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https://www.ademcetinkaya.com/2022/09/can-neural-networks-predict-stock_53.html | 1,713,252,228,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817073.16/warc/CC-MAIN-20240416062523-20240416092523-00329.warc.gz | 585,819,844 | 59,825 | Time series forecasting has been widely used to determine the future prices of stock, and the analysis and modeling of finance time series importantly guide investors' decisions and trades. In addition, in a dynamic environment such as the stock market, the nonlinearity of the time series is pronounced, immediately affecting the efficacy of stock price forecasts. Thus, this paper proposes an intelligent time series prediction system that uses sliding-window metaheuristic optimization for the purpose of predicting the stock prices. We evaluate Prism Johnson Limited prediction models with Reinforcement Machine Learning (ML) and Spearman Correlation1,2,3,4 and conclude that the NSE PRSMJOHNSN stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period: The dominant strategy among neural network is to Hold NSE PRSMJOHNSN stock.
Keywords: NSE PRSMJOHNSN, Prism Johnson Limited, stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures.
## Key Points
1. Prediction Modeling
2. Can statistics predict the future?
3. Technical Analysis with Algorithmic Trading
## NSE PRSMJOHNSN Target Price Prediction Modeling Methodology
This paper tries to address the problem of stock market prediction leveraging artificial intelligence (AI) strategies. The stock market prediction can be modeled based on two principal analyses called technical and fundamental. In the technical analysis approach, the regression machine learning (ML) algorithms are employed to predict the stock price trend at the end of a business day based on the historical price data. In contrast, in the fundamental analysis, the classification ML algorithms are applied to classify the public sentiment based on news and social media. We consider Prism Johnson Limited Stock Decision Process with Spearman Correlation where A is the set of discrete actions of NSE PRSMJOHNSN stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Spearman Correlation)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Reinforcement Machine Learning (ML)) X S(n):→ (n+4 weeks) $\begin{array}{l}\int {r}^{s}\mathrm{rs}\end{array}$
n:Time series to forecast
p:Price signals of NSE PRSMJOHNSN stock
j:Nash equilibria
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## NSE PRSMJOHNSN Stock Forecast (Buy or Sell) for (n+4 weeks)
Sample Set: Neural Network
Stock/Index: NSE PRSMJOHNSN Prism Johnson Limited
Time series to forecast n: 26 Sep 2022 for (n+4 weeks)
According to price forecasts for (n+4 weeks) period: The dominant strategy among neural network is to Hold NSE PRSMJOHNSN stock.
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Yellow to Green): *Technical Analysis%
## Conclusions
Prism Johnson Limited assigned short-term B2 & long-term Ba1 forecasted stock rating. We evaluate the prediction models Reinforcement Machine Learning (ML) with Spearman Correlation1,2,3,4 and conclude that the NSE PRSMJOHNSN stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period: The dominant strategy among neural network is to Hold NSE PRSMJOHNSN stock.
### Financial State Forecast for NSE PRSMJOHNSN Stock Options & Futures
Rating Short-Term Long-Term Senior
Outlook*B2Ba1
Operational Risk 7269
Market Risk3757
Technical Analysis8663
Fundamental Analysis3684
Risk Unsystematic4080
### Prediction Confidence Score
Trust metric by Neural Network: 75 out of 100 with 705 signals.
## References
1. Breiman L. 1993. Better subset selection using the non-negative garotte. Tech. Rep., Univ. Calif., Berkeley
2. E. Collins. Using Markov decision processes to optimize a nonlinear functional of the final distribution, with manufacturing applications. In Stochastic Modelling in Innovative Manufacturing, pages 30–45. Springer, 1997
3. Matzkin RL. 1994. Restrictions of economic theory in nonparametric methods. In Handbook of Econometrics, Vol. 4, ed. R Engle, D McFadden, pp. 2523–58. Amsterdam: Elsevier
4. Mnih A, Teh YW. 2012. A fast and simple algorithm for training neural probabilistic language models. In Proceedings of the 29th International Conference on Machine Learning, pp. 419–26. La Jolla, CA: Int. Mach. Learn. Soc.
5. E. Altman, K. Avrachenkov, and R. N ́u ̃nez-Queija. Perturbation analysis for denumerable Markov chains with application to queueing models. Advances in Applied Probability, pages 839–853, 2004
6. Efron B, Hastie T. 2016. Computer Age Statistical Inference, Vol. 5. Cambridge, UK: Cambridge Univ. Press
7. Keane MP. 2013. Panel data discrete choice models of consumer demand. In The Oxford Handbook of Panel Data, ed. BH Baltagi, pp. 54–102. Oxford, UK: Oxford Univ. Press
Frequently Asked QuestionsQ: What is the prediction methodology for NSE PRSMJOHNSN stock?
A: NSE PRSMJOHNSN stock prediction methodology: We evaluate the prediction models Reinforcement Machine Learning (ML) and Spearman Correlation
Q: Is NSE PRSMJOHNSN stock a buy or sell?
A: The dominant strategy among neural network is to Hold NSE PRSMJOHNSN Stock.
Q: Is Prism Johnson Limited stock a good investment?
A: The consensus rating for Prism Johnson Limited is Hold and assigned short-term B2 & long-term Ba1 forecasted stock rating.
Q: What is the consensus rating of NSE PRSMJOHNSN stock?
A: The consensus rating for NSE PRSMJOHNSN is Hold.
Q: What is the prediction period for NSE PRSMJOHNSN stock?
A: The prediction period for NSE PRSMJOHNSN is (n+4 weeks) | 1,571 | 6,160 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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-2024-18 | latest | en | 0.901182 |
http://research.stlouisfed.org/fred2/graph/?g=dgW | 1,429,343,373,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246633972.52/warc/CC-MAIN-20150417045713-00057-ip-10-235-10-82.ec2.internal.warc.gz | 218,620,727 | 18,687 | # FRED Graph
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
Release:
Restore defaults | Save settings | Apply saved settings
w h
Graph Background: Plot Background: Text:
Color:
(a) Civilian Labor Force Participation Rate, Percent, Seasonally Adjusted (CIVPART)
The series comes from the 'Current Population Survey (Household Survey)'
The source code is: LNS11300000
Civilian Labor Force Participation Rate
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Color:
(a) Civilian Employment-Population Ratio, Percent, Seasonally Adjusted (EMRATIO)
The series comes from the 'Current Population Survey (Household Survey)'
The source code is: LNS12300000
Civilian Employment-Population Ratio
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Graph Data
Graph Image
Suggested Citation
US. Bureau of Labor Statistics, Civilian Labor Force Participation Rate [CIVPART], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/CIVPART/, April 18, 2015.
US. Bureau of Labor Statistics, Civilian Employment-Population Ratio [EMRATIO], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/EMRATIO/, April 18, 2015.
Retrieving data.
Graph updated.
#### Recently Viewed Series
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Name: Email: | 725 | 2,857 | {"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} | 2.953125 | 3 | CC-MAIN-2015-18 | latest | en | 0.792666 |
http://macosx.gnu-darwin.org/ProgramDocuments/NTL/tour-ex6.html.body | 1,596,995,175,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738562.5/warc/CC-MAIN-20200809162458-20200809192458-00486.warc.gz | 70,028,953 | 1,646 | A Tour of NTL: Examples: Floating Point Classes
# A Tour of NTL: Examples: Floating Point Classes
NTL also supports arbitrary precision floating point with the class RR. Additionally, it supports two specialized classes: quad_float, which gives a form of quadruple precision, but without an extended exponent range, and xdouble, which gives double precision, but with an extended exponent range. The advantage of the latter two classes is efficiency.
Here again is a program that reads a list of numbers from the input, and outputs the sum of their squares, using the class RR.
```#include <NTL/RR.h>
int main()
{
RR acc, val;
acc = 0;
while (SkipWhiteSpace(cin)) {
cin >> val;
acc += val*val;
}
cout << acc << "\n";
}
```
The precision used for the computation can be set by executing
``` RR::SetPrecision(p);
```
which sets the effective precision to p bits. By default, p=150. All of the basic arithmetic operations compute their results by rounding to the nearest p-bit floating point number. The semantics of this are exactly the same as in the IEEE floating point standard (except that there are no special values, like "infinity" and "not a number").
The number of decimal digits of precision that are used when printing an RR can be set be executing
``` RR::SetOutputPrecision(d);
```
which sets the output precision to d. By default, d=10.
See RR.txt for details.
By replacing the occurences of RR by either quad_float or xdouble, one gets an equivalent program using one of the other floating point classes. The output precision for these two classes can be controlled just as with RR. See quad_float.txt and xdouble.txt for details. | 372 | 1,660 | {"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} | 2.78125 | 3 | CC-MAIN-2020-34 | latest | en | 0.856304 |
https://math.stackexchange.com/questions/1215095/for-what-values-of-alpha-does-1-alpha-does-1-alpha-1-complex-num | 1,566,264,967,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315174.57/warc/CC-MAIN-20190820003509-20190820025509-00367.warc.gz | 543,040,175 | 31,252 | # For what values of $\alpha$ does $1^{\alpha}$ does $1^{\alpha} = 1$. complex numbers
For what values of $\alpha$ does $1^{\alpha}$ does $1^{\alpha} = 1$. What are the possible values of $1^{\alpha}$? What are the values of $1^{\frac{1}{2}}$? (Hint: use the definition of $z^{\alpha}$.)
Attempt: Recall the definition of a complex $\alpha$ constant where $z \neq 0$, then $z^{\alpha} = e^{\alpha \log z}$.
Then, the possible values for $1^{\alpha}$ using the definition are: $1^{\alpha} = e^{\alpha \log 1} = e^{\alpha [\log 1 + i\arg 1]} = e^{\alpha i 2 k\pi }$.
And when $\alpha = \frac{1}{2}$ we have $1^ {\frac {1}{2}} = e^{\frac{1}{2} i 2 k\pi } = e^{ik\pi}$.
I don't know how to continue.
I dont know for what values of $\alpha$ does $1^{\alpha}$ does $1^{\alpha} = 1$. Can someone please help me? I would really appreciate it. Thank you in advance.
• How are you defining $\log z$? Is it a single-valued function or multi-valued? If single-valued with $\log 1=0$ then it is true for all values $\alpha.$ If multivalued, then are you asking when all values of $1^{\alpha}$ are $1$? Because if multivalued, then $1$ is always one of the values of $1^{\alpha}$. – Thomas Andrews Apr 1 '15 at 0:29
Hint: You want $e^{2\pi k \alpha i} = 1$. What values of $z$ give you $e^{iz} = 1$? You'll want to distinguish between the cases $k = 0$ and $k \ne 0$.
• when $z = \pi$ we have $e^{i\pi} = 1$ . And when $k = 0$ we have $e^{2\pi k\alpha i} = e^0 = 1$ When k is not equal to zero, then when $k = 2\pi$ or any multiple of $2\pi$ it will be equal to 1? – Mahidevran Apr 1 '15 at 0:27
• @Mahidevran $e^{i\pi}=-1$, not $1$. – AlexR Apr 1 '15 at 0:29
• When $z$ is an integer multiple of $2\pi$ you get $1$. So you want $k\alpha$ to be ... Here $k$ is an integer, it's not going to be $2\pi$. – Robert Israel Apr 1 '15 at 2:49
• No, it doesn't. Any integer will work if $1$ does. – Robert Israel Apr 1 '15 at 6:51 | 703 | 1,916 | {"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} | 4.09375 | 4 | CC-MAIN-2019-35 | latest | en | 0.768955 |
https://mathexamination.com/course/null-set,-negligible-set.php | 1,606,654,598,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141198409.43/warc/CC-MAIN-20201129123729-20201129153729-00345.warc.gz | 391,747,061 | 7,763 | ## Take My Null Set, Negligible Set Course
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https://studysoup.com/tsg/913194/engineering-mechanics-statics-8-edition-chapter-2-5-problem-2-61 | 1,604,194,761,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922746.99/warc/CC-MAIN-20201101001251-20201101031251-00070.warc.gz | 517,072,384 | 11,583 | ×
×
# The indicated forcecouple system is applied to a small shaft at the center of the plate
ISBN: 9781118807330 401
## Solution for problem 2/61 Chapter 2/5
Engineering Mechanics: Statics | 8th Edition
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Problem 2/61
The indicated forcecouple system is applied to a small shaft at the center of the plate. Replace this system by a single force and specify the coordinate of the point on the x-axis through which the line of action of this resultant force passes.
Step-by-Step Solution:
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##### ISBN: 9781118807330
This full solution covers the following key subjects: . This expansive textbook survival guide covers 33 chapters, and 919 solutions. The answer to “The indicated forcecouple system is applied to a small shaft at the center of the plate. Replace this system by a single force and specify the coordinate of the point on the x-axis through which the line of action of this resultant force passes.” is broken down into a number of easy to follow steps, and 44 words. Engineering Mechanics: Statics was written by and is associated to the ISBN: 9781118807330. Since the solution to 2/61 from 2/5 chapter was answered, more than 227 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 2/61 from chapter: 2/5 was answered by , our top Engineering and Tech solution expert on 03/14/18, 04:44PM. This textbook survival guide was created for the textbook: Engineering Mechanics: Statics, edition: 8.
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# If (ā7ā24i)1/2=x-iy, then x2+y2=
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Solution
## The correct option is B 25 √−7−24i=x-iy Squaring both sides, -7-24i=x2-y2-i(2xy) Equating real and imaginary parts,we get x2-y2=-7 And 2xy=24 ∴ x2+y2= √49+576=√625=25
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# Four women and three men must be seated in a row for a group photo
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?
A) 240
B) 480
C) 720
D) 1440
E) 5640
*kudos for all correct solutions
Hi
_ M _ M _ M _
" No two men can sit next to each other" - we need to place woman between each man. This can be done in 4C4 ways. In this particular case we can say that they need to alternate.
We can arrange 4 women in 4! ways and 3 men in 3! ways.
Total # of sitting arrangements:
4C4 * 4! * 3! = 24 * 6 = 144
Am I missing something?
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
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Top Contributor
vitaliyGMAT wrote:
GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?
A) 240
B) 480
C) 720
D) 1440
E) 5640
*kudos for all correct solutions
Hi
_ M _ M _ M _
" No two men can sit next to each other" - we need to place woman between each man. This can be done in 4C4 ways. In this particular case we can say that they need to alternate.
We can arrange 4 women in 4! ways and 3 men in 3! ways.
Total # of sitting arrangements:
4C4 * 4! * 3! = 24 * 6 = 144
Am I missing something?
Your solution only allows for one configuration: W M W M W M W
However, we can also have W W M W M W M or W M W W M W M, etc.
Cheers,
Brent
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
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Kudos
GMATPrepNow wrote:
vitaliyGMAT wrote:
GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?
A) 240
B) 480
C) 720
D) 1440
E) 5640
*kudos for all correct solutions
Hi
_ M _ M _ M _
" No two men can sit next to each other" - we need to place woman between each man. This can be done in 4C4 ways. In this particular case we can say that they need to alternate.
We can arrange 4 women in 4! ways and 3 men in 3! ways.
Total # of sitting arrangements:
4C4 * 4! * 3! = 24 * 6 = 144
Am I missing something?
Your solution only allows for one configuration: W M W M W M W
However, we can also have W W M W M W M or W M W W M W M, etc.
Cheers,
Brent
Agrrrr...
I've chosen wrong gaps! Thanks.
We should choose seats between women.
Final solution:
5C3*4!*3! = 1440.
Good question.
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
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Kudos
try arrange women first : 4 women could get arrange 4! ways and they will create 5 gaps among themselves. Now, we have 5 gaps and 3 males could be seated on those gaps in 5p3 ways. Therefore, total arrangements would be 4!*5*4*3=1440 -> option C
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
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Kudos
GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?
A) 240
B) 480
C) 720
D) 1440
E) 5640
*kudos for all correct solutions
no two men can sit together so, we have
_W_W_W_W_
SO Women can sit in 4! ways
3 men have 5 spaces to sit so: (5C3) * 3!(sort among themselves)
TOTAL = 4!* 3! * 5C2 = 1440
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
1
Kudos
_W_W_W_W_
4 women = 4! = 24
3 men = 3! = 6
24*6 = 144
now out of 5 spaces any 3 coz 3 men
5c3 = 5c2 = 10
144*10 = 1440
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
Can someone help me understand the assumption that there are 5 chairs available after the 4 women have been seated? Seems to me that this is an insufficiently detailed question prompt. Wouldn't a more logical assumption be that there are 3 remaining chairs after the 4 women have been seated for a total of 7 chairs - the same as the number of people?
Thanks.
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Four women and three men must be seated in a row for a group photo [#permalink]
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Kudos
Shazriki wrote:
Can someone help me understand the assumption that there are 5 chairs available after the 4 women have been seated? Seems to me that this is an insufficiently detailed question prompt. Wouldn't a more logical assumption be that there are 3 remaining chairs after the 4 women have been seated for a total of 7 chairs - the same as the number of people?
Thanks.
Hi there ,
Kindly visualise this W_W _W_W
let's name these 3 spaces 1,2,3 from left to right
i.e w 1 w 2 w 3 w
but what if the person chooses to sit like this
1ww2w3
1w2ww3w
though it satisfies the initial condition , what you have assumed fails in such cases
hence to avoid this confusion , we arrange 4 women _W_W_W_W_
please note that the _ represents a space and 3 spaces out of these 5 spaces can be chosen. w1w2W3w is not the only combination , there would be other possibilities as shown above
lets say you are seated and there is a possibility that a person can sit at either side of you viz. Left/ right
i hope this is clear
kudos if this helps
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?
A) 240
B) 480
C) 720
D) 1440
E) 5640
*kudos for all correct solutions
Let three men be A,B,C
total number of ways =
4 w & A = 5! x4
4 w & B =5! x 4
4 w & C = 5! x 4 = 3(5!x4)= 1440
multiplied by 4 : becaause the man can sit with any of the 4 women.
is my approach correct?
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
1
Kudos
first let sit 4 women.
This can be done in 4! ways = 4*3*2*1=24 ways
among 4 women , there are 5 spaces
3 men can sit in these 5 sits in 5C3* 3!ways = 10 *3*2*1=60 ways
total ways =24*60=1440
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
Top Contributor
hanyhamdani wrote:
GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?
A) 240
B) 480
C) 720
D) 1440
E) 5640
*kudos for all correct solutions
Let three men be A,B,C
total number of ways =
4 w & A = 5! x 4
4 w & B =5! x 4
4 w & C = 5! x 4 = 3(5!x4)= 1440
multiplied by 4 : becaause the man can sit with any of the 4 women.
is my approach correct?
It's hard to tell whether your approach is correct.
What does "4 w & A = 5! x 4" represent?
Cheers,
Brent
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
Someone please explain where I have gone wrong -
1) Unconstrained way of arranging 4W & 3M -> 7!
2) Now, to formulate scenarios when no men can sit together, I have grouped them. Hence now we have 5 entities 4W & 1M (group). Also the men can be arranged 3! ways.
3) So the final answer is -> 7! - 5!3!
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Four women and three men must be seated in a row for a group photo [#permalink]
GMATPrepNow wrote:
GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?
A) 240
B) 480
C) 720
D) 1440
E) 5640
*kudos for all correct solutions
Take the task of arranging the 7 peopl and break it into stages.
Stage 1: Arrange the 4 women in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 4 women in 4! ways (= 24 ways)
So, we can complete stage 1 in 24 ways
IMPORTANT: For each arrangement of 4 women, there are 5 spaces where the 3 men can be placed.
