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http://www.jiskha.com/display.cgi?id=1219442311 | 1,369,220,351,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701614932/warc/CC-MAIN-20130516105334-00078-ip-10-60-113-184.ec2.internal.warc.gz | 540,747,618 | 3,136 | Wednesday
May 22, 2013
# Homework Help: Math
Posted by Happyface on Friday, August 22, 2008 at 5:58pm.
What is the surface area and volume of a rectangular prism with a length of 10in, a width of 10in, and a height of 20in?
Thank you.
• Math - drwls, Friday, August 22, 2008 at 6:05pm
Use the same procedure I described in my answer to your other question.
• Math - Happyface, Friday, August 22, 2008 at 6:16pm
Thank you
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### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
There are several ways to turn a fraction into a decimal but one way is to divide the numerator by the denominator
In , you divide 8 by 10
8÷10=.80
Get the whole number 5 and add it to .80
5+.80=5.80 | 135 | 537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-04 | latest | en | 0.93132 |
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# Problem 18. Bullseye Matrix
Solution 1976051
Submitted on 14 Oct 2019 by nguyen thai son son
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 5; a = [3 3 3 3 3; 3 2 2 2 3; 3 2 1 2 3; 3 2 2 2 3; 3 3 3 3 3]; assert(isequal(bullseye(n),a));
2 Pass
n = 7; a = [4 4 4 4 4 4 4; 4 3 3 3 3 3 4; 4 3 2 2 2 3 4; 4 3 2 1 2 3 4; 4 3 2 2 2 3 4; 4 3 3 3 3 3 4; 4 4 4 4 4 4 4]; assert(isequal(bullseye(n),a))
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Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```surveyors rod = 5.0292 meter (length) ``` Related Measurements: Try converting from "rod" to actus (Roman actus), agate (typography agate), engineers chain, finger, furlong (surveyors furlong), Greek palm, Greek span, light yr (light year), line, m (meter), mile, nail (cloth nail), nautical mile, palm, parasang, skein, soccer field, span (cloth span), spindle, stadium (Roman stadium), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: rod = .14175258 actus (Roman actus), .08601142 arpentlin, 11 Biblical cubit, .1375 bolt (of cloth), .02291667 cable length, 19,800 caliber (gun barrel caliber), 1.31E-08 earth to moon (mean distance earth to moon), 28,618.92 en (typography en), .165 engineers chain, 2.75 fathom, 10.87 Greek cubit, 25 link (surveyors link), 88 nail (cloth nail), .00090518 nautical league, 14,309.46 point (typography point), 16.99 Roman foot, 22 span (cloth span), .02716795 stadia (Greek stadia), 165.97 sun (Japanese sun), 16.5 survey foot.
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 458 | 1,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-50 | latest | en | 0.714357 |
https://stat.ethz.ch/pipermail/r-help/2006-April/103004.html | 1,715,994,671,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057216.39/warc/CC-MAIN-20240517233122-20240518023122-00107.warc.gz | 479,428,250 | 2,284 | # [R] Help on computing a matrix from another matrix in R
Russ Branaghan rbranaghan at bigredrooster.com
Wed Apr 5 00:42:23 CEST 2006
```Hello All,
I am new to R and am having a problem.
I have the following 29 x 5 array called temp with a mode "list".
Sincere Exciting Competent Sophisticated Rugged
3.5 2.5 3.8 3.5 3.0
2.0 2.5 3.5 2.5 2.3
2.8 3.0 2.3 1.5 1.8
2.0 4.0 2.8 1.5 4.5
2.5 1.5 1.8 1.3 1.3
4.0 2.0 3.0 2.3 2.5
3.0 1.8 3.0 3.0 2.0
1.7 1.7 1.3 1.0 1.3
2.7 2.3 1.5 1.3 1.8
2.8 1.0 3.5 4.0 1.8
3.3 1.5 3.0 2.3 1.5
2.3 1.3 1.3 1.0 1.0
2.0 1.3 1.0 1.3 1.0
3.0 2.0 2.0 2.3 1.3
2.5 1.5 2.0 1.0 1.5
2.5 1.3 2.8 2.8 1.5
2.3 1.5 1.5 1.3 1.3
2.0 2.3 2.3 1.8 2.8
2.0 3.3 3.0 2.3 3.8
2.8 3.8 4.0 2.3 4.8
2.8 3.8 3.0 2.3 4.3
2.5 2.0 3.0 2.8 2.0
2.3 3.0 1.8 1.5 2.0
3.0 3.3 3.3 2.8 2.3
3.5 2.8 3.3 1.8 2.0
5.0 1.5 1.5 1.3 1.3
2.7 1.7 1.5 1.3 1.5
1.8 1.3 1.3 1.0 1.0
2.8 4.5 3.8 3.0 4.0
I am trying to compute the similarity of each row with each other row by
summing the absolute values of the differences between each item in each row
with each item in each other row. So for example since row 1 is 3.5, 2.5,
3.8, 3.5, 3.0 and row 2 is 2.0, 2.5, 3.5, 2.5, and 2.3, the sum of the
absolute values of the differences is 3.5. This is given by this code:
sum(abs(temp[1,]-temp[2,]))
My issue is that I want a 29 x 29 matrix of these similarity scores. I have
made some progress, but I cannot quite get it. The closest code I have is
the following (all on one line, of course):
ans<- for (firstrow in 1:29) { for (escrow in 1:29) {
print(sum(abs(temp[firstrow,] - temp[secrow,]) ) ) } }
I realize that print is not the right function to be using, but at least it
displays my results in a long list. Also, ans has the following
characteristics:
> ans
[1]
> dim(ans)
NULL
> mode(ans)
1] "numeric"
Can someone help me by telling me how to get this into a 29 x 29 matrix of
the right mode? | 1,100 | 2,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-22 | latest | en | 0.763975 |
https://aprove.informatik.rwth-aachen.de/eval/lowerbounds/html/proofs/103899.master.html | 1,723,380,795,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640997721.66/warc/CC-MAIN-20240811110531-20240811140531-00064.warc.gz | 80,311,488 | 20,592 | ### (0) Obligation:
Runtime Complexity TRS:
The TRS R consists of the following rules:
ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
id_inc(x) → x
id_inc(x) → s(x)
div(x, y) → if(ge(y, s(0)), ge(x, y), x, y)
if(false, b, x, y) → div_by_zero
if(true, false, x, y) → 0
if(true, true, x, y) → id_inc(div(minus(x, y), y))
Rewrite Strategy: INNERMOST
### (1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)
Converted Cpx (relative) TRS to CDT
### (2) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(z0, 0) → true
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
id_inc(z0) → z0
id_inc(z0) → s(z0)
div(z0, z1) → if(ge(z1, s(0)), ge(z0, z1), z0, z1)
if(false, z0, z1, z2) → div_by_zero
if(true, false, z0, z1) → 0
if(true, true, z0, z1) → id_inc(div(minus(z0, z1), z1))
Tuples:
GE(z0, 0) → c
GE(0, s(z0)) → c1
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(z0, 0) → c3
MINUS(0, z0) → c4
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
ID_INC(z0) → c6
ID_INC(z0) → c7
DIV(z0, z1) → c8(IF(ge(z1, s(0)), ge(z0, z1), z0, z1), GE(z1, s(0)), GE(z0, z1))
IF(false, z0, z1, z2) → c9
IF(true, false, z0, z1) → c10
IF(true, true, z0, z1) → c11(ID_INC(div(minus(z0, z1), z1)), DIV(minus(z0, z1), z1), MINUS(z0, z1))
S tuples:
GE(z0, 0) → c
GE(0, s(z0)) → c1
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(z0, 0) → c3
MINUS(0, z0) → c4
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
ID_INC(z0) → c6
ID_INC(z0) → c7
DIV(z0, z1) → c8(IF(ge(z1, s(0)), ge(z0, z1), z0, z1), GE(z1, s(0)), GE(z0, z1))
IF(false, z0, z1, z2) → c9
IF(true, false, z0, z1) → c10
IF(true, true, z0, z1) → c11(ID_INC(div(minus(z0, z1), z1)), DIV(minus(z0, z1), z1), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
ge, minus, id_inc, div, if
Defined Pair Symbols:
GE, MINUS, ID_INC, DIV, IF
Compound Symbols:
c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11
### (3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 8 trailing nodes:
IF(true, false, z0, z1) → c10
GE(z0, 0) → c
GE(0, s(z0)) → c1
MINUS(z0, 0) → c3
MINUS(0, z0) → c4
IF(false, z0, z1, z2) → c9
ID_INC(z0) → c6
ID_INC(z0) → c7
### (4) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(z0, 0) → true
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
id_inc(z0) → z0
id_inc(z0) → s(z0)
div(z0, z1) → if(ge(z1, s(0)), ge(z0, z1), z0, z1)
if(false, z0, z1, z2) → div_by_zero
if(true, false, z0, z1) → 0
if(true, true, z0, z1) → id_inc(div(minus(z0, z1), z1))
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
DIV(z0, z1) → c8(IF(ge(z1, s(0)), ge(z0, z1), z0, z1), GE(z1, s(0)), GE(z0, z1))
IF(true, true, z0, z1) → c11(ID_INC(div(minus(z0, z1), z1)), DIV(minus(z0, z1), z1), MINUS(z0, z1))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
DIV(z0, z1) → c8(IF(ge(z1, s(0)), ge(z0, z1), z0, z1), GE(z1, s(0)), GE(z0, z1))
IF(true, true, z0, z1) → c11(ID_INC(div(minus(z0, z1), z1)), DIV(minus(z0, z1), z1), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
ge, minus, id_inc, div, if
Defined Pair Symbols:
GE, MINUS, DIV, IF
Compound Symbols:
c2, c5, c8, c11
### (5) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
### (6) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(z0, 0) → true
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
id_inc(z0) → z0
id_inc(z0) → s(z0)
div(z0, z1) → if(ge(z1, s(0)), ge(z0, z1), z0, z1)
if(false, z0, z1, z2) → div_by_zero
if(true, false, z0, z1) → 0
if(true, true, z0, z1) → id_inc(div(minus(z0, z1), z1))
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
DIV(z0, z1) → c8(IF(ge(z1, s(0)), ge(z0, z1), z0, z1), GE(z1, s(0)), GE(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
DIV(z0, z1) → c8(IF(ge(z1, s(0)), ge(z0, z1), z0, z1), GE(z1, s(0)), GE(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
ge, minus, id_inc, div, if
Defined Pair Symbols:
GE, MINUS, DIV, IF
Compound Symbols:
c2, c5, c8, c11
### (7) CdtUsableRulesProof (EQUIVALENT transformation)
The following rules are not usable and were removed:
id_inc(z0) → z0
id_inc(z0) → s(z0)
div(z0, z1) → if(ge(z1, s(0)), ge(z0, z1), z0, z1)
if(false, z0, z1, z2) → div_by_zero
if(true, false, z0, z1) → 0
if(true, true, z0, z1) → id_inc(div(minus(z0, z1), z1))
### (8) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
DIV(z0, z1) → c8(IF(ge(z1, s(0)), ge(z0, z1), z0, z1), GE(z1, s(0)), GE(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
DIV(z0, z1) → c8(IF(ge(z1, s(0)), ge(z0, z1), z0, z1), GE(z1, s(0)), GE(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, DIV, IF
Compound Symbols:
c2, c5, c8, c11
### (9) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace DIV(z0, z1) → c8(IF(ge(z1, s(0)), ge(z0, z1), z0, z1), GE(z1, s(0)), GE(z0, z1)) by
DIV(0, s(z0)) → c8(IF(ge(s(z0), s(0)), false, 0, s(z0)), GE(s(z0), s(0)), GE(0, s(z0)))
DIV(s(z0), s(z1)) → c8(IF(ge(s(z1), s(0)), ge(z0, z1), s(z0), s(z1)), GE(s(z1), s(0)), GE(s(z0), s(z1)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0), GE(0, s(0)), GE(z0, 0))
DIV(x0, 0) → c8(IF(false, ge(x0, 0), x0, 0), GE(0, s(0)), GE(x0, 0))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
### (10) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(0, s(z0)) → c8(IF(ge(s(z0), s(0)), false, 0, s(z0)), GE(s(z0), s(0)), GE(0, s(z0)))
DIV(s(z0), s(z1)) → c8(IF(ge(s(z1), s(0)), ge(z0, z1), s(z0), s(z1)), GE(s(z1), s(0)), GE(s(z0), s(z1)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0), GE(0, s(0)), GE(z0, 0))
DIV(x0, 0) → c8(IF(false, ge(x0, 0), x0, 0), GE(0, s(0)), GE(x0, 0))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(0, s(z0)) → c8(IF(ge(s(z0), s(0)), false, 0, s(z0)), GE(s(z0), s(0)), GE(0, s(z0)))
DIV(s(z0), s(z1)) → c8(IF(ge(s(z1), s(0)), ge(z0, z1), s(z0), s(z1)), GE(s(z1), s(0)), GE(s(z0), s(z1)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0), GE(0, s(0)), GE(z0, 0))
DIV(x0, 0) → c8(IF(false, ge(x0, 0), x0, 0), GE(0, s(0)), GE(x0, 0))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
K tuples:none
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, IF, DIV
Compound Symbols:
c2, c5, c11, c8
### (11) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
DIV(x0, 0) → c8(IF(false, ge(x0, 0), x0, 0), GE(0, s(0)), GE(x0, 0))
### (12) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(0, s(z0)) → c8(IF(ge(s(z0), s(0)), false, 0, s(z0)), GE(s(z0), s(0)), GE(0, s(z0)))
DIV(s(z0), s(z1)) → c8(IF(ge(s(z1), s(0)), ge(z0, z1), s(z0), s(z1)), GE(s(z1), s(0)), GE(s(z0), s(z1)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0), GE(0, s(0)), GE(z0, 0))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(0, s(z0)) → c8(IF(ge(s(z0), s(0)), false, 0, s(z0)), GE(s(z0), s(0)), GE(0, s(z0)))
DIV(s(z0), s(z1)) → c8(IF(ge(s(z1), s(0)), ge(z0, z1), s(z0), s(z1)), GE(s(z1), s(0)), GE(s(z0), s(z1)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0), GE(0, s(0)), GE(z0, 0))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
K tuples:none
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, IF, DIV
Compound Symbols:
c2, c5, c11, c8
### (13) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 4 trailing tuple parts
### (14) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(z0), s(z1)) → c8(IF(ge(s(z1), s(0)), ge(z0, z1), s(z0), s(z1)), GE(s(z1), s(0)), GE(s(z0), s(z1)))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(z0), s(z1)) → c8(IF(ge(s(z1), s(0)), ge(z0, z1), s(z0), s(z1)), GE(s(z1), s(0)), GE(s(z0), s(z1)))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
K tuples:none
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, IF, DIV
Compound Symbols:
c2, c5, c11, c8, c8
### (15) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
We considered the (Usable) Rules:
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
minus(0, z0) → 0
And the Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(z0), s(z1)) → c8(IF(ge(s(z1), s(0)), ge(z0, z1), s(z0), s(z1)), GE(s(z1), s(0)), GE(s(z0), s(z1)))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [4]
POL(DIV(x1, x2)) = [4]x1
POL(GE(x1, x2)) = 0
POL(IF(x1, x2, x3, x4)) = [4]x3
POL(MINUS(x1, x2)) = 0
POL(c11(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(c8(x1)) = x1
POL(c8(x1, x2, x3)) = x1 + x2 + x3
POL(false) = [3]
POL(ge(x1, x2)) = [4]x1
POL(minus(x1, x2)) = x1
POL(s(x1)) = x1
POL(true) = 0
### (16) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(z0), s(z1)) → c8(IF(ge(s(z1), s(0)), ge(z0, z1), s(z0), s(z1)), GE(s(z1), s(0)), GE(s(z0), s(z1)))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(z0), s(z1)) → c8(IF(ge(s(z1), s(0)), ge(z0, z1), s(z0), s(z1)), GE(s(z1), s(0)), GE(s(z0), s(z1)))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
K tuples:
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, IF, DIV
Compound Symbols:
c2, c5, c11, c8, c8
### (17) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace DIV(s(z0), s(z1)) → c8(IF(ge(s(z1), s(0)), ge(z0, z1), s(z0), s(z1)), GE(s(z1), s(0)), GE(s(z0), s(z1))) by
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(0), s(s(z0))) → c8(IF(ge(s(s(z0)), s(0)), false, s(0), s(s(z0))), GE(s(s(z0)), s(0)), GE(s(0), s(s(z0))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
### (18) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(0), s(s(z0))) → c8(IF(ge(s(s(z0)), s(0)), false, s(0), s(s(z0))), GE(s(s(z0)), s(0)), GE(s(0), s(s(z0))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(0), s(s(z0))) → c8(IF(ge(s(s(z0)), s(0)), false, s(0), s(s(z0))), GE(s(s(z0)), s(0)), GE(s(0), s(s(z0))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
K tuples:
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, IF, DIV
Compound Symbols:
c2, c5, c11, c8, c8
### (19) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
### (20) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c8(GE(s(s(z0)), s(0)), GE(s(0), s(s(z0))))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c8(GE(s(s(z0)), s(0)), GE(s(0), s(s(z0))))
K tuples:
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, IF, DIV
Compound Symbols:
c2, c5, c11, c8, c8, c8
### (21) CdtGraphSplitRhsProof (BOTH BOUNDS(ID, ID) transformation)
Split RHS of tuples not part of any SCC
### (22) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
K tuples:
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, IF, DIV
Compound Symbols:
c2, c5, c11, c8, c8, c
### (23) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
We considered the (Usable) Rules:none
And the Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(DIV(x1, x2)) = [4]
POL(GE(x1, x2)) = 0
POL(IF(x1, x2, x3, x4)) = [4]
POL(MINUS(x1, x2)) = 0
POL(c(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(c8(x1)) = x1
POL(c8(x1, x2, x3)) = x1 + x2 + x3
POL(false) = [3]
POL(ge(x1, x2)) = [2] + x2
POL(minus(x1, x2)) = [3] + [3]x1 + [3]x2
POL(s(x1)) = 0
POL(true) = 0
### (24) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
K tuples:
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, IF, DIV
Compound Symbols:
c2, c5, c11, c8, c8, c
### (25) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace DIV(x0, s(z0)) → c8(IF(ge(z0, 0), ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0))) by
DIV(0, s(z0)) → c8(IF(ge(z0, 0), false, 0, s(z0)), GE(s(z0), s(0)), GE(0, s(z0)))
DIV(s(z0), s(z1)) → c8(IF(ge(z1, 0), ge(z0, z1), s(z0), s(z1)), GE(s(z1), s(0)), GE(s(z0), s(z1)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
### (26) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
DIV(0, s(z0)) → c8(IF(ge(z0, 0), false, 0, s(z0)), GE(s(z0), s(0)), GE(0, s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(0, s(z0)) → c8(IF(ge(z0, 0), false, 0, s(z0)), GE(s(z0), s(0)), GE(0, s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
K tuples:
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, IF, DIV
Compound Symbols:
c2, c5, c11, c8, c8, c
### (27) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing tuple parts
### (28) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
K tuples:
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, IF, DIV
Compound Symbols:
c2, c5, c11, c8, c8, c
### (29) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)
The following tuples could be moved from S to K by knowledge propagation:
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
### (30) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
K tuples:
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, IF, DIV
Compound Symbols:
c2, c5, c11, c8, c8, c
### (31) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace DIV(z0, 0) → c8(IF(ge(0, s(0)), true, z0, 0)) by
DIV(x0, 0) → c8(IF(false, true, x0, 0))
### (32) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(x0, 0) → c8(IF(false, true, x0, 0))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
DIV(x0, 0) → c8(IF(false, true, x0, 0))
K tuples:
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, IF, DIV
Compound Symbols:
c2, c5, c11, c8, c8, c
### (33) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing nodes:
DIV(x0, 0) → c8(IF(false, true, x0, 0))
### (34) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
S tuples:
GE(s(z0), s(z1)) → c2(GE(z0, z1))
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
K tuples:
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
GE, MINUS, IF, DIV
Compound Symbols:
c2, c5, c11, c8, c8, c
### (35) CdtForwardInstantiationProof (BOTH BOUNDS(ID, ID) transformation)
Use forward instantiation to replace GE(s(z0), s(z1)) → c2(GE(z0, z1)) by
GE(s(s(y0)), s(s(y1))) → c2(GE(s(y0), s(y1)))
### (36) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
GE(s(s(y0)), s(s(y1))) → c2(GE(s(y0), s(y1)))
S tuples:
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
GE(s(s(y0)), s(s(y1))) → c2(GE(s(y0), s(y1)))
K tuples:
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
MINUS, IF, DIV, GE
Compound Symbols:
c5, c11, c8, c8, c, c2
### (37) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 3 trailing nodes:
DIV(0, s(z0)) → c8(GE(s(z0), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(s(z0)), s(0)))
DIV(s(0), s(s(z0))) → c(GE(s(0), s(s(z0))))
### (38) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
GE(s(s(y0)), s(s(y1))) → c2(GE(s(y0), s(y1)))
S tuples:
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z1)), s(0)), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)), GE(s(0), s(0)), GE(s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(z0), s(0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(s(z0), s(0)), GE(x0, s(z0)))
GE(s(s(y0)), s(s(y1))) → c2(GE(s(y0), s(y1)))
K tuples:
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
MINUS, IF, DIV, GE
Compound Symbols:
c5, c11, c8, c8, c2
### (39) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 6 trailing tuple parts
### (40) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
GE(s(s(y0)), s(s(y1))) → c2(GE(s(y0), s(y1)))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
S tuples:
MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1))
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
GE(s(s(y0)), s(s(y1))) → c2(GE(s(y0), s(y1)))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
K tuples:
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
MINUS, IF, DIV, GE
Compound Symbols:
c5, c11, c8, c2, c8
### (41) CdtForwardInstantiationProof (BOTH BOUNDS(ID, ID) transformation)
Use forward instantiation to replace MINUS(s(z0), s(z1)) → c5(MINUS(z0, z1)) by
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
### (42) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
GE(s(s(y0)), s(s(y1))) → c2(GE(s(y0), s(y1)))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
S tuples:
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
GE(s(s(y0)), s(s(y1))) → c2(GE(s(y0), s(y1)))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
K tuples:
DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
IF, DIV, GE, MINUS
Compound Symbols:
c11, c8, c2, c8, c5
### (43) CdtForwardInstantiationProof (BOTH BOUNDS(ID, ID) transformation)
Use forward instantiation to replace DIV(s(x0), s(x1)) → c8(GE(s(x0), s(x1))) by
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
### (44) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
GE(s(s(y0)), s(s(y1))) → c2(GE(s(y0), s(y1)))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
S tuples:
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
GE(s(s(y0)), s(s(y1))) → c2(GE(s(y0), s(y1)))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
IF, GE, DIV, MINUS
Compound Symbols:
c11, c2, c8, c8, c5
### (45) CdtForwardInstantiationProof (BOTH BOUNDS(ID, ID) transformation)
Use forward instantiation to replace GE(s(s(y0)), s(s(y1))) → c2(GE(s(y0), s(y1))) by
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
### (46) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
S tuples:
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
IF, DIV, MINUS, GE
Compound Symbols:
c11, c8, c8, c5, c2
### (47) CdtRewritingProof (BOTH BOUNDS(ID, ID) transformation)
Used rewriting to replace DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(s(z1)), s(0)), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1)))) by DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
### (48) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
S tuples:
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
IF, DIV, MINUS, GE
Compound Symbols:
c11, c8, c8, c5, c2
### (49) CdtRewritingProof (BOTH BOUNDS(ID, ID) transformation)
Used rewriting to replace DIV(s(z0), s(0)) → c8(IF(ge(s(0), s(0)), true, s(z0), s(0))) by DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
### (50) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
S tuples:
IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
IF, DIV, MINUS, GE
Compound Symbols:
c11, c8, c5, c8, c2
### (51) CdtInstantiationProof (BOTH BOUNDS(ID, ID) transformation)
Use instantiation to replace IF(true, true, z0, z1) → c11(DIV(minus(z0, z1), z1), MINUS(z0, z1)) by
IF(true, true, s(x0), s(x1)) → c11(DIV(minus(s(x0), s(x1)), s(x1)), MINUS(s(x0), s(x1)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(s(x0)), s(s(x1))), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)), MINUS(s(x0), s(0)))
### (52) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(x0), s(x1)) → c11(DIV(minus(s(x0), s(x1)), s(x1)), MINUS(s(x0), s(x1)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(s(x0)), s(s(x1))), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)), MINUS(s(x0), s(0)))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(x0), s(x1)) → c11(DIV(minus(s(x0), s(x1)), s(x1)), MINUS(s(x0), s(x1)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(s(x0)), s(s(x1))), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)), MINUS(s(x0), s(0)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11
### (53) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 1 trailing tuple parts
### (54) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(x0), s(x1)) → c11(DIV(minus(s(x0), s(x1)), s(x1)), MINUS(s(x0), s(x1)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(s(x0)), s(s(x1))), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(x0), s(x1)) → c11(DIV(minus(s(x0), s(x1)), s(x1)), MINUS(s(x0), s(x1)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(s(x0)), s(s(x1))), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (55) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace IF(true, true, s(x0), s(x1)) → c11(DIV(minus(s(x0), s(x1)), s(x1)), MINUS(s(x0), s(x1))) by
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
### (56) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(s(x0)), s(s(x1))), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(s(x0)), s(s(x1))), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (57) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
We considered the (Usable) Rules:
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
minus(0, z0) → 0
And the Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(s(x0)), s(s(x1))), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(DIV(x1, x2)) = [4]x1
POL(GE(x1, x2)) = 0
POL(IF(x1, x2, x3, x4)) = [4]x3
POL(MINUS(x1, x2)) = 0
POL(c11(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(c8(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(false) = [3]
POL(ge(x1, x2)) = [5]x1
POL(minus(x1, x2)) = x1
POL(s(x1)) = [4] + x1
POL(true) = 0
### (58) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(s(x0)), s(s(x1))), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(s(x0)), s(s(x1))), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (59) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(s(x0)), s(s(x1))), s(s(x1))), MINUS(s(s(x0)), s(s(x1)))) by
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
### (60) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (61) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
We considered the (Usable) Rules:
ge(z0, 0) → true
And the Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = [3]
POL(DIV(x1, x2)) = [1] + [5]x2
POL(GE(x1, x2)) = [2]x2
POL(IF(x1, x2, x3, x4)) = x1
POL(MINUS(x1, x2)) = [2]x2
POL(c11(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(c8(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(false) = [5]
POL(ge(x1, x2)) = [1]
POL(minus(x1, x2)) = [2] + [2]x2
POL(s(x1)) = 0
POL(true) = [1]
### (62) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (63) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
We considered the (Usable) Rules:
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
minus(0, z0) → 0
And the Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(DIV(x1, x2)) = x1
POL(GE(x1, x2)) = 0
POL(IF(x1, x2, x3, x4)) = x3
POL(MINUS(x1, x2)) = 0
POL(c11(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(c8(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(false) = [2]
POL(ge(x1, x2)) = 0
POL(minus(x1, x2)) = x1
POL(s(x1)) = [2] + x1
POL(true) = 0
### (64) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (65) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace IF(true, true, s(x0), s(0)) → c11(DIV(minus(s(x0), s(0)), s(0))) by
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
### (66) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (67) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
We considered the (Usable) Rules:
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
minus(0, z0) → 0
And the Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(DIV(x1, x2)) = [2] + x1
POL(GE(x1, x2)) = 0
POL(IF(x1, x2, x3, x4)) = [2] + x3
POL(MINUS(x1, x2)) = 0
POL(c11(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(c8(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(false) = [3]
POL(ge(x1, x2)) = 0
POL(minus(x1, x2)) = x1
POL(s(x1)) = [2] + x1
POL(true) = 0
### (68) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (69) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1)))) by
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
### (70) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (71) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0))) by
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
IF(true, true, s(0), s(0)) → c11(DIV(0, s(0)))
### (72) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
IF(true, true, s(0), s(0)) → c11(DIV(0, s(0)))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (73) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace DIV(x0, s(z0)) → c8(IF(true, ge(x0, s(z0)), x0, s(z0)), GE(x0, s(z0))) by
DIV(0, s(z0)) → c8(IF(true, false, 0, s(z0)), GE(0, s(z0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
### (74) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
IF(true, true, s(0), s(0)) → c11(DIV(0, s(0)))
DIV(0, s(z0)) → c8(IF(true, false, 0, s(z0)), GE(0, s(z0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
DIV(0, s(z0)) → c8(IF(true, false, 0, s(z0)), GE(0, s(z0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (75) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)
Removed 2 trailing tuple parts
### (76) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
IF(true, true, s(0), s(0)) → c11(DIV(0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
DIV(0, s(z0)) → c8
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
DIV(0, s(z0)) → c8
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11, c8
### (77) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
DIV(0, s(z0)) → c8
We considered the (Usable) Rules:none
And the Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
IF(true, true, s(0), s(0)) → c11(DIV(0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
DIV(0, s(z0)) → c8
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(DIV(x1, x2)) = [1] + x2
POL(GE(x1, x2)) = 0
POL(IF(x1, x2, x3, x4)) = [1]
POL(MINUS(x1, x2)) = 0
POL(c11(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(c8) = 0
POL(c8(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(false) = [3]
POL(ge(x1, x2)) = [3]x1
POL(minus(x1, x2)) = [5] + x1 + [2]x2
POL(s(x1)) = 0
POL(true) = 0
### (78) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
IF(true, true, s(0), s(0)) → c11(DIV(0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
DIV(0, s(z0)) → c8
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
DIV(0, s(z0)) → c8
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11, c8
### (79) CdtNarrowingProof (BOTH BOUNDS(ID, ID) transformation)
Use narrowing to replace IF(true, true, x0, s(x1)) → c11(DIV(minus(x0, s(x1)), s(x1)), MINUS(x0, s(x1))) by
IF(true, true, 0, s(x1)) → c11(DIV(0, s(x1)), MINUS(0, s(x1)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
### (80) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
IF(true, true, s(0), s(0)) → c11(DIV(0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
DIV(0, s(z0)) → c8
IF(true, true, 0, s(x1)) → c11(DIV(0, s(x1)), MINUS(0, s(x1)))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
IF(true, true, 0, s(x1)) → c11(DIV(0, s(x1)), MINUS(0, s(x1)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(x0)), s(s(x1))) → c11(DIV(minus(s(x0), s(x1)), s(s(x1))), MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(z0), s(0)) → c11(DIV(minus(z0, 0), s(0)))
DIV(0, s(z0)) → c8
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11, c8
### (81) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)
Removed 3 trailing nodes:
IF(true, true, s(0), s(0)) → c11(DIV(0, s(0)))
DIV(0, s(z0)) → c8
IF(true, true, 0, s(x1)) → c11(DIV(0, s(x1)), MINUS(0, s(x1)))
### (82) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (83) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)
The following tuples could be moved from S to K by knowledge propagation:
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
### (84) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (85) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
We considered the (Usable) Rules:
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
minus(0, z0) → 0
And the Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(DIV(x1, x2)) = [4] + [4]x1
POL(GE(x1, x2)) = [4]
POL(IF(x1, x2, x3, x4)) = [4]x3
POL(MINUS(x1, x2)) = [3]
POL(c11(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(c8(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(false) = [2]
POL(ge(x1, x2)) = [4]x1
POL(minus(x1, x2)) = [1] + x1
POL(s(x1)) = [4] + x1
POL(true) = 0
### (86) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
S tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (87) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
We considered the (Usable) Rules:
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
minus(0, z0) → 0
And the Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(DIV(x1, x2)) = [2] + [3]x1 + x2
POL(GE(x1, x2)) = 0
POL(IF(x1, x2, x3, x4)) = [3]x3 + x4
POL(MINUS(x1, x2)) = [3]
POL(c11(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(c8(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(false) = [3]
POL(ge(x1, x2)) = [3]x1
POL(minus(x1, x2)) = x1
POL(s(x1)) = [4] + x1
POL(true) = 0
### (88) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
S tuples:
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (89) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^2))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
We considered the (Usable) Rules:
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
minus(0, z0) → 0
And the Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(DIV(x1, x2)) = [2]x12
POL(GE(x1, x2)) = 0
POL(IF(x1, x2, x3, x4)) = [2]x32
POL(MINUS(x1, x2)) = [2]x1
POL(c11(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(c8(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(false) = 0
POL(ge(x1, x2)) = 0
POL(minus(x1, x2)) = x1
POL(s(x1)) = [1] + x1
POL(true) = 0
### (90) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
S tuples:
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (91) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^2))) transformation)
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
We considered the (Usable) Rules:
minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
minus(0, z0) → 0
And the Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
The order we found is given by the following interpretation:
Polynomial interpretation :
POL(0) = 0
POL(DIV(x1, x2)) = [2]x1 + [2]x12
POL(GE(x1, x2)) = [2]x1
POL(IF(x1, x2, x3, x4)) = [2]x32
POL(MINUS(x1, x2)) = [1]
POL(c11(x1)) = x1
POL(c11(x1, x2)) = x1 + x2
POL(c2(x1)) = x1
POL(c5(x1)) = x1
POL(c8(x1)) = x1
POL(c8(x1, x2)) = x1 + x2
POL(false) = 0
POL(ge(x1, x2)) = 0
POL(minus(x1, x2)) = x1
POL(s(x1)) = [1] + x1
POL(true) = 0
### (92) Obligation:
Complexity Dependency Tuples Problem
Rules:
ge(0, s(z0)) → false
ge(s(z0), s(z1)) → ge(z0, z1)
ge(z0, 0) → true
minus(z0, 0) → z0
minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
IF(true, true, s(s(z0)), s(s(z1))) → c11(DIV(minus(z0, z1), s(s(z1))), MINUS(s(s(z0)), s(s(z1))))
IF(true, true, s(z0), s(0)) → c11(DIV(z0, s(0)))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
S tuples:none
K tuples:
DIV(s(s(y0)), s(s(y1))) → c8(GE(s(s(y0)), s(s(y1))))
IF(true, true, s(z0), s(z1)) → c11(DIV(minus(z0, z1), s(z1)), MINUS(s(z0), s(z1)))
IF(true, true, s(s(x0)), s(s(x1))) → c11(MINUS(s(s(x0)), s(s(x1))))
DIV(s(z0), s(0)) → c8(IF(ge(0, 0), true, s(z0), s(0)))
DIV(s(x0), s(z0)) → c8(IF(ge(z0, 0), ge(x0, z0), s(x0), s(z0)), GE(s(x0), s(z0)))
DIV(s(s(z0)), s(s(z1))) → c8(IF(ge(s(z1), 0), ge(z0, z1), s(s(z0)), s(s(z1))), GE(s(s(z0)), s(s(z1))))
DIV(s(z0), s(z1)) → c8(IF(true, ge(z0, z1), s(z0), s(z1)), GE(s(z0), s(z1)))
MINUS(s(s(y0)), s(s(y1))) → c5(MINUS(s(y0), s(y1)))
GE(s(s(s(y0))), s(s(s(y1)))) → c2(GE(s(s(y0)), s(s(y1))))
Defined Rule Symbols:
ge, minus
Defined Pair Symbols:
DIV, MINUS, GE, IF
Compound Symbols:
c8, c5, c8, c2, c11, c11
### (93) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)
The set S is empty | 51,813 | 107,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-33 | latest | en | 0.278145 |
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Written by Ines May 15, 2021 ยท 4 min read
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Multiplication Worksheets Fun. For example if you forget 82 you might remember 2816. 15052019 Then once kids have a solid understanding you can provide additional practice in fun non-threatening ways like the multiplication facts worksheets and games youll find below. From basics like multiplying by twos to complex concepts such as three-digit multiplication our multiplication worksheets help elementary school students of all ages improve this vital skill. Multiplication tables and charts given here help children to solve these problems quickly.
This Growing Bundle Of Multiplication Tables From 2 To 12 Is Designed To Help Students Practice Le Multiplication Facts Times Tables Worksheets Multiplication From pinterest.com
Find enjoyable Times Table Games and Fun multiplication practice to provide great multiplication resources for teacher students and parents. Cut in half then times 10. This worksheet therefore is available to improve your kids in the understanding of multiplication. 1 to 4. Practice row and column multiplication of simple and large-digit numbers. Have fun exploring these 50 multiplication and division activities.
### 15052019 Then once kids have a solid understanding you can provide additional practice in fun non-threatening ways like the multiplication facts worksheets and games youll find below.
These are Multiplication Worksheets. Multiplication cut across all aspect of life. This is why they must be sound in their multiplication topics to be sound in this. For younger students we offer printable multiplication tables and various puzzles like multiplication crosswords and fill-in-the-blanks. Basic Multiplication Worksheets This site has hundreds of basic multiplication activities. Free Printable Multiplication Worksheets.
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Each page includes a set of multiplication facts to solve. Easy Way To Learn Multiplication. These are great for multiplication practice at home if your school is. Cut in half then times 10. 2 to 6.
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Multiplication Free Printables Resources to get some serious training on times table 1-12. Using skip counting worksheets is a simple way to introduce the basic. This worksheet therefore is available to improve your kids in the understanding of multiplication. This is why they must be sound in their multiplication topics to be sound in this. 08072019 So this page has multiple resources that can help make learning multiplication and dividing fun.
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They can learn to skip count play games to work on memorizing their math facts or just practice their long multiplication and division skills through puzzles and other fun worksheets. Basic Multiplication Worksheets This site has hundreds of basic multiplication activities. Up To Ten. These times tables worksheets include single-digit and double-digit multiplication practice worksheets missing factor activities color by number activities and even a few multiplication games. These are Multiplication Worksheets.
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The ones are simply counting and. Multiplication Worksheets One Minute Timed Multiplication Worksheets. 3 to 15. For younger students we offer printable multiplication tables and various puzzles like multiplication crosswords and fill-in-the-blanks. Worksheet Number Range Online.
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27072019 MULTIPLICATION WORKSHEETS Free Printable Multiplication Worksheets provided here has numerous exercises to sharpen your childs multiplication skills. These basic math fact multiplication worksheets are similar to the. Print off the Bingo multiplication worksheets on Table Fables for a hilarious game of Bingo that gets children practising their Times Tables in a new immersive away. This set of practice pages is NOT your typical worksheet. Double then double again Example 49.
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Dec 20 . 4 min read | 1,134 | 5,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-21 | latest | en | 0.850129 |
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# physics lab 6 - not of the new position. 4.) Repeat step 3...
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Adebayo Adesina 106223646 PHYSICS 133 EXPERIMENT 7 SIMPLE HARMONIC MOTION Introduction: “In this lab, we study the phenomenon of simple harmonic motion for a mass-and-spring system and for a variety of pendulums” Lab 7 .We will experiment with a glider mounted on an air track and a pendulum that will be swung at various amplitudes according to our digression. We will also be able to practice the concept of a spring constant and other formulas that relate to oscillations. This lab will be broken down into various sections that allow us to analyze and begin to understand mass spring system and the simple pendulum. PROCEDURE: 1.) Measure the mass of the glider 2.) Attach a spring to the glider and take note of its equilibrium position. 3.) Attach a spring to the glider with a mass at the end of the spring and take
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Unformatted text preview: not of the new position. 4.) Repeat step 3 two more times with different masses. 5.) Remove the string and attach a spring to the other end of the glider connected to the air track. 6.) Use the computer to take pendulum measurements of the motion of the glider. 7.) Add a mass to the glider and repeat step 6 8.) Record information from computer in accordance with step 6 and 7 9.) Measure the length of the pendulum 10.) Swing the pendulum from a certain angle 11.) Time the amount of time it takes for the pendulum to swing ten times 12.) Repeat steps ten and eleven, three more times 13.) Measure the mass of the ruler 14.) Place the ruler on the pendulum 15.) Time the amount of time it takes for the ruler to swing ten times. DATA : Attached...
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## This note was uploaded on 04/06/2010 for the course WST 301 taught by Professor Kandi during the Spring '10 term at Suffolk.
Ask a homework question - tutors are online | 489 | 2,010 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-09 | longest | en | 0.852887 |
https://www.physicsforums.com/threads/integral-of-natural-log-part-ii.13954/ | 1,539,891,953,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512014.61/warc/CC-MAIN-20181018194005-20181018215505-00015.warc.gz | 1,025,045,591 | 13,025 | # Homework Help: Integral of Natural Log Part II
1. Feb 5, 2004
### wubie
Hi,
I am still having trouble with taking the integral of the following:
integral of ln(x+1) dx
I am trying to do it by parts but I end up getting stuck. I had no problem doing:
integral of ln x dx
But I can't seem to get
integral of ln (x+1) dx
Let u = ln (x+1) then du = 1/ (x+1)
Let dv = dx then v = x.
So then
integral of ln (x+1) dx =
ln(x+1)*x - integral of x/(x+1) dx
I then though that I would integrate the last part by parts again.
Let u = x then du = dx
Let dv = 1/(x+1) then v = ln(x+1)
So then
integral of x/(x+1) dx =
x*ln(x+1) - integral of ln(x+1) dx.
So far I now have
integral of ln (x+1) dx =
ln(x+1)*x - [x*ln(x+1) - integral of ln(x+1) dx]
But then all I end up getting is
integral of ln (x+1) dx = integral of ln (x+1) dx
So I am back to where I started. What am I doing incorrectly?
Any help is appreciated. Thankyou.
2. Feb 5, 2004
### stoffer
you know the integral of ln[x] and now u want integral of ln[x+1]?
Is it just me or do I hear x+1 begging to be substituted??
3. Feb 6, 2004
### wubie
!@#$!@#$!@#$!@#$%!@#\$!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
What is wrong with me tonight? I can't see anything. >8(
Thanks stoffer.
Cheers.
4. Feb 6, 2004
### stoffer
no prob. we all have our moments
5. Feb 6, 2004
### NateTG
$$\int \ln(x+1) dx$$
let $$u=x+1$$
then $$du=1 dx$$
so you have
$$\int \ln(u) du$$
which is:
now you can apply the formula I gave you in the other thread so you get:
$$u \ln u - u = (x+1) \ln (x+1) - (x+1)$$
FYI derivation of the formula:
$$\int \ln(x) dx = \int \ln(x) * 1 dx$$
now by parts we have:
$$u=\ln(x)$$
$$dv=1dx$$
so
$$du=\frac{1}{x} dx$$
$$v=x$$
so
$$\int 1 * \ln(x) dx=\int u dv = uv+C_1-\int v du =$$
$$x \ln x +C_1- \int x* \frac{1}{x} dx=x \ln x +C_1- \int 1dx = x \ln x +C_1- x+C_2=$$
$$x \ln x -x +C$$ | 713 | 1,875 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-43 | latest | en | 0.825047 |
https://philoid.com/question/16288-the-area-of-a-trapezium-is-960-cm-2-if-the-parallel-sides-are-34-cm-and-46-cm-find-the-distance-between-them | 1,716,233,295,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00119.warc.gz | 394,920,613 | 7,967 | ##### The area of a trapezium is 960 cm2. If the parallel sides are 34 cm and 46 cm, find the distance between them.
Area of trapezium =
Area of trapezium =
=
Therefore = 24 cm
8 | 57 | 183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-22 | latest | en | 0.85646 |
http://mathhelpforum.com/advanced-algebra/159271-coset-help.html | 1,529,605,960,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864256.26/warc/CC-MAIN-20180621172638-20180621192638-00424.warc.gz | 206,119,125 | 9,963 | 1. ## Coset help
Q: Suppose that $\displaystyle a$ has order 15. Find all of the left cosets of
$\displaystyle < a^{5} >$ in $\displaystyle < a >$.
Proof. First, lets list the elements of $\displaystyle < a >$
$\displaystyle < a >=\{e, a, a^{2}, a^{3} , a^{4} , a^{5} , a^{6} , a^{7} , a^{8} , a^{9} , a^{10} , a^{11} , a^{12} , a^{13} , a^{14}\}$
The cosets would be:
$\displaystyle < a^{5} >, a < a^{5} >, a^{2} < a^{5} >, a^{3} < a^{5} >, a^{4} < a^{5} >$
I don't understand where the choices $\displaystyle e, a, a^{2}, a^{3}, a^{4}$ came from. I see that there will be 5 cosets, since $\displaystyle |<a^{5}>|||<a>|=\frac{15}{3}=5$, but I don't understand the choices of the cosets.
As I understand it, there will be more cosets, but cosets parition the group into distinct sets. So, once you have found all the cosets whos union is the orginal group, you are done. Correct?
2. Originally Posted by Danneedshelp
Q: Suppose that $\displaystyle a$ has order 15. Find all of the left cosets of
$\displaystyle < a^{5} >$ in $\displaystyle < a >$.
Proof. First, lets list the elements of $\displaystyle < a >$
$\displaystyle < a >=\{e, a, a^{2}, a^{3} , a^{4} , a^{5} , a^{6} , a^{7} , a^{8} , a^{9} , a^{10} , a^{11} , a^{12} , a^{13} , a^{14}\}$
The cosets would be:
$\displaystyle < a^{5} >, a < a^{5} >, a^{2} < a^{5} >, a^{3} < a^{5} >, a^{4} < a^{5} >$
I don't understand where the choices $\displaystyle e, a, a^{2}, a^{3}, a^{4}$ came from. I see that there will be 5 cosets, since $\displaystyle |<a^{5}>|||<a>|=\frac{15}{3}=5$, but I don't understand the choices of the cosets.
As I understand it, there will be more cosets, but cosets parition the group into distinct sets. So, once you have found all the cosets whos union is the orginal group, you are done. Correct?
"So, once you have found all the cosets whos union is the orginal group, you are done. Correct?" Well, yes...obviously!
The question here is to understand that a subgroup of a group defines an equivalence relation on the group
whose equivalence classes are precisely the left (or right) cosets of the group wrt that sbgp.
Not only that: if $\displaystyle H\leq G\,,\,x,y,\in G\,,\,then\,\,xH=yH\Longleftrightarrow y^{-1}x\in H$ , so you could check easily whether
two of the sets you were given are or not different cosets.
For example, $\displaystyle a<a^5>=a^3<a^5>\Longleftrightarrow a^{-3}a\in<a^5>$ . But $\displaystyle a^{-3}a=a^{-2}=a^{13}\notin <a^5>\Longrightarrow$ the cosets
are different.
Tonio | 858 | 2,504 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-26 | latest | en | 0.816539 |
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X | 464 | 2,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.927785 |
http://mathhelpforum.com/differential-geometry/169553-every-0-there-such-z.html | 1,481,019,953,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541896.91/warc/CC-MAIN-20161202170901-00159-ip-10-31-129-80.ec2.internal.warc.gz | 176,025,757 | 12,907 | # Thread: For every ε>0 there is an a∈A such that a<z+ε
1. ## For every ε>0 there is an a∈A such that a<z+ε
So there are a few questions that go along with this but I've hopefully worked them out correctly so I am only posting the ones I definitely have no clue about.
Let A be a bounded set of reals and z a real number such that the following holds: For every ε>0 there is an a∈A such that a<z+ε (if it is true, prove it. If not, then show a counterexample.
(1) inf A $\leq$ z
I said it was TRUE but I don't know how I would prove it.
(2) inf A $\geq$ z
This one I have no clue about. I'm assuming its FALSE because I said the above statement is true.
(3) sup A $\leq$ z
I said True by trichotomy..but I know that isn't a proof so how would I prove this one as well.
Thanks!
2. Denote $m=\inf A\;,\;M=\sup A$
If $m>z$ choose $\epsilon=m-z$ .
(2) False. Choose $A=[0,1)\;,\;z=1$ .
(3) False. Choose $A=[0,1)\;,\;z=1/2$ .
Fernando Revilla
3. Thanks! Now if the same conditions hold For every $\epsilon$>0 there is an a $\in$A such that a<z+ $\epsilon$: what about..
1. A has an element a $\geq$z
I said false: [-1,1]=A , z=1 (or would it be the open set (-1,1)?)
2. Inf A=z
I said false: (0,1)=A, z=1
3. Sup A=z
I said false: [-2,-1] z=0 ( would this counterexample also work for sup A $\geq$z as well?)
4. Originally Posted by alice8675309
Thanks! Now if the same conditions hold For every $\epsilon$>0 there is an a $\in$A such that a<z+ $\epsilon$: what about..
1. A has an element a $\geq$z
I said false: [-1,1]=A , z=1 (or would it be the open set (-1,1)?)
The above, 1), is true. In #2 is a proof of that.
5. Originally Posted by Plato
The above, 1), is true. In #2 is a proof of that.
Sorry, I did read reply #2 but I was honestly checking the answers I had. I think I'll have to draw a picture to connect everything.
6. Originally Posted by Plato
The above, 1), is true. In #2 is a proof of that.
Now looking at my paper, the reason I had posted the question A has an element a $\geq$z and said it was false was because I had another question with the same conditions stated in the first post that said:
A has an element a $\leq$ z and I said True because if a=z then a<z+ $\epsilon$ holds for all $\epsilon$>0 and if a <z then a<z+ $^epsilon$ holds for all $\epsilon$>0.
Do I have them backwards? Which one is true and which one is false and what would be a counterexample for the false one?
7. Given that $\left( {\forall \varepsilon > 0} \right)\left( {\exists a \in A} \right)\left[ {z \leqslant a < z + \varepsilon } \right]$.
If $m=\text{inf}(A)$ suppose that it is true that $z. Then let $\varepsilon =(m-z)>0$.
So $\left( {\exists a \in A} \right)\left[ {z \leqslant a < z + (m - z)} \right]$.
But that says $a. What is wrong with that?
Thus we have that $m\le z.$
8. Originally Posted by Plato
Given that $\left( {\forall \varepsilon > 0} \right)\left( {\exists a \in A} \right)\left[ {z \leqslant a < z + \varepsilon } \right]$.
If $m=\text{inf}(A)$ suppose that it is true that $z. Then let $\varepsilon =(m-z)>0$.
So $\left( {\exists a \in A} \right)\left[ {z \leqslant a < z + (m - z)} \right]$.
But that says $a. What is wrong with that?
Thus we have that $m\le z.$
Ohhh I see it now! So for the one that states A has an element a $\leq$z does this work as a counterexample? A=[0,1) z=1?
9. Originally Posted by alice8675309
Let A be a bounded set of reals and z a real number such that the following holds: For every ε>0 there is an a∈A such that a<z+ε (if it is true, prove it. If not, then show a counterexample. (1) inf A $\leq$ z
(2) inf A $\geq$ z
(3) sup A $\leq$ z
Above I have quoted the OP.
I answered (1), proved it is true.
Both (2) & (3) are false.
To show that we need to find an example of both the set A and the value z.
Let $A=[0,1]$ then $0=\inf(A)~\&~1=\sup(A)$.
If we choose $z=\frac{1}{2}$ both are false.
10. Originally Posted by Plato
Above I have quoted the OP.
I answered (1), proved it is true.
Both (2) & (3) are false.
To show that we need to find an example of both the set A and the value z.
Let $A=[0,1]$ then $0=\inf(A)~\&~1=\sup(A)$.
If we choose $z=\frac{1}{2}$ both are false.
Thanks so much! I'm sorry if it got confusing because I did post three additional questions in reply #3 which I guess I should have started a new thread for. However, are the counterexamples in reply #3 for 2 and 3 by any chance correct? Again thanks you've been a big help! This topic isn't my strongest | 1,456 | 4,472 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 47, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2016-50 | longest | en | 0.961494 |
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If the numbers 5, 7 and 4 go into this function machine, what numbers will come out?
Shapes in a Grid
Stage: 2 Challenge Level:
Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?
Rabbits in the Pen
Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
How Do You Do It?
Stage: 2 Challenge Level:
This group activity will encourage you to share calculation strategies and to think about which strategy might be the most efficient.
Fingers and Hands
Stage: 2 Challenge Level:
How would you count the number of fingers in these pictures?
Sept03 Sept03 Sept03
Stage: 2 Challenge Level:
This number has 903 digits. What is the sum of all 903 digits?
Being Thoughtful - Primary Number
Stage: 1 and 2 Challenge Level:
Number problems at primary level that require careful consideration. | 1,957 | 8,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2017-39 | latest | en | 0.899621 |
https://interviewmania.com/discussion/27588-verbal-reasoning-matrix-arrangement | 1,721,690,659,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517927.60/warc/CC-MAIN-20240722220957-20240723010957-00459.warc.gz | 271,703,217 | 10,770 | Home » Verbal Reasoning » Matrix Arrangement » Question
#### Matrix Arrangement
Direction: Study the following information carefully to answer these questions.
A, B, C, D, E, F and G are members of a sports club and have liking for different games, viz Carrom, Table Tennis, Badminton, Bridge, Hockey, Football and Lawn Tennis but not necessarily in the same order. Each one of them has a liking for different musical instruments, viz Sitar, Guitar, Harmonium, Flute, Tabla, Banjo and Santoor, not necessarily in the same order. B likes Carrom and Banjo. E likes to play Bridge but not Harmonium or Tabla. The one who plays Hockey plays Sitar. F plays Guitar but not Table Tennis or Lawn Tennis. A plays badminton and Flute. The one who plays Lawn Tennis does not play Tabla. C plays Harmonium and G plays Hockey.
1. Which of the following combinations of gameperson-musical instrument is definitely correct?
2. Table Tennis-E-Santoor
3. Lawn Tennis-D-Tabla
4. Table Tennis-C-Tabla
5. None of these
##### Correct Option: E
As per given question, draw a table with Member, Sports and Instrument. Write the Member , Sports and Instrument in table as per given question.
As per Question,
A plays badminton and Flute. (Write in Table as shown in below figure.)
B likes Carrom and Banjo. (Write in Table as shown in below figure.)
C plays Harmonium.
(Write in Table as shown in below figure.)
E likes to play Bridge.
(Write in Table as shown in below figure.)
F plays Guitar.
(Write in Table as shown in below figure.)
G plays Hockey.
(Write in Table as shown in below figure.)
The one who plays Hockey plays Sitar.
(Write in Table for G as shown in below figure.)
Now Only two Instrument left which are Santoor and Tabla.
E likes to play Bridge but not Harmonium or Tabla.
So E likes to play Santoor.
(Write in Table for E as shown in below figure.)
Now only one Instrument is left which will be play by D.
(Write in Table for D as shown in below figure.)
F plays Guitar but not Table Tennis or Lawn Tennis.
It means Table Tennis or Lawn Tennis play only by C Or D.
As per one more condition,
The one who plays Lawn Tennis does not play Tabla.
D Plays Tabla so C will play only Lawn Tennis.
(Write in Table for D as shown in below figure.) | 538 | 2,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-30 | latest | en | 0.917744 |
https://www.physicsforums.com/threads/photonic-computer.249651/ | 1,527,470,547,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00008.warc.gz | 793,493,628 | 19,756 | # Photonic computer
1. Aug 12, 2008
### UltraViolet
Hello!
My name is Alexandru, i am from Romania student at Mechanics University. I am working on a project and i also try to confirm some theory'es. I recently developed build and test a mechanism using the inertial propulsion principle. It is working but keep it in test stage. Now the advantage of this mechanism is that the acceleration is constant so its only a mater of time until this engine would reach speed of light. What would happen when it reaches the speed of light? I thinked of mounting a laptop whit wireless technology incorporated and programed to connect automatically at any free wireless access point. Is it posible that this machine when reach the speed of light to act as light do? This machine let's say will weigh's 10 Kg. and Einstein's theory say when a body reaches speed of light his mass and gravity will get infinite values and some physicists say this machine will implode in itself because of this big forces... My question: how to implode? I mean when gravity, magnetism and the mass of a body becomes infinite... what is this mean? When you travel from A to B in no time... Einstein sayd time is 0 for a body when it reaches the speed of light... am i wrong?
Maybe there is someone who know's: How manny photons exist in this universe?
I will appreciate all comments, discussion, explanations and answers... anything u have in mind.
Thank you very much!
2. Aug 12, 2008
### Staff: Mentor
Constant according to whom?
How much time?
3. Aug 12, 2008
### tiny-tim
Welcome to PF!
Hello Alexandru! Welcome to PF!
You say that the acceleration is constant …
but have you included the increase in mass?
I suspect that your mechanism provides a constant rate of increase of momentum.
In Newtonian physics, that's d(mv)/dt = constant,
so v = constant x t.
But in relativity, it's:
d(mv)/dt = d(m0v/√(1 - v2/c2)/dt = constant,
so v/√(1 - v2/c2) = constant x t,
and so v/c must always be less than 1.
4. Aug 12, 2008
### UltraViolet
Doc Al
Constant according to whom?
The acceleration of the mechanism, engine or ship how should i call it. :)
Q: How much time?
I am sure i can get an acceleration of 100 Km/h every 3 seconds.
5. Aug 12, 2008
### Staff: Mentor
My point was: Are you talking about a constant proper acceleration or a constant acceleration with respect to some inertial frame? One is possible; the other, not.
Not with respect to an inertial frame. At least not forever.
As tiny-tim explained, it gets harder and harder to continue accelerating something with respect to some inertial frame. The speed will approach the speed of light as a limit as time goes to infinity. It will take forever.
6. Aug 12, 2008
### throng
With regards to your machine. if it were to reach infinite mass Volume=0 it wouldn't exist in a geometric "place."
If it exceeded light speed and time was 0 it would take no time to travel between geometric points (Place to place).
So the machine would actually be everywhere at once but nowhere in the universe.
I don't think this machine would come back - so dont invest too much money.
Ha Ha - Hope it enlivens the imagination
7. Aug 12, 2008
### tiny-tim
Hi throng!
ah … you're referring to the Infinite Improbability Drive!
No, it's ok … when the drive is disconnected from the random motion generator (generally at T-time), the wave-function collapses, and the machine can be observed in a particular position.
It can then supply telemetry, or other useful information, back to Earth.
8. Aug 14, 2008
### UltraViolet
throng! if u exceed speed of light... u mention this, from my little knowledge about Einstein's theory, i remember that Time reaches 0 value at the speed of light... so if u would go further shouldnt time take negative values? i mean this should be a good explanation to time travel... i believe (not convinced) for makin a trip into the past there should be a good physical explanation, and Einstein seem to explain it very well... at the speed of light the ship becomes somethin infinte... i think it will act exactly the same like a photon, light wave... are photons affected by friction? i belive they dont meet obstacles in theyr trajectory .. but light can reflect from objects ... i am curious about installing a wireless laptop on the ship and when it reaches speed of light ... infinite mass, gravity, time reach 0 value ... i think in that fraction of second this ship get's to be present in all time from the time it reaches until the big end :)... gets to be present in all places in universe is present in everything ... maybe.. but that laptop whit wireless technology setup to autoconnect to the first detected wireless access point ... so i mean .. in a moment that ship is present here there and everywhere in the same time and the time that comes ... we should detect manny manny connections .. the same computer gets connected... somethin like cloning .. ) im sure this sound very crazy but .. .every byte, bit from this universe we will know about it .. a supernetwork super computer .. there is a market in my town where there is free wireless internet access but also in that market there are buildings and a statue... made of some metal.. that statue contains a very big number of atoms wich i think they contain also manny photons... on a PC is very simple to delete and to empty trash ..
9. Aug 14, 2008
### UltraViolet
Doc Al
http://www.rexresearch.com/cookip/cookip.htm
"The Conversion of Centrifugal Force Into Linear Force and Motion"
by Robert Cook
lets spin an arm of 30 inch and a weight of 100 grams at the end of the arm, rotateing it whit 100 rpm we get a centrifugal force. Now if we duble the rpm we get some changes.. at 200 rpm the centrifugal force will grow too .. at 3000 rpm the arm may dont resist and broke and that weight if it hits the surface of a scale will show that something heavy is on it... the scale will show there is a weight of 10 kg on it .. until all the energy is disipated... its simple i think: the harder u throw the higher it goes... so let me get back to the arm that is spinning, when it spins it distributes centrifugal force on 360 degrees but if we somehow redirect 180 degrees in the same direction .. then we get a linear motion... it will be very helpfull for me if you know or someone can help me here how to estimate, i need to make some conversions wich i dont know, dont have documentation. How could i calculate the centrifugal force of that spining arm and compare it whit the force needed to lift a weight from the ground? at what RPM that weight of 100 grams will generate enough centrifugal force to equalize the force needed to lift 1 kg from the ground 1 inch.
10. Aug 14, 2008
### UltraViolet
tiny-tim, i wanted to answer you yesterday but didnt had time... yes youre right, but when this ship reaches speed of light dosnt apply Einstein's theory of relativity? the principle of my mechanism is inertial propulsion... its like haveing a metall indestructible ball in your front and whit a gun you fire that ball, now that ball starts to move on a direction given by the bullet you shot... now the shooter (you) will travel the same speed behind the ball and from the same distance you shoot again that ball... its easyer to imagine this in space where friction whit atmosphere is less..
11. Aug 14, 2008
### UltraViolet
tiny-tim
this is how my mechanism act's...
12. Aug 14, 2008
### UltraViolet
tiny-tim
"No, it's ok … when the drive is disconnected from the random motion generator (generally at T-time), the wave-function collapses, and the machine can be observed in a particular position.
It can then supply telemetry, or other useful information, back to Earth."
i strongly belive it needs only to touch only 1 milisecond or much more less time the speed of light because in that moment it becomes infinite mass, gravity and time 0, so it means it will be present in all our future... present in all things the whole solar system will have that ship at the base as a particle that ship.. the ship that reached speed of light .. what i am very curious .. would it be posible for the laptop from that ship to connect wirelessly to a free internet access point? Are photons subatomic particles? i mean do they exist in the composition of an electron or atom nuclei? even so .. there is so much light comming from the sun in day time and light half earth surface and manny wireless spots.. )
13. Aug 14, 2008
Staff Emeritus
Oh boy.
First, this has nothing to do with photonic computers.
Second, the "Cook Inertial Propulsion Engine" you link to is pure crackpottery. As Cook himself says, it violates conservation of momentum.
Third, I simply don't believe you have a working machine that works on the "inertial propulsion principle", more commonly called a reactionless drive. Much less one that will accelerate at ~1g until nearing the speed of light. I notice that you mention that you are "sure you can get it" - not that you've actually done this. The dustbin of history is filled with crackpots whose perpetual motion machines almost worked.
14. Aug 14, 2008
### George Jones
Staff Emeritus
This thread has degenerated too far, so I have closed it. | 2,127 | 9,245 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-22 | longest | en | 0.929481 |
https://wordpandit.com/tricks-for-divisibility-2/ | 1,718,348,663,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00414.warc.gz | 578,097,207 | 50,417 | Select Page
Previously, we have learned the basics of divisibility and various divisibility checks and rules. In this article, we will learn some shortcuts and tricks for divisibility that will help you solve questions quickly. Without wasting time, let’s get going with the tricks.
Some basics tips and tricks you studied in the previous lessons that you can use for divisibility:
• All whole numbers are divisible by 1.
• A number is divisible by 2 if it is even.
• A non-zero number is divisible by 5 if it ends in 0 or 5.
• In order to check the divisibility of a number by a composite number, divide the composite divisor into prime factors, which are co-prime and then check for its divisibility with each. For example, to check the divisibility of a number with 12, break down 12 into 3 and 4.
Tricks for Divisibility:
1. an – bn is always divisible by a-b
Example: 85–55 is divisible by 8-5= 3
Remember it by:
1. an – bn is divisible by a + b when n is even
Example: 710 – 510 is divisible by 7-5= 2
Remember it by:
a3 – b3 is not divisible by a + b
a2 – b2 is also divisible by a + b
a4 – b4 is also divisible by a + b
1. an + bn is divisible by a + b when n is odd
711 + 511 is divisible by 7 + 5 = 12
Remember it by:
a3 + b3 is divisible by a + b
a2 + b2 is NOT divisible by a + b
a4 + b4 is NOT divisible by a + b
1. an + bn + cn is divisible by a + b + c when n is odd.
73 + 53 + 23= 343 + 125 + 8 = 476 divisible by 7 + 5 + 2 = 14
Given below are some questions for you to apply these formulae:
Example 1: 3223 + 1723 is definitely divisible by:
a. 49
b. 15
c. 49 & 15
d. None of these.
As an + bn is divisible by a + b when n is odd
So 3223 + 1723 is divisible by 32 + 17 = 49.
Example 2: 3223 – 1723 is definitely divisible by:
a. 49
b. 15
c. 49 & 15
d. none of these.
As an – bn is always divisible by a-b.
So 3223 – 1723 is divisible by 32 – 17 = 15.
Example 3: 32232 – 17232 is definitely divisible by:
a. 49
b. 15
c. 49 & 15
d. none of these.
As an – bn is always divisible by a-b and an – bn is divisible by a + b when n is even
So 32232 – 17232 is divisible by both 32 – 17 = 15 and 32 + 17 = 49.
Example 4: 35 + 55+ 75 is definitely divisible by:
a. 8
b. 7
c. 15
d. all of these.
As an + bn + cn is divisible by a + b + c when n is odd.
So 35 + 55 + 75 is divisible by 3 + 5 + 7 = 15
Try out some more questions given in the exercise.
EXERCISE:
Question 1: (49)15 –1 is exactly divisible by:
(1) 50 (2) 51 (3) 29 (4) 8
Solution (4)
By property 1
an – bn is always divisible by a-b
(49)15 –(1)15 is exactly divisible by 49–1 = 48, that is a multiple of 8.
Question 2: If a and b are two odd positive integers, by which of the following integers I (a4 –b4) always divisible?
(1) 3 (2) 6 (3) 8 (4) 12
Solution (3)
an – bn is divisible by a + b when n is even
an – bn is always divisible by a-b
a4 – b4 = (a2 + b2) (a + b) (a – b)
put a=3 , b=1 ( because 1,3 are the smallest odd numbers)
Required number = (3 + 1) (3 – 1) = 8
Question 3: m and n are positive integers and (m–n) is an even number, then (m2–n2) will be always divisible by
(1) 4 (2) 6 (3) 8 (4) 12
Solution (1)
Let m – n = 2x
m + n = 2x
(m–n) (m+n) = 4x2
m2–n2 = 4x2
Question 4: It is given that (232 + 1) is exactly divisible by a certain number, which one of the following is also definitely divisible by the same number?
(1) 296 + 1 (2) 7 × 233 (3) 216 – 1 (4) 216 + 1
Solution (1)
Question 5: The greatest whole number, by which the expression n4 + 6n3 + 11n2 + 6n + 24 is divisible for every natural number n, is
(1) 6 (2) 24 (3) 12 (4) 48
Solution (4)
For n = 1
n4 + 6n3 + 11n2 + 6n + 24
=1 + 6 + 11 + 6 + 24 = 48
For, n = 2
n4 + 6n3 + 11n2 + 6n + 24
=16 + 48 + 44 + 12 + 24
=144 which is divisible by 48.
Clearly, 48 is the required number.
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No thanks, I don't want it. | 1,440 | 4,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2024-26 | latest | en | 0.938739 |
http://reference.wolfram.com/legacy/v7/ref/Replace.html | 1,511,389,598,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806676.57/warc/CC-MAIN-20171122213945-20171122233945-00269.warc.gz | 251,876,201 | 9,841 | This is documentation for Mathematica 7, which was
based on an earlier version of the Wolfram Language.
# Replace
Replace[expr, rules]applies a rule or list of rules in an attempt to transform the entire expression expr. Replace[expr, rules, levelspec]applies rules to parts of expr specified by levelspec.