If we let W represent each woman, we can add the spaces as follows: _W_W_W_W_
So, if we place the men in 3 of the available spaces, we can ENSURE that two men are never seated together.
Let's let A, B and C represent the 3 men.
Stage 2: Place man A in an available space.
There are 5 spaces, so we can complete stage 2 in 5 ways.
Stage 3: Place man B in an available space.
There are 4 spaces remaining, so we can complete stage 3 in 4 ways.
Stage 4: Place man C in an available space.
There are 3 spaces remaining, so we can complete stage 4 in 3 ways.
By the Fundamental Counting Principle (FCP), we can complete the 4 stages (and thus seat all 7 people) in (24)(5)(4)(3) ways (= 1440 ways)
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So be sure to learn this technique.
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Hi GMATPrepNow
So what i did was i did the total- not possible. Total being 7!=5040
Not possible : 3 Men M1,M2,M3
1st condition M1 and M2 not sit together : 1440
2nd condition M2 and M3:1440
3rd condition M3 and M1 :1440
Subtracting i got 720. Please tell me where i got my logic wrong
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
Consider the arrangement *W*W*W*W*
The three men can be placed in any 3 of the 5 *s to ensure that no 2 men sit next to each other
No of ways of choosing 3 * out of 5 = 5C3
The three men can arrange themselves in 3! ways and similarly the four women can arrange themselves in 4! ways
Therefore, the total number of ways in which the seven can sit with no two men next to each other = 5C3*3!*4! = 10*6*24 = 1440
Hit Kudos if this helped!
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
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AlN wrote:
Hi GMATPrepNow
So what i did was i did the total- not possible. Total being 7!=5040
Not possible : 3 Men M1,M2,M3
1st condition M1 and M2 not sit together : 1440
2nd condition M2 and M3:1440
3rd condition M3 and M1 :1440
Subtracting i got 720. Please tell me where i got my logic wrong
How are you calculating the 3 values (of 1440)?
Also, if 5040 = the total number of arrangements
And if 3(1440) = the number of arrangements in which the men are NOT together, then 5040 - 3(1440) = the number of arrangements in which the men ARE together (and the question says the men CANNOT be together)
Cheers,
Brent
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
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Hi,
First of all without restrictions we have 7! ways to arrange 7 people (5,040)
Now let’s find out the number of ways in which at least 2 men are sitting together and subtract it from 7!
1) Case in which all 3 men are sitting together : we glue them and consider them as one single entity we get 5! However in the glued entity we can arrange them in 3! Ways therefore all three can sit together in 5!*3! Ways (120*6 = 720 ways)
2) Case in which 2 of the men sit together : we glue them as well and we get 6! Ways, and because we can arrange the glued entity in 2! Ways ( AB or BA) we get 6!2! Ways just for one couple of men.
As we have 3 men, we can make 3 couples and therefore we will multiply 6!2! By 3 and we get 720*2*3= 4320 ways
But… Hold on! In the 4320 ways, the case in which all 3 men sit together is already accounted for so we subtract 720 from 4320 and we get 3600.
Finally, there are 3600 ways in which 2 or more men sit together therefore there are 5040-3600= 1440 ways in which no two men are sitting together.
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Re: Four women and three men must be seated in a row for a group photo [#permalink]
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# What is a Multiplication Table?
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There has been decline in use of the multiplication table since 1989.
Many primary grade math programs insist on memorization of the multiplication table.
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The multiplication table is the familiar grid that contains an x quadrant, ranging from 0-12, 0-10, or 0-9, and a y quadrant with numbers of the same range. The product of any two numbers can be found by looking at the intersection between x and y. For example, if you wanted to find 8 X 2, you would merely look at the 8, and move down to the 2 space, to find the number 16. Alternately, since order does not affect simple multiplication, you could find the number 2 and count over to reach its intersection with the number 8.
The products of the intersection of numbers would be listed in several spots on the table. In a simple 0-9 multiplication table, the number 16 would be listed 3 times. You would find it at the intersection of 8,2, 2,8 and 4,4.
Many primary grade math programs insist on memorization of the multiplication table, or knowing your “times tables.” Actually, though the reference is to the multiplication table, students may never use the real multiplication table. They may instead simply memorize multiplication facts in order. The multiplication table is sometimes seen as a crutch, because students can use it without memorizing the facts, or alternately, they can use a calculator. Early grades may display large multiplication tables in a classroom, but from grade 3 and up, most classes no longer display them, or teachers cover them during tests so students don’t use them for multiplication problems.
There has been decline in use of the multiplication table since 1989, when the US National Council of Teachers of Mathematics (NCTM) suggested that students should evolve their own methods for figuring out multiplication problems. This suggestion didn't entirely solve the problem of memorizing multiplication, since some students appear not to evolve their own methods. Some students appear better served by actual memorization, while others might be able to make observations about numbers that help them keep in mind how to solve each problem.
In truth, knowing your multiplication facts often makes more advanced math simpler. There is a direct correlation between declining grades in mathematics and failure to memorize the multiplication facts, especially from 1-9 X 1-9. Failure to understand such facts often makes advanced concepts like long division extremely challenging. Some teachers now try to employ some of the suggestions made by the NCTM and also reinforce memorization. The NCTM revised their own statements to stress the importance of memorization of basic math facts, after a large body of evidence suggested de-emphasis had led to greater problems with math.
If your child is having difficulty memorizing math facts, it doesn’t hurt to have a few examples of the multiplication table posted at home. The ceiling over a child’s bed may be an excellent place to put one, and kid’s bathrooms also are good places. This may help the child find patterns and become better at learning math facts, simply because they have greater exposure to the multiplication table. You can purchase inexpensive large tables at teacher supply stores, in bookstores and on the Internet. You can also make your own with a child, to reinforce learning math facts.
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### Making Maths: Rolypoly
##### Stage: 1 and 2 Challenge Level:
Paint a stripe on a cardboard roll. Can you predict what will happen when it is rolled across a sheet of paper?
##### Stage: 2 Challenge Level:
How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
### Regular Rings 1
##### Stage: 2 Challenge Level:
Can you work out what shape is made by folding in this way? Why not create some patterns using this shape but in different sizes?
### Folding Flowers 2
##### Stage: 2 Challenge Level:
Make a flower design using the same shape made out of different sizes of paper.
### World of Tan 2 - Little Ming
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming?
### Folding Flowers 1
##### Stage: 2 Challenge Level:
Can you visualise what shape this piece of paper will make when it is folded?
### World of Tan 1 - Granma T
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Granma T?
### World of Tan 27 - Sharing
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Fung at the table?
### World of Tan 28 - Concentrating on Coordinates
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming playing the board game?
### Triangles in the Middle
##### Stage: 3, 4 and 5 Challenge Level:
This task depends on groups working collaboratively, discussing and reasoning to agree a final product.
### L-ateral Thinking
##### Stage: 1 and 2 Challenge Level:
Try this interactive strategy game for 2
### Midpoint Triangle
##### Stage: 2 Challenge Level:
Can you cut up a square in the way shown and make the pieces into a triangle?
### Dicey
##### Stage: 2 Challenge Level:
A game has a special dice with a colour spot on each face. These three pictures show different views of the same dice. What colour is opposite blue?
### Three Squares
##### Stage: 1 and 2 Challenge Level:
What is the greatest number of squares you can make by overlapping three squares?
### Green Cube, Yellow Cube
##### Stage: 2 Challenge Level:
How can you paint the faces of these eight cubes so they can be put together to make a 2 x 2 cube that is green all over AND a 2 x 2 cube that is yellow all over?
### Jomista Mat
##### Stage: 2 Challenge Level:
Looking at the picture of this Jomista Mat, can you decribe what you see? Why not try and make one yourself?
### Nine-pin Triangles
##### Stage: 2 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
##### Stage: 2 Challenge Level:
How many DIFFERENT quadrilaterals can be made by joining the dots on the 8-point circle?
### World of Tan 29 - the Telephone
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this telephone?
### You Owe Me Five Farthings, Say the Bells of St Martin's
##### Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
### World of Tan 7 - Gat Marn
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this plaque design?
### Makeover
##### Stage: 1 and 2 Challenge Level:
Exchange the positions of the two sets of counters in the least possible number of moves
### Fractional Triangles
##### Stage: 2 Challenge Level:
Use the lines on this figure to show how the square can be divided into 2 halves, 3 thirds, 6 sixths and 9 ninths.
### Regular Rings 2
##### Stage: 2 Challenge Level:
What shape is made when you fold using this crease pattern? Can you make a ring design?
### Triangular Faces
##### Stage: 2 Challenge Level:
This problem invites you to build 3D shapes using two different triangles. Can you make the shapes from the pictures?
### World of Tan 12 - All in a Fluff
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these rabbits?
### Making Tangrams
##### Stage: 2 Challenge Level:
Here's a simple way to make a Tangram without any measuring or ruling lines.
### Turning Cogs
##### Stage: 2 Challenge Level:
What happens when you turn these cogs? Investigate the differences between turning two cogs of different sizes and two cogs which are the same.
### World of Tan 11 - the Past, Present and Future
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the telescope and microscope?
### World of Tan 14 - Celebrations
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Little Ming and Little Fung dancing?
### World of Tan 4 - Monday Morning
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of Wai Ping, Wah Ming and Chi Wing?
### World of Tan 15 - Millennia
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the workmen?
### The Development of Spatial and Geometric Thinking: the Importance of Instruction.
##### Stage: 1 and 2
This article looks at levels of geometric thinking and the types of activities required to develop this thinking.
### World of Tan 9 - Animals
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this goat and giraffe?
### World of Tan 6 - Junk
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this junk?
### World of Tan 5 - Rocket
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the rocket?
### Conway's Chequerboard Army
##### Stage: 3 Challenge Level:
Here is a solitaire type environment for you to experiment with. Which targets can you reach?
### World of Tan 13 - A Storm in a Tea Cup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these convex shapes?
### World of Tan 8 - Sports Car
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this sports car?
### Fred the Class Robot
##### Stage: 2 Challenge Level:
Billy's class had a robot called Fred who could draw with chalk held underneath him. What shapes did the pupils make Fred draw?
### Domino Numbers
##### Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be? | 1,875 | 7,959 | {"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} | 3.359375 | 3 | CC-MAIN-2017-43 | longest | en | 0.873311 |
http://mathgpselaboration.blogspot.com/2009/07/m5n3c.html | 1,508,462,636,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823605.33/warc/CC-MAIN-20171020010834-20171020030834-00850.warc.gz | 210,906,673 | 10,868 | ## Sunday, July 26, 2009
### M5N3c - Multiplying and dividing by numbers less than one
M5N3. Students will further develop their understanding of the meaning of multiplication and division with decimals and use them. c. Multiply and divide with decimal fractions including decimal fractions less than one and greater than one.
Consider a problem like the following:
A ribbon costs \$ 1.80 for one meter. If you want to buy 0.8 meter of the ribbon, how much will it cost?
When children are asked what operation they would use to solve this problem, many will pick division. Some might think those children simply do not understand multiplication and division. However, that is not the case. Those children pick division as the operation because they know that "multiplication make bigger and division makes smaller." Research has shown that many children, and adults, hold this misconception.
Actually, calling it a "misconception" may be inappropriate. Rather, it is an overgeneralization children make based on their experiences. While students are working only with whole numbers, the only exception to this generalization is multiplication by 0 and 1. However, once they go beyond the "basic facts" stage of multiplication learning, practically all experiences involve multiplying by a number greater than one. The same can be said of division. The only time division does not result in a number less than the dividend is when it is divided by one. However, once again, practically all children's experiences before this point are division by a number greater than one.
Once the range of numbers is expanded to include decimal numbers and fractions, however, there are many cases where we do multiply or divide by numbers less than one. Therefore, it is an important goal of mathematics teaching that our students overcome this overgeneralization. A potentially powerful tool for this purpose is double number lines. If we represent the ribbon problem, it will look like this:
From this diagram, you can easily see that if the multiplier (represented on the bottom number line) is less than 1, the product (? mark) will be on the left of the multiplicand (1.80, i.e., amount corresponding to 1). On the other hand, if the multiplier is greater than 1, the product will be to the right of the multiplicand. Therefore, we can generalize:
If multiplier is greater than 1, multiplicand < product.
If multiplier is less than 1, product < multiplicand.
Similarly, you can use double number lines to contrast the situations when the divisors are less than 1 and those cases where the divisors are greater than 1.
However, the most difficult part for students (thus for teachers) is to help them understand that these situations are indeed situations where multiplication is the appropriate operation. For some students, double number line may not be sufficient. Another possible tool is to write mathematical expressions using words to describe the relationship among the quantities involved. In the ribbon problem, there are three quantities: cost of 1 meter of ribbon, total length of ribbon, and the price. The relationship among these three quantities can be expressed as
Price = [Cost of 1 meter] x [Total Length].Thus, for this problem, ? = 1.80 x 0.8.
An implicit, yet very important, goal of teaching multiplication and division of fractions and decimal numbers is to expand students' understanding of these operations. In early elementary grades, these operations are considered in equal group situations. Thus, when the multiplier or the divisor (in the case of fair-sharing division, the quotient in the case of measurement division) becomes something other than whole numbers, students have difficulty interpreting what it means. Through teaching of multiplication and division of decimal numbers and fractions, we want students to develop more proportional understanding of these operations. For example, A x B = C, should be interpreted as "A is to 1, C is to B," or "C is B times as much as A." Although this is not an explicitly stated goal in the GPS, it is something all teachers must keep in mind. | 839 | 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} | 4.3125 | 4 | CC-MAIN-2017-43 | longest | en | 0.960383 |
https://theonlinewebtools.com/hex-to-binary | 1,718,886,137,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00104.warc.gz | 500,504,862 | 13,539 | Decode Hexadecimal into Binary with One Click - The Effortless Encoder
Hexadecimal numbers play an integral role in computing and programming languages despite binary being the foundational machine language. Hex allows abbreviating long binary sequences into shorter numeric representations utilizing base-16 digits from 0-9 and A-F.
This provides a more human-readable yet still highly compact format for coding and data. Assembly languages rely heavily on hex to represent instructions and memory addresses. Hex color codes concisely define web palette values and ARGB encodings. When inspecting file contents, bytes are displayed in hexadecimal rather than cumbersome binary. Behind the scenes, processors ultimately interpret instructions in binary and operating systems store files as binary data. Hexadecimal serves as an important intermediate shorthand to interface between human programmers and the underlying binary reality.
Hex to Binary Manual Conversion Steps
Performing hexadecimal to binary conversion manually requires referencing decode tables, carefully padding and concatenating binary sequences, and precision to avoid errors. Initially, you would need a table listing the binary equivalent of each hex digit from 0 to F. Then for each hex number in your input, find its matching binary representation from the table. Any binary groups less than 4 digits would need padding zeros added to maintain proper 4-digit spacing.
Finally, accurately concatenate all the binary groups together into a single output string while avoiding mistakes. This multi-step process becomes extremely tedious and time consuming for longer hex inputs, with ample opportunity for minor errors or miscounts that would invalidate the final binary output. Just one errant digit could undermine the translation. And repeating this manual workflow frequently when dealing with hex ultimately proves frustratingly inefficient.
Streamlined Online Hex to Binary Conversion
Modern online calculators automate hex to binary translation, eliminating the need for manual lookups and calculations. Our user-friendly Hex to Binary encoder gives instant trouble-free encoding in 3 easy steps:
1. Enter or paste your Hex values into the converter
2. Click the "Convert" button
3. Instantly receive the resulting binary number
By handling the translation internally, our tool converts any size hex input to binary quickly and accurately. The automation provides immense time savings over manual methods.
The Power of Hexadecimal and Binary Together
Hexadecimal and binary each offer distinct advantages that together provide a robust system for computing needs. Binary provides the ultra-compact base-2 foundation for efficiently storing data and executing processor instructions using discrete on/off electrical signals. Meanwhile, hexadecimal allows inputting coded instructions and numeric values using easy base-16 notation instead of long binary strings.
Programmers appreciate hexadecimal for more concise and readable representations. Yet binary remains the machine language driving computation underneath. Our hexadecimal to binary encoder bridges these two worlds perfectly. It enables effortlessly translating from succinct hex inputs into optimized binary outputs that computers can readily interpret. The tool also works in reverse for decoding binary back to hex. Together, interfacing between hexadecimal and binary provides the best of both - human usability meeting machine performance.
We care about your data and would love to use cookies to improve your experience. | 609 | 3,577 | {"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} | 2.640625 | 3 | CC-MAIN-2024-26 | latest | en | 0.849931 |
http://betterlesson.com/lesson/resource/2546967/websites-for-road-trip-docx | 1,477,275,715,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719463.40/warc/CC-MAIN-20161020183839-00258-ip-10-171-6-4.ec2.internal.warc.gz | 31,491,977 | 19,114 | ## Websites for Road Trip.docx - Section 1: Set the Stage
Unit 7: Building Functions
Lesson 2 of 12
## Big Idea: Combine functions to help determine which mode of travel is cheapest.
Print Lesson
3 teachers like this lesson
Standards:
Subject(s):
Math, Function Operations and Inverses, Algebra II, function, master teacher project, 11th Grade
50 minutes
### Merrie Rampy
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Algebra II » Cubic Functions
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Fort Collins, CO
Environment: Suburban
###### Parking and Pencils: Step Functions and Piecewise Functions
12th Grade Math » Functioning with Functions
Big Idea: Model situations using unconventional (step and piecewise) functions.
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Troy, MI
Environment: Suburban | 207 | 901 | {"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} | 2.78125 | 3 | CC-MAIN-2016-44 | longest | en | 0.824365 |
http://perplexus.info/show.php?pid=10958&cid=58962 | 1,591,029,498,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419056.73/warc/CC-MAIN-20200601145025-20200601175025-00556.warc.gz | 94,324,616 | 4,830 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
3 by 3 by 5 (Posted on 2017-09-22)
The 15 triplets listed below were created by dividing each
of 5 not-so-common 9-letter words - each into 3 parts:
QUA, SUN, UAC, LLE, QUA,
ALL, SQU, LOQ, ITY, ARE,
AQU, UNE, LED, VIA, DRI.
You are requested to
- reconstruct the original words
- provide a short comment on each
- discover one common feature, present in all five.
Have fun!
See The Solution Submitted by Ady TZIDON No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
computer finds few words, apparently not all compatible | Comment 3 of 6 |
loquacity
aquarelle
unequaled
unequaled
duplicates coming from duplication of QUA.
DefDbl A-Z
Dim crlf\$, wordPart\$(15)
Form1.Visible = True
Text1.Text = ""
crlf = Chr\$(13) + Chr\$(10)
For i = 1 To 15
wordPart(i) = Mid("quasunuacllequaallsquloqityareaquuneledviadri", 3 * i - 2, 3)
Next i
For a = 1 To 15
For b = 1 To 15
If b <> a Then
For c = 1 To 15
If c <> a And c <> b Then
w\$ = wordPart(a) + wordPart(b) + wordPart(c)
If isWord(w) Then
Text1.Text = Text1.Text & w & crlf
End If
End If
Next c
End If
Next b
Next a
Text1.Text = Text1.Text & crlf & " done"
End Sub
Function isWord(w\$)
n = Len(w\$)
w1\$ = Space\$(n)
Open "\words\words" + LTrim\$(Str\$(n)) + ".txt" For Binary As #2
l = LOF(2) / n
low = 1: high = l
Do
middle = Int((low + high) / 2)
Get #2, (middle - 1) * n + 1, w1\$
If w1\$ = w\$ Then isWord = 1: Close 2: Exit Function
If w1\$ < w\$ Then low = middle + 1 Else high = middle - 1
Loop Until low > high
isWord = 0
Close 2
End Function
Posted by Charlie on 2017-09-22 12:17:16
Search: Search body:
Forums (0) | 618 | 1,729 | {"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} | 3.28125 | 3 | CC-MAIN-2020-24 | latest | en | 0.682116 |
https://solomonegash.com/econometrics/rbook1/serial-correlation-and-heteroskedasticity-in-time-series-regressions.html | 1,716,035,375,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057422.43/warc/CC-MAIN-20240518121005-20240518151005-00578.warc.gz | 477,790,458 | 16,262 | # 12 Serial Correlation and Heteroskedasticity in Time Series Regressions
Also covered using Python and Stata
``````library(wooldridge)
library(lmtest)
library(stargazer)
library(car)
library(dynlm)
library(prais)
library(sandwich)
rm(list = ls())``````
## 12.1 Example 12.1. Testing for AR(1) Serial Correlation in the Phillips Curve
``````ts_phillips <- ts(phillips)
uhat1 <- resid(dynlm(inf ~ unem + 1, data=ts_phillips))
phill_inf1 <- dynlm(uhat1 ~ L(uhat1,1))
ts_phillips2 <-ts(subset(phillips, phillips\$year<1997))
uhat2 <- resid(dynlm(inf ~ unem + 1, data=ts_phillips2))
phill_inf2 <- dynlm(uhat2 ~ L(uhat2, 1))
uhat3<- resid(dynlm(cinf ~ unem + 1, data=ts_phillips2))
phill_cinf <- dynlm(uhat3 ~ L(uhat3, 1))
stargazer(phill_inf1, phill_inf2, phill_cinf, keep.stat=c("n","rsq", "adj.rsq"), no.space=TRUE, type="text")``````
``````##
## ==========================================
## Dependent variable:
## -----------------------------
## uhat1 uhat2 uhat3
## (1) (2) (3)
## ------------------------------------------
## L(uhat1, 1) 0.572***
## (0.108)
## L(uhat2, 1) 0.573***
## (0.116)
## L(uhat3, 1) -0.036
## (0.124)
## Constant -0.112 -0.113 0.194
## (0.318) (0.359) (0.300)
## ------------------------------------------
## Observations 55 48 47
## R2 0.345 0.346 0.002
## Adjusted R2 0.333 0.332 -0.020
## ==========================================
## Note: *p<0.1; **p<0.05; ***p<0.01``````
## 12.2 Example 12.2. Testing for AR(1) Serial Correlation in the Minimum Wage Equation
``````tsprminwge <- ts(prminwge)
minwg_reg <- dynlm(lprepop ~ lmincov + lprgnp + lusgnp + t, data=tsprminwge)
uhat <- resid(minwg_reg)
AR1c <- dynlm(uhat ~ lmincov + lprgnp + lusgnp + t + L(uhat, 1) , data=tsprminwge)
AR1s <- dynlm(uhat ~ L(uhat, 1))
stargazer(AR1c, AR1s, column.labels=c("AR1c", "AR1s"), no.space=TRUE, type="text")``````
``````##
## ===========================================================
## Dependent variable:
## ---------------------------------------
## uhat
## AR1c AR1s
## (1) (2)
## -----------------------------------------------------------
## lmincov 0.038
## (0.035)
## lprgnp -0.078
## (0.071)
## lusgnp 0.204
## (0.195)
## t -0.003
## (0.004)
## L(uhat, 1) 0.481*** 0.417**
## (0.166) (0.159)
## Constant -0.851 -0.001
## (1.093) (0.004)
## -----------------------------------------------------------
## Observations 37 37
## R2 0.242 0.165
## Residual Std. Error 0.028 (df = 31) 0.027 (df = 35)
## F Statistic 1.983 (df = 5; 31) 6.895** (df = 1; 35)
## ===========================================================
## Note: *p<0.1; **p<0.05; ***p<0.01``````
## 12.3 Example 12.3. Testing for AR(3) Serial Correlation
``````tsbarium<-ts(barium)
barium_reg <- dynlm(lchnimp ~ lchempi + lgas + lrtwex + befile6 + affile6 + afdec6, data=tsbarium)
u <- resid(barium_reg)
AR3 <- dynlm(u ~ lchempi + lgas + lrtwex + befile6 + affile6 + afdec6 + L(u, 1) + L(u, 2) + L(u, 3) + 1, data = tsbarium)
summary(AR3)``````
``````##
## Time series regression with "ts" data:
## Start = 4, End = 131
##
## Call:
## dynlm(formula = u ~ lchempi + lgas + lrtwex + befile6 + affile6 +
## afdec6 + L(u, 1) + L(u, 2) + L(u, 3) + 1, data = tsbarium)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.89072 -0.32250 0.05873 0.36376 1.19650
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -14.36915 20.65567 -0.696 0.4880
## lchempi -0.14316 0.47203 -0.303 0.7622
## lgas 0.62331 0.88597 0.704 0.4831
## lrtwex 0.17867 0.39103 0.457 0.6486
## befile6 -0.08592 0.25101 -0.342 0.7327
## affile6 -0.12212 0.25470 -0.479 0.6325
## afdec6 -0.06683 0.27437 -0.244 0.8080
## L(u, 1) 0.22149 0.09166 2.417 0.0172 *
## L(u, 2) 0.13404 0.09216 1.454 0.1485
## L(u, 3) 0.12554 0.09112 1.378 0.1709
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.5704 on 118 degrees of freedom
## Multiple R-squared: 0.1159, Adjusted R-squared: 0.04848
## F-statistic: 1.719 on 9 and 118 DF, p-value: 0.09202``````
``linearHypothesis(AR3, c("L(u, 1)=0 ", "L(u, 2)=0" , "L(u, 3)=0"))``
``````## Linear hypothesis test
##
## Hypothesis:
## L(u,0
## L(u, 2) = 0
## L(u, 3) = 0
##
## Model 1: restricted model
## Model 2: u ~ lchempi + lgas + lrtwex + befile6 + affile6 + afdec6 + L(u,
## 1) + L(u, 2) + L(u, 3) + 1
##
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 121 43.394
## 2 118 38.394 3 5.0005 5.1229 0.00229 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1``````
## 12.4 Example 12.4. Prais-Winsten Estimation in the Event Study
``````ols <- dynlm(lchnimp ~ lchempi + lgas + lrtwex + befile6 + affile6 + afdec6, data=tsbarium)
stargazer(ols, column.labels=c("OLS"), no.space=TRUE, single.row = TRUE, type="text")``````
``````##
## ===============================================
## Dependent variable:
## ---------------------------
## lchnimp
## OLS
## -----------------------------------------------
## lchempi 3.117*** (0.479)
## lgas 0.196 (0.907)
## lrtwex 0.983** (0.400)
## befile6 0.060 (0.261)
## affile6 -0.032 (0.264)
## afdec6 -0.565* (0.286)
## Constant -17.803 (21.045)
## -----------------------------------------------
## Observations 131
## R2 0.305
## Residual Std. Error 0.597 (df = 124)
## F Statistic 9.064*** (df = 6; 124)
## ===============================================
## Note: *p<0.1; **p<0.05; ***p<0.01``````
``pw <- prais_winsten(lchnimp ~ lchempi + lgas + lrtwex + befile6 + affile6 + afdec6, data=tsbarium, index = "t")``
``````## Iteration 0: rho = 0
## Iteration 1: rho = 0.2708
## Iteration 2: rho = 0.291
## Iteration 3: rho = 0.293
## Iteration 4: rho = 0.2932
## Iteration 5: rho = 0.2932
## Iteration 6: rho = 0.2932
## Iteration 7: rho = 0.2932``````
``summary(pw)``
``````##
## Call:
## prais_winsten(formula = lchnimp ~ lchempi + lgas + lrtwex + befile6 +
## affile6 + afdec6, data = tsbarium, index = "t")
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.99386 -0.32219 0.03747 0.40226 1.50281
##
## AR(1) coefficient rho after 7 iterations: 0.2932
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -37.07770 22.77830 -1.628 0.1061
## lchempi 2.94095 0.63284 4.647 8.46e-06 ***
## lgas 1.04638 0.97734 1.071 0.2864
## lrtwex 1.13279 0.50666 2.236 0.0272 *
## befile6 -0.01648 0.31938 -0.052 0.9589
## affile6 -0.03316 0.32181 -0.103 0.9181
## afdec6 -0.57681 0.34199 -1.687 0.0942 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.5733 on 124 degrees of freedom
## Multiple R-squared: 0.2021, Adjusted R-squared: 0.1635
## F-statistic: 5.235 on 6 and 124 DF, p-value: 7.764e-05
##
## Durbin-Watson statistic (original): 1.458
## Durbin-Watson statistic (transformed): 2.087``````
## 12.5 Example 12.5. Static Phillips Curve
``````phillips2 <- ts(subset(phillips, phillips\$year<1997))
ols <- lm(inf ~ unem, data=phillips2)
stargazer(ols, column.labels=c("OLS"), no.space=TRUE, single.row = TRUE, type="text")``````
``````##
## ===============================================
## Dependent variable:
## ---------------------------
## inf
## OLS
## -----------------------------------------------
## unem 0.468 (0.289)
## Constant 1.424 (1.719)
## -----------------------------------------------
## Observations 49
## R2 0.053
## Residual Std. Error 3.131 (df = 47)
## F Statistic 2.616 (df = 1; 47)
## ===============================================
## Note: *p<0.1; **p<0.05; ***p<0.01``````
``pw <- prais_winsten(inf ~ unem, data=phillips2, index = "year")``
``````## Iteration 0: rho = 0
## Iteration 1: rho = 0.5727
## Iteration 2: rho = 0.7307
## Iteration 3: rho = 0.7719
## Iteration 4: rho = 0.7792
## Iteration 5: rho = 0.7803
## Iteration 6: rho = 0.7805
## Iteration 7: rho = 0.7805
## Iteration 8: rho = 0.7805
## Iteration 9: rho = 0.7805``````
``summary(pw)``
``````##
## Call:
## prais_winsten(formula = inf ~ unem, data = phillips2, index = "year")
##
## Residuals:
## Min 1Q Median 3Q Max
## -5.5470 -2.6764 -0.3579 1.4108 10.2853
##
## AR(1) coefficient rho after 9 iterations: 0.7805
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 8.2959 2.2314 3.718 0.000535 ***
## unem -0.7157 0.3135 -2.283 0.026988 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2.267 on 47 degrees of freedom
## Multiple R-squared: 0.1358, Adjusted R-squared: 0.1174
## F-statistic: 7.386 on 1 and 47 DF, p-value: 0.009174
##
## Durbin-Watson statistic (original): 0.8027
## Durbin-Watson statistic (transformed): 1.91``````
## 12.6 Example 12.6. Differencing the Interest Rate Equation
``````tsintdef <- ts(intdef)
reg1 <- dynlm(i3 ~ inf + def, data=tsintdef)
u <- resid(reg1)
AR1u <- dynlm(u ~ L(u, 1), data=tsintdef)
reg2 <-dynlm(ci3 ~ cinf + cdef, data=tsintdef)
e <- resid(reg2)
AR1e <- dynlm(e ~ L(e, 1), data=tsintdef)
stargazer(reg1, reg2, AR1u, AR1e, column.labels=c("reg1", "reg2", "AR1u", "AR1e"), no.space=TRUE, keep.stat=c("n","rsq", "adj.rsq"), type="text")``````
``````##
## ==============================================
## Dependent variable:
## ---------------------------------
## i3 ci3 u e
## reg1 reg2 AR1u AR1e
## (1) (2) (3) (4)
## ----------------------------------------------
## inf 0.606***
## (0.082)
## def 0.513***
## (0.118)
## cinf 0.149
## (0.092)
## cdef -0.181
## (0.148)
## L(u, 1) 0.623***
## (0.110)
## L(e, 1) 0.072
## (0.134)
## Constant 1.733*** 0.042 0.015 -0.041
## (0.432) (0.171) (0.190) (0.166)
## ----------------------------------------------
## Observations 56 55 55 54
## R2 0.602 0.176 0.377 0.005
## Adjusted R2 0.587 0.145 0.366 -0.014
## ==============================================
## Note: *p<0.1; **p<0.05; ***p<0.01``````
## 12.7 Example 12.7. The Puerto Rican Minimum Wage
``````OLS <- lm(lprepop ~ lmincov + lprgnp + lusgnp + t, data=prminwge)
Newey <- coeftest(OLS, vcov.=NeweyWest(OLS, lag=2, prewhite=FALSE, adjust=TRUE, verbose=TRUE))``````
``````##
## Lag truncation parameter chosen: 2``````
``stargazer(OLS, Newey, column.labels=c("OLS", "Newey"), no.space=TRUE, keep.stat=c("n","rsq", "adj.rsq"), type="text")``
``````##
## =========================================
## Dependent variable:
## ----------------------------
## lprepop
## OLS coefficient
## test
## OLS Newey
## (1) (2)
## -----------------------------------------
## lmincov -0.212*** -0.212***
## (0.040) (0.046)
## lprgnp 0.285*** 0.285***
## (0.080) (0.100)
## lusgnp 0.486** 0.486*
## (0.222) (0.279)
## t -0.027*** -0.027***
## (0.005) (0.006)
## Constant -6.663*** -6.663***
## (1.258) (1.536)
## -----------------------------------------
## Observations 38
## R2 0.889
## =========================================
## Note: *p<0.1; **p<0.05; ***p<0.01``````
``summary(prais_winsten(lprepop ~ lmincov + lprgnp + lusgnp + t , data=tsprminwge, index = "year"))``
``````## Iteration 0: rho = 0
## Iteration 1: rho = 0.4197
## Iteration 2: rho = 0.5325
## Iteration 3: rho = 0.5796
## Iteration 4: rho = 0.5999
## Iteration 5: rho = 0.6086
## Iteration 6: rho = 0.6123
## Iteration 7: rho = 0.6139
## Iteration 8: rho = 0.6146
## Iteration 9: rho = 0.6149
## Iteration 10: rho = 0.615
## Iteration 11: rho = 0.6151
## Iteration 12: rho = 0.6151
## Iteration 13: rho = 0.6151
## Iteration 14: rho = 0.6151
## Iteration 15: rho = 0.6151
## Iteration 16: rho = 0.6151``````
``````##
## Call:
## prais_winsten(formula = lprepop ~ lmincov + lprgnp + lusgnp +
## t, data = tsprminwge, index = "year")
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.079574 -0.028121 -0.004816 0.005918 0.075978
##
## AR(1) coefficient rho after 16 iterations: 0.6151
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -4.652853 1.376470 -3.380 0.00188 **
## lmincov -0.147711 0.045842 -3.222 0.00286 **
## lprgnp 0.251383 0.116462 2.158 0.03826 *
## lusgnp 0.255711 0.231750 1.103 0.27784
## t -0.020502 0.005856 -3.501 0.00135 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.02852 on 33 degrees of freedom
## Multiple R-squared: 0.7509, Adjusted R-squared: 0.7207
## F-statistic: 24.87 on 4 and 33 DF, p-value: 1.47e-09
##
## Durbin-Watson statistic (original): 1.014
## Durbin-Watson statistic (transformed): 1.736``````
## 12.8 Example 12.8. Heteroskedasticity and the Efficient Markets Hypothesis
``````tsnyse <-ts(nyse)
u2 <- resid(dynlm(return ~ return_1, data=tsnyse))**2
summary(dynlm(u2 ~ return_1 + 1, data = tsnyse))``````
``````##
## Time series regression with "ts" data:
## Start = 3, End = 691
##
## Call:
## dynlm(formula = u2 ~ return_1 + 1, data = tsnyse)
##
## Residuals:
## Min 1Q Median 3Q Max
## -9.689 -3.929 -2.021 0.960 223.730
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.6565 0.4277 10.888 < 2e-16 ***
## return_1 -1.1041 0.2014 -5.482 5.9e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 11.18 on 687 degrees of freedom
## Multiple R-squared: 0.04191, Adjusted R-squared: 0.04052
## F-statistic: 30.05 on 1 and 687 DF, p-value: 5.905e-08``````
## 12.9 Example 12.9. ARCH in Stock Returns
``````tsnyse <-ts(nyse)
u <- resid(dynlm(return ~ return_1, data=tsnyse))
u2 <- u*u
u2_reg <- dynlm(u2 ~ L(u2, 1) + 1, data = tsnyse)
u_reg <- dynlm(u ~ L(u, 1) + 1)
stargazer(u2_reg, u_reg, no.space=TRUE, type="text")``````
``````##
## ===========================================================
## Dependent variable:
## ----------------------------
## u2 u
## (1) (2)
## -----------------------------------------------------------
## L(u2, 1) 0.337***
## (0.036)
## L(u, 1) 0.001
## (0.038)
## Constant 2.947*** -0.001
## (0.440) (0.081)
## -----------------------------------------------------------
## Observations 688 688
## R2 0.114 0.00000 | 5,889 | 16,992 | {"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-2024-22 | latest | en | 0.405924 |
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• 0:04 Partitioning a Line Segment
• 0:52 Using Slope
• 2:34 Example
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Lesson Transcript
Instructor: Laura Pennington
Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.
Partitioning a line segment involves breaking the line segment up in a particular way. This lesson will show how to use the slope of a line segment to find a point that partitions the line segment into a given ratio.
## Partitioning a Line Segment
At the circus, Mike the magnificent is walking the tight rope. It takes him 10 equal size steps to get across the rope. He takes seven steps flawlessly, then wobbles a bit, and quickly takes the last three steps to land safely on the end platform.
It just so happens that Mike just performed a mathematical feat called partitioning a line segment. Partitioning a directed line segment, AB, into a ratio a/b involves dividing the line segment into a + b equal parts and finding a point that is a equal parts from A and b equal parts from B.
Consider Mike again. The point at which he wobbled on the tight rope (a line segment of 10 equal parts) is seven equal parts from the start and three equal parts from the end, so the point at which Mike wobbled partitioned the line segment into the ratio 7/3.
## Using Slope
Partitioning a directed line segment seems simple enough, but what if we are given the two endpoints of a directed line segment, and want to find the point that partitions the line segment into the ratio a/b?
Thankfully, we can do this fairly easily using parts of the slope of the line segment. The slope of the line segment with endpoints (x1, y1) and (x2, y2) gives us the rate at which y is changing with respect to x, and we can find it using the slope formula:
Slope = Rise/Run = (Change in y)/(Change in x) = (y2 - y1)/(x2 - x1)
If we are given a line segment AB, where A = (x1, y1) and B = (x2, y2), and we want to partition it into the ratio a/b, then we want to find a point P that falls a equal parts from point A and b equal parts from point B on the line segment. We can do this using the following steps:
1. Determine the ratio, call it c, comparing a to the entire length of the line segment using the formula c = a/(a + b). This ratio gives the fraction of the way that P is from A to B.
2. Find the rise (y2 - y1) and run (x2 - x1) of the slope of the line segment.
3. Add câ‹…(run) to the x1, and add câ‹…(rise) to y1. This takes point A and moves it a/(a + b) of the way to point B, which is exactly the point P that we want.
These steps also give way to a nice easy formula for P:
P = (x1 + c(x2 - x1), y1 + c(y2 - y1))
Hmmm…that seems to make sense, but don't you think an example will make things even more clear?
## Example
Suppose we have a directed line segment AB, where A = (1,2) and B = (8,7), and we want to partition it with the ratio 3/5. In other words, we want to find a point, P, that is three equal parts from A and is five equal parts from B. Let's take it through our steps, and then we'll verify our answer with our formula.