• The rules must be of the form lhs->rhs or lhs:>rhs.
• A list of rules can be given. The rules are tried in order. The result of the first one that applies is returned. If none of the rules apply, the original expr is returned.
• If the rules are given in nested lists, Replace is effectively mapped onto the inner lists. Thus Replace[expr, {{r11, r12}, {r21, ...}, ...}] is equivalent to {Replace[expr, {r11, r12}], Replace[expr, {r21, ...}], ...}.
• Delayed rules defined with :> can contain /; conditions.
• Replace uses standard level specifications:
n levels 1 through n Infinity levels 1 through Infinity {n} level n only {n1,n2} levels n1 through n2
• The default value for levelspec in Replace is {0}, corresponding to the whole expression.
• A positive level n consists of all parts of expr specified by n indices.
• A negative level -n consists of all parts of expr with depth n.
• Level -1 consists of numbers, symbols and other objects that do not have subparts.
• With the option setting , Replace includes heads of expressions, and their parts.
• Replacements are performed to parts specified by levelspec even when those parts have Hold or related wrappers.
Replace by default applies rules only to complete expressions:
It does not map down to subparts:
A list of rules gives a list of results:
Replace by default applies rules only to complete expressions:
Out[1]=
It does not map down to subparts:
Out[2]=
A list of rules gives a list of results:
Out[1]=
Scope (2)
Replace at level 1:
Replace also works with RuleDelayed: | 444 | 1,858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2017-47 | latest | en | 0.870259 |
http://mathematica.stackexchange.com/questions/46214/why-my-differential-equations-become-true | 1,469,286,386,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823072.2/warc/CC-MAIN-20160723071023-00219-ip-10-185-27-174.ec2.internal.warc.gz | 160,958,577 | 19,421 | # Why my differential equations become True? [duplicate]
This question already has an answer here:
I've been trying to solve a system of nonlinear differential equation, but the conditions are a bit weird.
Two of the differentials equate to the same equation, but have different boundary conditions. The third differential equate to something else.
For the background on the problem I am trying to solve, it is related to concentration as a function of time for a reaction.
Here is the code I have put, which was more or less straightforward, but I keep getting True. I've tried using DSolve and NDsolve, but I am unable to get a graph for my three concentration profiles (I do not need an analytical solution, just a graphical one).
NDSolve[{
A'[t] == -kf A[t] *B[t] + kb * c[t],
B'[t] == -kf A[t] *B[t] + kb * c[t],
c'[t] == kb * c[t] + kf A[t] B[t],
A[0] == 1, B[0] == 2, c[0] == 0},
{A, B, c}, {t, 100}]
kf and kb are constants I've already equated into my code.
But I keep resulting with:
NDSolve[{True, True, True, A[0] == 1, B[0] == 2, c[0] == 0}, {A, B, c}, {t, 100}]
I'm thinking it's because I'm multiplying two functions together, I haven't solved ODEs this way. I'd appreciate any input, or hints/guides/direction. Thanks!
If there is something unclear, or if I am lacking proper question asking etiquette please let me know!
-
## marked as duplicate by Michael E2 plotting StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 17 '15 at 15:04
Then you must have once mixed up = and ==. You need Clear[Derivative]. Values stored in Derivative can't be cleared by Clear["Global*"] because it actually means clearing all the variables under context "Global" while Context[Derivative] gives "System". – xzczd Apr 16 '14 at 7:15
@xzczd New users might fall in to this Context trap. I haven't found another post to mark this as a duplicate. Perhaps your comment is worthy of an answer just so it is in the books? – bobthechemist Apr 16 '14 at 13:12
@bobthechemist OK, let me elaborate it into an answer. – xzczd Apr 16 '14 at 13:42
Apparently, I got to close this because it was originally tagged "plotting", since removed. I meant to just make it simple vote. If there's disagreement, respond or flag and maybe a moderator can help. – Michael E2 May 17 '15 at 15:08
Clear[Derivative] first.
OK, it's surprising that there seems to be no regular answer to this common problem for beginners, let me elaborate my comment into an answer. If you restart your Mathematica and run your code again then you'll find your problem no longer exists anymore! Then, why? Because Mathematica is unstable?
Of course not. Just recall what you've done before meeting the error, you may find the following scene in the corner of memory:
The value of kf and kb is chosen at random.
It's another common mistake of Mathematica newbies, that is, mix up = and ==.
Your code was in a mess now, but you didn't feel worried because you've already learned from some material that you can use
Clear["Global*"]
to fix this. You cheerfully placed this line at the beginning of your code and ran it again, only to find:
What happened? Why does the magic of Clear["Global*"] lose its power?
To answer this I'd like to first explain why you need to clear the variables after you mistyped == to = since I presumingly guess you may still be vague about this. Try the following code:
A[0] = 0
A[0]
A[0] == 0
Clear[A]
A[0] == 0
0
0
True
A[0] == 0
BTW, though I've execute A[0] for several times to show the variation, you can judge whether A owns a value just by observing its color: it's black when it does, otherwise it'll be blue.
As you see, Set (=) will give value to its left hand side. (Of course there exist cases that the left hand side is Protected, for example a + b = 4 won't give the left side a value, but it's another story and I'd like not to talk about it here. ) If you don't Clear it, it'll be always there and break your equations. When introducing this issue, many materials will tell readers that you can simply use Clear["Global*"] to "clear all the variables" at once, but it doesn't work for your case, yeah, you already know that, but do you know what's the exact meaning of Clear["Global*"]?
If you check "Details" of document of Clear, you'll find the following description:
Clear["context*"] clears all symbols in a particular context.
What's "context"? "context" is something that every symbol in Mathematica owns. You can check it by function Context, for example:
Context[a]
Context[NDSolve]
"Global"
"System"
So Clear["Global*"] is clearing the values of symbols under context Global, which is the stronghold for most symbols that'll have values (at least for beginners), but, what you can Clear is only "most", not "all". Try the following code:
A'[0] = 0
A'[0]
Clear["Global*"]
A'[0]
Clear[A]
A'[0]
HoldForm[FullForm[A'[0]]]
0
0
0
0
Derivative[1][A][0]
Aha, A'[0] is one of the exceptions, the value of A'[0] isn't stored in A, in fact it's stored as SubValues (this is another story that I'd like to omit here, you can search it in this site or have a look at Leonid Shifrin's excellent book ) of Derivative, then what's the context of Derivative?:
Context[Derivative]
"System"
So values in it can't be cleared by Clear["Global*] (BTW 2, for most cases it can be shortened as Clear["*"], meaning clear all the values of symbols under the current context, which is usually Global), what you need to clear it is
Clear[Derivative]
Though some warnings generated, Clear["System*"] can be used too if you like.
BTW 3, another symbol that is likely to trig this problem is Subscript.
-
@xzcxd: I must say this an excellent answer, one of the most helpful I have read. – David Nov 21 '14 at 17:33
I think it is a Mathematica design error to store derivative of a user symbol in the system context. Everything relating to user symbols/variables should be kept outside system context. Nothing should leak outside like with this example. – Nasser Nov 27 '14 at 5:29 | 1,687 | 6,467 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2016-30 | latest | en | 0.852017 |
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Liberty University ECON 213 Problem Set 3 complete solutions correct answers A+ work
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Liberty University ECON 213 Problem Set 3 complete solutions correct answers A+ work
1. Data for the market for graham crackers is shown below. Calculate the elasticity of demand between the following prices.
Price of crackers
Quantity Demanded (per month)
\$1.00–\$1.50: ___________________________________
\$1.50–\$2.00: ___________________________________
\$2.00–\$2.50: ___________________________________
\$2.50–\$3.00: ___________________________________
Now, assume the price of graham crackers is \$2.75. Should firms raise or lower their prices if they want to increase revenue? Explain this in terms of elasticity.
2. Assume the competitive market shown below faces a short run price of \$10. Using the graph below, identify the following:
Profit-maximizing output: _______________________
In the long run, the price falls to \$7.50. Why does this happen?
What is the new profit-maximizing output? _______________________
3. A local hardware store is trying to decide whether to stay open. They have found that their industry is extremely competitive and profits have shrunk considerably. Knowing that you have taken an economics course, the owners have asked for your opinion. Draw 2 completely labeled graphs to help you explain the shutdown decision. One graph must be for the market as a whole, and the other must be for this store in particular. Assume that the store is losing money; however, explain why they may want to stay open for a little while longer. (Note: Your answer should include a written explanation of your graph.)
4. Use the production function below to answer the following questions:
Units of Labor
Total Output
MP
a.) Calculate marginal productivity (MP) and put this in the table.
b.) At what level of employment does diminishing marginal productivity begin?
c.) At what level of employment does marginal productivity become negative?
d.) Why does marginal product become negative?
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Liberty University ECON 213 Problem Set 3 complete solutions correct answers A+ work
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Liberty University ECON 213 Problem Set 3 complete solutions correct answers A+ work 1. Data for the market for graham crackers is shown below. Calculate the elasticity of demand between the following prices. Price of crackers Quantity Demanded (per month) \$1.00–\$1.50: ___________________________________ \$1.50–\$2.00: ___________________________________ \$2.00–\$2.50: ___________________________________ \$2.50–\$3.00: ___________________________________ Now,...
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Plot the following two lines: Y = 35 – X and Y = 5 + 2X Find the point where the two lines intersect each other, and find the points of intersection with the x and y axes. Solution: To plot the lines, select a value for x and solve the equation for y to yield point (x,y). Repeat the process to find another point, and draw a line through the two points. NOTE: These are linear equations (equations representing straight lines) in the form of Y= b + mX , where b is the y-intercept and m is the slope of the line. Ex: Let X = 1; then Y = 35 – 1; Y = 34; (X,Y) : (1, 34) (A) and Let X = 20; then Y = 35 – 20; Y = 15; (X,Y) : (20, 15) (B) The line Y = 35 – X passes through points (1, 34) and (20, 15). Let X = 1; then Y = 5 + 2(1); Y = 7 (X,Y) : (1, 7) (C) Let X = 20; then Y = 5 + 2(20); Y = 45 (X,Y) : (20, 45) (D) The line Y = 5 + 2X passes through points (1, 7) and (20, 45) NOTE: It is often easiest to plug 0 in for x and then for y, but the above example shows that this technique will work with any values you choose. A
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# A test $$1.6 \times 10^{-19} Cb$$ is moving with velocity $$\overrightarrow v = ( 2 \hat i +3 \hat j)$$ m/sec is magnetic field $$\overrightarrow B = ( 2 \hat i + 3 \hat j ) wb /m^2$$ .the magnetic force on the test charge:-
A
$$6 \hat k T$$
B
$$( 4 \hat i + 6 \hat j ) T$$
C
$$( 4 \hat i + 6 \hat j ) \times 10^{-19} T$$
D
zero
Solution
Verified by Toppr
#### Correct option is D. zeroGivenCharge=$$1.6*10^{-19}$$v=(2i+3j) m/sB=(2i+3j) wb/mSolutionForce=q(vxB)Cross product of v and B is 0Therefore F is 0The correct option is D
13
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Q1
A test $$1.6 \times 10^{-19} Cb$$ is moving with velocity $$\overrightarrow v = ( 2 \hat i +3 \hat j)$$ m/sec is magnetic field $$\overrightarrow B = ( 2 \hat i + 3 \hat j ) wb /m^2$$ .the magnetic force on the test charge:-
View Solution
Q2
The magnetic force acting on a charged particle of charge2μC in a magnetic field of 2T acting in y-direction, when the particle velocity is (2^i+3^j) × 106 ms1 is :
View Solution
Q3
A charge q=4μC has an instantaneous velocity v=(2^i3^j+^k)×106m/s in a uniform magnetic field B=(2^i+5^j3^k)×102T.What is the force on the charge?
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If an electron moves with a velocity, $$\vec{v} = (2\hat{i} - 3\hat{j} + \hat{k}) \,m/s$$ in a magnetic field, $$\vec{B} = (2\hat{i} + \hat{j} - \hat{k})T.$$ then, calculate the Lorentz force.
View Solution
Q5
A particle of mass 1 mg and having charge 1 μC is moving in a magnetic field, B=(2^i+3^j+^k) T with velocity v=(2^i+^j^k) km/s. Find the magnitude of acceleration.
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# Re:Re: st: confidence intervals of predicted risk after stcox
From petretta@unina.it From Steve Samuels To statalist@hsphsun2.harvard.edu Subject Re:Re: st: confidence intervals of predicted risk after stcox Date Wed, 31 Oct 2012 10:21:49 +0100 Date Tue, 30 Oct 2012 16:09:05 -0400
```Many thanks to Steve Samuels for the explanation.
I apologize for the previous private replay.
Mario
Mario wrote to me privately. When this happens I refer the writer to the
FAQ paragraph that starts "Do not request private replies...".
My problems are:
1. If i gen survival probability and not risk, the upper and lower
limits of ci seem inverted:
```
They should be inverted. You previously computed CIs for the failure ("risk") curve, now you are computing the CIs for the survival curve. A high risk for failure implies a low risk for survival and vice-versa. This is mathematically so because F = 1 -S, as you
```previously calculated. Thus the upper and lower limits for the two curves are
switched.
[Show Quoted Text - 33 lines][Nascondi Testo quotato]
Gen survlb15 = base15^exp(lb)
gen survub15 = base15^exp(ub)
gen surv15 = base15^exp(xb)
sum survlb15 surv15 survub15
variable | obs mean std. Dev. Min max
-------------+--------------------------------------------------------
survlb15 | 48 .7716674 .2110915 .381534 .9887756
surv15 | 48 .5276728 .343838 .0012602 .9593767
survub15 | 48 .2910488 .3057171 8.11e-21 .8586743
2. To compare the survival result at 15 months obtained in subjects
with age40 = 20 and drug = 3, i used the stata command survci
(-survci- sj11-4: St0217_1 ? Initially written by yulia marchenko of
statacorp and thereafter adjourned by matthew cefalu (the stata
journal (2011), number 1, pp. 64?81 pointwise confidence intervals
for the covariate-adjusted survivor function in the cox model).
The code i used is:
Survci, at(age40=20 drug=3) outfile(age20_drug3).
The results of survci at 15 months are:
_t _d _surv _se _lb _ub
15 1 .85324285 .07846486 .6101214 .95029744
those of my code are the same as regard surv15 but very different as
regard the confidence intervals (lower and upper)
surv15 survlb15 survub15
.8532429 .9722255 4089079
There are two mistakes here.
1. The minor one is that the -survci- gives CIs for someone
with with age = 20 and drug = 3, whereas the *15 values
you show are averages over the entire sample. There's no
reason to expect similarity.
2. The major mistake is that your original CIs reflect only variability of
the estimated betas, whereas -survci- properly takes into account the fact
that the baseline survival probability is also an estimate.
Steve
[Show Quoted Text - 59 lines][Nascondi Testo quotato]
-----------------------------------------
date: Fri, 26 oct 2012 17:15:04 -0400
from: Steve samuels <sjsamuels@gmail.Com>
subject: Re: St: Confidence intervals of predicted risk after stcox
mario, when i run your code, and sum the new variables, i get:
Variable | obs mean std. Dev. Min max
- -------------+--------------------------------------------------------
lb15 | 48 .2283326 .2110915 .0112244 .6184661
risk15 | 48 .4723272 .343838 .0406233 .9987398
ub15 | 48 .7089512 .3057171 .1413257 1
i see nothing strange about them. What is it that does not make sense to you?
Steve
on oct 26, 2012, at 6:45 am, petretta@unina.It wrote:
Hi all,
i use stata/ic 12.1 for windows (32-bit).
I run a cox model to calculate for each patient the predicted probability
of event at 15 months, using predict xb, predict basesurv and adjiusting
the baseline risk
webuse cancer, clear
gen age40=age-40
stcox age40 i.Drug
predict xb, xb
predict basesurv, basesurv
sum basesurv if _t<16
scalar base15 = r(min)
gen risk15 = 1 - base15^exp(xb)
i ask if it is possible to have also the 95% confidence intervals of these
estimate.
I try:
Predict se_xb, stdp
gen lb = xb - invnormal(0.975)*se_xb
gen ub = xb + invnormal(0.975)*se_xb
gen lb15 = 1 - base15^exp(lb)
gen ub15 = 1 - base15^exp(ub)
but the results seems do not have sense.
Thanks for your consideration.
Mario petretta
dpt. of internal medicine, cardiology and heart surgery
naples university federico ii -italy
Mario, when I run your code, and sum the new variables, I get:
Variable | Obs Mean Std. Dev. Min Max
- -------------+--------------------------------------------------------
lb15 | 48 .2283326 .2110915 .0112244 .6184661
risk15 | 48 .4723272 .343838 .0406233 .9987398
ub15 | 48 .7089512 .3057171 .1413257 1
I see nothing strange about them. What is it that does not make sense to you?
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/faqs/resources/statalist-faq/
* http://www.ats.ucla.edu/stat/stata/
Mario Petretta
Dipartimento di Medicina Clinica Scienze Cardiovascolari e Immunologiche
Facoltà di Medicina e Chirurgia
Università di Napoli Federico II
081 - 7462233
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/faqs/resources/statalist-faq/
* http://www.ats.ucla.edu/stat/stata/
```
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# 43895276 What is the missing number in the 3×3 magic square?
Open in App
Solution
## The sum of numbers in any row or column or diagonal of the magic square is the same. The sum of numbers in 1st row = 4 + 3 + 8 = 15. It is same for all rows, columns and diagonals. So in 2nd row, 9 + 5 is 14, the missing number is 15−14=1.
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### Volume and Surface Area of a Box
```Date: 04/06/2008 at 18:04:46
From: Mary
Subject: volume: why doesn't a sheet of notebook paper, folded into..
Why doesn't a sheet of notebook paper folded into a square tube give
the same volume if folded into a square tube "the other way" ??
If I took a piece of paper 10' x 20' and folded it into a square tube
10' long, with four sides measuring 5' across (width) the volume = 10'
x 5' x 5' = 250 cuft.
If I took the same paper and made a square tube 20' long, with four
sides measuring 2.5' across (width) the volume = 20' x 2.5' x 2.5' =
62.5 cuft. Why don't I get the same volume ?
Ok--to assume it would be the same volume was a bad premise, as the
formula proves...but wouldn't a box-maker, using the commodity of
cardboard, get the same volume if he used the same surface area of
cardboard IN ANY BOX CONFIGURATION HE DESIRED ?
So if I am trying to find an equality between two different shaped
boxes, and am now willing to include the surface area consideration
of the ends (I didn't consider the ends at all in the square tube),
shouldn't I be able to say that the surface area of one box = the
surface area of the other box, and then they would have the same
volume ?? So that a box-maker could use the same amount of cardboard
any which way they wanted and get the same volume ?
But that didn't work either--why not ? Shouldn't the invested
cardboard material, when shaped into a rectangular shape, gain the
same volume no matter the side and length dimensions chosen ? Can
you explain that ?
Here's the work telling me my second assumption (equal material
investment produces equal volume) is also wrong:
sa = 2(lw) + 2(hw) + 2(lh)
box measures l x w x h = 6' x 4' x 3'.
sa = 2(6' x 4') + 2(3' x 4') + 2(6' x 3') = 108 sqft
(box volume = 3' x 4' x 6' = 72 cuft)
make a second box using 108 sqft of cardboard
make box measure l x w x h = 6' x 2' x h > > solve for h
108 sqft = sa = 2(6' x 2') + 2(h' x 2') + 2(6' x h')
108 sqft = 24 sqft + 4'h' + 12'h'
108-24 sqft = 16'h
84/16 = h = 5.25'
for the investment in 108 sqft cardboard, this volume will be less,
at 6' x 2' x 5.25' = 63 cuft
Why doesn't that work ?
How does the box maker know how to maximize volume per cardboard
investment ?
I never made this comparison before and always assumed if you folded
a paper "the other way", you would get the same volume.
Can you explain ?
Thank you.
Mary
```
```
Date: 04/06/2008 at 22:50:07
From: Doctor Peterson
Subject: Re: volume: why doesn't a sheet of notebook paper, folded into..
Hi, Mary.
Thanks for writing to Dr. Math. You wrote:
>..but wouldn't a box-maker, using the commodity of cardboard, get the
>same volume if he used the same surface area of cardboard IN ANY BOX
>CONFIGURATION HE DESIRED ?
This is why math is needed; what SEEMS correct on the surface is often
wrong, so we can't trust our intuition. We need to be able to check it.
It simply is not true that when the surface area is the same, the
volume will be the same. The same is true of plane figures--you can
have figures with the same perimeter, but different areas.
Let's take a simple example. Suppose I have a sealed "zip-lock"
plastic bag. At first, the way it comes, it has no air in it; its
volume is zero. But if you fill it with something--let's say,
leftover beans--it puffs out to a new shape with THE SAME SURFACE
AREA (since it's still made of the same plastic, which didn't stretch,
just bent). But its volume is now significantly greater than zero!
So the volume depends on the shape, not just the surface area.
>here's the work telling me my second assumption (equal material
>investment produces equal volume) is also wrong:
>
>sa = 2(lw) + 2(hw) + 2(lh)
> box measures l x w x h = 6' x 4' x 3'.
>
>sa = 2(6' x 4') + 2(3' x 4') + 2(6' x 3') = 108 sqft
> (box volume = 3' x 4' x 6' = 72 cuft)
>
>make a second box using 108 sqft of cardboard
> make box measure l x w x h = 6' x 2' x h > > solve for h
>
>108 sqft = sa = 2(6' x 2') +2(h' x 2') + 2(6' x h')
>108 sqft = 24 sqft + 4'h' + 12'h'
>108-24 sqft = 16'h'
>84/16 = h = 5.25'
>for the investment in 108 sqft cardboard, this volume will be less,
>at 6' x 2' x 5.25' = 63 cuft
>
>why doesn't that work ?
You can see even more easily that it won't work. At the extreme, you
could have h=0, and SA = 2lw can still be 108 (with l=6 and w=9, for
example); but the volume is now zero! That's a simpler version of
what I did with the zip-lock bag. If it can be either zero or
non-zero with different heights, it clearly isn't constant! (This is
a standard way to test a conjecture quickly, because the extremes make
the difference more obvious.)
>How does the box maker know how to maximize volume per cardboard
>investment ?
There are some standard problems in calculus that involve maximizing
the volume of a container. You do this by writing an equation that
says the surface area is a certain value (or whatever constraint the
problem involves--a variant is that a box is being made out of a
certain size piece of cardboard with some parts cut out) and use that
to relate the variables involved (e.g. the length, width, and height),
and then use that relationship to express the volume as a function of
all but one of the variables. Then you can use differentiation to
find the maximum of this function.
>I never made this comparison before and always assumed if you folded
>a paper "the other way", you would get the same volume.
As an engineer friend of mine used to say constantly, NEVER ASSUME
ANYTHING! Now you know.
If you have any further questions, feel free to write back.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
```
Date: 04/07/2008 at 10:51:06
From: Mary
Subject: volume: why doesn't a sheet of notebook paper, folded into..
Thank you for your answer. Today, I will make a contribution.
Your zip-lock example is a good image for me--I can accept zero volume
of the flattened bag, so that works for me.
And the reminder to use the extreme examples (set one parameter to
zero) is good to know, too. THANK YOU for being there!
Mary
```
Associated Topics:
Middle School Measurement
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https://developer.apple.com/library/mac/documentation/GraphicsImaging/Reference/CGAffineTransform/index.html | 1,464,359,228,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049276780.5/warc/CC-MAIN-20160524002116-00169-ip-10-185-217-139.ec2.internal.warc.gz | 940,551,290 | 11,583 | Quartz 2D Reference Collection CGAffineTransform Reference
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# CGAffineTransform Reference
The `CGAffineTransform` data structure represents a matrix used for affine transformations. A transformation specifies how points in one coordinate system map to points in another coordinate system. An affine transformation is a special type of mapping that preserves parallel lines in a path but does not necessarily preserve lengths or angles. Scaling, rotation, and translation are the most commonly used manipulations supported by affine transforms, but skewing is also possible.
Quartz provides functions that create, concatenate, and apply affine transformations using the `CGAffineTransform` data structure. For information on how to use affine transformation functions, see Quartz 2D Programming Guide.
You typically do not need to create an affine transform directly—CGContext Reference describes functions that modify the current affine transform. If you don’t plan to reuse an affine transform, you may want to use `CGContextScaleCTM`, `CGContextRotateCTM`, `CGContextTranslateCTM`, or `CGContextConcatCTM`.
### Functions
• ``` CGAffineTransformMake ```
Returns an affine transformation matrix constructed from values you provide.
#### Declaration
Objective-C
``` CGAffineTransform CGAffineTransformMake ( CGFloat a, CGFloat b, CGFloat c, CGFloat d, CGFloat tx, CGFloat ty ); ```
#### Parameters
``` a ``` The value at position [1,1] in the matrix. ``` b ``` The value at position [1,2] in the matrix. ``` c ``` The value at position [2,1] in the matrix. ``` d ``` The value at position [2,2] in the matrix. ``` tx ``` The value at position [3,1] in the matrix. ``` ty ``` The value at position [3,2] in the matrix.
#### Return Value
A new affine transform matrix constructed from the values you specify.
#### Discussion
This function creates a `CGAffineTransform` structure that represents a new affine transformation matrix, which you can use (and reuse, if you want) to transform a coordinate system. The matrix takes the following form:
Because the third column is always `(0,0,1)`, the `CGAffineTransform` data structure returned by this function contains values for only the first two columns.
If you want only to transform an object to be drawn, it is not necessary to construct an affine transform to do so. The most direct way to transform your drawing is by calling the appropriate `CGContext` function to adjust the current transformation matrix. For a list of functions, see CGContext Reference.
#### Availability
Available in OS X v10.0 and later.
• ``` CGAffineTransformMakeRotation(_:) CGAffineTransformMakeRotation ```
Returns an affine transformation matrix constructed from a rotation value you provide.
#### Declaration
Swift
``` func CGAffineTransformMakeRotation(_ angle: CGFloat) -> CGAffineTransform ```
Objective-C
``` CGAffineTransform CGAffineTransformMakeRotation ( CGFloat angle ); ```
#### Parameters
``` angle ``` The angle, in radians, by which this matrix rotates the coordinate system axes. In iOS, a positive value specifies counterclockwise rotation and a negative value specifies clockwise rotation. In OS X, a positive value specifies clockwise rotation and a negative value specifies counterclockwise rotation.
#### Return Value
A new affine transformation matrix.
#### Discussion
This function creates a `CGAffineTransform` structure, which you can use (and reuse, if you want) to rotate a coordinate system. The matrix takes the following form:
The actual direction of rotation is dependent on the coordinate system orientation of the target platform, which is different in iOS and OS X. Because the third column is always `(0,0,1)`, the `CGAffineTransform` data structure returned by this function contains values for only the first two columns.
These are the resulting equations that Quartz uses to apply the rotation to a point (x, y):
If you want only to rotate an object to be drawn, it is not necessary to construct an affine transform to do so. The most direct way to rotate your drawing is by calling the function `CGContextRotateCTM`.
#### Availability
Available in OS X v10.0 and later.
• ``` CGAffineTransformMakeScale(_:_:) CGAffineTransformMakeScale ```
Returns an affine transformation matrix constructed from scaling values you provide.
#### Declaration
Swift
``` func CGAffineTransformMakeScale(_ sx: CGFloat, _ sy: CGFloat) -> CGAffineTransform ```
Objective-C
``` CGAffineTransform CGAffineTransformMakeScale ( CGFloat sx, CGFloat sy ); ```
#### Parameters
``` sx ``` The factor by which to scale the x-axis of the coordinate system. ``` sy ``` The factor by which to scale the y-axis of the coordinate system.
#### Return Value
A new affine transformation matrix.
#### Discussion
This function creates a `CGAffineTransform` structure, which you can use (and reuse, if you want) to scale a coordinate system. The matrix takes the following form:
Because the third column is always `(0,0,1)`, the `CGAffineTransform` data structure returned by this function contains values for only the first two columns.
These are the resulting equations that Quartz uses to scale the coordinates of a point (x,y):
If you want only to scale an object to be drawn, it is not necessary to construct an affine transform to do so. The most direct way to scale your drawing is by calling the function `CGContextScaleCTM`.
#### Availability
Available in OS X v10.0 and later.
• ``` CGAffineTransformMakeTranslation(_:_:) CGAffineTransformMakeTranslation ```
Returns an affine transformation matrix constructed from translation values you provide.
#### Declaration
Swift
``` func CGAffineTransformMakeTranslation(_ tx: CGFloat, _ ty: CGFloat) -> CGAffineTransform ```
Objective-C
``` CGAffineTransform CGAffineTransformMakeTranslation ( CGFloat tx, CGFloat ty ); ```
#### Parameters
``` tx ``` The value by which to move the x-axis of the coordinate system. ``` ty ``` The value by which to move the y-axis of the coordinate system.
#### Return Value
A new affine transform matrix.
#### Discussion
This function creates a `CGAffineTransform` structure. which you can use (and reuse, if you want) to move a coordinate system. The matrix takes the following form:
Because the third column is always `(0,0,1)`, the `CGAffineTransform` data structure returned by this function contains values for only the first two columns.
These are the resulting equations Quartz uses to apply the translation to a point (x,y):
If you want only to move the location where an object is drawn, it is not necessary to construct an affine transform to do so. The most direct way to move your drawing is by calling the function `CGContextTranslateCTM`.
#### Availability
Available in OS X v10.0 and later.
• ``` CGAffineTransformTranslate(_:_:_:) CGAffineTransformTranslate ```
Returns an affine transformation matrix constructed by translating an existing affine transform.
#### Declaration
Swift
``` func CGAffineTransformTranslate(_ t: CGAffineTransform, _ tx: CGFloat, _ ty: CGFloat) -> CGAffineTransform ```
Objective-C
``` CGAffineTransform CGAffineTransformTranslate ( CGAffineTransform t, CGFloat tx, CGFloat ty ); ```
#### Parameters
``` t ``` An existing affine transform. ``` tx ``` The value by which to move x values with the affine transform. ``` ty ``` The value by which to move y values with the affine transform.
#### Return Value
A new affine transformation matrix.
#### Discussion
You use this function to create a new affine transform by adding translation values to an existing affine transform. The resulting structure represents a new affine transform, which you can use (and reuse, if you want) to move a coordinate system.
#### Availability
Available in OS X v10.0 and later.
• ``` CGAffineTransformScale(_:_:_:) CGAffineTransformScale ```
Returns an affine transformation matrix constructed by scaling an existing affine transform.
#### Declaration
Swift
``` func CGAffineTransformScale(_ t: CGAffineTransform, _ sx: CGFloat, _ sy: CGFloat) -> CGAffineTransform ```
Objective-C
``` CGAffineTransform CGAffineTransformScale ( CGAffineTransform t, CGFloat sx, CGFloat sy ); ```
#### Parameters
``` t ``` An existing affine transform. ``` sx ``` The value by which to scale x values of the affine transform. ``` sy ``` The value by which to scale y values of the affine transform.
#### Return Value
A new affine transformation matrix.
#### Discussion
You use this function to create a new affine transformation matrix by adding scaling values to an existing affine transform. The resulting structure represents a new affine transform, which you can use (and reuse, if you want) to scale a coordinate system.
#### Availability
Available in OS X v10.0 and later.
• ``` CGAffineTransformRotate(_:_:) CGAffineTransformRotate ```
Returns an affine transformation matrix constructed by rotating an existing affine transform.
#### Declaration
Swift
``` func CGAffineTransformRotate(_ t: CGAffineTransform, _ angle: CGFloat) -> CGAffineTransform ```
Objective-C
``` CGAffineTransform CGAffineTransformRotate ( CGAffineTransform t, CGFloat angle ); ```
#### Parameters
``` t ``` An existing affine transform. ``` angle ``` The angle, in radians, by which to rotate the affine transform. In iOS, a positive value specifies counterclockwise rotation and a negative value specifies clockwise rotation. In OS X, a positive value specifies clockwise rotation and a negative value specifies counterclockwise rotation.
#### Return Value
A new affine transformation matrix.
#### Discussion
You use this function to create a new affine transformation matrix by adding a rotation value to an existing affine transform. The resulting structure represents a new affine transform, which you can use (and reuse, if you want) to rotate a coordinate system.
The actual direction of rotation is dependent on the coordinate system orientation of the target platform, which is different in iOS and OS X.
#### Availability
Available in OS X v10.0 and later.
• ``` CGAffineTransformInvert(_:) CGAffineTransformInvert ```
Returns an affine transformation matrix constructed by inverting an existing affine transform.
#### Declaration
Swift
``` func CGAffineTransformInvert(_ t: CGAffineTransform) -> CGAffineTransform ```
Objective-C
``` CGAffineTransform CGAffineTransformInvert ( CGAffineTransform t ); ```
#### Parameters
``` t ``` An existing affine transform.
#### Return Value
A new affine transformation matrix. If the affine transform passed in parameter `t` cannot be inverted, Quartz returns the affine transform unchanged.
#### Discussion
Inversion is generally used to provide reverse transformation of points within transformed objects. Given the coordinates (x,y), which have been transformed by a given matrix to new coordinates (x’,y’), transforming the coordinates (x’,y’) by the inverse matrix produces the original coordinates (x,y).
#### Availability
Available in OS X v10.0 and later.
• ``` CGAffineTransformConcat(_:_:) CGAffineTransformConcat ```
Returns an affine transformation matrix constructed by combining two existing affine transforms.
#### Declaration
Swift
``` func CGAffineTransformConcat(_ t1: CGAffineTransform, _ t2: CGAffineTransform) -> CGAffineTransform ```
Objective-C
``` CGAffineTransform CGAffineTransformConcat ( CGAffineTransform t1, CGAffineTransform t2 ); ```
#### Parameters
``` t1 ``` The first affine transform. ``` t2 ``` The second affine transform. This affine transform is concatenated to the first affine transform.
#### Return Value
A new affine transformation matrix. That is, t’ = t1*t2.
#### Discussion
Concatenation combines two affine transformation matrices by multiplying them together. You might perform several concatenations in order to create a single affine transform that contains the cumulative effects of several transformations.
Note that matrix operations are not commutative—the order in which you concatenate matrices is important. That is, the result of multiplying matrix `t1` by matrix `t2` does not necessarily equal the result of multiplying matrix `t2` by matrix `t1`.
#### Availability
Available in OS X v10.0 and later.
• ``` CGPointApplyAffineTransform ```
Returns the point resulting from an affine transformation of an existing point.
#### Declaration
Objective-C
``` CGPoint CGPointApplyAffineTransform ( CGPoint point, CGAffineTransform t ); ```
#### Parameters
``` point ``` A point that specifies the x- and y-coordinates to transform. ``` t ``` The affine transform to apply.
#### Return Value
A new point resulting from applying the specified affine transform to the existing point.
#### Availability
Available in OS X v10.0 and later.
• ``` CGSizeApplyAffineTransform ```
Returns the height and width resulting from a transformation of an existing height and width.
#### Declaration
Objective-C
``` CGSize CGSizeApplyAffineTransform ( CGSize size, CGAffineTransform t ); ```
#### Parameters
``` size ``` A size that specifies the height and width to transform. ``` t ``` The affine transform to apply.
#### Return Value
A new size resulting from applying the specified affine transform to the existing size.
#### Availability
Available in OS X v10.0 and later.
• ``` CGRectApplyAffineTransform(_:_:) CGRectApplyAffineTransform ```
Applies an affine transform to a rectangle.
#### Declaration
Swift
``` func CGRectApplyAffineTransform(_ rect: CGRect, _ t: CGAffineTransform) -> CGRect ```
Objective-C
``` CGRect CGRectApplyAffineTransform ( CGRect rect, CGAffineTransform t ); ```
#### Parameters
``` rect ``` The rectangle whose corner points you want to transform. ``` t ``` The affine transform to apply to the `rect` parameter.
#### Return Value
The transformed rectangle.
#### Discussion
Because affine transforms do not preserve rectangles in general, the function `CGRectApplyAffineTransform` returns the smallest rectangle that contains the transformed corner points of the `rect` parameter. If the affine transform `t` consists solely of scaling and translation operations, then the returned rectangle coincides with the rectangle constructed from the four transformed corners.
#### Availability
Available in OS X v10.4 and later.
• ``` CGAffineTransformIsIdentity(_:) CGAffineTransformIsIdentity ```
Checks whether an affine transform is the identity transform.
#### Declaration
Swift
``` func CGAffineTransformIsIdentity(_ t: CGAffineTransform) -> Bool ```
Objective-C
``` bool CGAffineTransformIsIdentity ( CGAffineTransform t ); ```
#### Parameters
``` t ``` The affine transform to check.
#### Return Value
Returns `true` if `t` is the identity transform, `false` otherwise.
#### Availability
Available in OS X v10.4 and later.
• ``` CGAffineTransformEqualToTransform(_:_:) CGAffineTransformEqualToTransform ```
Checks whether two affine transforms are equal.
#### Declaration
Swift
``` func CGAffineTransformEqualToTransform(_ t1: CGAffineTransform, _ t2: CGAffineTransform) -> Bool ```
Objective-C
``` bool CGAffineTransformEqualToTransform ( CGAffineTransform t1, CGAffineTransform t2 ); ```
#### Parameters
``` t1 ``` An affine transform. ``` t2 ``` An affine transform.
#### Return Value
Returns `true` if `t1` and `t2` are equal, `false` otherwise.
#### Availability
Available in OS X v10.4 and later.
### Data Types
• ``` CGAffineTransform ```
A structure for holding an affine transformation matrix.
#### Declaration
Swift
``` struct CGAffineTransform { var a: CGFloat var b: CGFloat var c: CGFloat var d: CGFloat var tx: CGFloat var ty: CGFloat init() init(a a: CGFloat, b b: CGFloat, c c: CGFloat, d d: CGFloat, tx tx: CGFloat, ty ty: CGFloat) } ```
Objective-C
``` struct CGAffineTransform { CGFloat a; CGFloat b; CGFloat c; CGFloat d; CGFloat tx; CGFloat ty; }; typedef struct CGAffineTransform CGAffineTransform; ```
#### Fields
``` a ```
The entry at position [1,1] in the matrix.
``` b ```
The entry at position [1,2] in the matrix.
``` c ```
The entry at position [2,1] in the matrix.
``` d ```
The entry at position [2,2] in the matrix.
``` tx ```
The entry at position [3,1] in the matrix.
``` ty ```
The entry at position [3,2] in the matrix.
#### Discussion
In Quartz 2D, an affine transformation matrix is used to rotate, scale, translate, or skew the objects you draw in a graphics context. The `CGAffineTransform` type provides functions for creating, concatenating, and applying affine transformations.
In Quartz, affine transforms are represented by a 3 by 3 matrix:
Because the third column is always `(0,0,1)`, the `CGAffineTransform` data structure contains values for only the first two columns.
Conceptually, a Quartz affine transform multiplies a row vector representing each point (x,y) in your drawing by this matrix, producing a vector that represents the corresponding point (x’,y’):
Given the 3 by 3 matrix, Quartz uses the following equations to transform a point (x, y) in one coordinate system into a resultant point (x’,y’) in another coordinate system.
The matrix thereby “links” two coordinate systems—it specifies how points in one coordinate system map to points in another.
Note that you do not typically need to create affine transforms directly. If you want only to draw an object that is scaled or rotated, for example, it is not necessary to construct an affine transform to do so. The most direct way to manipulate your drawing—whether by movement, scaling, or rotation—is to call the functions `CGContextTranslateCTM`, `CGContextScaleCTM`, or `CGContextRotateCTM`, respectively. You should generally only create an affine transform if you want to reuse it later.
#### Availability
Available in OS X v10.0 and later.
### Constants
• ``` CGAffineTransformIdentity ```
The identity transform.
#### Declaration
Swift
`let CGAffineTransformIdentity: CGAffineTransform`
Objective-C
```const CGAffineTransform CGAffineTransformIdentity; ```
#### Constants
• `CGAffineTransformIdentity`
`CGAffineTransformIdentity`
The identity transform:
Available in OS X v10.0 and later. | 3,617 | 18,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2016-22 | latest | en | 0.660241 |
http://mathhelpforum.com/calculus/73689-finding-sequence-functions-difficult-problem-about-integration.html | 1,529,859,266,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866984.71/warc/CC-MAIN-20180624160817-20180624180817-00026.warc.gz | 195,600,148 | 10,231 | 1. ## finding sequence of functions! difficult problem about integration..
Let $\displaystyle f,g$ are ANY non-negative and measurable. $\displaystyle \mu$ is measure
Find two function $f_n,g_n$ such that HAVE ONLY FINITELY MANY VALUE and
for any natural number $n$,
$\displaystyle |\int^{\infty}_0 \mu(\{ x : f(x) > t \}) dt - \int^{\infty}_0 \mu(\{ x : f_n(x) > t \}) | \leq \frac {C}{n}$ and
$\displaystyle |\int^{\infty}_0 \mu(\{ x : g(x) > t \}) dt - \int^{\infty}_0 \mu(\{ x : g_n(x) > t \}) | \leq \frac {C}{n}$ and
$\displaystyle |\int^{\infty}_0 \mu(\{ x : f(x) + g(x) > t \}) dt - \int^{\infty}_0 \mu(\{ x : f_n(x) + g_n(x) > t \}) | \leq \frac {C}{n}$.
And C is independent of n.
*Important* above integral is (improper) riemann integral !!
(In fact, this problem is from ANALYSIS by Lieb and Loss.. so definition of integral is different from other books
it defines integral by $\displaystyle \int f d\mu : = \int^{\infty}_0 \mu({x: f(x) > t}) dt$ so i wrote the problem as above..)
P.S. sorry for poor english..
-I correct some mistakes -
2. Originally Posted by ramsey88
Let f,g are non-negative and measurable. $\displaystyle \mu$ is measure
Find two function $\displaystyle f_n,g_n$ such that HAVE ONLY FINITELY MANY VALUE and
for any natural number n,
$\displaystyle |\int^{\infty}_0 \mu(\{ x : f(x) > t \}) dt - \int^{\infty}_0 \mu(\{ x : f_n(x) > t \}) | \leq \frac {C}{n}$ and
$\displaystyle |\int^{\infty}_0 \mu(\{ x : g(x) > t \}) dt - \int^{\infty}_0 \mu(\{ x : g_n(x) > t \}) | \leq \frac {C}{n}$ and
$\displaystyle |\int^{\infty}_0 \mu(\{ x : f(x) + g(x) > t \}) dt - \int^{\infty}_0 \mu(\{ x : f_n(x) + g(x) > t \}) | \leq \frac {C}{n}$.
And C is independent of n.
*Important* above integral is (improper) riemann integral !!
(In fact, this problem is from ANALYSIS by Lieb and Loss.. so definition of integral is different from other books
it defines integral by $\displaystyle \int f d\mu : = \int^{\infty}_0 \mu({x: f(x) > t}) dt$ so i wrote the problem as above..)
P.S. sorry for poor english..
I'm not clear on what the problem requires. Is there any requirement that would prevent a trivial example like $\displaystyle f_n(1)= 1$ and undefined for other x, for all n, $\displaystyle g_n(0)= 1$ and undefined for other x, for all n?
In that case, all integrals are 0 so you can take C to be any positive number.
3. Originally Posted by HallsofIvy
I'm not clear on what the problem requires. Is there any requirement that would prevent a trivial example like $\displaystyle f_n(1)= 1$ and undefined for other x, for all n, $\displaystyle g_n(0)= 1$ and undefined for other x, for all n?
In that case, all integrals are 0 so you can take C to be any positive number.
$\displaystyle f_n, g_n$ should satisfy above condition for any non-negative,measurable function f,g. | 909 | 2,805 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-26 | latest | en | 0.700859 |
https://cloud.tencent.com/developer/article/1068068 | 1,571,670,756,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987779528.82/warc/CC-MAIN-20191021143945-20191021171445-00419.warc.gz | 424,757,180 | 14,385 | # Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 113031 Accepted Submission(s): 26130
Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
Sample Output
Case 1: 14 1 4
Case 2: 7 1 6
Author
Ignatius.L
http://acm.hdu.edu.cn/showproblem.php?pid=1003
1 #include<iostream>
2 #include<vector>
3 using namespace std;
4 int main()
5 {
6 int t,n,i,k,st,en,maxn,sum,flag=1;
7 cin>>t;
8 while(t--)
9 {
10 cin>>n;
11 vector<int>arr(n);
12 for(i=0;i<n;i++)
13 cin>>arr[i];
14 st=en=sum=0;
15 for(i=0,k=0;i<n;i++)
16 {
17 sum+=arr[i];
18 if(i==0||sum>maxn)
19 {
20 maxn=sum;
21 en=i;
22 st=k;
23 }
24 if(sum<0)
25 {
26 sum=0;
27 k=i+1;
28 }
29 }
30 cout<<"Case "<<flag++<<":"<<endl;
31 cout<<maxn<<" "<<st+1<<" "<<en+1<<endl;
32 if(t!=0)cout<<endl;
33 }
34 return 0;
35 }
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37170 | 1,028 | 3,109 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-43 | latest | en | 0.578349 |
https://www.loomsolar.com/blogs/collections/how-to-calculate-solar-panel-battery-and-inverter | 1,716,645,152,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058829.24/warc/CC-MAIN-20240525131056-20240525161056-00111.warc.gz | 729,157,682 | 43,347 | When you plan to install solar panel, battery and inverter, then you must be wondering about how to decide the capacity of these components. On the basis of our practical experience, below guide will help you.
## Step 1: Load Calculation
The best way to calculate load calculation is to use best quality clamp meter. Let’s assume for this calculation we will use a clamp meter that will support 600A . Using this clamp meter, we can calculate power consumption of home, office, shop, school, medical clinic, & petrol pump, etc. Follow the below steps to Calculate the power consumption:
1. Turn on all the appliances
2. Use clamp meter in phase wire of electric meter
While you use clamp meter, it shows power consumption in Amp. Such as 5Amp, 10Amp, & 20Amp, etc. Indian grid voltage range starts from 220V-240V. But you must check voltage as well using this clamp meter.
Load Calculation = 220V * 5 A
= 1100W
## Step 2: Backup Time
In above steps, I have calculated the power consumption. On the basis of this calculated power consumption, we should know how many hours you may want to run appliances after power cut. If your answer will be 4 to 5 hrs. That means:
Battery Storage = Total Load * Backup Time
= 1100W * 4 hrs.
= 4400W
## Step 3: Battery Capacity
In above steps, You will know about your required battery storage. There are two types of battery technology in India – lead acid battery & lithium battery. In lead acid battery, 150Ah lead acid battery is the most popular battery for homes and businesses. One 150Ah battery stores 1300 Watt. That means, you need
Battery Capacity = Required Battery Storage / Battery Storage
= 4400W / 1300W
= 4 Batteries
Generally, 4 batteries of 150Ah comes in 48V.
In case of lithium battery, you need only one battery that comes with 5kWh, 48V.
## Step 4: Inverter Capacity
In above steps, you will know both the battery capacity and voltage. According to battery voltage, capacity, and power consumption. In simple language, I need here a 48V & 5kW inverter.
Inverter Capacity = Load + Load * 20%
= 1100W + 1100W * 20%
= 1100W + 220W
= 1320W
That means, you need around 1.3kW inverter capacity.
## Step 5: Solar Panel Capacity
Finally, you need to calculate the solar panel capacity. We always know that solar panels generates DC voltage (22V to 50V). In simple terms,
Solar Panel Capacity = 3 * Battery Capacity
= 3 * 600Ah
= 1800Watt
That means, you need 1.8kW capacity of solar panels and the highest wattages of solar panels in India is around 540W. If you choose these solar panels, then you will need around 4 solar panels for charging your battery as well as run your home loads.
Youngbuin jean martial
How to calculate solar batteries and inverter
John Whyte
My load is 2000w run for 24hrs
I need Sola panels, batteries, inverter
Jyoti Sagar Nayak
I want totally details knowledge
Jan Kim
please help to compute full set up solar panel system for the load of 1hp window type aircon.
thank you ao much
Abdullahi Abdulfatayi
Please help me out to calculate of numbers of solar panels, battery and inverter to used in a shop that has a eight CCTV camera, 2 refrigerator, 15 LED bulb,four ceiling fan, 2 standing fan and 2 external light
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Muzzamil
i like to install solar system at my home.
4 fans
10 led bulb
washing machine
freigde
iron
can you please send me quotation of this system
thanks
muzzamil
cell: 0324-2636303
Jivan Joshi
We have two AC of total 2.5 ton cap. , Four ceiling fans , three tube lights , one hp motor – single phase.
Calculate load and number of panels to be used and battery required . Thanks .
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My brother recommended I might like this website. He was entirely right. This post actually made my day. You can not imagine just how much time I had spent for this info! Thanks!
Ishaq Yunusa Ishaq
I need more explanation with exact example that goes direct to the point,examle Soler selection to meet up battery charging
Ali hossain
Dear.i want to know that. Calculated size of solar panel.battery bank & inverter 3 phase industrial system. My load 33 Amp 3 phase system. Plz.calculated size of inverter Battery & solar panel.
Uduma Ndubuisi Ogbonnaya
I need to know more about inverter, battery and solar calculation
abed Nazari
Hello to all this information about pv system.
Is very useful for me.
Ram Naresh Sharma
Please share me capacity will be in kw or .KVA.rating.
Ileni
Helloo ! Help me on specific ,that how many lead acid battery’s required 1575 watts. And how can I get the supplier if wanna start a business of order ing the equipments of renewable energy.
Ileni
Helloo ! Help me on specific ,that how many lead acid battery’s required 1575 watts. And how can I get the supplier if wanna start a business of order ing the equipments of renewable energy.
myo hlaing
I want a solar inverter and battery cables size calculation excel software files link.please send me email.
Irfan ullah
Your calculation is not correct plz correct
Using website name electrical technology
Azhaan
Hii
In 96 volt inverter how many solar panels or what power can we use
Nonnyblack
Hi!
What’s the advantage of half cut solar panel over traditional single-face solar panel?
Nonnyblack
Hi!
I’ve got a question: how many 200ah lead-acid batteries will 24(440w) half cut solar panels charge?
Abdifatah Ahmed
What are the main reasons a hybrid or off-grid solar system cannot operate without batteries or a startup generator that operate with the PV system during sunshine not only for storage and backup during outages?
Abdifatah Ahmed
What are the main reasons a hybrid or off-grid solar system cannot operate without batteries or a startup generator that operate with the PV system during sunshine not only for storage and backup during outages?
Kabir Elyakub Zakirai
Thanks Mr India for your input. Thanks alot we learned from your post. Please keep it up
Mehmood
I wish to learn more about design and installation of solar system. Please, kindly give a elaborate examples on calculations on sizing of control charger, solar panels, batteries and battery back up, inverter, Thanks alot
otsile dijo
i want to leran more about solar
otsile dijo
i want to leran more about solar
what is the calculated panels?
Q1 what is the cost of 540watt pennal
Q2 How much batteries we install for1080watt pennals
Q3 is there any install ment scheme
Q4 is there any dealership in jandkashmir
Q5 i want only such solar who charge my battery in cold and snow weather
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Calculators | 1,853 | 7,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-22 | latest | en | 0.8912 |
http://stackoverflow.com/questions/5810502/manhattan-distance-algorithm-job-interview-question | 1,455,487,182,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454702032759.79/warc/CC-MAIN-20160205195352-00008-ip-10-236-182-209.ec2.internal.warc.gz | 207,557,828 | 19,429 | # Manhattan distance algorithm job interview question
Given a set of n points in an X-Y plane, how can I determine if every point is at least separated by every other point by a Manhattan distance of 5 units in time less than O(n^2)?
What is the best algorithm to implement this?
Thank you.
-
This is not a question but an assignment. – Bart Kiers Apr 27 '11 at 21:02
How do you know that's an assignment? This easily is an interview question. – arasmussen Apr 27 '11 at 21:03
@arasmussen, "assignment" does not necessarily mean homework. My point is that the OP did not ask a question. – Bart Kiers Apr 27 '11 at 21:06
Good point. Zaya, what work have you done on this question so far? What are your thoughts? What does it mean for the algorithm that it is less than O(n^2)? – arasmussen Apr 27 '11 at 21:08
@arasmussen I totally agree with Bart. I think you didn't get his point. What he meant is that the sentence given by zaya is an assignment (order). It is written in imperative form. It is not a question asking something. – sawa Apr 27 '11 at 21:09
1. Sort the points by `x`. This takes time 'O(n log(n))'.
2. Divide the range into strips of width 10. (You'll need an obvious bit of care here for the pathological case where one point has x-coordinate 1 and the next has x-coordinate 1020.) `O(n)`
3. For each strip:
1. Take the set of points within that strip, or within an x of 5 to either side, and sort them by y. This is `O(n log(n))` across all strips.
2. For each point in the strip, find the Manhattan distance to all other points in that slightly wider strip whose y coordinate is within 5 of their own. If you find any within distance 5, exit and report false. This is `O(n)` across all strips.
4. Report true.
This algorithm is `O(n log(n))`. I strongly advise that you demonstrate for yourself that the pointwise Manhattan comparison in 1.2 takes `O(n)` operations, even if the answer is false.
For true it is simple - it follows from the fact that there is a maximum number of other points that can be squeezed in a 20x10 box without 2 getting within 5. For false it is trickier, you can have a lot of other points in that box, but by the time you have compared a fixed number of them to the rest, you must have found two within distance 5. Either way a given point participates in a fixed maximum number of point to point comparisons before you have your answer.
-
how is sorting by y for all strips O(n)? Using radix sort? – DShook Apr 27 '11 at 21:34
@DShook: Because I had a brain fart. Fixed. – btilly Apr 27 '11 at 22:20
I might be wrong here, but since you have up to n strips, sorting each of those strips by y results in n * (n log(n)) operations, meaning n^2 log(n). Am I wrong? I recognize that in practice it will be a much diminished set (i.e. if there are many strips, then the values in each are limited, so the internal sorting is smaller; if there are few strips then there are few sorts) so it should be fast, but it still technically seems n^2 log(n) to me... – DRobinson Feb 19 '13 at 15:00
@DRobinson If there are `n` strips, then each sort is `O(1)`. In general if you have `k` lists and `n` things between them, sorting all `k` lists cannot be worse than sorting `n` things with up to `k` labels by label and then by thing, so no matter how many lists there are it is no worse than `O(n log(n))` to sort them all. – btilly Feb 21 '13 at 2:28
Yes, but take for example a situation where there is n/2 strips, two of which each have about n/4 elements (the remaining n/2 elements are distributed among the remaining (n/2 - 2) strips). Is this not, technically, going over O(n) strips (because there are n/2 of them), and in each sorting O(n) [because worst case has n/4 -> O(n)] values? That would be O(n/2 * O(n/4 log(n/4)) = O(n * nlog(n)). Obviously it will be much faster but in pure big O that still seems like n^2log(n) to me. – DRobinson Feb 21 '13 at 19:37 | 1,062 | 3,916 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2016-07 | latest | en | 0.946872 |
https://math.stackexchange.com/questions/3719079/prove-that-sqrt32-sqrt34-is-irrational | 1,610,923,670,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513194.17/warc/CC-MAIN-20210117205246-20210117235246-00697.warc.gz | 439,297,008 | 28,813 | # Prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational. [duplicate]
Prove that $$\sqrt[3]{2} + \sqrt[3]{4}$$ is irrational.
My steps so far: I found that the polynomial $$y^3-6y-6=0$$ has roots $$\sqrt[3]{2} + \sqrt[3]{4}.$$ Can I use this to prove that $$\sqrt[3]{2} + \sqrt[3]{4}$$ is irrational? If so, how? I was thinking of using Proof by Contradiction, but I'm not so sure.
• The given expression can be written as $\sqrt[3] 2(\sqrt[3] 2+1)$. Knowing that $\sqrt[3] 2$ is irrational, the product of two irrational numbers is irrational, too. – SarGe Jun 14 '20 at 8:49
• @Doubtnut "the product of two irrational numbers is irrational, too" Not so, e.g. $\sqrt{2}\cdot\sqrt{2}=2$. – J.G. Jun 14 '20 at 8:50
• @Doubtnut No you are not right, for example $\pi$ and $e$ are both irrationals but it is still an open problem that $\pi\cdot e$ is whether irrational or not. – mertunsal Jun 14 '20 at 8:51
• Maybe it might be said that (it is a rough conjecture feel free to refute me as you want) product of algebraic irrational numbers with different degrees are irrational. – mertunsal Jun 14 '20 at 8:53
• In this case, $\sqrt[3]2$ and $\sqrt[3]2+1$ both have the same degree. @mertunsal22 – Angina Seng Jun 14 '20 at 9:13
Actually, $$\sqrt[3]2+\sqrt[3]4$$ is not a root of that polynomial. But it is a root of $$x^3-6 x-6$$. By the rational roots theorem, the only rational roots that that polynomial can have are $$\pm1$$, $$\pm2$$, $$\pm3$$, and $$\pm6$$. Since none of them is actually a root, your number is irrational.
• @MathIsFun123 Use Factor Theorem and calculate $p(\sqrt[3]{2} + \sqrt[3]{4})$. – Landuros Jun 14 '20 at 8:52
• Did you read the edited version of my answer? I wrote there that it is a root of $x^3-6x-6$. – José Carlos Santos Jun 14 '20 at 8:56 | 607 | 1,776 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-04 | latest | en | 0.86398 |
https://www.mycoursehelp.com/QA/To-invest-you-want-to-create-a-portfolio/7450/1 | 1,620,633,687,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989115.2/warc/CC-MAIN-20210510064318-20210510094318-00163.warc.gz | 971,058,204 | 8,260 | Create an Account
Home / Questions / To invest you want to create a portfolio of three companies
To invest you want to create a portfolio of three companies
To invest, you want to create a portfolio of three companies (A, B and C) with stocks. The expected monthly returns of these shares are 1%, 2% and 3% respectively. Standard deviations are 1%, 1.5% and 2%, respectively. There is a positive correlation of 0.5 between the returns of shares of companies A and B. There is no correlation between A and C and between B and C.
(a) If you only want to create a portfolio consisting of A and B shares, find the portfolio weights that minimize the risk (standard deviation) of this portfolio. Calculate the expected the result of this portfolio.
(b) Find the weight of the active portfolio which provides the expected monthly return of 5%. Calculate the standard deviation of the portfolio. Compare it with the risk of the portfolio in (a).
(c) If you are investing \$ 100,000 in the portfolio in (b), calculate how much you want to buy or sell from each individual.
May 14 2018 View more View Less | 249 | 1,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-21 | latest | en | 0.925588 |
https://amara.org/en/videos/CcxW75CLF10D/en/6546/239488/ | 1,638,077,683,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358469.34/warc/CC-MAIN-20211128043743-20211128073743-00407.warc.gz | 171,425,584 | 26,769 | ## Multiplication 2: The Multiplication Tables
• 0:01 - 0:03
At this point I think you
know a little bit about
• 0:03 - 0:05
what multiplication is.
• 0:09 - 0:12
What we're going to do in this
video is to give you just a ton
• 0:12 - 0:17
of more practice and start you
on your memorization of the
• 0:17 - 0:18
multiplication tables.
• 0:18 - 0:20
And if you watch enough Khan
Academy videos, and hopefully
• 0:20 - 0:23
you will in the future, you'll
realize that I'm normally not
• 0:23 - 0:25
a big fan of memorization.
• 0:25 - 0:27
But the one thing about
multiplication is if you
• 0:27 - 0:30
tables that we'll start to do
• 0:30 - 0:34
in this video, it'll pay huge
benefits the rest of your life.
• 0:34 - 0:38
So I promise you, do it now,
you'll never forget it, and the
• 0:38 - 0:40
rest of your life everything
will be-- well, I don't want to
• 0:40 - 0:43
make false promises to you, but
they'll be better than if you
• 0:43 - 0:46
didn't memorize your
multiplication tables.
• 0:46 - 0:47
So what are the
multiplication tables?
• 0:47 - 0:50
Well that's all of the
different numbers
• 0:50 - 0:51
times each other.
• 0:51 - 0:54
So let's actually do a
little bit review.
• 0:54 - 0:59
So if I say what is 2 times 1?
• 0:59 - 1:02
That is equal to 2
plus itself one time.
• 1:02 - 1:05
So this is equal to just 2.
• 1:05 - 1:07
That's 2 plus itself one time.
• 1:07 - 1:08
I don't have to say plus
anything because there's
• 1:08 - 1:09
only one 2 there.
• 1:09 - 1:13
I could also write this as
1 plus itself two times.
• 1:13 - 1:15
So that's also 1 plus 1.
• 1:15 - 1:18
Well that also equals 2.
• 1:18 - 1:18
Fair enough.
• 1:18 - 1:20
So 2 times 1 is 2.
• 1:20 - 1:22
And if you watched the last
video, what's 2 times 0?
• 1:22 - 1:23
Well that's 0.
• 1:23 - 1:27
So you don't have to memorize
your 0 multiplication tables
• 1:27 - 1:31
because everything times 0 is
0, or 0 times anything is 0.
• 1:31 - 1:31
So let's see.
• 1:31 - 1:32
What's 2 times 2?
• 1:36 - 1:38
Well, this is equal to--
we're going to add 2
• 1:38 - 1:39
to itself two times.
• 1:39 - 1:42
So that's 2 plus 2.
• 1:42 - 1:43
And there's only a
way to do that.
• 1:43 - 1:45
I could say take this 2 and
add it to itself two times,
• 1:45 - 1:47
but it's the same thing.
• 1:47 - 1:48
And what's 2 plus 2?
• 1:48 - 1:49
That's equal to 4.
• 1:49 - 1:51
What's 2 times 3?
• 1:51 - 1:58
2 times 3 is equal
to 2 plus 2 plus 2.
• 1:58 - 2:03
It can also be
equal to 3 plus 3.
• 2:03 - 2:06
Right, we learned in the previous video
this statement can be written
• 2:06 - 2:07
either of these ways.
• 2:07 - 2:09
And in either case,
what's it equal to?
• 2:09 - 2:12
Well, 3 plus 3 is the same
thing as 2 plus 2 plus 2,
• 2:12 - 2:15
and that's equal to 6.
• 2:15 - 2:16
All right.
• 2:16 - 2:17
Now what is 2 times 4?
• 2:21 - 2:26
Well that's equal to 2
plus 2 plus 2 plus 2.
• 2:26 - 2:30
And notice, it's exactly
what 2 times 3 was.
• 2:30 - 2:33
2 times 3 was that.
• 2:33 - 2:36
I have that here, but now I'm
just adding another 2 to it.
• 2:36 - 2:40
So if we're too lazy to sit
here and add 2 plus 2 is 4.
• 2:40 - 2:41
4 plus 2 is 6.
• 2:41 - 2:43
Instead of doing that, we could
say, hey look, we already know
• 2:43 - 2:46
that this thing over
here, this was 6.
• 2:46 - 2:48
We figured it out in the
previous line right there.
• 2:48 - 2:52
We figured out this is 6, so we
could just say, oh, 2 times 4
• 2:52 - 2:56
is going to be 2 more than
that, which is equal to 8.
• 2:56 - 2:57
And you should hopefully
see that pattern.
• 2:57 - 3:03
As we go from 2 times 1,
to 2 times 2, to 2 times
• 3:03 - 3:04
3, what's happening?
• 3:04 - 3:06
How much are we going up by?
• 3:06 - 3:08
From 2 to 4 we're going plus 2.
• 3:08 - 3:11
From 4 to 6 we're going plus 2.
• 3:11 - 3:13
And then from 6 to 8
we're going plus 2.
• 3:13 - 3:16
So you could figure out what
2 times 5 is, even without
• 3:16 - 3:17
• 3:17 - 3:23
2 times 5 is equal to 2 plus
2 plus 2 plus 2 plus 2.
• 3:23 - 3:26
It could also be
written as 5 plus 5.
• 3:26 - 3:29
2 times 4 could've been
written as 4 plus 4 as well.
• 3:29 - 3:30
And what's that equal to?
• 3:30 - 3:33
We could add all of these up
or we could add these two up.
• 3:33 - 3:36
Or we could just say it's going
to be two more than 2 times 4.
• 3:36 - 3:39
So it's going to be 10.
• 3:39 - 3:42
I'll finish the 2 times tables.
• 3:42 - 3:45
And I think you see all of the
patterns that emerge from it.
• 3:45 - 3:48
So 2 times 6.
• 3:48 - 3:52
Well, that's going to be 2
plus itself six times.
• 3:52 - 3:52
Let's see.
• 3:52 - 3:56
1, 2, 3, 4, 5, 6, which
should also be equal to
• 3:56 - 3:59
6 plus itself two times.
• 3:59 - 4:01
This could be
interpreted either way.
• 4:01 - 4:03
And that's going to
be equal to 12.
• 4:03 - 4:08
Once again, two more than 2
times 5 right, because we're adding
• 4:08 - 4:10
2 to itself one more time.
• 4:10 - 4:12
So it's going to be two more.
• 4:12 - 4:14
Let's keep going.
• 4:14 - 4:17
So 2 times 7.
• 4:17 - 4:23
2 times 7 is equal to-- well, instead of, I
could write 2 plus 2 plus 2
• 4:23 - 4:27
plus 2-- this is getting
tiring-- plus 2 plus 2.
• 4:27 - 4:28
Is that 7?
• 4:28 - 4:31
1, 2, 3, 4, 5, 6, 7.
• 4:31 - 4:34
And that's the same thing as 7
plus 7, which you may or may
• 4:34 - 4:37
not know is equal to 14.
• 4:37 - 4:40
You could just say hey, that's
going to be two more than 12.
• 4:40 - 4:44
So 12 plus 1 plus 2
is-- 12 plus 1 is 13.
• 4:44 - 4:46
12 plus 2 is 14.
• 4:46 - 4:48
All right, let's
just keep going.
• 4:48 - 4:51
2 times 8.
• 4:51 - 4:54
I could do all of this business
here where I add the 2's or I
• 4:54 - 4:57
could say look, it's just going
to be two more than 2 times 7.
• 4:57 - 4:59
So I could say it's
going to be 14 plus 2.
• 4:59 - 5:00
two to that one.
• 5:00 - 5:02
So I could say it's 16.
• 5:02 - 5:06
Or I could also say
that's 8 plus 8.
• 5:06 - 5:07
That's also 16.
• 5:07 - 5:09
I could have done all the 2's
out, but if you like you
• 5:09 - 5:15
could do that for your
own benefit and learning.
• 5:15 - 5:18
Alright, we're almost-- well, we could
go forever because there
• 5:18 - 5:19
is no largest number.
• 5:19 - 5:22
I can keep going.
• 5:22 - 5:25
2 times 9 times 10 times 100
times 1,000 times 1,000,000.
• 5:25 - 5:28
But I'm going to stop at 12
because that tends to be what
• 5:28 - 5:29
people need to memorize.
• 5:29 - 5:32
But if you really want
to be a mathlete you
• 5:32 - 5:34
want to go up to 20.
• 5:34 - 5:37
Lets let's go to 2 times 9.
• 5:37 - 5:39
That's going to be two
more than 2 times 8.
• 5:39 - 5:41
It's going to be 18.
• 5:41 - 5:43
Or that's 9 plus 9.