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## Electrical Engineering Objective Questions MCQ For Graduate Engineers Fully Solved
Electrical Engineering Objective Questions MCQ For Graduate Engineers Fully Solved
## Electrical Engineering objective questions MCQ for graduate Engineers Fully Solved
1.If the applied voltage of a certain transformer is increased hr 50% and the frequency is reduced to 50% (assuming that the magnetic circuit remains unsaturated), the maximum core flax density will
(a) change to three times the original value
(b) change to 1.5 times the original value
(c) change to 0.5 times as the original value
(d) remain the same as the original value
2. The low—voltage winding of a core type transformer is subdivided into two equal halves, each of half the original width of the single winding will the high voltage winding in between (instead of having the usual construction of low-voltage winding adjacent to the core and surrounded by the high-voltage winding). Such an interlacing of coils would make the combined primary and secondary leakage reactance (in terms of primary) nearly.
(a)twice
(b) equal
(c)half
(d) one -fourth
3. Two 3-limb, 3-phase delta-star connected transformers are supplied from the same source. One of the transformers is of. Dy I and the other is of Dy II / connection. The phase difference between the corresponding phase voltages of the secondaries would he
(a) 0°
(b) 30°
(c) 60°
(d) 120°
4. In a transformer fed from a fundamental frequency voltage source, the source of harmonics k the
(b) poor insulation
(c) iron loss
(d) saturation of core
5. A 40 kVA transformer/ a core loss of400 W and a full load copper of of 800 W. The proportion of full-load at maximum efficiency is
(a)50%
(b) 62.3%
(c) 70.7%
(d)100 %
6.A single phase transformer has rating of 15 KVA 600 / 120 V. It is recommended as an auto transformer to supply at 720 V from a 600 V primary source. The maximum load it can supply is ?
(a)90 kVA
(b)18 kVA
(c)15 kVA
(d) 12 kVA
7. Equalizing pulses in TV are placed during the
(a) vertical blanking period
(b) horizontal blanking period
(c)serration
(d) horizontal retrace
8. The most useful approach to radar system for monitoring the speed of moving vehicles is:
(b) Monopulse
9. A. dc shunt generator, when driven at its rated speed, is found to he not generating any voltage. Which of the following would account for this?
I. There is no residual magnetism
2. The connection of the field winding is not proper with respect to the armature terminals.
3. The resistance of the field circuit is greater than the critical field resistance.
4. The load resistance is less than the critical armature resistance.
Select the correct answer using the codes given below:
Codes:
(a)3 and 4
(b)l,2 and 4
(c)1,2 and 3
(d)l.2,3 and 4
10. To have spark/as commutation, the armature reaction effect in a dc machine is neutralised by
(a) using compensating winding and commutating poles
(b) Shifting the brush axis from the geometrical neutral axis to the magnetic neutral axis
(c) fixing the brush axis in line with the pole axis
(d) increasing the field excitation
11.In a dc shunt general or working on load, the brushes are moved forward in the direction of rotation, as a result of this, commutation will
(a)improve but terminal voltage will fall -
(b)worsen and terminal voltage will fall
(c) improve and terminal voltage will rise
(d) worsen and terminal voltage will rise
12. Consider the following statements:
The maximum range of radar can be increased by
1. increasing the peak transmitted power
2. increasing the gain of the receiver
3. increasing the diameter of the antenna
4. reducing the wavelength used
Of these statements
(a) 1, 3 and 4 are correct
(b) 1, 2, 3 and 4 are correct
(c) 2 and 4 are correct
(d)1 and 3 are correct
1. A submarine cable repeater contains filters for the two directions of transmits
2. Armored submarine cable is used for the shallow-shore ends of the cable.
3. Fibre optic submarine cable is used to prevent inadvertent ploughing in of the cable
Of these statements
(a) 1 and 2 are correct
(b) 2 and 3 are correct
(c) 1 and 3 are correct
(d) 1, 2 and 3 are correct
14. A dc overcompounded generator was operating satisfactorily and supplying power to an infinite bus when the prime mover failed to supply any mechanical power. The machine would then run as a ?
(a) cumulatively compounded motor with speed reversed
(b) cumulatively compounded motor with direction of rotation as before
(c) differentially compounded motor with speed reversed
(d) differentially compounded motor with direction of speed as before.
15 To couple a coaxial line to a parallel wire line, It is best to use a
(a) slotted line
(b) balun
(c)directional coupler
(a) quarter-wave transformer
16. When a synchronous motor is running at synchronous speed, the damper winding produces
(a) damping toruqe
(b) eddy current torque
(c)torques aiding the developing torque
(d) no torque
17. For a given developed power, a synchronous motor operating from a constant voltage and constant frequency supply, will draw the minimum and maximum armature currents, Imin and Imax respectively, corresponding to?
(a)Imin at unity pf, but Imax at zero pf
(b) Imax at unity p, but Imin at zero p1
(c) both Imin and Imax at unity pf
(d )both Imin and Imax at zero pf
18. Consider the folio wing statements regarding an RC phase-shift oscillator:
1. The amplifier gain is positive
2. The amplifier gain is negative.
3. The phase shift introduced by the feedback network is 180
4. The phase shift introduced by the feedback network is 360
Of these statements
(a) 1 and 3 are correct
(b) 2 and 3 are correct
(c) 2 and 4 are correct
(d) 1 and 4 are correct
19 While conducting a “slip “ test for determination of direct and quadrature axis synchronous reactance ‘Xd’ and ‘Xq’ of salient pole synchronous machine, the rotor of the machine is run with a slip ‘s’ and stator supply frequency ‘f’. The frequency of
1. voltage induced across open field terminals,
2. envelope of the armature terminal voltage.
3. envelope of the armature current, and
4. armature current
will be respectively
(a) sf, sf,sf and f
(b) sf, f,sf and f
(c) f, sf,f and af
(d) f, (1-s)f,(2-s)f and sf
20. If two induction motors A and B are identical except that the air-gap of motor ‘A ‘ is 59% greater than that of motor ‘B’, then
(a) the no-load power factor of A will be better than that of B.
(b) the no-load power factor of A will be poorer than that of B.
(c) the core losses of A will be more than those of B
(d) the operating flux of A will be smaller than that of B.
21 A 6-pole, 50 Hz, 3-phase synchronous motor and an 8-pole, 50 Hz. 3—phase slipring induction motor are mechanically coupled and operate on the some 3-phase, 50 Hz supply system If they are left open-circuited, then the frequency of the voltage produced across any two slip rings would be?
(a) 12.5 Hz
(b) 25.0 Hz
(c) 37.5 Hz
(d) 50.0 Hz
22. Which of the following statements regarding skewing of motor bars in a squirrel-cage induction motor are correct?
I. It prevents cogging;
2. It produces more uniform torque.
3. it increases starting torque.
4. It reduces motor ‘hum’ during its operation.
Select the correct answer using the codes given below.
Codes:
(a)2,3 and 4
(b) l,2 and 3
(c)1,3 and 4
(d) l,2 and 4
23. The rotor power output of a 3-phase induction motor is 15kW and the corresponding slip is 4%. The rotor copper loss will be
(a) 600 W
(b) 625 W
(c) 650 W
(d) 700 W
24. If an input signal with non-zero dc component is applied to a low pass RC network, the dc component in the output will be ?
(a) the same as that in the input
(b) less than that in the input
(c) more than that in the input
(d) zero
25. A 3 phase wound rotor induction motor, when started with load connected to its shaft, was found to start but settle down at about half synchronous speed. If the rotor winding as well as the stator winding were star connected, then the cause of the malfunction could be attributed to
(a) one of the stator phase windings being short–circuited
(b) one of the supply fuses being blown
(c) one of the rotor phases being open-circuited
(d) two of the rotor phases being open circuited
26. Consider the following statements regarding fractional horse power shaded-pole motor:
1. Its direction of rotation is from unshaded to shaded portion of the poles.
2.. Its direction of rotation is from shaded to unshaded portion of the poles.
3. It can remain stalled for short periods without any harm.
4. It has a very poor power factor.
Of these statements
(a) 1, 3 and 4 are correct
(b) 2, 3 and 4 are correct
(c) 2 and 4 are correct
(d) 1 and 3 are correct
27. In the case of a converter-inverter speed control arrangement f an induction motor operating with v/f constant and with negligible stator impedance.
(a) the maximum torque is independent of frequency
(b) the maximum torque is proportional to frequency
(c) the slip at maximum torque is proportional to frequency
(d) the starting torque is proportional to frequency
28 An amplifier with mid-band gain A = 500 has negative feedback beta = 1/100 the upper cut-off without feedback were at 60 kHz, then with feedback it would become
(a) 10kHz
(b) 12kHz
(c) 300 kHz
(d) 360 kHz
29. If the discharge is 1 m cube and the head of water is 1 m then the power generated by the alternator in one hour (assume 100% efficiency of generator and turbine) will be
(a) 10kW
(b) 73/75kW
(c) 746/75kW
(d) 100kw
30. Control rods used In unclear reactors are made of
(a) zirconium
(b) boron
(c) beryllium
31. if alpha = 0.98, Ico=6 microA I beta =100 microA for a transistor, Then the value of Ic will he
(a) 2.3mA
(b) 3.lmA
(c) 4.6mA
(d) 5.2mA
32. In a 3-core extra-high voltage cable, a metallic screen around each core-insulation is provided to
(a) facilitate heat dissipation
(h) give mechanic strength
(d) obtain longitudinal electric stress
33. Galloping In transmission line conductors arises generally due to
(a) asymmetrical layers of ice formation
(b) vortex phenomenon in light winds
(c) heavy weight of the line conductors
(d) adoption of horizontal conductor configurations
34.. In a 3- phase rectifier circuit, thyristor number 1, 2 and 3 are connected respectively to R, Y and B phases of the star-connected transformer secondary. When the current is being commutated front thyristor No. 1 to No. 2, the effect of the transformer leakage and the ac system Inductance will be such that it will?
(a) Prolong the conduction in No. 1 and delay the turn on of No. 2 correspondingly.
(b) stop the conduction in No. 1 at the scheduled time, but delay the turn on of No. 2
(c) produce conduction in both No. 1 and No. 2 in parallel for an overlapping period through a transient
(d) double the voltage output through a commutation transient
35. The incremental generating costs of two generating units are given by
IC1=0.10 X + 20 Rs /MWhr
IC1=0.15 Y + 18 Rs /MWhr
where X and V are power generated by the two units in MW.
For a total demand of 300 MW, the value (in MW) of X and Y will be respectively
(a) 172 and 123
(b) 123 and 172
(c) 175 and 125
(d) 200 and 100
36. Consider the following statements:
To provide reliable protection for a distribution transformer against over voltages using lightning arrestors, it is essential that the
2. distance between the transformer and the arrestor is small
3. transformer and the arrestor have a common inter-connecting ground.
4. Spark over voltage of the arrestor is greater than the residual voltage.
Of these statements
(a) 1, 3 and 4 are correct
(b) 2 and 3 are correct
(c)2, 3 and 4 are correct
(d) 1 and 4 are correct
37. The reflection coefficient of a short-circuited line is
(a) -1
(b) 1
(c) 0.5
(d) zero
38.Iif an intrinsic semiconductors is doped with a very small amount of born then in the extrinsic semiconductor so farmed, the number of electrons and holes will
(a)decrease
(b) increase and decrease respectively
(c)increase
(d) decrease and increase respectively
39. Hollow conductors are used in transmission lines to
(a) reduce weight of copper
(b) improve stability
(c) reduce corona
(d) increase power transmission capacity
40. In the solution of load-flow equation, Newton-Raphson (NR) method is superior to the Gauss- Seidel (GS) method, because the
(a) time taken to perform one iteration in the NR method is less When compared to the time taken in the OS method
(b) number of iterations required in the NR method is more when compared to that in the GS method
(c) number of iterations required is not independent of the size of the system in the NR method
(d) convergence characteristics of the NR method are not affected by the selection of slack bus
41. In a synchronous generator, a divided winding rotor is preferable to a conventional winding rotor because of
(a) higher efficiency
(c) higher short-circuit ratio
(d) better damping
42. Consider the following statements regarding speed control of induction motors by means of external rotor resistors:
1. Reduction in speed is accompanied by reduced efficiency.
2. With a large resistance in the rotor circuit, the speed would vary considerably with variation in torque
3. The method is very complicated
The Disadvantages of such a method of speed control would include
(a) l and 2
(b) 2 and 3
(c) l and 3
(d) 1,2 and 3
43. Zero sequence currents can flow from a line into a transformer bank if the windings are in ?
(a) grounded star/delta
(b) delta/star
(c) star/grounded star
(d) delta/delta
44. When a line to ground fault occurs, the current in a faulted phase is 100 A. The Zero Sequence current in this case will be
(a) zero
(b) 33.3 A
(c) 66.6A
(d) 100 A
45. Consider the following statements:
Switched mode power supplies are preferred over the continuous types, because they are
1. suitable for use in both ac and dc.
2; more efficient,
3. suitable for low-power circuits.
4. suitable for high-power circuits.
Of these statements
(a) 1 and 2 are correct
(b) 1 and 3 are correct
(c) 2 and 3 are correct
(d) 2 and 4 are correct
46. The power generated by two plants are P1 = 50MW, P2= 40 MW.
If the loss coefficients are B11 = 0.001, B22 = 0.0025 and B12 = -0.0005, the power loss will be
(a)5.5 MW
(b) 6.5 MW
(c) 4.5Mw
(d) 8.5 MW
47. In dc choppers, per unit ripple is maximum when then duty cycle is?
(a) 0.2
(b) 0.5
(c) 0.7
(d) 0.9
48. The following data pertain to two alternators working in parallel and supplying a total load of 80 MW:
Machine 1 : 40 MVA with 5% speed regulation
Machine 2: 60 MVA with 5% speed regulation
The load sharing between machines 1 and 2 will be
(a)P1/48MW, P2/32MW
(b) 40MW. 40MW
(c)30MW,50MW
(d) 32 MW. 48MW
49. The per unit impedance of a synchronous machine is 0.242. If the base voltage is Increased by 1.1 times, the per unit value will be
(a)0.266
(b) 0.242
(c)0.220
(d) 0.200
50. A 3-pulse converter feeds a pure resistive load at a firing angle of alpha = 60°. The average value of current flowing in the load is 10 A, If a very large inductance is connected in the load circuit, then the
(a) average value of current will remain as 10 A
(b) average value of current wilt become greater than 10 A
(c) average value of current will become less than 10 A
(d) trend of variation of current cannot be predicted unless the exact value of the inductance connected is known | 4,432 | 16,102 | {"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-2020-10 | longest | en | 0.409542 |
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# Although people in France consume fatty foods
Author Message
Intern
Joined: 20 Jun 2018
Posts: 12
Although people in France consume fatty foods [#permalink]
### Show Tags
26 Jun 2018, 00:46
00:00
Difficulty:
45% (medium)
Question Stats:
33% (01:40) correct 67% (01:35) wrong based on 6 sessions
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Although people in France consume fatty foods at a rate comparable to the United States, their death rates from heart disease are far lower in France.
A) people in France consume fatty foods at a rate comparable to the United States, their
B) people in France and the United States consume fatty foods at about the same rate, the
C) fatty foods are consumed by people in France at a comparable rate to the United States's, their
D) the rate of fatty foods consumed in France and the United States is about the same, the
E) the rate of people consuming fatty foods is about the same in France and the United States, the
Math Expert
Joined: 02 Sep 2009
Posts: 49206
Re: Although people in France consume fatty foods [#permalink]
### Show Tags
26 Jun 2018, 00:49
Mike03 wrote:
Although people in France consume fatty foods at a rate comparable to the United States, their death rates from heart disease are far lower in France.
A) people in France consume fatty foods at a rate comparable to the United States, their
B) people in France and the United States consume fatty foods at about the same rate, the
C) fatty foods are consumed by people in France at a comparable rate to the United States's, their
D) the rate of fatty foods consumed in France and the United States is about the same, the
E) the rate of people consuming fatty foods is about the same in France and the United States, the
Please search before posting: https://gmatclub.com/forum/although-peo ... 52828.html
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Re: Although people in France consume fatty foods &nbs [#permalink] 26 Jun 2018, 00:49
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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®. | 657 | 2,810 | {"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-2018-39 | latest | en | 0.93442 |
https://www.folkstalk.com/2022/10/how-to-make-a-grading-system-in-python-with-code-examples.html | 1,709,636,147,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948234904.99/warc/CC-MAIN-20240305092259-20240305122259-00394.warc.gz | 739,955,431 | 14,581 | # How To Make A Grading System In Python With Code Examples
How To Make A Grading System In Python With Code Examples
We will use programming in this lesson to attempt to solve the How To Make A Grading System In Python puzzle. This is demonstrated by the following code.
if scores >= 90 and <= 100:
return 'A'
elif scores >= 80 and <= 89:
return 'B'
elif scores >= 70 and <= 79:
return 'C'
elif scores >= 60 and <= 69:
return 'D'
elif scores >= 50 and <= 59:
return 'E'
else:
return 'F'
We were able to fix the How To Make A Grading System In Python problemcode by looking at a number of different examples.
## How do you create a grading system?
9.12 Guidelines for Creating an Effective Grading System
• Keep your eyes on the prize.
• An effective grading system fosters communication.
• Grades should reflect a nonjudgmental posture.
• Intentional imprecision.
• Use points only when necessary.
• No surprises.
• Find a balance that works for you.
• Valuing the learning process.
## How does Python calculate average mark of students?
Python Average The formula to calculate average in Python is done by calculating the sum of the numbers in the list divided by the count of numbers in the list.26-Aug-2022
## How do you get a percentage in Python?
Use the division / operator to divide one number by another. Multiply the quotient by 100 to get the percentage. The result shows what percent the first number is of the second.09-Jul-2022
## How do you create a grading program in C?
First of all we will take input a mark of subject from the candidate and according to following condition we will calculate the grade.
• If marks <50 then Grade is F.
• if marks >=50 <60 then Grade is D.
• if marks >=60 <70 then Grade is C.
• if marks >=70 <80 then Grade is B.
• if marks >=80 <90 then Grade is A.
## What is the 7 point grading scale?
The 7-point grading scale consists of five marks designating a passing level (12, 10, 7, 4 and 02) as well as two marks designating a non-passing level (00 and -3).
## What are alternatives to grading?
• Gamification.
• Live Feedback.
• A Continuous 'Climate of Assessment'
• Standards-Based Reporting.
• “So?
• Metacognitive Action/Reflection/Narrative/Anecdotal.
• Digital Portfolios.
## How does Python calculate total marks?
Python: How to calculate total, average and percentage of marks of five subjects
• Read marks of five subjects in some variables say S1, S2, S3, S4 and S5.
• Apply the formula for finding sum i.e. total = S1 + S2 + S3 + S4 + S5.
• Apply the formula to find average using above total i.e. average = total / 5.