• 5:43 - 5:44
Also 18.
• 5:44 - 5:46
What's 2 times 10?
• 5:46 - 5:48
And 10 times tables
are interesting.
• 5:48 - 5:50
And we're going to see a
pattern there in a second
• 5:50 - 5:53
when we try to complete
an entire times tables.
• 5:53 - 5:55
So 2 times 10?
• 5:55 - 5:57
Two more than 2 times 9.
• 5:57 - 5:59
It's 20.
• 5:59 - 6:01
Or we could also say
that's 10 plus 10.
• 6:01 - 6:03
10 plus itself two times.
• 6:03 - 6:05
Now what's interesting
• 6:05 - 6:09
This looks just like
a 2 with a 0 added.
• 6:09 - 6:11
And you're going to see
that anything times 10.
• 6:11 - 6:12
You just put a 0 on the right.
• 6:12 - 6:14
And you can think
about why that is.
• 6:14 - 6:16
You can view this
as two 10's is 20.
• 6:16 - 6:18
That's what 20 is.
• 6:18 - 6:20
We're almost done.
• 6:20 - 6:22
Let's do 2 times 11.
• 6:22 - 6:26
2 times 11 is going to be 2
more than this right here.
• 6:26 - 6:28
It's going to be 22.
• 6:28 - 6:30
Another interesting pattern.
• 6:30 - 6:32
I have the number repeated
twice-- a 2 and a 2.
• 6:32 - 6:33
Interesting.
• 6:33 - 6:38
Something to watch out
for as we look at other
• 6:38 - 6:39
multiplication tables.
• 6:39 - 6:42
And then finally-- it's not
finally, we could keep going.
• 6:42 - 6:45
2 times-- that's too
dark of a color.
• 6:45 - 6:47
2 times 12.
• 6:47 - 6:51
2 times 12 is going to be
two more than 2 times 11.
• 6:51 - 6:51
That's 24.
• 6:51 - 6:54
We could have also written
that as 12 plus 12.
• 6:54 - 6:56
Or we could've said 2 plus
2 plus 2 plus 2 plus
• 6:56 - 6:58
2 twelve times.
• 6:58 - 7:00
It all gets you to 24.
• 7:00 - 7:01
So that's the 2 times
tables and I think
• 7:01 - 7:02
you see the pattern.
• 7:02 - 7:05
Every time you multiply it by
one higher number you just
• 7:05 - 7:07
add 2 to that number.
• 7:07 - 7:09
So now that we see that
pattern, let's see
• 7:09 - 7:12
if we can complete a
multiplication table.
• 7:12 - 7:16
So what I want to do, I'm going
to write all the numbers.
• 7:16 - 7:18
Let's see. One.
• 7:18 - 7:19
I hope I have space for this.
• 7:19 - 7:29
1, 2, 3, 4, 5, 6, 7, 8, 9.
• 7:29 - 7:31
Actually, I'll just
do it till 9.
• 7:31 - 7:32
I'll just keep going.
• 7:32 - 7:33
9.
• 7:33 - 7:34
Actually I won't have space to
do that because I want you
• 7:34 - 7:36
to see the entire table.
• 7:36 - 7:38
So I'm just going up till
9 here, but I encourage
• 7:38 - 7:40
you after this video to
complete it on your own.
• 7:40 - 7:43
Maybe if we have time I'll
complete it here as well.
• 7:43 - 7:45
So these are the first numbers
that I'm going to multiply.
• 7:45 - 7:52
And I'm going to multiply
it times 1, 2, 3, 4,
• 7:55 - 7:57
5, 6, 7, 8, and 9.
• 7:57 - 8:00
What I' gonna do is, I'm gonna
• 8:00 - 8:02
So first of all
• 8:02 - 8:03
Actually I should have
written this 1 under-
• 8:03 - 8:05
well, what's 1 times 1?
• 8:05 - 8:06
So this is the way I'm
going to view it.
• 8:06 - 8:09
Whatever is 1 times 1 I'm
going to write here.
• 8:09 - 8:10
Well that's 1.
• 8:10 - 8:12
What's 1 times 2?
• 8:12 - 8:12
That's 2.
• 8:12 - 8:14
What's 1 times 3?
• 8:14 - 8:14
That's 3.
• 8:14 - 8:16
1 times anything is that
number, so I can just
• 8:16 - 8:21
write 4, 5, 6, 7, 8, 9.
• 8:21 - 8:24
1 times 9 is 9.
• 8:24 - 8:25
Fair enough.
• 8:25 - 8:26
Now let's do the
2 times tables.
• 8:26 - 8:28
I'll do that in a blue.
• 8:28 - 8:32
Actually, let me do 1 in that
color and now in maybe a
• 8:32 - 8:34
darker blue I'll do
the 2 times tables.
• 8:34 - 8:35
What's 2 times 1?
• 8:35 - 8:36
That's 2.
• 8:36 - 8:38
It's the same thing
as 1 times 2.
• 8:38 - 8:40
Notice, these two numbers
are the same thing.
• 8:40 - 8:42
What's 2 times 2?
• 8:42 - 8:43
That's 4.
• 8:43 - 8:45
2 times 3 is 6.
• 8:45 - 8:46
We just did this.
• 8:46 - 8:50
Every time you increment or you go
you multiply by a higher
• 8:50 - 8:51
number, you just add by 2.
• 8:51 - 8:53
2 times 4 is 8.
• 8:53 - 8:55
Same thing as 4 times 2.
• 8:55 - 8:57
2 times 5 is 10.
• 8:57 - 8:59
2 times 6 is 12.
• 8:59 - 9:01
I'm just adding 2 every time.
• 9:01 - 9:04
Up here I added 1 from every
step, here I'm adding 2.
• 9:04 - 9:07
2 times 7, 14.
• 9:07 - 9:09
2 times 8, 16.
• 9:09 - 9:13
2 two times 9, 18.
• 9:13 - 9:18
All right, let's do
our 3 times tables.
• 9:18 - 9:18
I'll do it in yellow.
• 9:22 - 9:24
3 times 1 is 3.
• 9:24 - 9:25
Notice, 3 times 1 is 3.
• 9:25 - 9:27
1 times 3 is 3.
• 9:27 - 9:29
These are the same values.
• 9:29 - 9:33
3 times 2 is the same
thing as 2 times 3.
• 9:33 - 9:38
3 times 2 should be the
same thing as 2 times 3.
• 9:38 - 9:40
So it's 6.
• 9:40 - 9:40
And that makes sense.
• 9:40 - 9:46
3 plus 3 is 6 or 2
plus 2 plus 2 is 6.
• 9:46 - 9:48
So every time here we're
going to increase by 3.
• 9:48 - 9:49
See the pattern.
• 9:49 - 9:51
3 times 3 is 9.
• 9:51 - 9:53
3 plus 3 plus 3.
• 9:53 - 9:55
So we went from 3 to 6 to 9.
• 9:55 - 9:57
So 3 times 4 is going to be 12.
• 9:57 - 9:59
I'm just adding 3 every time.
• 9:59 - 10:01
12 plus 3 is 15.
• 10:01 - 10:03
15 plus 3 is 18.
• 10:03 - 10:06
18 plus 3 is 21.
• 10:06 - 10:08
21 plus 3 is 24.
• 10:08 - 10:11
24 plus 3 is 27.
• 10:11 - 10:13
So 3 times 9 is 27.
• 10:13 - 10:15
3 times 8 is 24.
• 10:15 - 10:20
So if you were to say 8 plus
8 plus 8, it would be 24.
• 10:20 - 10:22
So now I'm going to speed it
up a little bit now that
• 10:22 - 10:23
we see the pattern.
• 10:23 - 10:25
And you should do this on your
own and you really should
• 10:25 - 10:27
memorize everything
we're doing.
• 10:27 - 10:29
You should actually go
all the way up to 12
• 10:29 - 10:31
in both directions.
• 10:31 - 10:31
So let's see.
• 10:31 - 10:35
4 times 1 is 4.
• 10:35 - 10:38
I'm just going to go up
by increments of 4.
• 10:38 - 10:40
So 4 plus 4 is 8.
• 10:40 - 10:42
8 plus 4 is 12.
• 10:42 - 10:44
12 plus 4 is 16.
• 10:44 - 10:46
16 plus 4 is 20.
• 10:46 - 10:48
20 plus 4 is 24.
• 10:48 - 10:51
4 times 6 is 24.
• 10:51 - 10:53
4 times 7, 28.
• 10:53 - 10:54
I'm just going up by 4.
• 10:54 - 10:59
32 and 36.
• 10:59 - 11:01
All right, 5 times 1.
• 11:01 - 11:07
5 times 1 is going to be 5.
• 11:07 - 11:09
Actually, we know that
anything-- well, I want to keep
• 11:09 - 11:11
changing colors, so I'll just
do it in rows like this.
• 11:11 - 11:13
5 times 1 is 5.
• 11:13 - 11:16
5 times 2 is 10.
• 11:16 - 11:17
5 times 3 is 15.
• 11:17 - 11:18
I'm just going to
increase by 5.
• 11:18 - 11:21
5 times tables are very fun as
well because every number
• 11:21 - 11:24
you're going to add-- if we
multiply 5 times-- well, we'll
• 11:24 - 11:26
learn about even and
odd in the future.
• 11:26 - 11:30
But every other number in its
times tables is going to end
• 11:30 - 11:32
with a 5, and then every other
one's going to end with a 0.
• 11:32 - 11:35
Because if you add 5
to 15 you get 20.
• 11:35 - 11:41
You get 25, 30, 35, 40, 45.
• 11:41 - 11:43
Fair enough.
• 11:43 - 11:47
6 times tables, We'll do it in green.
• 11:47 - 11:48
6 times 1 is 6.
• 11:48 - 11:49
That's easy.
• 11:49 - 11:51
You add 6 to that, you get 12.
• 11:51 - 11:52
You add 6 to that, you get 18.
• 11:52 - 11:54
You add 6 to that, you get 24.
• 11:54 - 11:56
You add 6 to that, you get 30.
• 11:56 - 12:01
Then you go 6 more, 36, 42, 48.
• 12:01 - 12:05
48 plus 6 is 54.
• 12:05 - 12:08
6 times 9 is 54.
• 12:08 - 12:09
All right, we're almost there.
• 12:09 - 12:12
7 times 1, that's 7.
• 12:12 - 12:14
7 times 1 is 7.
• 12:14 - 12:16
7 times 2 is 14.
• 12:16 - 12:18
7 times 3, 21.
• 12:18 - 12:20
7 times 4, 28.
• 12:20 - 12:24
7 times 5, what's 28 plus 7?
• 12:24 - 12:24
Let's see.
• 12:24 - 12:25
If you add 2 you get to 30.
• 12:25 - 12:28
Then you have 5, 35.
• 12:28 - 12:29
7 times 6, 42.
• 12:29 - 12:33
7 times 7, 49.
• 12:33 - 12:35
7 times 8.
• 12:35 - 12:38
Seven times is going to be
7 plus this, so it's 56.
• 12:38 - 12:42
I always used to get confused
between 7 times 8 being 56
• 12:42 - 12:44
and 6 times 9 being 54.
• 12:44 - 12:46
So now that I pointed out to
you that I always got confused
• 12:46 - 12:48
between those two, it's your
job not to be confused
• 12:48 - 12:49
by those two.
• 12:49 - 12:53
7 times 8 you could
say has the 6 in it.
• 12:53 - 12:55
6 times 9 doesn't
have the 6 in it.
• 12:55 - 12:56
That's the way I think of it.
• 12:56 - 12:58
Anyway, 7 times 9.
• 12:58 - 12:59
We're going to add
another 7 here.
• 12:59 - 13:01
It's going to be 63.
• 13:01 - 13:05
I'll do it in the same color.
• 13:05 - 13:08
All right, we're at
our 8 times tables.
• 13:08 - 13:11
8 times 1 is 8.
• 13:11 - 13:13
8 times 2 is 16.
• 13:13 - 13:14
24.
• 13:14 - 13:16
Alright, 8 times 3 is 24.
• 13:16 - 13:18
And if we go to 3 times 8
we should also see the 24.
• 13:18 - 13:20
Yep, it's there.
• 13:20 - 13:21
These values are the same.
• 13:21 - 13:23
So we're actually
doing things twice.
• 13:23 - 13:25
We're doing it when you do
8 times 3 and we're doing
• 13:25 - 13:27
it when we did 3 times 8.
• 13:27 - 13:28
Let's see.
• 13:28 - 13:31
8 times 4, you're going
to add 8 to it-- 32.
• 13:31 - 13:32
40.
• 13:32 - 13:35
Add another 8, 48.
• 13:35 - 13:37
Notice, 8 times 6, 48.
• 13:37 - 13:40
6 times 8, 48.
• 13:40 - 13:42
All right, 8 times 7.
• 13:42 - 13:46
Well, we already pointed
that one out, that was 56.
• 13:46 - 13:48
8 times 8, 64.
• 13:48 - 13:52
8 times 9, add 8
to this, is 72.
• 13:52 - 13:55
Now we're at the
9 times tables.
• 13:55 - 13:57
I'm running out of colors.
• 13:57 - 14:00
Maybe I'll reuse
a color or two.
• 14:00 - 14:01
I'll use the blue again.
• 14:01 - 14:03
9 times 1 is 9.
• 14:03 - 14:07
9 times 2, 18 9 times 3-- we
actually know all of these.
• 14:07 - 14:09
We could look it up in the rest
of the table because 9 times 3
• 14:09 - 14:11
is the same thing as 3 times 9.
• 14:11 - 14:13
It's 27.
• 14:13 - 14:15
Add 9 to that.
• 14:15 - 14:18
27 plus 9 is 36.
• 14:18 - 14:22
36 plus 9 is 45.
• 14:22 - 14:25
Notice, every time you add
9, you go almost up by
• 14:25 - 14:26
10, but 1 less than that.
• 14:26 - 14:30
So up by 10 would be 46, and
then one less than that is 45.
• 14:30 - 14:33
But anyway, we'll talk about, notice, the
1's-- well, I'll talk more
• 14:33 - 14:34
about it in the future.
• 14:34 - 14:38
But we go from a 9, 8,
7, 6, 5 on this digit,
• 14:38 - 14:39
on the second digit.
• 14:39 - 14:43
And on this digit here
you go 1, 2, 3, 4.
• 14:43 - 14:44
So it's an interesting pattern.
• 14:44 - 14:47
Another interesting pattern is
the digits will add up to 9.
• 14:47 - 14:49
3 plus 6 is 9, 2 plus 7 is 9.
• 14:49 - 14:51
We'll talk more about that
in the future and maybe
• 14:51 - 14:53
prove that to you.
• 14:53 - 14:56
9 times 6, 54.
• 14:56 - 14:58
• 14:58 - 15:02
9 times 7, 63.
• 15:02 - 15:04
9 times 8, 72.
• 15:04 - 15:06
9 times 9 is 81.
• 15:06 - 15:07
I don't know if
you can see that.
• 15:07 - 15:08
81.
• 15:08 - 15:09
There you go.
• 15:09 - 15:11
Now, I could keep going.
• 15:11 - 15:14
Actually, I should keep going.
• 15:14 - 15:18
Well, I realize this video
is already pretty long.
• 15:18 - 15:20
I want you to memorize this
right now because this is
• 15:20 - 15:21
going to get you pretty far.
• 15:21 - 15:25
In the next video I'm going to
do the times tables past 9.
• 15:25 - 15:27
See you soon.
Title:
Multiplication 2: The Multiplication Tables
Description:
Introduction to the multiplication tables from 2-9.
more » « less
Video Language:
English
Duration:
15:27
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• Follow the directions to make a print of the Olympics rings logo. Part of our Sports and Olympics Theme Unit
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• Each chart below lists the swimmers and times for a swimming event. In the blank, write the place that each swimmer finished.
• Cut out the nine pairs of summer Olympic theme pictures and make a memory match game.
• Determine your own resting and active heart rate with this learning center.
• Read the graph of the gold medal winners from the 2004 Olympics and interpret the results. Students could also create further charts for other Olympics games and for silver and bronze winners.
• "Color the torches with the answer 5 to find the path from Greece to the Olympic Flame."
• Add this activity to you Olympic Games Theme Unit. Students will learn about the Food Pyramid and Nutrition.
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• Use the Venn diagram to compare slalom and flatwater kayaking.
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• Cut out the pictures and sort them into the correct classes on the next page. Glue them in place.
• Color the chart to show the statistics for estimated worldwide population for these endangered animals.
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• Set of bulletin board sized, wood design letters to spell "Round-Up."
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Algebra -> Algebra -> Exponents-negative-and-fractional -> SOLUTION: solve (b^2-4b+4)^12 Log On
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Algebra: Negative and Fractional exponents Solvers Lessons Answers archive Quiz In Depth
Question 611506: solve (b^2-4b+4)^12Answer by lwsshak3(6522) (Show Source): You can put this solution on YOUR website!solve (b^2-4b+4)^12 =[(b-2)^2]^12 =(b-2)^24 | 176 | 575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2013-20 | latest | en | 0.715993 |
https://math4finance.com/general/ten-identical-slips-of-paper-each-contain-one-number-from-one-to-ten-inclusive-the-papers-are-put-into-a-bag-and-then-mixed-around-which-statements-about-the-situation-are-true-check-all-that-apply-p-6-p-1-p-5-p-gt-10-0p-1-lt-x-lt | 1,718,503,523,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00722.warc.gz | 343,590,344 | 7,347 | Q:
# Ten identical slips of paper each contain one number from one to ten, inclusive. The papers are put into a bag and then mixed around. Which statements about the situation are true? Check all that apply.P(6) = P(1)P(5) = P(>10) = 0P(1 < x < 10) = 100%S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}If A β S; A could be {1, 3, 5, 7, 9}
Accepted Solution
A:
For E2020 users, the correct answer is A, C, E, F. | 163 | 412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-26 | latest | en | 0.864731 |
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## The Franck-Hertz ice cream truck
Imagine that there’s an ice cream truck parked near a school. It’s 3pm; class is dismissed and a steady stream of kids approach the truck. For simplicity, let’s stipulate the following rules:
1. If a kid has enough money, he or she will buy an ice cream cone;
2. A kid will only buy one ice cream cone;
3. A kid will buy the most expensive ice cream cone he or she can.
After the kids buy ice cream (or not) they continue on their way, having whatever leftover money they might have in their pocket.
My question is this: by seeing how much money kids have left, what can you determine about the prices of ice cream cones? (Astute observers will note that this toy model has many similarities to the Franck-Hertz experiment of 1914, but we’re not there yet.)
Let’s look first at some hypothetical data. Here we have plotted L (money left over in a kid’s pocket) as a function of M (how much money a kid started with) for a school with 11 kids:
There doesn’t seem to be a pattern. Maybe we don’t have enough data? Let’s increase the number of kids to 42:
The sawtooth form of the function is characteristic of problems of this type, and can be understood intuitively. First of all, notice that the function is linear to begin with, with a slope of exactly one; this means that below a certain threshold value of M, a kid can’t afford any ice cream, so he/she ends up with the same amount of money he/she started with. Eventually, if M ≥ \$1, a kid can buy a cone; L then plummets because the kid has spent a dollar.
It seems obvious from the graph, then, that ice cream cones are priced at \$1, \$2, and \$3; beyond that we don’t have any data so can’t draw more conclusions.
Now, if we relax the stipulation that kids only buy one cone, then our data is ambiguous. Maybe the cones are priced \$1/\$2/\$3, or maybe cones are always just \$1, and kids are buying more than one cone if they can. We can’t distinguish between these cases; that’s why I put the original stipulation there to begin with.
Let’s try a more challenging graph:
This graph is much harder to interpret. The first peak is much bigger than the others; there is also a very tiny peak on the right-hand side. It helps if you know where the jumps are: L jumps down to zero whenever M is equal to \$0.67, \$1.02, \$1.34, \$1.69, \$2.01, and \$2.04. If you want to work out the ice cream cone prices for yourself, feel free. I’ll wait.
The solution? A little trial and error will give you two different prices for ice cream cones: A= \$0.67 and B=\$1.02. Then the jumps occur at A, B, 2A, A+B, 3A, and 2B. If we were doing physics, we’d make a prediction: we would expect the next jump to occur at \$2.36, which is B+2A. Observing this would support our hypothesis. But if the next jump occurred at \$2.22, say, then we’d have to revise our theory and posit a new ice cream C priced at \$2.22.
What does any of this have to do with physics?
I use this example when I introduce the Franck-Hertz experiment to my students. This experiment was first performed in 1914 (an auspicious year!) and provided support for Bohr’s idea that atoms have specific (quantized) energy levels. Electrons are accelerated and shot towards mercury atoms (in a vapor). The electrons may then give energy to the atoms (exciting them) or they may not, bouncing off without loss of kinetic energy. We look at how much energy the electrons have to start with, and how much they end up with, and thereby deduce the energy levels of the mercury atoms.
How is that possible? In terms of the analogy, make the following transformations:
kids –> electrons
ice cream truck –> Hg atoms
kids buying ice cream at certain prices –> electrons giving certain amount of energy to Hg atoms
prices of ice cream cones –> energy levels of Hg
M (initial amount of money) –> V (proportional to initial kinetic energy of electron)
L (leftover money) –> I (proportional to final kinetic energy of electron)
If energy levels of an atom are truly quantized, then we would expect a graph of I vs. V to look like our graphs above, with an increasing sawtooth pattern. Where the drops occur will then tell us the specific energy levels of the Hg atom. (Incidentally, why do we use V and I? Well, in the actual experiment the easiest way to measure initial kinetic energy is by measuring the voltage used to accelerate the electrons; the easiest way to measure final kinetic energy is to measure the current of the electrons after they have passed through the Hg vapor.)
How did the experiment go? Here are the results:
It should be obvious that atomic energy levels have specific “prices”, and that there’s a minimum amount of energy that an electron must have in order to “buy” an atomic transition (exciting the atom at the expense of the electron’s kinetic energy). It remains for the experimenter to do some elementary unit conversions, to translate I and V into final and initial kinetic energy.
[Note for advanced students and/or physicists: it is interesting to ponder the following questions, which highlight the fact that the ice cream analogy is not perfect: (1) why is the I vs.V graph not linear at low voltage? (2) After “spending” kinetic energy to cause the first excited state, why does the electron’s energy not drop all the way down to zero?]
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https://www.brainkart.com/article/Solved-Example-Problems-for-Physics--Nature-of-Physical-World-and-Measurement_34448/ | 1,680,198,202,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00523.warc.gz | 740,748,080 | 17,971 | Home | Solved Example Problems for Physics: Nature of Physical World and Measurement
# Solved Example Problems for Physics: Nature of Physical World and Measurement
Physics : Nature of Physical World and Measurement
## SOLVED EXAMPLE
1. Find the dimensions of a and b in the formula [P+ a/V2][V-b]=RT where P is pressure and V is the volume of the gas
Solution:
By the principle of homogeneity, a / V2 is of the dimensions of pressure and b is of the dimensions of volume.
2. Show that (P-5/6 ρ1/2 E1/3 ) is of the dimension of time. Here P is the pressure, is the density and E is the energy of a bubble)
Solution:
Dimension of Pressure = [ML-1 T -2]
Dimension of density = [ML-3 ]
Dimension of Energy = [ML2 T -2 ]
By substituting in the given equation,
3. Find the dimensions of mass in terms of Energy, length and time
Solution:
Let the dimensions of Energy, Length and Time be [E ],[L],[T] respectively.
We know that Force = mass x acceleration
4. A physical quantity Q is found to depend on quantities x,y,z obeying relation Q =x1y3/z1. The percentage errors in x, y and z are 2%, 3% and 1% respectively. Find the percentage error in Q.
Solution:
5. The mass and volume of a body are found to be 4±.03 kg and 5±.01 m3 respectively. Then find the maximum possible percentage error in density.
Solution:
Error in density = error in mass + error in volume
6. Using a Vernier Callipers, the length of a cylinder in different measurements is found to be 2.36 cm, 2.27 cm, 2.26 cm, 2.28 cm, 2.31 cm, 2.28 cm and 2.29 cm. Find the mean value, absolute error, the relative error and the percentage error of the cylinder.
Solution:
The given readings are 2.36 cm, 2.27 cm, 2.26 cm, 2.28 cm, 2.31 cm, 2.28 cm and 2.29 cm
Absolute errors in the measurement are,
7. The shadow of a pole standing on a level ground is found to be 45 m longer when the sun’s altitude is 30° than when it was 60°. Determine the height of the pole. [Given √3 = 1.73]
Solution:
Let the height of the pole be h.
Substituting the values of x in the above equation
8. Calculate the number of times a human heart beats in the life of 100 years old man. Time of one heart beat = 0.8s.
Solution:
Life of the man = 100 years
100 years includes 76 normal years and 24 leap years
Total no of days = 76 × 365 + 24 × 366 = 36524 days
Number of seconds = 36524 × 24 × 3600 = 3.155 x 109 second
9. The parallax of a heavenly body measured from two points diametrically opposite on equator of earth is 2. Calculate the distance of the heavenly body. [Given radius of the earth = 6400km] [1″ = 4.85 10-6 rad]
Solution:
Angle θ = 21 = 2 x 60″ = 120″ = 120 × 4.85 × 10-6 rad
The distance of heavenly body
10. Convert a velocity of 72 kmh-1 into ms-1 with the help of dimensional analysis.
Solution:
The dimensional formula for velocity is [L T-1]
11. Check the correctness of the following equation using dimensional analysis. Make a comment on it.
= ut + 1/4 at2 where s is the displacement, u is the initial velocity, t is the time and a is the acceleration produced.
Solution:
Dimension for distance s = [L]
Dimension for initial velocity
Dimension for time t = [T]
Dimension for acceleration
According to the principle of homogeneity,
Dimensions of LHS = Dimensions of RHS
Substituting the dimensions in the given formula
= ut + 1/4 at 1/4 is a number. It has no dimensions
As the dimensional formula of LHS is same as that of RHS, the equation is dimensionally correct.
Comment:
But actually it is a wrong equation. We know that the equation of motion is
So, a dimensionally correct equation need not be the true (or) actual equation S = ut + 1/4 at2
But a true equation is always dimensionally correct.
12. Round - off the following numbers as indicated.
a) 17.234 to 3 digits
b) 3.996 × 105 to 3 digits
c) 3.6925 × 10-3 to 2 digits
d) 124783 to 5 digits.
Solution:
a) 17.2
b) 4.00 × 105
c) 3.7 × 10-3
d) 124780
13. Solve the following with regard to significant figures.
a) √(4.5-3.31)
b) 5.9 × 105 - 2 .3 × 104
c) 7.18 + 4.3
d) 6.5 + .0136
Solution:
a) Among the two, the least number of significant after decimal is one
b) The number of minimum significant figures is 2
c) The lowest least number of significant digit after decimal is one
7.18 + 4.3 = 11.48 Rounding off we get 11.5
The lowest least number of significant digit after decimal is one
14. Arrive at Einstein’s mass-energy relation by dimensional method (E = mc2)
Solution:
Let us assume that the Energy E depends on mass m and velocity of light c.
E =kma cb where K a constant
Dimensions of E = [ML2T -2 ]
Dimensions of m = [M]
Dimensions of c = [LT-1]
Substituting the values in the above equation
The value of constant k = 1
E = mc2 This is Einstein’s mass energy relation
15. The velocity of a body is given by the equation v = b/t + c t2+dt3. Find the dimensional formula for b.
Solution:
(b/t) should have the dimensions of velocity
b has the dimensions of (velocity time)
16. The initial and final temperatures of liquid in a container are observed to be 75.4 ± 0.5°C and 56.8 ± 0.2°C. Find the fall in the temperature of the liquid.
Solution:
t1 = (75.4 ± 0.5)°C
t2 = (56.8 ± 0.2)°C
Fall in temperature = (75.4 ± 0.5°C) – (56.8 ± 0.2°C)
17. Two resistors of resistances R1 = 150 ± 2 Ohm and R2 = 220 ± 6 Ohm are connected in parallel combination. Calculate the equivalent resistance.
Solution:
The equivalent resistance of a parallel combination
Substituting the value,
18. A capacitor of capacitance C = 3.0 ± 0.1 µF is charged to a voltage of V = 18 ± 0.4Volt . Calculate the charge Q [Use Q= CV]
Solution:
(C + C) = (3.0 ± 0.1) µf
(V + V) = (18 ± 0.4) V
Q = CV
## Solved Example Problems for Nature of Physical World and Measurement
Example 1.1
From a point on the ground, the top of a tree is seen to have an angle of elevation 60°. The distance between the tree and a point is 50 m. Calculate the height of the tree?
## Solution
Angle θ = 60°
The distance between the tree and a point = 50 m
Height of the tree (h) = ?
For triangulation method tan
h = x tan θ
= 50 × tan 60°
= 50 × 1.732
h = 86.6 m
The height of the tree is 86.6 m.
## Parallax method
Very large distances, such as the distance of a planet or a star from the Earth can be measured by the parallax method. Parallax is the name given to the apparent change in the position of an object with respect to the background, when the object is seen from two different positions. The distance between the two positions (i.e., points of observation) is called the basis (b). For example, consider Figure 1.4., an observer is specified by the position O. The observer is holding a pen before him, against the background of a wall. When the pen is looked at first by our left eye L (closing the right eye) and then by our right eye R (closing the left eye), the position of the pen changes with respect to the back ground of the wall. The shift in the position of an object (say, a pen) when viewed with two eyes, keeping one eye closed at a time is known as Parallax. The distance between the left eye (L) and the right eye (R) in this case is the basis.
LOR is called the parallax angle or parallactic angle.
Taking LR as an arc of length b and radius LO = RO = x
we get θ = b/x, b-basis, x-unknown distance
Knowing ‘b’ and measuring θ, we can calculate x.
## Determination of distance of Moon from Earth
In Figure 1.5, C is the centre of the Earth. A and B are two diametrically opposite places on the surface of the Earth. From A and B, the parallaxes θ1 and θ2 respectively of Moon M with respect to some distant star are determined with the help of an astronomical telescope. Thus, the total parallax of the Moon subtended on Earth AMB = θ1 + θ2 = θ.
If θ is measured in radians, then
Knowing the values of AB and θ,
we can calculate the distance MC of Moon from the Earth.
Example 1.2
The Moon subtends an angle of 1° 55 at the base line equal to the diameter of the Earth. What is the distance of the Moon from the Earth? (Radius of the Earth is 6.4 × 106 m)
Solution
Radius of the Earth = 6.4 × 106 m
From the Figure 1.5 AB is the diameter of the Earth (b)= 2 × 6.4 × 106 m Distance of the Moon from the Earth x = ?
The word RADAR stands for radio detection and ranging. A radar can be used to measure accurately the distance of a nearby planet such as Mars. In this method, radio waves are sent from transmitters which, after reflection from the planet, are detected by the receiver. By measuring, the time interval (t) between the instants the radio waves are sent and received, the distance of the planet can be determined as
where v is the speed of the radio wave. As the time taken (t) is for the distance covered during the forward and backward path of the radio waves, it is divided by 2 to get the actual distance of the object. This method can also be used to determine the height, at which an aeroplane flies from the ground.
Example 1.3
A RADAR signal is beamed towards a planet and its echo is received 7 minutes later. If the distance between the planet and the Earth is 6.3 × 1010 m. Calculate the speed of the signal?
Solution
The distance of the planet from the Earth d = 6.3 × 1010 m
The speed of signal
Example 1.4
In a series of successive measurements in an experiment, the readings of the period of oscillation of a simple pendulum were found to be 2.63s, 2.56 s, 2.42s, 2.71s and 2.80s. Calculate (i) the mean value of the period of oscillation (ii) the absolute error in each measurement (iii) the mean absolute error (iv) the relative error (v) the percentage error. Express the result in proper form.
Solution
Example 1.5
Two resistances R1 = (100 ± 3) Ω, R2 = (150 ± 2) Ω, are connected in series. What is their equivalent resistance?
Solution
Equivalent resistance R = ?
Equivalent resistance R = R1 + R2
Example 1.6
The temperatures of two bodies measured by a thermometer are t1 = (20 + 0.5)°C, t2 = (50 ± 0.5)°C. Calculate the temperature difference and the error therein.
Solution
Example 1.7
The length and breadth of a rectangle are (5.7 ± 0.1) cm and (3.4 ± 0.2) cm respectively. Calculate the area of the rectangle with error limits.
Solution
Example 1.8
The voltage across a wire is (100 ± 5)V and the current passing through it is (10±0.2) A. Find the resistance of the wire.
Solution
Example 1.9
A physical quantity x is given by x
If the percentage errors of measurement in a, b, c and d are 4%, 2%, 3% and 1% respectively then calculate the percentage error in the calculation of x.
Solution
The percentage error in x is given by
The percentage error is x = 17.5%
Example 1.10
State the number of significant figures in the following
i. 600800
ii. 400
iii. 0.007
iv. 5213.0x
v. 2.65 × 1024 m
vi. 0.0006032
Solution
Solution: i) four ii) one iii) one iv) five v) three vi) four
Example 1.11
Round off the following numbers as indicated
i) 18.35 up to 3 digits
ii) 19.45 up to 3 digits
iii) 101.55 × 106 up to 4 digits
iv) 248337 up to digits 3 digits
v) 12.653 up to 3 digits.
Solution
i) 18.4 ii) 19.4 iii) 101.6 × 106 iv) 248000 v) 12.7
Example 1.12
Convert 76 cm of mercury pressure into Nm−2 using the method of dimensions.
Solution
In cgs system 76 cm of mercury pressure = 76 × 13.6 × 980 dyne cm−2
The dimensional formula of pressure P is [ML−1T−2]
Example 1.13
If the value of universal gravitational constant in SI is 6.6x10−11 Nm2 kg−2, then find its value in CGS System?
Solution
Let GSI be the gravitational constant in the SI system and Gcgs in the cgs system. Then
The dimensional formula for G is M−1 L3T −2
Example 1.14
Check the correctness of the equation
using dimensional analysis method
Solution
Both sides are dimensionally the same, hence the equations
is dimensionally correct.
Example 1.15
Obtain an expression for the time period T of a simple pendulum. The time period T depends on (i) mass ‘m’ of the bob (ii) length ‘l’ of the pendulum and (iii) acceleration due to gravity g at the place where the pendulum is suspended. (Constant k = 2π) i.e
Solution
Here k is the dimensionless constant. Rewriting the above equation with dimensions
Comparing the powers of M, L and T on both sides, a=0, b+c=0, -2c=1
Solving for a,b and c a = 0, b = 1/2, and c = −1/2
From the above equation T = k. m0 l1/2 g−1/2
Example 1.16
The force F acting on a body moving in a circular path depends on mass of the body (m), velocity (v) and radius (r) of the circular path. Obtain the expression for the force by dimensional analysis method. (Take the value of k=1)
where k is a dimensionless constant of proportionality. Rewriting above equation in terms of dimensions and taking k = 1, we have
Comparing the powers of M, L and T on both sides
From the above equation we get
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
11th Physics : UNIT 1 : Nature of Physical World and Measurement : Solved Example Problems for Physics: Nature of Physical World and Measurement | | 3,885 | 13,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-14 | longest | en | 0.788253 |
http://jean-pierre.moreau.pagesperso-orange.fr/Cplus/multpol_cpp.txt | 1,607,138,827,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-50/segments/1606141746033.87/warc/CC-MAIN-20201205013617-20201205043617-00474.warc.gz | 35,523,799 | 1,660 | /**************************************************** * This program multiplies a polynomial P(x) by a * * polynomial Q(x) * * ------------------------------------------------- * * Ref.: "Mathématiques en Turbo-Pascal By M. Ducamp * * and A. Reverchon (vol 2), Eyrolles, Paris, 1988" * * [BIBLI 05]. * * ------------------------------------------------- * * SAMPLE RUN: * * * * MULTIPLY TWO POLYNOMIALS: * * * * P(X) = x3 - 6x + 7 * * Q(x) = 5x5 -3x4 +x2 -3 * * * * +5 X8 -3 X7 -30 X6 +54 X5 -21 X4 -9 X3 +7 X2 * * * * +18 X -21 * * * * ------------------------------------------------- * * Functions used (of unit Polynoms): * * * * AddNumber(), EnterPolynom(), DisplayPolynom() * * and MultNumber(). * * * * C++ version by J-P Moreau. * * (To be linked with polynoms.cpp).* * (www.jpmoreau.fr) * ****************************************************/ #include #include #include #include "polynoms.h" ar_polynom *P, *Q, *R; // P(X) * Q(X) = R(X) - R can be either P or Q bool MultPolynom(ar_polynom *P, ar_polynom *Q, ar_polynom *R) { int i,j, n; ar_number u; ar_polynom *nr; nr = (ar_polynom *) calloc(1,sizeof(ar_polynom)); //verify that P and Q are not void if (P->degree==0 && P->coeff[0].value==0) return FALSE; if (Q->degree==0 && Q->coeff[0].value==0) return FALSE; nr->degree=P->degree+Q->degree; if (nr->degree>AR_MAXPOL) return FALSE; // R degree is too big for (n=0; n<=nr->degree; n++) { if (!SetNumber(&nr->coeff[n],"0")) return FALSE; for (i=0; i<=P->degree; i++) { j=n-i; if (j>=0 && j<=Q->degree) { if (!MultNumber(P->coeff[i],Q->coeff[j],&u)) return FALSE; if (!AddNumber(nr->coeff[n],u,&nr->coeff[n])) return FALSE; } } } //copy nr in R R->degree=nr->degree; for (i=0; i<=nr->degree; i++) R->coeff[i]=nr->coeff[i]; free(nr); return TRUE; } void main() { //dynamic memory allocation of polynomials P = (ar_polynom *) calloc(1,sizeof(ar_polynom)); Q = (ar_polynom *) calloc(1,sizeof(ar_polynom)); R = (ar_polynom *) calloc(1,sizeof(ar_polynom)); printf("\n MULTIPLY TWO POLYNOMIALS:\n\n"); if (!EnterPolynom(" P(X) = ", P)) return; if (!EnterPolynom(" Q(X) = ", Q)) return; printf("\n"); if (MultPolynom(P,Q,R)) DisplayPolynom(R); else printf(" Error in multiplication.\n"); printf("\n\n"); free(P); free(Q); free(R); } //end of file multpol.cpp | 728 | 2,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-50 | longest | en | 0.330412 |
https://oscar-system.github.io/Singular.jl/dev/alghom/ | 1,721,058,221,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00277.warc.gz | 404,061,959 | 5,036 | Algebra Homomorphisms
Singular.jl allows the creation of algebra homomorphisms of Singular polynomial rings over Nemo/Singular coefficient rings.
The default algebra homomorphism type in Singular.jl is the Singular SAlgHom type.
Additionally, a special type for the identity homomorphism has been implemented. The type in Singular.jl for the latter is SIdAlgHom.
All algebra homomorphism types belong directly to the abstract type AbstractAlgebraHomomorphism{T}.
Algebra Homomorphism functionality
Constructors
Given two Singular polynomial rings $R$ and $S$ over the same base ring, the following constructors are available for creating algebra homomorphisms.
Singular.AlgebraHomomorphismMethod
AlgebraHomomorphism(D::PolyRing, C::PolyRing, V::Vector)
Constructs an algebra homomorphism $f: D \to C$, where the $i$-th variable of $D$ is mapped to the $i$-th entry of $V$. $D$ and $C$ must be polynomial rings over the same base ring.
source
Examples
julia> L = FiniteField(3, 2, "a")
(Finite field of characteristic 3 and degree 2, a)
julia> R, (x, y, z, w) = polynomial_ring(L[1], ["x", "y", "z", "w"];
ordering=:negdegrevlex)
(Singular polynomial ring (9,a),(x,y,z,w),(ds(4),C), spoly{n_GF}[x, y, z, w])
julia> S, (a, b, c) = polynomial_ring(L[1], ["a", "b", "c"];
ordering=:degrevlex)
(Singular polynomial ring (9,a),(a@1,b,c),(dp(3),C), spoly{n_GF}[a, b, c])
julia> V = [a, a + b^2, b - c, c + b]
4-element Vector{spoly{n_GF}}:
a
b^2 + a
b + a^4*c
b + c
julia> f = AlgebraHomomorphism(R, S, V)
Algebra homomorphism
from Singular polynomial ring (9,a),(x,y,z,w),(ds(4),C)
to Singular polynomial ring (9,a),(a@1,b,c),(dp(3),C)
Defining equations: spoly{n_GF}[a, b^2 + a, b + a^4*c, b + c]
Operating on objects
It is possible to act on polynomials and ideals via algebra homomorphisms.
Examples
R, (x, y, z, w) = polynomial_ring(Nemo.ZZ, ["x", "y", "z", "w"];
ordering=:negdegrevlex)
S, (a, b, c) = polynomial_ring(Nemo.ZZ, ["a", "b", "c"];
ordering=:degrevlex)
V = [a, a + b^2, b - c, c + b]
f = AlgebraHomomorphism(R, S, V)
id = IdentityAlgebraHomomorphism(S)
J = Ideal(R, [x, y^3])
p = x + y^3 + z*w
K = f(J)
q = f(p)
Composition
AbstractAlgebra.composeMethod
compose(f::AbstractAlgebra.Map(Singular.SAlgHom),
g::AbstractAlgebra.Map(Singular.SAlgHom))
Return an algebra homomorphism $h: domain(f) \to codomain(g)$, where $h = g(f)$.
source
A short command for the composition of $f$ and $g$ is f*g, which is the same as compose(f, g).
Examples
julia> R, (x, y, z, w) = polynomial_ring(QQ, ["x", "y", "z", "w"];
ordering=:negdegrevlex)
(Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C), spoly{n_Q}[x, y, z, w])
julia> S, (a, b, c) = polynomial_ring(QQ, ["a", "b", "c"];
ordering=:degrevlex)
(Singular polynomial ring (QQ),(a,b,c),(dp(3),C), spoly{n_Q}[a, b, c])
julia> V = [a, a + b^2, b - c, c + b]
4-element Vector{spoly{n_Q}}:
a
b^2 + a
b - c
b + c
julia> W = [x^2, x + y + z, z*y]
3-element Vector{spoly{n_Q}}:
x^2
x + y + z
y*z
julia> f = AlgebraHomomorphism(R, S, V)
Algebra homomorphism
from Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C)
to Singular polynomial ring (QQ),(a,b,c),(dp(3),C)
Defining equations: spoly{n_Q}[a, b^2 + a, b - c, b + c]
julia> g = AlgebraHomomorphism(S, R, W)
Algebra homomorphism
from Singular polynomial ring (QQ),(a,b,c),(dp(3),C)
to Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C)
Defining equations: spoly{n_Q}[x^2, x + y + z, y*z]
julia> idR = IdentityAlgebraHomomorphism(R)
Identity algebra homomorphism
from Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C)
to Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C)
Defining equations: spoly{n_Q}[x, y, z, w]
julia> h1 = f*g
Algebra homomorphism
from Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C)
to Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C)
Defining equations: spoly[x^2, 2*x^2 + 2*x*y + y^2 + 2*x*z + 2*y*z + z^2, x + y + z - y*z, x + y + z + y*z]
julia> h2 = idR*f
Algebra homomorphism
from Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C)
to Singular polynomial ring (QQ),(a,b,c),(dp(3),C)
Defining equations: spoly{n_Q}[a, b^2 + a, b - c, b + c]
julia> h3 = g*idR
Algebra homomorphism
from Singular polynomial ring (QQ),(a,b,c),(dp(3),C)
to Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C)
Defining equations: spoly{n_Q}[x^2, x + y + z, y*z]
julia> h4 = idR*idR
Identity algebra homomorphism
from Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C)
to Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C)
Defining equations: spoly{n_Q}[x, y, z, w]
Preimages
Examples
julia> R, (x, y, z, w) = polynomial_ring(QQ, ["x", "y", "z", "w"];
ordering=:negdegrevlex)
(Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C), spoly{n_Q}[x, y, z, w])
julia> S, (a, b, c) = polynomial_ring(QQ, ["a", "b", "c"];
ordering=:degrevlex)
(Singular polynomial ring (QQ),(a,b,c),(dp(3),C), spoly{n_Q}[a, b, c])
julia> I = Ideal(S, [a, a + b^2, b - c, c + b])
Singular ideal over Singular polynomial ring (QQ),(a,b,c),(dp(3),C) with generators (a, b^2 + a, b - c, b + c)
julia> f = AlgebraHomomorphism(R, S, gens(I))
Algebra homomorphism
from Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C)
to Singular polynomial ring (QQ),(a,b,c),(dp(3),C)
Defining equations: spoly{n_Q}[a, b^2 + a, b - c, b + c]
julia> idS = IdentityAlgebraHomomorphism(S)
Identity algebra homomorphism
from Singular polynomial ring (QQ),(a,b,c),(dp(3),C)
to Singular polynomial ring (QQ),(a,b,c),(dp(3),C)
Defining equations: spoly{n_Q}[a, b, c]
julia> P1 = preimage(f, I)
Singular ideal over Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C) with generators (x, y, z, w)
julia> P2 = preimage(idS, I)
Singular ideal over Singular polynomial ring (QQ),(a,b,c),(dp(3),C) with generators (a, b^2 + a, b - c, b + c)
julia> K1 = kernel(f)
Singular ideal over Singular polynomial ring (QQ),(x,y,z,w),(ds(4),C) with generators (4*x - 4*y + z^2 + 2*z*w + w^2)
julia> K2 = kernel(idS)
Singular ideal over Singular polynomial ring (QQ),(a,b,c),(dp(3),C) with generators (0) | 2,119 | 5,979 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-30 | latest | en | 0.702876 |
https://www.rocketmath.com/2020/11/12/data-proven-approach-to-math-facts-fluency/ | 1,713,083,822,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816875.61/warc/CC-MAIN-20240414064633-20240414094633-00334.warc.gz | 895,765,649 | 54,757 | ## A Data-Proven Approach to Measuring Math Facts Fluency (That Kids Love!)
You know your student is building math facts fluency, but can you quantify it? With Rocket Math’s Online Game, you can track your student’s progress learning addition, subtraction, multiplication, and division (and 12 other Learning Tracks) from level A to Z. Progress is tracked numerically through 1-Minute Races so that you can compare results over time, between students, and across classrooms.
## How Rocket Math’s 1-Minute Races Measure Math Fact Fluency
Starting in the Fall of 2020, the Rocket Math Online Game began scheduling 1-Minute Races for students as they worked through the 16 different Learning Tracks taught in the game.
The 1-Minute Races present a random selection of all the facts in each Learning Track. Students are encouraged to skip facts they haven’t learned, but the scores shown here are the correct answers. When a student completes sets A, i, R, and Z, the 1-Minute Races (also known as math facts fluency tests) are scheduled. We wait until after students complete set A to give the first pre-test, so we know the students know how to use the interface and answer questions.
As the chart below shows, Set A represents the average student’s starting score. Set i and R show intermediate progress. And finally, Set Z represents the average fluency students have achieved by playing through levels 26 A to Z in that Learning Track.
#### Average 1-Minute Race Scores of Students Playing the Rocket Math Online Game
The more correct problems per minute, the more fluent the learner. On average, by the end of a Learning Track students should aim for 20 to 30 correct problems per minute in the Online Game races. Want to learn more about how fast students should be with math facts? Read here.
## Data show improved math facts fluency as students work through each Learning Track
As you can see above, there is a steady rise in fluency as students work through each Learning Track. Because fluency scores go up consistently at each measuring point, and they do so in every Learning Track, we can be sure that the Online Game is effectively developing fluency.
## Data shows improvement based on how many math facts answered per minute
The fact that on each teststudents can correctly answer more problems shows that they are learning, and the data proves it. That is very important, as you don’t purchase this game simply to entertain children. The goal is to help students achieve math facts fluency, and we are seeing that here. Individual results will, of course, vary.
## Not only an excellent online math speed test but also an excellent tool for research
Teachers can also assign these 1-Minute Races at any time, or even on a weekly basis. The scheduled tests, however, show the best effect of the Online Game because the amount of work done to achieve the tests is equal among students. Both scheduled and teacher-assigned results are kept, and the results of these tests can be exported in excel files, so they are available for research purposes. We invite you to study your students’ math fact fluency using the Rocket Math Online Game and 1-Minute Races. One thing is certain from our data, the Rocket Math Online Game works! | 728 | 3,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-18 | longest | en | 0.928978 |
http://2011.isiproceedings.org/Abstracts/STS048.01.html | 1,519,431,592,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814872.58/warc/CC-MAIN-20180223233832-20180224013832-00036.warc.gz | 1,768,078 | 2,113 | Uses of Parametric Modelling of Income and Direct Estimates of Aggregate Income Indicators from EU-SILC Data and the Derived Synthetic Dataset AMELIA with Focus on the Generalized Beta Distribution of the Second Kind (GB2). Derivation of Design-Based Estimation
Monique Graf, Desislava Nedyalkova
Statistical Methods Unit, Swiss Federal Statistical Office, Neuchâtel, Switzerland
The reason why parametric estimation may be useful, when empirical data and estimators are available is threefold: 1. to stabilize estimation; 2. to get insight into the relationships between the characteristics of the theoretical distribution and a set of indicators, e.g. by sensitivity plots; 3. to deduce the whole distribution from known empirical indicators, when the raw data are not available. The talk will address these points and convey the experiences done within the AMELI project on the parametric estimation of the EU-SILC monetary indicators.
Special emphasis will be laid on the Generalized Beta distribution of the second kind (GB2), derived by McDonald (1984). Apart from the scale parameter, this distribution has three shape parameters: the first governing the overall shape, the second the lower tail and the third the upper tail of the distribution. These characteristics give to the GB2 a large flexibility for fitting a wide range of empirical distributions and it has been established that it outperforms other four-parameter distributions for income data (Kleiber and Kotz, 2003). We have studied different types of estimation methods, taking into account the design features of the EU-SILC surveys. Pseudo-maximum likelihood estimation of the parameters is compared with a nonlinear fit from the indicators. Variance estimation is done by linearization and different types of simplified formulas for the variance proposed in the litterature are evaluated by simulation. The computations are made on the synthetic universe AMELIA constructed from the EU-SILC data (Muennich et al., 2010) and the simulation is done with the R package SimFrame (Templ et al., 2010). Both AMELIA and SimFrame are developped in the context of the AMELI project. The parametric methods we have developped are made available in the R package GB2, which is part of the output of the AMELI project.
Keywords: Income data; Monetary indicators; Parametric distribution; generalized beta distribution of the second kind; Design-based variance estimation
Biography: Monique Graf leads a group of experts in statistical methods at the Swiss Federal Statistical Office. The group is part of the Statistical methods unit and is involved in the statistical consulting for the Office, as well as for other offices at the federal and regional level. Research and development are also duties of the statistical methods unit. Monique is responsible for several Work packages in the FP7 AMELI project and will speak about the research conducted by the statistical methods team. | 595 | 2,951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-09 | latest | en | 0.876667 |
imilon.medium.com | 1,627,687,052,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154032.75/warc/CC-MAIN-20210730220317-20210731010317-00589.warc.gz | 307,852,942 | 33,864 | Logistic Regression: A Detailed Overview from Scratch
Fundamentals of logistic regression and concepts like maximum likelihood function, cross-entropy are covered briefly
Logistic regression is one of the most commonly used algorithms for machine learning and is a building block for neural networks. It is important to understand the fundamentals and maths behind logistic regression. In this post, we will explore the fundamentals of logistic regression, and concepts like maximum likelihood function, cross-entropy/log-likelihood are covered briefly. In the end, we have provided codes for Logistic Regression in Python from scratch.
Previously, we have covered linear classifier models and we have described the concept of decision boundaries. In linear classifier models, we take some features xᵢ, some random weights wᵢ and calculate the…
Descriptive Statistics: A Brief Introduction
Descriptive and summary statistics associated with histogram, boxplot, and scatter plot are covered briefly.
Statistics is very important in our day to day life. There are so many applications of statistics in sectors like predictive analytics, machine learning, etc. It is really important to understand statistics well in order to understand these high-level concepts. In this post, we will introduce different parts of statistics, and then in later posts, we will describe them briefly.
Statistics
Statistics is the discipline that concerns data collection, organization, analysis, interpretation, and presentation. So what is data? Data is simply defined as distinct pieces of information that can be in many forms e.g. text, video, databases, spreadsheet, audio, images, etc…
Outlier Detection: A Comprehensive overview of different methods
Mathematical and statistical background of Various outlier detection methods with a detailed comparison between the use cases
Most real-world data sets have outliers that have unusually small or big values than the typical values of the data set. Outliers can have a negative impact on data analysis if not detected correctly and also outliers can reveal significant information and characteristics of data. There are several methods to detect an outlier in the data set. It is important to understand the characteristics of these methods properly since they might be quite powerful with large normal data but it is problematic to apply them to nonnormal data or small sample sizes without proper knowledge of their characteristics. …
Overfitting and Regularization in Machine Learning
Why regularization is used in machine learning and how it reduces overfitting balancing bias-variance
In the earlier era of machine learning, there used to be a lot of discussion on the bias-variance trade-off as was we could increase bias and reduce variance, or reduce bias and increase variance. But back in the pre-deep learning era, we didn’t have many tools that just reduce bias or just reduce variance without hurting the other one.
But in the modern deep learning, neural network or big data era getting a bigger network almost always just reduces bias without necessarily hurting variance, so long as we regularize appropriately. Also, getting more data always reduces variance and doesn’t hurt…
Neural Network: A Complete Beginners Guide from Scratch
A detailed explanation of Mathematics and concepts behind Neural Network
Neural Network has become a crucial part of modern technology. It has influenced our daily life in a way that we have never imagined. From e-commerce and solving classification problems to autonomous driving, it has touched everything. In this article, we are gonna discuss all the important aspects of neural networks in the simplest way possible and at the end of the tutorial, we have also provided code implemented in python for all the described parts.
Motivation
Animal brains even the small ones like the brain of a pigeon was more capable of than digital computers with huge processing power and…
Linear Classifiers: An Introduction to Classification
Linear Classifiers are one of the most commonly used classifiers and Logistic Regression is one of the most commonly used linear classifiers. The concepts we are going to learn here will actually extend a lot of other classification methods beyond linear classifiers. We’re going to learn the fundamental concepts, but also the underlying algorithms that let us optimize the parameters of this model to fit your training data.
Multiple Regression from Scratch in Python
Previously, we have discussed briefly the simple linear regression. Here we will discuss multiple regression or multivariable regression and how to get the solution of the multivariable regression. At the end of the post, we will provide the python code from scratch for multivariable regression.
Motivation
A single variable linear regression model can learn to predict an output variable y when there is only one input variable, x and there is a linear relationship between y and x, that is, y≈w₀+w₁*x. Well, that might not be a great predictive model for most cases. For example, let’s assume we are going to…
Simple Linear Regression: An Introduction to Regression from scratch
Regression is a very fundamental concept in statistics, machine learning, and Neural networks. Imagine plotting the correlation between rainfall frequency and agriculture production in high school. Increase in rainfall generally increases agriculture production. Fitting a line to those points enables us to predict the production rate under different rain conditions. It was actually the very simplest form of linear regression.
In simple words, regression is a study of how to best fit a curve to summarize a collection of data. It’s one of the most powerful and well-studied types of supervised learning algorithms. In regression, we try to understand the…
Bias-Variance Tradeoff: A Comprehensive Graphical Representation
To build an accurate machine learning model we need to have a proper understanding of the error. In forming predictions of a model there are three sources of error: noise, bias, and variance. Having proper knowledge of error and bias-variance would help us building accurate models and avoiding mistakes of overfitting and underfitting.
In this tutorial, our case study will be how to predict house prices. We have a dataset which consists of house prices with the square feet of the house associated with it. | 1,188 | 6,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2021-31 | latest | en | 0.929888 |
https://stackoverflow.com/questions/3887063/weighted-average-calculation-in-mysql | 1,652,907,876,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522309.14/warc/CC-MAIN-20220518183254-20220518213254-00154.warc.gz | 629,155,200 | 65,052 | # Weighted average calculation in MySQL?
I am currently using the following query to get some numbers:
SELECT gid, count(gid), (SELECT cou FROM size WHERE gid = infor.gid)
FROM infor
WHERE id==4325
GROUP BY gid;
The output I am getting at my current stage is the following:
+----------+-----------------+---------------------------------------------------------------+
| gid | count(gid) | (SELECT gid FROM size WHERE gid=infor.gid) |
+----------+-----------------+---------------------------------------------------------------+
| 19 | 1 | 19 |
| 27 | 4 | 27 |
| 556 | 1 | 556 |
+----------+-----------------+---------------------------------------------------------------+
I am trying to calculate the weighted average i.e.
(1*19+4*27+1*556)/(19+27+556)
Is there a way to do this using a single query?
Use:
SELECT SUM(x.num * x.gid) / SUM(x.cou)
FROM (SELECT i.gid,
COUNT(i.gid) AS num,
s.cou
FROM infor i
LEFT JOIN SIZE s ON s.gid = i.gid
WHERE i.id = 4325
GROUP BY i.gid) x
• Awesome... Thank you very much for this. I was about to write a nested loop inside a procedure but then came across an article that said "If you require a nested loop, then you did not look at a JOIN" :) Oct 8, 2010 at 2:32
• @legend: The advice is correct, but JOINs also risk inflating records if there's more than one child associated to the parent. If you want distinct rows from the parent, it is better to use a subquery (EXISTS would be my recommendation). Oct 8, 2010 at 2:35
• I see. In my case, there's exactly one element but I will keep your advice in mind. Oct 8, 2010 at 2:45
• @Legend: Excellent - the better you know your data, the better your queries will be. Oct 8, 2010 at 2:47
You could place your original query as a sub-query and SUM the records. I could not test this as I don't have the dataset you do, but it should work in theory ;)
SELECT SUM(gid)/SUM(weights) AS calculated_average FROM (
SELECT gid, (COUNT(gid) * gid) AS weights
FROM infor
WHERE id = 4325
GROUP BY gid);
• +1 for this approach. Thank You. Just curious to know which one is more efficient. Will run to EXPLAIN and see. Oct 8, 2010 at 2:33
• @Legend, no problem. You may in fact need the JOIN, but from the OP, I didn't see its necessity. Oct 8, 2010 at 2:37 | 624 | 2,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-21 | longest | en | 0.829387 |
https://www.excelforum.com/excel-formulas-and-functions/507493-time-elapsed-formula.html | 1,652,901,724,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522309.14/warc/CC-MAIN-20220518183254-20220518213254-00178.warc.gz | 882,555,423 | 14,035 | # time elapsed formula
1. ## time elapsed formula
I need to calculate two formulas.
1. Need to create a formula for % time elapsed (based on 365 days)
2. Need to create a formula for % to Forecast vs time elapsed .
2. ## Re: time elapsed formula
For 1, are you saying you have a start date, an end date, and a current date,
and you want to know what % of the period has elapsed from start to current? If
so, try:
=(current - start) / (end - start), format as percent.
For 2, you might be able to compare this calculated % with the forecast %. If
that's not what you are looking for, post back with more details.
--
Regards,
Fred
"sue" <sue@discussions.microsoft.com> wrote in message
>I need to calculate two formulas.
>
> 1. Need to create a formula for % time elapsed (based on 365 days)
> 2. Need to create a formula for % to Forecast vs time elapsed .
3. ## Re: time elapsed formula
Let's say the spread sheet measures sales volume for 365 days (06 calendar
year). So, some of the columns on the page going across are:
\$ sales month to date; \$ sales ytd; Forecasted volume; % to forecast ytd.
I think the formula is year to date \$ sales / total forecast = % to yearly
forecast..
but we will be comparing that to total time elapsed for the year.. I don't
know how to do do it. This should be a running formula that "ticks" away
the year in %.. so we know how much time we have left to make the number vs %
to actual \$'s achieved to date. Make sense?
"Fred Smith" wrote:
>
> For 1, are you saying you have a start date, an end date, and a current date,
> and you want to know what % of the period has elapsed from start to current? If
> so, try:
>
> =(current - start) / (end - start), format as percent.
>
> For 2, you might be able to compare this calculated % with the forecast %. If
> that's not what you are looking for, post back with more details.
>
> --
> Regards,
> Fred
>
>
> "sue" <sue@discussions.microsoft.com> wrote in message
> >I need to calculate two formulas.
> >
> > 1. Need to create a formula for % time elapsed (based on 365 days)
> > 2. Need to create a formula for % to Forecast vs time elapsed .
>
>
>
4. ## Re: time elapsed formula
Make sense? No.
Try posting a specific example. I'm sure that will help us figure out what
you're trying to do.
--
Regards,
Fred
"sue" <sue@discussions.microsoft.com> wrote in message
news:D3EC7EE4-E6D5-4943-B2A2-5021DD9A0235@microsoft.com...
> Let's say the spread sheet measures sales volume for 365 days (06 calendar
> year). So, some of the columns on the page going across are:
>
> \$ sales month to date; \$ sales ytd; Forecasted volume; % to forecast ytd.
>
> I think the formula is year to date \$ sales / total forecast = % to yearly
> forecast..
>
> but we will be comparing that to total time elapsed for the year.. I don't
> know how to do do it. This should be a running formula that "ticks" away
> the year in %.. so we know how much time we have left to make the number vs %
> to actual \$'s achieved to date. Make sense?
>
> "Fred Smith" wrote:
>
>>
>> For 1, are you saying you have a start date, an end date, and a current date,
>> and you want to know what % of the period has elapsed from start to current?
>> If
>> so, try:
>>
>> =(current - start) / (end - start), format as percent.
>>
>> For 2, you might be able to compare this calculated % with the forecast %. If
>> that's not what you are looking for, post back with more details.
>>
>> --
>> Regards,
>> Fred
>>
>>
>> "sue" <sue@discussions.microsoft.com> wrote in message
>> >I need to calculate two formulas.
>> >
>> > 1. Need to create a formula for % time elapsed (based on 365 days)
>> > 2. Need to create a formula for % to Forecast vs time elapsed .
>>
>>
>>
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Search Engine Friendly URLs by vBSEO 3.6.0 RC 1 | 1,070 | 4,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-21 | latest | en | 0.887115 |
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# Ang bansang saudi arabia lanmang sa mga bansa sa mundo ang hindi nagpapahintulot sa kababaihan na magmaneho ng sasakyan. ito ay katangian ng
### Another question on Physics
Physics, 26.11.2019 11:28
Aroller coaster car rapidly pick up speed as it rolls down a slope. as it start down the slope, its speed is 4 minute/second. but 3 second later at the bottom of the slope, its speed 22minute/second. what is average acceleration
Physics, 27.11.2019 05:28
Identify the variables in this hypothesis. if the pressure of a gas is increased, then the volume will decrease because the particles of the gas will be forced closer together. the independent variable is. the dependent variable is. | 201 | 719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-27 | latest | en | 0.795819 |
https://studyandanswers.com/mathematics/question12027258 | 1,627,782,283,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154127.53/warc/CC-MAIN-20210731234924-20210801024924-00677.warc.gz | 549,741,835 | 15,165 | , 22.06.2019 21:00 cia029587691
Answer the question with shiloh/ the jellybeans
right
step-by-step explanation:
22 and 22
step-by-step explanation:
10 plus 12 equals 22.
12 plus 10 also equals 22 because regardless of what order they're in, they're gonna equal the same thing.
step-by-step explanation:
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The cross-sectional areas of a right triangular prism and a right cylinder are congruent. the right triangular prism has a height of 6 units, and the right cylinder has a height of 6 units. which conclusion can be made from the given information? the volume of the triangular prism is half the volume of the cylinder. the volume of the triangular prism is twice the volume of the cylinder. the volume of the triangular prism is equal to the volume of the cylinder. the volume of the triangular prism is not equal to the volume of the cylinder. | 286 | 1,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-31 | latest | en | 0.862018 |
http://ir.lib.ncu.edu.tw:88/thesis/view_etd.asp?URN=90221004&fileName=GC90221004.pdf | 1,571,582,204,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986710773.68/warc/CC-MAIN-20191020132840-20191020160340-00284.warc.gz | 98,564,984 | 8,816 | ### 博碩士論文 90221004 詳細資訊
以作者查詢圖書館館藏 、以作者查詢臺灣博碩士 、以作者查詢全國書目 、勘誤回報 、線上人數：9 、訪客IP：3.231.228.109
(Structure of Traveling Waves in Delayed Cellular Neural Networks)
★ 遲滯型細胞神經網路之行進波 ★ 網格型微分方程的行進波的數值解 ★ 某類網格型微分方程行波解的存在性，唯一性及穩定性 ★ 某類週期性網格型微分方程行波解之研究 ★ 網格型動態系統行波解之研究 ★ 矩陣值勢能上的sofic測度 ★ 在Sofic Shift上的多重碎型分析 ★ 某類傳染病模型微分方程行波解之研究 ★ 某類三維癌症模型之整體穩定性分析 ★ 三種競爭合作系統之行波解的存在性 ★ 離散型Lotka-Volterra競爭系統之行波解的穩定性
1. 本電子論文使用權限為同意立即開放。
2. 已達開放權限電子全文僅授權使用者為學術研究之目的，進行個人非營利性質之檢索、閱讀、列印。
3. 請遵守中華民國著作權法之相關規定，切勿任意重製、散佈、改作、轉貼、播送，以免觸法。
(CNN)行進波解的結構。利用Monotone Iteration 及
Shooting的方法我們可以證明行進波之解結構隨著速度的改變而有不同的行為。
for one-dimensional cellular neural networks with distributed delayed signal
transmission. By using the monotone iteration method and shooting
method, we describe the transition of wave profiles from monotonicity,
damped oscillation, periodicity, unboundedness and back to monotonicity
as the wave speed is varied.
★ 行進波
★ traveling waves
1. Introduction.............................................................2
2. Properties of Characteristic Equation.....................5
3. Existence of Monotonic Traveling Waves................7
3.1. Construction of Upper and Lower Solutions.....7
3.2. Monotone Iteration Method.............................11
3.3. Proof of Main Theorem (I)..............................14
4. Structure of Non-Monotonic Traveling Waves.......15
4.1. Basic Properties of Asymptotic Initial Value
Problem...................................................................15
4.2. Proof of Main Theorem (II)............................26
References.....................................................................28
systems, J. Di . Eqns., 149 (1998), pp. 248-291.
[2] L. O. Chua, CNN: A Paradigm for Complexity, World Scientific Series on
Nonlinear Science, Series A, Vol. 31, World Scientific, Singapore, 1998.
[3] L. O. Chua and T. Roska, The CNN paradigm, IEEE Trans. Circuits Syst.,
40 (1993), pp. 147-156.
[4] L. O. Chua and L. Yang, Cellular neural networks: Theory, IEEE Trans.
Circuits Syst., 35 (1988), pp. 1257-1272.
[5] L. O. Chua and L. Yang, Cellular neural networks: Applications, IEEE
Trans. Circuits Syst., 35 (1988), pp. 1273-1290.
[6] T. Erneux and G. Nicolis, Propagation waves in discrete bistable reactiondi
usion systems, Physica D, 67 (1993), pp. 237-244.
[7] I. Gyori and G. Ladas, Oscillating Theory of Delay Di erential Equations
with Applications, Oxford University Press, Oxford, 1991.
[8] C.-H. Hsu and S.-S. Lin, Existence and multiplicity of traveling waves in a
lattice dynamical system, J. Di . Eqns., 164 (2000), pp. 431-450.
[9] C.-H. Hsu, S.-S. Lin and W. Shen, Traveling waves in cellular neural networks,
Internat. J. Bifur. and Chaos, 9 (1999), pp. 1307-1319.
[10] C.-H. Hsu and S.-Y. Yang, On camel-like traveling wave solutions in cellular
neural networks, preprint, 2002.
[11] H. Hudson and B. Zinner, Existence of traveling waves for a generalized
discrete Fisher’s equations, Comm. Appl. Nonlinear Anal., 1 (1994), pp.
23-46.
[12] J. Juang and S.-S. Lin, Cellular neural networks: mosaic pattern and spatial
chaos, SIAM J. Appl. Math., 60 (2000), pp. 891-915.
[13] J. P. Keener, Propagation and its failure in coupled systems of discrete
excitable cells, SIAM J. Appl. Math., 47 (1987), pp. 556-572.
[14] J. Mallet-Paret, The global structure of traveling waves in spatial discrete
dynamical systems, J. Dyn. Di . Eqns., 11 (1999), pp. 49-127.
[15] L. Orzo, Z. Vidnyanszky, J. Hamori and T. Roska, CNN model of the
feature linked synchronized activities in the visual thalamo-cortical system,
Proc. 1996 Fourth IEEE Int. Workshop on CNN and Their Applications,
pp. 291-296, Seville, Spain, June 24-26, 1996.[16] P. Thiran, K. R. Crounse, L.O. Chua, and M. Hasler, Pattern formation
properties of autonomous cellular neural networks, IEEE Trans. Circuit
Syst., 42 (1995), pp. 757-774.
[17] P. Thiran, Dynamics and Self-Organization of Locally Coupled Neural
Networks, Presses Polytechniques et Universitaires Romandes, Lausanne,
Switzerland, 1997.
[18] F. Werblin, T. Roska and L.O. Chua, The analogic cellular neural network
as a bionic eye, Internat. J. Circuit Theory Appl., 23 (1994), pp. 541-569.
[19] P. Weng and J. Wu, Deformation of traveling waves in delayed cellular
neural networks, preprint, 2001.
[20] J. Wu and X. Zou, Asymptotical and periodic boundary value problems
of mixed FDEs and wave solutions of lattice di erential equations, J. Di .
Eqns., 135 (1997), pp. 315-357.
[21] B. Zinner, Existence of traveling wavefront solutions for discrete Nagumo
equation, J. Di . Eqns., 96 (1992), pp. 1-27. | 1,613 | 4,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-43 | latest | en | 0.295727 |
https://algotree.org/algorithms/backtracking/word_search/ | 1,701,887,585,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100602.36/warc/CC-MAIN-20231206162528-20231206192528-00072.warc.gz | 121,862,988 | 16,848 | # Search a word in a given 2 dimensional matrix of letters
Backtracking : Program to search a word in a given 2 dimensional matrix of letters
``````#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
using namespace std;
class WordSearch {
public:
string sofar;
vector<string> board;
string word;
// The size of the board would always be less than or equal to 8
bool visited[8][8] = {{ false }};
bool Search (int x, int y, int pos) {
if (pos == word.size()) {
return true;
} else {
// Co-ordinates difference of the adjacent cells in order Up, Right, Down and Left
int dx[4] = { -1, 0, 1, 0 };
int dy[4] = { 0, 1, 0, -1 };
int rows = board.size();
int cols = board[0].size();
// Visit adjacent cells in order Up, Right, Down & Left
for (int i=0; i<4; i++) {
int newx = x + dx[i];
int newy = y + dy[i];
if (newx < 0 or newx >= rows or
newy < 0 or newy >= cols or visited[newx][newy] == true)
continue;
if (board[newx][newy] == word[pos]) {
pos++;
visited[newx][newy] = true;
if (Search (newx, newy, pos) == true) // Word has been found so return true.
return true;
visited[newx][newy] = false; // Backtrack and try other adjacent letter first.
pos--;
}
}
}
return false;
}
// Below function returns true if the word [word] exist on the board
bool Exist (vector<string> arg_board, string arg_word) {
int pos; // Tracks the position of the matching letters of the word
board = arg_board;
word = arg_word;
int rows = board.size();
int cols = board[0].size();
memset(visited, false, sizeof(bool) * 8 * 8);
for (int r=0; r<rows; r++) {
for (int c=0; c<cols; c++) {
pos = 0;
if (board[r][c] == word[pos]) {
pos++;
visited[r][c] = true;
if (Search(r, c, pos))
return true;
visited[r][c] = false;
}
}
}
return false;
}
};
int main() {
WordSearch W;
//====================================
vector<string> vec_1 = {{"ABCE"},
{"SFCS"},
if (W.Exist(vec_1, "ABCCED") == 1)
cout << "Yes" << endl;
else
cout << "No" << endl;
//====================================
vector<string> vec_2 = {{"AAAAA"},
{"AAAAA"},
{"AAAAA"},
{"AAAAA"}};
if (W.Exist(vec_2, "AAB") == 1)
cout << "Yes" << endl;
else
cout << "No" << endl;
//====================================
vector<string> vec_3 = {{"FGABG"},
{"EFAEC"},
{"GGFBD"},
{"GGDCD"}};
if (W.Exist(vec_3, "FEGGGDCDDCGBAG") == 1)
cout << "Yes" << endl;
else
cout << "No" << endl;
//====================================
vector<string> vec_4 = {{"FCDFFD"},
{"BGCBFC"}};
if (W.Exist(vec_4, "FBGCDCBFFF") == 1)
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
``````
Output
``````Yes
No
Yes
Yes
`````` | 824 | 2,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-50 | latest | en | 0.435608 |
https://encyclopedia2.thefreedictionary.com/Personal+space | 1,571,453,085,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986688674.52/warc/CC-MAIN-20191019013909-20191019041409-00032.warc.gz | 471,124,703 | 15,184 | # space
(redirected from Personal space)
Also found in: Dictionary, Thesaurus, Medical, Legal, Acronyms, Wikipedia.
## space
1.
a. the region beyond the earth's atmosphere containing the other planets of the solar system, stars, galaxies, etc.; universe
b. (as modifier): a space probe
2.
a. the region beyond the earth's atmosphere occurring between the celestial bodies of the universe. The density is normally negligible although cosmic rays, meteorites, gas clouds, etc., can occur. It can be divided into cislunar space (between the earth and moon), interplanetary space, interstellar space, and intergalactic space
b. (as modifier): a space station
3. Music any of the gaps between the lines that make up the staff
4. Maths a collection of unspecified points having properties that obey a specified set of axioms
## space
The near-vacuum existing beyond the atmospheres of all bodies in the Universe. The extent of space, i.e. whether it is finite or infinite, is as yet unresolved. See intergalactic medium; interplanetary medium; interstellar medium.
## Space
The unlimited continuous three-dimensional expanse in which all material objects exist; all the area in and around a structure, or volume between specified boundaries, and the interval between two objects.
## Space
in mathematics, a logically conceivable form or structure that is used as a setting in which other forms and various constructions are realized. For example, in elementary geometry the plane or space is the setting in which various figures are constructed. In most spaces, we introduce relations whose formal properties are similar to those of ordinary spatial relations, such as distance between points or congruence of figures. Consequently, such spaces may be said to represent logically conceivable spacelike forms.
Historically, the first mathematical space was three-dimensional Euclidean space, which is an approximate abstract image of physical space; it has remained a very important space in mathematics. The general concept of space took shape in mathematics as a result of the gradual, increasingly broad generalization and modification of the concepts of the geometry of Euclidean space. The first spaces differing from three-dimensional Euclidean space were introduced in the first half of the 19th century. These spaces were Lobachevskian space and n-dimensional Euclidean space. The general concept of mathematical space was advanced in 1854 by B. Riemann. The process of generalizing, refining and concretely defining the concept followed various directions; for example, such concepts as vector space, Hilbert space, Riemannian space, function space, and topological space were developed.
In contemporary mathematics a space is defined as a set of objects, which are called the points of the space. These objects may be. for example, geometric figures, functions, or the states of a physical system. When we consider a set of objects as a space, we deal not with the individual properties of the objects but with only those properties of the set that are determined by relations that we wish to take into account or that we introduce by definition. These relations between points and various configurations, or sets of points, determine the geometry of the space. When the geometry is constructed axiomatically, the basic properties of these relations are expressed in the corresponding axioms.
Three examples of spaces are metric spaces, spaces of events, and phase spaces. In a metric space, the distance between points is defined. Thus, the functions f(x) continuous on an interval [a, b] form a metric space—whose points are the functions f (x)— when the distance between f1(x) and f2(x) is defined as the maximum of the absolute value of the difference between the two functions:
r = max ǀf1(x) –f2(x
The concept of space of events plays an important role in the geometric interpretation of the theory of relativity. Every event is characterized by its position—the coordinates x, y, and z— and the time t of its occurrence. The set of all possible events is thus a four-dimensional space, in which an event or point is defined by the four coordinates x, y, z, and t.
Phase spaces are studied in theoretical physics and mechanics. The phase space of a physical system is the set of all the possible states of the system. The states are the points of the space.
The spaces in these examples are of significance in the actual universe since the set of possible states of a physical system or the set of events with space and time coordinates has real existence. Consequently, we are dealing with forms of reality that, although not spatial in the ordinary sense, are spacelike in structure. The question of which mathematical space reflects most accurately the general properties of physical space is answered experimentally. Thus, it has been established that in describing physical space Euclidean geometry is not always sufficiently accurate, and Riemannian geometry is used in the present-day theory of physical space (seeRELATIVITY, THEORY OF). The concept of space in mathematics is also discussed in the articles GEOMETRY, MATHEMATICS, and MULTIDIMENSIONAL SPACE.
A. D. ALEKSANDROV
## space
[spās]
(astronomy)
Specifically, the part of the universe lying outside the limits of the earth's atmosphere.
More generally, the volume in which all celestial bodies, including the earth, move.
(communications)
The open-circuit condition or the signal causing the open-circuit condition in telegraphic communication; the closed-circuit condition is called the mark.
(mathematics)
In context, usually a set with a topology on it or some other type of structure.
## space
(character)
The space character, ASCII 32.
See octal forty.
## space
(1) In digital electronics, a 0 bit. Contrast with mark.
(2) The trendy word that started in the 1990s for area or field of endeavor. For example, the phrase "we are involved in the videoconferencing space" refers simply to the videoconferencing industry. To many, this sounds more chic than using a word such as "field," "arena" or "industry."
References in periodicals archive ?
Exploring the changes in blood pressure values during the violation of personal space, a study reported that shorter interpersonal space led to changes, though insignificant, in blood pressure levels as well as increases in anxiety and tension.7 Similarly, in some studies found that people experienced significant increases in their systolic blood pressure values when they suffered from psychological stress.5,6,16 In the present study, the participants might have regarded it as a threat when their blood pressure was measured by the male student, for they did not know him.
Your personal space will mirror your loves and dreams just as much as it will reflect your life.
How do we stop this, what is a safe distance to maintain to avoid personal space invasion?
Key words: Personal space, Invasion, Confederate, Library
A It is perfectly normal behaviour for many young people not to have any concept of personal space.
Both genders mostly used personal space, but more female students used intimate space, and more male students used social and public space.
The proxemics of personal space applies to social media as well, based upon the user's cultural background their divulgence of personal information varies.
For Sketchbooks: The Hidden Art of Designers, Illustrators & Creatives (Laurence King, \$30), editor Richard Brereton explores the freedom a sketchbook offers the professional creative-a personal space where the imagination can run free without deadlines or budgetary constraints.
3M Privacy Filters prevent invasion of personal space by darkening laptop screens when viewed from the side.
The research, among 800 adults, found that a growing number of under-35s wanted to live alone to have greater individual freedom and personal space.
Choose comfortable furniture arranged to ensure adequate personal space and avoid crowding.
Mum Hayley, 29, said: "It wasn't locked up, but you think in your back garden, which is your own personal space, that it would be OK."
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Open / Close | 1,681 | 8,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-43 | latest | en | 0.900113 |
https://www.alphacodingskills.com/sql-server/notes/sql-server-func-asin.php | 1,643,367,083,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305494.6/warc/CC-MAIN-20220128104113-20220128134113-00449.warc.gz | 682,637,480 | 6,345 | # SQL Server - ASIN() Function
The SQL Server (Transact-SQL) ASIN() function returns arc sine of a value. The returned value will be in the range -𝜋/2 through 𝜋/2. In special cases it returns the following:
• If the number is not within the range of -1 to 1, then an error is returned.
Note: ASIN() is the inverse of SIN().
### Syntax
```ASIN(x)
```
### Parameters
`x` `Required. `Specify the value.
### Return Value
Returns the arc sine of the value.
### Example 1:
The example below shows the usage of ASIN() function.
```SELECT ASIN(0.2);
Result: 0.2013579207903308
SELECT ASIN(0.8);
Result: 0.9272952180016123
SELECT ASIN(1);
Result: 1.5707963267948966
SELECT ASIN(-1);
Result: -1.5707963267948966
SELECT ASIN(0);
Result: 0.0
SELECT ASIN(-0.2);
Result: -0.2013579207903308
```
### Example 2:
Consider a database table called Sample with the following records:
Datax
Data 1-1
Data 2-0.5
Data 30
Data 40.5
Data 51
The query given below can be used to calculate the arc sine of records of column x.
```SELECT *, ASIN(x) AS ASIN_Value FROM Sample;
```
This will produce the result as shown below:
DataxASIN_Value
Data 1-1-1.5707963267948966
Data 2-0.5-0.52359877559829893
Data 300.0
Data 40.50.52359877559829893
Data 511.5707963267948966
❮ SQL Server Functions
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# Chapter 3
Time Value of
Money
3-1
## 2001 Prentice-Hall, Inc.
Fundamentals of Financial Management, 11/e
Created by: Gregory A. Kuhlemeyer, Ph.D.
Carroll College, Waukesha, WI
3-2
## The Interest Rate
Simple Interest
Compound Interest
Amortizing a Loan
## The Interest Rate
Which would you prefer -- \$10,000
today or \$10,000 in 5 years?
Obviously, \$10,000 today.
You already recognize that there is
TIME VALUE TO MONEY!!
3-3
Why TIME?
Why is TIME such an important
TIME allows you the opportunity to
postpone consumption and earn
INTEREST.
3-4
postulate
## All operations with money must be
compared between alternatives
to find the best result.
3-5
## Interest rate is a simple but
prominent equivalent of any
change of time value of money.
5
Types of Interest
Simple
Interest
## Interest paid (earned) on only the original
amount, or principal borrowed (lent).
Compound
Interest
## Interest paid (earned) on any previous
interest earned, as well as on the
principal borrowed (lent).
3-6
## Future value and present
value
Changing in time value of money
gets future and present
nomination
Getting from present value to
future value is called
compounding.
3-7
## Getting from future value to
present value is called
Formula
3-8
SI = P0(i)(n)
SI:
Simple Interest
P0:
i:
n:
Assume
## that you deposit \$1,000 in an
account earning 7% simple interest for
2 years. What is the accumulated
interest at the end of the 2nd year?
SI
3-9
= P0(i)(n)
= \$1,000(.07)(2)
= \$140
What
deposit?
FV
Future
= P0 + SI
= \$1,000 + \$140
= \$1,140
## Value is the value at some future
time of a present amount of money, or a
series of payments, evaluated at a given
interest rate.
3-10
What
## is the Present Value (PV) of the
previous problem?
The Present Value is simply the
\$1,000 you originally deposited.
That is the value today!
Present
3-11
## Value is the current value of a
future amount of money, or a series of
payments, evaluated at a given interest
rate.
Power of Time
Figure 5.1 Future Value and Compound Interest Illustrated
## Future value of original investment increases with time, unless
interest rate is zero.
12
3-12
FIN3000,
## Power of Interest Rate
Figure 5.1 Future Value and Compound Interest Illustrated
value.
13
3-13
FIN3000,
## Where to use simple
interest
Money
market instruments
Treasury
Local
bills (T-bill)
bills
Certificate
of deposit (CD)
Commercial
3-14
Bill
paper (CP)
of exchange
14
Money market
## Pure discount securities
Contracts up to 1 year
## Huge volume and vigorous competition
No physical place
## Essentially for professionals ( banks,institutional
investors, brokerage firms, companies)
## Liquidity ( fine spreads based on interest rate of
lending and borrowing)
3-15
Creditworthiness
15
3-16
T-bills
## Domestic instruments issued by governments to raise short
term finance balancing cashflow
auction
Negotiable
## Generally 13,26,52 weeks
Certificate of deposit - CD
deposit
16
## Money market securities 2
Commercial paper- CP
Issued
## by large, safe and wellknown companies bypassing
banks to achieve lower
borrowing rates (sometimes
below the banks prime rate)
Very
3-17
## short term (max 270 days,
most 60days or less)
17
## Trade bill, bills of exchange,
bankersacceptance
Used
purposes
The
## seller draws up a bill to
sign it
3-18
Could
be sold at a discount
18to
## Future value and present value
(1 + r) is a future value factor (FVF)
To simplify calculations of FV use table of FVF.
Years
1%
2%
3%
4%
5%
6%
7%
8%
9%
10%
1,01
1,02
1,03
1,04
1,05
1,06
1,07
1,08
1,09
1,1
1,02
1,04
1,06
1,08
1,10
1,12
1,14
1,17
1,19
1,21
1,03
1,06
1,09
1,12
1,16
1,19
1,23
1,26
1,295
1,33
1,04
1,08
1,13
1,17
1,22
1,26
1,31
1,36
1,41
1,46
1,05
1,1
1,16
1,22
1,28
1,34
1,40
1,47
1,54
1,61
1,06
1,13
1,19
1,27
1,34
1,42
1,50
1,59
1,68
1,77
1,07
1,15
1,23
1,32
1,41
1,50
1,61
1,71
1,83
1,94
1,08
1,17
1,27
1,37
1,48
1,59
1,72
1,85
1,99
2,14
1,09
1,20
1,30
1,42
1,55
1,69
1,84
1,999
2,17
2,36
10
1,1
1,22
1,34
1,48
1,63
1,79
1,97
2,16
2,37
2,59
3-19
19
## Why Compound Interest?
Future Value (U.S. Dollars)
3-20
20000
10% Simple
Interest
7% Compound
Interest
10% Compound
Interest
15000
10000
5000
0
1st Year 10th
Year
20th
Year
30th
Year
Capital market
Instruments
3-21
Bonds
Government bonds
## Mortgage or other assets backed bonds
Corporate
Foreign
Junk
Shares
Preferred
Normal
Innovations
Convertibles
21
Future Value
Single Deposit (Graphic)
Assume that you deposit \$1,000 at
a compound interest rate of 7% for
2 years.
7%
\$1,000
FV2
3-22
Future Value
Single Deposit (Formula)
FV1 = P0 (1+i)1
= \$1,000 (1.07)
= \$1,070
Compound Interest
## You earned \$70 interest on your \$1,000
deposit over the first year.
This is the same amount of interest you
would earn under simple interest.
3-23
Future Value
Single Deposit (Formula)
FV1
= P0 (1+i)1
FV2
= FV1 (1+i)1
= P0 (1+i)(1+i) = \$1,000(1.07)(1.07)
= P0 (1+i)2
= \$1,000(1.07)2
= \$1,144.90
= \$1,000 (1.07)
= \$1,070
## You earned an EXTRA \$4.90 in Year 2 with
compound over simple interest.
3-24
General Future
Value Formula
FV1 = P0(1+i)1
FV2 = P0(1+i)2
etc.
## General Future Value Formula:
FVn = P0 (1+i)n
or FVn = P0 (FVIFi,n) -- See Table I
3-25
## Valuation Using Table I
FVIFi,n is found on Table I at the end
of the book or on the card insert.
3-26
Period
1
2
3
4
5
6%
1.060
1.124
1.191
1.262
1.338
7%
1.070
1.145
1.225
1.311
1.403
8%
1.080
1.166
1.260
1.360
1.469
## Using Future Value Tables
FV2
= \$1,000 (FVIF7%,2)
= \$1,000 (1.145)
= \$1,145 [Due to Rounding]
Period
6%
7%
8%
1
1.060
1.070
1.080
2
1.124
1.166
1.145
3
1.191
1.225
1.260
4
1.262
1.311
1.360
5
1.338
1.403
1.469
3-27
## Use the highlighted row
of keys for solving any
of the FV, PV, FVA,
problems
N:
Number of periods
I/Y:Interest rate per period
PV:
Present value
PMT:
Payment per period
FV:
Future value
CLR TVM: Clears all of the inputs
into the above TVM keys
3-28
Inputs
I/Y
PV
PMT
FV
Compute
3-29
## Focus on 3rd row of keys (will be
displayed in slides as shown above)
Press:
2nd
3-30
CLR TVM
I/Y
-1000
PV
PMT
CPT
FV
## Solving the FV Problem
Inputs
Compute
N:
I/Y:
PV:
PMT:
FV:
3-31
-1,000
I/Y
PV
PMT
FV
1,144.90
2 periods (enter as 2)
7% interest rate per period (enter as 7 NOT .07)
\$1,000 (enter as negative as you have less)
Not relevant in this situation (enter as 0)
## Story Problem Example
Julie Miller wants to know how large her deposit
of \$10,000 today will become at a compound
annual interest rate of 10% for 5 years.
10%
\$10,000
FV5
3-32
## Calculation based on general formula:
FVn = P0 (1+i)n
FV5 = \$10,000 (1+ 0.10)5
= \$16,105.10
Calculation
based on Table I:
FV5 = \$10,000 (FVIF10%, 5)
= \$10,000 (1.611)
= \$16,110 [Due to Rounding]
3-33
Press:
2nd
3-34
CLR TVM
10
I/Y
-10000
PV
PMT
CPT
FV
Inputs
Compute
10
-10,000
I/Y
PV
PMT
FV
16,105.10
## The result indicates that a \$10,000
investment that earns 10% annually
for 5 years will result in a future value
of \$16,105.10.
3-35
Quick! How long does it take to
double \$5,000 at a compound rate
of 12% per year (approx.)?
We will use the Rule-of-72.
3-36
The Rule-of-72
Quick! How long does it take to
double \$5,000 at a compound rate
of 12% per year (approx.)?
Approx. Years to Double = 72 / i%
72 / 12% = 6 Years
[Actual Time is 6.12 Years]
3-37
Inputs
N
Compute
12
-1,000
+2,000
I/Y
PV
PMT
FV
6.12 years
## The result indicates that a \$1,000
investment that earns 12% annually
will double to \$2,000 in 6.12 years.
Note: 72/12% = approx. 6 years
3-38
## Basic principles of finance
3-39
Time value of
money - a dollar
today worth more
than a dollar
tomorrow
A safe dollar is
worth more than a
risky one
39
## Theory of Present Value
3-40
Theory by John B.
Williams
Based on :
dividends and
assumes long-term
decisions
Compares actual
value and real
value
40
Basics
Yield
Rate
of return
Rate
of interest
Income
3-41
Maturity
## Future and present value
Simple interest
Compound interest
41
Present Value
Single Deposit (Graphic)
Assume that you need \$1,000 in 2 years.
Lets examine the process to determine
how much you need to deposit today at a
discount rate of 7% compounded annually.
7%
\$1,000
PV0
3-42
PV1
Present Value
Single Deposit (Formula)
PV0 = FV2 / (1+i)2
= FV2 / (1+i)2
0
7%
= \$1,000 / (1.07)2
= \$873.44
1
\$1,000
PV0
3-43
General Present
Value Formula
PV0 = FV1 / (1+i)1
PV0 = FV2 / (1+i)2
etc.
## General Present Value Formula:
PV0 = FVn / (1+i)n
or
3-44
## Valuation Using Table II
PVIFi,n is found on Table II at the end
of the book or on the card insert.
Period
1
2
3
4
5
3-45
6%
.943
.890
.840
.792
.747
7%
.935
.873
.816
.763
.713
8%
.926
.857
.794
.735
.681
## Using Present Value Tables
PV2
3-46
= \$1,000 (PVIF7%,2)
= \$1,000 (.873)
= \$873 [Due to Rounding]
Period
6%
7%
8%
1
.943
.935
.926
2
.890
.873
.857
3
.840
.816
.794
4
.792
.763
.735
5
.747
.713
.681
## 2. Future value and present value
Table of present value factor
3-47
Years
1%
2%
3%
4%
5%
6%
7%
8%
9%
10%
0,99
0,98
0,97
0,96
0,95
0,94
0,935
0,93
0,92
0,91
0,98
0,96
0,94
0,92
0,91
0,89
0,87
0,86
0,84
0,83
0,97
0,94
0,92
0,89
0,86
0,84
0,82
0,79
0,77
0,75
0,96
0,92
0,89
0,85
0,82
0,79
0,76
0,74
0,71
0,68
0,95
0,91
0,87
0,82
0,78
0,75
0,71
0,68
0,65
0,62
0,94
0,89
0,84
0,79
0,75
0,70
0,67
0,63
0,596
0,56
0,93
0,87
0,81
0,76
0,71
0,67
0,62
0,58
0,55
0,51
0,92
0,85
0,79
0,73
0,68
0,63
0,58
0,54
0,50
0,47
0,914
0,84
0,77
0,70
0,64
0,59
0,54
0,50
0,46
0,42
10
0,905
0,82
0,74
0,68
0,61
0,56
0,51
0,46
0,42
0,39
47
## Solving the PV Problem
Inputs
Compute
N:
I/Y:
PV:
PMT:
FV:
3-48
I/Y
PV
+1,000
PMT
FV
-873.44
2 periods (enter as 2)
7% interest rate per period (enter as 7 NOT .07)
Compute (Resulting answer is negative deposit)
Not relevant in this situation (enter as 0)
\$1,000 (enter as positive as you receive \$)
## Story Problem Example
Julie Miller wants to know how large of a
deposit to make so that the money will
grow to \$10,000 in 5 years at a discount
rate of 10%.
10%
\$10,000
PV0
3-49
## Calculation based on general formula:
PV0 = FVn / (1+i)n
PV0 = \$10,000 / (1+ 0.10)5
= \$6,209.21
## Calculation based on Table I:
PV0 = \$10,000 (PVIF10%, 5)
= \$10,000 (.621)
= \$6,210.00 [Due to Rounding]
3-50
Inputs
Compute
10
I/Y
PV
+10,000
PMT
FV
-6,209.21
## The result indicates that a \$10,000
future value that will earn 10% annually
for 5 years requires a \$6,209.21 deposit
today (present value).
3-51
Types of Annuities
An
## Annuity represents a series of equal
payments (or receipts) occurring over a
specified number of equidistant periods.
Ordinary
## Annuity: Payments or receipts
occur at the end of each period.
Annuity
## Due: Payments or receipts
occur at the beginning of each period.
3-52
Examples of Annuities
3-53
## Car Loan Payments
Mortgage Payments
Retirement Savings
Parts of an Annuity
(Ordinary Annuity)
End of
Period 1
Today
3-54
End of
Period 2
End of
Period 3
\$100
\$100
\$100
## Equal Cash Flows
Each 1 Period Apart
Parts of an Annuity
(Annuity Due)
Beginning of
Period 1
\$100
\$100
\$100
Today
3-55
Beginning of
Period 2
Beginning of
Period 3
## Equal Cash Flows
Each 1 Period Apart
Overview of an
Ordinary Annuity -- FVA
Cash flows occur at the end of the period
i%
. . .
R
R = Periodic
Cash Flow
FVAn =
R(1+i)n-1 +
R(1+i)n-2 +
## ... + R(1+i)1 + R(1+i)0
3-56
FVAn
n+1
Example of an
Ordinary Annuity -- FVA
Cash flows occur at the end of the period
\$1,000
\$1,000
7%
\$1,000
\$1,070
\$1,145
FVA3 = \$1,000(1.07)2 +
\$1,000(1.07)1 + \$1,000(1.07)0 \$3,215 = FVA3
= \$1,145 + \$1,070 + \$1,000
= \$3,215
3-57
## Hint on Annuity Valuation
The future value of an ordinary
annuity can be viewed as
occurring at the end of the last
cash flow period, whereas the
future value of an annuity due
can be viewed as occurring at
the beginning of the last cash
flow period.
3-58
## Valuation Using Table III
FVAn
FVA3
= R (FVIFAi%,n)
= \$1,000 (FVIFA7%,3)
= \$1,000 (3.215) = \$3,215
Period
6%
7%
8%
1
1.000
1.000
1.000
2
2.060
2.070
2.080
3
3.184
3.246
3.215
4
4.375
4.440
4.506
5
5.637
5.751
5.867
3-59
Inputs
Compute
N:
I/Y:
PV:
PMT:
FV:
3-60
-1,000
I/Y
PV
PMT
FV
3,214.90
## 3 periods (enter as 3 year-end deposits)
7% interest rate per period (enter as 7 NOT .07)
Not relevant in this situation (no beg value)
\$1,000 (negative as you deposit annually)
Overview View of an
Cash flows occur at the beginning of the period
## FVADn = R(1+i)n + R(1+i)n-1 +
... + R(1+i)2 + R(1+i)1
= FVAn (1+i)
3-61
. . .
i%
R
n-1
R
Example of an
Cash flows occur at the beginning of the period
\$1,000
\$1,000
\$1,070
7%
\$1,000
\$1,145
\$1,225
2
1
\$1,000(1.07) + \$1,000(1.07)
= \$1,225 + \$1,145 + \$1,070
= \$3,440
3-62
## Valuation Using Table III
= R (FVIFAi%,n)(1+i)
= \$1,000 (FVIFA7%,3)(1.07)
= \$1,000 (3.215)(1.07) = \$3,440
Period
6%
7%
8%
1
1.000
1.000
1.000
2
2.060
2.070
2.080
3
3.184
3.246
3.215
4
4.375
4.440
4.506
5
5.637
5.751
5.867
3-63
Inputs
-1,000
I/Y
PV
PMT
FV
3,439.94
Compute
## Complete the problem the same as an ordinary annuity
problem, except you must change the calculator setting
to BGN first. Dont forget to change back!
Step 1:
Press
2nd
BGN
keys
3-64
Step 2:
Press
2nd
SET
keys
Step 3:
Press
2nd
QUIT
keys
Overview of an
Ordinary Annuity -- PVA
Cash flows occur at the end of the period
i%
n+1
. . .
R
R = Periodic
Cash Flow
PVAn
## PVAn = R/(1+i)1 + R/(1+i)2
+ ... + R/(1+i)n
3-65
Example of an
Ordinary Annuity -- PVA
Cash flows occur at the end of the period
\$1,000
\$1,000
7%
\$1,000
\$ 934.58
\$ 873.44
\$ 816.30
\$2,624.32 = PVA3
3-66
PVA3 =
\$1,000/(1.07)1 +
\$1,000/(1.07)2 +
\$1,000/(1.07)3
= \$2,624.32
## Hint on Annuity Valuation
The present value of an ordinary
annuity can be viewed as
occurring at the beginning of the
first cash flow period, whereas
the present value of an annuity
due can be viewed as occurring
at the end of the first cash flow
period.
3-67
## Valuation Using Table IV
PVAn
PVA3
= R (PVIFAi%,n)
= \$1,000 (PVIFA7%,3)
= \$1,000 (2.624) = \$2,624
Period
6%
7%
8%
1
0.943
0.935
0.926
2
1.833
1.808
1.783
3
2.673
2.577
2.624
4
3.465
3.387
3.312
5
4.212
4.100
3.993
3-68
## Future and present value of
stream of cash flow
Table of future value factor of
annuity
Years
1%
2%
3%
4%
5%
6%
7%
8%
9%
10%
1,00
1,00
1,00
1,00
1,00
1,00
1,00
1,00
1,00
1,00
2,01
2,02
2,03
2,04
2,05
2,06
2,07
2,08
2,09
2,1
3,03
3,06
3,09
3,12
3,15
3,18
3,22
3,25
3,28
3,31
4,06
4,12
4,2
4,25
4,31
4,38
4,44
4,51
4,57
4,64
5,1
5,2
5,3
5,42
5,53
5,64
5,75
5,87
5,99
6,11
6,2
6,3
6,5
6,63
6,8
6,98
7,15
7,34
7,52
7,72
7,2
7,4
7,7
7,898
8,14
8,39
8,65
8,92
9,2
9,49
8,3
8,6
8,9
9,21
9,55
9,897
10,26
10,64
11,03
11,45
9,4
9,8
10,16
10,58
11,03
11,49
11,98
12,49
13,02
13,58
10
10,5
10,95
11,46
12,01
12,58
13,18
13,82
14,49
15,19
15,94
3-69
69
annuity factor
Years
1%
2%
3%
4%
5%
6%
7%
8%
9%
10%
0,99
0,98
0,97
0,96
0,95
0,94
0,93
0,925
0,917
0,91
1,97
1,94
1,91
1,89
1,86
1,83
1,81
1,78
1,76
1,74
2,94
2,88
2,83
2,76
2,72
2,67
2,62
2,58
2,53
2,49
3,90
3,81
3,72
3,63
3,55
3,47
3,39
3,31
3,24
3,17
4,85
4,71
4,58
4,45
4,33
4,21
4,10
3,99
3,89
3,79
5,796
5,60
5,42
5,24
5,08
4,91
4,77
4,62
4,49
4,36
6,73
6,47
6,23
6,00
5,79
5,58
5,39
5,21
5,03
4,87
7,65
7,33
7,02
6,73
6,46
6,21
5,97
5,75
5,53
5,33
8,57
8,16
7,79
7,44
7,11
6,8
6,52
6,25
5,99
5,76
10
9,47
8,98
8,73
8,11
7,72
7,36
7,02
6,71
6,42
6,14
3-70
70
Inputs
Compute
N:
I/Y:
PV:
PMT:
FV:
3-71
I/Y
PV
-1,000
PMT
FV
2,624.32
## 3 periods (enter as 3 year-end deposits)
7% interest rate per period (enter as 7 NOT .07)
\$1,000 (negative as you deposit annually)
Not relevant in this situation (no ending value)
Overview of an
Cash flows occur at the beginning of the period
i%
R
n-1
. . .
R
R: Periodic
Cash Flow
## PVADn = R/(1+i)0 + R/(1+i)1 + ... + R/(1+i)n-1
= PVAn (1+i)
3-72
Example of an
Cash flows occur at the beginning of the period
\$1,000
\$1,000
7%
\$1,000.00
\$ 934.58
\$ 873.44
## PVADn = \$1,000/(1.07)0 + \$1,000/(1.07)1 +
\$1,000/(1.07)2 = \$2,808.02
3-73
## Valuation Using Table IV
= \$1,000 (2.624)(1.07) = \$2,808
Period
6%
7%
8%
1
0.943
0.935
0.926
2
1.833
1.808
1.783
3
2.673
2.577
2.624
4
3.465
3.387
3.312
5
4.212
4.100
3.993
3-74
Inputs
I/Y
PV
-1,000
PMT
FV
2,808.02
Compute
## Complete the problem the same as an ordinary annuity
problem, except you must change the calculator setting
to BGN first. Dont forget to change back!
Step 1:
Press
2nd
BGN
keys
3-75
Step 2:
Press
2nd
SET
keys
Step 3:
Press
2nd
QUIT
keys
## Steps to Solve Time Value
of Money Problems
2. Determine if it is a PV or FV problem
3. Create a time line
4. Put cash flows and arrows on time line
5. Determine if solution involves a single
CF, annuity stream(s), or mixed flow
## 6. Solve the problem
7. Check with financial calculator (optional)
3-76
## Mixed Flows Example
Julie Miller will receive the set of cash
flows below. What is the Present Value
at a discount rate of 10%?
1
10%
\$600
PV0
3-77
## \$600 \$400 \$400 \$100
How to Solve?
1. Solve a piece-at-a-time by
discounting each piece back to t=0.
2. Solve a group-at-a-time by first
breaking problem into groups of
annuity streams and any single
cash flow group. Then discount
each group back to t=0.
3-78
Piece-At-A-Time
0
1
10%
\$600
\$545.45
\$495.87
\$300.53
\$273.21
\$ 62.09
## \$1677.15 = PV0 of the Mixed Flow
3-79
Group-At-A-Time (#1)
0
10%
\$600
## \$600 \$400 \$400 \$100
\$1,041.60
\$ 573.57
\$ 62.10
\$1,677.27 = PV0 of Mixed Flow [Using Tables]
\$600(PVIFA10%,2) =
\$600(1.736) = \$1,041.60
\$400(PVIFA10%,2)(PVIF10%,2) = \$400(1.736)(0.826) = \$573.57
\$100 (PVIF10%,5) =
\$100 (0.621) =
\$62.10
3-80
Group-At-A-Time (#2)
0
\$400
\$400
\$400
\$200
\$200
4
\$400
\$1,268.00
Plus
PV0 equals
\$1677.30.
\$347.20
Plus
5
\$100
\$62.10
3-81
## Solving the Mixed Flows
Problem using CF Registry
Use
the highlighted
key for starting the
process of solving a
mixed cash flow
problem
Press
3-82
the CF key
and down arrow key
through a few of the
keys as you look at
the definitions on
the next slide
## Solving the Mixed Flows
Problem using CF Registry
Defining the calculator variables:
For CF0:
This is ALWAYS the cash flow occurring
at time t=0 (usually 0 for these problems)
For Cnn:* This is the cash flow SIZE of the nth
group of cash flows. Note that a group may only
contain a single cash flow (e.g., \$351.76).
For Fnn:*
This is the cash flow FREQUENCY of the
nth group of cash flows. Note that this is always a
positive whole number (e.g., 1, 2, 20, etc.).
3-83
## * nn represents the nth cash flow or frequency. Thus, the
first cash flow is C01, while the tenth cash flow is C10.
## Solving the Mixed Flows
Problem using CF Registry
Steps in the Process
3-84
Step 1:
Press
Step 2:
Press
Step 3: For CF0 Press
CF
2nd
0
CLR Work
Enter
Step 4:
Step 5:
Step 6:
Step 7:
600
2
400
2
Enter
Enter
Enter
Enter
For F01 Press
For C02 Press
For F02 Press
key
keys
keys
keys
keys
keys
keys
## Solving the Mixed Flows
Problem using CF Registry
Steps in the Process
3-85
100
Enter
keys
Enter
keys
Step 10:
Step 11:
Press
Press
keys
key
NPV
Enter
10
Step 13:
Press
CPT
Result:
## Present Value = \$1,677.15
keys
key
Frequency of
Compounding
General Formula:
FVn = PV0(1 + [i/m])mn
3-86
n:
m:
i:
FVn,m:
Number of Years
Compounding Periods per Year
Annual Interest Rate
FV at the end of Year n
PV0:
## PV of the Cash Flow today
Impact of Frequency
Julie Miller has \$1,000 to invest for 2
years at an annual interest rate of
12%.
Annual
FV2
= 1,000(1+ [.12/1])(1)(2)
= 1,254.40
Semi
FV2
= 1,000(1+ [.12/2])(2)(2)
= 1,262.48
3-87
Impact of Frequency
Qrtly
FV2
= 1,000(1+ [.12/4])(4)(2)
= 1,266.77
Monthly
FV2
= 1,000(1+ [.12/12])(12)(2)
= 1,269.73
Daily
FV2
= 1,000(1+[.12/365])(365)(2)
= 1,271.20
3-88
## Solving the Frequency
Problem (Quarterly)
Inputs
Compute
2(4)
12/4
-1,000
I/Y
PV
PMT
FV
1266.77
## The result indicates that a \$1,000
investment that earns a 12% annual
rate compounded quarterly for 2 years
will earn a future value of \$1,266.77.
3-89
## Solving the Frequency
Problem (Quarterly Altern.)
Press:
2nd P/Y
2nd
QUIT
12
I/Y
-1000
PV
PMT
2
3-90
CPT
ENTER
2nd xP/Y N
FV
Problem (Daily)
Inputs
N
Compute
I/Y
PV
PMT
FV
1271.20
## The result indicates that a \$1,000
investment that earns a 12% annual
rate compounded daily for 2 years will
earn a future value of \$1,271.20.
3-91
## Solving the Frequency
Problem (Daily Alternative)
Press:
## 2nd P/Y 365 ENTER
2nd
QUIT
12
I/Y
-1000
PV
PMT
2
3-92
CPT
2nd xP/Y N
FV
Effective Annual
Interest Rate
Effective Annual Interest Rate
The actual rate of interest earned
rate for factors such as the number
of compounding periods per year.
(1 + [ i / m ] )m - 1
3-93
BWs Effective
Annual Interest Rate
Basket Wonders (BW) has a \$1,000
CD at the bank. The interest rate
is 6% compounded quarterly for 1
year. What is the Effective Annual
Interest Rate (EAR)?
EAR = ( 1 + 6% / 4 )4 - 1
= 1.0614 - 1 = .0614 or 6.14%!
3-94
Converting to an EAR
Press:
3-95
2nd
I Conv
ENTER
ENTER
CPT
2nd
QUIT
1.
2.
## Determine the interest in Period t.
(Loan balance at t-1) x (i% / m)
3.
## Compute principal payment in Period t.
(Payment - interest from Step 2)
4.
## Determine ending balance in Period t.
(Balance - principal payment from Step 3)
5.
3-96
## Amortizing a Loan Example
Julie Miller is borrowing \$10,000 at a
compound annual interest rate of 12%.
Amortize the loan if annual payments are
Step 1: Payment
PV0
= R (PVIFA i%,n)
\$10,000
= R (PVIFA 12%,5)
\$10,000
= R (3.605)
R = \$10,000 / 3.605 = \$2,774
3-97
End of
Year
0
Payment
Interest
Principal
---
---
---
Ending
Balance
\$10,000
\$2,774
\$1,200
\$1,574
8,426
2,774
1,011
1,763
6,663
2,774
800
1,974
4,689
2,774
563
2,211
2,478
2,775
297
2,478
\$13,871
\$3,871
\$10,000
3-98
Inputs
Compute
12
10,000
I/Y
PV
PMT
FV
-2774.10
## The result indicates that a \$10,000 loan
that costs 12% annually for 5 years and
will be completely paid off at that time will
require \$2,774.10 in annual payments.
3-99
## Using the Amortization
Functions of the Calculator
Press:
2nd
Amort
ENTER
ENTER
Results:
BAL = 8,425.90
PRN = -1,574.10
INT =
-1,200.00
3-100
## Using the Amortization
Functions of the Calculator
Press:
2nd
Amort
ENTER
ENTER
Results:
BAL = 6,662.91
PRN = -1,763.99
INT =
-1,011.11
3-101
## Using the Amortization
Functions of the Calculator
Press:
2nd
Amort
ENTER
ENTER
Results:
0.00
PRN =-10,000.00
INT =
BAL =
-3,870.49
## Entire 5 Years of loan information
3-102
Usefulness of Amortization
3-103
1.
## Determine Interest Expense -Interest expenses may reduce
taxable income of the firm.
2.
## Calculate Debt Outstanding -- The
quantity of outstanding debt
may be used in financing the
day-to-day activities of the firm. | 9,208 | 23,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-35 | latest | en | 0.90721 |
https://www.setzeus.com/public-blog-post/the-fourier-series | 1,675,608,171,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00336.warc.gz | 976,570,245 | 12,835 | # The Fourier Series
### A Basic Overview & Visual Introduction To The Magic Of Waves
##### Beyond Calculus
The Deep Study Of Nature Is The Most Fruitful Source Of Knowledge
It’s warmly appropriate that Jean-Baptiste Fourier left us with the poignant quote above as a stark reminder to continuously turn to our connection with nature as a source of inspiration for knowledge. It’s appropriate because, well, Fourier’s greatest contribution, the Fourier Series, both literally & figuratively, stems from a deep study of nature.
Literally, his key contribution to the annals of math history, the sole purpose of this exposition, came from the solution to a question posed by nature — mainly, how does the temperature on a metallic plate distribute over time? How about at any given point on the plate? Figuratively, the search for the solution stemmed from a long-spanning tradition: our innate need to make sense of the world around us by describing it in terms of a circle.
Since ancient times, the circle was placed on a pedestal as the simplest shape for abstract comprehension. A simple center point & a fixed-length radius/string was all needed—every point on the perimeter perfectly equidistant from the center. The key to understanding the Fourier series (thus the Fourier transform & finally the Discrete Fourier Transform) is our ancient desire to express everything in terms of circles. The genius connection the rest of this piece revolves around, the heart of Fourier’s observation, stems from the elegantly-captivating realization below: from simple rotations in a circle we can create trigonometric functions of sine & cosine.
# Introduction
As the term “ancient” in the previous sentence implies, Jean-Baptiste Joseph Fourier (1768–1830 A.D.) was far from the first person to realize this. However, he was the first to cleverly note that multiple simple waves, either sine or cosine, could be added to perfectly duplicate any type of periodic function. More importantly, the series bears his name because he derived the ingenious method that reverse-engineered his observation: the Fourier Series setup & the required Fourier analysis is the process necessary to uncover all the sine & cosine waves that converge to a targeted function. Specifically, once setup, the analysis consists of deriving the coefficients (radius of) & frequencies (“speed” of rotation on) of the many circles whose summation mimics that of any generic periodic function.
The Fourier Series is the circle & wave-equivalent of the Taylor Series. Assuming you’re unfamiliar with that, the Fourier Series is simply a long, intimidating function that breaks down any periodic function into a simple series of sine & cosine waves. It’s a baffling concept to wrap your mind around, but almost any function can be expressed as a series of sine & cosine waves created from rotating circles. To give you an idea of just how pervasive this new perspective can be, take a look at the example below, where we outline a bird strictly using attached circles:
The larger implications of the Fourier Series, it’s application to non-periodic functions through the Fourier Transform, have long provided one of the principal methods of analysis for mathematical physics, engineering, & signal processing. The Fourier Series a key underpinning to any & all digital signal processing — take a moment realize the breadth of this. Fourier’s work has spurred generalizations & applications that continue to develop right up to the present. As we’ll learn below, while the original theory of Fourier Series applies to periodic functions occurring in a natural wave motion, such as with light & sound, its generalizations, relate to significantly wider settings, such as the time-frequency analysis underlying the recent theories of wavelet analysis & local trigonometric analysis.
## Post-Heat Equation
Baron Jean Baptiste Joseph Fourier (1768−1830) first introduced the idea that any periodic function can be represented by a series of sines & cosines waves in 1828; published in his dissertation Théorie analytique de la chaleur, which loosely translates to The Analytical Theory of Heat, Fourier’s work is a result of arriving at the answer to a particular heat equation. Panda the Red beautifully recounts this particular journey, as a result, we’re mainly focusing on everything post-heat equation discovery.
In short, from the heat equation, Fourier evolved his findings to develop the Fourier Series; since then, the Fourier Series has only increased in importance (though more through the Fourier Transform), particularly in the digital age. From creating the base for physics such as Brownian motion, for finance such as in the Black-Scholes equations, or for electrical engineering such as in digital processing, Fourier’s work has only grown in both theoretical & practical applications.
Since we’re covering Fourier Series here, however, our scope of work is slightly narrowed. Despite casually mentioning that the Fourier Series is only applicable to periodic function, the truth is a bit more nuanced.
## Dirichlet Conditions
First, it must be noted that unlike the Fourier Transform, a Fourier Series cannot be applied to general functions — they can only converge to periodic functions. Yet that’s not all, to guarantee a convergence of simple sine & cosine waves, three specific criteria must be met. Known as the Dirichlet conditions, named after one Peter Gustav Lejeune Dirichlet, all three conditions must be met for a periodic function with some period-length 2L:
1. It has a finite number of discontinuities within the period 2L
2. It has a finite average value in the period 2L
3. it has a finite number of positive & negative maxima & minima
The three criteria above mainly ask: “does the function have bounded variation?” If f(x) is periodic over the length of some period 2L, & checks-off each condition list above, then the Fourier Series guarantees that some mix of cosine & sine waves can replicate f(x). Next, we’ll dive into the Fourier Series itself — starting from a very high level to working our way down to calculating exact coefficients.
# Modern Fourier Series
An infinite series of numbers either goes to infinite or converges to a number, the same way an infinite series of expressions (either polynomials or trigonometric) either goes to infinite or converges to a function (or shape). Conversely, if we’re given a shape, we can approximate its function by creating an infinite series of varying sin & cosine waves.
The Fourier Series is simply a function that’s described & derived by a literal summation of waves & constants.
## Formula Overview
We start with a high-level overview of the Fourier Series. Below, f(x) (left-side) is the target function we’re attempting to approximate through the Fourier Series (right-side):
The “Fourier Analysis” is simply the actual process of reverse-engineering, or constructing from scratch (sin & cos) a period function with the setup above — the goal is to solve for coefficients a0, an & bn. The most commonly-seen notation for the Fourier Series looks like the above. Before we dive into the coefficients, let’s re-frame the above by explaining the two different parts.
f(x) = Avg. Function Value + Sine/Cosine Waves Series
The first part of the Fourier Series, the leading division that includes the coefficient a0 is simply the average value of the function; more specifically, it’s the net area between −L & L, divided by 2L (the period of the function).
The second part of the equation, notated with a sigma/series symbol, represents the literal summation of different cosine & sines waves that should converge to target function; as one can tell, both trigonometric functions are carried to the nth-degree in the series. For this second half of the equation, the challenge is to solve for an & bn.
## Fourier Analysis
As we head into the weeds of Fourier Analysis & we start to solve for our target coefficients (a0,an, & bn), there’s good news & bad news. The good news: there’s a standard setup for deriving all three coefficients as well as multiple short-cuts that we’ll introduce later on. The bad news: solving for a0,an & bn is straight-forward, yet far from simple. All three coefficients are solved through the following integrals:
Solving For a0 — The Average Value
The first term on the left, a0 is at times referred to as the “average value” coefficient for that very reason — it’s simply an integral of the function we’re attempting to replicate over it’s fixed period.
Solving For aN— Summation of Cosine Waves
aN is the leading coefficient for the cosine waves in our series; our goal is to figure out how this coefficient behaves at different values in the series.
Solving For bN— Summation of Sine Waves
Conversely, bN is the leading coefficient for the sine waves in our series; our goal here is to again figure out how this coefficient behaves at different values in the series.
aN & bN are essentially the varying “weights” of their respective waves — they provide us with an approximation of which wave we’re “mixing” in most for any given series.
## Shortcuts — Even & Odd Functions
Now, thankfully, most Fourier Series are drastically reduced in their complexity early-on; based on the symmetry of the target function f(x), whether the function is even or odd, we can usually eliminate at least one of the coefficients. For a review, it’s worth remembering that a function, relative to its symmetry across the origin or the y-axis, can be considered even or odd:
• F(x) is even if F(-x)=F(x), such as Cosine(x), F(x)*Cosine(x), etc…
• F(x) is odd if F(-x)=-F(x), such as Sine(x), F(x)*Sine(x), etc…
This section can make our lives a lot easier because it reduces the work required. The key shortcut here is to always start a Fourier analysis by first checking whether F(x), the function or shape we’re approximating, is odd, even or neither. If a function is odd or even, we’re in luck — to recall some basic calculus, let’s remind ourselves what happens when we integrate either of the two trigonometric functions over some fixed period:
• The integral of Cos(x) from -L to L is 0
• The integral of Sin(x) from -L to L is also 0
Excellent — with both sets of facts above, it’s clear now how a function’s symmetry drastically reduces its complexity for a Fourier Analysis; basically, in most, not all, problems that we encounter, the Fourier coefficients a0, aN or bN become zero after integration. With knowledge of even & odd functions, a zero coefficient is predicted without performing the integration, leading us to, essentially, a powerful shortcut. Let’s closely inspect both cases further.
# Even Functions: Half-Range Fourier Cosine Series
A function F(x) is said to be even if F(-x) = F(x) for all values of x; therefore, the graph of an even function is always symmetrical about the y-axis (aka — it is a mirror image). For example, take a look at the graph of the function below, F(x) = cos(πx):
Clearly, the above is symmetrical across the y-axis. If a function is even, then it follows that the integral part of solving for bN, no matter the nth-term of bN, is also equal to zero. Therefore, we can safely eliminate the bN part of our original series, leaving us with the truncated Fourier Series of an even function. Known as a Half-Range Fourier Cosine Series, it looks like the following:
An even function has only cosine terms in its Fourier expansion: the key to understanding this & the following shortcut is the simple reminder that every Fourier Series setup starts with both a sine & a cosine function.
# Odd Functions: Half-Range Fourier Sine Series
A function F(x) is said to be odd if F(-x) = -F(x) for all values of x; therefore, the graph of an odd function is always symmetrical about the origin (aka — it’s unchanged when flipped over the x-axis & y-axis). For example, take a look at the graph of the function below, F(x) = sin(πx):
It’s a bit harder to tell, but the above is symmetrical across the origin. If a function is odd, then it follows that the integral of the series terms including aN , no matter the nth-term of aN, is also equal to zero. Therefore, we can safely eliminate the aN part of our original series, leaving us with the truncated Fourier Series of an odd function; known as a Half-Range Fourier Sine Series. That’s not all, however, odd functions include extraneous information that helps us eliminate an additional term: a0. Think this through — if a function is symmetrical across the origin, then this means that the area above the x-axis is equal to the area below the x-axis; which means that the average value of the function, our a0 term, is also equal to zero. Therefore, for a Half-Range Fourier Sine Series, we can safely eliminate both our first time a0 & our cosine term as such:
Much of the original setup now truncated, an odd function has only sine terms in its Fourier expansion; clearly, this is a significantly-simpler setup than our starting Fourier Series.
## Examples
It’s now time to walk through an actual Fourier Series example! For this example, we’re going to replicate a square-wave that oscillates from troughs of -1 to crests of 1 with a period of 2π; we’re going to analyze the function from -π to π-. This takes the following form (picture on the left/below).
The very first step to setting up a Fourier Series is not to jump into the setup, but rather to check if the target function displays either type of symmetry; looking at the graph, it’s pretty clear that it is indeed symmetrical around the origin. Therefore, the function we’re working with is odd. That tiny piece of analysis drastically reduces the complexity & required steps to complete our Fourier Series. Since we know it’s an odd-function, this means we can treat it as a Half-Range Fourier Sine Series (described above). We start our actual journey through this example with the substantially-simpler setup:
Reading left-to-right, f(t) is the function we’re approximating with our Fourier series. As you can tell, we’ve already eliminated both the a0 & aN terms, we only have a series-sum of sine waves left to engineer. Past the semi-colon, to the right, we have the remaining coefficient that we need to solve.
Here’s the part the tripped me up the first time: the f(t) on the right side is simply the value of the shape / function we’re approximating. In this particular example, as shown in the shape above, the value of the function f(t) is piecewise: from -π to 0, f(t) = -1; from 0 to π, f(t) = 1. Therefore, if we split bN to two different integrations, (-π,0) to (0,π), we can simply substitute the f(t) variable with either -1 or 1:
Next, we work out a few sample values of n to analyze patterns that’ll hint to the convergence of our coefficient bN. Let’s start by writing out n = 1:
The above isn’t too complicated — feel free to plug into wolfram alpha to double-check. It tells us though, that for the first value of n = 1, our coefficient of bN converges to the fraction 4/π. We’ll now repeat this process for four additional values of n in hopes of noticing a pattern:
Is there a discernible pattern? Yes. Again, please double-check these piece-wise integrations with Wolfram Alpha or another advanced calculator. Looking at the above, it’s notable that all even values of bN converge to zero, while all the odd values of bN converge to: 4 / n*π.
With bN solved, we can now plug the coefficient back into our Half-Range Fourier Sine Series that we setup above. Let’s now write out the first few terms of our series below:
This is a bit convoluted, however, it’s already perfectly accurate: the Fourier Series on the right indeed converges to our target square-wave. We can further confirm this by simplifying & animating exactly how this convergence happens over time:
With our Fourier Series now properly solved, let’s take a quick moment to visually confirm what we solved. The animation below shows exactly how each of the terms above corresponds to a circle with a specific radius & frequency that, in summation, draw out our intended square graph:
Each circle has a different radius & frequency. Observable in the third column of the GIF above, by appending each circle at the end of the radius of the preceding circle, our wave gradually approaches a square wave. For a last & final check, we’re going to overlay the series as we approach infinity on top of the very initial opening graph:
I can’t think of a word more accurate to capture this than: beautiful. It’s simply captivating to watch in action & nothing short of rewarding to fully comprehend the underlying mechanics.
# Onto Fourier Transforms
The Fourier Series is a way of representing periodic functions as an infinite sum of simpler sine & cosine waves. From signal processing to approximation theory to partial differential equations, it’s hard to overstate just how intricately the Fourier Series is tied with physics phenomena — anything with an identifiable pattern can be described with varying sin & cosine waves.
Yet…that’s not the end of the story. As it’d turn out a few decades afterward, the scope of our Fourier Series is quite limited compared to its successor, the Fourier Transform. The Fourier Series is used to represent a periodic function by a discrete sum, while the Fourier Transform is used to represent a general, non-periodic function. The Fourier transform is essentially the limit of the Fourier series of a function as the period approaches infinity. At the heart of all digital-based technology, it’s the next stop on our journey for those curious to understand the nature of our everyday objects just a tad bit more.
###### Learning Sources
Fourier Series
An Introduction To Fourier Series & Integrals | 3,814 | 17,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-06 | longest | en | 0.893537 |
https://karpacz.cz/36256/7uCJ23/about_crushing_aggregate_crushing_value.html | 1,632,838,037,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060803.2/warc/CC-MAIN-20210928122846-20210928152846-00470.warc.gz | 378,686,114 | 8,100 | #### Aggregate crushing value test Step by Step
2021-2-1 The average crushing value in the different tests is the actual crushing value of aggregate. Important points to remember. It is recommended that the crushing value of aggregate should be less than 30%. If it is more than 30 % then 10% finer aggregate is suggested to use. The aggregate whose crushing value is less than 30% can be used for
#### DETERMINATION OF AGGREGATE CRUSHING VALUE
2017-5-29 The aggregate crushing value gives a relative measure of the resistance of an aggregate to crushing under a gradually applied compressive load. Crushing value is a measure of the strength of the aggregate. The aggregates should therefore have minimum crushing value. 3. APPARATUS
#### Aggregate Crushing Value Test Significance & Test
2021-1-29 Calculation of Aggregate Crushing Value. The crushing value of aggregates is measured in the ratio between the crushed aggregates, obtained from the IS sieve 2.36mm and the weight of the dried aggregate sample. At least three samples should be tested from the same batch and average value
#### Aggregate Crushing Value Civil Engineering
AGGREGATE CRUSHING VALUE This test helps to determine the aggregate crushing value of coarse aggregates as per IS: 2386 (Part IV) 1963. The apparatus used is Cylindrical measure and plunger, Compression testing machine, IS Sieves of sizes 12.5mm, 10mm and 2.36mm
#### Aggregate Crushing Value Test Determine
2021-1-31 Concept and significance of the Aggregate crushing value test. The ‘ aggregate crushing value test ` gives a relative measure of the resistance of an aggregate to crushing under a gradually applied compressive load. Aggregate crushing value is defined as the percentage by weight of the crushed (or finer) material obtained when the test aggregates are subjected to a specified load under
#### AGGREGATE CRUSHING VALUE iricen.gov.in
2019-3-11 AGGREGATE CRUSHING VALUE 1.Objective The aggregate crushing value gives a relative measure of the resistance of an aggregate to crushing under a gradually applied compressive load. With aggregate of aggregate crushing value 30 or higher, the result may be anomalous, and in such cases the ten percent fines value should be determined instead.
#### How to determine the crushing value of aggregate?
2020-5-22 The aggregate crushing value= B/A*100 . Where, B is weight of fraction passing 2.36mm sieve ; A is weight of surface dry sample taken in mould. Note. The aggregate crushing value should not be more than 45% for aggregate used for concrete other than the wearing surfaces. 30% for concrete used the surfaces such as runway, roads and air field
#### What is the Crushing value of Aggregates..??
2020-5-27 The crushing value of an aggregate indicates its resistance against crushing under gradually applied compressive loads. The standard aggregate crushing test is made on aggregate passing 12.5mm IS sieve and retained on 10 mm IS sieve. Procedure. About 6.5 Kg material consisting of aggregates passing 12.5mm IS sieve and retained on 10mm sieve is
#### Aggregate Crushing Value Test L & T Learning
2019-8-27 The aggregate crushing value is a numerical index of the resistance of an aggregate to crushing under a gradually applied compressive load. It is expressed as the percentage by mass of the crushed (or finer) material obtained when the test sample is subjected to a specified load under standard conditions. Aggregates with lower crushing value
#### How to determine the crushing value of aggregate?
2020-5-22 The aggregate crushing value= B/A*100 . Where, B is weight of fraction passing 2.36mm sieve ; A is weight of surface dry sample taken in mould. Note. The aggregate crushing value should not be more than 45% for aggregate used for concrete other than the wearing surfaces. 30% for concrete used the surfaces such as runway, roads and air field
#### DETERMINATION OF AGGREGATE CRUSHING VALUE
2017-5-29 The aggregate crushing value gives a relative measure of the resistance of an aggregate to crushing under a gradually applied compressive load. Crushing value is a measure of the strength of the aggregate. The aggregates should therefore have minimum crushing value. 3. APPARATUS
#### How to Determine the Aggregate Crushing Strength
2021-1-28 The aggregate crushing value is defined as the ratio of the weight of fines passing through specified IS (here 2.36mm sieve) to the total weight of sample expressed in percentage. W1= Weigh of the total aggregates (Wa-We) W2= Weigh of aggregates passing through 2.36mm IS sieve.
#### Aggregate Crushing Value Civil4M
2017-10-16 Significance : The aggregate crushing value gives relative measure of the resistance to crush under a gradually applied compressive load. Definition : It is defined as the ratio of the weight of fines to the total sample weight expressed as a percent when tested as per the IS procedure.
#### Aggregate Crushing Value Test Vidyarthiplus
aggregate crushing value indicates strong aggregates, as the crushed fraction is low. ~hus the test can be used to assess the suitability of aggregates with reference to the crushing strength for various types of pavement components. The aggregates used for the surface course of
#### aggregate crushing value procedure proves-projekt.de
7 7 Aggregate Crushing Value and 10 Fines Value . Number report the mean as the aggregate crushing value, unless the individual results differ by more than times the mean value in this case repeat the test on two further specimens, calculate the median of the four results to the nearest whole number and report the median as the aggregate crushing value a data sheet is given as form
#### Aggregate Crushing Value Calculator, Test on course
Calculate Aggregate Crushing Value. The ratio of the weight of fines formed to the total sample weight in each test shall be expressed as a percentage, the result being recorded to the first decimal place: Aggregate Crushing Value = W B W A × 100. Where, W A is weight of surface-dry sample
#### Aggregate Crushing Value Apparatus Crushing Value
Our Aggregate Crushing Value Apparatus is used to measure the aggregate resistance of Crushing. We have designed our Aggregate Crushing Value Apparatus as per IS 2376 (PART-IV) & BS 812. In addition, the company is a prominent Aggregate Crushing Value Apparatus Manufacturer, Expo more...
#### Importance Of Aggregate Crushing Value Test
Aggregate Properties The Idiots' Guide to Highways Maintenance. The UK test to determine the aggregate crushing value is,BS 812 : Testing Aggregates : Part 110 Method of...
#### Aggregate Crushing Value Test L & T Learning
The aggregate crushing value is a numerical index of the resistance of an aggregate to crushing under a gradually applied compressive load. It is expressed as the percentage by mass of the crushed (or finer) material obtained when the test sample is subjected to a specified load under standard conditions. Aggregates with lower crushing value
#### Aggregate Crushing Value Test Vidyarthiplus
aggregate crushing value indicates strong aggregates, as the crushed fraction is low. ~hus the test can be used to assess the suitability of aggregates with reference to the crushing strength for various types of pavement components. The aggregates used for the surface course of
#### Aggregate Crushing Value Civil4M
2017-10-16 Significance : The aggregate crushing value gives relative measure of the resistance to crush under a gradually applied compressive load. Definition : It is defined as the ratio of the weight of fines to the total sample weight expressed as a percent when tested as per the IS procedure.
#### Aggregate Crushing Value Calculator, Test on course
Calculate Aggregate Crushing Value. The ratio of the weight of fines formed to the total sample weight in each test shall be expressed as a percentage, the result being recorded to the first decimal place: Aggregate Crushing Value = W B W A × 100. Where, W A is weight of surface-dry sample
#### Aggregate Crushing Value (ACV) Sets Mechanical
The Aggregate Crushing Value (ACV) Test Set provides a relative measure of the resistance of an aggregate to crushing under a gradually applied compressive load. Each set consists of steel cylinder, plunger, base plate, cylindrical measure and tamping rod. All parts of the apparatus are powder coated or galvanized steel, heat treated and ground
#### AGGREGATE CRUSHING VALUE (IS:2386-Part 4-1963)
2013-5-8 Aggregate crushing Value =(W B /W A) × 100 Record. The mean of the two results is reported to the nearest whole number as the aggregate crushing value of the tested material. Safety & Precautions: Use hand gloves while removing containers from oven after switching off the oven. To wear safety shoes & helmet during the time of test. | 1,876 | 8,832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-39 | latest | en | 0.857446 |
https://www.vanessabenedict.com/what-is-a-gram/ | 1,685,424,037,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645089.3/warc/CC-MAIN-20230530032334-20230530062334-00638.warc.gz | 1,160,473,616 | 15,114 | # What makes up 1 gram?
#### ByVanessa
Jun 9, 2022
The gram is a unit of mass in the metric system.
Originally defined as of 1795 as “the absolute weight of a volume of pure water equal to the cube of the hundredth part of a metre [1 cm³], and at the … Wikipedia
Ingredient Weight Chart
gram (n.) also gramme, metric unit of weight, 1797, from French gramme (18c.), from Late Latin gramma “small weight,” from Greek gramma “small weight,” a special use of the classical word meaning “a letter of the alphabet” (see -gram).
Metric Units
Untitled Document
## What makes up 1 gram
A gram is a model of mass in the metric computer system, defined as one thousandth (1 in 10 – 3) of a given kilogram. The gram was originally known as a unit equal to the mass of one cubic centimeter associated with pure water at 4 °C (the cold at which water becomes dense), the maximum possible.
## What is 1 gram approximately equal to
comparisons. 1 gram is approximately equal to 1 small paper or paperclip cap.
## What is gram in food
Gram Besan flour or heartbeat flour made from a variety of chickpea rue called Gram chickpeas. It is the main ingredient in the major cuisines of the Indian subcontinent, including Indian, Bangladeshi, Burmese, Nepalese, Pakistani, Sri Lankan and Caribbean cuisines. grams of flour.
## Is gram a mass or weight
Gram (g), also gram, a written unit of mass or weight, where it is used, in particular in some units of centimeters and gram-seconds (see International System of Units).
## What is the difference between Gram positive and Gram negative organisms when referring to Gram staining ie what makes Gram positive purple and Gram negative pink
Cells with a thick cell surface appear blue (gram positive) because crystal violet is retained in our cells, so red absorption dyes are not visible. Cells that develop a thin cell wall and subsequently become discolored turn red (gram-negative).