## How do you average a list in Python?
Using Python sum() function len() function is used to calculate the length of the list i.e. the count of data items present in the list. Further, statistics. sum() function is used to calculate the sum of all the data items in the list. Note: average = (sum)/(count).03-Aug-2022
## How do you print total marks in Python?
Sum all subjects marks using the arithmetic operator. Calculate average using this formula average = total / 5 . And calculate percentage using this formula: percentage = (total / 500) * 100. Print result.15-Jun-2022
## How do you calculate 5% in Python?
To calculate a percentage in Python, use the division operator (/) to get the quotient from two numbers and then multiply this quotient by 100 using the multiplication operator (*) to get the percentage.23-Jun-2022 | 833 | 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} | 3.546875 | 4 | CC-MAIN-2024-10 | latest | en | 0.857827 |
https://www.futurestarr.com/blog/mathematics/a12-out-of-30-as-a-percentage | 1,656,200,937,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103036176.7/warc/CC-MAIN-20220625220543-20220626010543-00221.warc.gz | 834,631,957 | 23,325 | FutureStarr
A12 out of 30 as a percentage
## 12 out of 30 as a percentage
We all want to succeed as entrepreneurs. But what does that mean? The road to success is never easy and there’s not just one path. So, what should you do if you want to succeed in a niche where you don’t know a lot about it?
### Percentage
via GIPHY
This percentage calculator is a tool that lets you do a simple calculation: what percent of X is Y? The tool is pretty straightforward. All you need to do is fill in two fields, and the third one will be calculated for you automatically. This method will allow you to answer the question of how to find a percentage of two numbers. Furthermore, our percentage calculator also allows you to perform calculations in the opposite way, i.e., how to find a percentage of a number. Try entering various values into the different fields and see how quick and easy-to-use this handy tool is. Is only knowing how to get a percentage of a number is not enough for you? If you are looking for more extensive calculations, hit the advanced mode button under the calculator. other than being helpful with learning percentages and fractions, this tool is useful in many different situations. You can find percentages in almost every aspect of your life! Anyone who has ever been to the shopping mall has surely seen dozens of signs with a large percentage symbol saying "discount!". And this is only one of many other examples of percentages. They frequently appear, e.g., in finance where we used them to find an amount of income tax or sales tax, or in health to express what is your body fat. Keep reading if you would like to see how to find a percentage of something, what the percentage formula is, and the applications of percentages in other areas of life, like statistics or physi.
Other than being helpful with learning percentages and fractions, this tool is useful in many different situations. You can find percentages in almost every aspect of your life! Anyone who has ever been to the shopping mall has surely seen dozens of signs with a large percentage symbol saying "discount!". And this is only one of many other examples of percentages. They frequently appear, e.g., in finance where we used them to find an amount of income tax or sales tax, or in health to express what is your body fat. Keep reading if you would like to see how to find a percentage of something, what the percentage formula is, and the applications of percentages in other areas of life, like statistics or physics. (Source: www.omnicalculator.com)
### Fraction
via GIPHY
Percentages are sometimes better at expressing various quantities than decimal fractions in chemistry or physics. For example, it is much convenient to say that percentage concentration of a specific substance is 15.7% than that there are 18.66 grams of substance in 118.66 grams of solution (like in an example in percentage concentration calculator). Another example is efficiency (or its special case - Carnot efficiency). Is it better to say that a car engine works with an efficiency of 20% or that it produces an energy output of 0.2 kWh from the input energy of 1 kWh? What do you think? We are sure that you're already well aware that knowing how to get a percentage of a number is a valuable ability.
Although Ancient Romans used Roman numerals I, V, X, L, and so on, calculations were often performed in fractions that were divided by 100. It was equivalent to the computing of percentages that we know today. Computations with a denominator of 100 became more standard after the introduction of the decimal system. Many medieval arithmetic texts applied this method to describe finances, e.g., interest rates. However, the percent sign % we know today only became popular a little while ago, in the 20th century, after years of constant evolution. (Source: www.omnicalculator.com)
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June 25, 2022 | Jamshaid Aslam | 1,199 | 4,998 | {"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} | 2.84375 | 3 | CC-MAIN-2022-27 | latest | en | 0.958838 |
https://www.bevfitchett.us/machine-gun-v4/step-1.html | 1,563,369,576,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525187.9/warc/CC-MAIN-20190717121559-20190717143559-00329.warc.gz | 644,415,380 | 7,566 | ## Step
2. The loss in velocity due to the initial compression of the springs is equal to:
Determine the velocity loss for various values of t, substract each from the corresponding ordinate of the free recoil velocity curve and draw a curve through the resulting points. If the effect of the spring constant proves to be negligible, this curve is the retarded velocity curve.
3. Integrate under the curve drawn in step 2 to obtain the displacement curve.
4. Assume that the curve drawn in step 3 represents the actual time-travel curve and use this curve to determine the retardation due to the spring constant. (Use the combined spring constant for the barrel spring and bolt spring, K1 + K2.) Ordinarily, it will be found that this retardation is so small that it will not have any effect worthy of consideration.
5. In the event that the retardation determined in step 4 is sufficient to affect the velocity, use it to modify the curve drawn in step 2 and then integrate under the new curve to obtain a corrected displacement curve.
6. Steps \ and 5 can be repeated as often as is necessary until no significant change occurs in the displacement curve. Actually, this process of successive approximation should never be necessary and satisfactory results should be achieved in the first three steps or at least in the first five steps.
Fig. 2-10 shows the curves obtained for the gun of the example. The total loss in velocity due to the combined effect of the initial compressions of the springs during the first 0.010 second is:
The loss due to the combined effect of the spring constants as determined by the method of step 4 is only about 0.345 foot per second. The final curves shown in fig. 2-10 are the result of performing step 5. Since the velocity loss due to the cffcct of the spring constant is so small, step 6 need not be taken.
The remainder of the displacement curve for the recoil stroke can now be determined analytically by using equation 2-10:
. KD + F, . T / K • _, F„ 1 Fo d- K *in LVKTr KD + F0K
0 0 | 455 | 2,037 | {"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} | 3.484375 | 3 | CC-MAIN-2019-30 | latest | en | 0.921015 |
http://math.stackexchange.com/questions/337782/pseudo-random-number-generation-on-the-gpu?answertab=active | 1,448,653,118,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398450559.94/warc/CC-MAIN-20151124205410-00035-ip-10-71-132-137.ec2.internal.warc.gz | 148,969,384 | 20,084 | # Pseudo Random Number Generation on the GPU
Idea
For a Path Tracing application I need to generate good quality pseudo random numbers in the closed range [0~1]. Because I'm doing this on the GPU (HLSL Shader Model 5) I can only use 32bit variables. My initial approach is the following set-up:
1. Ever Frame the pixel shader receive a good pseudo random number ([0~1]) from the CPU using C++'s std::mt19937 generator and std::uniform_real_distribution.
2. Because for each pixel this number is the same I also use the screen coordinates u and v of each pixel these are also in [0~1].
3. I then call the Multiply With Carry method like below.
Algorithm
// seed is the value given from the cpu
float3 random = Random(seed + u, seed + v)l
// Multiply With Carry, returns 3 floating point values {x, y, z}
// x: the random number
// y, z: new seeds for the next time we need a random number
float3 Random(float seed_a, float seed_b)
{
uint m_z = asuint(seed_a);
uint m_w = asuint(seed_b);
m_z = 36969 * (m_z & 65535) + (m_z >> 16);
m_w = 18000 * (m_w & 65535) + (m_w >> 16);
float r = ((m_z << 16) + m_w) / (float)0xFFFFFFFF;
return float3(r, asfloat(m_z), asfloat(m_w));
}
This produces the following output. (The left part is the random number obtained from the Random method for this pixel, the right part is the visualization of u and v as Red and Green.
Result
As you can see there is clearly a pattern, so the randomness is not good at all. Which hurts the performance of my algorithm tremendously. This is probably due to the fact that the original Multiply with Carry method assumes m_z and m_w are 64 bit integers, not 32 bit ones.
What I want
What I'm looking for is a way to fix my implementation of the Multiply with Carry method so that it produces reasonably good pseudo random numbers and works in the closed [0~1] interval instead of the open [0~1] interval. However since it is quite possible that this method can only work right using 64 bits integers I'm would also be really glad if someone can suggest another pseudo random number generator algorithm that:
• Works with 32 bit numbers
• Produces uniformly distributed results in the closed [0~1] interval
• Does not require too much state information, (this is why I chose MwC since it only needs to store 2 variables) since that is hard on the GPU. 16 32 bit variables would be the maximum I think since I can store that in 4x4 matrix which is easy to pass around.
-
It's badly wrong to re-seed a PRNG each time you need a random number -- that means that the randomness you end up with does not really come from the PRNG itself, but just from how random you can make each new seed (which, in this case, is "not very"). With SIMD parallelism you'll probably want to seed a PRNG for each processing unit once and for all, and then keep carrying that seed forwards through the entire computation. (But I don't know how easy it is to express this if you otherwise let the compiler parallelize your loops for you). – Henning Makholm Mar 22 '13 at 10:35
Unfortunately I cant share information between pixels using a shader so I need to seed each pixel individually. Shaders also do not retain information between frames so that makes everything a lot harder. Still I do believe that the performance I'm getting now is really abysmal. Note that for this example I only take one random number from the PRNG but usually I take hundreds per pixel. – Roy T. Mar 22 '13 at 10:44
So the code you're showing is not what actually runs? Your comment says "new seeds for the next time we need a random number". If you're using "hundreds" of random numbers for each seed, you should have plenty of time to run the initial seed through, say, one of the integer hashes from burtleburtle.net, which ought to get rid of the most visually obvious patterns in the first outputs. – Henning Makholm Mar 22 '13 at 10:51
No wait, that is how MwC works, the two seeds are its state. I mean I only seed them using the number from the CPU and the U and V variables once each frame. – Roy T. Mar 22 '13 at 11:00
You might be interested in this site proposal – Daniel Pendergast Nov 18 '13 at 18:49
I queried fellow students and they brought my attention to this paper: http://http.developer.nvidia.com/GPUGems3/gpugems3_ch37.html
Since the paper is quite long and might not be online forever below is the general idea:
Linear Congruent Generators are ideal for use on the GPU because they are simple and do not require much state (only the previously generated number). but they are not 'random enough' for, for example, Monte Carlo based simulation. A generator like a Mersenne Twister would be better but requires way too much state to be stored.
The solution proposed by the paper is combining several LCGs using a combined Tausworthe Generator (as used by the Mersenne Twister) this guarantees a much better randomness without having to store as much state as the Mersenne Twister. The final algorithm looks like this:
struct RandomResult
{
uint4 state;
float value;
};
uint TausStep(uint z, int S1, int S2, int S3, uint M)
{
uint b = (((z << S1) ^ z) >> S2);
return (((z & M) << S3) ^ b);
}
uint LCGStep(uint z, uint A, uint C)
{
return (A * z + C);
}
RandomResult Random(uint4 state)
{
state.x = TausStep(state.x, 13, 19, 12, 4294967294);
state.y = TausStep(state.y, 2, 25, 4, 4294967288);
state.z = TausStep(state.z, 3, 11, 17, 4294967280);
state.w = LCGStep(state.w, 1664525, 1013904223);
RandomResult result;
result.state = state;
result.value = 2.3283064365387e-10 * (state.x ^ state.y ^ state.z ^ state.w);
return result;
}
Note that the initial seed values for state should be larger than 128! (For background information reed the paper) and that you should fill the seed with 4 good random numbers from the CPU + four values unique for that pixel to get a nice result.
-
There's some Perlin noise code on the following site that produces a number between -1 and +1 that I have used in the past. I am sure it could be changed to produce 0 to 1. http://www.gamedev.net/page/resources/_/technical/game-programming/simple-clouds-part-1-r2085
-
Hey MistyManor, I tried those links, and they indeed give me nice noisy image but it has a really low period which in the end was insufficient for the Monte Carlo simulation which is why I kept searching for something better. I answered my own question now after some searching, but I wish I could accept your answer as well since it is certainly not wrong and will be sufficient for most uses. – Roy T. Mar 24 '13 at 21:02 | 1,667 | 6,546 | {"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} | 3.125 | 3 | CC-MAIN-2015-48 | longest | en | 0.843484 |
http://home.wangjianshuo.com/20060806_the_world_is_not_created_by_genius.htm | 1,563,351,819,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525133.20/warc/CC-MAIN-20190717081450-20190717103450-00098.warc.gz | 74,007,991 | 20,159 | # The World is Not Created by Genius
Lao Hua has an interesting blog about a game he facilitates. The game is like this:
• Gather a group of people (the more, the better)
• Ask each person to write down a number between 0 – 100
• Calculate the average of all these numbers
• The one who guessed closest to 2/3 of the average number wins
Which number would you guess if you were in the game?
My Guess
I played the game for the first time yesterday. My guess was 8.2
I thought if everyone randomly chooses a number, the average should be 50, and 2/3 of it is 33.3.
I assume everyone should know this. If everyone knows it, people will try 33.3333, and the average is 33.33. 2/3 of it is 22.22
If everyone is as smart and think of this, people will irratate, and the number is becoming smaller and smaller, like this.
50
33.33333333
22.22222222
14.81481481
9.87654321
6.58436214
4.38957476
2.926383173
1.950922116
My brain started to hurt, and finally, I thought, at least it should be within 1-10, so I randomly chose 8.2
The Final Result
This is the guess from 12 people:
30
98.16
32
50
12
33.3
22
8
8.2
18
28.68
37
The average is 31.445, and 2/3 of it is 20.96333333. Finally, the person who guessed 22 won the game. I lose miserably.
I agree that world is not created by genius now.
## 17 thoughts on “The World is Not Created by Genius”
1. Shrek7 says:
let me get my abacus out to quickly calculate the correct average.
2. RWM says:
Reminds me of the ‘lowest unique bid’ type of auction – where people pay some low fee for placing a bid, with the lowest unique bidder being the winner.
People trying to second guess people.
3. Post this year’s FIRST comment on Wang Jian Shuo’s website.
Have been such a long period.
4. The result may vary if you told them like this: Hey, if you write down the most accurate guess, you would win a car/ thousands dollars something… ‘coz in that way, people will take it more seriously, and numbers as 98.16 will hardly be chosen…
5. Realy Interesting Game.
6. SSC says:
Interesting game. I am not sure if this one of the classical games from game theory.
It seems people rarely get further than two rounds of iterations. I guess the smarter a group is, the smaller the number one should guess.
What if everyone guesses zero, who will be the winner?
7. haha, interesting~~
8. earflying says:
The result is volatile as the plays change,but it still worths trying~~~
9. drunkpriest says:
–I thought if everyone randomly chooses a number, the average should be 50, and 2/3 of it is 33.3. — I thought if everyone randomly chose a number, the average would be 50, …[to keep the consistency of tense in a the sentence]
–I assume ….– I assumed…. [to keep the consistency of tense in the context]
The similar mistakes can be found in your post.
10. so funny game.if sb can guess the accurate number,he must be a man who can peep all the numbers written on the paper.but it’s impossible. hahaha…………
11. Jie Lun says:
Interesting game, it really makes one think about his/her chances :)
btw, drunkpriest, I bet you could do something better with your life instead of telling people how to write proper English (and no, my English is not that good either). Give the man a break, his English is good enough for me, I wish more people wrote like he did :)
12. very interesting guess.
13. icePhoenix says:
I guess 22,and I’m right!?What a coincidence!
14. drunkpriest says:
Hi Jielun,
Did I offend you? I’m sorry if I did.
Drunkpriest
15. Helen says:
What an interesting game! Well, as to me, the result depends on the biggest number. Maybe the one who guessed 98.16 hadn’t known how to play the game.
16. At the Money says:
Interesting, indeed. People reveal more things about themselves in this little game than, say, when they are drunk. Here is my take on several game participants.
98.16 – What can I say? He is either so full of himself or just completely stupidÂ… or both.
50 – He is one of those students who never get that “100” because he writes down his answer before reading the question.
33.3 – Self-centered bastard!
8 – Cousin of “18”.
18 – He thinks that he is always the favored child by some unexplainable external force. Really he is just a person who willingly hand his fate over to someone else.
8.2 – He is calculated but believes the randomness of life. Well, if I’m right on this one, it is just because I’ve been reading your blog. | 1,147 | 4,453 | {"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} | 3.5 | 4 | CC-MAIN-2019-30 | longest | en | 0.952555 |
http://www.jiskha.com/display.cgi?id=1291636274 | 1,496,136,452,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463614620.98/warc/CC-MAIN-20170530085905-20170530105905-00327.warc.gz | 667,422,931 | 3,906 | # Physics
posted by on .
if a ball is thrown vertically into the air, leaving the ground at a height of 1.8m, with a speed of 9.7m s -1. When it rises and falls to the ground what will the balls speed be as it hits the ground? please help with an explanation as i am so stuck, my exam is in two days and i need help.
thanx
Becky
• Physics - ,
I would do this with energy.
Final total energy=initial total energy
1/2 m vf^2+zeroPE=mg(hi)+1/2 mvi^2
Vf^2=2g*1+vi^2
you may have had this formula presented to you for memory as..
• Physics - ,
Thank you Bob Pursley :)
• Physics - ,
Penejoldo
### Answer This Question
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### Related Questions
More Related Questions
Post a New Question | 206 | 731 | {"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-2017-22 | latest | en | 0.918215 |
http://www.ask.com/question/what-is-a-180-degree-angle-called | 1,394,361,386,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999677213/warc/CC-MAIN-20140305060757-00016-ip-10-183-142-35.ec2.internal.warc.gz | 225,230,097 | 16,490 | # What Is a 180 Degree Angle Called?
A 180 degree angle is also referred to as a straight angle or a half circle, while 360 degrees is a full rotation. An angle is the space within two or more planes that diverge from a common point. Other examples of geometrical angles are acute angles, obtuse angles, complementary angles, supplementary angles and others. | 78 | 359 | {"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} | 2.640625 | 3 | CC-MAIN-2014-10 | longest | en | 0.916803 |
https://www.econverter.net/force/pound-force/gram-force/ | 1,601,576,281,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402131986.91/warc/CC-MAIN-20201001174918-20201001204918-00403.warc.gz | 755,929,580 | 7,083 | 1 Pound-force = 453.59237 Gram-force
## How to convert from Pound-force to Gram-force?
### Pound-force to Gram-force?
To convert Pound-force to Gram-force: Every 1 Pound-force equals 453.59237 Gram-force. For example, 100 Pound-force equal 100 * 453.59237 = 45359.237 Gram-force and so on..
Pound-force to Newton | Pound-force to Dyne | Pound-force to Poundal | Pound-force to Kilogram-force | Pound-force to Kip-force | Pound-force to Ton-force (metric) |
## Pound-force to Gram-force Conversions Table
1 Pound-force = 453.59237 Gram-force
2 Pound-force = 907.18474 Gram-force
4 Pound-force = 1814.36948 Gram-force
5 Pound-force = 2267.96185 Gram-force
10 Pound-force = 4535.9237 Gram-force
20 Pound-force = 9071.8474 Gram-force
25 Pound-force = 11339.80925 Gram-force
50 Pound-force = 22679.6185 Gram-force
100 Pound-force = 45359.237 Gram-force
200 Pound-force = 90718.474 Gram-force
250 Pound-force = 113398.0925 Gram-force
500 Pound-force = 226796.185 Gram-force
1000 Pound-force = 453592.37 Gram-force
2000 Pound-force = 907184.74 Gram-force
2500 Pound-force = 1133980.925 Gram-force
5000 Pound-force = 2267961.85 Gram-force
10000 Pound-force = 4535923.69999 Gram-force
20000 Pound-force = 9071847.39999 Gram-force
25000 Pound-force = 11339809.24999 Gram-force
50000 Pound-force = 22679618.49997 Gram-force
100000 Pound-force = 45359236.99994 Gram-force
200000 Pound-force = 90718473.99988 Gram-force
500000 Pound-force = 226796184.9997 Gram-force
1000000 Pound-force = 453592369.9994 Gram-force
1 Gram-force = 0.0022 Pound-force
2 Gram-force = 0.00441 Pound-force
4 Gram-force = 0.00882 Pound-force
5 Gram-force = 0.01102 Pound-force
10 Gram-force = 0.02205 Pound-force
20 Gram-force = 0.04409 Pound-force
25 Gram-force = 0.05512 Pound-force
50 Gram-force = 0.11023 Pound-force
100 Gram-force = 0.22046 Pound-force
200 Gram-force = 0.44092 Pound-force
250 Gram-force = 0.55116 Pound-force
500 Gram-force = 1.10231 Pound-force
1000 Gram-force = 2.20462 Pound-force
2000 Gram-force = 4.40925 Pound-force
2500 Gram-force = 5.51156 Pound-force
5000 Gram-force = 11.02311 Pound-force
10000 Gram-force = 22.04623 Pound-force
20000 Gram-force = 44.09245 Pound-force
25000 Gram-force = 55.11557 Pound-force
50000 Gram-force = 110.23113 Pound-force
100000 Gram-force = 220.46226 Pound-force
200000 Gram-force = 440.92452 Pound-force
500000 Gram-force = 1102.31131 Pound-force
1000000 Gram-force = 2204.62262 Pound-force | 760 | 2,423 | {"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} | 3.578125 | 4 | CC-MAIN-2020-40 | latest | en | 0.407661 |
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Power series or polynomial with infinitely many terms
The sum of a power series is a function
Maclaurin and Taylor series
The radius of convergence or the interval of convergence
Power series or polynomial with infinitely many terms
A real power series in x around the origin (or centered at the origin) is a series of functions of the form
and the power series around a given point x = x0 (or centered at x0) is a series of the form
where the coefficients an are fixed real numbers and x is a real variable.