## Which is are true regarding features of PESA Act 1996 1 Gram Sabha shall identify beneficiaries under poverty alleviation programs 2 the recommendations of the Gram Sabha is mandatory prior to grant of prospecting license for minor minerals 3 Gram Sabha
1) Gram Shall sabha defines the beneficiaries of anti-poverty programs. 2) Gram Sabha recommendations are mandatory prior to issuing planning permission for mining of small minerals downstream. 4) Every village level panchayat must receive a certificate of use from my Gram Sabha.
Untitled Document
## How does the Gram staining procedure differentiate between gram negative and Gram-positive bacteria quizlet
Gram-positive bacteria have a large amount of peptidoglycan in their wall segment, which allows them to retain Ravenscroft’s violet dye, which allows them to label violet-blue. Gram-negative bacteria have much less peptidoglycan in their cell wall and therefore definitely do not retain the crystal violet dye that gives them their pink-red color.
## How does the Gram staining procedure differentiate between Gram negative and gram positive bacteria
Gram-positive and unhealthy cells have walls containing hard layers of peptidoglycan (90% smart walls). They turn purple. Gram-negatives accept bacterial walls with thin layers behind the peptidoglycan (10% wall) and higher lipid content. This keeps the pink color.
Untitled Document | 735 | 3,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-23 | latest | en | 0.918647 |
https://www.mathworks.com/matlabcentral/answers/335544-matlab-play-notes-for-different-duration-s-as-specified-by | 1,660,480,782,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572033.91/warc/CC-MAIN-20220814113403-20220814143403-00572.warc.gz | 775,233,035 | 25,670 | # MATLAB play notes for different duration's as specified by
10 views (last 30 days)
Commented: sanbadgerdude on 18 Apr 2017
I am trying to play a string of notes, each for a different duration. The following is what I have come up with but can not seem to get it to work properly. I am new to this and trying to understand. My intent is to have an array of notes then an array of times that each note will play. I may play the same note but hold it for a different duration of time.
if true
% code
end
clc
clear all
fs = 16000;
t1=[0:1/fs:0.3];
t2=[0:1/fs:0.6];
t3=[0:1/fs:0.9];
C2 = 65.4064;
D2 = 73.4162;
F2 = 87.3071;
staff1 = [C2 D2 F2];
time1 = [t1 t2 t3];
play1 = sin(2*pi*staff1*time1);
sound(play1,fs);
Geoff Hayes on 18 Apr 2017
sanbadgerdude - when I run your above code, I observe the following error
Error using *
Inner matrix dimensions must agree.
play1 = sin(2*pi*staff1*time1);
This is because the dimensions of staff1 is 1x3 whereas the dimensions for time1 is 1x28803, and so multiplying the two together is going to fail. What you can do instead is multiply each time intervals by a frequency prior to passing this in to the sine function. Try
ft = [t1*C2 t2*D2 t3*F2];
play1 = sin(2*pi*ft);
Thank you! I have been beating myself up over this. I had started to write out a long drawed out code when I saw your answer. Very much appreciated! Such a simple overlooked resolution. | 422 | 1,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-33 | latest | en | 0.930231 |
https://gmatclub.com/forum/in-1850-lucretia-mott-published-her-discourse-on-women-48661.html?sort_by_oldest=true | 1,513,588,397,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948612570.86/warc/CC-MAIN-20171218083356-20171218105356-00073.warc.gz | 592,162,880 | 58,269 | It is currently 18 Dec 2017, 01:13
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# In 1850 Lucretia Mott published her Discourse on Women,
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11 Jul 2007, 18:05
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In 1850 Lucretia Mott published her Discourse on Women, arguing in a treatise for women to have equal political and legal rights and for changes in the married women’s property laws.
A. arguing in a treatise for women to have equal political and legal rights
B. arguing in a treatise for equal political and legal rights for women
C. a treatise that advocates women’s equal political and legal rights
D. a treatise advocating women’s equal political and legal rights
E. a treatise that argued for equal political and legal rights for women
[Reveal] Spoiler: OA
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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11 Jul 2007, 19:28
Well, C, D and E all have sort of modification problems.
The correct answer should be between A and B. I like B. It is shorter and more concise.
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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11 Jul 2007, 21:08
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It's E!
The "for..." agrees with the latter "...and for"
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Re: SC: Discourse on Women [#permalink]
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31 Jan 2008, 06:35
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In 1850 Lucretia Mott published her Discourse on Women, arguing in a treatise for
women to have equal political and legal rights
and for changes in the married women’s
property laws.
A. arguing in a treatise for women to have equal political and legal rights
X published a book, arguing ... something. This is wrong for me as the meaning can stand as book was published while she was arguing !
B. arguing in a treatise for equal political and legal rights for women
Same as A. Guys ( please any Guru ) please advice for the same, I don't think I am confident in my explanation
C. a treatise that advocates women’s equal political and legal rights
[ Not parallel ]
D. a treatise advocating women’s equal political and legal rights
[ Not parallel ]
E. a treatise that argued for equal political and legal rights for women
Correct. Happy to see a for in both sides.
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Re: SC: Discourse on Women [#permalink]
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31 Jan 2008, 08:35
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I am with B.
C , D , E are out (can't assign human traits to objects)
B says she argues in a treatise for (equal political and legal rights) for woman
CORRECT
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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04 Apr 2011, 09:00
nevergiveup wrote:
Well, C, D and E all have sort of modification problems.
The correct answer should be between A and B. I like B. It is shorter and more concise.
It is E. In B, "arguing in a treatise" doesn't make any sense.
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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04 Apr 2011, 17:36
Am with E for two reasons -
1) does not change the original sentence, C,D use the word ''advocates and advocating'' instead of the original ''arguing''.
2)maintains the clarity by defining what the Discourse was all about by starting with '' a treatise...''
Hope that helps !
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03 May 2011, 03:24
38. In 1850, Lucretia Mott published her Discourse on Women, arguing in a treatise for women to have equal political and legal rights and for changes in the married women’s property laws.
(A) arguing in a treatise for women to have equal political and legal rights
(B) arguing in a treatise for equal political and legal rights for women
(C) a treatise that advocates women’s equal political and legal rights
(D) a treatise advocating women’s equal political and legal rights
(E) a treatise that argued for equal political and legal rights for women
why is option
[Reveal] Spoiler:
B
Is it probably because it is wrong to introduce a prep. phrase "in a ..." within the idiom "argue for"
Otherwise,
[Reveal] Spoiler:
B
can be correct given that the present participle arguing modifies the first clause to Lucretia...
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03 May 2011, 08:36
I narrowed down my answer choices to B and E and incorrectly selected B.
I am guessing between B and E, it is tense which should become reason to elimate B. Published is simple past and so preferred would be argued. E clearly modifies treatise and is clear.
I will also wait for more expert comments?
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04 May 2011, 02:21
wininblue wrote:
38. In 1850, Lucretia Mott published her Discourse on Women, arguing in a treatise for women to have equal political and legal rights and for changes in the married women’s property laws.
(A) arguing in a treatise for women to have equal political and legal rights
(B) arguing in a treatise for equal political and legal rights for women
(C) a treatise that advocates women’s equal political and legal rights
(D) a treatise advocating women’s equal political and legal rights
(E) a treatise that argued for equal political and legal rights for women
why is option
[Reveal] Spoiler:
B
Is it probably because it is wrong to introduce a prep. phrase "in a ..." within the idiom "argue for"
Otherwise,
[Reveal] Spoiler:
B
can be correct given that the present participle arguing modifies the first clause to Lucretia...
First of all a treatise is a formal and systematic written discourse on some subject. So our sentence is:
LM published her discourse on women,__________
So A and B are wrong as I expect treatise to be used just after comma.
(C) a treatise that advocates women’s equal political and legal rights..wrong as treatise cannot adovcate.
(D) a treatise advocating women’s equal political and legal rights...wrong
(E) a treatise that argued for equal political and legal rights for women ...better. Also argued for is correct.
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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01 Mar 2012, 15:06
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Here , arguing should modify the whole clause "Lucretia Mott published her Discourse on Women", but it is incorrectly associating "arguing" with "Mott".
It would have been correct , if the sentence is as follows :
In 1850 Lucretia Mott published her Discourse on Women, arguing equal political and legal rights for women and for changes in the married women’s property laws.
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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10 Jul 2012, 17:28
eyunni wrote:
In 1850 Lucretia Mott published her Discourse on Women, arguing in a treatise for
women to have equal political and legal rights
and for changes in the married women’s
property laws.
A. arguing in a treatise for women to have equal political and legal rights
B. arguing in a treatise for equal political and legal rights for women
C. a treatise that advocates women’s equal political and legal rights
D. a treatise advocating women’s equal political and legal rights
E. a treatise that argued for equal political and legal rights for women
Can someone please explain, what is structure of sentence above.
IS it : Time , clause , modifier. Then how modifier is modifying discourse.
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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10 Jul 2012, 20:24
‘Discourse on women’ is the title of Mott’s publication comprising her speeches. This publication is a called a treatise that gives information about what the publication contains and as such, it is an appositive modifier of ‘Discourse on Women
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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11 Jul 2012, 17:02
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@raphardu
Correct Sentence:
In 1850 Lucretia Mott published her Discourse on Women, a treatise that argued for equal political and legal rights for women
The structure of the correct sentence, which is (E), is as follows:
The phrase beginning with 'a treatise...' is an adjectival phrase modifying Discourse on Women.
Also, a 'treaty', or any document for that matter, can 'argue'.
e.g. The Declaration of Independence argued for the preeminence of men's freedom over the whimsies of tyranny.
Hope that helps .
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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01 Nov 2012, 04:17
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sachindia wrote:
Experts pls explain why A and B are wrong..
I am not the expert, but i will try to give my 2cents.
A and B : arguing..... is a verb-ing modifier modifying the preceding clause. Verb-ing modifier after the clause must tell additional information about the clause or present the result of the preceding clause. However, in A and B arguing... is not providing any information about the action of publishing. Also no result of the preceding clause is presented. hence option A and B are wrong. C and D are out due to parallelism issue. option E corrects the anomaly
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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05 May 2013, 21:59
1
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Hi folks ,
I would like to have more clarity on why option (B) is not in the RACE.
-Ing forms can be used in 4 ways :
Noun(Gerund)
Verb(Prog. tense)
ADJ ( The CAT SLEEPING on floor is ill)
ADVERB(X published article,mentioning all important facts)
So LM PUBLISHED her discourse, and in the act of publish "ARGUING"......
Rgds,
TGC
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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10 May 2013, 13:30
4
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targetgmatchotu wrote:
ADVERB(X published article,mentioning all important facts)
So LM PUBLISHED her discourse, and in the act of publish "ARGUING"......
Rgds,
TGC
Hi TGC,
You are perfectly right, ING form over here is actually used to modify LM, however that is not the error here.
Error is what happens after that.
"in a treatise for women" a treatise is a formal document for some cause(here for women rights) it doesn't make sense saying "arguing in a treatise for women to have " it could have been arguing in her treatise, the construction is similar to say..for eg, "i am arguing in a book to have more rights for poor kids." I am not actually(Physically) arguing in a book. I could have said, arguing for poor kids in my book, I sought rights for poor kids. or some construction similar to that.
Let me know if that helps..
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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13 May 2013, 03:52
4
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In 1850 Lucretia Mott published her Discourse on Women, arguing in a treatise for women to have equal political and legal rights and for changes in the married women’s property laws.
A. arguing in a treatise for women to have equal political and legal rights
its seems that Mott is arguing
B. arguing in a treatise for equal political and legal rights for women
its seems that Mott is arguing
C. a treatise that advocates women’s equal political and legal rights
Parallelism error
D. a treatise advocating women’s equal political and legal rights
Parallelism error
E. a treatise that argued for equal political and legal rights for women
Correct
For equal political and legal rights is parallel with for changes in the ...
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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09 Jul 2013, 11:57
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A> The construction is “arguing”+”in a treatise…” or participle + prepositional phrase. Or simply, it’s a participial phrase. Now, the participial phrase most likely modifies the subject, which is “Lucretia Mott”. Now the question is - Is Lucretia Mott arguing or is the argument in her book? When we look at the phrase “arguing in a treatise” , it is apparent that Lucretia is not arguing. Also, to+be construction is wordy. Be watchful
B> Same as above. The participial phrase as a modifier is not doing the justice.
C> An appositive “a treatise that..” is fixing the above problem. But it is introducing another problem. “a treatise that advocates women’s equal political and legal rights and for changes in the married women's property laws”. It is important to understand the construction of appositive. An appositive could be simply noun or noun+modifier. Here, we can write - a treatise (noun) that that advocates women’s equal political and legal rights (relative or adjective clause) and (a treatise) for changes in the married women's property laws (prepositional phrase). As you can see, a treatise for changes in the married women's property laws (prepositional phrase) does not make any sense. To correct this, we need (a treatise) that advocates for changes in the married women's property laws (adjective clause)
D> It is having the same problem as C. A treatise for changes in the married women's property laws (prepositional phrase) does not make any sense.
E> This answer choice fixed both problem. It introduced appositive correctly. Also, a parallelism is established with two prepositional phrases “for equal political and legal rights for women” and “for changes in the married women's property laws” that are modifying verb “argue” and act as a adverbial modifier
Consider a Kudos if you find this explanation helpful
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink]
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26 Jul 2013, 05:20
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eyunni wrote:
In 1850 Lucretia Mott published her Discourse on Women, arguing in a treatise for
women to have equal political and legal rights
and for changes in the married women’s
property laws.
A. arguing in a treatise for women to have equal political and legal rights
B. arguing in a treatise for equal political and legal rights for women
C. a treatise that advocates women’s equal political and legal rights
D. a treatise advocating women’s equal political and legal rights
E. a treatise that argued for equal political and legal rights for women
This question had bothered me long before I figured out why it is being said that Lucretia Mott is doing two things, although with casual read one feels its one thing.
I think the most important thing to notice here is the presence of article "a" before treatise. If we were to refer to Discourse on Women then the article should have been "the", which would mean we are referring to Discourse on Women. So the sentence then should have been "arguing in the treatise for women to have equal political and legal rights
Hope this helps....
Kudos if my post has helped you.
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Re: In 1850 Lucretia Mott published her Discourse on Women, [#permalink] 26 Jul 2013, 05:20
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https://byjus.com/questions/abcd-is-a-quadrilateral-in-which-ad-bc/ | 1,631,985,522,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00615.warc.gz | 209,422,020 | 37,403 | # ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that ∠ABD = ∠BAC.
$\mathrm{ABCD} \text { is a quadrilateral, where } \mathrm{AD}=\mathrm{BC} \text { and } \angle \mathrm{DAB}=\angle \mathrm{CBA}\\ \text { In } \triangle \mathbf{A B D} \text { and } \triangle \mathbf{B A C} \text { , }\\ \Rightarrow \mathbf{A D}=\mathbf{B C}\\ \Rightarrow \angle \mathrm{DAB}=\angle \mathrm{CBA} \quad[\text { Given }]\\ \Rightarrow \mathbf{A B}=\mathbf{B A} \text { [Common side] }\\ \triangle \mathbf{A B D} \cong \triangle \mathbf{B A C}\\ \angle \mathrm{ABD}=\angle \mathrm{BAC} [\mathrm{CPCT}]$ | 233 | 602 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-39 | latest | en | 0.342575 |
https://inneka.com/programming/js/how-can-i-do-an-asc-and-desc-sort-using-underscore-js/ | 1,566,387,793,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315936.22/warc/CC-MAIN-20190821110541-20190821132541-00499.warc.gz | 497,617,891 | 9,608 | # How can I do an asc and desc sort using underscore.js?
## How can I do an asc and desc sort using underscore.js?
I am currently using underscorejs for sort my json sorting. Now I have asked to do an ascending and descending sorting using underscore.js. I do not see anything regarding the same in the documentation. How can I achieve this?
### Solution 1:
You can use `.sortBy`, it will always return an ascending list:
``````_.sortBy([2, 3, 1], function(num) {
return num;
}); // [1, 2, 3]
``````
But you can use the .reverse method to get it descending:
``````var array = _.sortBy([2, 3, 1], function(num) {
return num;
});
console.log(array); // [1, 2, 3]
console.log(array.reverse()); // [3, 2, 1]
``````
Or when dealing with numbers add a negative sign to the return to descend the list:
``````_.sortBy([-3, -2, 2, 3, 1, 0, -1], function(num) {
return -num;
}); // [3, 2, 1, 0, -1, -2, -3]
``````
Under the hood `.sortBy` uses the built in `.sort([handler])`:
``````// Default is ascending:
[2, 3, 1].sort(); // [1, 2, 3]
// But can be descending if you provide a sort handler:
[2, 3, 1].sort(function(a, b) {
// a = current item in array
// b = next item in array
return b - a;
});
``````
### Solution 2:
Descending order using underscore can be done by multiplying the return value by -1.
``````//Ascending Order:
_.sortBy([2, 3, 1], function(num){
return num;
}); // [1, 2, 3]
//Descending Order:
_.sortBy([2, 3, 1], function(num){
return num * -1;
}); // [3, 2, 1]
``````
If you’re sorting by strings not numbers, you can use the charCodeAt() method to get the unicode value.
``````//Descending Order Strings:
_.sortBy(['a', 'b', 'c'], function(s){
return s.charCodeAt() * -1;
});
``````
### Solution 3:
The Array prototype’s reverse method modifies the array and returns a reference to it, which means you can do this:
``````var sortedAsc = _.sortBy(collection, 'propertyName');
var sortedDesc = _.sortBy(collection, 'propertyName').reverse();
``````
Also, the underscore documentation reads:
In addition, the Array prototype’s methods are proxied through the chained Underscore object, so you can slip a `reverse` or a `push` into your chain, and continue to modify the array.
which means you can also use `.reverse()` while chaining:
``````var sortedDescAndFiltered = _.chain(collection)
.sortBy('propertyName')
.reverse()
.filter(_.property('isGood'))
.value();
``````
### Solution 4:
Similar to Underscore library there is another library called as ‘lodash’ that has one method “orderBy” which takes in the parameter to determine in which order to sort it. You can use it like
``````_.orderBy('collection', 'propertyName', 'desc')
``````
For some reason, it’s not documented on the website docs.
Related: Smoothing arcs/plot points in D3.js/GeoJSON/TopoJSON/Shapefile (somewhere along the way)
### Solution 5:
Underscore Mixins
Extending on @emil_lundberg’s answer, you can also write a “mixin” if you’re using Underscore to make a custom function for sorting if it’s a kind of sorting you might repeat in an application somewhere.
For example, maybe you have a controller or view sorting results with sort order of “ASC” or “DESC”, and you want to toggle between that sort, you could do something like this:
Mixin.js
``````_.mixin({
sortByOrder: function(stooges, prop, order) {
if (String(order) === "desc") {
return _.sortBy(stooges, prop).reverse();
} else if (String(order) === "asc") {
return _.sortBy(stooges, prop);
} else {
return stooges;
}
}
})
``````
Usage Example
``````var sort_order = "asc";
var stooges = [
{name: 'moe', age: 40},
{name: 'larry', age: 50},
{name: 'curly', age: 60},
{name: 'July', age: 35},
{name: 'mel', age: 38}
];
_.mixin({
sortByOrder: function(stooges, prop, order) {
if (String(order) === "desc") {
return _.sortBy(stooges, prop).reverse();
} else if (String(order) === "asc") {
return _.sortBy(stooges, prop);
} else {
return stooges;
}
}
})
// find elements
var banner = \$("#banner-message");
var sort_name_btn = \$("button.sort-name");
var sort_age_btn = \$("button.sort-age");
function showSortedResults(results, sort_order, prop) {
banner.empty();
banner.append("<p>Sorting: " + prop + ', ' + sort_order + "</p><hr>")
_.each(results, function(r) {
banner.append('<li>' + r.name + ' is '+ r.age + ' years old.</li>');
})
}
// handle click and add class
sort_name_btn.on("click", function() {
sort_order = (sort_order === "asc") ? "desc" : "asc";
var sortedResults = _.sortByOrder(stooges, 'name', sort_order);
showSortedResults(sortedResults, sort_order, 'name');
})
sort_age_btn.on('click', function() {
sort_order = (sort_order === "asc") ? "desc" : "asc";
var sortedResults = _.sortByOrder(stooges, 'age', sort_order);
showSortedResults(sortedResults, sort_order, 'age');
})
``````
Here’s a JSFiddle demonstrating this: JSFiddle for SortBy Mixin
Related: How to change the text of a label? | 1,375 | 4,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-35 | latest | en | 0.538149 |
https://fivethirtyeight.com/features/can-you-unravel-these-number-strings/amp/ | 1,701,623,498,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00482.warc.gz | 297,401,588 | 16,578 | Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-sized and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, and you may get a shoutout in next week’s column. If you need a hint, or if you have a favorite puzzle collecting dust in your attic, find me on Twitter.
## Riddler Express
From Peter Neuhaus, a puzzling sequence:
A mysterious figure emerges from the shadows and hands you a note with the following list of six numbers:
1
11
21
1,211
111,221
312,211
He wants to know one thing: What number comes next?
## Riddler Classic
From Brad Ramirez, a second shrewd sequence:
Take a look at this string of numbers:
333 2 333 2 333 2 33 2 333 2 333 2 333 2 33 2 333 2 333 2 …
At first it looks like someone fell asleep on a keyboard. But there’s an inner logic to the sequence:
This sequence creates itself, number by number, as explained in the image above. Each digit refers to the number of consecutive 3s before a certain 2 appears. Specifically, the first digit refers to the number of consecutive 3s that appear before the first 2, the second digit refers to the number of 3s that appear consecutively before the second 2, and so on toward infinity.
The sequence never ends, but that won’t stop us from asking us questions about it. What is the ratio of 3s to 2s in the entire sequence?
## Solution to last week’s Riddler Express
Congratulations to 👏 Kari Matthews-Vaughn 👏 of Tulsa, Oklahoma, winner of last week’s Express puzzle!
You’re the president of the United States and you have a problem: Someone on your staff keeps leaking stories to the press. So you and your new chief of staff devise a plan. You will give different pieces of information to different staffers so that you’ll learn who the leakers are by seeing what information ends up in the newspaper or on TV. (You know you have only one leaker, and you know he or she leaks any story he or she is given.) If there are 100 people on your staff, how many different stories do you need to be able to identify your leaker for sure?
You’ll need seven stories.
The trick is to release stories sequentially. Here’s your strategy: Tell half your staff (50 people) one story and withhold it from the other half (also 50 people). If it leaks, you know the leaker is in the first half, and if doesn’t, you know the leaker is in the second half. Do the same thing again with the 50 remaining suspects: Give a second story to 25 of them and withhold it from the other 25. Once you see whether it leaks or not, you’ll have narrowed it down to 25 suspects. Rinse and repeat.
With this strategy, you’ll narrow down the staff, in the worst-case scenario, in a sequence like so: 100, 50, 25, 13, 7, 4, 2, 1. (If you get lucky, the leaker will be in one of the smaller groups created when you split the odd numbers in half, which might let you find them sooner.) So in seven steps, you’ll find the leaker for sure.
This is an example of a binary search. In the worst-case scenario, this type of search will take (rounded up to the nearest whole number) steps. The number equals about 6.6, so we’ll need 7 steps to guarantee we find the leaker.
## Solution to last week’s Riddler Classic
Congratulations to 👏 Seth Colbert-Pollack 👏 of Minneapolis, winner of last week’s Classic puzzle!
Fans of “Dungeons & Dragons” will have fond feelings for four-sided dice, which are shaped like regular tetrahedrons. Some of you might have noticed, in those long hours of fantasy battle, that if you touch five of these pyramids face-to-face-to-face, they come agonizingly close to forming a closed pentagon. Alas, there remains a tiny angle of empty space left between two of the pyramids. What is the measure of that angle?
If you knew what to look for, you could pretty quickly work out the answer to this problem via Wikipedia. The site lists a tetrahedron’s dihedral angle — the angle between two intersecting planes — as 70.528779 degrees. There are five dice, so five such angles. There are 360 degrees in all, so a gap of 7.356105 degrees.
But if you eschewed help from the internet, or if you just wanted to start from scratch, here’s how you could get there, adapted from a tidy solution submitted by Matthew Baron.
Assume the sides of the triangular faces of the dice have a length of 1. Now, begin with a single face and find the height of the triangle.
We know from the Pythagorean theorem that , so
Now we want to find the angle between two faces — the dihedral angle. Let’s call it
We can find the angle using the law of cosines:
Simplifying, , so degrees. And, again, subtracting five such angles from a total of 360 degrees ( ) gives our answer: about 7.4 degrees.
Solver Amy Leblang happened to have some tetrahedral dice on hand, and she verified the result:
## Want to submit a riddle?
Email me at oliver.roeder@fivethirtyeight.com.
Filed under | 1,235 | 5,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-50 | latest | en | 0.92797 |
https://www.numbersaplenty.com/5703658 | 1,701,909,608,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100626.1/warc/CC-MAIN-20231206230347-20231207020347-00176.warc.gz | 1,025,469,910 | 2,965 | Search a number
5703658 = 22851829
BaseRepresentation
bin10101110000011111101010
3101201202221121
4111300133222
52430004113
6322125454
766323512
oct25603752
911652847
105703658
113246274
121ab088a
13124915c
14a86842
15779e8d
hex5707ea
5703658 has 4 divisors (see below), whose sum is σ = 8555490. Its totient is φ = 2851828.
The previous prime is 5703641. The next prime is 5703679. The reversal of 5703658 is 8563075.
It is a happy number.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in only one way, i.e., 3621409 + 2082249 = 1903^2 + 1443^2 .
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1425913 + ... + 1425916.
Almost surely, 25703658 is an apocalyptic number.
5703658 is a deficient number, since it is larger than the sum of its proper divisors (2851832).
5703658 is a wasteful number, since it uses less digits than its factorization.
5703658 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 2851831.
The product of its (nonzero) digits is 25200, while the sum is 34.
The square root of 5703658 is about 2388.2332381909. The cubic root of 5703658 is about 178.6698031367.
The spelling of 5703658 in words is "five million, seven hundred three thousand, six hundred fifty-eight".
Divisors: 1 2 2851829 5703658 | 471 | 1,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-50 | latest | en | 0.853557 |
https://metanumbers.com/24459 | 1,656,462,040,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103619185.32/warc/CC-MAIN-20220628233925-20220629023925-00751.warc.gz | 450,601,148 | 7,385 | 24459 (number)
24,459 (twenty-four thousand four hundred fifty-nine) is an odd five-digits composite number following 24458 and preceding 24460. In scientific notation, it is written as 2.4459 × 104. The sum of its digits is 24. It has a total of 3 prime factors and 8 positive divisors. There are 15,720 positive integers (up to 24459) that are relatively prime to 24459.
Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 24
• Digital Root 6
Name
Short name 24 thousand 459 twenty-four thousand four hundred fifty-nine
Notation
Scientific notation 2.4459 × 104 24.459 × 103
Prime Factorization of 24459
Prime Factorization 3 × 31 × 263
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 24459 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 24,459 is 3 × 31 × 263. Since it has a total of 3 prime factors, 24,459 is a composite number.
Divisors of 24459
1, 3, 31, 93, 263, 789, 8153, 24459
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 33792 Sum of all the positive divisors of n s(n) 9333 Sum of the proper positive divisors of n A(n) 4224 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 156.394 Returns the nth root of the product of n divisors H(n) 5.79048 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 24,459 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 24,459) is 33,792, the average is 4,224.
Other Arithmetic Functions (n = 24459)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 15720 Total number of positive integers not greater than n that are coprime to n λ(n) 3930 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2709 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 15,720 positive integers (less than 24,459) that are coprime with 24,459. And there are approximately 2,709 prime numbers less than or equal to 24,459.
Divisibility of 24459
m n mod m 2 3 4 5 6 7 8 9 1 0 3 4 3 1 3 6
The number 24,459 is divisible by 3.
Classification of 24459
• Arithmetic
• Deficient
Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
Other numbers
• LucasCarmichael
• Sphenic
Base conversion (24459)
Base System Value
2 Binary 101111110001011
3 Ternary 1020112220
4 Quaternary 11332023
5 Quinary 1240314
6 Senary 305123
8 Octal 57613
10 Decimal 24459
12 Duodecimal 121a3
20 Vigesimal 312j
36 Base36 ivf
Basic calculations (n = 24459)
Multiplication
n×y
n×2 48918 73377 97836 122295
Division
n÷y
n÷2 12229.5 8153 6114.75 4891.8
Exponentiation
ny
n2 598242681 14632417734579 357894305370067761 8753736815046487366299
Nth Root
y√n
2√n 156.394 29.0277 12.5057 7.5455
24459 as geometric shapes
Circle
Diameter 48918 153680 1.87943e+09
Sphere
Volume 6.12921e+13 7.51774e+09 153680
Square
Length = n
Perimeter 97836 5.98243e+08 34590.2
Cube
Length = n
Surface area 3.58946e+09 1.46324e+13 42364.2
Equilateral Triangle
Length = n
Perimeter 73377 2.59047e+08 21182.1
Triangular Pyramid
Length = n
Surface area 1.03619e+09 1.72445e+12 19970.7
Cryptographic Hash Functions
md5 6320267b4965e7a50b2996cbd2d6c60f 51fe2c8ea7c615597b4111e3b07b274d8c2578e5 c724c617cbd2d3566ee00cba1e9b81a0934e3163954fb9eed879263362d65489 3865ab11e91b4264bfaf1e4fda4fc95cd8347e5deb2d9fcb29993f101a4fe4d12a4853a1be7767c0860e6b26ef85534eae46c43e54a35fade04877d26194ec53 bcff9aa63fd77ed7523e5f07cb43f5bfe5d0860b | 1,456 | 4,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-27 | latest | en | 0.828983 |
http://www.algebra.com/algebra/homework/Exponents/Exponents.faq.question.283561.html | 1,444,191,131,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736682102.57/warc/CC-MAIN-20151001215802-00119-ip-10-137-6-227.ec2.internal.warc.gz | 361,485,809 | 4,980 | # SOLUTION: Rewrite the following expression with positive exponents {{{(46xy)^(-2/7)}}} Choose the correct equivalent expression. A) {{{(46xy)^(7/2)}}} B) {{{1/(46xy)^(-2/7)}}} C)
Algebra -> Exponents -> SOLUTION: Rewrite the following expression with positive exponents {{{(46xy)^(-2/7)}}} Choose the correct equivalent expression. A) {{{(46xy)^(7/2)}}} B) {{{1/(46xy)^(-2/7)}}} C) Log On
Algebra: Exponents and operations on exponentsSection SolversSolvers LessonsLessons Answers archiveAnswers
Click here to see ALL problems on Exponents Question 283561: Rewrite the following expression with positive exponents Choose the correct equivalent expression. A) B) C) D) Answer by oberobic(2304) (Show Source): | 196 | 722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2015-40 | latest | en | 0.536392 |
https://sfepy.org/doc-devel/_modules/sfepy/linalg/utils.html | 1,718,256,737,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.11/warc/CC-MAIN-20240613025523-20240613055523-00267.warc.gz | 483,875,588 | 10,170 | # Source code for sfepy.linalg.utils
```from itertools import product
import numpy as nm
from numpy.lib.stride_tricks import as_strided
import numpy.linalg as nla
import scipy as sc
from sfepy.base.base import assert_, insert_method, output, Struct
[docs]def norm_l2_along_axis(ar, axis=1, n_item=None, squared=False):
"""Compute l2 norm of rows (axis=1) or columns (axis=0) of a 2D array.
n_item ... use only the first n_item columns/rows
squared ... if True, return the norm squared
"""
assert_(axis in [0, 1])
assert_(ar.ndim == 2)
other = 1 - axis
vec = nm.zeros((ar.shape[other],), dtype=nm.float64)
if n_item is None:
n_item = ar.shape[axis]
else:
n_item = min( n_item, ar.shape[axis] )
if axis == 1:
for ii in range( n_item ):
vec += ar[:,ii]**2
else:
for ii in range( n_item ):
vec += ar[ii,:]**2
if not squared:
vec = nm.sqrt( vec )
return vec
[docs]def normalize_vectors(vecs, eps=1e-8):
"""
Normalize an array of vectors in place.
Parameters
----------
vecs : array
The 2D array of vectors in rows.
eps : float
The tolerance for considering a vector to have zero norm. Such
vectors are left unchanged.
"""
norms = norm_l2_along_axis(vecs, axis=1)
ii = norms > eps
vecs[ii] = vecs[ii] / norms[ii][:, None]
[docs]def dets_fast(a):
"""
Fast determinant calculation of 3-dimensional array.
Parameters
----------
a : array
The input array with shape (m, n, n).
Returns
-------
out : array
The output array with shape (m,): out[i] = det(a[i, :, :]).
"""
from packaging import version
if version.parse(nm.__version__) >= version.parse('1.8'):
return nm.linalg.det(a)
else:
from numpy.linalg import lapack_lite
from numpy.core import intc
a = a.copy()
m = a.shape[0]
n = a.shape[1]
lapack_routine = lapack_lite.dgetrf
pivots = nm.zeros((m, n), intc)
flags = nm.arange(1, n + 1).reshape(1, -1)
for i in range(m):
tmp = a[i]
lapack_routine(n, n, tmp, n, pivots[i], 0)
sign = 1. - 2. * (nm.add.reduce(pivots != flags, axis=1) % 2)
idx = nm.arange(n)
d = a[:, idx, idx]
absd = nm.absolute(d)
sign *= nm.multiply.reduce(d / absd, axis=1)
nm.log(absd, absd)
return sign * nm.exp(logdet)
[docs]def invs_fast(a, det=None):
"""
Fast inversion calculation of 4-dimensional array.
Parameters
----------
a : array
The input array with shape (c, q, n, n).
det: array
To speed up the calculation, enter the already calculated determinant.
Returns
-------
out : array
The output array with shape (c, q, n, n):
out[c, q] = inv(a[c, q, :, :]).
"""
ax = nm.einsum("ij...->...ij", a)
dim = a.shape[-1]
inv_ax = nm.empty_like(ax)
if det is None:
det_a = dets_fast(a)[..., None, None]
else:
det_a = det.reshape(a.shape[:2] + (1, 1))
if dim == 3:
inv_ax[0, 0] = -ax[1, 2] * ax[2, 1] + ax[1, 1] * ax[2, 2]
inv_ax[1, 0] = ax[1, 2] * ax[2, 0] - ax[1, 0] * ax[2, 2]
inv_ax[2, 0] = -ax[1, 1] * ax[2, 0] + ax[1, 0] * ax[2, 1]
inv_ax[0, 1] = ax[0, 2] * ax[2, 1] - ax[0, 1] * ax[2, 2]
inv_ax[1, 1] = -ax[0, 2] * ax[2, 0] + ax[0, 0] * ax[2, 2]
inv_ax[2, 1] = ax[0, 1] * ax[2, 0] - ax[0, 0] * ax[2, 1]
inv_ax[0, 2] = -ax[0, 2] * ax[1, 1] + ax[0, 1] * ax[1, 2]
inv_ax[1, 2] = ax[0, 2] * ax[1, 0] - ax[0, 0] * ax[1, 2]
inv_ax[2, 2] = -ax[0, 1] * ax[1, 0] + ax[0, 0] * ax[1, 1]
elif dim == 2:
inv_ax[0, 0] = ax[1, 1]
inv_ax[1, 0] = -ax[1, 0]
inv_ax[0, 1] = -ax[0, 1]
inv_ax[1, 1] = ax[0, 0]
elif dim == 1:
inv_ax[0, 0] = 1.
else:
raise NotImplementedError(f'matrix dimension {dim}x{dim}')
return nm.einsum("...ij->ij...", inv_ax) / det_a
[docs]def get_blocks_stats(blocks, *args):
"""
Return statistics of array/matrix `blocks` defined by indices in `args`.
Returns
-------
stats: structured array
The array with 'shape', 'min', 'mean' and 'max' fields at positions of
each matrix block.
Examples
--------
>>> import numpy as nm
>>> from sfepy.linalg.utils import get_blocks_stats
>>>
>>> A = nm.eye(3)
>>> B = nm.full((3,2), 2)
>>> C = nm.full((1,3), 3)
>>> D = nm.full((1,2), 4)
>>> M = nm.block([[A, B], [C, D]])
>>>
>>> sr = [slice(0, 3), slice(3, 5)]
>>> sc = [slice(0, 3), slice(3, 4)]
>>> stats = get_blocks_stats(M, sr, sc)
>>>
>>> print(stats['shape'])
[[(3, 3) (3, 1)]
[(1, 3) (1, 1)]]
>>>
>>> print(stats['min'])
[[0. 2.]
[3. 4.]]
"""
bindices = args
idim = len(args)
bdim = blocks.ndim
bshape = [len(indices) for indices in bindices]
if idim == 1:
bshape = bshape * blocks.ndim
bindices *= blocks.ndim
elif idim != bdim:
raise ValueError('wrong number of dimensions of block indices!'
f' (can be 1 or {bdim}, is {idim})')
dt = blocks.dtype
sizes = nm.empty(bshape,
dtype=[('shape', tuple), ('min', dt), ('mean', dt),
('max', dt), ('maxabs', dt)])
for iflat, ii in enumerate(product(*bindices)):
key = nm.unravel_index(iflat, bshape)
block = blocks[ii]
sizes[key] = ((block.shape, block.min(), block.mean(), block.max(),
nm.abs(block).max()))
return sizes
[docs]def print_array_info(ar):
"""
Print array shape and other basic information.
"""
ar = nm.asanyarray(ar)
print(ar.shape, 'c_contiguous:', ar.flags.c_contiguous, \
'f_contiguous:', ar.flags.f_contiguous)
print('min:', ar.min(), 'mean:', ar.mean(), 'max:', ar.max())
[docs]def output_array_stats(ar, name, verbose=True):
ar = nm.asarray(ar)
if not len(ar):
output('%s: empty' % name)
elif nm.isrealobj(ar):
output('%s\nmin: % .6e mean: % .6e median: % .6e max: % .6e'
% (name, ar.min(), ar.mean(), nm.median(ar), ar.max()),
verbose=verbose)
else:
output_array_stats(ar.real, 'Re(%s)' % name , verbose=verbose)
output_array_stats(ar.imag, 'Im(%s)' % name, verbose=verbose)
[docs]def max_diff_csr(mtx1, mtx2):
aux = nm.abs((mtx1 - mtx2).data)
return aux.max() if len(aux) else 0.0
##
# 21.11.2005, c
[docs]def split_range( n_item, step ):
num = n_item / step
out = [step] * num
aux = sum( out )
if aux < n_item:
out.append( n_item - aux )
return out
##
# Inspired on net (ASPN Recipec).
# 14.12.2005, c
[docs]def permutations( seq ):
ls = len( seq )
if ls <= 1:
yield seq
else:
for ii in range( ls ):
for perm in permutations( seq[:ii] + seq[ii+1:] ):
yield [seq[ii]] + perm
##
# 14.12.2005, c
[docs]def cycle( bounds ):
"""
Cycles through all combinations of bounds, returns a generator.
More specifically, let bounds=[a, b, c, ...], so cycle returns all
combinations of lists [0<=i<a, 0<=j<b, 0<=k<c, ...] for all i,j,k,...
Examples:
In [9]: list(cycle([3, 2]))
Out[9]: [[0, 0], [0, 1], [1, 0], [1, 1], [2, 0], [2, 1]]
In [14]: list(cycle([3, 4]))
[[0, 0], [0, 1], [0, 2], [0, 3], [1, 0], [1, 1], [1, 2], [1, 3], [2, 0],
[2, 1], [2, 2], [2, 3]]
"""
nb = len( bounds )
if nb == 1:
for ii in range( bounds[0] ):
yield [ii]
else:
for ii in range( bounds[0] ):
for perm in cycle( bounds[1:] ):
yield [ii] + perm
[docs]def combine( seqs ):
"""Same as cycle, but with general sequences.
Example:
In [19]: c = combine( [['a', 'x'], ['b', 'c'], ['dd']] )
In [20]: list(c)
Out[20]: [['a', 'b', 'dd'], ['a', 'c', 'dd'], ['x', 'b', 'dd'],
['x', 'c', 'dd']]
"""
nb = len( seqs )
if nb == 1:
for ii in seqs[0]:
yield [ii]
else:
for ii in seqs[0]:
for perm in combine( seqs[1:] ):
yield [ii] + perm
[docs]def assemble1d(ar_out, indx, ar_in):
"""
Perform `ar_out[indx] += ar_in`, where items of `ar_in`
corresponding to duplicate indices in `indx` are summed together.
"""
if len(indx) > 0:
zz = nm.zeros_like(indx)
aux = sc.sparse.coo_matrix((ar_in, (indx, zz)), dtype=ar_in.dtype)
aux = aux.tocsr().tocoo() # This sums the duplicates.
ar_out[aux.row] += aux.data
[docs]def unique_rows(ar, return_index=False, return_inverse=False):
"""
Return unique rows of a two-dimensional array `ar`. The arguments follow
`numpy.unique()`.
"""
ar = nm.ascontiguousarray(ar)
# View the rows as a 1D structured array.
arv = ar.view(ar.shape[1] * [('', ar.dtype)])
out = nm.unique(arv, return_index=return_index,
return_inverse=return_inverse)
if isinstance(out, tuple):
uarv = out[0]
else:
uarv = out
# Restore the original dimensions.
uar = uarv.view(ar.dtype).reshape((-1, ar.shape[1]))
if isinstance(out, tuple):
out = (uar,) + out[1:]
else:
out = uar
return out
[docs]def argsort_rows(seq):
"""
Returns an index array that sorts the sequence `seq`. Works along
rows if `seq` is two-dimensional.
"""
seq = nm.asanyarray(seq)
if seq.ndim == 1:
ii = nm.argsort(seq)
else:
ii = nm.lexsort(seq.T[::-1])
return ii
[docs]def map_permutations(seq1, seq2, check_same_items=False):
"""
Returns an index array `imap` such that `seq1[imap] == seq2`, if
both sequences have the same items - this is not checked by default!
In other words, finds the indices of items of `seq2` in `seq1`.
"""
assert_(len(seq1) == len(seq2))
seq1 = nm.asanyarray(seq1)
seq2 = nm.asanyarray(seq2)
i1 = argsort_rows(seq1)
i2 = argsort_rows(seq2)
if check_same_items:
assert_(seq1.shape == seq2.shape)
assert_((seq1[i1] == seq2[i2]).all())
ii = nm.argsort(i2)
imap = i1[ii]
return imap
[docs]def mini_newton( fun, x0, dfun, i_max = 100, eps = 1e-8 ):
x = x0
ii = 0
while ii < i_max:
r = fun( x )
err = nla.norm( r )
## print ii, x, r, err
if err < eps: break
mtx = dfun( x )
try:
dx = nm.dot( nla.inv( mtx.T ), r )
except:
break
x = x - dx
ii += 1
return x
[docs]def insert_strided_axis(ar, axis, length):
"""
Insert a new axis of given length into an array using numpy stride
tricks, i.e. no copy is made.
Parameters
----------
ar : array
The input array.
axis : int
The axis before which the new axis will be inserted.
length : int
The length of the inserted axis.
Returns
-------
out : array
The output array sharing data with `ar`.
Examples
--------
>>> import numpy as nm
>>> from sfepy.linalg import insert_strided_axis
>>> ar = nm.random.rand(2, 1, 2)
>>> ar
array([[[ 0.18905119, 0.44552425]],
[[ 0.78593989, 0.71852473]]])
>>> ar.shape
(2, 1, 2)
>>> ar2 = insert_strided_axis(ar, 1, 3)
>>> ar2
array([[[[ 0.18905119, 0.44552425]],
[[ 0.18905119, 0.44552425]],
[[ 0.18905119, 0.44552425]]],
[[[ 0.78593989, 0.71852473]],
[[ 0.78593989, 0.71852473]],
[[ 0.78593989, 0.71852473]]]])
>>> ar2.shape
(2, 3, 1, 2)
"""
shape = list(ar.shape)
shape.insert(axis, length)
strides = list(ar.strides)
strides.insert(axis, 0)
out = as_strided(ar, shape=shape, strides=strides)
return out
[docs]def dot_sequences(mtx, vec, mode='AB'):
"""
Computes dot product for each pair of items in the two sequences.
Equivalent to
>>> out = nm.empty((vec.shape[0], mtx.shape[1], vec.shape[2]),
>>> dtype=vec.dtype)
>>> for ir in range(mtx.shape[0]):
>>> out[ir] = nm.dot(mtx[ir], vec[ir])
Parameters
----------
mtx : array
The array of matrices with shape `(n_item, m, n)`.
vec : array
The array of vectors with shape `(n_item, a)` or matrices with shape
`(n_item, a, b)`.
mode : one of 'AB', 'ATB', 'ABT', 'ATBT'
The mode of the dot product - the corresponding axes are dotted
together:
'AB' : `a = n`
'ATB' : `a = m`
'ABT' : `b = n` (*)
'ATBT' : `b = m` (*)
(*) The 'BT' part is ignored for the vector second argument.
Returns
-------
out : array
The resulting array.
Notes
-----
Uses `numpy.matmul()` via the `@` operator.
"""
if vec.ndim == mtx.ndim:
squeeze = False
else:
squeeze = True
vec = vec[..., None]
if 'BT' in mode:
ax = list(range(vec.ndim))
vec = vec.transpose((ax[:-2]) + [ax[-1], ax[-2]])
if 'AT' in mode:
ax = list(range(mtx.ndim))
mtx = mtx.transpose((ax[:-2]) + [ax[-1], ax[-2]])
out = mtx @ vec
if squeeze:
out = out[..., 0]
return out
[docs]def apply_to_sequence(seq, fun, ndim, out_item_shape):
"""
Applies function `fun()` to each item of the sequence `seq`. An item
corresponds to the last `ndim` dimensions of `seq`.
Parameters
----------
seq : array
The sequence array with shape `(n_1, ..., n_r, m_1, ..., m_{ndim})`.
fun : function
The function taking an array argument of shape of length `ndim`.
ndim : int
The number of dimensions of an item in `seq`.
out_item_shape : tuple
The shape an output item.
Returns
-------
out : array
The resulting array of shape `(n_1, ..., n_r) + out_item_shape`. The
`out_item_shape` must be compatible with the `fun`.
"""
n_seq = nm.prod(seq.shape[0:-ndim], dtype=int)
aux = nm.reshape(seq, (n_seq,) + seq.shape[-ndim:])
out = nm.empty((n_seq,) + out_item_shape, dtype=seq.dtype)
for ii, item in enumerate(aux):
out[ii,:] = fun(item)
out = nm.reshape(out, seq.shape[0:-ndim] + out_item_shape)
return out
##
# 30.08.2007, c
[docs]class MatrixAction( Struct ):
##
# 30.08.2007, c
[docs] def from_function( fun, expected_shape, dtype ):
def call( self, vec ):
aux = fun( vec )
assert_( aux.shape[0] == self.shape[0] )
return nm.asanyarray( aux, dtype = self.dtype )
obj = MatrixAction( shape = expected_shape,
dtype = dtype,
kind = 'function' )
insert_method( obj, call )
return obj
from_function = staticmethod( from_function )
##
# 30.08.2007, c
[docs] def from_array( arr ):
def call( self, vec ):
return nm.asarray( sc.dot( self.arr, vec ) )
obj = MatrixAction( shape = arr.shape,
dtype = arr.dtype,
arr = arr,
kind = 'array' )
insert_method( obj, call )
return obj
from_array = staticmethod( from_array )
##
# 30.08.2007, c
def __call__( self, vec ):
return self.call( vec )
##
# 30.08.2007, c
[docs] def to_array( self ):
if self.kind == 'array':
return self.arr
else:
print('cannot make array from MatrixAction of kind %s!' % self.kind)
raise ValueError
``` | 4,483 | 13,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-26 | latest | en | 0.462451 |
http://www.questionotd.com/2011/ | 1,723,584,175,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641085898.84/warc/CC-MAIN-20240813204036-20240813234036-00734.warc.gz | 46,012,627 | 87,666 | ## Wednesday, December 21, 2011
### Snowball's Chance
0 = S C in H
2 = G for E B (BBS)
3 = H on the D C
4 = F on a P
26 = M in a M
76 = T in the M M
97 = P W
600 = C in the L B
## Tuesday, December 20, 2011
### 200 D for P G in M
1 H in a D
3 V in a T
4 S on a V
8 P of S in the E L
8 D a W in the B S
9 J on the S C
9 S in T T T
## Monday, December 19, 2011
### 90 Degrees in a Right Angle
3 B M (S H T R)
5 T on a F
9 L of a C
15 M on a D M C
23 P of C in the H B
26 L of the A
27 B in the N T
39 B of the O T
90 D in a R A
## Friday, December 16, 2011
### 24 Hours in a Day
24 = B. B. in a P.
24 = B. B. to a C.
24 = D. P. to the I. for M. I.
24 = H. in a D.
24 = S. Z. of I. T.
## Wednesday, December 14, 2011
### 451 D F
435 = S. in the H. of R.
451 = D. F. at which B. B.
500 = F. F. H. C.
500 = M. in the I. F. H.
500 = S. in a R.
## Tuesday, December 13, 2011
### Lucky 13
13 = U for S
13 L in a B D
13 = S on the A F
54 = C in a D (with the J)
9 = P in the S S
## Monday, December 12, 2011
### Been A While Since I Did Some Statistics
How many 2 digit numbers can you make using the digits 1, 2, 3 and 4 without repeating the digits?
## Friday, December 09, 2011
### Imagine What They Would Do to the Linens
What always goes to bed with shoes on?
## Thursday, December 08, 2011
### Commuters Nightmare
What kind of running means walking?
## Wednesday, December 07, 2011
### Can Climb Mountains
What can climb mountains,
cross streams,
handle hundreds of feet each day,
yet never moves?
I've been reading the Dresden series by taking the books out of the library. It's been saving me a lot of money given how many books Butcher puts out.
## Monday, December 05, 2011
### Middle You Can See With
The middle you can see with,
The end is end's start,
The beginning I won't tell you,
But the whole stops your heart.
A Jeff Original (TM)
Have your own riddle to post? Send them in (use my e-mail in the About Me section in the sidebar).
## Friday, December 02, 2011
### For Others I'm Slow
For some I go fast
for others I'm slow.
To most people, I'm an obsession
relying on me is a well practiced lesson.
This question was posted by an anonymous reader who signed it : ~A.S.T. & Heidelberg High School~
## Wednesday, November 30, 2011
### Dagger Thrust to the Heart
A dagger thrust at my own heart,
Dictates the way I’m swayed.
Left I stand, and right I yield,
To the twisting of the blade.
What am I?
This one reminds me of Kingdom Hearts.
## Tuesday, November 29, 2011
### Five Women and Five Household Items
Five women each purchase a household item for use in a different room. Can you work out the full name of each woman, her item and where she keeps it?
1. Mrs. Simpson does not keep her item in the bedroom.
2. Amy has a television; Mrs. Griggs has a radio.
3. Kylie does not keep her item in the bedroom.
4. Clara does not have a telephone.
5. Mrs. Williams does not keep her item in the kitchen.
6. Kylie keeps hers in the conservatory.
7. Michelle has a bookcase; Mrs. Dingle has a computer.
8. Michelle does not keep her in the living room.
9. Mrs. Pringle keeps hers in the study; Roxanne keeps hers in the kitchen.
First Names: Amy, Clara, Kylie, Michelle, Roxanne
Last Names: Dingle, Griggs, Pringle, Simpson, Williams
Items: Bookcase, Computer, Radio, Telephone, Television
Locations: Bedroom, Conservatory, Kitchen, Living Room, Study
Speaking of WOT series, anyone have an update on when the last book is due out? I can't believe I've been reading this series for 20 years, now.
## Monday, November 28, 2011
### First Is Often at the Front Door
My first is often at the front door.
My second is found in the cereal family.
My third is what greed leads you to acquire.
My whole is one of the united states.
Seriously, is Angry Birds getting to be everywhere?
## Wednesday, November 23, 2011
### Pilgrims Trivia
Last year, I left you with some Thanksgiving trivia. This year I think I'll post some trivia questions on the Mayflower and the Pilgrims. Since I'll be gone until next week, the answers are posted below the questions.
Questions
1. You know about the Mayflower, but what was the name of the other ship the Pilgrims had planned on carrying then to the New World?
2. What country did the Pilgrims leave from?
3. How many children were born on the Mayflower during the voyage?
4. How many days did it take before the first land fall?
5. John Carver was the first governor of the Plymouth Colony, how many terms did he serve?
6. 102 passengers set sail, 102 passengers made landing (one birth and one death). But how many survived that first winter?
7. What was the name of the tribe that helped the Pilgrims and later celebrated that first Thanksgiving with them?
8. Who was the captain of the Mayflower?
1. Speedwell was the name of the other ship. After having trouble and having to turn around twice, the Pilgrims decided to travel on the Mayflower alone.
2. The Pilgrims had been living in the Netherlands but left from England on Sept 6, 1620.
3. One, his name was Oceanus.
4. 66 days, for an average of 2 mph.
5. He was elected twice, the first time he was elected in November 1620, while aboard the Mayflower. He was again elected in March 1621. He died just one month later in April 1621.
6. Only 51 survived the first winter in Plymouth.
7. The Wampanoag tribe, which was led by Massassoit.
8. The captain of this ship was Christopher Jones. During that first winter, the Mayflower crew also suffered there in Plymouth, with almost half dying. It did not leave until April 5, 1621.
## Tuesday, November 22, 2011
### Magnetic Bar of Iron
There are two bars of iron. One bar is magnetized along its length, while the other is not. Using just the two bars, without any other items, how can you tell which bar is magnetized and which bar is not?
I do like the Geronimo Stilton series. The Kingdom of Fantasy series in particular has been enjoyable.
## Monday, November 21, 2011
### Counting Odd One Out
Which word is the odd one out: First Second Third Forth Fifth Sixth Seventh Eighth.
Has anyone tried the Kindle Fire? I'm wondering if it's as good as all the press.
## Tuesday, November 15, 2011
### Missing Gallows
An old parchment describes the location of buried treasure:”On the island there are only two trees, A and B, and the remains of a gallows. Start at the gallows and count the steps required to walk in a straight line to tree A. At the tree turn 90 degrees to the left and then walk forward the same number of steps. At the point where you top drive a spike into the ground.
Now return to the gallows and walk in a straight line, counting your steps, to tree B. When you reach the tree, turn 90 degrees to the right and take the same number of steps forward, placing another spike at the point where you stop. Dig at the point exactly halfway between the spikes and you will find the treasure.”
However, our hero when he gets to the island finds the gallows missing. Is there any way he can still get to the treasure?
## Monday, November 14, 2011
### Quick Logic on Bees and Flowers
In a pond there are some flowers with some bees hovering over the flowers. How many flowers and bees are there if both the following statements are true:
1. If each bee lands on a flower, one bee doesn’t get a flower.
2. If two bees share each flower, there is one flower left out.
## Thursday, November 10, 2011
### Which Man Died?
There are five acquaintances. One of them shot and killed one of the other five. Which man is the murderer?
1. Dan ran in the NY City marathon yesterday with one of the innocent men.
2. Mike considered being a farmer before he moved to the city.
3. Jeff is a top-notch computer consultant and wants to install Ben's new computer next week.
4. The murderer had his leg amputated last month and has not recovered fully.
5. Ben met Jack for the first time six months ago.
6. Jack has been in seclusion since the crime.
7. Dan used to drink heavily.
8. Ben and Jeff built their last computers together.
9. The murderer is Jack’s brother; they grew up together in Seattle.
## Tuesday, November 08, 2011
### Five Gates
A joy to hold or hold for joy
Windows out or maybe in
Never filled but too much may lessen
And sometimes to be, lost in the cold
Whilst another finds the spice of life
But in truth.
Gates to the world?
## Monday, November 07, 2011
### A Time When They're Green
A time when they're green,
A time when they're brown.
But both of these times cause me to frown.
But just in between for a very short while,
They're perfect and yellow and cause me to smile!
What are they?
His Dark Materials is good so far. I'm thrown off by the start of the second book, though. To start off with a whole new main character is disturbing after focusing in so much on Lyra in the first.
## Thursday, November 03, 2011
### Steps Into It
What does a dog do, that a man steps into?
And no, I'm not going for the obvious answer!
## Wednesday, November 02, 2011
### One is Sitting Down But Will Never Get Up
I give you a group of three,
One is sitting down but will never get up,
The second eats as much as is given to him,
yet is always hungry.
The third goes away and never returns...
What are they?
No one seems to be responding to my questions about what they're reading?
## Tuesday, November 01, 2011
### All About But Cannot Be Seen
All about but cannot be seen,
can be captured,
cannot be held,
no throat but can be heard.
What am I?
Exploding stones? Who would have thought?
## Friday, October 28, 2011
### Run to the Water
Fifty people are standing on the beach. One of them pulls out a gun and says "I will shoot the last one that gets in the water."
There are two people left on the beach, one of them is running to the water, the other one is you.
What would you do?
Frustrated with my iRoomba lately. I think it needs a new battery, I hope it works.
## Thursday, October 27, 2011
### Five Hundred, Five and One
Five hundred begins it, five hundred ends it,
Five in the middle is seen;
First of all figures, the first of all letters,
Take up their stations between.
Join all together, and then you will bring
Before you the name of an eminent king
## Wednesday, October 26, 2011
### Safe and Secure
As a whole I am both safe and secure,
Behead me, and I become a place of meeting.
Restore me and I become the domain of beasts.
What am I?
I do like to watch Biggest Lose but I've never tried their cookbooks. Have you?
## Tuesday, October 25, 2011
### Work Smarter, Not Harder
What work is it that the faster you work,
the longer it is before your done
and the slower you work the sooner your finished?
Just started The Secret of Zoom. Good starting chapter and the girl is a funny observer.
## Monday, October 24, 2011
### Feeling Cold?
What traps warmth, sprays forth ice and water and slips through your fingers with ease?
Got George R. R. Martin's A Game of Thrones 4-Book Bundle on sale, but have never watched the show.
## Thursday, October 20, 2011
### Long or Short, Round or Square
I can be long, or I can be short.
I can be grown, and I can be bought.
I can be painted, or left bare.
I can be round, or square.
What am I?
Still reading How Firm a Foundation (Safehold) by David Weber, how about you?
## Wednesday, October 19, 2011
### Bent or Broken
What can be made, bent, broken, changed and followed but not twisted?
Book I'm reading right now: The Secret of Platform 13
## Tuesday, October 18, 2011
### Three Riddles for the Price of One
Which three things all have eyes but none can see?
Which two things have mouths but cannot speak?
What has ears but cannot hear?
Currently, I'm reading How Firm a Foundation (Safehold)
## Monday, October 17, 2011
### Couple of Rebus
A couple of rebus, can you find the phrases hidden inside these two clues?
1.
Peace
Earth
2.
Jumping
Joy
Joy
Joy
Joy
From the book: Everything Brain Strain Book: Over 400 Puzzles, Riddles, And Mind-Benders To Flex Your Mental Muscles (Everything: Sports and Hobbies)
## Thursday, October 13, 2011
### Prime Time Soccer
At the end of the soccer season, every player had scored a prime number of goals and the average for the eleven was also a prime number. No player's tally was the same as anyone else's, and neither was it the same as the average.
Given that nobody had scored more than forty-five goals, how many goals did each player score?
Like yesterday, this puzzle comes out of The World's Biggest Book of Brainteasers & Logic Puzzles.
## Wednesday, October 12, 2011
### Shepherds on Hillsides
In every group of shepherds in Hyperborea, at least one is a Sororean, who always speaks truthfully, and at least one is a Nororean, who always speaks falsely.
A visitor approached four shepherds on a hillside and asked each how many of the four were Sororeans. These answers were given:
• Three of us are Sororeans.
• One of us is.
• There are two of us.
• None of us are Sororeans.
The visitor approached four more shepherds on another hillside and asked how many were Nororeans. Their answers follow:
• We are all Nororeans.
• One of us is.
• Three of us are.
• The fourth shepherd declined to speak.
How many of the shepherds on the two hillsides were Sororeans?
This puzzle comes out of The World's Biggest Book of Brainteasers & Logic Puzzles
## Tuesday, October 11, 2011
### Letter Equation Competition
3 S and Y O
6 P for a T D
20 M in a P
4 R for a G S H R
## Friday, October 07, 2011
### Fly to Clouds of Heaven
I am a wingless bird, flying even to the clouds of heaven. I give birth to tears of mourning in pupils that meet me, even though there is no cause for grief, and at once on my birth I am dissolved into air. What am I?
## Thursday, October 06, 2011
### Shift to Enter
You may enter, but you may not come in, I have space, but no room, I have keys, but open no lock. What am I?
## Tuesday, October 04, 2011
### Modern Day Riddle
I am always at your side. To a slab my tail is tied And my eyes are both inside my belly. Tickle my back, one/two/three! Wisdom and folly are yours to see. -http://catb.org/~esr/riddle-poems.html
## Monday, October 03, 2011
### Moon and Sea
The Moon is my father,
the Sea is my mother;
I have a million brothers,
I die when I reach land.
## Friday, September 30, 2011
### Old Time Riddle of Rings
A hoard of rings am I,
but no fit gift for a bride;
I await a sword's kiss.
## Thursday, September 29, 2011
### Riddle on the Wave
A wonder on the wave / water became bone.
- Book of Exeter
## Wednesday, September 28, 2011
### Shoulder a Gargantuan Wooden Rail
1. Perambulate in moccasins, and shoulder a gargantuan wooden rail. 2. Allow somnolent quadrupeds that are homo sapien's greatest comrades to remain reclining. 3. Lack of what is required is the matriarch of inspiration. 4. A maximum amount of purposeful activity and a minimum amount of disport and dalliance cause Jack to become a dim-witted, stagnant dunce of the male species. 5. That which is acquired without difficulty is dispersed with equal facility. 6. It is more desirable to arrive in the medium of time which constitutes a later than desirable hour or date than not to arrive at all.
## Tuesday, September 27, 2011
### Say What?
He who locks himself into the arms of Morpheus promptly at eventide, and starts the day before it is officially announced by the rising sun, excels in physical fitness, increases his economic assets and celebrates with remarkable efficiency. What common phrase is hidden in the complicated sentence above?
## Friday, September 23, 2011
### Friday Night Calls
The sun bakes them, the hand breaks them, the foot treads on them, and the mouth tastes them. What are they?
## Thursday, September 22, 2011
### A Logic Master Party
I go to a party with my wife. When we get there, four other couples arrive at the same time. We all know each other, so we all greet each other. A greeting can be a handshake, a kiss, a hug, or whatever. When everyone is done I ask everyone how many times they shook another person's hand. All answers I get are different. Given that nobody greets their own spouse, how many hands did my wife shake?
## Wednesday, September 21, 2011
### Travel Over Cobblestone or Gravel
Used left or right, I get to travel, over cobblestone or gravel. Used up, I vie for sweet success, used down, I cause men great duress. What am I?
## Tuesday, September 20, 2011
### Is it Possible
Find a 5 digit number, as big as possible, that when you multiply it by a single digit number, you get a six digit number, in which all digits are identical.
## Friday, September 16, 2011
### Three Darts in One Hemisphere
You throw three darts onto the surface of a globe, each from a randomly chosen direction. What is the probability that all three darts are in one hemisphere?
## Wednesday, September 14, 2011
### A Bit More Complicated
You have 55 matches arranged in some number of piles of different sizes. You now do the following operation: pick one match from each pile, and form a new pile. You repeat this ad infinitum. What is the steady state? Is it unique? By steady state I mean the number of piles will remain unchanged or you create an unending loop.
## Tuesday, September 13, 2011
### Six Digits
You have a number that consists of 6 different digits. This number multiplied by 2, 3, 4, 5, and 6 yields, in all cases, a new 6-digit number, which, in all cases, is a permutation of the original 6 different digits. What's the number?
## Monday, September 12, 2011
### Create Twenty-Four
Create the number 24 using (all of) 1, 3, 4, and 6. You may add, subtract, multiply, and divide. Parentheses are free. You must use each digit only once. Note that you may not "glue" digits together. (14 - 6) * 3 is not a solution. 13 * 4 * 6 is not a solution either (powers not allowed).
## Wednesday, September 07, 2011
### Feels Like Punctuation is Missing
Two friends mr and mr saw see one day mr see saw sea and mr saw didn't see sea see saw sea and jumped in sea saw didn't see sea but jumped in in sea See saw saw in sea and saw saw see in sea. see saw?
## Tuesday, September 06, 2011
### Eight Lab Rats Work on Their Logic
Eight lab rats (Nolan, Shorty, Spike, Evelyn, Herman, Dottie, Ruth, and George) are dispersed among an arrangement of ten boxes as represented above by the letters A to J. The boxes connect vertically and horizontally, but not diagonally. Eight boxes each have one rat, and two boxes are vacant. From the clues given, determine where each rat is and which boxes are vacant.
1. Herman is in the same horizontal row as either or both vacant boxes.
2. Shorty connects with Ruth.
3. Nolan connects with a vacant box but not with Dottie.
4. Shorty connects with a vacant box.
5. Evelyn is not in B or I but does connect with Spike.
6. Nolan has a corner box.
7. A vacant box connects only with George and Evelyn.
8. Dottie is in G.
Found this one at Crad Kilodney.
## Friday, September 02, 2011
My first is foremost legaly. My second circles outwardly. My third leads all in victory. My fourth twice ends a nominee. What am I?
## Tuesday, August 30, 2011
### Is Never Falls Located in Never Land?
Iron roof, glass walls, burns and burns and never falls.
## Monday, August 29, 2011
### Feet Do Not Touch the Ground
No mouth, no eyes, yet a nose, two arms, two feet, and as it goes, the feet don’t touch the ground, but all the way the head runs round. What is it?
## Thursday, August 25, 2011
### Steamed, Not Shaken
I run, though I have no legs to be seen. I possess no heat, yet I do have steam. I have no voice to let words out, but from far away you can still hear me shout. What am I?
## Wednesday, August 24, 2011
### I Take My Time
I move very slowly at an imperceptible rate, although I take my time, I am never late. I accompany life, and survive past demise, I am viewed with esteem in many women’s eyes. What am I?
## Tuesday, August 23, 2011
### Second is Performed
My second is performed by my first, and, it is thought, a thief by the marks of my whole might be caught. What am I?
## Monday, August 22, 2011
### Window and Lamp
I am a window, I am a lamp, I am clouded, I am shining, and I am coloured; set in white, I fill with water and overflow. I say much, but I have no words. What am I?
## Friday, August 19, 2011
### Always Feeling Old
Always old, sometimes new, never sad, sometimes blue. Never empty, sometimes full, never pushes, always pulls.
## Wednesday, August 17, 2011
### More is Less
The more you have of it, the less you see. What is it?
## Tuesday, August 16, 2011
### Green is the Color of Death
High born, my touch is gentle, purest white is my lace; silence is my kingdom, green is the colour of my death. What am I?
## Monday, August 15, 2011
### Bitterly Pressed
Two brothers we are, great burdens we bear, all day we are bitterly pressed; Yet this I will say – we are full all the day, and empty when we go to rest. What are we?
## Friday, August 12, 2011
### Blue as Blue Can Be
What is always blue?
## Thursday, August 11, 2011
### Not an Egg
What falls but never breaks, what breaks but never falls?