A power series with real coefficients is said to be real or complex according as both x and x0 are real or complex numbers.
Therefore, the nth partial sum of a power series is a polynomial of degree n,
The sum of a power series is a function
The sum of a power series is a function
the domain of which is the set of those values of x for which the series converges to the value of the function.
Maclaurin and Taylor series
Consider the polynomial function
f(x) = anxn + an - 1xn - - 1 + · · · a3x3 + a2x2 + a1x + a0.
If we write the value of the function and the values of its successive derivatives, at the origin, then
f(0) = a0, f '(0) = 1· a1, f ''(0) = 1· 2a2, f '''(0) = 1· 2· 3a3, . . . , f (n)(0) = n!an
so we get the coefficients;
Then, the polynomial f(x) with infinitely many terms, written as the power series
and
where 0! = 1, f (0)(x0) = f(x0) and f (n)(x0) is the nth derivative of f at x0,
represents an infinitely differentiable function and is called Maclaurin series and Taylor series respectively.
The radius of convergence or the interval of convergence
- If the power series anxn converges when x = x1, then it converges for every x that is closer to the
origin than x1, that is, whenever | x | < | x1 |.
- If the power series anxn diverges when x = x1, then it diverges for every x that is further from the origin
than x1, that is, whenever | x | > | x1 |.
The real power series converges for all absolute value of x that are less than a number r, called the radius of convergence or the interval of convergence, written | x | < that is, the open interval - r < x < r.
We apply the root test or the ratio test to find the interval of convergence. Thus,
the power series converges,
the power series diverges.
Therefore the inequality,
defines the interval in which the power series is absolutely convergent.
Thus, denoting the right side of the above inequality by r, we get the interval of convergence | x | < r saying, for every x between - r and r the series converges absolutely while, for every x outside that interval the series diverges.
Therefore, the radius of convergence of the power series,
The power series anxn converges absolutely at every point x from the open interval - r < x < r and diverges for all x outside this interval. At the endpoints - r and r, the series may converge or diverge so these points must be checked for convergence, individually.
The power series an(x - x0)n converges absolutely at every point x from the open interval |x - x0| < r or
x0 - r < x < x0 + r and diverges for all x outside this interval.
Note that the series may or may not converge when | x - x0 | = r that is, when x = x0 - r or x = x0 + r.
To determine whether the power series converge or diverge at endpoints we should plug each endpoint into the given series and apply appropriate test for convergence.
Functions contents E | 898 | 3,423 | {"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} | 4.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.889061 |
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4个回答
event对象没有做兼容性处理
function inp(){
var event = arguments.callee.caller.arguments[0] || window.event;
var code= event.which || event.keyCode; | 1,294 | 4,395 | {"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-2020-16 | latest | en | 0.107146 |
https://robertlovespi.net/tag/truncated-cuboctahedron/ | 1,685,494,784,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646181.29/warc/CC-MAIN-20230530230622-20230531020622-00304.warc.gz | 565,984,879 | 39,375 | # Packing Space with Great Rhombcuboctahedra and Octagonal Prisms
…And so on….
[Software credit: I made these images using Stella 4d, which you can try for free right here.]
# Three Archimedean Solids Which Fill Space Together: The Great Rhombcuboctahedron, the Truncated Tetrahedron, and the Truncated Cube
To start building this space-filling honeycomb of three Archimedean solids, I begin with a great rhombcuboctahedron. This polyhedron is also called the great rhombicuboctahedron, as well as the truncated cuboctahedron.
Next, I augment the hexagonal faces with truncated tetrahedra.
The next polyhedra to be added are truncated cubes.
Now it’s time for another layer of great rhombcuboctahedra.
Now more truncated tetrahedra are added.
Now it’s time for a few more great rhombcuboctahedra.
Next come more truncated cubes.
More great rhombcuboctahedra come next.
More augmentations using these three Archimedean solids can be continued, in this manner, indefinitely. The images above were created with Stella 4d: Polyhedron Navigator, a program you may try for yourself at http://www.software3d.com/Stella.php.
# ZigZag: A Faceting of the Great Rhombcuboctahedron
I made this using Stella 4d: Polyhedron Navigator. You can try this program for free at http://www.software3d.com/Stella.php.
# A Faceted Great Rhombcuboctahedron
Some prefer to call the great rhombcuboctahedron the “truncated cuboctahedron,” instead. Whichever term you prefer, this is a faceted version of that Archimedean solid. I made it using Stella 4d: Polyhedron Navigator, software you may find here.
# 494 Circles, Each, Adorning Two Great Rhombcuboctahedra, with Different (Apparent) Levels of Anxiety
The design on each face of these great rhombcuboctahedra is made from 19 circles, and was created using both Geometer’s Sketchpad and MS-Paint. I then used a third program, Stella 4d (available here), to project this image on each of a great rhombcuboctahedron’s 26 faces, creating the image above.
If you watch carefully, you should notice an odd “jumping” effect on the red, octagonal faces in the polyhedron above, almost as if this polyhedron is suffering from an anxiety disorder, but trying to conceal it. Since I like that effect, I’m leaving it in the picture above, and then creating a new image, below, with no “jumpiness.” Bragging rights go to the first person who, in a comment to this post, figures out how I eliminated this anxiety-mimicking effect, and what caused it in the first place.
Your first hint is that no anti-anxiety medications were used. After all, these polyhedra do not have prescriptions for anything. How does one “calm down” an “anxious” great rhombcuboctahedron, then?
On a related note, it is amazing, to me, that simply writing about anxiety serves the purpose of reducing my own anxiety-levels. It is an effect I’ve noticed before, so I call it “therapeutic writing.” That helped me, as it has helped me before. (It is, of course, no substitute for getting therapy from a licensed therapist, and following that therapist.) However, therapeutic writing can’t explain how this “anxious polyhedron” was helped, for polyhedra can’t write.
For a second hint, see below.
.
.
[Scroll down….]
.
.
Second hint: the second image uses approximately twice as much memory-storage space as the first image used.
# Unsquashing the Squashed Meta-Great-Rhombcuboctahedron
I noticed that I could arrange eight great rhombcuboctahedra into a ring, but that ring, rather than being regular, resembled an ellipse.
I then made a ring of four of these elliptical rings.
After that, I added a few more great rhombcuboctahedra to make a meta-rhombcuboctahedron — that is, a great rhombcuboctahedron made of rhombcuboctahedra. However, it’s squashed. (I believe the official term for this is “oblate,” but “squashed” also works, at least for me.)
So now I’m wondering if I can make this more regular. In other words, can I “unsquash” it? I notice that even this squashed metapolyhedron has regular rings on two opposite sides, so I make such a ring, and start anew.
I then make a ring of those . . .
. . . And, with two more ring-additions, I complete the now-unsquashed meta-great-rhombcuboctahedron. Success!
To celebrate my victory, I make one more picture, in “rainbow color mode.”
[All images made using Stella 4d, available here: http://www.software3d.com/Stella.php.]
# A Non-Convex Variant of the Great Rhombcuboctahedron
I made this with Stella 4d, software you can try here.
# A Great Rhombcuboctahedron, Decorated with Circles and Hexagons
The images on the faces of this polyhedron were created with Geometer’s Sketchpad and MS-Paint. Projecting these images onto these faces, and then creating this rotating image, was accomplished using Stella 4d: Polyhedron Navigator— a program you can try for yourself, for free, at http://www.software3d.com/Stella.php.
## Five of the Thirteen Archimedean Solids Have Multiple English Names
### Image
I call the polyhedron above the rhombcuboctahedron. Other names for it are the rhombicuboctahedron (note the “i”), the small rhombcuboctahedron, and the small rhombicuboctahedron. Sometimes, the word “small,” when it appears, is put in parentheses. Of these multiple names, all of which I have seen in print, the second one given above is the most common, but I prefer to leave the “i” out, simply to make the word look and sound less like “rhombicosidodecahedron,” one of the polyhedra coming later in this post.
My preferred name for this polyhedron is the great rhombcuboctahedron, and it is also called the great rhombicuboctahedron. The only difference there is the “i,” and my reasoning for preferring the first name is the same as with its “little brother,” above. However, as with the first polyhedron in this post, the “i”-included version is more common than the name I prefer.
Unfortunately, this second polyhedron has another name, one I intensely dislike, but probably the most popular one of all — the truncated cuboctahedron. Johannes Kepler came up with this name, centuries ago, but there’s a big problem with it: if you truncate a cuboctahedron, you don’t get square faces where the truncated parts are removed. Instead, you get rectangles, and then have to deform the result to turn the rectangles into squares. Other names for this same polyhedron include the rhombitruncated cuboctahedron (given it by Magnus Wenninger) and the omnitruncated cube or cantitruncated cube (both of these names originated with Norman Johnson). My source for the named originators of these names is the Wikipedia article for this polyhedron, and, of course, the sources cited there.
This third polyhedron (which, incidentally, is the one of the thirteen Archimedean solids I find most attractive) is most commonly called the rhombicosidodecahedron. To my knowledge, no one intentionally leaves out the “i” after “rhomb-” in this name, and, for once, the most popular name is also the one I prefer. However, it also has a “big brother,” just like the polyhedron at the top of this post. For that reason, this polyhedron is sometimes called the small rhombicosidodecahedron, or even the (small) rhombicosidodecahedron, parentheses included.
I call this polyhedron the great rhombicosidodecahedron, and many others do as well — that is its second-most-popular name, and identifies it as the “big brother” of the third polyhedron shown in this post. Less frequently, you will find it referred to as the rhombitruncated icosidodecahedron (coined by Wenninger) or the omnitruncated dodecahedron or icosahedron (names given it by Johnson). Again, Wikipedia, and the sources cited there, are my sources for these attributions.
While I don’t use Wenninger’s nor Johnson’s names for this polyhedron, their terms for it don’t bother me, either, for they represent attempts to reduce confusion, rather than increase it. As with the second polyhedron shown above, this confusion started with Kepler, who, in his finite wisdom, called this polyhedron the truncated icosidodecahedron — a name which has “stuck” through the centuries, and is still its most popular name. However, it’s a bad name, unlike the others given it by Wenninger and Johnson. Here’s why: if you truncate an icosidodecahedron (just as with the truncation of a cuboctahedron, described in the commentary about the second polyhedron pictured above), you don’t get the square faces you see here. Instead, the squares come out of the truncation as rectangles, and then edge lengths must be adjusted in order to make all the faces regular, once more. I see that as cheating, and that’s why I wish the name “truncated icosidodecahedron,” along with “truncated cuboctahedron” for the great rhombcuboctahedron, would simply go away.
Here’s the last of the Archimedean solids with more than one English name:
Most who recognize this shape, including myself, call it the truncated cube. A few people, though, are extreme purists when it comes to Greek-derived words — worse than me, and I take that pretty far sometimes — and they won’t even call an ordinary (Platonic) cube a cube, preferring “hexahedron,” instead. These same people, predictably, call this Archimedean solid the truncated hexahedron. They are, technically, correct, I must admit. However, with the cube being, easily, the polyhedron most familiar to the general public, almost none of whom know, let alone use, the word “hexahedron,” this alternate term for the truncated cube will, I am certain, never gain much popularity.
It is unfortunate that five of the thirteen Archimedean solids have multiple names, for learning to spell and pronounce just one name for each of them would be task enough. Unlike in the field of chemistry, however, geometricians have no equivalent to the IUPAC (International Union of Pure and Applied Chemists), the folks who, among other things, select official, permanent names and symbols for newly-synthesized elements. For this reason, the multiple-name problem for certain polyhedra isn’t going away, any time soon.
(Image credit: a program called Stella 4d, available at www.software3d.com/Stella.php, was used to create all of the pictures in this post.) | 2,569 | 10,205 | {"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-2023-23 | latest | en | 0.883711 |
https://www.jiskha.com/display.cgi?id=1268151030 | 1,516,413,146,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888341.28/warc/CC-MAIN-20180120004001-20180120024001-00250.warc.gz | 939,630,067 | 4,023 | # physics
posted by .
A 380 turn solenoid of length 32.0 cm and radius 3.10 cm carries a current of 4.80 A. Find the following.
(a) the magnetic field strength inside the coil at its midpoint
mT
(b) the magnetic flux through a circular cross-sectional area of the solenoid at its midpoint
T ยท m2
• physics -
(4pi*10^-7)(4.8*380/.32)
=____mT
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More Similar Questions | 766 | 2,860 | {"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-2018-05 | latest | en | 0.894007 |
http://petersonbiology.com/math230Notes/introHypothesisTestsAndPvalues.html | 1,679,625,516,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945242.64/warc/CC-MAIN-20230324020038-20230324050038-00742.warc.gz | 37,134,081 | 10,799 | # An introduction to statistics in R
### A series of tutorials by Mark Peterson for working in R
There are no data sets for today. We will be inputting a few examples of our own instead.
## 14.2 Overview
This chapter introduces the concept of hypothesis testing and the language associated with it. We have spent a lot of time working on how to generate a sample from a population. Here, we begin the process of using a sample to infer things about a population.
To start, we want to focus our attention a little more closely on how to generate a hypothetical sampling distribution, given an idea about what we think the should look like.
## 14.3 Definitions
I can not stress enough how important it is that you read this section.
The concepts here will be vital to where we go next. We will discuss them in context as they arise, but there is no substitute for understanding them now.
### 14.3.1 Hypotheses
#### 14.3.1.1 Null Hypothesis
The null hypothesis, written in the notation H0, (that is an “H” with a zero subscript; pronounce “H nought”) states that there is no effect or difference, or that the mean is some set level. The null hypothesis is always a specific value (often 0), and cannot be a range of values. It must be a specific value, because we are going to use that specific value to imagine what we expect would happen if it were true.
One way to think about the null hypothesis is to think of it as the most boring thing that could be true about the data. It is the status quo, the default, the complete lack of an effect or difference. We will see these in action many times, but make sure to think through carefully.
#### 14.3.1.2 Alternative Hypothesis
The alternative hypothesis, written Ha, is the claim that we are seeking to support, namely, that there is an effect or differnce. Said another way, the alternative hypothesis is simply that the null is not true. The alternative hypothesis is not a point estimate, but rather encompasses a range of possible values that are different than the point estimate made by H0.
#### 14.3.1.3 Put together
It is important to remember that we always need both a null and an alternative hypothesis. If we want to know if NFL linemen are heavier than skill position players, we need a null (that they are not different weights) as well as an alternative (that they are different). It is also crucial to remember that what we are actually testing (see significance below) is how unlikely it would be to get our data (or more extreme) if the null hypothesis is true.
### 14.3.2 Statistical test
A statistical test is the use of a sample to assess a statement about a population.
We can know the value of a sample statistic exactly (e.g., we know that exactly 7 of the 8 beads were red), but we often don’t (and can’t) know the value of a population parameter exactly (e.g., we can’t weigh every person in the U.S). A statistical test is a way of deciding if our estimate from a sample tells us something about the population as a whole.
Specifically, for most of this course, we are going to ask how likely the data were to occur if the null hypothesis were true. We will use this information to decide about the “significance” of our result.
### 14.3.3 Statistical significance
When results are so extreme that we determine it is unlikely that they would have occurred if the null hypothesis is true, we say that the result it statistically significant. In this case, we “reject” H0. If we reject H0, it gives evidence to support our Ha. However, this is not a guarantee that Ha is true.
An important point here is that we never “accept” H0. Even data that perfectly match our expectations if H0 were true could have occured under some other true population parameter. A traditional null-hypothesis significance test can only reject or not reject H0 (the null).
### 14.3.4 P-value
Some of you may have encountered the idea of p-values before, for others, this may be your first exposure to the term. In either case, we will spend a substantial amount of time, this chapter and throughout the semester, discussing and interpretting p-values This definition is so important to everything we are doing that I am highlighting it here:
Definition of a P-value
A p-value is the probability of generating data that are as extreme as, or more extreme than, your collected data, given that the null hypothesis was true.
Put another way, a p-value is the probability that your data (or more extreme data) would have been generated by chance if the null hypothesis is an accurate description of the world. In this way, we are essentially asking how rare our data would be if the null hypothesis is true.
We care about this because it allows us to say how unexpected our data are. If they are very unlikely (have a low p-value), we can reject the null hypothesis and say that it is unlikely to be true because it is unlikely to have produced these data. We will see more about how to decide how low is low enough to make this decision in the coming chapters.
Many books and materials focus on the idea of p-values long before they show you how to generate these values. I think it is instructive to start here. So, here, we are going to focus on how to generate p-values for some simple cases, then we will spend time on interpretation throughout the next chapters.
Pay attention to these worked examples (and your assigned problems), as they are going to be very helpful to your understanding of the underlying concepts. I will endeavor to spend class time focusing on applications.
## 14.4 A coin-flipping example
If we think that someone might have a heads-biased coin (a coin that comes up heads more often than tails), we might want to test that hypothesis. For this, we can set our null hypothesis as a “fair” coin (one that is equally likely to come up heads or tails), and our alternative hypothesis as a probablity of heads greater than 0.5. In math notation:
So, if we flip the coin 10 times and get 7 heads, do we think the coin is biased?
One question we can ask is how likely that result is by chance. To do that, lets flip 10,000 fair coins ten times each and see how often we get seven (or more) heads.
This might take a while. Let’s try R instead.
First, save the data that we collected. (Note that, if you have data in a data.frame already, this would likely mean just saving a column or subset of one.)
# Store our data
theData <- rep( c("H","T"),
c( 7 , 3))
# Save the number of heads
# View the data
table(theData)
## theData
## H T
## 7 3
Next, we need to draw many, many samples from our null distribution (the distribution that matches H0). The loop we make to do that should look very familiar. It is very, very similar to a bootstrap or sampling loop. The only difference is that, instead of re-sampling from our collected samples or a known population, we instead take samples from a distribution that matches our H0.
Here, that means from a coin that comes up heads and tails equally often. As with our other sampling, when we generate our samples from H0, it is very important that we use the same size as our collected data. We want to match everything we can about our sample collection method.
First, it is always a good idea to take one simple sample. This is a good way to get a sense of what is happening, and is an opportunity to catch errors that might creep into the proccess.
# Create a fair coin for sampling
fairCoin <- c("H","T")
# Generate a "test" sample
tempData <- sample( fairCoin,
size = length(theData),
replace = TRUE)
# Look at the output
tempData
## [1] "T" "T" "H" "H" "T" "T" "T" "H" "H" "H"
# Count the number of heads
sum(tempData == "H")
## [1] 5
So, we see that we got a series of “H” and “T”, and that we can count up that we got 5 heads. All of this looks reasonable, so we can proceed with making our loop. If you were doing this yourself (instead of copying what I have below), you would likely copy your sampling lines above, and just make sure that you store the sum(tempData == "H") into a variable (that’s how I made this one).
# Initialize our variable
# Note that all we need to save is the number of heads
for(i in 1: 10000){
# Generate a sample
tempData <- sample( fairCoin,
size = length(theData),
replace = TRUE)
# Count the number of heads
}
# Visualize our results
xlab = "Number of heads in sample",
main = "Null distribution of 10 coin flips")
This histogram shows us a nice, unimodal, symmetric distribution. As expected, it is centered at 5 heads and 5 tails. Because the values can only range from 0 to 10, we can look at the data directly with a table. Note, however, that for many (most) data sets that is a bad idea, and that the histogram will be much more informative.
# Make a table (Note: only do this with small samples)
table(nHeadsSamples)
0 1 2 3 4 5 6 7 8 9 10
11 101 465 1157 2051 2518 2024 1137 421 103 12
The first thing this shows us is that, when we flip a fair coin 10 times, it will occasionally come up all heads or all tails, just by chance.
Now that we have this, we want to calculate how likely it was to get 7 heads in 10 flips of a coin we know is fair. First, let’s add a line to the plot to show where we are looking:
abline(v = nHeadsData,
col = "red3", lwd =3)
Note here that, because of the way the histogram is built, the values right at the line (i.e. 7 heads) appear in the bar just left of the line. That is ok, just make sure to keep that in mind as we consider the value we are about to calculate. An alternative plot might help in this case, though it will apply in so few situations that I do not want you to try to make this plot[*] For those of you that want to do it anyway: I did this using barplot, but barplot doesn’t put things quite where you would expect. If you save the output of barplot (e.g. spots <- barplot(table(1:3))) you can see where the bars are actually centered. This is more than you need to do for class, but if you are interested, I at least wanted to give you an explanation for why your likely first attempts would have given odd results.
What we want to figure out is how many of our samples from the null distribution gave us 7 or more heads. Recall that the p-value asks about data as extreme as, or more extreme than, the data collected. So, we get to introduce a new logical test operator, the >= for “greater than or equal to”. We count up the number that are as or more extreme, and divide by the total number. Because that is the same as calculating a mean, we can do it in one short step. However, don’t confuse this with calculating the mean of a numeric variable.
# Test likelihood of our data
propAsExtreme
## [1] 0.1673
This suggests that we are likely to get 7 or more heads in 10 flips roughly 17% of the time we flip a fair coin. This seems fairly common (about 1 time in 6, so this probably doesn’t give us enough reason to reject the null hypothesis that the coin is fair. This does not mean that the coin is neccesarily fair; it only means that we cannot say it is not fair.
### 14.4.1 Try it out
Imagine that instead of getting 7 heads in 10 flips, we had gotten 10 heads in our 10 flips. How often is that likely to have occurred by chance, if the coin is fair? What does that suggest to you? Our null and alternative hypotheses remain the same.
You should get a value of about 0.0011, though your answer will vary a bit because of the random sampling. Note that your value may be printed in scientific notation. If you get something like 11e-04, it means $$11*10^{-4}$$. The number after the e is the exponent on the 10. It is just a short hand way of showing scientific notation (and is common in many programs). Don’t forget to plot your null distribution either; you should get something like this
## 14.5 (Sample) size matters
Just like in our other distributions (bootstrapping and sampling) e once again find that sample size matters in these tests What if, instead of flipping the coin 10 times, we flipped it 100? Imagine if we got the same proportion of heads, giving us 70 heads in 100 flips. Now, how unlikely is this.
Here, you will (hopefully) see why I used some of the variables I did above. Copy and paste all of what you just did. All we need to change is our initial data theData to have 70 heads and 30 tails, instead of 7 and 10. If instead of using length() in the loop I had typed in 10, we’d now have to change that as well.
# Store our data, for ease of access
theData <- rep( c("H","T"),
c( 70, 30))
# Save the number of heads
# Create a fair coin for sampling
# Note, we use one of each face
fairCoin <- c("H","T")
#####################
# Run the full loop #
#####################
# Initialize our variable
# Note that all we need to save is the number of heads
for(i in 1: 10000){
# Copy the test sample into our loop
# (or build the loop around it)
# Generate a sample
tempData <- sample( fairCoin,
size = length(theData),
replace = TRUE)
# Count the number of heads
}
# Make a table (Note: only do this with small samples)
# Now it is a bad idea
# Visualize our results
col = "red3", lwd =3)
In all of this code, the only things I changed were the original data, and commenting out the table (it would be quite large). Now, again with exactly the same code, we can again test how many of our samples had as many (or more) heads
than our original sample (70).
# Test likelihood of our data
propAsExtreme
## [1] 0
Here, I got no samples with 70 or more heads from the null distribution. In a large class, at least a few of you will get one or two. Still, with that few occurrences, we can rather comfortably say that the data (70 heads in 100 flips) were unlikely to have happened by chance if the coin were fair. So, we reject the null hypothesis, and say that we think the coin is unlikely to be fair.
### 14.5.1 Try it out
How likely is it to get 110 heads in 200 flips? Calculate and interpret a p-value. Don’t forget a plot.
You should get a p-value of about 0.09.
## 14.6 Two tails?
What if, however, didn’t suspect that the coin was heads biased, but just that it might be biased? Would only checking things greater than 7 or 70 make sense?
Instead, we usually use a two-tailed test. This way, we will include the biases in both directions, rather than only one. For almost every hypothesis you test, a two-tailed test will be more appropriate. Only if you have a strong reason to expect a particular direction of difference (e.g., if Ha is greater than or less than, rather than just not equal), is a one-tailed test appropriate.
There are two ways of getting this value: doubling the one-tailed value and actually measuring both. The former is a bit simpler, but the latter is more robust to problems caused by non-symmetrical data.
For you, go back and run the example with only 10 flips, as the effects are easier to see. In most circumstances, you would want to use different variable names, so that you can keep testing both things.
And a table:
0 1 2 3 4 5 6 7 8 9 10
11 104 435 1197 2102 2417 2002 1185 442 96 9
To test the first, we can just regenerate the proportion we calculated before, and double it.
# Test likelihood of our data
propAsExtreme * 2
## [1] 0.3464
Now, we see that such a bias is likely to occur about 1 time in 3, quite likely.
To test more formally, we can add a second value, one for below our expected values.
To do this, we first calculate how many heads we expected if the coin were fair.
# how many heads we expect
nHeadsExpected
## [1] 5
As your intution told you, this is the middle of the null distribution. Next, we figure out how far away from that value our data were. For this, it is often best to use the absolute value, calculated using abs() in R. This way, the value is always positive, and we don’t need to worr as much about getting our directions right.
# How extreme our data are
ourDiff
## [1] 2
Finally, we calculate how many values are as (or more extreme) as our original sample. That is, how many were as far away, or further, from the expected value in either direction.
# Calculate the propotion that are that extreme
# in the other direction
propAsExtremeLess + propAsExtremeMore
## [1] 0.3479
This gives us a value very similar to the above, but it is a bit safer. We can look at the values separately to see why this might be the case:
propAsExtremeLess propAsExtremeMore
0.1747 0.1732
While the values are very similar, there are slight differences. In more complicated analyses, these differences may be meaningful.
### 14.6.1 Try it out
Calculate the two-tailed p-value for the example above with 105 head in 200 flips.
## 14.7 A note on significance thresholds
We can all probably agree that 7 out of 10 heads is not extreme enough to reject our null, but that 70 out of 100 is. Somewhere between these two extremes lies a dividing line of acceptability. We will spend this week discussing and finding that line. | 4,110 | 16,936 | {"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": 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} | 4 | 4 | CC-MAIN-2023-14 | latest | en | 0.951485 |
https://gromacs.bioexcel.eu/t/integration-of-g-r/739 | 1,652,923,488,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522741.25/warc/CC-MAIN-20220519010618-20220519040618-00462.warc.gz | 341,159,700 | 4,774 | # Integration of g(r)
GROMACS version: 2018
GROMACS modification: No
Hi everyone,
Recently, I am confused by the result of integration of g®. In the manul, the definition of g®_AB is g®_AB = pho®/ pho_local, the r_max for pho_local is the half of box.
Based on the definition, the integration from zero to rmax of g®_AB*pho_local will be the number of B (smaller than total number of B) in the sphere with radius r_max.
However, when I calculated the integration, I found that the result is bigger than the total number of B.
For example, if there are 26 A monomers and 26 B monomers in the system, the integration of g_ABpho_local will smaller than 26. But I got the value larger than 46.
The pho_local = average_number/(4.0/3
pirmaxrmax*rmax), here average_number is accumulate number of B at distance rmax got from the output file using gmx rdf -cn
## the integration script i used:
double dr = rr[1]-rr[0];
sumb = 0;
double psv = 0.0,rri;
for(i = 0; i < ri; i ++){
rri = (i + 0.5) * dr;
double sv = (4.0 / 3.0) * PI * rri * rri * rri;
double dsv = sv-psv;
sumb += (gr[i])*(dsv);
psv = sv;
cout<<"sumb = "<<sumb<<endl;
}
double pho_local = 24.8/((4.0 / 3.0) * PI * rri * rri * rri);
cout<<"total monomer number ~: "<<sumb*pho_local<<endl;;
I do not find anything wrong in my script. So could anyone help me to explain why the result is unexpected. Is there some misunderstanding of the definition of g® that GROMACS used to analyze data?
Best
isimuly
Hi isimuly,
purely following the math of (5.388) in the manual (http://manual.gromacs.org/2021-beta1/manual-2021-beta1.pdf) and using A=B you would obtain the number of particles from the distribution function by integrating the following way:
\langle \rho_{local}\rangle \int dr 4\pi r^2 g_{AA}(r) = 1/N_A \int d r \sum_{i\in A}\sum_{j\in A}\delta(r_{ij}-r) = N_A^2/N_A = N_A
(the integral on the right hand side over the delta function just counts the A,A pairs)
I agree that it’s a bit counter-intuitive to use 4\pi r^2 instead of the difference of the sphere volumes; it might help to think in terms of the continuous integral first and then approximate. There, you see that you obtain the factor by integrating in spherical coordinates over the complete angle range for \phi, \psi for a volume element r^2 \sin\phi \,\mathrm{d}\phi \,\mathrm{d}\psi \,\mathrm{d}r | 701 | 2,337 | {"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-2022-21 | latest | en | 0.777347 |
https://math.snu.edu.in/node/8279 | 1,657,022,336,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104576719.83/warc/CC-MAIN-20220705113756-20220705143756-00545.warc.gz | 422,676,183 | 10,445 | Geometry of Curves & Surfaces | Department of Mathematics
Geometry of Curves & Surfaces
A Major Elective for B.Sc. (Research) Mathematics.
Credits (Lec:Tut:Lab)= 3:1:0 (3 lectures and 1 tutorial weekly)
Prerequisites: MAT 102 Calculus II or MAT 103 Mathematical Methods I. And MAT 160 Linear Algebra I.
Overview: This course combines the traditional approach to learn the basic concepts of curves and surfaces with the symbolic manipulative abilities of Mathematica. Students will learn and study the classical curves/surfaces as well as more interesting curves/surfaces using computer methods. For example, to see the effect of change of parameter, the student will explore and observe with the help of Mathematica and then the mathematical proof of the observation will be developed in the class.
Detailed Syllabus:
1- Curves in the plane: Length of a curve, Vector fields along curves. Famous plane curves: cycloids, lemniscates of Bernoulli, cardioids, catenary, cissoid of Diocles, tractrix, clothoids, pursuit curves.
2- Regular curve, curvature of a curve in a plane, curvature and torsion of a curve in R3. Determining a plane curve from given curvature.
3- Global properties of plane curves: Four vertex theorem, Isoperimetric inequality.
4- Curves on Sphere. Loxodromes on spheres, animation of curves on a sphere.
5- Review of calculus in Euclidean space.
6- Surfaces in Euclidean spaces: Patches in R3, local Gauss map, Regular surface, Tangent vectors.
7- Example of surfaces: Graphs of a function of two variables, ellipsoid, stereographic ellipsoid, tori, paraboloid, seashells.
8- Orientable and Non-orientable surfaces. Mobius strip, Klein Bottle.
9- The shape operator, normal curvature, Gaussian and mean curvature, fundamental forms.
10- Surfaces of revolution
References:
1. Modern Differential Geometry of Curves and Surfaces with Mathematica, Third Edition by Elsa Abbena, Simon Salamon, Alfred Gray.
2. Elementary Differential Geometry by A.N. Pressley, Springer Undergraduate Mathematics Series.
3. Differential Geometry of Curves and Surfaces by Manfredo DoCarmo. | 489 | 2,107 | {"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-2022-27 | latest | en | 0.842681 |
http://www.algonotes.com/en/minimal-memory-dp/ | 1,558,731,696,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257767.49/warc/CC-MAIN-20190524204559-20190524230559-00371.warc.gz | 234,910,363 | 25,094 | # Dynamic programming on trees with minimal memory
Dynamic programming on trees has always been a beloved topic of mine. I think it nicely teaches fundamental principles of DP and recurrence, at least on a theoretical level. But in practice, when one has to implement a concrete program using DP on trees, there are some traps which have to be avoided. For instance, one has to be aware of the fact that usually such a program consumes pretty decent amount of memory. Asymptotically it is linear on the number of vertices in the tree, but the constant could be quite big. In this article we will show how much of this memory is really necessary when programming in C++, the most used language in programming competitions.
We will illustrate it with a classic problem of counting the number of maximum-size matchings in an undirected tree. There is nothing special about this particular problem, apart from the fact that I like it and it has nice educational properties – it is not very complicated, but at the same time no trivial and it nicely exemplifies what we can expect from DP on trees. Therefore most of the things we say here can be applied to other DP on trees (maybe except the very last bits, in which we will use some features of this particular problem). Especially the part dealing with reading a tree from the input to the memory is universal and interesting all by itself.
This article features task Matchings that I have prepared for Polish Olympiad in Informatics Training Camp in 2010 and used once more at Brazilian ICPC Summer School in 2015. It is significantly expanded version of a lecture I gave at the latter camp.
## Definition of the problem and DP recurrences
You can try solving problem Matchings from Polish Olympiad in Informatics Training Camp 2010.
We are given an undirected tree which has $n$ vertices and $n-1$ edges. A matching is a subset of edges in which every vertex is adjacent to at most one edge from the subset. The size of a matching is the number of edges in this subset. We want to calculate the maximum size of a matching and also the number of matchings of this size. The size of a matching is limited by $\frac{n}{2}$, since each edge in a matching “uses” two vertices. However, the number of matchings can be exponential (in a star-shaped tree where each ray has three vertices, there are $\frac{n}{3} 2^{n/3}$ matchings), so we use a standard trick and ask for this number modulo given integer $M$.
The DP solution for this problem is not very hard. As usual, we pick an arbitrary vertex $v_\star$ and make it a root of the tree, then we direct every edge to make a directed tree. Finally we traverse this tree from leaves to the root, performing some calculations for every vertex $v$ based on calculations done previously for its children $u_0, \ldots, u_{m-1}$.
### Finding the maximum size
First, let's concentrate on finding the maximum size. We will proceed bottom-up and for every vertex $v$ we will calculate the maximum size of a matching in a subtree rooted in vertex $v$. In fact we will calculate two values: $dt[v]$ – the maximum size of a matching in the subtree which uses vertex $v$, i.e. there is an edge $vu_i$ for some $i$ in the matching, and $dn[v]$ – the maximum size of a matching in the subtree which does not use vertex $v$, i.e. there is no such edge.
The recurrence for these values is as follows. When the matching does not use vertex $v$, all its children $u_i$ are free to be used (or not) in their respective subtrees which can be considered independently. Thus it is:
$dn[v] = \sum_{i} \max(dt[u_i], dn[u_i])$
For case when the matching uses vertex $v$, we have to decide which edge $vu_i$ goes into the matching. Then in the subtree rooted in $u_i$ we cannot use this vertex, since it is already matched with its parent. Of course we use the edge which results in a matching of the largest size. Therefore
$dt[v] = \max_{i} (1 + dn[u_i] + \sum_{j \neq i} \max(dt[u_j], dn[u_j]))$
We assume that sum over empty set is $0$, and max over empty set is $-\infty$, thus for a leaf $v$ we have $dn[v] = 0$ and $dt[v] = -\infty$. The above formula for $dt[v]$ is not very efficient, since it calculates sums over vertices many times. It is better to calculate the sum once (and in fact this is exactly what we have done in $dn[v]$), and then remove one element from it:
$dt[v] = dn[v] + \max_{i} (1 + dn[u_i] - \max(dt[u_i], dn[u_i]))$
Now calculating $dn[v]$ and $dt[v]$ takes linear time in number of children of vertex $v$, therefore calculating these values for all vertices takes $O(n)$.
We can simplify these formulas a little bit. Assume that for some vertex $v$ we have $dn[v] \geq 0$, therefore there is a non-empty matching which does not use $v$. If we consider a matching of maximum size and a top-most vertex which is matched in it by some edge, we can (since vertices above it are not used) move this edge to the top, so vertex $v$ will be used by it. Thus we have $dt[v] \geq dn[v]$. On the other hand if $dn[v] = 0$, then either $v$ is a leaf (and then $dt[v] = -\infty$) or $dt[v] \geq 1$. So let's fix the definition of $dt[v]$ to be $0$ in case when $v$ is a leaf. Thanks to that change we will always have $dt[v] \geq dn[v]$, thus $dt[v] = \max(dt[v], dn[v])$ and the formulas become:
$dn[v] = \sum_{i} dt[u_i]$
$dt[v] = dn[v] + \max_{i} (1 + dn[u_i] - dt[u_i])$
And for leaves we have $dn[v] = dt[v] = 0$.
### Finding the number of maximum-size matchings
Now let's proceed to counting the matchings. Again, for every vertex $v$ we introduce two variables $cn[v]$ and $ct[v]$ – the numbers of maximum-size matchings in subtree $v$ which (do not) use vertex $v$. We also use notation $c[v]$ to denote a number of maximum-size matchings in the subtree, regardless of situation of vertex $v$. Since $dt[v] \geq dn[v]$, then $ct[v]$ is always included, but we include $cn[v]$ only if $dn[v]$ is equal to $dt[v]$:
$c[v] = ct[v] + cn[v] \cdot [ dn[v] = dt[v] ]$
For $cn[v]$, that is when calculating number of maximum-size matchings which do not use vertex $v$, we independently consider all subtrees rooted in children of vertex $v$, so we just take a product of numbers of all maximum-size matchings in these subtrees:
$cn[v] = \prod_{i} c[u_i]$
For calculating $ct[v]$ we must iterate over each possible edge $vu_i$, and see whether the maximum size of a matching using this edge equals to $dt[v]$. In case the answer is positive, we replace $c[u_i]$ with $cn[u_i]$ in already calculated product $cn[v]$ (because we do not want to recalculate the product again), and finally take a sum of such products:
$ct[v] = \sum_{i} ( cn[v] / c[u_i] \cdot cn[u_i] ) \cdot [ dt[v] = dn[v] + 1 + dn[u_i] - dt[u_i] ]$
Unfortunately, there is a slight problem with the above formula – we cannot perform division. It would be perfectly fine, if we had worked with numbers in full precision, since $c[v]$ is never equal to $0$ and we can do normal division. But since we are working with numbers modulo $M$, we have to multiply by inversion, and $c[u_i]$ could simply evaluate to $0$ modulo $M$. (And if $M$ was not a prime number, even more numbers wouldn't have their inversions.)