## Tuesday, August 09, 2011
### Cannot Get Enough Number Equations
21 = D on a D 29 = D in F in a L Y 64 = S on a C B 5 = D in a Z C 90 = D in a R A
## Monday, August 08, 2011
### Equations or Songs?
9 N on the T C 21 N on the G S 76 T L the B P 99 B OF B on the W
## Thursday, August 04, 2011
### A Little Bit Less Historic
31 = D in O =
31 = F of BRIC
31 = ICFABR
32 = DF at which WF
32 = T counting WT
33 = CQ for the IFH
36 = I on a YS
36 = IIAY
36 = N on a RW
36 = RM
37 = NBTDC
38 = N on a RW
39 = L in ML
## Tuesday, August 02, 2011
### How High Can You Go?
1805 = SFS
1854 = FNS in the CW
1863 = ALD the GA
1865 = PALA by JWB
1901 = FNPA
1943 = FECB
1945 = B of N and H
1953 = EHCE
1963 = PKA
1969 = NA and EAW on the M
1977 = EP, K of R, D
1989 = TSP in B
## Monday, August 01, 2011
### Letter Equations of 50 or More
50 = C in a HD
50 = R to T
50 = S on the AF
50 = S in the U
50 = W to LYL
50 = Y in a J
52 = C in a D
52 = W in a Y
54 = C in a D with the J
54 = S on a RC
55 = D at P
56 = SOTDOI
57 = HV
## Thursday, July 28, 2011
### Teenagers
Mary, John and Pete have red, brown, and blonde hair, and are 13, 14, and 15 years old . Using the following clues determine the hair color, and age of each child.
1. The youngest has blonde hair.
2. John is older than Pete.
3. John does not have red hair and Pete does not have blonde hair.
## Tuesday, July 26, 2011
### Felis Catus Logic Puzzle
Five cats lived with their owners on Felis Street. The cats, including Ratchett, caused the local fire brigade to be called out five times one night last week. They all did that mad cat thing of becoming stuck up a tree.
Trees: Ash, Beech, Elm, Oak, Willow
Firemen: Ken, Jock, Boris, Dirk, Rick
Cats: Fluffy, Slater, Percy, Tiddles, Ratchett
Which fireman rescued which cat from which tree?
1. Fluffy, the giant ginger tom, became stuck up the willow tree; his rescuer was not Ken.
2. Jock did not rescue a cat from the willow tree, did Boris?
3. Slater, the tortoiseshell moggy, was rescued by Dirk.
4. Rick rescued the cat from the ash tree, this was not the adorable black and white kitten called Percy.
5. Tiddles was not stuck in the ash tree.
6. Ken did not rescue a black and white cat, but he climbed a tree whose first letter was directly after the initial of the name of the fireman that climbed the beech tree to rescue one defenceless little kitty.
## Monday, July 25, 2011
### Single Living Logic
An apartment building contains six apartments with three upstairs (201,202,203) and three downstairs (101,102,103), each with a single occupant.
Three of them are female - Sally, Denise, and Jenny. Three of them are male - Bob, John, and Tom.
Each person owns a vehicle - two U.S. cars - a Ford Escort, and a Chevy Impala, 3 foreign cars - a Zastava Yugo, a Honda Civic, and a Toyota Corrolla, and a bicycle.
With the clues below can you show which person lives in each apartment and what vehicle they own?
Bob lives on the lower floor.
Sally lives between John and Denise.
John owns an American car.
The person in Apartment 203 owns a bicycle.
Jenny lives directly below John.
The Yugo owner lives in Apartment 103.
Bob lives between Tom and Jenny.
Tom owns a foreign built car.
The person in apartment 101 drives a foreign built car.
The Impala owner lives on the lower floor.
The Escort owner lives upstairs.
A female owns the bicycle.
A male owns the Escort.
The person in apartment 101 owns the Toyota.
## Friday, July 22, 2011
### Starts with ne
What starts with "Ne,"
And is one of nine,
And ends with "ne,"
And is eighth in line?
## Thursday, July 21, 2011
### Last Carrol Puzzle
(a) No interesting poems are unpopular among people of real taste.
(b) No modern poetry is free from affectation.
(c) All your poems are on the subject of soap-bubbles.
(d) No affected poetry is popular among people of real taste.
(e) No ancient poem is on the subject of soap-bubbles.
I pulled the three Lewis Carrol puzzles from here.
## Tuesday, July 19, 2011
### More Lewis Carrol Logic
What can you conclude from the following four statements?
(a) None of the unnoticed things, met with at sea, are mermaids.
(b) Things entered in the log, as met with at sea, are sure to be worth remembering.
(c) I have never met with anything worth remembering, when on a voyage.
(d) Things met with at sea, that are noticed, are sure to be recorded in the log.
## Monday, July 18, 2011
### Lewis Carrol, Three Statements
What can you conclude from Lewis Carrol's logic puzzle:
(a) All babies are illogical.
(b) Nobody is despised who can manage a crocodile.
(c) Illogical persons are despised.
1.
123456789
US
2.
t___i___m___e
abde
3.
learning <-----
learning
learning
learning
4.
UWIN+
ULOSE+
1.
hi way
pass
2.
chawhowhorge
3.
camping
night
4.
playshort
5.
1,000,000 air
## Friday, July 01, 2011
### Does this make sense to you?
There is a field with sheep and cows.
Each sheep can see twice as many cows as it can see sheep.
Each cow can see the same number of sheep as it can see cows.
How many cows and how many sheep are there?
## Thursday, June 30, 2011
### Longfellow
Half-way up the hill, I see thee at last, lying beneath me with thy sounds and sights -- A city in the twilight, dim and vast, with smoking roofs, soft bells, and gleaming lights.
## Wednesday, June 29, 2011
### Two Bodies in One
Two bodies have I
Though both joined in one.
The more still I stand,
The quicker I run...
## Tuesday, June 28, 2011
### Take as Much as You Like
I am nothingness, and the more you take from me, the bigger I get.
## Monday, June 27, 2011
### A Counting Problem
How many terms are contained in (a + b + c)20 .
1.
DAYDAYOUT
2.
THOUGHT AN
3.
SEARCH
AND
4.
222222222day
5.
VIT_MIN
## Thursday, June 23, 2011
### 1 END 3 END 5 END 7 END 9 END
Can you tell what each of these five word puzzles mean?
1.
soiudte
2.
S M
E U
O S
G T
T C
A O
H M
W E
3.
ABCDEFGHIJKLMNOPQRSTUVWYZ
QQQQQQQQQQQQQQQQQQQQQQQQQ
4.
oz
e
5.
1 END 3 END 5 END 7 END 9 END
## Wednesday, June 22, 2011
### Tick-Tack-Toe Game
Tick-Tack-Toe
David and Angela play a game of tick-tack-toe. In this game, the players try to get three circles or three crosses in a row (horizontal, vertical, or diagonal).
A player always tries to win: if a player can place his own symbol (X or O) in a row which already contains two of his own symbols, he will do so.
A player always tries to avoid that his opponent wins: if a player can place his own symbol (X or O) in a row which already contains two of the symbols of his opponent, he will do so.
Of course, the first rule has precedence over the second rule, because the game can be won in this way.
In the game shown on the right, 6 moves have been done. David plays with crosses (X) and Angela plays with circles (0). However, we don't know who started the game.
Who will win this game?
If you're like me, you may be having trouble seeing the image of the board I tried to create. So here's a more basic image:
O | O |
----------
O | X |
----------
X | X |
## Tuesday, June 21, 2011
### Moon Cable
A large space agency has decided to build a base on the moon. For this purpose, a cable must be laid around the moon's equator. When the cable is laid, it turns out to be 1 meter short. In a quickly arranged meeting, it is decided to investigate the possibility to lay the whole cable in a groove.
How deep does the groove need to be to make this work?
The agency's director considers digging a groove, no matter how deep, around the entire circumference is too expensive. He suggests to lay the whole cable just a bit north of the equator. How many meters north of the moon's equator should the cable be laid to settle the problem of the lacking 1 meter of cable?
## Monday, June 20, 2011
### Golden Opportunity
It has a golden tail
but it hasn't got a body.
What is it?
## Saturday, June 18, 2011
Hi all,
I never do this, but I'd like to make a personal appeal. My daughter was entered into a contest to win an iPad2. She painted a pottery piece at PYOP for her teacher at school. The lady who runs it liked it enough to enter it into her summer contest to win an iPad2. At six years old, that would be quite a prize.
If you have a facebook account, please head on over to the PYOP album (https://www.facebook.com/media/set/?set=a.10150194490771739.307688.40310146738) and like here photo (the one with the butterfly). In case you're wondering, it's a dunk mug. You can put cookies down at the bottom, for dunking!
Thanks for helping out! We'll be back to our puzzles on Monday. Happy father's day to all the dads out there!
## Friday, June 17, 2011
### Ants on a Board
There are 100 ants on a board that is 1 meter long, each facing either left or right and walking at a pace of 1 meter per minute.
The board is so narrow that the ants cannot pass each other; when two ants walk into each other, they each instantly turn around and continue walking in the opposite direction. When an ant reaches the end of the board, it falls off the edge.
From the moment the ants start walking, what is the longest amount of time that could pass before all the ants have fallen off the plank? You can assume that each ant has infinitely small length.
## Thursday, June 16, 2011
### Reindeer Line
There are nine reindeer, all in a line. Can you figure out the order?
Comet behind Rudolph, Prancer and Cupid. Blitzen behind Cupid and in front of Donder, Vixen and Dancer. Cupid in front of Comet, Blitzen and Vixen. Donder behind Vixen, Dasher and Prancer. Rudolph behind Prancer and in front of Donder, Dancer and Dasher. Vixen in front of Dancer and Comet. Dancer behind Donder, Rudolph and Blitzen. Prancer in front of Cupid, Donder and Blitzen. Dasher behind Prancer and in front of Vixen, Dancer and Blitzen. Donder behind Comet and Cupid. Cupid in front of Rudolph and Dancer. Vixen behind Rudolph, Prancer and Dasher.
This one is from Kevin.
## Wednesday, June 15, 2011
### Generous Buzz
When Buzz gathered his 2001 records for his CPA, he found that he had made six donations, each for a different amount of money and totaling \$1500, to six local youth sports teams, including a softball team. Given the data below, can you find how much in sports deductions Buzz has for 2001: the contribution to each team and the sport the team plays?
1. Buzz donated twice as much to the basketball team as he did to the Knights.
2. The contribution to the Comets was less than that to the football team.
3. For his donation to the soccer team, which was \$50 more than he gave the Rovers, the soccer team advertised Buzz's gas station on the back of their jerseys.
4. The Lions got \$150 more from the service station operator than the baseball team did.
5. The smallest donation of the six was for \$50.
6. The Hawks received twice as much money from Buzz as the soccer team.
7. Before actually seeing the receipts again, Buzz thought that his largest donation had been \$600 and that he had given the hockey team \$250; he found that the largest contribution was for less than \$600 and that the hockey donation was for less than \$250.
8. Buzz gave the Knights \$50 more than he gave the Devils.
9. The largest donation wasn't the one to the youth basketball team.
## Tuesday, June 14, 2011
### Six of One
The expression, "Six of one, half a dozen of another," is commonly used to indicate that two alternatives are essentially equivalent, because six and a half dozen are equal quantities. But are "six dozen dozen dozen" and "a half dozen dozen dozen" equal?
## Monday, June 13, 2011
### There are 5 ships in a port
The greek ship leaves at six and carries coffee.
The ship in the middle has a black chimney
The English ship leaves at nine.
The French ship with a blue chimney is to the left of a ship that carries coffee.
Right to the ship carrying cocoa is a ship going to Marseille.
The Brazilian ship is heading for Manila.
Next to the ship carrying rice is a ship with a green chimney.
A ship going to Genoa leaves at five.
The Spanish ship leaves at seven and is to the right of the ship going to Marseille.
The ship with a red chimney goes to Hamburg.
Next to the ship leaving at seven is a ship with a white chimney.
The ship on the border carries corn.
The ship with a black chimney leaves at eight.
The ship carrying corn is anchored next to the ship carrying rice.
The ship to Hamburg leaves at six.
Which ship is going to Port Said?
Which ship is carrying the tea?
## Thursday, June 09, 2011
### Teacher is so Smart
After studying roman numerals for a short time, the teacher had given the class some free time. Sally's teacher walked up to her desk and saw that she had written:
XI + I = X
on her paper.
He was quick to point out her mistake. Sally protested, saying she had it right. After demonstrating what she meant, the teacher apologized and moved on.
What did Sally point out?
## Wednesday, June 08, 2011
### Helen and Who Are Not Very Close
Yesterday evening, Helen and her husband invited their neighbours (two couples) for a dinner at home. The six of them sat at a round table. Helen tells you the following:
1. Victor sat on the left of the woman who sat on the left of the man who sat on the left of Anna.
2. Esther sat on the left of the man who sat on the left of the woman who sat on the left of the man who sat on the left of the woman who sat on the left of my husband.
3. Jim sat on the left of the woman who sat on the left of Roger.
4. I did not sit beside my husband.
What is the name of Helen's husband?
## Tuesday, June 07, 2011
### Random Walk
A drunk leaves a bar (rather forcefully). After standing up and dusting himself off (rather unsuccessfully), he decides to walk home. Unfortunately for him, he has no idea where home is and has no way of deciding which way to go. He is on a street traveling east-west only, making things simpler for him.
With each step, he's just as likely to turn around and walk back in the direction he came from. After 20 steps, how likely is it he will be back where he started from (right in front of the bar)?
## Monday, June 06, 2011
### Time to Play
It goes up, but at the same time goes down. Up toward the sky, and down toward the ground. It's present tense and past tense too, come for a ride, just me and you.
What is it?
## Friday, June 03, 2011
### Split the Tab
You've been out to a restaurant with a group of people from work. The bill is now on it's way. Time for either a bit of philosophy/game theory. What's the best way to handle the bill? You have three choices: Split the bill evenly, pay for what you ordered, or got to the trouble of asking for separate checks?
For a good write up on the problem, try mindyourdecisions.
## Thursday, June 02, 2011
### Not Exactly Chicken Nuggets
On a nice summer day, two tourists visit the Dutch city of Gouda. During their tour through the center they spot a cosy terrace. They decide to have a drink and, as an appetizer, a portion of hot bitterballs. The waiter tells them that the bitterballs can be served in portions of 6, 9, or 20.
What is the largest number of bitterballs that cannot be ordered in these portions?
## Wednesday, June 01, 2011
### Class is Almost Over
The gentlemen Dutch, English, Painter, and Writer are all teachers at the same secondary school. Each teacher teaches two different subjects. Furthermore:
• Three teachers teach Dutch language
• There is only one math teacher
• There are two teachers for chemistry
• Two teachers, Simon and mister English, teach history
• Peter doesn't teach Dutch language
• Steven is chemistry teacher
• Mister Dutch doesn't teach any course that is taught by Karl or mister Painter.
The Question: What is the full name of each teacher and which two subjects does each one teach?
In case anything has been missed:
The four first names are Karl, Peter, Simon, and Steven
The last four names are Dutch, English, Painter and Writer.
The classes are Chemistry, Dutch, History and Math.
## Tuesday, May 31, 2011
### Usual Suspects
Four suspects; A, B, C and D have been arrested for possibly stealing the car of the chief of police. They were all interviewed personally by the chief, and were hooked up to the lie detector.
Each suspect gave three statements during the examinations, that are listed below:
Suspect A:
In high-school I was in the same class as suspect C.
Suspect B has no driving license.
The thief didn't know that it was the car of the chief of police.
Suspect B:
Suspect C is the guilty one.
Suspect A is not guilty.
I never sat behind the wheel of a car.
Suspect C:
I never met suspect A until today.
Suspect B is innocent.
Suspect D is the guilty one.
Suspect D:
Suspect C is innocent.
I didn't do it.
Suspect A is the guilty one.
Unfortunately, there was a lightning strike and the machine went haywire. After the recovery process, the machine only reported that exactly four of the twelve statements were true, but not which ones. And now all four suspects have gotten lawyers and are refusing to be re-interviewed.
Can you figure out which suspect did it?
## Wednesday, May 25, 2011
There are four cards lying on the table. In order, the cards show: E V 2 7. Each card has a digit on one side and a letter on the other side.
Which cards should you turn around to test the following statement: "when there is a vowel on one side of a card, then there is an even digit on the other side"?
## Tuesday, May 24, 2011
There is only one true answer.
## Monday, May 23, 2011
Which of the following statements are true?
1. Precisely one of these statements is untrue.
2. Precisely two of these statements are untrue.
3. Precisely three of these statements are untrue.
4. Precisely four of these statements are untrue.
5. Precisely five of these statements are untrue.
6. Precisely six of these statements are untrue.
7. Precisely seven of these statements are untrue.
8. Precisely eight of these statements are untrue.
9. Precisely nine of these statements are untrue.
10. Precisely ten of these statements are untrue.
## Thursday, May 19, 2011
### How Good Are You at Adding
ABCABA
BBDCAA
ABEABB
+ ABDBAA
--------------
AAFGBDH
Each letter represents one digit. No two letters represent the same digit.
What's the sum equal to?
## Wednesday, May 18, 2011
### Socks Again
So, here you are, packing your socks up again in the dark. You would think you would have learned your lesson from the whole 'cat incident', but no. You still have 2 red, 4 yellow, 6 purple, 8 brown, 10 white, 12 green, 14 black, 16 blue, 18 grey, and 20 orange socks (see yesterdays post on sock packing). But on your last trip you were embarrassed to find you had ended up packing three pairs of orange socks. You certainly stood out in that crowd. And when your boss asked you on the third day if you even bothered to change your socks... well!
This time, you want to guarantee you have three different pairs, to make it clear, without having to say so, that yes... you do indeed change your socks once a day.
So how many socks do you have to pull out before you know for sure you have three different colored pairs?
## Tuesday, May 17, 2011
### Packing Socks in the Dark
Image via Wikipedia
In your bedroom you have a drawer with 2 red, 4 yellow, 6 purple, 8 brown, 10 white, 12 green, 14 black, 16 blue, 18 grey, and 20 orange socks. The socks are all jumbled together in the drawer. It is dark in your bedroom, your partner is still asleep and so you don't want to turn on the light to see. So you cannot see which sock is which in the dark.
You are packing for a trip that will take you three days, so you need three pairs of socks.
How many socks do you need to take out to be sure that you have at least three pairs of socks of the same color?
## Monday, May 16, 2011
### Lucky Or Not
Image by cliff1066™ via Flickr
You are a card shark, thrown in jail after you were caught dealing from the bottom of a deck. The sheriff, who has a reputation for being honest, has told you the swinging judge is on his way to town for the trial. You plead for him to let you out promising never to return to his territory again.
He says, "Sure, but let's see how lucky you truly are."
He shows you a special deck with 23 cards in it. One side of each cards has an X, while the other side has on O. He throws the cards behind you (you cannot turn around and look, yet). He tells you 14 of them have landed with the X side up (which means the other nine landed with the O up).
He pushes you into the cell and closes the door, leaving you without any light at all. Calling through the door, he sets the challenge. "Put the cards into two stacks, so that both stacks have the same number of X's facing up! I'll be back in one hour."
You can hear him laughing as he walks away.
You spend the next 55 minutes gathering the cards together. One had fallen behind the cot's leg, making it especially hard to find in the dark. You made sure not to flip any of them over so that there are still 14 cards with the X side up, assuming the sheriff wasn't lying to you about that.
You have five minutes left to meet the challenge. What do you do?
## Thursday, May 12, 2011
### No One Volunteered to Put an Apple on Their Head
Image via Wikipedia
William is playing archery blindfolded. The first arrow he shoots, unfortunately misses the bulls eye. The second arrow misses the bulls eye by an even wider margin. William shoots his third and final arrow.
How likely is it William's third shot is also worse than his first shot?
## Wednesday, May 11, 2011
What can bring back the dead,
make us cry,
make us laugh,
make us young,
born in an instant,
## Monday, May 09, 2011
### Boy Wakes Up
A man wakes up in the morning, but it is still dark out. He lights 10 candles throughout his house as he gets ready to go out for the day. A strong wind comes through the front window, blowing out two of the candles. So he shuts the window. He notices a wind blows through the back door, blowing out another candle, so he shuts the door.
He goes out for the day and comes back that evening. How many candles does the man still have?
## Friday, May 06, 2011
### Can You Trust the Delivery Guy
Image via Wikipedia
An undercover cop was shocked by what he had overheard. The gang he had infiltrated was planning on murdering his old partner Jake. He knew the gang had bribed someone else at the police dept, but he didn't know who. He had to get a message out to his partner so they could meet.
Quickly, he wrote out a message and got it delivered. Since he didn't know who to trust at the department, it was in code.
At the dept, Jake opened up the letter and read: "Flame mate weighty soak shave. comedy debut stake scared.", glanced at the clock on the wall and realized he still had time to make the meet.
How did Jake know when and where to meet?
## Thursday, May 05, 2011
### Seen But Not Heard
What can be seen even though you cannot see it?
## Wednesday, May 04, 2011
### Twist and Shout
You can spin, wheel and twist, but this thing can turn without moving.
What is it?
## Monday, May 02, 2011
### Need Want a Some Hint Help
2 9 3 1 8 4 3 6 5 7 ?
What's the next number in this sequence?
## Friday, April 29, 2011
### Why Money is Important
A horse is worth as much as two bulls and one sheep.
A bull is worth as much as two cows.
Two cows are worth as much as five donkeys.
A donkey is worth as much as four sheep.
How many sheep is a horse worth?
## Thursday, April 28, 2011
### Net Figures
The numbers 1 up to and including 8 must be put in the circles of the depicted net. However, numbers in neighbouring circles must differ more than 1. So, for example, circles connected to a circle with a 4 may not contain a 3 or a 5.
Think that one was easy? Try this one.
The numbers 1 up to and including 9 must be put in the circles of the second figure in such a way that the sum of the corners of each of the 7 triangles (4 small ones and 3 large ones) is equal.
To answer in the comments, give the numbers in order for each row of the two figures.
## Wednesday, April 27, 2011
### Make Just One Move
26 - 63 = 1
The above equation is wrong, but can you fix it with just one move?
## Tuesday, April 26, 2011
### Anyone Else Back From Vacation?
Image via Wikipedia
In a contest, four fruits (an apple, a banana, an orange, and a pear) have been placed in four closed boxes (one fruit per box). People may guess which fruit is in which box. 123 people participate in the contest. When the boxes are opened, it turns out that 43 people have guessed none of the fruits correctly, 39 people have guessed one fruit correctly, and 31 people have guessed two fruits correctly
How many people have guessed three fruits correctly, and how many people have guessed four fruits correctly?
## Monday, April 25, 2011
### In My Opinion, This is Hard to Do
This is a famous problem from 1882, to which a prize of \$1000 was awarded for the best solution. The task is to arrange the seven numbers 4, 5, 6, 7, 8, 9, and 0, and eight dots in such a way that an addition approximates the number 82 as close as possible. Each of the numbers can be used only once. The dots can be used in two ways: as decimal point and as symbol for a recurring decimal. For example, the fraction 1/3 can be written as
.
. 3
The dot on top of the three denotes that this number is repeated infinitely. If a group of numbers needs to be repeated, two dots are used: one to denote the beginning of the recurring part and one to denote the end of it. For example, the fraction 1/7 can be written as
. .
. 1 4 2 8 5 7
Note that '0.5' is written as '.5'.
How close can you get to the number 82?
## Thursday, April 14, 2011
### The Tax Man Cometh
(1) Everyone is afraid of the tax man.
(2) The tax man is afraid of only me.
Who am I?
## Wednesday, April 13, 2011
### Twice the Fun at Half the Price
Someone shows you two boxes and he tells you that one of these boxes contains two times as much as the other one, but he does not tell you which one this is. He lets you choose one of these boxes, and opens it. It turns out to be filled with \$10. Now he gives you the opportunity to choose the other box instead of the current one (and give back the \$10 from the first box), because the second box could contain twice as much (i.e. \$20).
The Question: Should you choose the second box, or should you stick to your first choice to maximize the expected amount of money?
## Tuesday, April 12, 2011
### Letter Blocks
Molly has a set of four alphabet blocks. Each side of these blocks is printed with a different letter, making 24 in total. Molly notices that by rearranging the blocks, she can spell each of the following words:
BOXY, BUCK, CHAW, DIGS, EXAM, FLIT,
GIRL, JUMP, OGRE, OKAY, PAWN, ZEST
What letters are on each block?
## Monday, April 11, 2011
### Riddle of Bilbo
An eye in a blue face
Saw an eye in a green face.
"That eye is like to this eye"
Said the first eye,
"But in low place
Not in high place."
## Wednesday, April 06, 2011
### So Here On Every Stop
Summer days make me want to play.
Hello, weather! What a beautiful day!
On the ground I see my bare feet.
Everyone knows it's time for a treat!
So, what is the answer to this riddle?
## Tuesday, April 05, 2011
### Riddle Behind You
The more you take, the more you leave behind.
What is it?
## Friday, April 01, 2011
### Not Trying to Fool You, But Who Turned on the Snow?
I am used to bat with, yet I never get a hit.
I am near a ball, yet it is never thrown.
What am I?
## Thursday, March 31, 2011
### Feasting Bright Upon the Eyes
I know a thousand faces, and count the tailed heads, feasting bright upon the eyes, of many who have died. Wielding well a mighty power, who hath but humble stature.
What am I?
## Tuesday, March 29, 2011
### Riddle Without Weapons
I defend without weapons, stand without legs, wound without force, and am harder to fight than to kill. What am I?
## Monday, March 28, 2011
Image via Wikipedia
Mary and Linda were two of the first five people who came into Jake's Hardware yesterday. Each of the five, including Jones and Silkwood, wanted to have a key duplicated. From this information and the clues, can you determine the full names of Jake's first five customers, the order in which they visited the store, and what type of key each had copied?
1. None of the five has the same first and last initial.
2. The first customer duplicated a key to her boat and the fifth duplicated her car key.
3. Sam was Jake's customer before John, but not just before.
4. One customer, not John, duplicated a house key.
5. Smith is not the one who got a key to a garage duplicated.
6. Susan got a copy of her office key made.
7. Mills got a key after the one who got a garage key made.
8. Larkin isn't the one who got an office key.
In no particular order:
Order: 1, 2, 3, 4, 5
First Name: Sam, John, Susan, Mary, Linda
Last Name: Jones, Silkwood, Smith, Mills, Larkin
Type of Key: Boat, Car, House, Garage, Office
## Friday, March 25, 2011
### A Plumber, a Carpenter and an Electrician Walk Into a Bar
A plumber, a carpenter, and an electrician applied for jobs with a famous construction firm. From the information given below can you identify each worker's full name, job, and age?
1.
Surnames: Tim, John, Bjorn.
Ages: 25, 30, 35 (but not respectively).
2. Tim, the electrician, is older than Federer.
4. The plumber is 30 years old.
5. The youngest worker of the three is not called Bjorn.
## Wednesday, March 23, 2011
The king explained to the prisoner that in Room 1, if a lady is in it, then the sign on the door is true. But if a tiger is in it, the sign is false. In the Room II, the situation is the opposite: a lady in the room means the sign on the door is false, and a tiger in the room means the sign is true. It is possible that both rooms contain ladies or both rooms contain tigers, or that one room contains a lady and the other a tiger.
Room 1 Sign: Both rooms contain ladies
Room 2 Sign: Both rooms contain ladies
Pointing to the signs, the king asked the prisoner which room did he wish to choose?
This puzzle comes from the Lady or the Tiger
## Tuesday, March 22, 2011
### Dancing Partners
Image via Wikipedia
At the annual dancer's ball a number of very experienced dancers performed their favourite dance with their favourite partners. Alan danced the tango, whilst Becky watched the waltz. James and Charlotte were fantastic together. Keith was magnificent during his foxtrot and Simon excelled at the rumba. Jessica danced with Alan, but Laura did not dance with Simon.
Can you determine who danced with whom and which dance they each enjoyed?
## Monday, March 21, 2011
### Which Spy Did the Deed
- First Name - Surname - Code Name
1. Nick - Hing - Hadenov
2. Choo - Anfit - Fingsarov
3. Hans - Errs - Nodapov
4. Baz - Teale - Tikelikov
At the Spy-of-the-Year competition, the results had just been announced. The problem was, someone had tampered with the paper they were written on, altering the order, so that although each item was in the correct column, only one item in each column was correctly positioned. The judges could only remember the following facts about the correct order.
1) Hing was not second
2) Nodapov was one place above Teale
3) Neither Fingsarov nor Nodapov was first
4) Hing was one place below Nick
5) Neither Hans nor Nick were third.
## Friday, March 18, 2011
### I Can See Clearly
What builds up castles, tears down mountains, makes some blind, helps others to see?
## Wednesday, March 16, 2011
### Emeralds and Diamonds
I am emeralds and diamonds, lost by the moon. I am found by the sun and picked up soon. What am I?
## Tuesday, March 15, 2011
### One Two Four
Some have one. Some have two. Some have four. None have three.
What are they?
EDIT: Sorry for those who tried to answer this as it was originally written. I've fixed it so that it says none have three.
## Monday, March 14, 2011
### A Little Scrabble Help, Please
What 5 letter word can be rearranged 3 different times to get 3 different words each containing 1 more syllable than the last?
Hint: The word has no duplicates of letters.
## Friday, March 11, 2011
A woman owns a pie shop and the first day she had 13 customers, the second day she had 14 customers, the third 95, and the fourth 62. Following the sequence, how many customers will she have tomorrow?
## Wednesday, March 09, 2011
### How to Jog
Jenn picked a book off the highest shelf in her room. On the spine she read "How to Jog". She ran out of the room and opened the book but found it had absolutely nothing to do with jogging.
## Tuesday, March 08, 2011
### Liar!
Image via Wikipedia
A boy and a girl are talking.
"I am a boy" - said the child with black hair.
"I am a girl" - said the child with white hair.
At least one of them lied. Who is the boy and who is the girl?
## Monday, March 07, 2011
### Five Gears
Image via Wikipedia
There are five gears connected in a row, the first one is connected to the second one, the second one is connected to the third one, and so on.
How much faster would the last gear be if the second gear was twice the size of the first gear, and all the other gears were the same size as the first gear?
## Thursday, March 03, 2011
### Words in Common
What do these words have in common: age, blame, curb, dance, evidence, fence, gleam, harm, interest, jam, kiss, latch, motion, nest, order, part, quiz, rest, signal, trust, use, view, win, x-ray, yield, zone?
## Wednesday, March 02, 2011
### Hiking Through the Forest
I got it in a forest but didn't want it. Once I had it, I couldn't see it. The more I searched for it, the less I liked it. I took it home in my hand because I could not find it. What was it?
1.
Rest
2.
Passed
Again
1.
otwone
2.
r
g rose e i
n
3.
e
k
a
m
4.
podpodpod
5.
take pets
## Wednesday, February 23, 2011
### Two Riddles to Enjoy
1. I have 24 keys, but they can't open any locks.
2. Many people own a copy of me. Without me the world would fall. Who am I?
## Tuesday, February 22, 2011
### Does Anyone Buy Analog Clocks Anymore?
If we follow the hour and minute hands of a clock for 24 hours how many times will they form a right angle?
1.
N N N N N N N
A A A A A A A
C C C C C C C
2.
AALLLL
3.
i4i
4.
Math The
## Thursday, February 17, 2011
### Ring Around the Marbles
A boy has four red marbles and eight blue marbles. He arranges his twelve marbles randomly, in a ring. What is the probability that no two red marbles are adjacent?
## Tuesday, February 15, 2011
### Free Tickets
At a movie theater, the manager announces that a free ticket will be given to the first person in line whose birthday is the same as someone in line who has already bought a ticket. You have the option of getting in line at any time. Assuming that you don't know anyone else's birthday, and that birthdays are uniformly distributed throughout a 365 day year, what position in line gives you the best chance of being the first duplicate birthday?
## Monday, February 14, 2011
### DDDWESTDDD
The rebus are back again. If you look carefully, you can see the words or common phrases hidden in the arrangement of letters and words below. For example, the title represents West Indies (West in D's).
1.
lem
2.
job I'm job
3.
Eye e
See Except
4.
dox
dox
## Friday, February 11, 2011
### Biggest Loser
In the most recent Biggest Loser competition, the contestants had to arrange the order of five dishes from least calories to most. If they did it blind-folded, how many different ways could the food be arranged?
But let's say they have some knowledge about a couple of the dishes. They know the plate of hamburger (loaded with everything) and chips should be near the top, while the plate of broccoli should be near the bottom of the scale. How does that affect the number of choices?
1.
belt
hitting
2.
3.
settle
4.
----------
________
## Wednesday, February 09, 2011
### Some Title is Better Than No Title
Rebus are phrases hidden inside word pictures. Can you solve them all?
1.
heasicknessandlth
2.
on
it all
3.
stafornce
4.
way
wrong way
## Thursday, February 03, 2011
### I Am So Tired of Getting Snowed In
5 children in this puzzle had a bowl of cereal for breakfast with one drink and another bowl of cereal (different than breakfast) as an after-school snack.
The children were:
Dean, Emma, Julia, Patrick, Thomas
The cereals were:
Brownies, Cococrunch, Crumbler, Krispo & Wheeties
The drinks were:
Coffee, Milk, Orange Juice(OJ), Tea & Water.
Hints:
1.The child who ate Wheeties for breakfast had a bowl of Krispo after school.
2.Emma wasn’t the child who drank coffee in the morning & ate a bowl of Brownies in the afternoon
3.The child who drank a glass of water with his/her Brownies at breakfast time is either Patrick of Thomas
4.Emma didn’t eat Cococrunch or Crumbler for breakfast
5.The child who had a glass of milk ate Cococrunch for breakfast, but not Crumbler as an after-school snack
6.Patrick drank either milk or tea for breakfast.
7.The child who drank coffee (decaf!) didn’t eat a bowl of Crumbler for breakfast.
8.Dean (who never drinks coffee) didn’t have a bowl of Crumbler for his breakfast.
## Tuesday, February 01, 2011
### Feeding Time
Joe Stevens was in charge of feeding all of the animals in the morning. He had a regular schedule that he followed every day.
• The giraffes were fed before the zebras but after the monkeys.
• The bears were fed 15 minutes after the monkeys.
• The lions were fed after the zebras.
• Joe always began feeding the animals at 6:30 AM.
• It takes 15 minutes for him to finish feeding and move on to the next cage.
• The five types of animals are bears, giraffes, lions, monkeys, and zebras.
## Monday, January 31, 2011
### Digging Away
It takes one day for ten workers to dig ten holes.
It takes two days for five workers to dig ten holes.
How much does it take for one man to dig HALF a hole?
## Wednesday, January 26, 2011
### Another Carroll Riddle
'First, the fish must be caught.'
That is easy: a baby, I think, could have
caught it.
'Next, the fish must be bought.' That is easy: a penny, I think, would have
bought it.
'Now cook me the fish!'
That is easy, and will not take more than a minute.
'Let it lie in a dish!'
That is easy, because it already is in it.
'Bring it here! Let me sup!'
It is easy to set such a dish on the table.
'Take the dish-cover up!'
Ah, that is so hard that I fear I'm unable!
For it holds like glue-
Holds the lid to the dish, while it lies in the
middle:
Which is easiest to do,
Un-dish-cover the fish, or dishcover the
riddle?
## Tuesday, January 25, 2011
### Ravens and Desks
Why is a raven like a writing desk?
- Lewis Carroll
## Monday, January 24, 2011
### Lewis Carroll Box
John gave his brother James a box:
About it there were many locks.
James woke and said it gave him pain;
So gave it back to John again.
The box was not with lid supplied
Yet caused two lids to open wide:
And all these locks had never a key
What kind of box, then, could it be?
## Friday, January 21, 2011
### Back in School Counting
Image via Wikipedia
Of 28 students taking at least one subject, the number taking Math and English but not History equals the number taking Math but not History or English. No student takes English only or History only, and six students take Math and History but not English. The number taking English and History but not Math is 5 times the number taking all three subjects.
If the number taking all three subjects is even and non-zero, how many are taking English and Math but not History?
## Thursday, January 20, 2011
### Can You Solve the Pattern
748
397
6510
?84
Can you figure out the pattern? What is the missing number?
## Tuesday, January 18, 2011
### River Crossing
I have posted more complicated crossing problems before, but with another six inches of snow to shovel outside, this seems good enough to me.
A dad and two sons needed to cross the river. They have a canoe and paddles, but no other equipment. The canoe has a weight limit of 150 lbs. Luckily, dad weighs 150 lbs. The two sons weigh 75 lbs each. How can the all get across the river using the canoe?
## Thursday, January 13, 2011
### Snowed In
Image by Getty Images via @daylife
It snowed last night! Four couples woke up to a winter wonderland this morning. Due to the high winds during the storm, the snow drifted as it fell. The plows were out early but still, it took awhile to get all the streets in town cleared. Determine the full name of each couple, how much snow they got form the storm (between 6 and 10 inches), and what time the road they lived on was plowed out (between 8:30 am and 10:00 am).
1. Rebecca, whose last name isn’t Heart, was plowed out before Greg but after the couple who got eight inches of snow.
2. Mr. and Mrs. Shephard have two inches more snow than the couple who were plowed out at 8:30 am.
3. Karl and Debbie weren’t plowed out at 9:00 am though they were plowed out before Peter.
4. Stan Grant, who isn’t married to Martha, got one inch more snow than Greg did.
5. The couples were plowed out, from earliest to latest, in the following order: the couple who got eight inches of snow, Alice, Peter, Mr. and Mrs. White.
## Tuesday, January 11, 2011
### (ice)^3
Rebus are puzzles you need to solve by looking at the words, their relationships to each others and trying to imagine how it represents a common phrase.
For instance, BIRD can mean Big bird or pa per can mean paper cut.
1.
dinner dinner
table
2.
wosteplf
3.
ware
mholey
4.
(ice)^3
## Friday, January 07, 2011
### I'M you
1.
Dribble Dribble
2.
long
do
3.
Math The
4.
my own heart a person
5.
Must get here
Must get here
Must get here
1.
search
and
2.
injury + insult
3.
welieight
4.
belt
hitting
5.
EZ
iii
## Wednesday, January 05, 2011
Image via Wikipedia
During the recent Fast Track Cup, the first four horses did well to finish ahead of the field. After Dinner was not last and the horse wearing blue was not third. Bitter Twist was before the horse in green. Dare Devil came first. Bitter Twist finished before After Dinner. Catch Me wore red but Bitter Twist did not wear yellow.
Can you determine where each horse finished and the colours they wore?
## Tuesday, January 04, 2011
### Corn Can Be Fractionalized
Image by kairoinfo4u via Flickr
Ahmes's Papyrus
About 1650 B. C., Egyptian scribe Ahmes, made a transcript of even more ancient mathematical scriptures dating to the reign of the Pharaoh Amenemhat III. In 1858 Scottish antiquarian, Henry Rhind came into possession of Ahmes's papyrus. The papyrus is a scroll 33 cm wide and about 5.25 m long filled with funny math riddles. One of the problems is as follows:
100 measures of corn must be divided among 5 workers, so that the second worker gets as many measures more than the first worker, as the third gets more than the second, as the fourth gets more than the third, and as the fifth gets more than the fourth. The first two workers shall get seven times less measures of corn than the three others.
How many measures of corn shall each worker get? (You can have fractional measures of corn.)
## Monday, January 03, 2011
### May Not See it For a While
I am around long before dawn.
But by lunch I am usually gone.
You can see me summer, fall, and spring.
I like to get on everything.
But when winter winds start to blow;
Burr, then it's time for me to go!
What am I?
Oh, and Happy New Year everyone!
Enter your Email and join hundreds of others who get their Question of the Day sent right to their mailbox | 17,037 | 65,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-33 | latest | en | 0.929621 |
http://resources.schoolscience.co.uk/cda/14-16/physics/copch3pg1.html | 1,571,477,244,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986692723.54/warc/CC-MAIN-20191019090937-20191019114437-00063.warc.gz | 161,000,660 | 5,455 | page 2
What makes it turn?
You can take a motor apart to see what it’s made of.
What's inside an electric motor?
Coil The coil is made of copper wire - because it is such an excellent conductor (see conducting properties). It is wound onto an armature. The coil becomes an electromagnet when a current flows through it. Armature The armature supports the coil and can help make the electromagnet stronger. This makes the motor more efficient. Permanent magnets There are two permanent magnets. They produce a steady magentic field so that the coil will turn when a current flows in it. Some motors have electromagnets instead of permanent magnets. These are made from more coils of copper wire. Commutator Each end of the coil is connected to one of the two halves of the commutator. The commutator swaps the contacts over every half turn. Brushes The brushes press on the commutator. They keep contact with the commutator even though it is spinning round. The current flows in and out of the motor through the brushes. Steel former The former made of magnetic material links the two permanent magnets and, in effect, makes them into a single horseshoe shaped magnet. Commercial motors often use a horseshoe magnet.
Picture 2. A simple electric motor. Use the highlighted words in the text to find out how it works.
How does it work? The motor is connected to a battery. When the switch is closed, the current starts to flow and the coil becomes an electromagnet. In this case the current is flowing anticlockwise in the top of the coil. This makes the top a north pole. This north pole is attracted to the south pole on the left. So the top of the coil turns towards the left. Notice that the bottom of the coil is a south pole and is attracted to the magnet on the right. Once the coil gets to the upright position, there is no turning force on it because the electromagnet of the coil is lined up with the permanent magnets. If the current in the coil were constant, the coil would stop in this position. However, to keep it spinning, the commutator breaks contact in this position. So the current stops for an instant. The momentum of the coil keeps it going and the contacts are reconnected. However, they are now the other way around. So, the side of the coil that used to be a south pole is now a north pole. The commutator will keep swapping the contacts every half turn (when the coil is in the upright position). In this way, the motor keeps spinning.
Question 1
Look at the descriptions below. Each one describes a part of an electric motor. In each case, type in the name of the part. Choose from the list at the bottom.
Use only small letters and make sure you spell the names correctly.
Provides a steady magnetic field. Automatically reverses the current in the coil. Becomes an electromagnet when a current flows. Feeds the current into or out of the motor. Supports the coil and can strengthen the magnetic field. commutator, brushes, coil, armature, magnets. | 646 | 2,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-43 | latest | en | 0.951646 |
http://www.clawpack.org/gallery/pyclaw/gallery/quadrants.html | 1,548,136,379,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583829665.84/warc/CC-MAIN-20190122054634-20190122080634-00432.warc.gz | 294,037,683 | 4,976 | # 2-dimensional Euler equations¶
Simple example solving the Euler equations of compressible fluid dynamics:
$\begin{split}\rho_t + (\rho u)_x + (\rho v)_y & = 0 \\ (\rho u)_t + (\rho u^2 + p)_x + (\rho uv)_y & = 0 \\ (\rho v)_t + (\rho uv)_x + (\rho v^2 + p)_y & = 0 \\ E_t + (u (E + p) )_x + (v (E + p))_y & = 0.\end{split}$
Here $$\rho$$ is the density, (u,v) is the velocity, and E is the total energy. The initial condition is one of the 2D Riemann problems from the paper of Liska and Wendroff.
## Source:¶
#!/usr/bin/env python
# encoding: utf-8
r"""
==========================
Simple example solving the Euler equations of compressible fluid dynamics:
.. math::
\rho_t + (\rho u)_x + (\rho v)_y & = 0 \\
(\rho u)_t + (\rho u^2 + p)_x + (\rho uv)_y & = 0 \\
(\rho v)_t + (\rho uv)_x + (\rho v^2 + p)_y & = 0 \\
E_t + (u (E + p) )_x + (v (E + p))_y & = 0.
Here :math:\rho is the density, (u,v) is the velocity, and E is the total energy.
The initial condition is one of the 2D Riemann problems from the paper of
Liska and Wendroff.
"""
from __future__ import absolute_import
from clawpack import riemann
from clawpack.riemann.euler_4wave_2D_constants import density, x_momentum, y_momentum, \
energy, num_eqn
from clawpack.visclaw import colormaps
def setplot(plotdata):
r"""Plotting settings
Should plot two figures both of density.
"""
plotdata.clearfigures() # clear any old figures,axes,items data
# Figure for density - pcolor
plotfigure = plotdata.new_plotfigure(name='Density', figno=0)
# Set up for axes in this figure:
plotaxes = plotfigure.new_plotaxes()
plotaxes.xlimits = 'auto'
plotaxes.ylimits = 'auto'
plotaxes.scaled = True
plotaxes.title = 'Density'
# Set up for item on these axes:
plotitem = plotaxes.new_plotitem(plot_type='2d_pcolor')
plotitem.plot_var = density
plotitem.pcolor_cmap = colormaps.yellow_red_blue
plotitem.pcolor_cmin = 0.
plotitem.pcolor_cmax = 2.
# Figure for density - Schlieren
plotfigure = plotdata.new_plotfigure(name='Schlieren', figno=1)
# Set up for axes in this figure:
plotaxes = plotfigure.new_plotaxes()
plotaxes.xlimits = 'auto'
plotaxes.ylimits = 'auto'
plotaxes.title = 'Density'
plotaxes.scaled = True # so aspect ratio is 1
# Set up for item on these axes:
plotitem = plotaxes.new_plotitem(plot_type='2d_schlieren')
plotitem.schlieren_cmin = 0.0
plotitem.schlieren_cmax = 1.0
plotitem.plot_var = density
return plotdata
def setup(use_petsc=False):
if use_petsc:
import clawpack.petclaw as pyclaw
else:
from clawpack import pyclaw
solver = pyclaw.ClawSolver2D(riemann.euler_4wave_2D)
solver.all_bcs = pyclaw.BC.extrap
domain = pyclaw.Domain([0.,0.],[1.,1.],[100,100])
solution = pyclaw.Solution(num_eqn,domain)
gamma = 1.4
solution.problem_data['gamma'] = gamma
solver.dimensional_split = False
solver.transverse_waves = 2
# Set initial data
xx, yy = domain.grid.p_centers
l = xx < 0.8
r = xx >= 0.8
b = yy < 0.8
t = yy >= 0.8
solution.q[density,...] = 1.5 * r * t + 0.532258064516129 * l * t \
+ 0.137992831541219 * l * b \
+ 0.532258064516129 * r * b
u = 0.0 * r * t + 1.206045378311055 * l * t \
+ 1.206045378311055 * l * b \
+ 0.0 * r * b
v = 0.0 * r * t + 0.0 * l * t \
+ 1.206045378311055 * l * b \
+ 1.206045378311055 * r * b
p = 1.5 * r * t + 0.3 * l * t + 0.029032258064516 * l * b + 0.3 * r * b
solution.q[x_momentum,...] = solution.q[density, ...] * u
solution.q[y_momentum,...] = solution.q[density, ...] * v
solution.q[energy,...] = 0.5 * solution.q[density,...]*(u**2 + v**2) + p / (gamma - 1.0)
claw = pyclaw.Controller()
claw.tfinal = 0.8
claw.solution = solution
claw.solver = solver
claw.output_format = 'ascii'
claw.outdir = "./_output"
claw.setplot = setplot
return claw
if __name__ == "__main__":
from clawpack.pyclaw.util import run_app_from_main
output = run_app_from_main(setup, setplot) | 1,309 | 3,960 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2019-04 | latest | en | 0.516557 |
http://www.ifa.hawaii.edu/~barnes/ast626_95/tm.html | 1,516,116,531,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886437.0/warc/CC-MAIN-20180116144951-20180116164951-00698.warc.gz | 469,878,493 | 5,145 | # Triaxial Models
## Astronomy 626: Spring 1995
The general problem of modeling triaxial galaxies is illustrated with a couple of special cases. In separable models the allowed orbits are fairly simple, and the job of populating orbits so as to produce the density distribution is well-understood. In scale-free models the allowed orbits are more complex, and it is not clear if such models can be in true equilibrium. If not, secular evolution over many tens of dynamical times may play a role in shaping elliptical galaxies.
## Schwarzschild's Method
Schwarzschild (1979) invented a powerful method for constructing equilibrium models of galaxies without explicit knowledge of the integrals of motion. To use this method,
1. Specify the mass model rho(x) and find the corresponding potential;
2. Construct a grid of K cells in position space;
3. Chose initial conditions for a set of N orbits, and for each one,
• integrate the equations of motion for many orbital periods, and
• keep track of how much time the orbit spends in each cell, which is a measure of how much mass the orbit contributes to that cell;
4. Determine non-negative weights for each orbit such that the summed mass in each cell is equal to the mass implied by the original rho(x).
The last step is the most difficult. Formally, let P_i(c) be the mass contributed to cell c by orbit i; the task is then to find N non-negative quantities Q_i such that
``` -- N
\
(1) M(c) = | Q_i P_i(c)
/
-- i = 1
```
simultaneously for all cells, where M(c) is the integral of rho(x) over cell c. This may be accomplished by taking N > K, so as to obtain a reasonably rich set of `basis functions', and using any one of a number of numerical techniques, including linear programming (Schwarzschild 1979), non-negative least squares (Pfenniger 1984), Lucy's method (Newton & Binney 1984), or maximum entropy (Richstone & Tremaine 1988).
In general Eq. 1 has many solutions, reflecting the fact that many different distribution functions are consistent with a given mass model. Some methods allow one to specify additional constraints so as to select solutions with special properties (maximum rotation, radial anisotropy, etc.).
## Separable Potentials
Very roughly speaking, a separable potential is one in which the orbit of a star may be decomposed into three separate motions (or two separate motions, in two dimensions). Motion in separable potentials is tractable analytically, and exact expressions for the integrals of motion are known. Separable potentials are rather special, mathematically speaking; it is highly unlikely that real galaxies have such potentials. However, numerical experiments indicate that most triaxial density distributions with cores generate potentials which share certain key features with separable potentials.
### Orbits
The orbits in a separable potential may be classified into distinct families. Each family is associated with a set of closed and stable orbits. In two dimensions, for example, there are two types of closed, stable orbits; one type (i) oscillates back and forth along the major axis, and the other type (ii) loops around the center. Because these orbits are stable, other orbits which start nearby will remain nearby at later times. The families associated with types (i) and (ii) are known as box and loop orbits, respectively (see BT87, Chapter 3.3.1).
In three dimensions, a separable potential permits four distinct orbit families:
• box orbits,
• short-axis tube orbits,
• inner long-axis tube orbits, and
• outer long-axis tube orbits.
The short-axis tubes are orbits which loop around the short (minor) axis, while long-axis tubes loop around the long (major) axis. The two families of long-axis tube orbits arise from different closed stable orbits and explore different regions of space (see BT87, Figure 3-20). There are no `intermediate-axis tube' orbits since closed orbits looping around the intermediate axis are unstable. In general, triaxial potentials with cores have orbit families much like those in separable potentials.
The time-averaged angular momentum of a star on a box orbit vanishes; such orbits therefore do not contribute to the net rotation of the system. Short-axis and long-axis tube orbits, on the other hand, preserve a definite sense of rotation about their respective axes; consequently, their time-averaged angular momenta do not vanish. The total angular momentum vector of a non-rotating triaxial galaxy may lie anywhere in the plane containing the short and long axes (Levison 1987).
### Models
Using Schwarzschild's method, it is possible to numerically determine distribution functions f(E,I_2,I_3) corresponding to separable triaxial models (Statler 1987). A somewhat more restricted set of models can be constructed exactly; these models make use of all available box orbits, but only those tube orbits with zero radial thickness (Hunter & de Zeeuw 1992). Apart from the choice of streaming motion, thin tube models are unique. One use of such models is to illustrate the effects of streaming motion by giving all tube orbits the same sense of rotation; the predicted velocity fields display a wide range of possibilities (Arnold et al. 1994). Non-zero streaming on the projected minor axis is a generic feature of such models; a number of real galaxies exhibit such motions and thus must be triaxial.
## Scale-Free Potentials
In scale-free models all properties have either power-law or logarithmic dependence on radius. In particular, scale-free models with density profiles scaling as r^-2 have logarithmic potentials and flat rotation curves. While real galaxies are not entirely scale-free, power-law density distributions are reasonable approximations to the central regions of some elliptical galaxies and to the halos of galaxies in general.
### Orbits
If the density falls as r^-2 or faster, then all box orbits are replaced by minor orbital families called boxlets (Gerhard & Binney 1985, Miralda-Escude & Schwarzschild 1989). Each boxlet family is associated with a closed and stable orbit arising from a resonance between the motions in the x and y directions.
Moreover, some scale-free potentials have irregular orbits; these have no integrals of motion apart from the energy E. In principle, such an orbit can wander everywhere on the phase-space hypersurface of constant E, but in actuality such orbits show a complicated and often fractal-like structure.
### Models
The appearance of boxlets instead of boxes poses a problem for model builders because boxlets are `centrophobic' (meaning that they avoid the center) and consequently do not provide the elongated density distributions of the box orbits they replace. As a result, the very existence of scale-free triaxial systems is open to doubt (de Zeeuw 1995).
The scale-free elliptic disk is a relatively simple two-dimensional analog of a scale-free triaxial system. Because the model is scale-free, the radial dimension can be folded out when applying Schwarzschild's method; thus the calculations are fast (Kuijken 1993). The result is that self-consistent models can be built using the available boxlets, loops, and irregular orbits, but the minimum possible axial ratio b/a increases as the numerical resolution of the calculation is improved.
Similar results hold for scale-free models in three dimensions. Models have been constructed for triaxial logarithmic potentials with a range of axial ratios b/a and c/a (Schwarzschild 1993). Tubes and regular boxlets provide sufficient variety to produce models with c/a > 0.5, but not flatter. However, over intervals of ~50 dynamical times, irregular orbits behave like `fuzzy regular orbits', and by including them it becomes possible to build near-equilibrium models as flat as c/a = 0.3.
## Rotation, Chaos, & Secular Evolution
Figure rotation adds a new level of complexity to the orbit structure of triaxial systems. A few models have been constructed using Schwarzschild's method, but little is known about the existence and stability of such systems. N-body experiments indicate that at least some such systems are viable models of elliptical galaxies. Rotation tends to steer orbits away from the center and so may lessen the effects of central density cusps.
Realistic potentials are likely to have some irregular or chaotic orbits, and there is no reason to think that such orbits are systematically avoided by processes of galaxy formation. Over ~10^2 or more dynamical times, such orbits tend to produce nearly round density distributions.
Consequently, it is likely that secular evolution over timescales of ~10^2 dynamical times may be changing the structures of elliptical galaxies (Binney 1982, Gerhard 1986). The outer regions are not likely to be affected since dynamical times are long at large radii, but significant changes may occur in the central cusps where dynamical times are ~10^6 years (de Zeeuw 1995).
## References
• Arnold, R.A., de Zeeuw, P.T., & Hunter, C. 1994, M.N.R.A.S., submitted.
• Binney, J.J. 1982, M.N.R.A.S. 201, 15.
• de Zeeuw, P.T. 1995, in The Formation and Evolution of Galaxies, ed. C. Munoz-Tunon & F. Sanchez, p. 231.
• Gerhard, O.E. 1986, M.N.R.A.S. 219, 373.
• Gerhard, O.E. & Binney, J.J. 1985, M.N.R.A.S. 216, 467.
• Hunter, C. & de Zeeuw, P.T. 1992, Ap.J 398, 79.
• Kuijken, K. 1993, Ap.J. 409, 68.
• Levison, H.F. 1987, Ap.J. 320, L93.
• Miralda-Escude, J. & Schwarzschild, M. 1989, Ap.J. 339, 752.
• Newton, A. & Binney, J.J. 1984, M.N.R.A.S. 210, 711.
• Pfenniger, D. 1984, Astr.Ap. 141, 171.
• Richstone, D.O. & Tremaine, S.D. 1988, Ap.J. 327, 82.
• Schwarzschild, M. 1979, Ap.J. 232, 236.
• Schwarzschild, M. 1993, Ap.J. 409, 563.
• Statler, T.S. 1987, Ap.J. 321, 113.
Joshua E. Barnes (barnes@zeno.ifa.hawaii.edu) | 2,309 | 9,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-05 | latest | en | 0.887314 |
https://worldbuilding.stackexchange.com/questions/136518/how-to-make-variable-super-soldier-size-the-norm-versus-a-single-size | 1,718,325,636,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861517.98/warc/CC-MAIN-20240613221354-20240614011354-00116.warc.gz | 568,606,465 | 47,489 | # How to make variable super-soldier size the norm versus a single size
Imagine you can build your soldiers in a lab somewhere. You can control everything, and you've found a the ideal humanoid form (ignoring potentially better forms here). You could build them in all sizes big and small, but this would also mean that production of weapons and equipment would need to account for various sizes and be more intensive, so using a single size for your super-soldiers would be more ideal.
You could have a few set sizes. Pilots of vehicles and aircraft would likely benefit from being small in size as it would mean they need less space inside the vehicle, but then you would simply have two set standard sizes: "normal" soldiers and "pilot" soldiers that are smaller but all still the same size within their category. For clarity let's say all normal soldiers are 1.8m in size and all pilots 1m in size without any further variation. You can possibly imagine a few more size variants like a heavy duty variant that is larger but that too would be set at a specific height without further variation.
What I want is to have these soldiers greatly differ in size, fir example from 1.5 to 2.5m, but I can't find a good reason. Especially for the kind of large-scale warfare I'm imagining the industrial advantage of building them all one size would be so immense it can hardly be ignored. It would likely need to be in part an evolutionary reason that allows a variable size army to succeed more than one with set sizes.
These soldiers rely for the most part on equipment, so building it into their bodies matching their size requirements is not going to be an option.
• Do you want, say, all "standard soldiers" to have different sizes? And also, how much variation do you expect - if a standard soldier is 1.8m, then can they go to 1m minimum or is it, say, 1.7m- 1.9m?
– VLAZ
Commented Jan 14, 2019 at 12:11
• @vlaz I want all standard soldiers to have different sizes yes. From for example 1.5m to 2.5m and vary constantly. Commented Jan 14, 2019 at 13:55
• OK, so that's a big variation. Initially I was thinking something smaller. It's useful to add this clarification to the question.
– VLAZ
Commented Jan 14, 2019 at 14:12
• Do you have a good in-lore reason for why you need to produce the equipment like old school industry would in a single size fits all? furthermore modern armies already vary 1.5 meters to 2 meters. The more advanced your civilization, the less advantage there is from producing just single sized jackets and it's easier to produce to individual fit. Commented Jan 14, 2019 at 15:05
• @Lassikinnunen the in-lore reason is the same as the real-life one: individual fit or adaptable fit production requires more costly factories and more resources to produce. Additionally it is a more complex supply line to maintain. Stockpiling replacement parts and equipment would become harder as you would need to stockpile per individual or needs to be requested and transported to the exact person each time rather than "We'll have this many soldiers who need a replacement part and they can grab it off the pile". Commented Jan 14, 2019 at 15:37
Weapons come in different size, usually called calibers for fire weapons, specialized on different use cases. I.e. no sane soldier would use a naval cannon to shot an opposing soldier.
Weapons are already produced on industrial scale, yet there are different sizes.
Your soldiers are nothing more than a new type of weapon. Just find the optimal use case for each size.
• Similar to ammo production, they come in specific sizes. For army purposes you have a few common types for specific purposes with as much commonality as possible between weapons. This is also why many countries separately came to more or less the same bullet sizes for their armies. Yet when looking at the entire market theres many different ammo types and even within one caliber size the length can differ for different purposes. But you could be on to something: what if there are so many ideal sizes for specific tasks that you might as well have variable sizes? Commented Jan 14, 2019 at 13:51
• @Demigan, if it's ideal it logically must be unique.
– L.Dutch
Commented Jan 14, 2019 at 13:56
• @Demigan: if that’s the case then it might simply look to an outside observer like the sizes are variable, when in actual fact you have thousands of different clones designed for very specific ‘ideal’ jobs assembled into units with enough overlap in utility to cover any shortfalls. To me it looks like a diverse unit, to you it looks like a Brute mk.B3, a Brute mk.Q5, a Scout mk.I1, a Sniper custom grown for this exact mission and a first-generation Officer model. Commented Jan 15, 2019 at 7:34
The problem with identical clones is they lack diversity. One disease will wipe them all out, there won't be small people able to wriggle through cracks and large people able to apply huge brute strength or tall people able to see further.
If you standardize everything into conformity then you reduce the range of different tasks your teams can perform. Instead you want a mix of different physical types across the range and then let each take the role they are best suited for.
• I assume "disease" in this case could be replaced with "one situation designed against these highly specific soldiers". It would be a rather stupid organization that builds soldiers to specifications but gives them all the exact same genes (even though this is exactly what happens in geneticly modified food, but that is because of patents and economy being more efficiënt if you dont create genetic diversity). You can still have variable genes and the same height. Commented Jan 14, 2019 at 16:52
• @Demigan: it would be a rather stupid society that risks elimination of most of the yield of an important crop that way, yet here we are. Commented Jan 14, 2019 at 17:37
• @mikolak and that's the exact point I alluded to yes. But an answer based on "what if they are (still) stupid" only answers the question in the scenario that they are stupid. A reason that is useful even if they arent is far better an answer. Commented Jan 14, 2019 at 18:55
• @Demigan : I was pointing out that a) these kinds of mistakes have happened before modern GMOs, b) they are being repeated again - therefore, it's not so incredible to believe they will be made in the future. And it doesn't have to originate from "stupidity", it could be necessity (a certain level of tweaking is lost technology), it could be greed ("cost optimization" for a megacorp-style entity). The point is all it takes is for one side of a conflict making this kind of mistake to have an in-universe object lesson. It's your story of course, I'm merely showing that this solution is viable. Commented Jan 14, 2019 at 19:38
Your methods are helpless against mutation.
After the initial injection of DNA into the egg, one can no longer regulate their biological development outside of nourishment and hormone treatment. This allows possibilities of mutation during cell growth.
A particularly dominant mutation could spread throughout the fetus/embryo and result in a soldier with their own quirks. Every now and then one is born with different eye or hair colours, a slightly greater intellect or, coincidentally, an unusual height.
The mutations should not be too radical 'by default', but they could be amplified as a side-effect of the chemical cocktail used to accelerate growth.
Alternatively, mutation inhibitors may have been invented to reduce the frequency or severity of mutations. These chemicals may be too expensive for extensive use and are reserved for an elite force (an interesting concept).
It may simply not be worth ejecting all non-conforming soldiers due to the cost of fertilising an egg/monitoring it constantly/feeding it until it is born. Therefore the soldiers will have to adapt and be trained for their standard equipment.
I hope this provides some inspiration.
Edit:
Having soldiers of various heights could aid remaining unsuspicious when executing covert operations, since otherwise a soldier of exactly e.g. 1.836m in height would be suspicious and taken aside for inspection at border controls.
In large groups, having soldiers of various heights would aid visibility of a singular position, similar to stages.
Varying heights could aid in combat specialisation, with lower centers of gravity allowing for more athletic maneuvers while taller soldiers could perhaps have more strength or leverage.
Short soldiers would take less space, less food, less material for uniforms and less weight among other things, making them ideal for mechanical deployment, sieges or other forms of long-term semi-isolation.
Taller beings would be better suited as look-outs or for transport purposes, being able to march quicker, wade through deep marshes and step/climb over obstacles more efficiently, but having a high center of gravity would not necessarily be advantageous as far as I can tell.
• Although it could be an option, it goes against the "you can control everything" part of the question. I would like a solid enough reason to actively encourage variable sizes despite the higher cost in industrial capacity this causes. Commented Jan 14, 2019 at 14:18
• @Demigan Your question did not exclude the possibility of your control over everything being imperfect, but I now understand your question better. I shall edit my answer to include more possibilities. Commented Jan 14, 2019 at 14:31
# Different models for different jobs
The ideal infantryman looks nothing like the ideal pilot. The ideal pilot doesn’t necessarily look much like the ideal tank operator. The ideal tank operator certainly doesn’t look much like the ideal medic. An army is full of people who play different roles, and there’s no reason to make them all look the same.
# Like father, like son
If you want a fun excuse for this, say that each job role has a ‘blueprint’, based on the most outstanding representative of that role at the time when this super soldier was begun. All the medics are the perfected ‘sons’ of a highly decorated combat medic, etc. This gives you diverse looking troops who are incredibly proud of their lineage.
• Is that you, Boba Fett?? Commented Jan 15, 2019 at 7:35
Unification gives great economical bonus, thus specialization is about great efficiency.
Let take a look back: two centuries ago, grenadiers used to be tallest soldiers as they had longer arm and could do bigger swing movement while throwing a grenade (resulting a longer throw distance). Contrary, hussar cavalry was about light body weight to improve horse attack speed (naturally leading to low-stature dexterous recruits).
So, your war economy model is way more complex than checkers vs. chess dilemma and should include to-be-born cost + manufacture associated equipment as well as war field efficiency metrics.
Something about the manufacturing process produces different results. Maybe the farther you are from the center of the incubator the taller you are. Or maybe there's some sort of layered, onion-like ingredient and the "strips" closer the center are more potent.
Anyway, there is this unavoidable consequence that each batch produces an array of sizes.
This has another advantage in that it enforces societal roles. Pilot's are more important because very few are produced per batch.
## Chip Binning Anecdote
Interestingly enough, this type of manufacturing already exists! Have you ever wondered how Intel makes so many different models of processors? They actually only make a few (i3, i5, i7, etc.). All the variants, like i7 8700k, happen because the manufacturing process does not produce exact results. One wafer (batch) produces CPU's of all different specs. Here is an article that explains it (scroll down to step 10).
Just like chip binning, your manufacturing process could produce people in different sizes all in one go. These different sized people could then be "binned" into their proper positions.
In the past, supersoldiers were produced that way but it turned out failure to conform with the strict requirements lead to many being abandoned part way through production. In contrast, the requirements were found to be far too strict and with advancements in supersoldier manufacturing processes many previously discarded soldiers were found to be perfectly capable of less demanding tasks. Most supersoldiers are now binned to maximize yield.
The best supersoldier are the ones who win wars. Unfortunately, that doesn't tell us exactly which characteristics to give our supersoldiers.
A possible solution to this issue is to use a method similar to natural selection: To produce new supersoldiers, randomly select two existing supersoldiers, mix their DNA, apply a slight mutation and grow the new, and hopefully better supersoldier from the result. The random selection should be biaised toward the fittest soldiers according to some metric of your choice.
That way, the soldiers become gradually more efficient at war, which will ensure the dominance of that army for decades to come.
Using this method would mean that the army is made of genetically diverse soldiers, which implies varied sizes.
There are a few problems with this method (completely random DNA mutation produces a lot of waste, random DNA mixing may produce suboptimal results, etc...), but they can be fixed with sufficiently advanced knowledge and technology, but that should not be a problem for a supersoldier-producing civilisation. | 2,897 | 13,557 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-26 | latest | en | 0.968672 |
https://brilliant.org/discussions/thread/actual-taste-of-calculus/?sort=top | 1,582,548,103,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145941.55/warc/CC-MAIN-20200224102135-20200224132135-00160.warc.gz | 300,882,674 | 16,501 | # Actual Taste of Calculus
Hi Brilliant!
Recently many users have started posting great problems in Calculus section (level 3-5). All of them include variety of functions, new ideas, skill, etc. to solve them.
But I feel just being able to solve an integral or summation doesn't make one good at calculus. There is no use of solving complicated integrals/summations when you are not able to apply them properly. For example the problem, Looks Like Gabriel's Horn To Me is an application of functions which are frequently used in the Calculus section. But very few users have solved it.
I asked a user, who recently posted some good problems, whether he was able to apply them or not? He replied he barely knows application of calculus. I was stunned. What's the point of studying so much when you are not able to apply them? I've seen many people eagerly preparing for Brilliant Integration Contest. But only a few of them have covered all the basics.
Some users have not at all gone into the depth of Differential Calculus and have directly done Integral Calculus. This leads to Nothing. I'll give an example: There is a Slam Dunk Contest in NBA. Many eagerly prepare themselves for that contest. But is that contest as valuable as the NBA title? The answer is no. Just being able to dunk perfectly when no one is standing in front of a player will be of no use in the game. Similar is the case in Calculus.
So please make your basics very strong then start applying them. Later move into high level. No one has set a time limit to calculus. So take your own time and become a master.
I hope everyone liked this note.
Thanks.
3 years, 11 months ago
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This is something that I have echoed often, which is that your ability to apply your knowledge is much more important than the depth your knowledge. Someone who knows how to analyze / problem-solve / research, can easily be taught the new knowledge that he needs, and applies his skills in this area. This is especially true in university / workplace, where they do not expect you to have had all of the knowledge going in, but you learn and apply it along the way.
This is why we set up the chapters with as many levels of challenge quizzes as possible (and we're still working on it with the help of the community). For someone who only knows quadratic equations, how far can he push himself to apply it to novel situations?
Staff - 3 years, 11 months ago
Thanks for sharing your views :)
- 3 years, 11 months ago
Great Note. But my basics are clear. I spent 1 year on it.
- 3 years, 11 months ago
Cheers you don't come under that category then.
- 3 years, 11 months ago
I learnt how to apply calculus before I actually dive into things like how to integrate and differentiate, which is also a reason why many of my previous problems require computational methods (Because I did not know how to evaluate it, so I just used Mr wolf.) I think that learning both is important, or it would just seem useless to learn.
- 3 years, 11 months ago
So early you gave your opinion :3
- 3 years, 11 months ago
.__.
- 3 years, 11 months ago
Frog with medium mouth.
- 3 years, 11 months ago
Differential and Integral calculus have great applications in engineering. Many times you have to deal with the situations which can be solved using transforms which is an important part of Integral calculus. There are still many more things which I can't tell here in this short description.
Use of differential calculus is most common. And we all know most probably.
- 3 years, 11 months ago
Yeah that's what I wanted to say. People ignore differential calculus thinking it is easy and of no use.
- 3 years, 11 months ago
Your Hot integral problems are really too tough but yet awesome !!! are they your self constructed problems ? @Aman Rajput
- 3 years, 11 months ago
Thank you for the advice , I have started applications :)
- 3 years, 11 months ago
That's very nice!
- 3 years, 11 months ago
@Aditya Kumar I didn't get your views clearly. By application, do you mean applications of already known concepts to solve problems that are generally require knowledge of advance concepts, or do you mean applying (advance or elementary) concepts to real life problems?
- 3 years, 11 months ago
The latter one :). See the link. It uses one of advanced concept.
- 3 years, 11 months ago
I have posted some problems which will be really helpful for others to clear their basics...For example , The series of Ignited Integrals can really be helpful to clear the basics of Integral Calc. Btw can you suggest me some good books for differencial equations?
- 3 years, 11 months ago
Piskunov volume 2 is good for diffy
- 3 years, 11 months ago
Gr8 thanks...
- 3 years, 11 months ago
Is it E-book?
- 3 years, 10 months ago
You'll get a book.
- 3 years, 10 months ago | 1,599 | 6,494 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 9, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-10 | latest | en | 0.960035 |
https://oeis.org/A226731 | 1,721,669,470,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00423.warc.gz | 371,676,096 | 4,538 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A226731 a(n) = (2n - 1)!/(2n). 1
20, 630, 36288, 3326400, 444787200, 81729648000, 19760412672000, 6082255020441600, 2322315553259520000, 1077167364120207360000, 596585001666576384000000, 388888194657798291456000000 (list; graph; refs; listen; history; text; internal format)
OFFSET 3,1 COMMENTS For n < 3, the formula does not produce an integer. The ratio of the product of the partition parts of 2n into exactly two parts to the sum of the partition parts of 2n into exactly two parts. For example, a(3) = 20, and 2*3 = 6 has 3 partitions into exactly two parts: (5,1), (4,2), (3,3). Forming the ratio of product to sum (of parts), we have (5*1*4*2*3*3)/(5+1+4+2+3+3) = 360/18 = 20. - Wesley Ivan Hurt, Jun 24 2013 LINKS Seiichi Manyama, Table of n, a(n) for n = 3..225 Index entries for sequences related to partitions. FORMULA a(n) = A009445(n-1)/A005843(n) = A002674(n)/A001105(n). - Wesley Ivan Hurt, Jun 24 2013 a(n) ~ sqrt(Pi)*2^(2*n-1)*n^(2*n-3/2)/exp(2*n). - Ilya Gutkovskiy, Nov 01 2016 From Amiram Eldar, Mar 10 2022: (Start) Sum_{n>=3} 1/a(n) = e - 8/3. Sum_{n>=3} (-1)^(n+1)/a(n) = cos(1) + sin(1) - 4/3. (End) EXAMPLE a(3) = (2*3 - 1)!/(2*3) = 5!/6 = 120/6 = 20. MAPLE seq((2*k-1)!/(2*k), k=1..20); # Wesley Ivan Hurt, Jun 24 2013 MATHEMATICA Table[(2n-1)!/(2n), {n, 3, 20}] (* Harvey P. Dale, Jun 19 2013 *) PROG (PARI) a(n) = (2*n-1)!/(2*n); \\ Michel Marcus, Nov 01 2016 CROSSREFS Cf. A001105, A002674, A005843, A009445, A093353, A143623, A211374. Sequence in context: A129906 A125722 A336412 * A201724 A006410 A159874 Adjacent sequences: A226728 A226729 A226730 * A226732 A226733 A226734 KEYWORD nonn,easy AUTHOR Wesley Ivan Hurt, Jun 15 2013 STATUS approved
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Last modified July 22 13:14 EDT 2024. Contains 374499 sequences. (Running on oeis4.) | 806 | 2,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-30 | latest | en | 0.61863 |
http://forums.cgsociety.org/archive/index.php?t-54574.html | 1,526,984,736,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864657.58/warc/CC-MAIN-20180522092655-20180522112655-00271.warc.gz | 113,553,096 | 7,742 | View Full Version : between
(B-HOLM)04 April 2003, 06:05 PMuhm I was wondering if there was a kind of between function...I have been looking for some stuff in the help docs, but haven't found anything :shrug: I wanna do like this if (frame == between(1, 3) { do stuff } hope your understand :/
mark_wilkins
04 April 2003, 06:26 PM
proc int between (float \$test, float \$low, float \$high)
{
int \$return_val = ((\$test >= \$low) && (\$test <= \$high));
return \$return_val;
}
\$frame = frame;
if (between(\$frame, 1, 3))
{
do stuff
}
This between procedure is inclusive, meaning that if \$test equals \$low or \$high it comes back true.
-- Mark
(B-HOLM)
04 April 2003, 06:53 PM
sorry I wasn't clear enough...I meant that I wanted an interval function...
so that when the fram is equal to 1,2 and 3 then stuff would happen...did that clear it up?
mark_wilkins
04 April 2003, 07:01 PM
How is what you are requesting different than what I gave you?
In the example I posted, if the frame is 1, 2, 3, or anything in between, the contents of the if statement will execute. I'm not sure what else you want.
To be more clear, it is not possible to make the == operator test your frame number to determine whether it is within a range, but it is possible to test whether it is within an interval by using a function, and I have implemented a function to do this, which you will see above.
-- Mark
(B-HOLM)
04 April 2003, 07:04 PM
well, as I understand it then the thing you gave me is how to make a function that returnes the number between to numbers...
but what I want is something that like when :
if( frame == 1 - 5 ) //from 1 to 5
{
then do stuff
}
mark_wilkins
04 April 2003, 07:06 PM
Huh???
Using the function I defined above:
\$frame = frame;
if (between(\$frame, 1, 5))
{
do whatever you want done when
the frame number is between 1 and 5,
inclusive.
}
The function I posted returns one (1) if the first argument \$test is between \$low and \$high, and returns zero (0) if the first argument \$test is not between \$low and \$high.
The if statement executes its contents if the between function returns 1 and skips the contents if the between function returns 0.
-- Mark
(B-HOLM)
04 April 2003, 07:09 PM
ok, well I'm sorry, guess I just misread your script I guess :cry: :thumbsdow
mark_wilkins
04 April 2003, 07:10 PM
it's OK. :D
Is it clear now or would more explanation help?
-- Mark
(B-HOLM)
04 April 2003, 07:15 PM
well, the only things I can't understand is:
1. why floats in the parameter?
2. what is the return for? been so long time ago since I've done anything with return :(
3. and well...I can't understand what the \$test is for...is it what the function equals or is the return what the function equals....
I now I'm not very clear, 'cause my programming knowledge isn't great :/
mark_wilkins
04 April 2003, 12:53 AM
1) The values in the parameters are floats (short for floating-point numbers, or numbers with fractional parts) because frame numbers can be fractional. For example, you can set the timeline to 1.2 or 1.4. (Try it!)
2) When you define a function like this
proc int test (float \$a, float \$b, float \$c)
{
(whatever)
}
that means that the procedure "test" will make a calculation using the values of \$a, \$b, and \$c, and that the result will be an integer. To figure out whether one number is between two other numbers, you need three numbers: \$test is the number you're checking, \$low is the lower bound of the range, and \$high is the higher bound.
3) If a procedure passes back a value to the expression that contains it (which the "int" in the declaration above says it will) then you need to use a return statement to tell it what to return. For example, if you use the procedure above like this:
print (test(1, 2, 4) + 3);
this print statement executes the test procedure. Whatever the return statement in that test procedure says to return becomes the value that gets added to 3, then the result of that addition gets printed out.
-- Mark
(B-HOLM)
04 April 2003, 01:55 PM
hehe, I just took a look at it today, and I understand it all now. Must have been very tired yesterday hehe.
But Mark, once again thanks for all your help :)
mark_wilkins
04 April 2003, 04:08 AM
No problem!!!
-- Mark
CGTalk Moderation
01 January 2006, 09:00 PM
This thread has been automatically closed as it remained inactive for 12 months. If you wish to continue the discussion, please create a new thread in the appropriate forum.
1 | 1,209 | 4,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-22 | latest | en | 0.916757 |
https://www.onlinemath4all.com/integration-practice-questions-with-solutions.html | 1,571,725,908,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987803441.95/warc/CC-MAIN-20191022053647-20191022081147-00042.warc.gz | 1,006,579,952 | 13,499 | # INTEGRATION PRACTICE QUESTIONS WITH SOLUTIONS
## About "Integration Practice Questions With Solutions"
Integration Practice Questions With Solutions :
Here we are going to see some example problems in integration.
To find the formulas used in integration, please visit the page "Integration Formulas for Class 12"
## Integration Practice Questions With Solutions - Questions
Question 1 :
Integrate the following with respect to x
∫ (x + 5)6 dx
Solution :
∫ (x + 5)dx = (x + 5)(6+1)/(6 +1) + c
= (x + 5)7/7 + c
Question 2 :
Integrate the following with respect to x
∫ 1/(2 - 3x)4 dx
Solution :
∫ 1/(2 - 3x)4 dx = ∫ (2 - 3x)-4 dx
= (2 - 3x)(-4 + 1) / (-4 + 1) ⋅ (-3) + c
= (2 - 3x)-3 / (-3) (-3) + c
= (1/9) [1/(2 - 3x)3] + c
Question 3 :
Integrate the following with respect to x
√(3x + 2) dx
Solution :
∫ √(3x + 2) dx = ∫ (3x + 2)1/2 dx
= (3x + 2)3/2 / (3/2) (3) + c
= (3x + 2)3/2 / (9/2) + c
= (2/9)(3x + 2)3/2 + c
Question 4 :
Integrate the following with respect to x
sin 3x dx
Solution :
sin 3x dx = (- cos 3x/3) + c
Question 5 :
Integrate the following with respect to x
cos (5 - 11x) dx
Solution :
cos (5 - 11x) dx = sin (5 - 11x) / (-11) + c
= (-1/11) sin (5 - 11x) + c
Question 6 :
Integrate the following with respect to x
cosec2(5x - 7) dx
Solution :
cosec2(5x - 7) dx = -cot (5x - 7) (1/5) + c
= (-1/5) cot (5x - 7) + c
Question 7 :
Integrate the following with respect to x
e3x- 6 dx
Solution :
e3x- 6 dx = e3x- 6/3 + c
= (1/3)e3x- 6 + c
Question 8 :
Integrate the following with respect to x
e8 - 7x dx
Solution :
e8 - 7x dx = e8 - 7x /(-7) + c
= (-1/7)e8-7x + c
Question 9 :
Integrate the following with respect to x
1/(6 - 4x) dx
Solution :
1/(6 - 4x) dx = (log (6 - 4x))/-4 + c
= (-1/4) (log (6 - 4x)) + c
Question 10 :
Integrate the following with respect to x
sec2 x/5 dx
Solution :
sec2 x/5 dx = tan (x/5)/(1/5) + c
= 5 tan (x/5) + c
Question 11 :
Integrate the following with respect to x
cosec (5x + 3) cot (5x + 3) dx
Solution :
cosec (5x + 3) cot (5x + 3) dx = [- cosec (5x + 3)]/5 + c
= (-1/5) cosec (5x + 3) + c
Question 12 :
Integrate the following with respect to x
30 sec (2 - 15x) tan (2 - 15x) dx
Solution :
30 sec (2 - 15x) tan (2 - 15x) dx
= 30 sec (2 - 15x)/(-15) + c
= -2 sec (2 - 15x) + c
Question 13 :
Integrate the following with respect to x
1/√(1 - (4x)2) dx
Solution :
1/√(1 - (4x)2) dx
= sin-1(4x)/4 + c
= (1/4)sin-1(4x) + c
Question 14 :
Integrate the following with respect to x
1/√(1 - 81x2) dx
Solution :
1/√(1 - 81x2) dx
= 1/√(1 - (9x)2) dx
= sin-1 (9x) / 9 + c
(1/9) sin-1(9x) + c
Question 15 :
Integrate the following with respect to x
1/(1 + 36x2) dx
Solution :
1/(1 + 36x2) dx
= 1/(1 + (6x)2) dx
= tan-1(6x)/6 + c
= (1/6) tan-1(6x) + c
After having gone through the stuff given above, we hope that the students would have understood, "Integration Practice Questions With Solutions"
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Sum of all three four digit numbers formed using 1, 2, 5, 6 | 1,795 | 5,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-43 | longest | en | 0.571886 |
https://www.albert.io/ie/multivariable-calculus/find-product-of-first-four-nonzero-coefficients-of-power-series | 1,481,436,072,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544140.93/warc/CC-MAIN-20161202170904-00024-ip-10-31-129-80.ec2.internal.warc.gz | 901,513,078 | 19,722 | ?
Free Version
Moderate
# Find Product of First Four Nonzero Coefficients of Power Series
MVCALC-1GOEMO
Find the product of the first four nonzero coefficients of the power series solution to:
$$(2x+1) y'' - y' + xy = 0, \quad y(0) = 1 \quad y'(0) = 2$$
...centered at $x_0 = 0$.
A
$-2$
B
$-1$
C
$0$
D
$1$
E
None of the Above | 128 | 341 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2016-50 | longest | en | 0.787296 |
https://a5theory.com/dijkstras-algorithm-shortest-path-routing-algorithm/ | 1,653,529,592,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662595559.80/warc/CC-MAIN-20220526004200-20220526034200-00621.warc.gz | 113,423,172 | 21,695 | # Dijkstra’s Algorithm: Shortest Path Routing Algorithm/ How do you solve Dijkstra’s shortest path algorithm?
Hello Friends, In this blog post, I am going to let you know about an interesting shortest path algorithm which is also known as the Dijkstra’s Algorithm.
Inside this blog post, we will discuss a few important questions related to Dijkstra Algorithm like
What is Dijkstra’s algorithm for example? How do you use Dijkstra’s algorithm? Does Dijkstra’s algorithm always work? Is Dijkstra a BF? Why the Dijkstra algorithm is greedy? Which shortest path algorithm is best?….
…How does the Kruskal algorithm work? Is Dijkstra greedy or dynamic programming? How do you find the shortest path algorithm? How do you solve Dijkstra’s problem? How does Prim’s algorithm work?…
…What is the time complexity of the Dijkstra algorithm? Why does Dijkstra fail negative weights? Does Google Maps use Dijkstra? What is the finiteness of the algorithm? What do you mean by shortest path algorithm?…
…What is a single-source shortest path algorithm? Is Bellman-Ford greedy? What algorithm does GPS use?
### How do you find the shortest path between two nodes?/ Dijkstra’s algorithm steps?/ What is Dijkstra’s algorithm for example?
This ‘Dijkstra algorithm’ technique is widely used in many forms because it is simple and easy to understand. The idea is to make a graph of the subnet,…
… with each node of the graph representing a router and each arc of the graph representing a communication line(often known as a link).
For selecting a route between a given pair of routers, the algorithm just finds the shortest path between them on the graph. One method of measuring path length is the number of hops.
Using this metric, the paths ABC and ABE are equally long as shown in fig1. Another metric is the geographic distance in kilometers, in which case ABC is much longer than ABE.
Several other metrics are also possible besides hops and physical distance. As an example, each arc could be labeled with the mean queueing and transmission delay for some standard test packets determined by hourly test runs.
The shortest path is the fastest path with this graph labeling, rather than the path with the fewest arcs or kilometers.
Generally, the labels on the arcs could be computed as a function of the distance, bandwidth, average traffic, communication cost, mean queue length, measured delay, and other factors.
Then the algorithm would compute the shortest path by changing the weighting function, measured according to any one of a number of criteria or a combination of criteria.
Many algorithms for computing the shortest path between two nodes of a graph are known. every node is labeled with its distance from the source node along the best-known path.
In starting, the paths are not known, therefore all nodes are labeled with infinity. As the algorithm proceeds and paths are obtained, the labels may change, reflecting better paths.
A label may be either permanent or tentative. All labels are tentative in starting. When it is discovered that a label tells the shortest possible path from the source to that node, it is made permanent and never changed thereafter.
For illustrating the working of the labeling algorithm see the weighted undirected graph of fig1(a) where the weights represent distance. For finding the shortest path from A to D.
We start out by marking node A as permanent, denoted by a filled-in circle. Then, each of the nodes adjacent to A (the working node), relabels each one with the distance to A.
Whenever a node is relabeled, it is also labeled with the node from which the probe was made, therefore the final path can be constructed later. After examining each of the nodes adjacent to A,…
all the tentatively labeled nodes are examined in the whole graph and made the one with the smallest label permanent as shown in fig1(b). This one is the new working node.
Now, starting is done from B and all nodes adjacent to it are examined. If the sum of the label on B and the distance from B to the node being considered is less than the label on that node, a shorter path is obtained, therefore the label is relabeled.
When all the nodes adjacent to the working node have been inspected and the tentative labels changed if possible, the whole graph is searched for the tentatively labeled node with the smallest value.
This node is made permanent and is the working node for the next round. The first five steps of the algorithm are shown in fig1.
### How do you use Dijkstra’s algorithm?
There are a few simple steps of the Dijkstra algorithm which are given below to find the shortest distance.
Very first you need to initialize the distance as per the Dijkstra algorithm.
After initializing the node pick the first node and calculate the distance to the adjacent nodes.
Here pick the shortest distance node out of all the adjacent nodes.
Now again repeat the process and calculate all the adjacent nodes calculation with the new node which we have found in the previous search.
And doing this you will find the shortest path or tree between two vertex or nodes in the graph using Dijkstra Algorithm.
### Does Dijkstra’s algorithm always work?
Dijkstra algorithm works correctly for all the weighted and directed graphs with non-negative weights and is easily able to find the shortest path between two given vertex.
### Is Dijkstra a BF?
We can say that the Dijkstra algorithm does the same as BFS does. Dijkstra algorithm relies on the property of the shortest path as s to t and also between any pair of vertices along the path.
So we can conclude that the Dijkstra algorithm can be implemented as the BFS with the priority queue however it is not the only implementation.
### Why the Dijkstra algorithm is greedy?
There are both scenarios with the Dijkstra algorithm. It can be considered as greedy as we choose the closest vertex. And on the other end, it can be also considered as dynamic as we update the value with the help of previously calculated values.
But we would like to consider it closer to the dynamic algorithm than the greedy algorithm. As we can’t predict the step-by-step process to calculate the shortest distance between A to B.
### Which shortest path algorithm is best?
With the Dijkstra Algorithm’s help, we are able to find the shortest path from one vertex whereas the Floyd-Warshall algorithm is able to find out the shortest distance between each pair of vertices as it is a higher running time than the Dijkstra algorithm.
### How does the Kruskal algorithm work?
Kruskal algorithm is used to find the minimum spanning tree from an undirected connected weighted graph. In each step, we have to find out the lowest weight node that should not form a cycle in the spanning tree.
### How do you find the shortest path algorithm? What are the algorithms for finding the shortest path?
For finding the shortest path Dijkstra algorithm is not only the single option rather you can find it with the help of some other shortest path algorithm which is given below:
BFS(Breath-first-search)
DFS(Depth-first-search)
Bidirectional search
Dijkstra Algorithm
Bell-Man Ford algorithm
### How does Prim’s algorithm work?
Prim’s algorithm is also known as Jarnik’s algorithm and it is a greedy algorithm that is used to find the minimum spanning tree without forming any cycle out of an undirected connected and weighted graph.
### What is the time complexity of the Dijkstra algorithm?
O(V 2) is the time complexity of the Dijkstra algorithm. But when it is the case of the min-priority queue it is down to O(V + E log V).
### Why does Dijkstra fail negative weights?
The main aim of the Dijkstra algorithm is to find the optimal path instead of just finding any path. So with the negative weights,…
… it can not find the optimal path as it can not form any loop with the nodes that it has visited before. So with the negative weights, it will end to any random path but not the perfect optimal path.
### Does Google Maps use Dijkstra?
This is the fact that the Dijkstra algorithm power the shortest pathfinding in an effective way. And we can say that all navigation applications like Google Map,…
… and Apple Map are using the Dijkstra algorithm or another search algorithm maybe a little extended and updated form but surely include the basics of these search algorithms like Dijkstra algorithm.
### Can Dijkstra find the longest path?
The basic use of the Dijkstra algorithm is to find the shortest route from one starting vertex say ‘i’ to an end vertex say j. But Dijkstra algorithm can also be used to find the longest distance or route if some simple modification update is made.
### What is the difference between Bellman-Ford and Dijkstra?
Both are used to find the optimal path or shortest path between two pairs of vertices but the only main difference between them is that Bellman is capable to handle…
… the negative weights within the edges and the Dijkstra algorithm is not capable to handle the negative weights, it can only work perfectly with the positive weights.
### What is the difference between BFS and Dijkstra’s algorithms?
Both are quite similar but have some minor differences. BFS is used to calculate the shortest distance or path in an undirected graph whereas in the Dijkstra algorithm we find the shortest path in the directed graph.
## Conclusion:
So, in this blog post, we have learned about the Dijkstra algorithm and other search algorithms for finding the shortest path or optimal path. We have also seen the process step by step to solve the shortest path problem using the Dijkstra algorithm.
Within this blog post, we have gone through What is Dijkstra’s algorithm with an example, How do you use Dijkstra’s algorithm, Does Dijkstra’s algorithm always work, Is Dijkstra a BF, Why Dijkstra algorithm is greedy, Which shortest path algorithm is best, How does Kruskal algorithm work, Is Dijkstra greedy or dynamic programming, How do you find shortest path algorithm, How do you solve Dijkstra’s problem, How does Prim’s algorithm work, What is the time complexity of Dijkstra algorithm, Why does Dijkstra fail negative weights, Does Google Maps use Dijkstra, What is the finiteness in an algorithm, What do you mean by shortest path algorithm, What is single-source shortest path algorithm, Is Bellman-Ford greedy, What algorithm does GPS use.
In the case of any queries, you can write to us at [email protected] we will get back to you ASAP.
Hope! you would have enjoyed this post about ‘Dijkstra’s Algorithm for the shortest path‘.
Please feel free to give your important feedback in the comment section below.
Have a great time! Sayonara!
#### Anurag
I am a blogger by passion, a software engineer by profession, a singer by consideration and rest of things that I do is for my destination. | 2,207 | 10,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2022-21 | latest | en | 0.904949 |
https://encyclopedia2.thefreedictionary.com/Eosinophil+differentiation+factor | 1,653,765,856,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00111.warc.gz | 293,296,624 | 11,369 | # factor
(redirected from Eosinophil differentiation factor)
Also found in: Dictionary, Thesaurus, Medical, Legal, Financial.
## factor
factor, in arithmetic, any number that divides a given number evenly, i.e., without any remainder. The factors of 12 are 1, 2, 3, 4, 6, and 12. Similarly in algebra, any one of the algebraic expressions multiplied by another to form a product is a factor of that product, e.g., a+b and ab are factors of a2b2, since (a+b)(ab)=a2b2. In general, if r is a root of a polynomial equation f(x)=0, then (xr) is a factor of the polynomial f(x).
## factor
[′fak·tər]
(mathematics)
For an integer n, any integer which gives n when multiplied by another integer.
For a polynomial p, any polynomial which gives p when multiplied by another polynomial.
For a graph G, a spanning subgraph of G with at least one edge.
(statistics)
A quantity or a variable being studied in an experiment as a possible cause of variation.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
## factor
1. Maths
a. one of two or more integers or polynomials whose product is a given integer or polynomial
b. an integer or polynomial that can be exactly divided into another integer or polynomial
2. Med any of several substances that participate in the clotting of blood
3. Law, Commerce a person who acts on another's behalf, esp one who transacts business for another
4. former name for a gene
5. Commercial law a person to whom goods are consigned for sale and who is paid a factorage
6. (in Scotland) the manager of an estate
Collins Discovery Encyclopedia, 1st edition © HarperCollins Publishers 2005
## factor
A quantity which is multiplied by another quantity. | 439 | 1,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-21 | latest | en | 0.873262 |
https://www.hackmath.net/en/example/3951?tag_id=61,78 | 1,563,326,350,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525004.24/warc/CC-MAIN-20190717001433-20190717023433-00103.warc.gz | 716,032,607 | 6,647 | # Angle in RT
Determine the size of the smallest internal angle of a right triangle whose sides constitutes sizes consecutive members of arithmetic progressions.
Result
x = -90 °
#### Solution:
Checkout calculation with our calculator of quadratic equations.
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this example are needed these knowledge from mathematics:
Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? See also our right triangle calculator. See also our trigonometric triangle calculator.
## Next similar examples:
1. Angle
Determine the size of the smallest internal angle of a right triangle which angles forming the successive members of the arithmetic sequence.
2. Alfa, beta, gama
In the triangle ABC is the size of the internal angle BETA 8 degrees larger than the size of the internal angle ALFA and size of the internal angle GAMA is twice the size of the angle BETA. Determine the size of the interior angles of the triangle ABC.
3. Alfa beta gama
The triangle's an interior angle beta is 10 degrees greater than the angle alpha and gamma angle is three times larger than the beta. Determine the size of the interior angles.
4. Angles of the triangle
ABC is a triangle. The size of the angles alpha, beta are in a ratio 4: 7. The angle gamma is greater than the angle alpha by a quarter of a straight angle. Determine angles of the triangle ABC.
5. Internal angles IST
Determine internal angles of isosceles trapezium ABCD /a, c are the bases/ and if: alpha:gamma = 1:3
6. Internal angles
One internal angle of the triangle JAR is 25 degrees. The difference is the size of the two other is 15°. Identify the size of these angles.
7. Triangle angles
The angles α, β, γ in triangle ABC are in the ratio 6:2:6. Calculate size of angles.
8. Angles in triangle
The triangle is ratio of the angles β:γ = 6:8. Angle α is 40° greater than β. What are the size of angles of the triangle?
9. Angles in a triangle
The angles of the triangle ABC make an arithmetic sequence with the largest angle γ=83°. What sizes have other angles in a triangle?
10. Angles ratio
The internal angles of a triangle are in ratio 1:4:5 What kind of triangle is it? (solve internal angles and write down and discuss)
11. Triangle angles
In a triangle ABC the interior angle at the vertex C is twice as the internal angle at the point A. Outer angle at the point B measured 117 degrees. How big is the outer angle at the vertex A?
12. Clouds
From two points A and B on the horizontal plane was observed forehead cloud above the two points under elevation angle 73°20' and 64°40'. Points A , B are separated by 2830 m. How high is the cloud?
13. Largest angle of the triangle
What is the largest angle of the triangle if the second angle is 10° greater than twice the first and the third is 30° smaller than the second?
14. Circular sector
I have a circular sector with a length 15 cm with an unknown central angle. It is inscribed by a circle with radius 5 cm. What is the central angle alpha in the circular sector?
15. Garage
There are two laths in the garage opposite one another: one 2 meters long and the second 3 meters long. They fall against each other and stay against the opposite walls of the garage and both laths cross 70 cm above the garage floor. How wide is the garag
16. Angles of a triangle
In the triangle ABC, the angle beta is 15° greater than the angle alpha. The remaining angle is 30° greater than the sum of the angles alpha and beta. Calculate the angles of a triangle.
17. Height of the arc - formula
Calculate the height of the arc if the length of the arc is 77 and chord length 40. Does exist a formula to solve this? | 898 | 3,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2019-30 | latest | en | 0.858332 |
https://discourse.mcneel.com/t/intersections-between-circles/84992 | 1,656,746,432,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103989282.58/warc/CC-MAIN-20220702071223-20220702101223-00564.warc.gz | 270,422,352 | 5,767 | # Intersections between circles
Hello all,
Please, I’m trying to write a script that checks for the intersection between inclined circles on Rhino and calculates the length. I tried using curvecurveintersection, but it’s always returning no intersection, wvwn when the circles are intersected.
NB: my circles are guid objects I inserted in a box, then saved the GUIDs in a list, so I want to use the list to find intersection between the circles
Can you post some code with a sample file that’s not working?
Blockquote
``````> def LengthOfIntersection(circle_list):
> length = 0
> print circle_list
> for i in range(len(circle_list)):
> curveA = circle_list[i]
> for j in range(len(circle_list)-1):
> if j != i:
> intersection = rs.CurveCurveIntersection(curveA,circle_list[j])
> #print intersection
> if intersection != None:
>
> length += math.sqrt((intersection[1][1][0]-intersection[0][1][0])**2 + (intersection[1][1][1]-intersection[0][1][1])**2)``````
The circle_list is a list of GUIDs
I have now converted the circles into surfaces for the intersections to be detected. But I am getting a weird error
``````def LengthOfIntersection(surface_list):
length = 0
for i in range(len(surface_list)):
curveA = surface_list[i]
for j in range(len(surface_list)):
if j != i:
intersection = rs.CurveCurveIntersection(curveA,surface_list[j])
#print intersection
if intersection != None:
``````
`````` length += math.sqrt((intersection[1][1][0]-intersection[0][1][0])**2 + (intersection[1][1][1]-intersection[0][1][1])**2)
else: continue
print length
``````
Message: unable to convert ef49509a-0bb4-46d9-b03d-e966960c84e2 into Curve geometry
Traceback:
line 1447, in CurveCurveIntersection, “C:\Users\falol\AppData\Roaming\McNeel\Rhinoceros\6.0\Plug-ins\IronPython (814d908a-e25c-493d-97e9-ee3861957f49)\settings\lib\rhinoscript\curve.py”
line 1002, in coercecurve, “C:\Users\falol\AppData\Roaming\McNeel\Rhinoceros\6.0\Plug-ins\IronPython (814d908a-e25c-493d-97e9-ee3861957f49)\settings\lib\rhinoscript\utility.py”
line 171, in LengthOfIntersection, "C:\Users\falol\AppData\Roaming\McNeel\Rhinoceros\6.0\scripts\sec.
This is the function generating and converting the circles to surfaces.
``````def AddCircle(n,radius,boxlength):
"""
A function to add a fixed number of circles in a cube
"""
#initialize a list to store circle GUIDs
circle_list = []
surface_list = []
#a loop to insert the fixed number of circles
for i in range(n):
#create a center point for the circle
x = random.uniform(0+radius,boxlength-radius)
y = random.uniform(0+radius,boxlength-radius)
z = random.uniform(0+radius,boxlength-radius)
#store the random origin
origin = [x,y,z]
#convert the origin to a plane
plane = InclinePlane(origin)
#insert the circle in the cube
my_circle = rs.AddCircle(plane,radius)
#append the circle GUID to the list
circle_list.append(my_circle)
surf = rs.AddPlanarSrf(my_circle)
surface_list.append(surf[0])
print surface_list
return surface_list`````` | 839 | 3,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-27 | latest | en | 0.675855 |
https://www-verimag.imag.fr/~boulme/videos/poster-oopsla20.srt | 1,624,400,224,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-25/segments/1623488525399.79/warc/CC-MAIN-20210622220817-20210623010817-00257.warc.gz | 559,590,028 | 3,338 | 1 00:00:00,080 --> 00:00:04,720 Hello, I will briefly present you here two breakthroughs that we did for formally verified 2 00:00:04,720 --> 00:00:08,020 compilers in Coq. A formally verified compiler is a compiler 3 00:00:08,020 --> 00:00:12,179 with a proof of correctness. So, all the transformations have to be proven 4 00:00:12,179 --> 00:00:14,860 correct. Our first breakthrough is to provide scalable 5 00:00:14,860 --> 00:00:19,000 scheduling optimizations. A scheduling optimization is one where the 6 00:00:19,000 --> 00:00:22,430 instructions are reordered. This kind of transformation is very hard to 7 00:00:22,430 --> 00:00:27,160 prove structurally, so we split it into an untrusted oracle which gives us a schedule 8 00:00:27,160 --> 00:00:33,180 proposition, and then we execute symbolically the returned schedule to check its correctness. 9 00:00:33,180 --> 00:00:37,740 Our second contribution is that we are the first to target an interlocked VLIW processor, 10 00:00:37,740 --> 00:00:42,150 that is, a processor with explicit instruction parallelism. 11 00:00:42,150 --> 00:00:46,030 So here is a tiny example to introduce you the impact of scheduling. 12 00:00:46,030 --> 00:00:50,750 In this example, we assume that the ADD and SUB have a latency of 1, and the LOAD has 13 00:00:50,750 --> 00:00:54,820 a latency of 2. So we consider here a simple model, a 3 stage 14 00:00:54,820 --> 00:00:59,170 pipeline. There is the DECODE stage, the EXEC1 stage 15 00:00:59,170 --> 00:01:04,760 where ADD and SUB finish, and EXEC2 where LOAD finishes. 16 00:01:04,760 --> 00:01:10,280 In this example, if we run the instructions in order, there will be a stall because of 17 00:01:10,280 --> 00:01:16,190 the R3 register not being available. However, if we permute the instructions 2 18 00:01:16,190 --> 00:01:21,420 and 1, we see that we win one cycle but there is no more stall. 19 00:01:21,420 --> 00:01:27,140 In practice, our untrusted oracle has a bunch of resource constraints, and also latency 20 00:01:27,140 --> 00:01:32,290 constraints which depend on both the processor latencies and the original program. 21 00:01:32,290 --> 00:01:37,600 Then, to ensure that this scheduling is actually correct, we execute symbolically the original 22 00:01:37,600 --> 00:01:41,780 code and the scheduled code, and we check that they are the same. 23 00:01:41,780 --> 00:01:47,700 For architectures such as the Kalray processor, a bundle is a group of instructions that are 24 00:01:47,700 --> 00:01:52,090 to be executed in parallel. In our work, we defined semantics for bundle 25 00:01:52,090 --> 00:01:55,229 execution. And then we used a scheduling pass in order 26 00:01:55,229 --> 00:01:59,350 to generate bundles into CompCert. So, our verifier works in two steps. 27 00:01:59,350 --> 00:02:03,620 As mentionned earlier, the first step verifies the equivalence of the unordered and ordered 28 00:02:03,620 --> 00:02:07,190 basic block. And then the last step verifies that each 29 00:02:07,190 --> 00:02:11,210 bundle has the same execution between sequential and parallel semantics. 30 00:02:11,210 --> 00:02:15,569 Finally, we give some performance comparisons. The top chart compares the performance of 31 00:02:15,569 --> 00:02:19,840 the original code to the scheduler. Compared to the original version without any 32 00:02:19,840 --> 00:02:27,420 scheduling we gain 40% in average, and compared to the greedy bundler we gain 20% in average. 33 00:02:27,420 --> 00:02:31,180 The bottom chart compares CompCert to the GCC of Kalray. 34 00:02:31,180 --> 00:02:34,650 Most safety critical applications use GCC without any optimization. 35 00:02:34,650 --> 00:02:38,900 We can see that we always beat GCC -O0 by a large margin. 36 00:02:38,900 --> 00:02:43,810 We also beat GCC -O1 most of the time. As for -O2 and -O3, we are not there yet, 37 00:02:43,810 --> 00:02:47,459 but we are getting closer. In order to further bridge the gap, we are 38 00:02:47,459 --> 00:02:53,069 currently working on superblock scheduling before register allocation. 39 00:02:53,069 --> 00:03:00,040 I'm going to demonstrate the CompCert compiler and our optimizations on a simple example. 40 00:03:00,040 --> 00:03:06,430 We are going here to see an example of a matrix multiplication. 41 00:03:06,430 --> 00:03:09,300 So this is the very simple code that performs the matrix multiplication. 42 00:03:09,300 --> 00:03:15,940 We are going to compile it, and we are then going to execute it and measure the amount 43 00:03:15,940 --> 00:03:20,700 of cycles. Now we will run CompCert on the example without 44 00:03:20,700 --> 00:03:25,330 any postpass optimization. So you will see the actual original code. 45 00:03:25,330 --> 00:03:29,130 So here is the code from the innermost loop - that one. 46 00:03:29,130 --> 00:03:34,310 That code isn't optimal at all because we only have one instruction per bundle. 47 00:03:34,310 --> 00:03:40,580 And also we have some stalls in the pipeline. For example we have this multiply-add here 48 00:03:40,580 --> 00:03:45,091 that takes two cycles, except that we issue a load on the cycle right after. 49 00:03:45,091 --> 00:03:55,239 So that makes for a very suboptimal performance. If we try to run it, we get this amount of 50 00:03:55,239 --> 00:03:57,069 cycles. 51 00:03:57,069 --> 00:04:01,350 Now we will try to run CompCert with the first version of the scheduler, which will just 52 00:04:01,350 --> 00:04:06,290 pack instructions together into bundles. On this example, the oracle packed together 53 00:04:06,290 --> 00:04:11,110 some instructions, and the verificator verified that this is actually correct. 54 00:04:11,110 --> 00:04:15,440 This will give a better performance, however we still have issues with stalls as you can 55 00:04:15,440 --> 00:04:21,739 see here. We can try to run it. 56 00:04:21,739 --> 00:04:28,169 And we see we are slightly better, but it could still be improved. 57 00:04:28,169 --> 00:04:31,159 So now let's try to re-order the instructions with our list scheduler. 58 00:04:31,159 --> 00:04:35,229 Here is the result. We can see that a lot of instructions have been moved, which result 59 00:04:35,229 --> 00:04:41,550 in a much more efficient code. In particular, we can execute four instructions 60 00:04:41,550 --> 00:04:48,889 at the same time on the same bundle. Now if you run that example, we can see we 61 00:04:48,889 --> 00:04:52,449 have a much better performance than before. 62 00:04:52,449 --> 00:04:53,449 Thanks for your attention. | 2,110 | 6,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-25 | latest | en | 0.904609 |
https://codegolf.stackexchange.com/questions/163272/matrix-in-slash-order | 1,653,341,532,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00276.warc.gz | 216,477,442 | 91,599 | # Matrix in "slash" order
Given two positive numbers N >= 2 and N <= 100 create a matrix which follows the following rules:
• First Number starts at position [0,0]
• Second Number starts at position [0,1]
• Third number goes below First Number (position [1,0])
• Following numbers goes in "slash" direction
• Range of numbers used is [1, N1 * N2]. So, numbers goes from starting 1 to the result of the multiplication of both inputs.
Input
• Two numbers N >= 2 and N <= 100. First number is the amount of rows, Second number the amount of columns.
Output
• Matrix. (Can be outputted as a multidimensional array or a string with line breaks)
Example:
Given numbers 3 and 5 output:
1 2 4 7 10
3 5 8 11 13
6 9 12 14 15
Given numbers 2 and 2
1 2
3 4
Given Numbers 5 and 5
1 2 4 7 11
3 5 8 12 16
6 9 13 17 20
10 14 18 21 23
15 19 22 24 25
The shortest code in bytes wins.
• Can we use 0 indexing for any of the numbers?
– Jo King
Apr 24, 2018 at 12:41
• @JoKing No. Must start at 1. Apr 24, 2018 at 12:42
• Apr 24, 2018 at 12:49
• @LuisfelipeDejesusMunoz Perhaps a better term for the order is "diagonals"? Personally, I'd call it a "zig-zag", because it reminds me of Cantor's Zig-Zag proof, but that might confusing. Apr 24, 2018 at 14:12
• @LuisfelipeDejesusMunoz anti-diagonal is the term for the other diagonal.
– qwr
Apr 25, 2018 at 2:45
# Jelly, 6 5 bytes
pSÞỤs
Try it online!
### How it works
pSÞỤs Main link. Left argument: n. Right argument: k
p Take the Cartesian product of [1, ..., n] and [1, ..., k], yielding
[[1, 1], [1, 2], ..., [n, k-1], [n, k]].
SÞ Sort the pairs by their sums.
Note that index sums are constant on antidiagonals.
Ụ Grade up, sorting the indices of the sorted array of pairs by their values.
s Split the result into chunks of length k.
• Damn. Mine is 200+ bytes. Can you add some explanation pls? Apr 24, 2018 at 13:21
• God damn it, Dennis. Also, good job. Apr 24, 2018 at 13:35
• Wow, it is too "closely related". That's identical to the first link in miles' answer. Consider upvoting both. :) Apr 24, 2018 at 13:56
• I think it might be possible to do this with <atom><atom>¥þ but I can't find the right combination. oþ++þ is close but doesn't quite get there Apr 24, 2018 at 17:38
• @akozi So far, so good. The indices of the sorted array are [1, 2, 3, 4, 5, 6]. Ụ sorts this array, using the key that maps 1 to [1, 1], 2 to [1, 2], 3 to [2, 1], etc. Essentially, this finds the index of each pair from the sorted-by-sums array in the sorted-lexicographically array Apr 26, 2018 at 14:51
# Python 3, 91 bytes
def f(n,k):M=[(t//k+t%k,t)for t in range(n*k)];return zip(*k*[map([M,*sorted(M)].index,M)])
Try it online!
# R, 10160 54 bytes
function(M,N)matrix(rank(outer(1:M,1:N,"+"),,"l"),M,N)
Try it online!
Thanks to @nwellnhof for the suggestion of rank
function(M,N)matrix(unsplit(lapply(split(1:(M*N),unlist(split(x,x))),rev),x<-outer(1:M,1:N,"+")),M,N)
Try it online!
split is doing most of the work here; possibly there's a golfier algorithm but this definitely works.
Explanation:
function(M,N){
x <- outer(1:M,1:N,"+") # create matrix with distinct indices for the antidiagonals
idx <- split(x,x) # split into factor groups
items <- split(1:(M*N),unlist(idx)) # now split 1:(M*N) into factor groups using the groupings from idx
items <- lapply(items,rev) # except that the factor groups are
# $2:1,$3:2,3, (etc.) but we need
# $2:1,$3:3,2, so we reverse each sublist
matrix(unsplit(items,x),M,N) # now unsplit to rearrange the vector to the right order
# and construct a matrix, returning the value
}
Try it online! -- you can use wrap a print around any of the right-hand sides of the assignments <- to see the intermediate results without changing the final outcome, as print returns its input.
• Can you add some explanation pls? Apr 24, 2018 at 13:36
• @LuisfelipeDejesusMunoz added. If there's anything unclear, let me know and I'll try and clarify. Apr 24, 2018 at 13:56
• rank(x,1,"f") is 2 bytes shorter than order(order(x)). Apr 25, 2018 at 11:19
• @nwellnhof oh, very nice, but using rank(x,,"l") will get rid of the t as well. Apr 25, 2018 at 11:26
# Java 10, 121120109 105 bytes
m->n->{var R=new int[m][n];for(int i=0,j,v=0;i<m+n;)for(j=++i<n?0:i-n;j<i&j<m;)R[j][i-++j]=++v;return R;}
-11 bytes thanks to @OlivierGrégoire.
-4 bytes thanks to @ceilingcat.
Try it online.
Explanation:
m->n->{ // Method with two integer parameters and integer-matrix return-type
var R=new int[m][n]; // Result-matrix of size m by n
for(int i=0,j, // Index integers, starting at 0
v=0; // Count integer, starting at 0
i<m+n;) // Loop as long as i is smaller than m+n
for(j=++i<n?0 // Set j to 0 if i+1 is smaller than n
:i-n; // or to the difference between i and n otherwise
j<i&j<m;) // Inner loop j until it's equal to either i or m,
// so basically check if it's still within bounds:
R[j][i-++j]=++v; // Add the current number to cell j, i-(j+1)
return R;} // Return the result-matrix
• I realized this takes columns first and then rows. Apr 24, 2018 at 13:33
• @Luis I think it's convention to take coordinates as x,y/width,height
– Jo King
Apr 24, 2018 at 13:46
• 109 bytes Apr 25, 2018 at 15:23
$1(+/:@;)</.@i. -4 more bytes for this solution by miles. Thanks! Try it online! # J, 22 19 bytes -3 bytes thanks to FrownyFrog! ,$[:>:@/:@/:@,+/&i.
Try it online!
An implementation of Dennis' fantastic Jelly solution in J.
## Explanation:
Dyadic verb, takes left and right argument (m f n)
+/&i. creates lists 0..m-1 and 0..n-1 and makes an addition table for them:
3 +/&i. 5
0 1 2 3 4
1 2 3 4 5
2 3 4 5 6
[:>:@/:@/:@, flattens the table and grades the list twice and adds 1 to it:
3 ([:>:@/:@/:@,+/&i.) 5
1 2 4 7 10 3 5 8 11 13 6 9 12 14 15
,$ reshapes the list back into mxn table: 3 (-@],\[:>:@/:@/:@,+/&i.) 5 1 2 4 7 10 3 5 8 11 13 6 9 12 14 15 • -@],\,$ for −3 bytes. Apr 24, 2018 at 14:18
• @FrownyFrog - Of course, I feel stupid, it's so obvous now. Thank you! Apr 24, 2018 at 14:28
• 15 bytes $1(+/:@;)</.@i. with input as an array [r, c] Apr 24, 2018 at 14:38 • @miles: Very cool, thanks! I tried /. but could't achieve your result :) Apr 24, 2018 at 18:22 # APL+WIN, 38 or 22 bytes Prompts for integer input column then row: m[⍋+⌿1+(r,c)⊤m-1]←m←⍳(c←⎕)×r←⎕⋄(r,c)⍴m or: (r,c)⍴⍋⍋,(⍳r←⎕)∘.+⍳c←⎕ based on Dennis's double application of grade up. Missed that :( • Sorry for the question but is there somewhere I can test it? Apr 24, 2018 at 14:08 • @Luis felipe De jesus Munoz No problem. APL+WIN is not available on line but you can test it on the Dyalog website at tryapl.org if you replace the ⎕ characters with the integers of your choice. Apr 24, 2018 at 14:35 # Wolfram Language (Mathematica), 73 67 bytes Count elements in rows above: Min[j+k,#2]~Sum~{k,i-1} Count elements on the current row and below: Max[j-k+i-1,0]~Sum~{k,i,#} Put into a table and add 1. Voila: 1+Table[Min[j+k,#2]~Sum~{k,i-1}+Max[j-k+i-1,0]~Sum~{k,i,#},{i,#},{j,#2}]& Update: I realized there is a shorter way to count all the positions ahead of a normally specified position in the matrix with just one sum over two dimensions: Table[1+Sum[Boole[s-i<j-t||s-i==j-t<0],{s,#},{t,#2}],{i,#},{j,#2}]& Try it online! Try it online! # APL (Dyalog Unicode), 14 12 bytes {⍵⍴⍋⍋∊+/↑⍳⍵} Try it online! -2 thanks to ngn, due to his clever usage of ↑⍳. Based off of Dennis's 5-byte Jelly solution. • ∘.+⌿⍳¨⍵ -> +/↑⍳⍵ – ngn May 4, 2018 at 9:20 • @ngn Wow, that's a clever usage of ⍳ combined with ↑. May 4, 2018 at 12:53 # 05AB1E, 23 bytes *L<DΣ¹LILâOsè}UΣXsè}Á>ô Try it online! # Python 3, 164 bytes from numpy import* r=range def h(x,y): a,i,k,j=-array([i//y+i%y for i in r(x*y)]),1,2,0 while j<x+y:a[a==-j],i,k,j=r(i,k),k,k+sum(a==~j),j+1 a.shape=x,y;return a Try it online! This is definitely not the shortest solution, but I thought it was a fun one. • from numpy import* and dropping both n. is slightly shorter. Also, you can drop the space at ) for. And changing to Python 2 allows you to change return a to print a (in Python 3 it would be the same byte-count print(a)). Apr 24, 2018 at 14:25 • Thanks! I should have thought of import*. I'll never beat Dennis' answer, so I'll stick to Python 3. – maxb Apr 24, 2018 at 14:36 # Python 2, 93 bytes def f(b,a):i=1;o=[];exec"if b:o+=[],;b-=1\nfor l in o:k=len(l)<a;l+=[i]*k;i+=k\n"*a*b;print o Try it online! Semi-Ungolfed version: def f(b,a): i=1 o=[] for _ in range(a*b) if b: o+=[[]] b-=1 for l in o: if len(l)<a: l+=[i] i+=1 print o # Japt, 25 24 bytes Hardly elegant, but gets the job done. Working with 2D data in Japt is tricky. ;N×Ç<U©Ap[] A®Ê<V©Zp°T A ; // Set alternative default vars where A is an empty array. N×Ç // Multiply the inputs and map the range [0..U*V). <U // If the current item is less than the second input, ©Ap[] // add a new empty subarray into A. A® // Then, for each item in A, Ê<V // if its length is less than the first input, ©Zp°T // Add the next number in the sequence to it. A // Output the results, stored in A. I added the -Q flag in TIO for easier visualization of the results, it doesn't affect the solution. Bit off one byte thanks to Oliver. Try it online! • Speaking of ×, you can replace *V with N×. Apr 25, 2018 at 0:30 • @Oliver And here I was, thinking that shortcut is handy, but not a common use case. Thanks a lot! Apr 25, 2018 at 1:03 # JavaScript (Node.js), 103 bytes (a,b,e=[...Array(b)].map(_=>[]))=>f=(x=0,y=i=0)=>(x<a&y<b&&(e[y][x]=++i),x?f(--x,++y):y>a+b?e:f(++y,0)) Try it online! # TI-Basic, 76 bytes Prompt A,B {A,B🡒dim([A] 1🡒X For(E,1,B+A For(D,1,E If D≤A and E-D<B Then X🡒[A](D,E-D+1 X+1🡒X End End End [A] Prompts for user input and returns the matrix in Ans and prints it. TI-Basic is a tokenized language; all tokens used here are one byte, other than [A] which is 2 bytes. Note: TI-Basic (at least on the TI-84 Plus CE) only supports matrices up to 99x99, and so does this program. Explanation: Prompt A,B # 5 bytes, prompt for user input {A,B🡒dim([A] # 9 bytes, make the matrix the right size 1🡒X # 4 bytes, counter variable starts at 1 For(E,1,B+A # 9 bytes, Diagonal counter, 1 to A+B-1, but we can over-estimate since we have to check later anyway. For(D,1,E # 7 bytes, Row counter, 1 to diagonal count If D≤A and E-D<B # 10 bytes, Check if we are currently on a valid point in the matrix Then # 2 bytes, If so, X🡒[A](D,E-D+1 # 13 bytes, Store the current number in the current point in the matrix X+1🡒X # 6 bytes, Increment counter End # 2 bytes, End dimension check if statement End # 2 bytes, End row for loop End # 2 bytes, End dimension for loop [A] # 2 bytes, Implicitly return the matrix in Ans and print it # Perl 6, 61 59 bytes {($!={sort($_ Z=>1..*)>>.{*}})($!([X+] ^<<$_)).rotor(.[1])} Try it online! Another port of Dennis' Jelly solution. # Java (JDK 10), 142 131 bytes X->Y->{var A=new int[X][Y];int E=1;for(int y=0;y<Y+X-1;y++)for(int x=0;x<X;x++){if(y-x<0|y-x>Y-1)continue;A[x][y-x]=E++;}return A;} Try it online! Explanation: X->Y->{ // Method with two integer parameters and integer-matrix return-type var A=new int[X][Y]; // The Matrix with the size of X and Y int E=1; // It's a counter for(int y=0;y<Y+X-1;y++) // For each column plus the number of rows minus one so it will run as long as the bottom right corner will be reached for(int x=0;x<X;x++){ // For each row if(y-x<0|y-x>Y-1) // If the cell does not exist becouse it's out of range continue; // Skip this loop cycle A[x][y-x]=E++; // Set the cell to the counter plus 1 } return A; // Return the filled Array } Big thank to Kevin Cruijssen because I didn't know how to run my code on tio. Some code like the header and footer are stolen from him. -> His answer • 119 bytes: tio.run/… Feb 10, 2019 at 21:00 • 105 bytes Jul 22, 2021 at 6:54 # Husk, 9 bytes ṠCoηÖ¤×,ḣ Try it online! Most of the heavy lifting is done by ×, which creates pairs in the desired order so that it can support infinite lists. ### Explanation CηÖ×,ḣ²ḣ⁰⁰ (Expanded; let N and K denote the arguments.) ηÖ Take the list of indices of the sorted values of ×, the list of all pairs taken from ḣ² the range from 1 to N and ḣ⁰ the range from 1 to K, and C ⁰ cut it into lengths of K. # PHP, 115 bytes a pretty lazy approach; probably not the shortest possible. function($w,$h){for(;$i++<$h*$w;$r[+$y][+$x]=$i,$x--&&++$y<$h||$x=++$d+$y=0)while($x>=$w|$y<0)$y+=!!$x--;return$r;}
anonymous function, takes width and height as parameters, returns 2d matrix
try it online
# JavaScript (Node.js), 108105101 100 bytes
n=>(l=>{for(r=[];i<n*n;n*~-n/2+2>i?l++:l--*y++)for(T=y,t=l;t--;)r[T]=[...r[T++]||[],++i]})(y=i=0)||r
Try it online!
# Attache, 45 bytes
{Chop[Grade//2<|Flat!Table[+,1:_2,1:_],_]+1}
Try it online!
Anonymous lambda, where paramaters are switched. This can be fixed for +1 byte, by prepending ~ to the program. The test suite does this already.
## Explanation
This approach is similar to the J answer and the Jelly answer.
The first idea is to generate a table of values:
Table[+,1:_2,1:_]
This generates an addition table using ranges of both input parameters. For input [5, 3], this gives:
A> Table[+,1:3,1:5]
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
Then, we flatten this with Flat!:
A> Flat!Table[+,1:3,1:5]
[2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8]
Using the approach in the J answer, we can grade the array (that is, return indices of sorted values) twice, with Grade//2:
A> Grade//2<|Flat!Table[+,1:3,1:5]
[0, 1, 3, 6, 9, 2, 4, 7, 10, 12, 5, 8, 11, 13, 14]
Then, we need to chop the values up correctly, as in the Jelly answer. We can cut every _ elements to do this:
A> Chop[Grade//2<|Flat!Table[+,1:3,1:5],5]
0 1 3 6 9
2 4 7 10 12
5 8 11 13 14
Then, we just need to compensate for the 0-indexing of Attache with +1:
A> Chop[Grade//2<|Flat!Table[+,1:3,1:5],5]+1
1 2 4 7 10
3 5 8 11 13
6 9 12 14 15
And thus we have the result.
# Python 3, 259 bytes
So I did this a weird way. I noticed that there were two patterns in the way the array forms.
The first is how the top rows pattern has the difference between each term increasing from 1 -> h where h is the height and l is the length. So I construct the top row based on that pattern
For a matrix of dim(3,4) giving a max RoC = 3 We will see the top row of the form
1, (1+1), (2+2), (4+3) = 1, 2, 4, 7
Suppose instead that the dim(3,9) giving a max RoC = 3 we will instead see a top row of
1, (1+1), (2+2), (4+3), (7+3), (10+3), (13+3), (16+3), (19+3) = 1, 2, 4, 7, 10, 13, 16, 19, 22
The second pattern is how the rows change from one another. If we consider the matrix:
1 2 4 7 11
3 5 8 12 16
6 9 13 17 20
10 14 18 21 23
15 19 22 24 25
and subtract each row from the row below (ignoring the extra row) we get
2 3 4 5 5
3 4 5 5 4
4 5 5 4 3
5 5 4 3 2
Upon seeing this matrix we can notice this matrix is the sequence 2 3 4 5 5 4 3 2 where by each row is 5 terms of this pattern shifted by 1 for each row. See below for visual.
|2 3 4 5 5| 4 3 2
2 |3 4 5 5 4| 3 2
2 3 |4 5 5 4 3| 2
2 3 4 |5 5 4 3 2|
So to get the final matrix we take our first row we created and output that row added with the 5 needed terms of this pattern.
This pattern will always have the characteristics of beginning 2-> max value and ending max value -> 2 where the max value = min(h+1, l) and the number of times that the max value will appear is appearances of max = h + l -2*c -2 where c = min(h+1, l) - 2
So in whole my method of creating new rows looks like
1 2 3 7 11 + |2 3 4 5 5|4 3 2 = 3 5 8 12 16
3 5 8 12 16 + 2|3 4 5 5 4|3 4 2 = 6 9 13 17 20
6 9 13 17 20 + 2 3|4 5 5 4 3|4 2 = 10 14 18 21 23
10 14 18 21 23 + 2 3 4|5 5 4 3 2| = 15 19 22 24 25
Relevant code below. It didn't end up being short but I still like the method.
o,r=len,range
def m(l,h):
a,t=[1+sum(([0]+[x for x in r(1,h)]+[h]*(l-h))[:x+1]) for x in r(l)],min(l,h+1);s,c=[x for x in r(2,t)],[a[:]]
for i in r(h-1):
for j in r(o(a)):
a[j]+=(s+[t]*(l+h-2*(t-2)-2)+s[::-1])[0+i:l+i][j]
c+=[a[:]]
for l in c:print(l)
Try it online!
# Japt, 20 bytes
õ ïVõ)ñx
£bYgUñ¹ÄÃòV
Try it
r=[1..]
n!k=[[sum$lookup(i,j)$zip[(i,j)|s<-r,i<-n#r,j<-k#r,i+j==s]r|j<-k#r]|i<-n#r] | 6,076 | 16,325 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-21 | longest | en | 0.825516 |
http://frontend.turing.io/lessons/module-2/whiteboarding-and-pseudocoding.html | 1,553,168,203,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202523.0/warc/CC-MAIN-20190321112407-20190321134407-00128.warc.gz | 80,266,079 | 7,245 | title: Whiteboarding and Pseudocoding length: 120 tags: whiteboarding, interviews, pseudocoding module: 2 —
### Learning Goals
• Understand the relationship between problem-solving, programming, and algorithmic thinking
• Identify and implement different strategies for solving problems
• Identify common roadblocks to effective problem-solving
### Warm Up
• When you are presented with a problem, how do you solve it?
• In your journal, draw out the process of how you make toast. You must start with a clean sheet of paper and you are not allowed to write/use any words.
## Problem Solving
When you are presented with a problem… whether it is a complex mathematical problem or a broken printer, how do you solve it? Generally speaking, the first step in finding a solution to a problem is to be sure that you have clearly identified the problem itself. To solve problems, we can utilize the UPS strategy:
• Understand: Read the problem. Restate the problem by writing it down in your own words. What do you know? What do you not know?
• Plan: What problem strategy will you try? You might also draw a picture/representation of how you plan to approach the problem
#### Possible Problem Solving Strategies
Method Description Example
Trial and Error Continue trying different solutions until problem is solved Restarting phone, turning off WiFi, turning off bluetooth in order to determine why your phone is malfunctioning
Heuristic General problem-solving framework Working backwards, breaking a task into steps
Algorithm Step-by-step problem solving formula Instruction manual for installing new software on your computer
#### Journal/Reflection
Think about how you currently solve problems when you are programming. Has that changed since you were in Mod 1? What are some of the possible pros and cons of the methods listed above?
## Whiteboarding
In the context of building out programs, whiteboarding is a useful tool to ulitize in solving complex problems. If we rush into coding out the solution immediately, we completely skip over the `P`lanning part of the process… and its likely that we don’t fully `U`nderstand the problem at hand.
In the context of interviewing for a job, being able to whiteboard is important because it shows you can think through/discuss a problem without having to actually code it first - which can be a difficult skill to learn.
Things to remember
• Talk as you write
• Pay attention to the amount of space you have to write
• DON’T WIPE THE WHITEBOARD WITH YOUR HAND WHEN YOU ERASE THINGS
• High level vs low level
• Develop a shorthand for writing code:
``````if (foo === 'bar') {
this.changeFoo('baz')
} else if (foo === 'baz') {
this.changeFoo('bar');
} else {
this.changeFoo('caw caw');
}
``````
Could be written as shorthand:
``````foo = baz?
changeFoo(bar)
foo = bar?
changeFoo(baz)
else
changeFoo(caw caw)
``````
## Pseudocoding
Pseudocode is an artificial and informal language that helps programmers develop algorithms to solve problems. A key thing to remember is that psuedocode is not code. It is a way to force yourself to slow down and think about how to solve problems at a high-level, using natural language.
Pseudocode:
• gives you a fallback
• gives people 👀 the chance to intervene
• takes syntax errors out of the equation
### Basic Structure for Pseudocoding
Sequences: Sequences are the most straightforward part of structuring pseudocode. At the most basic level, a sequence is made up of commands or steps that must be followed. Any commands in a sequence are generally written as a list.
Here’s an IRL example of a sequence for feeding my dogs:
``````// Pick up dog bowls from the floor
// Bring bowls to pantry
// Open up big dog food bag
// Fill big bowl
// Open up small dog food bag
// Fill small bowl
// Call dogs by name
// Place both bowls on the floor
``````
Choices: Choices give us the option to take different paths based on certain conditions. You may find that you have just one path based on a condition, two paths, or multiple paths.
Here’s another IRL example of how I may choose a certain path (based on the weather) when dressing myself in the morning:
``````// Check the weather forecast
// If snow is expected, put on boots
// Otherwise, put on flip flops
``````
Iterations: Iterations allow us to repeat steps until a certain condition is met. They always begin with a choice. Here’s another IRL example of preheating my oven:
``````// While the oven is not 350 degrees, wait
``````
Which could also be written like this:
``````// Until the oven is 350 degrees, wait
``````
## PIA (Psuedocoding in action)
Task: Create your own `filter` function (on `Array.prototype`) to really understand/see how `filter` is working under the hood. You should implement a filter function that is like the `<Array>.filter()` in JavaScript.
``````[1,2,3,4].filter((num)=>{ return num > 3}) // Should return [4]
``````
#### Turn and Code:
In pairs, pull up and fork this codepen. You should be completing both phases for each individual test. Be sure that you are actively switching between being the driver and the navigator.
Phase I => (Understand and Plan) Break out the way that you would solve this problem in pseudocode. Remember, psuedocode is not code… but a way to work through the logic of the problem without worrying about syntax.
Person A: Driver
Person B: Navigator
Phase II => (Solve) Using your psuedocode as a guide (feeling free to make adjustments as necessary) to start to implement your solution in code.
Person B: Navigator
Person A: Driver
### Some possible pitfalls to problem solving
• Mental set - Persist in approaching a problem in a way that has worked in the past but is clearly not working now
• Anchoring bias - Tendency to focus on one particular piece of information when making decisions or problem-solving
• Availability bias - Decision is based upon either an available precedent or an example that may be faulty | 1,338 | 5,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-13 | longest | en | 0.919325 |
https://www.coursehero.com/file/6481188/Calc-3-Exam-2-Sin-2011/ | 1,516,156,334,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886792.7/warc/CC-MAIN-20180117003801-20180117023801-00413.warc.gz | 917,370,674 | 23,731 | Calc 3 Exam 2 Sin 2011
# Calc 3 Exam 2 Sin 2011 - MAC2313 Exam 2 Dr Sin No...
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Unformatted text preview: MAC2313 Exam 2 Dr Sin No Calculators. Answer the questions in the spaces provided on the question sheets. Please write your answers in full detail. If you run out of room for an answer, continue on the back of the page or on other paper. Name: 1. (7 points) Let f ( x,y ) = ( yx 2 x 4 + y 2 if ( x,y ) 6 = (0 , 0) if ( x,y ) = (0 , 0) . (a) (1 point) At which points of the plane is the function f ( x,y ) defined? (b) (4 points) At which points of the plane does the function f ( x,y ) have a limit? (c) (2 points) At which points of the plane is the function f ( x,y ) continuous? Solution: (a)The function is defined at all points of the plane. (b)At every point ( x,y ) 6 = (0 , 0) the function f ( x,y ) is defined by a rational function (whose denominator is not zero at ( x,y )), so f ( x,y ) has a limit there. At (0 , 0), we have lim x → x 4 +0 = 0 along x = 0, whereas lim x → x 4 x 4 + x 4 = 1 / 2 along y = x 2 , so f ( x,y ) has no limit at (0 , 0)....
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Calc 3 Exam 2 Sin 2011 - MAC2313 Exam 2 Dr Sin No...
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Ask a homework question - tutors are online | 484 | 1,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-05 | latest | en | 0.834983 |
https://mathoverflow.net/questions/95837/examples-of-theorems-with-proofs-that-have-dramatically-improved-over-time/152500 | 1,643,205,380,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304954.18/warc/CC-MAIN-20220126131707-20220126161707-00422.warc.gz | 454,048,250 | 36,135 | # Examples of theorems with proofs that have dramatically improved over time
I am looking for examples of theorems that may have originally had a clunky, or rather technical, or in some way non-illuminating proof, but that eventually came to have a proof that people consider to be particularly nice. In other words, I'm looking for examples of theorems for which have some early proof for which you'd say "ok that works but I'm sure this could be improved", and then some later proof for which you'd say "YES! That is exactly how you should do it!"
• Seems related to the earlier MO question mathoverflow.net/questions/43820/extremely-messy-proofs. May 3 '12 at 10:13
• I don't know the original proof, but I heard that the trick of Rabinovich provided a drastic improvement of the proof of Hilbert's Nullstellensatz. May 3 '12 at 21:42
• It would be also interesting to hear of theorems where people didn't think that the proof could be much improved, but then were proven wrong. Jul 9 '12 at 6:02
• Many answers here so far seem to have the form "The original proof of X was very complicated, but now one can prove it as a simple consequence of Y." -- But if the proof of Y is not itself simpler than the original proof of X, should this qualify? Jun 28 '13 at 2:57
• @LouisDeaett: Good question. There is something, though, still nicer about having a proof follow as a simple consequence of something complicated. It shows there is some larger (unifying?) idea. Dec 3 '13 at 6:50
The global (or homology) version of Cauchy’s theorem was given an elementary proof by John Dixon. I believe this is mentioned in Rudin's Real and Complex Analysis. A proof is available online at http://www.math.uiuc.edu/~r-ash/CV/CV3.pdf. This states "The elementary proof to be presented below is due to John Dixon, and appeared in Proc. Amer. Math. Soc. 29 (1971), pp. 625-626, but the theorem as stated is originally due to E.Artin."
• Very good example because it also shows the prize to pay for elegance: Dixon's proof is so simple because it defines a cycle in $\Omega$ as homologically trivial if the index of every point outside $\Omega$ is zero. But this hides the fact that homology is an internal property (invariant under homeomorphisms). May 16 '21 at 8:53
• @JochenWengenroth, there's no elegance bought either. Cauchy's theorem is trivial once one defines homologically trivial cycles as boundaries (subdivide a simplex into smaller ones such that each it contained in a ball in $\Omega$). It's then also obvious that a trivial cycle is also "extrinsically" trivial - simply apply the result to $1/(z-a)$. So, what's at stake in Dixon's proof is the purely topological converse statement. This also has a clean and truly elementary (contrary to Dixon's) proof by Artin. Oct 5 '21 at 14:28
One historical example that should probably be on this list is the Abel-Ruffini Theorem, which states that there is no general solution in radicals to polynomials of degree 5 and higher. Attempting to clarify the bigger picture of why the proof works may have been one of Galois's motivations for his development of what is now Galois Theory- and the proof we have now is quite insightful and illuminating.
an example from number theory (where such simplifications are not uncommon), Bertrands postulate
for any integer $$n > 3$$, there always exists at least one prime number $$p$$ with $$n < p < 2n − 2$$
was first simplified by Ramanujan and then later by Erdos who also proved a more general case.
another interesting case study here may be Lindemanns proof of transcendence of PI which is subsumed by later more general results. as Wikipedia states "Weierstrass proved the above more general statement in 1885. The theorem, along with the Gelfond–Schneider theorem, is extended by Baker's theorem, and all of these are further generalized by Schanuel's conjecture."
another "possible/controversial" famous/legendary case study here is Fermats Last Theorem; Fermat scribbled in the margin of his book that he had a remarkable proof, but modern consensus is that he must have been mistaken based on the 2020-hindsight of Wiles complex proof. however, strictly speaking, it has not been proven impossible that there exists a short proof.
it seems that later simplifications of proofs is a natural process of the historical/evolutionary progress of mathematics so that results once thought more arcane/inscrutable/complex become more accessible with the polishing/systematization of ideas/techniques.
• "However, strictly speaking, it has not been proven impossible that there exists a short proof." Well, sure, but for which theorem has it been proven impossible that there exists a short proof?? Dec 24 '13 at 11:30
• agreed. the question of short proofs is closely related to Kolmogorov complexity theory & Chaitin's incompleteness both tied up with undecidability. there do exist some results or so-called "no-go/barrier theorems" in the form that "a proof for [x] cannot use theory [y]".
– vzn
Dec 24 '13 at 16:11
The original proof of Muller-Schupp theorem saying that finitely generated groups with context-free word problem are exactly the virtually free groups, is really involved (though nice) and uses accessibility and Stallings results on ends.
But there is a short and elementary way to prove Muller-Schupp theorem using rewriting systems, as was recently done by Volker Diekert.
The Amitsur-Levitski Theorem (the standard non-commutative polynomial of order $2n$ vanishes identically on $M_n(k)$) qualifies. The original proof (1950) is messy, with no clear logical structure and takes 17 pages. The natural proof was given in 1976 by S. Rosset in a 2-pages article.
• Where can I find the 60-pages long original proof? "Minimal identities for algebras" is just 15 pages long ( ams.org/journals/proc/1950-001-04/S0002-9939-1950-0036751-9/… ), and proves AL (its Theorem 1) on its page 7. Does it rely on lots of other references? Sep 11 '15 at 11:28
• Also, it is question whether Rosset's proof is really "natural". It requires a lift to a characteristic-$0$ ring, which is suggested by nothing in the statement; and from there it still proceeds to apply some interesting and highly surprising tricks. I think there is much to be learned from that proof, but "natural" looks different. Sep 11 '15 at 11:35
• Darij, yes 17 pages indeed ! Sep 11 '15 at 15:53
• The paper is 15 pages long, and the theorem is proven on page 7. Now I am really curious how this adds up to 17 ;) Sep 11 '15 at 18:31
• One thing I can confirm, though: the original proof does look like a mess! Sep 11 '15 at 18:32
Godel's original proof of his Incompleteness Theorem was immensely more complicated Aaronson's one-paragraph derivation of Incompleteness from Undecidability of Halting Problem. Even including a proof of Undecidability of Halting it would be much shorter and clearer than the original proof.
• I agree that the concept of computability does give a nicer proof of Goedel's theorem, but I wouldn't describe the situation in quite this way. Aaronson's sketch glosses over the issue of expressing the halting of a Turing machine as an arithmetical statement. A large fraction of Goedel's proof is devoted to proving the expressibility of syntactical facts in the language of arithmetic, and these complexities are not avoided in Aaronson's sketch; they're just swept under the rug. So I don't think it's fair to say that Goedel's original proof was "immensely more complicated." Jan 10 '16 at 20:32
1. Chirka's proof ("On the propagation of holomorphic motions", 2004) of Slodkowski's theorem ("Holomorphic motions and polynomial hulls", 1991) is much simpler. (Slodkowski's paper is not that long, but uses a lot more difficult mathematics. A number of people had previously attempted to give alternative proofs, but these turned out to contain gaps.)
2. The original proof by Baker that repelling periodic points are dense in Julia sets of transcendental entire functions used the Ahlfors five islands theorem (a very deep result). The proof by Duval and Berteloot (Une démonstration directe de la densité des cycles répulsifs dans l’ensemble de Julia), building on work of Schwick, takes less than a page, and uses only very elementary results (notably, Zalcman's rescaling lemma for normal families). Even for rational functions, this is probably the simplest proof (simpler than the original ones of Fatou and Julia) currently in existence.
The first proof of the Hopf invariant one theorem due to Adams is very technical. It involves decomposing $sq^{2^n}$ as a composite of secondary cohomology operations when $n\geq 4$. Then Atiyah and Adams came up with a proof that uses $K$-theory which both admired for its elegance and simplicity.
Some results by Donaldson were simplified via the Seiberg-Witten invariants.
From Wikipedeia: Seiberg–Witten invariants are similar to Donaldson invariants and can be used to prove similar (but sometimes slightly stronger) results about smooth 4-manifolds. They are technically much easier to work with than Donaldson invariants; for example, the moduli spaces of solutions of the Seiberg–Witten equations tend to be compact, so one avoids the hard problems involved in compactifying the moduli spaces in Donaldson theory.
(Added Jan 7, '16) An additional piece of information I learned today from the Zabrodsky's lecture delivered by Peter Ozsváth is about the existence of exotic smooth structure on $$\mathbb R^4$$. The original proof was based on Freedman theorem and on Donaldson theorem. Results based on new knot invariants such as the Knot Floer homology (and simpler combinatorial descriptions of these invariants) can replace the "Donaldson side" of the proof by a much simpler argument.
Faltings' theorem (aka Mordell conjecture) can be taken as such an example. Different methods have been used so far with various difficulties.
Manjul Bhargava's proof of the 15 theorem was dramatically simpler than Conway and Schneeberger's original proof.
• The bad stuff were given to a computer in MB's proof while CS proof was all by hand. Is there a link to CS' proof? Nov 8 '13 at 18:49
• JAS, I think you are confusing the 15 theorem and the 290 theorem. CS never published a full proof of the 15 theorem because it was so messy (a sketch appears in Schneeberger's Ph.D. thesis), and did not think that the 290 theorem was provable at all. MB's proof of the 15 theorem is short and the computer calculations are pretty modest by modern standards. Granted, the proof of the 290 theorem by MB and Hanke is indeed highly computational, but again, CS didn't have a proof of it at all. Nov 8 '13 at 19:24
• ah ic. I think MB proved something more general. For every subset $S\subseteq\Bbb N$, there is a finite subset $S_f\subsetneq S$ such that if a quadratic form represents $S_f$, then it represents $S$. Is there a link to the proof? How constructive is the set $S_f$? Nov 8 '13 at 19:30
• There are several possible links; I think it's easier for you to Google it yourself than for me to try to post the links here. Nov 9 '13 at 0:47
• @Turbo: For the 15 theorem, see Manjul Bhargava, "On the Conway-Schneeberger fifteen theorem," in Quadratic forms and their applications, Contemp. Math. 272 (1999), 27–37. For the 290 theorem, try math.stanford.edu/~vakil/files/290-Theorem-preprint.pdf Jan 21 '16 at 21:22
DeMoivre's theorem. The pre-calc version of the proof relies on a lot of triangle geometry to establish trigonometric sum formulas and then uses induction. If you use Euler's Identity, it's a one line proof. (Of course, then there's a large amount of analysis implicit in the background.)
In machine learning and statistics, the technique of Rademacher complexities has a way of radically simplifying otherwise complicated proofs -- in particular those involving margin-based generalization bounds.
Some examples include Support Vector Machine margin bounds, which used to be proved via intricate combinatorial fat-shattering techniques of https://homes.di.unimi.it/~cesabian/Pubblicazioni/jacm-97b.pdf http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.42.6950&rep=rep1&type=pdf http://www.sciencedirect.com/science/article/pii/S0304397500001341 and are now proved in 3 lines (see the Thm. 4.3 in http://www.cs.nyu.edu/~mohri/mlbook/ ).
Another example is margin-based boosting bounds, which were originally proved via an involved covering and sampling argument, http://projecteuclid.org/euclid.aos/1024691352 and now has a very simple Rademacher-based proof (Thm. 5.7 in https://mitpress.mit.edu/books/boosting ).
See Ostrowski's proof of Luroth theorem in Schinzel's book "Polynomials with special regard to reducibility"
The four color theorem’s first proof by Appel and Haken, which actually contained mistakes, was proved again much more succinctly and successfully by Robertson, Sanders, Seymour, and Thomas (see for instance here). As far as I know, there is no better proof as yet, and unlikely there will ever be one, but who knows.
Liouville's Theorem that there exists a transcendental number had its proof greatly improved by Cantor who showed that a mere counting argument suffices.
Liouville's argument needs facts about how rationals interact with polynomials, and makes use of the metric structure on the reals.
Cantor's argument merely uses the fact that each integer polynomial can be described in finitely many symbols and has only finitely many roots. And it uses this to deduce a stronger result: that almost no reals are algebraic.
• I think the downvotes are excessive and have upvoted to compensate, but I also don't think this is fully an answer to the question: Liouvill's proof, while more complicated than the cardinality argument, is by no means "clunky, or rather technical, or in some way non-illuminating." May 10 '20 at 20:30
• @NoahSchweber I think it's both clunky and nonilluminating. The cardinality proof shows that the entire issue really has nothing to do with approximations, rationals or polynomials. You could extend the definition of 'algebraic' in many ways (for example to the closure of the algebraics under $\exp$ and $\log$) and Cantor's proof would show with no extra effort that there were still reals outside your set, whereas Liouville's proof would fall apart. Cantor's proof uses less to do more, and in a way which is more generalizable. May 10 '20 at 21:34
• In any case I added some of these justifications to my answer, so hopefully that will stem the flow of downvotes. May 10 '20 at 21:35
• I think the key point, though, is that Liouville's argument proves something stronger, and with respect to that conclusion it's a very illuminating and efficient proof. So it's less "Liouville proved X clunkily, then Cantor proved X efficiently" and more "Liouville proved X, then Cantor proved X efficiently, and in retrospect Liouville actually proved Y - which Cantor didn't." I don't think this quite counts; if anything it's an instance of nuking a mosquito, which isn't really the same in my opinion. (Of course, that's objective.) May 10 '20 at 21:50
• Doesn’t Liouville not only show existence of a transcendental number, but actually exhibit an explicit transcendental number? May 10 '20 at 21:50 | 3,718 | 15,251 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-05 | latest | en | 0.97608 |
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# If 10^50 74 is written as an integer in base 10 notation,
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Joined: 06 Nov 2008
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If 10^50 74 is written as an integer in base 10 notation, [#permalink]
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11 Dec 2008, 15:09
If 10^50 – 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?
A. 424
B. 433
C. 440
D. 449
E. 467
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11 Dec 2008, 15:11
C. 440.
Whenever there are large exponents like this, I like to use a smaller exponent and extract a rule from that simply because it's easier to use smaller numbers.
if we use 1000 (which is 10^3) we have 1000 - 74 = 926. So we know we'll have 2+6 for 8, but how many 9's will we have with the exponent being 50? well, if used 100 (10^2), we get 100 - 74 = 26, so here there are zero 9's. So let x represent the exponent value, we know that x - 2 will be the number of 9's. so 9(x-2) + 2 + 6 where x = the value of the exponent.
If x = 50, then 50 -2 = 48...48*9 = 432 + 2 + 6 = 440
haichao wrote:
If 10^50 – 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?
A. 424
B. 433
C. 440
D. 449
E. 467
--== Message from GMAT Club Team ==--
This is not a quality discussion. It has been retired.
If you would like to discuss this question please re-post it in the respective forum. Thank you!
To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
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Re: integer [#permalink] 11 Dec 2008, 15:11
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# If 10^50 74 is written as an integer in base 10 notation,
Moderator: chetan2u
# Events & Promotions
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 886 | 3,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-30 | latest | en | 0.902165 |
https://notebook.community/AllenDowney/ThinkBayes2/solutions/cookie_soln | 1,618,860,990,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038916163.70/warc/CC-MAIN-20210419173508-20210419203508-00409.warc.gz | 529,157,833 | 8,703 | # Think Bayes
This notebook presents example code and exercise solutions for Think Bayes.
Copyright 2018 Allen B. Downey
``````
In [1]:
# Configure Jupyter so figures appear in the notebook
%matplotlib inline
# Configure Jupyter to display the assigned value after an assignment
%config InteractiveShell.ast_node_interactivity='last_expr_or_assign'
# import classes from thinkbayes2
from thinkbayes2 import Hist, Pmf, Suite
``````
Here's the original statement of the cookie problem:
Suppose there are two bowls of cookies. Bowl 1 contains 30 vanilla cookies and 10 chocolate cookies. Bowl 2 contains 20 of each.
Now suppose you choose one of the bowls at random and, without looking, select a cookie at random. The cookie is vanilla. What is the probability that it came from Bowl 1?
If we only draw one cookie, this problem is simple, but if we draw more than one cookie, there is a complication: do we replace the cookie after each draw, or not?
If we replace the cookie, the proportion of vanilla and chocolate cookies stays the same, and we can perform multiple updates with the same likelihood function.
If we don't replace the cookie, the proportions change and we have to keep track of the number of cookies in each bowl.
Exercise:
Modify the solution from the book to handle selection without replacement.
Hint: Add instance variables to the `Cookie` class to represent the hypothetical state of the bowls, and modify the `Likelihood` function accordingly.
To represent the state of a Bowl, you might want to use the `Hist` class from `thinkbayes2`.
``````
In [2]:
# Solution
# We'll need an object to keep track of the number of cookies in each bowl.
# I use a Hist object, defined in thinkbayes2:
bowl1 = Hist(dict(vanilla=30, chocolate=10))
bowl2 = Hist(dict(vanilla=20, chocolate=20))
bowl1.Print()
``````
``````
chocolate 10
vanilla 30
``````
``````
In [3]:
# Solution
# Now I'll make a Pmf that contains the two bowls, giving them equal probability.
pmf = Pmf([bowl1, bowl2])
pmf.Print()
``````
``````
Hist({'vanilla': 30, 'chocolate': 10}) 0.5
Hist({'vanilla': 20, 'chocolate': 20}) 0.5
``````
``````
In [4]:
# Solution
# Here's a likelihood function that takes `hypo`, which is one of
# the Hist objects that represents a bowl, and `data`, which is either
# 'vanilla' or 'chocolate'.
# `likelihood` computes the likelihood of the data under the hypothesis,
# and as a side effect, it removes one of the cookies from `hypo`
def likelihood(hypo, data):
like = hypo[data] / hypo.Total()
if like:
hypo[data] -= 1
return like
``````
``````
In [5]:
# Solution
# Now for the update. We have to loop through the hypotheses and
# compute the likelihood of the data under each hypothesis.
def update(pmf, data):
for hypo in pmf:
pmf[hypo] *= likelihood(hypo, data)
return pmf.Normalize()
``````
``````
In [6]:
# Solution
# Here's the first update. The posterior probabilities are the
# same as what we got before, but notice that the number of cookies
# in each Hist has been updated.
update(pmf, 'vanilla')
pmf.Print()
``````
``````
Hist({'vanilla': 29, 'chocolate': 10}) 0.6000000000000001
Hist({'vanilla': 19, 'chocolate': 20}) 0.4
``````
``````
In [7]:
# Solution
# So when we update again with a chocolate cookies, we get different
# likelihoods, and different posteriors.
update(pmf, 'chocolate')
pmf.Print()
``````
``````
Hist({'vanilla': 29, 'chocolate': 9}) 0.4285714285714286
Hist({'vanilla': 19, 'chocolate': 19}) 0.5714285714285714
``````
``````
In [8]:
# Solution
# If we get 10 more chocolate cookies, that eliminates Bowl 1 completely
for i in range(10):
update(pmf, 'chocolate')
print(pmf[bowl1])
``````
``````
0.2621359223300971
0.13636363636363635
0.06104651162790699
0.023800528900642243
0.008061420345489446
0.002316602316602318
0.0005355548943766738
8.929900282780178e-05
8.118750253710948e-06
0.0
`````` | 1,116 | 3,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-17 | longest | en | 0.866358 |
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Ratio Formula Cheat Sheet
Prerequisite Formulas
Used in the various ratio formulas below
Measures the dollar amount of profit remaining
after cost of goods sold and is used as the
denominator in the Gross Margin calculation
Used as the denominator in the Return on Assets
calculation
Used as the denominator in the Return on Equity
calculation
Used as the denominator in the Receivables
Turnover calculation
Used as the denominator in the Inventory
Turnover calculation
Profitability Ratio Formulas
Used to measure how effective an organization
is at generating profits
Measures the percentage of each sales dollar
remaining after cost of sales that contributes to
funding the organization’s SG&A and other costs
Measures the percentage of each sales dollar that
remains after cost of goods sold and SG&A
expenses
Measures the percentage of each sales dollar that
remains after all expenses
Measures how effectively an organization uses its
assets to generate profits
Measures how effectively an organization uses
the capital contributed by equity investors
(excludes preferred stock)
Measures the percentage of net income paid out
as dividends
Asset Utilization Ratio Formulas
Used to measure the amount of time it takes or
how efficiently assets are turned into cash
Measures the number of times per year the
balance of receivables is collected and shows
how efficient an organization is in collections
Measures the average number of days it takes
from the time a sale is made until cash is
collected
Measures the number of times per year the
balance of inventory is output as cost of sales and
shows how efficient an organization is in
managing its inventory
Measures the average number of days it takes
from the time raw material is received until a
finished good is output
Measures how effective an organization is at
using its fixed assets to generate sales (note that
this formula uses net fixed assets, not gross)
Measures how effective an organization is at
using all of its assets to generate sales
Liquidity Ratio Formulas
Used to measure an organization’s ability to pay
off its obligations due within 1 year
Measures how total current assets compares to
total current liabilities – it is the most common
measure of short-term liquidity
Measures how liquid current assets (i.e. excluding
inventory) compare to total current liabilities – it
is more conservative than the current ratio
Debt Utilization Ratio Formulas
Used to measure whether an organization
carries an appropriate amount of debt
Measures how much of an organization’s assets
are financed through debt – a ratio over 1 means
100% or more are financed with debt
Measures whether an organization uses more
debt or more equity to finance operations – a
ratio of over 1 means debt is used more
Measures an organization’s ability to make its
interest payments by showing how much EBIT
exceeds or falls short of interest expense (over
100% means EBIT exceeds interest expense)
Market Ratio Formulas
Used by investors to evaluate how cheap or
expensive a stock is
Measures the net income per share of common
stock
(
)
Measures the number of years of profits implicit
in the share price (most commonly used market
ratio)
Measures the accounting book value per share of
common stock
Measures how the market price of the stock
compares with the common shareholder equity
(a value over 1 indicates the market price
exceeds the accounting book value)
Copyright 2011 | 742 | 3,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-24 | latest | en | 0.921319 |
https://calculator.academy/rc-circuit-calculator/ | 1,701,226,725,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100047.66/warc/CC-MAIN-20231129010302-20231129040302-00604.warc.gz | 191,068,410 | 51,317 | Enter the total capacitance and total resistance into the calculator below to calculate the frequency of an RC Circuit.
## RC Circuit Formula
The following equation can be used to calculate the frequency of an RC circuit.
f = 1/(2π * R * C)
• Where f is the frequency (Hz)
• R is the resistance (Ohms)
• C is the capacitance (Farads)
This equation is used only for RC circuits. As the name suggests, an RC circuit consists of a resistor with resistance r and a capacitor with capacitance c. The resistor or resistors are combined in a series in the case of using this formula There are two major uses for RC Circuit. One, to filter using the frequency, or to store energy using the capacitor. The total energy that can be stored by the circuit is only dictated by the capacitance, however, the rate of charge is affected by the resistor.
## RC Circuit Defintion
An RC circuit is defined as a circuit consisting of only a resistor and capacitor.
## How to calculate frequency of an RC Circuit?
How to calculate the frequency of an RC circuit?
1. First, measure the resistance.
Measure the resistance of the resistor.
2. Next, measure the capacitance.
Measure the capacitance of the capacitor.
3. Finally, calculate the frequency.
Use the equation above to calculate the frequency.
## FAQ
What is an RC Circuit?
An RC circuit is defined as a circuit consisting of only a resistor and capacitor. | 303 | 1,409 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-50 | latest | en | 0.906464 |
https://mathexamination.com/lab/squaring-the-circle.php | 1,624,489,966,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488544264.91/warc/CC-MAIN-20210623225535-20210624015535-00356.warc.gz | 356,585,738 | 8,594 | ## Take My Squaring The Circle Lab
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It is clear that Squaring The Circle as well as Modular Formulas are tough to find out, as well as it does take method to establish your very own feeling of intuition. But when you wish to address a mathematics problem, you have to utilize a tool, and also the devices for discovering are used when you are stuck, as well as they are not utilized when you make the wrong relocation. This is where lab Help Solution comes in.
For instance, what is wrong with the question is incorrect concepts, such as getting a partial worth when you do not have enough working parts to finish the entire task. There is a great reason that this was wrong, and also it is a matter of logic, not instinct. Logic permits you to comply with a detailed procedure that makes sense, as well as when you make an incorrect move, you are normally forced to either try to move forward and also deal with the mistake, or try to step and also do an in reverse step.
Another instance is when the student does not recognize a step of a procedure. These are both sensible failings, as well as there is no other way around them. Also when you are stuck in an area that does not allow you to make any type of kind of relocation, such as a triangular, it is still important to comprehend why you are stuck, to make sure that you can make a far better action and also go from the step you are stuck at to the following location.
With this in mind, the best means to address a stuck circumstance is to just take the step forward, instead of attempting to go backward. Both processes are various in their technique, yet they have some basic resemblances. However, when they are attempted with each other, you can swiftly inform which one is much better at addressing the issue, and also you can additionally tell which one is much more effective.
Allow's discuss the very first instance, which associates with the Squaring The Circle mathematics lab. This is not too challenging, so let's initial review exactly how to start. Take the adhering to procedure of connecting a component to a panel to be utilized as a body. This would certainly need three measurements, as well as would be something you would certainly need to connect as part of the panel.
Currently, you would have an extra dimension, however that doesn't imply that you can just keep that dimension as well as go from there. When you made your primary step, you can easily forget the dimension, and after that you would have to go back and also retrace your steps.
Nevertheless, rather than keeping in mind the additional dimension, you can utilize what is called a "mental faster way" to assist you keep in mind that added measurement. As you make your primary step, picture yourself taking the dimension as well as connecting it to the component you wish to connect to, and after that see just how that makes you really feel when you repeat the procedure.
Visualisation is an extremely powerful strategy, as well as is something that you need to not avoid over. Envision what it would seem like to actually attach the component and also have the ability to go from there, without the dimension.
Now, allow's take a look at the second instance. Allow's take the same procedure as before, and now the pupil needs to keep in mind that they are going to return one step. If you tell them that they need to return one action, yet then you get rid of the suggestion of needing to return one step, after that they will not recognize exactly how to proceed with the issue, they won't know where to try to find that step, as well as the procedure will be a mess.
Rather, make use of a psychological faster way like the psychological representation to psychologically show them that they are mosting likely to return one step. as well as put them in a setting where they can move forward from there. without needing to think of the missing out on a step.
## Hire Someone To Do Your Squaring The Circle Lab
" Squaring The Circle - Required Assist With a Math lab?" Regrettably, lots of trainees have had a trouble grasping the principles of direct Squaring The Circle. Thankfully, there is a new style for direct Squaring The Circle that can be made use of to teach straight Squaring The Circle to pupils that battle with this principle. Trainees can make use of the lab Aid Solution to help them learn new techniques in direct Squaring The Circle without facing a mountain of problems as well as without needing to take an examination on their ideas.
The lab Help Solution was created in order to assist struggling trainees as they move from college and also secondary school to the college and also task market. Lots of trainees are not able to take care of the stress of the knowing process as well as can have extremely little success in realizing the ideas of linear Squaring The Circle.
The lab Help Solution was developed by the Educational Screening Solution, that offers a range of various online tests that trainees can take and practice. The Examination Aid Service has actually aided numerous pupils enhance their ratings and can assist you improve your ratings too. As pupils move from university and also secondary school to the college as well as job market, the TTS will aid make your trainees' change simpler.
There are a couple of various manner ins which you can make use of the lab Aid Solution. The main way that students make use of the lab Help Service is via the Answer Managers, which can help trainees discover techniques in straight Squaring The Circle, which they can make use of to help them do well in their programs.
There are a number of issues that pupils experience when they initially use the lab Help Solution. Students are commonly overloaded as well as don't understand how much time they will require to commit to the Solution. The Response Managers can aid the trainees evaluate their idea learning and also help them to assess every one of the material that they have actually currently discovered in order to be gotten ready for their following course work.
The lab Help Solution works the same way that a professor performs in regards to assisting trainees grasp the ideas of linear Squaring The Circle. By offering your students with the tools that they require to find out the vital principles of straight Squaring The Circle, you can make your students more successful throughout their research studies. As a matter of fact, the lab Help Solution is so effective that numerous pupils have switched from typical math course to the lab Help Service.
The Task Manager is designed to aid pupils manage their research. The Task Manager can be set up to arrange how much time the student has offered to finish their appointed research. You can likewise establish a customized period, which is a wonderful function for students who have an active routine or a really hectic secondary school. This function can aid students stay clear of feeling bewildered with math jobs.
One more beneficial function of the lab Assist Service is the Student Aide. The Pupil Aide aids trainees handle their job and also gives them a place to upload their research. The Pupil Aide is useful for trainees that don't intend to get overwhelmed with responding to multiple inquiries.
As trainees get more comfy with their assignments, they are encouraged to connect with the Job Supervisor as well as the Student Aide to obtain an online support system. The on the internet support group can assist students keep their emphasis as they address their projects.
Every one of the assignments for the lab Assist Solution are consisted of in the bundle. Trainees can login and complete their assigned work while having the trainee help readily available behind-the-scenes to help them. The lab Aid Service can be a great aid for your trainees as they begin to navigate the tough college admissions as well as job hunting waters.
Pupils need to be prepared to obtain used to their jobs as swiftly as possible in order to reach their major objective of getting involved in the college. They need to strive sufficient to see outcomes that will allow them to walk on at the following degree of their research studies. Getting made use of to the process of finishing their projects is extremely important.
Trainees have the ability to locate different means to help them learn how to make use of the lab Help Service. Discovering just how to utilize the lab Help Service is important to students' success in university as well as task application.
## Pay Someone To Take My Squaring The Circle Lab
Squaring The Circle is utilized in a great deal of institutions. Some teachers, nevertheless, do not utilize it really efficiently or utilize it improperly. This can have a negative impact on the trainee's learning.
So, when designating tasks, make use of an excellent Squaring The Circle aid solution to aid you with each lab. These services give a selection of practical services, consisting of:
Tasks might require a great deal of assessing and also browsing on the computer system. This is when utilizing a help service can be a fantastic benefit. It permits you to obtain more job done, boost your understanding, and stay clear of a lot of tension.
These sorts of research services are a superb means to start dealing with the very best sort of help for your demands. Squaring The Circle is one of one of the most hard based on grasp for pupils. Dealing with a solution, you can ensure that your needs are fulfilled, you are instructed properly, and also you recognize the material appropriately.
There are many ways that you can show yourself to function well with the course and succeed. Use an appropriate Squaring The Circle help solution to assist you and also obtain the job done. Squaring The Circle is among the hardest courses to discover yet it can be easily mastered with the ideal aid.
Having a homework solution also assists to boost the student's qualities. It enables you to include added credit along with raise your Grade Point Average. Getting additional credit score is commonly a massive advantage in several universities.
Trainees that don't maximize their Squaring The Circle class will wind up moving ahead of the rest of the class. Fortunately is that you can do it with a fast as well as easy service. So, if you wish to continue in your class, use a good aid service. Something to bear in mind is that if you actually wish to boost your grade level, your program job needs to get done. As long as feasible, you require to understand and also work with all your troubles. You can do this with a good help service.
One benefit of having a research service is that you can aid yourself. If you don't feel great in your capacity to do so, after that a great tutor will be able to aid you. They will certainly have the ability to solve the troubles you face as well as help you recognize them to get a far better grade.
When you finish from secondary school and also get in college, you will certainly need to work hard in order to stay ahead of the various other pupils. That means that you will certainly need to strive on your homework. Using an Squaring The Circle service can help you get it done.
Keeping your qualities up can be difficult since you usually need to study a lot as well as take a lot of tests. You do not have time to work with your grades alone. Having a good tutor can be an excellent assistance due to the fact that they can help you and also your homework out.
An aid solution can make it simpler for you to handle your Squaring The Circle course. In addition, you can learn more regarding on your own and help you prosper. Locate the most effective tutoring solution as well as you will certainly be able to take your research abilities to the following degree. | 2,530 | 12,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-25 | latest | en | 0.978939 |
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# I WAS JUST WONDERING.............
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posted on May, 8 2003 @ 03:37 PM
I WAS wondering If there SUCH thing AS 4-d OR even 5-d.I KNOW THERES 3-d but still i think that has to be more dimensions we haven't discoverd yet
posted on May, 8 2003 @ 03:40 PM
is time considered the 4th dimension?
posted on May, 8 2003 @ 04:15 PM
I always thought Time was the 4th dimension....as for the 5th....
posted on May, 8 2003 @ 04:24 PM
the infinity of space?
posted on May, 8 2003 @ 11:01 PM
Who really knows??
posted on May, 8 2003 @ 11:43 PM
Time is just a manmade unit of measurement for the distance between two actions at any one given point in space.
posted on May, 8 2003 @ 11:45 PM
1,2,&3 are simple. you can realize these.
think of 4&5 as one, rather just space-time
they say light travels in waves, photons etc. but to be a way you need a medium. vacuum of space wont work. in order to travel it vibrates upon the 5th dimension itself.
6-10
dont ask me, its in the string. infact dont even try to fathom these
dimension n - the plane of infinite dimensions, the plane where god resides.
posted on May, 9 2003 @ 02:05 AM
There is somthing else that has to be looked at here. When it comes to dimentions this has to play into effect with what we call mathmatics. Sometimes things that seems so complex can have the simplest riddle to it. Call this just speculation.
If 1+1=2 then
a 1 dimentional object + a 1 dimentional object it would seem would have to = a 2 dimentional object or show the propertys of a 2 dimentional object.
The same could be said taking a 2 dimentional object + a 2 dimentional object would have to = a 4 dimentional object by mathmatics,
But dont ask me the physics as to how such a thing can occur. By all standards life and and all things have some way of forming that in one fashion or another can be measured. This however does not suggest that using a theory like this can define dimentions. In order for that to happen a commen thread of what a cirtin dimention would look like would have to be identifed before any kind of foundation could be taken into effect for measureing dimentions.
Not to mention I would imigine the laws of physics when taken into concideration the law for every action there is a equal or oppisite reaction. There is a possiblity that we live a what appears to be a 4th dimentional universe may be that only by what we can identify as being a 4th dimentional looking universe threw a observation standpoint. From a cirtin stand point when you look at how energy ceases to exist. This also has to be fully identifed before conculsions can be drawn on what exactly from every stand point is what the universe not man conciders to be 1 single dimention
In conclusion we may never know. Who knows if we will live that long as a species to find out
Also from the for every action there is a equal but oppisite reaction define then what is a negitive dimention? Is it a dimention that exists and doesnt exist at the same time or is it a dimentional gate way to a universe of oppisite energy feilds dependent on what energy one comes from to begin with. Yet again we may never know.
Is self awareness a curse
Falcon
posted on May, 9 2003 @ 02:08 AM
Originally posted by W_O_W
I WAS wondering If there SUCH thing AS 4-d OR even 5-d.I KNOW THERES 3-d but still i think that has to be more dimensions we haven't discoverd yet
Wait until you find a stash od '___' from the 60s, you'll know what 4d and 5d are!
posted on May, 9 2003 @ 10:43 AM
Mathematically speaking, yes.
posted on May, 9 2003 @ 11:04 AM
as time does not exist, it is not the 4th dimension, however, there ARE up to 11 dimensions. the last 6 of them however (6-11) are micro-dimensions that are wrapped around a super-string (and you can go find out what that is for yourself explaining string theory in full would take a while)
posted on May, 11 2003 @ 02:20 AM
Time is the forth dimension. When scientists tried to explain things about the universe they discovered that there would have to be some thing like 13 dimensions . There could be many more but they would have to be very small since we can’t see them (only a theory).
posted on May, 11 2003 @ 03:24 AM
Imagine a plane (2 dimensional space). Imagine in this plane lives 2 dimensional creatures, all they have is a X-Axis and a Y-Axis (in our space we have an X, Y , and Z axis). In short these creatures have only length and breadth, they have no thickness. The intersection of our thrid dimension Z axis with the plane of their existence would only be a point. These creatures would not be able to visualize a third dimension since they have only two axis of orientation. In a two dimensional system, each and every point in that system can be defined by setting up a coordinate system of two axis which intersect at right angles (X,Y coordinates). In a three dimensional system, each and every point in that system can be defined by setting up a coordinate system of three axis which intersect at right angles (X,Y,Z coordinates). The creatures of our imaginary two dimensional universe cannot visually conceive of a third dimension because the third dimensional axis would extend outside the plane of their dimension.
Now imagine our two dimensional creatures' universe was mapped on a surface of a sphere. There existence is confined soley to the surface of that sphere. Their two dimensional universe is not curved in a third dimension. They cannot visualize a third dimension but can infer its physical existence by certain effects in their universe. One is the expansion of their universe (like a ballon being blown up). Another effect would be that if they traveled in one direction long enough they would return to their starting point. (I only assuming a "closed universe".) By the same token we exist in a three dimensional universe that is curved in the fourth dimension. We cannot visualize this, but can mathematically infer the properties of this universe with N-dimensional geometry. Like the creatures in our imaginary two dimensional sphere universe, if we were to travel in a space ship in one direction long enough, we would return to our starting point. Curvature of the universe in the fourth dimension is the only explanation of the expansion of the universe discovered by Hubble. The question now is the universe open or closed. I think now the current theory is that we live in a closed universe because of the amount of dark matter.
posted on May, 11 2003 @ 03:57 AM
The dimensions of space could almost be endless when one considers the phenomena called brane's. Currently a very large particle accelerator called the Large Hadron Collinder is being constructed in Geneva. This with observations of cosmic microwave backgroud radiation, may determine weather we live on a brane.
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0 | 1,682 | 6,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-34 | latest | en | 0.954885 |
http://betterlesson.com/lesson/554026/finding-solutions-to-equations | 1,477,299,292,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719547.73/warc/CC-MAIN-20161020183839-00023-ip-10-171-6-4.ec2.internal.warc.gz | 30,149,485 | 20,849 | # Finding Solutions to Equations
Unit 6: Expressions, Equations, & Inequalities
Lesson 10 of 20
## Big Idea: Are the solutions to 30 – n = 15 and n – 30 = 15 the same? Why or why not? Students apply their knowledge to finding solutions to one-step equations.
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Math, Algebra, Expressions (Algebra), substitution, 6th grade, master teacher project, equation
50 minutes
### Andrea Palmer
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Nothing to upload details close | 309 | 1,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2016-44 | longest | en | 0.845946 |
http://facebook.stackoverflow.com/questions/4057513/levenshtein-distance-algorithm-better-than-onm/4068606 | 1,369,183,852,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700984410/warc/CC-MAIN-20130516104304-00057-ip-10-60-113-184.ec2.internal.warc.gz | 96,576,202 | 14,500 | # Levenshtein Distance Algorithm better than O(n*m)?
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I have been looking for an advanced levenshtein distance algorithm, and the best I have found so far is O(n*m) where n and m are the lengths of the two strings. The reason why the algorithm is at this scale is because of space, not time, with the creation of a matrix of the two strings such as this one:
Is there a publicly-available levenshtein algorithm which is better than O(n*m)? I am not averse to looking at advanced computer science papers & research, but haven't been able to find anything. I have found one company, Exorbyte, which supposedly has built a super-advanced and super-fast Levenshtein algorithm but of course that is a trade secret. I am building an iPhone app which I would like to use the Levenshtein distance calculation. There is an objective-c implementation available, but with the limited amount of memory on iPods and iPhones, I'd like to find a better algorithm if possible.
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Are you interested in reducing the time complexity or the space complexity ? The average time complexity can be reduced O(n + d^2), where n is the length of the longer string and d is the edit distance. If you are only interested in the edit distance and not interested in reconstructing the edit sequence, you only need to keep the last two rows of the matrix in memory, so that will be order(n).
If you can afford to approximate, there are poly-logarithmic approximations.
For the O(n +d^2) algorithm look for Ukkonen's optimization or its enhancement Enhanced Ukkonen. The best approximation that I know of is this one by Andoni, Krauthgamer, Onak
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This is very helpful, thanks! – Jason Nov 1 '10 at 10:08
I use this for DNA alignment; We check for the length of the sequences first since the logic for updating the Ukkonen barrier is heavier then just calculating the whole array. Also, take a look at "Time Warps, String Edits, and Macromolecules: The Theory and Practice of Sequence Comparison" for some more details. – nlucaroni Dec 13 '10 at 22:05
The original paper for the Ukkonen Approximate String Matching Algorithm is, cs.helsinki.fi/u/ukkonen/InfCont85.PDF. – nlucaroni Dec 13 '10 at 22:07
Actually, you don't need the last two rows of the matrix. The last row, plus the previous number in the current row, is enough. Also note that implementing Levenshtein this way is significantly faster than using the full matrix, probably due to CPU caching. – larsga Jan 4 at 11:26
If you only want the threshold function - eg, to test if the distance is under a certain threshold - you can reduce the time and space complexity by only calculating the n values either side of the main diagonal in the array. You can also use Levenshtein Automata to evaluate many words against a single base word in O(n) time - and the construction of the automatons can be done in O(m) time, too.
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Look in Wiki - they have some ideas to improve this algorithm to better space complexity: | 738 | 3,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2013-20 | latest | en | 0.946541 |
https://www.quizzes.cc/metric/percentof.php?percent=674&of=614 | 1,571,294,650,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672723.50/warc/CC-MAIN-20191017045957-20191017073457-00121.warc.gz | 1,036,819,361 | 3,176 | #### What is 674 percent of 614?
How much is 674 percent of 614? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 674% of 614 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 674% of 614 = 4138.36
Calculate another percentage below. Type into inputs
Find number based on percentage
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Find percentage based on 2 numbers
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Calculating six hundred and seventy-four of six hundred and fourteen How to calculate 674% of 614? Simply divide the percent by 100 and multiply by the number. For example, 674 /100 x 614 = 4138.36 or 6.74 x 614 = 4138.36
#### How much is 674 percent of the following numbers?
674 percent of 614.01 = 413842.74 674 percent of 614.02 = 413849.48 674 percent of 614.03 = 413856.22 674 percent of 614.04 = 413862.96 674 percent of 614.05 = 413869.7 674 percent of 614.06 = 413876.44 674 percent of 614.07 = 413883.18 674 percent of 614.08 = 413889.92 674 percent of 614.09 = 413896.66 674 percent of 614.1 = 413903.4 674 percent of 614.11 = 413910.14 674 percent of 614.12 = 413916.88 674 percent of 614.13 = 413923.62 674 percent of 614.14 = 413930.36 674 percent of 614.15 = 413937.1 674 percent of 614.16 = 413943.84 674 percent of 614.17 = 413950.58 674 percent of 614.18 = 413957.32 674 percent of 614.19 = 413964.06 674 percent of 614.2 = 413970.8 674 percent of 614.21 = 413977.54 674 percent of 614.22 = 413984.28 674 percent of 614.23 = 413991.02 674 percent of 614.24 = 413997.76 674 percent of 614.25 = 414004.5
674 percent of 614.26 = 414011.24 674 percent of 614.27 = 414017.98 674 percent of 614.28 = 414024.72 674 percent of 614.29 = 414031.46 674 percent of 614.3 = 414038.2 674 percent of 614.31 = 414044.94 674 percent of 614.32 = 414051.68 674 percent of 614.33 = 414058.42 674 percent of 614.34 = 414065.16 674 percent of 614.35 = 414071.9 674 percent of 614.36 = 414078.64 674 percent of 614.37 = 414085.38 674 percent of 614.38 = 414092.12 674 percent of 614.39 = 414098.86 674 percent of 614.4 = 414105.6 674 percent of 614.41 = 414112.34 674 percent of 614.42 = 414119.08 674 percent of 614.43 = 414125.82 674 percent of 614.44 = 414132.56 674 percent of 614.45 = 414139.3 674 percent of 614.46 = 414146.04 674 percent of 614.47 = 414152.78 674 percent of 614.48 = 414159.52 674 percent of 614.49 = 414166.26 674 percent of 614.5 = 414173
674 percent of 614.51 = 414179.74 674 percent of 614.52 = 414186.48 674 percent of 614.53 = 414193.22 674 percent of 614.54 = 414199.96 674 percent of 614.55 = 414206.7 674 percent of 614.56 = 414213.44 674 percent of 614.57 = 414220.18 674 percent of 614.58 = 414226.92 674 percent of 614.59 = 414233.66 674 percent of 614.6 = 414240.4 674 percent of 614.61 = 414247.14 674 percent of 614.62 = 414253.88 674 percent of 614.63 = 414260.62 674 percent of 614.64 = 414267.36 674 percent of 614.65 = 414274.1 674 percent of 614.66 = 414280.84 674 percent of 614.67 = 414287.58 674 percent of 614.68 = 414294.32 674 percent of 614.69 = 414301.06 674 percent of 614.7 = 414307.8 674 percent of 614.71 = 414314.54 674 percent of 614.72 = 414321.28 674 percent of 614.73 = 414328.02 674 percent of 614.74 = 414334.76 674 percent of 614.75 = 414341.5
674 percent of 614.76 = 414348.24 674 percent of 614.77 = 414354.98 674 percent of 614.78 = 414361.72 674 percent of 614.79 = 414368.46 674 percent of 614.8 = 414375.2 674 percent of 614.81 = 414381.94 674 percent of 614.82 = 414388.68 674 percent of 614.83 = 414395.42 674 percent of 614.84 = 414402.16 674 percent of 614.85 = 414408.9 674 percent of 614.86 = 414415.64 674 percent of 614.87 = 414422.38 674 percent of 614.88 = 414429.12 674 percent of 614.89 = 414435.86 674 percent of 614.9 = 414442.6 674 percent of 614.91 = 414449.34 674 percent of 614.92 = 414456.08 674 percent of 614.93 = 414462.82 674 percent of 614.94 = 414469.56 674 percent of 614.95 = 414476.3 674 percent of 614.96 = 414483.04 674 percent of 614.97 = 414489.78 674 percent of 614.98 = 414496.52 674 percent of 614.99 = 414503.26 674 percent of 615 = 414510 | 1,593 | 4,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2019-43 | latest | en | 0.792419 |
http://www.mathworks.com/matlabcentral/fileexchange/34143-meshing-thin-shells-using-four-noded-elements/content/Shell%20Mesh/MeshCylindricalShell.m | 1,427,427,499,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131294307.1/warc/CC-MAIN-20150323172134-00165-ip-10-168-14-71.ec2.internal.warc.gz | 634,868,099 | 7,719 | Code covered by the BSD License
# Meshing thin shells using four noded elements
### Siva Srinivas Kolukula (view profile)
To mesh thin shells using four noded elements to use in Finite Element Analysis
```function [coordinates,nodes] = MeshCylindricalShell(Radius,theta,Height,NH,NT)
% To Mesh a thin cylindrical shell with 4 noded Elements
%--------------------------------------------------------------------------
% Code written by : Siva Srinivas Kolukula |
% Senior Research Fellow |
% Structural Mechanics Laboratory |
% Indira Gandhi Center for Atomic Research |
% India |
% E-mail : allwayzitzme@gmail.com |
%--------------------------------------------------------------------------
%--------------------------------------------------------------------------
% Purpose:
% To Mesh a thin cylindrical shell to use in FEM Analysis
% Variable Description:
% Input :
% theta - Angle of the sector to which Shell needed
% Height - Height of the Cylindrical shell
% NH - Number of Elements along Height (Number of Rings)
% NT - Number of Angular sectors
% Output :
% coordinates - The nodal coordinates of the mesh
% -----> coordinates = [node X Y Z]
% nodes - The nodal connectivity of the elements
% -----> nodes = [node1 node2......]
%--------------------------------------------------------------------------
nel = NH*NT ; % Total Number of Elements in the Mesh
nnel = 4 ; % Number of nodes per Element
% Number of points (nodes) on the Height and Angluar discretization
npH = NH+1 ;
npT = NT+1 ;
nnode = npH*npT ; % Number of nodes
% Discretizing the Height and Angle of the cylinder
nH = linspace(0,Height,npH) ;
nT = linspace(0,theta,npT)*pi/180 ;
[H T] = meshgrid(nH,nT) ;
% Convert grid to cylindrical coordintes
ZZ = H ;
%mesh(XX,YY,ZZ) ;
% To get the Nodal Connectivity Matrix
coordinates = [XX(:) YY(:) ZZ(:)] ; % Coordinates of (X,Y,Z)
NodeNo = 1:nnode ;
nodes = zeros(nel,nnel) ;
% If elements along the Height and Angular discretization are equal
if npH==npT
NodeNo = reshape(NodeNo,npT,npH);
nodes(:,1) = reshape(NodeNo(1:npH-1,1:npT-1),nel,1);
nodes(:,2) = reshape(NodeNo(2:npH,1:npT-1),nel,1);
nodes(:,3) = reshape(NodeNo(2:npH,2:npT),nel,1);
nodes(:,4) = reshape(NodeNo(1:npH-1,2:npT),nel,1);
% If the elements along the axes are different
else%if npH>npT
NodeNo = reshape(NodeNo,npT,npH);
nodes(:,1) = reshape(NodeNo(1:npT-1,1:npH-1),nel,1);
nodes(:,2) = reshape(NodeNo(2:npT,1:npH-1),nel,1);
nodes(:,3) = reshape(NodeNo(2:npT,2:npH),nel,1);
nodes(:,4) = reshape(NodeNo(1:npT-1,2:npH),nel,1);
end
%
% Plotting the Finite Element Mesh
% Initialization of the required matrices
X = zeros(nnel,nel) ;
Y = zeros(nnel,nel) ;
Z = zeros(nnel,nel) ;
% Extract X,Y,Z coordinates for the (iel)-th element
for iel = 1:nel
X(:,iel) = coordinates(nodes(iel,:),1) ;
Y(:,iel) = coordinates(nodes(iel,:),2) ;
Z(:,iel) = coordinates(nodes(iel,:),3) ;
end
% Figure
fh = figure ;
set(fh,'name','Preprocessing for FEA','numbertitle','off','color','w') ;
fill3(X,Y,Z,'w') ;
title('Finite Element Mesh of thin Cylindrical shell','HandleVisibility','off') ; | 901 | 3,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2015-14 | longest | en | 0.47089 |
https://www.chemistry-dictionary.com/definition/fahrenheit.php | 1,716,713,965,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058872.68/warc/CC-MAIN-20240526074314-20240526104314-00290.warc.gz | 579,418,391 | 4,676 | # Definition of fahrenheit
1) The temperature scale where 32 degrees is the freezing point of water and 212 degrees at 760mm Hg (sea level) is the boiling point of water. To convert from Fahrenheit to centigrade, subtract 32 and then divide by 1.8 ((F-32)/1.8). To convert from centigrade into Fahrenheit, multiply the centigrade temperature by 1.8 then add 32 to the product (C*1.8+32=F). This scale is named for G. D. Fahrenheit (1686-1736), the German physicist, who invented it and was the individual who introduced the use of mercury (Hg) instead of alcohol in thermometers. Because of the health dangers related with mercury, the use of mercury in thermometers is losing favor and once again alcohol thermometers are becoming more popular. | 184 | 746 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-22 | latest | en | 0.853034 |
https://www.varsitytutors.com/ap_physics_c_mechanics-help/interpreting-rotational-motion-and-torque-diagrams | 1,726,706,034,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651944.55/warc/CC-MAIN-20240918233405-20240919023405-00726.warc.gz | 946,079,649 | 40,889 | # AP Physics C: Mechanics : Interpreting Rotational Motion and Torque Diagrams
## Example Questions
### Example Question #1 : Interpreting Rotational Motion And Torque Diagrams
A mechanic is using a wrench to loosen and tighten screws on an engine block and wants to increase the amount of torque he puts on the screws to adjust them more easily. Which of the following steps will not help him to do so?
Using a longer wrench
Using a wrench of a different material
Pulling or pushing at an angle that is more perpendicular to the wrench
Increasing the magnitude of the force he puts on the wrench
Pulling or pushing at an angle that is less perpendicular to the wrench | 144 | 676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-38 | latest | en | 0.880295 |
http://oeis.org/A032099 | 1,619,147,227,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039626288.96/warc/CC-MAIN-20210423011010-20210423041010-00058.warc.gz | 69,977,261 | 5,054 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A032099 "BHK" (reversible, identity, unlabeled) transform of 1,2,3,4,... 0
1, 2, 5, 10, 27, 66, 181, 470, 1263, 3310, 8767, 22980, 60449, 158354, 415353, 1087690, 2849675, 7461318, 19539429, 51156950, 133944931, 350677822, 918123935, 2403693960, 6293050657, 16475457986, 43133566061 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS From Petros Hadjicostas, Aug 21 2018: (Start) Using the formulae in C. B. Bower's web link below about transforms, it can be proved that, for k >= 2, the BHK[k] transform of sequence (c(n): n >= 1), which has g.f. C(x) = Sum_{n >= 1} c(n)*x^n, has generating function B_k(x) = (1/2)*(C(x)^k - C(x^2)^{k/2}) if k is even, and B_k(x) = C(x)*B_{k-1}(x) = (C(x)/2)*(C(x)^{k-1} - C(x^2)^{(k-1)/2}) if k is odd. For k=1, Bower assumes that the BHK[k=1] transform of (c(n): n >= 1) is itself, which means that the g.f. of the output sequence is C(x). (This assumption is not accepted by all mathematicians because a sequence of length 1 not only is reversible but palindromic as well.) Since a(m) = BHK(c(n): n >= 1)(m) = Sum_{k=1..m} BHK[k](c(n): n >= 1)(m) for m = 1,2,3,..., it can be easily proved (using sums of infinite geometric series) that the g.f. of (a(m): m >= 1) = BHK(c(n): n >= 1) is A(x) = (C(x)^2 - C(x^2))/(2*(1-C(x))*(1-C(x^2))) + C(x). (The extra C(x) is due of course to the special assumption made for the BHK[k=1] transform.) Here, BHK(c(n): n >= 1)(m) indicates the m-th element of the output sequence when the transform is BHK and the input sequence is (c(n): n >= 1). Similarly, BHK[k](c(n): n >= 1)(m) indicates the m-th element of the output sequence when the transform is BHK[k] (i.e., with k boxes) and the input sequence is (c(n): n >= 1). For the current sequence, c(n) = n for all n >= 1, and thus, C(x) = x/(1-x)^2. Substituting into the above formula for A(x), and doing the algebra, we get A(x) = x*(1-3*x+5*x^3-3*x^5+x^6)/((1-x)^2*(1-x-x^2)*(1+x-x^2)*(1-3*x+x^2)), which is the g.f. given by Christian G. Bower below. Using partial fraction decomposition, we get A(x) = (x/(1-x)^2)+ (1/2)*(x/(x^2-3*x+1) + (2*x+1)/(1+x-x^2) - (x+1)/(1-x-x^2)), from which we can prove the formula for a(n) below. (End) LINKS C. G. Bower, Transforms (2) FORMULA G.f.: x*(1-3x+5x^3-3x^5+x^6)/((1-x)^2(1-x-x^2)(1+x-x^2)(1-3x+x^2)). a(n) = n + (1/2)*(Fibonacci(2*n) + (-1)^(n+1)*Fibonacci(n-2) - Fibonacci(n+2)) for n >= 2. - Petros Hadjicostas, Aug 21 2018 EXAMPLE From Petros Hadjicostas, Aug 21 2018: (Start) According to C. G. Bower, in his website above, we have boxes of different colors and sizes (the size of the box is determined by the number of balls it can hold). Since c(n) = n, each box of size n can have one of n colors, say A, B, etc. (The colors for a box of size n are not necessarily the same as the colors for a box of size m <> n, but, for simplicity, we denote the colors by A, B, etc. for all box sizes.) Then a(n) = BHK(c(n): n >= 1)(n) = number of ways we can have boxes on a line such that the total number of balls is n and the array of boxes is reversible but not palindromic (with the exception when having only one box on the line). Hence, for n=1, there is only a(1) = 1 possible array, which is 1_A (since no other colors are allowed for a box of size 1). For n=2, the a(2) = 2 possible arrays for the boxes are: 2_A and 2_B (since a single box by itself is not consider a palindrome by Bower). For n=3, the a(3) = 5 possible arrays for the boxes are: 3_A, 3_B, 3_C, 1+2_A, 1+2_B. (Here, 1+2_A = 2_A+1 and 1+2_B = 2_B+1 since the arrays are reversible. Also, by writing 1, we mean 1_A.) For n=4, the a(4) = 10 possible arrays for the boxes are: 4_A, 4_B, 4_C, 4_D, 1+3_A, 1+3_B, 1+3_C, 2_A+2_B, 1+1+2_A, 1+1+2_B. (Note that 2_A+2_B = 2_B+2_A, 1+1+2_A = 2_A+1+1, and 1+1+2_B = 2_B+1+1. The arrays 1+2_A+1 and 1+2_B+1 are not allowed because they are palindromic.) (End) CROSSREFS Cf. A000045, A001906. Sequence in context: A053391 A183956 A221843 * A243797 A120896 A074801 Adjacent sequences: A032096 A032097 A032098 * A032100 A032101 A032102 KEYWORD nonn AUTHOR STATUS approved
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Last modified April 22 22:19 EDT 2021. Contains 343197 sequences. (Running on oeis4.) | 1,694 | 4,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-17 | latest | en | 0.784543 |
https://www.sidefx.com/docs/houdini/vellum/mass_thickness.html | 1,716,488,784,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058653.47/warc/CC-MAIN-20240523173456-20240523203456-00489.warc.gz | 873,300,461 | 4,623 | # Calculating mass and thickness
In Houdini 18 and later, the default behavior is to automatically calculate the mass and thickness of the Vellum geometry. This is means that as you change the resolution of your Vellum geometry, the behavior of the simulation shouldn’t change very much. This is particularly relevant to bend stiffness, since a heavier cloth will bend and wrinkle much differently than a very light cloth. This new method uses the length of a polyline, the area of a triangle, or the volume of a tetrahedron and the density of the material to compute the mass. It will also use the edge length to compute a thickness value, so that the points won’t overlap. Both of these values can be scaled by point attributes.
In Houdini 17.5 and earlier, the Mass was Set Uniform by default. In Houdini 18, you now have the option to Calculate Uniform, which will find the mass of the entire object and split it evenly among all of the points. However if you have an object that has a lot of varying mass per point (areas of higher resolution), splitting mass evenly could create artifacts. The new default behavior is to Calculate Varying, which will give each point a mass comparative to how much area is around it. Points in very dense areas become lighter, and points in really sparse areas become heavier.
Thickness has the same options as the Mass parameters. Calculate Uniform determines the thickness based on the average length of the edges in each piece, scaled by the Edge Length Scale parameter. This setting generally avoids excess self-collisions while maintaining uniform thickness. This is the default, because when thickness is the same over the whole object it will lie flat, which is what cloth is expected to do. If there is varying thickness in a piece of cloth, where densely sampled areas are thinner, the cloth wouldn’t lie flat. You can still choose Calculate Varying which automatically shrinks the `pscale` in areas with dense triangles and is larger in places with less triangles. This is useful for characters since it is likely that you will have highly varying resolutions where the pscales will overlap. For example, the area around the eyes. There is a checkbox called Visualize Thickness, which allows you to visualize the `pscale` without sending it through the solver.
Visualization of `pscale` when Thickness is set to Calculate Uniform in a dense area.
Visualization of `pscale` when Thickness is set to Calculate Varying in a dense area. | 511 | 2,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.943741 |
http://docscn.studygolang.com/src/math/gamma.go | 1,702,274,814,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103558.93/warc/CC-MAIN-20231211045204-20231211075204-00675.warc.gz | 11,542,387 | 6,312 | Black Lives Matter. Support the Equal Justice Initiative.
# Source file src/math/gamma.go
## Documentation: math
``` 1 // Copyright 2010 The Go Authors. All rights reserved.
2 // Use of this source code is governed by a BSD-style
4
5 package math
6
7 // The original C code, the long comment, and the constants
8 // below are from http://netlib.sandia.gov/cephes/cprob/gamma.c.
9 // The go code is a simplified version of the original C.
10 //
11 // tgamma.c
12 //
13 // Gamma function
14 //
15 // SYNOPSIS:
16 //
17 // double x, y, tgamma();
18 // extern int signgam;
19 //
20 // y = tgamma( x );
21 //
22 // DESCRIPTION:
23 //
24 // Returns gamma function of the argument. The result is
25 // correctly signed, and the sign (+1 or -1) is also
26 // returned in a global (extern) variable named signgam.
27 // This variable is also filled in by the logarithmic gamma
28 // function lgamma().
29 //
30 // Arguments |x| <= 34 are reduced by recurrence and the function
31 // approximated by a rational function of degree 6/7 in the
32 // interval (2,3). Large arguments are handled by Stirling's
33 // formula. Large negative arguments are made positive using
34 // a reflection formula.
35 //
36 // ACCURACY:
37 //
38 // Relative error:
39 // arithmetic domain # trials peak rms
40 // DEC -34, 34 10000 1.3e-16 2.5e-17
41 // IEEE -170,-33 20000 2.3e-15 3.3e-16
42 // IEEE -33, 33 20000 9.4e-16 2.2e-16
43 // IEEE 33, 171.6 20000 2.3e-15 3.2e-16
44 //
45 // Error for arguments outside the test range will be larger
46 // owing to error amplification by the exponential function.
47 //
48 // Cephes Math Library Release 2.8: June, 2000
49 // Copyright 1984, 1987, 1989, 1992, 2000 by Stephen L. Moshier
50 //
51 // The readme file at http://netlib.sandia.gov/cephes/ says:
52 // Some software in this archive may be from the book _Methods and
53 // Programs for Mathematical Functions_ (Prentice-Hall or Simon & Schuster
54 // International, 1989) or from the Cephes Mathematical Library, a
55 // commercial product. In either event, it is copyrighted by the author.
56 // What you see here may be used freely but it comes with no support or
57 // guarantee.
58 //
59 // The two known misprints in the book are repaired here in the
60 // source listings for the gamma function and the incomplete beta
61 // integral.
62 //
63 // Stephen L. Moshier
64 // moshier@na-net.ornl.gov
65
66 var _gamP = [...]float64{
67 1.60119522476751861407e-04,
68 1.19135147006586384913e-03,
69 1.04213797561761569935e-02,
70 4.76367800457137231464e-02,
71 2.07448227648435975150e-01,
72 4.94214826801497100753e-01,
73 9.99999999999999996796e-01,
74 }
75 var _gamQ = [...]float64{
76 -2.31581873324120129819e-05,
77 5.39605580493303397842e-04,
78 -4.45641913851797240494e-03,
79 1.18139785222060435552e-02,
80 3.58236398605498653373e-02,
81 -2.34591795718243348568e-01,
82 7.14304917030273074085e-02,
83 1.00000000000000000320e+00,
84 }
85 var _gamS = [...]float64{
86 7.87311395793093628397e-04,
87 -2.29549961613378126380e-04,
88 -2.68132617805781232825e-03,
89 3.47222221605458667310e-03,
90 8.33333333333482257126e-02,
91 }
92
93 // Gamma function computed by Stirling's formula.
94 // The pair of results must be multiplied together to get the actual answer.
95 // The multiplication is left to the caller so that, if careful, the caller can avoid
96 // infinity for 172 <= x <= 180.
97 // The polynomial is valid for 33 <= x <= 172; larger values are only used
98 // in reciprocal and produce denormalized floats. The lower precision there
99 // masks any imprecision in the polynomial.
100 func stirling(x float64) (float64, float64) {
101 if x > 200 {
102 return Inf(1), 1
103 }
104 const (
105 SqrtTwoPi = 2.506628274631000502417
106 MaxStirling = 143.01608
107 )
108 w := 1 / x
109 w = 1 + w*((((_gamS[0]*w+_gamS[1])*w+_gamS[2])*w+_gamS[3])*w+_gamS[4])
110 y1 := Exp(x)
111 y2 := 1.0
112 if x > MaxStirling { // avoid Pow() overflow
113 v := Pow(x, 0.5*x-0.25)
114 y1, y2 = v, v/y1
115 } else {
116 y1 = Pow(x, x-0.5) / y1
117 }
118 return y1, SqrtTwoPi * w * y2
119 }
120
121 // Gamma returns the Gamma function of x.
122 //
123 // Special cases are:
124 // Gamma(+Inf) = +Inf
125 // Gamma(+0) = +Inf
126 // Gamma(-0) = -Inf
127 // Gamma(x) = NaN for integer x < 0
128 // Gamma(-Inf) = NaN
129 // Gamma(NaN) = NaN
130 func Gamma(x float64) float64 {
131 const Euler = 0.57721566490153286060651209008240243104215933593992 // A001620
132 // special cases
133 switch {
134 case isNegInt(x) || IsInf(x, -1) || IsNaN(x):
135 return NaN()
136 case IsInf(x, 1):
137 return Inf(1)
138 case x == 0:
139 if Signbit(x) {
140 return Inf(-1)
141 }
142 return Inf(1)
143 }
144 q := Abs(x)
145 p := Floor(q)
146 if q > 33 {
147 if x >= 0 {
148 y1, y2 := stirling(x)
149 return y1 * y2
150 }
151 // Note: x is negative but (checked above) not a negative integer,
152 // so x must be small enough to be in range for conversion to int64.
153 // If |x| were >= 2⁶³ it would have to be an integer.
154 signgam := 1
155 if ip := int64(p); ip&1 == 0 {
156 signgam = -1
157 }
158 z := q - p
159 if z > 0.5 {
160 p = p + 1
161 z = q - p
162 }
163 z = q * Sin(Pi*z)
164 if z == 0 {
165 return Inf(signgam)
166 }
167 sq1, sq2 := stirling(q)
168 absz := Abs(z)
169 d := absz * sq1 * sq2
170 if IsInf(d, 0) {
171 z = Pi / absz / sq1 / sq2
172 } else {
173 z = Pi / d
174 }
175 return float64(signgam) * z
176 }
177
178 // Reduce argument
179 z := 1.0
180 for x >= 3 {
181 x = x - 1
182 z = z * x
183 }
184 for x < 0 {
185 if x > -1e-09 {
186 goto small
187 }
188 z = z / x
189 x = x + 1
190 }
191 for x < 2 {
192 if x < 1e-09 {
193 goto small
194 }
195 z = z / x
196 x = x + 1
197 }
198
199 if x == 2 {
200 return z
201 }
202
203 x = x - 2
204 p = (((((x*_gamP[0]+_gamP[1])*x+_gamP[2])*x+_gamP[3])*x+_gamP[4])*x+_gamP[5])*x + _gamP[6]
205 q = ((((((x*_gamQ[0]+_gamQ[1])*x+_gamQ[2])*x+_gamQ[3])*x+_gamQ[4])*x+_gamQ[5])*x+_gamQ[6])*x + _gamQ[7]
206 return z * p / q
207
208 small:
209 if x == 0 {
210 return Inf(1)
211 }
212 return z / ((1 + Euler*x) * x)
213 }
214
215 func isNegInt(x float64) bool {
216 if x < 0 {
217 _, xf := Modf(x)
218 return xf == 0
219 }
220 return false
221 }
222
```
View as plain text | 2,607 | 6,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-50 | latest | en | 0.462315 |
https://puzzle.queryhome.com/1127/why-did-the-woman-get-the-loan | 1,506,242,246,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689900.91/warc/CC-MAIN-20170924081752-20170924101752-00498.warc.gz | 698,436,573 | 31,684 | # Why did the woman get the loan?
0 votes
108 views
A woman goes into a bank in Dallas and asks for a \$2,000 loan for an upcoming trip she has planned to Europe. The loan officer tells her she will need some thing to deposit if she wants the loan.
She tells the officer "I'll leave my car, it's worth \$130,000." The bank takes the car for the loan, laughing at her for leaving such expensive thing.
A month later when the woman comes back she pays off her loan plus interest, costing her \$2,040. The bank manager smirks at her and asks "While you were gone we found out you're very wealthy. Why would you get such a small loan and leave such expensive thing?"
She tells him why and than he realizes she's not as dumb as they thought. Why did she get the loan?
posted Jun 5, 2014
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## 2 Solutions
+1 vote
She was able to park her car for 1 month in a safe place with only 40\$ cost.
solution Jun 17, 2014
0 votes
She parked her car in safe place with only \$40
solution Jun 17, 2014 by anonymous
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What is the maximum number of coins that pirate A might get? | 578 | 2,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-39 | latest | en | 0.983572 |
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