Therefore we need to use a slightly different approach, but since it is a little bit technical, we will postpone it to a later part of the article. For now just suppose that it can be done, in linear time and with no additional memory.
If vertex $v_\star$ is the root of the tree, then numbers $dt[v_\star]$ and $c[v_\star]$ are the final answer to the problem. Therefore, we have an algorithm which solves the problem in optimal time $O(n)$ and memory complexity $O(n)$. To be more precise, dynamic programming calculations use four variables per vertex, making in a total of $4n$ variables.
## The standard approach for implementation
We will make some assumptions on implementation. We assume that the standard integer type int is big enough to hold numbers $2n$ and $M-1$, but not necessarily bigger. We also assume that all pointers to the memory are of the same size as this int. The codes below will be written in C++ with the help of one macro:
#include <algorithm> #include <cstdio> #include <vector> using namespace std; #define REP(i,n) for(int i=0;i<(n);++i)
We assume that all global variables and arrays are initialized to $0$.
The implementation consists of two phases: reading the description of the tree from the standard input, and traversing the tree and performing DP calculations.
First we need to know how the tree is given in the input. The most common way for specifying graphs (and trees in particular) is by listing their edges. More precisely, first we number vertices of the tree (e.g. using integers from $0$ to $n-1$), and then we put number $n$ in the input, followed by $n-1$ pairs $a, b$ which mean that vertex number $a$ is adjacent to vertex number $b$. Note that we number vertices only for convenience of such representation. We do not care that a particular vertex has some particular number.
The easiest way to input such a tree is to create for each vertex a vector $adj[v]$ which will store all vertices adjacent to $v$. The code is as follows:
const int N = 1000000; int n; vector<int> adj[N]; void load_tree() { scanf("%d", &n); REP(i, n-1) { int a, b; scanf("%d%d", &a, &b); adj[a].push_back(b); adj[b].push_back(a); } }
The whole tree is defined in the input using about $2n$ ints (to be precise one int to describe the size and $2n-2$ ints to describe edges, but we will ignore additive constants here). Let's calculate how much memory does it need.
We have $n$ vectors, each consists of three pointers: start of the allocated buffer (pointer begin), one place after the last element in the buffer (pointer end), and one place after the allocated buffer (pointer capacity). Thus the length of the allocated buffer and the amount of filled the buffer are represented as differences between pointers capacity and end with the pointer begin. As we have assumed in the beginning, the pointers and integers have the same size, so we need $3n$ ints just for storing vectors (even when they are empty).
So what happens if we read a tree? On the first push_back to each vector, it will allocate a buffer of size $1$, and every time if current buffer is completely filled, it will reallocate it to a buffer twice as big. So sum of these buffers will need at least $2n$ int to store all values, but most probably more, because of unused reserved space. In worst-case it could be as big as $4n$ ints, e.g. for a star-shaped tree with one vertex $v$ of degree $n-1$ and $n-2$ being a power of two. The last push_back to $adj[v]$ will result in a vector of size $2n-4$ which during reallocating must coexist in memory with the old vector of size $n-2$, and of course we need $n-1$ ints for the one-element vectors for all the other vertices.
But that is not everything. What could be not obvious is that vector when allocating a buffer calls system function malloc which needs some memory to somehow keep track of all these allocated buffers. How much it depends on the implementation (and other factors such as memory alignment), but for sure it needs at least one pointer or pointer-size int, so let's assume that. That gives us another $n$ ints.
In summary: just to read the tree in the memory using vectors we need $6n$ ints, possibly more (even up to $8n$ ints).
### Calculating DP
Having read the tree we want to calculate the variables defined by DP recurrences. Again, the easiest way to do this is to make a DFS search through the tree starting from an arbitrary root, let's say $v_\star = 0$. We should track from which vertex we entered vertex $v$, so we will know which vertex from $adj[v]$ is the parent of $v$; the rest of them are its children. During the search we can calculate all DP arrays:
/* Recurse on node v with parent par (-1 if v is the root). */ void recursive_dp(int v, int par) { for (vector<int>::iterator it = adj[v].begin(); it != adj[v].end(); ++it) { const int u = *it; if (u != par) { recursive_dp(u, v); /* DP arrays for node u has been updated. */ /* Do something with edge from v to u, if needed. */ } } /* Update DP arrays for node v. */ update_v(v, par); } recursive_dp(root, -1);
The above code is quite general, and below is a function which calculates arrays dn and dt for our specific problem (we won't show yet how to calculate arrays cn and ct due to a problem with division mentioned before).
int dn[N], dt[N], cn[N], ct[N]; void update_v(int v, int par) { for (vector<int>::iterator it = adj[v].begin(); it != adj[v].end(); ++it) { const int u = *it; if (u == par) continue; dn[v] += dt[u]; dt[v] = max(dt[v], 1 + dn[u] - dt[u]); } dt[v] += dn[v]; }
This is the second phase: dynamic programming. Memory used by this phase can be divided into two parts: memory used by the DP arrays and memory used in order to traverse the tree. Let's calculate the latter. Since we are using recursion, we need to know how it's implemented by the compiler. There will be frames on the stack, each contains at least: return address, function arguments, and local variables. Return address is a pointer and in our case we have two arguments (v, par) and one local variable (it, which is iterator, thus can be implemented as a simple pointer). That's four ints in one frame. In the worst case our tree could be just a long path (so will be of depth $n$), thus the stack could contain $n$ frames, and will need $4n$ ints.
Actually it could be worse, since we do not know exactly what is inside the stack frame, until we dive into assembly. It is not uncommon to have there a pointer to the previous stack frame, temporary local variables (used to calculate values of subexpressions in complex expressions). Moreover, on certain architectures stack frames must be aligned to certain boundaries. So we state that we need at least $4n$ ints, but the reality could be much worse.
To sum up both phases of the algorithm, we need at least $6n$ ints (probably more) for storing the tree, $4n$ ints (probably more) for traversing it, and additional $4n$ ints for storing DP arrays. That is at least $14n$ ints for solving the task. Can you guess now how much this can be improved?
I just say: a lot. But I won't disclose precisely how much, I think it is more fun to unwind it step by step.
So let's begin.
## Replacing things we cannot control
The saddest part of the above estimates is that we said we need at least but probably more. First, we try to get rid of sources of this uncertainty: vectors with malloc (which were used for reading and storing the tree) and recursion (which was used for DP calculations).
### Replacing vectors
One clear redundancy with malloced vectors is that both malloc and vector store the length of allocated buffers. Since we cannot change the behavior of vector class (at least not without creating custom allocators which is not what I would like to do here), perhaps we can somehow manage without it. Which leads to a question: why do we need vectors in the first place? The problem is that we need vectors to store adjacency lists for vertices, but since we are loading edges one by one, we do not know beforehand, how much vertices will go to each list. Thus we need some kind of structure which will be resizable, and vectors look like the natural choice here.
But maybe vectors are not the best choice to store adjacency lists? Maybe real lists would fit better? Since list support push_back operation, then we only have to change declaration to list<int> adj[N];.
Memory calculation: each element of the list, apart from the value (int) stores two pointers to the previous and to the next element on the list (list is doubly-linked) – three ints in total. Moreover, each time we insert a new element, new memory for this single element is malloced, which results in at least one more int for malloc bookkeeping. There will be $2n$ elements in all lists, so the total memory consumption would be at least $8n$ ints. It's even bigger than the lower bound for vectors.
Of course, we don't need doubly-linked lists. We can use just fine forward_list<int> class, and replace push_back operations with push_front (since the actual order of vertices on lists is not important for us). That will leave us with at least $6n$ ints, and part of this memory still goes to malloc bookkeeping. It's some improvement over vectors, but not much (only over their upper bound).
Let's try a different approach. We wouldn't need vectors if we know the size of each adjacency list. Actually we could get this information, if we could go twice through the input:
int n, deg[N], ptr[N]; int* adj[N]; void load_tree_2() { scanf("%d", &n); REP(i, n-1) { scanf("%d%d", &a, &b); deg[a]++; deg[b]++; } REP(i, n) { adj[i] = new int[deg[i]]; ptr[i] = 0; } /* Rewind the input file */ lseek(0); scanf("%d", &n); REP(i, n-1) { scanf("%d%d", &a, &b); adj[a][ptr[a]++] = b; adj[b][ptr[b]++] = a; } }
For each vertex $v$ we first calculate its degree $deg[v]$, which is exactly the number of vertices in its adjacency list. After that we malloc a correctly-sized buffer $adj[v]$. Then we rewind the file, and read it once more, and now we store vertices in the arrays. Index $ptr[v]$ will keep track of where to put the next value in this array.
That gives us $3n$ ints for arrays deg, ptr and adj and at least another $n$ ints for malloc bookkeeping. And sum of buffer lengths is $2n$ ints, thus we need $6n$ ints here, same as lists. But we can make a clever trick here. Each time we put a vertex number during the second phase, we can advance pointer adj, so we do not need array ptr. If we need original beginnings of buffers, we can just subtract their lengths. Thus the code for the part after rewinding the input file is:
scanf("%d", &n); REP(i, n-1) { scanf("%d%d", &a, &b); *(adj[a]++) = b; *(adj[b]++) = a; } REP(i, n) { adj[i] -= deg[i]; }
Then the code to iterate over adjacency list of $v$ is as follows:
for (int it = 0; it < deg[v]; ++it) { const int u = adj[v][it]; /* Do something with edge from v to u. */ }
This uses only $5n$ ints for reading the tree. Unfortunately in many contests we load the data from the standard input, so we can only read it once. Thus this approach will not work, but it's good to know it anyhow. But don't throw away above code. If we cannot read list of edges twice, we can just store it using additional $2n$ ints:
int n, a[N], b[N], deg[N]; int* adj[N]; void load_tree_2() { scanf("%d", &n); REP(i, n-1) { scanf("%d%d", &a[i], &b[i]); deg[a[i]]++; deg[b[i]]++; } REP(i, n) { adj[i] = new int[deg[i]]; } REP(i, n-1) { *(adj[a[i]]++) = b[i]; *(adj[b[i]]++) = a[i]; } REP(i, n) { adj[i] -= deg[i]; } }
This approach needs $7n$ ints for reading the tree. It's certainly better than upper bound for vectors, but approach with lists gave us $6n$ ints. But the above code has one big advantage: we do need $7n$ ints for the reading phase, but after that phase we only need $5n$ ints, since arrays a and b will no longer be needed and can be reused for different purposes. That is why, from now on, we will specify memory requirements of the first phase using two numbers: memory needed to store the tree after the phase plus additional memory needed for actually performing the phase. In this notation the last approach needs $(5+2)n$ ints.
### Replacing malloc
The above approach still uses malloc. It is wasteful, since we do not control memory used by malloc, but moreover we know that it at least uses $n$ ints for keeping track of lengths of reserved buffers, and we also redundantly do it in array deg. Instead of malloc we will keep all adjacency lists in one array Adj of length $2n$:
int n, a[N], b[N], deg[N+1], Adj[2*N]; void load_tree_2() { scanf("%d", &n); REP(i, n-1) { scanf("%d%d", &a[i], &b[i]); deg[a[i]]++; deg[b[i]]++; } REP(i, n) { deg[i+1] += deg[i]; } REP(i, n-1) { Adj[--deg[a[i]]] = b[i]; Adj[--deg[b[i]]] = a[i]; } }
The idea is that the subsequent lists are stored after each other in Adj. After the second loop value $deg[v]$ stores the index after the last element of adjacency list of $v$. Therefore after the third loop each $deg[v]$ stores the beginning. That is why the better name for this array after the load code is begin. Since it will be quite common for us to reuse memory for different purposes, we will use aliases for clarity:
int* begin = deg;
Note that array begin has length $n+1$, since it has to store the end of the array Adj in the last element. That is how to iterate over adjacency list of $v$:
for (int it = begin[v]; it < begin[v+1]; ++it) { const int u = Adj[it]; /* Do something with edge from v to u. */ }
In this approach we need $5n$ ints for all arrays, but after that only deg and Adj are used, so it's just $3n$ ints. In total memory consumption for this phase is exactly $(3+2)n$ ints tops, which is clearly better than the first approach with vectors (even if we do not care to reuse arrays a and b).
### Replacing recursion
As we said before, recursion costs us at least $4n$ ints, but in practice it could be much worse. We pay all this just for the convenience that recursive programs are slightly easier to write, but for just a little more programming work we can rewrite the function in iterative way, and get rid of the most of this overhead. Let's see what data from the stack frame is essential: actually it's only vertex number v and iterator it. Observe that the second argument par is just a copy of v from the previous frame, and since recursive_dp function (apart from the call in main) is only called in one place, thus return addresses in all frames (except the first) will be the same. Clearly this is a lot of wasted space, which would not be wasted if we write it by hand.
So as a start, let's just translate this function in an iterative way:
int qe, q[N]; vector<int>::iterator it[N]; void iterative_dp(int root) { qe = 0; q[qe++] = -1; /* Sentinel */ q[qe++] = root; REP(i, n) { it[i] = adj[i].begin(); } while (qe > 1) { const int v = q[qe-1], par = q[qe-2]; if (it[v] == adj[v].end()) { /* Update DP arrays for node v. */ update_v(v, par); /* Do something with edge from par to v. */ qe--; } else { const int u = *it[v]++; if (u != par) { q[qe++] = u; } } } }
As we can see, we only use LIFO queue q (which replaces stack) and array it, so indeed we have managed to do DP traversal in memory of $2n$ ints. Of course, in the previous section we replaced vectors with array Adj. We can therefore rewrite this:
int ptr[N]; ... ptr[i] = begin[i]; ... if (ptr[v] == begin[v+1]) { ... const int u = Adj[ptr[v]++];
Finally we have a new solution without vectors, malloc and recursion. Since the first phase used $(3+2)n$ ints of memory and in the second phase we use $2n$ ints for tree traversal, in total we use $5n$ ints. Adding $4n$ ints for DP arrays we got $9n$ ints, which is slightly less than two-thirds of memory usage for the first solution. But this is just a warm-up.
## Compressing the tree
In the previous section we took a standard implementation of a standard approach and got rid of unnecessary memory allocated by parts of the code we didn't have control over (library and compiler code). Now when dust have settled after we squeezed the standard implementation, let's take a closer look at the standard approach we are using. Maybe here we also have some room for improvement?
We specifically focus on a format in which we store the tree in memory in order to make a DP traversal over it.
### Remove pointers to parents
For every vertex $v$ we store its adjacency list, that is a list of all vertices incident to it. But the DP calculations are done over a rooted tree in which one of this incident vertex becomes a parent of vertex $v$. Luckily, we enter $v$ from this parent, so we can skip it while iterating over adjacency list of $v$. But since the existence of these parents in adjacency lists is only a nuisance, why don't we remove all of them before calculating DP? Since there are $n-1$ parent elements in all lists, this removal would shorten array Adj by half!
This can be accomplished by another DFS traversal over tree (performed before DP calculations) during which for each non-root vertex $v$ we copy last element of its adjacency list in the position of parent element on this list:
void remove_parents(int root) { qe = 0; q[qe++] = -1; q[qe++] = root; REP(i, n) { ptr[i] = begin[i]; } while (qe > 1) { const int v = q[qe-1], par = q[qe-2]; if (ptr[v] == begin[v+1] - (par != -1)) { qe--; } else { const int u = Adj[ptr[v]]; if (u == par) { Adj[ptr[v]] = Adj[begin[v+1] - 1]; } else { q[qe++] = u; ptr[v]++; } } }
Next we can safely remove the last elements from all non-root adjacency lists. That means that we must move elements on adjacency list of vertex $v$ by $v-1$ or $v$ positions to the left, depending whether number $v$ is greater than $v_\star$ or not:
REP(v, n) { const int shift = v - (v > root); for (int it = begin[v]; it < begin[v+1]; ++it) { Adj[it - shift] = Adj[it]; } begin[v] -= shift; } begin[n] -= n-1; }
After such transformation we no longer have to check for parents during DP tree traversal. But what's even more important, the second half of array Adj is no longer needed and its $n$ ints can be reused. To be able to reuse only part of some array, we can make one big array array holding all the memory we need to reuse, and make aliases inside this array, e.g. like this:
int array[5*N+1], dt[N], cn[N], ct[N]; int* a = array; int* b = array + N; int* Adj = array + 2*N; int* deg = array + 4*N; int* q = array; int* ptr = array + N; int* dn = array + 3*N;
Just remember to initialize the array with zeros, if it comes from reused space.
Thanks to that the first phase uses memory of $(2+3)n$ ints. Adding $2n+4n$ ints for the second phase, we get $8n$ ints for the whole program.
### Changing order of DP computations
Observe that we can perform DP calculations in any order, as long as calculations for every child of given vertex $v$ is performed before calculations for vertex $v$. In functions recurrence_dp and iterative_dp we have done it using DFS traversal, since we at the same time needed to know the parents of vertices. But now, after removing parents, we can also traverse vertices in BFS manner if we please:
int qb, qe, q[N]; q[qe++] = root; while (qb != qe) { const int v = q[qb++]; for (int it = begin[v]; it < begin[v+1]; ++it) { q[qe++] = Adj[it]; /* Fill array q */ } }
This code fills the array $q$ of size $n$ with BFS order of visiting vertices in the tree. Note that in the loop we no longer have access to the parent of $v$, but we don't need it. One important property of this order is that vertex $v$ is in array q before each of its children. Therefore we can make DP traversal on the tree using only arrays Adj, begin and q. We iterate over vertices from q in reversed order:
REP(i, n) { const int v = q[n-1 - i]; for (int it = begin[v]; it < begin[v+1]; ++it) { const int u = Adj[it]; /* Do something with edge from v to u. */ } }
Therefore in order to traverse the tree in the second phase we can use BFS order in array q, instead of stack in array q and array ptr. Thanks to that we only need $n+4n$ ints for the second phase, thus $7n$ ints for the whole solution (half of memory usage of the standard approach).
### Remove vertex numbers
All of this work we have done during reading the tree was about finding the correct order of vertices. We hinted before that the actual number of vertices do not matter – the only thing that matters is the shape of the tree. So wouldn't it be nice if the vertices were numbered in the input in some sensible order, for example in the BFS order we calculated in the previous section? But since they aren't, maybe we can renumber them?
The idea is as follows. The array q contains numbers of vertices in the BFS ordering. Therefore the $i$-th node (zero-based) in this ordering has number $q[i]$. We would like to change number of this node to simply $i$. We store node numbers in array Adj and they are also used as indices to array begin.
First, let's look how would array begin look like after doing such renumbering. We can do it by restoring degrees of all vertices in BFS order (and storing them in array newdeg), and once more transforming it into array begin:
int* newdeg = array + N; REP(i, n) { const int v = q[i]; newdeg[i] = begin[v+1] - begin[v]; } begin[0] = 0; REP(i, n) { begin[i+1] = begin[i] + newdeg[i]; }
It looks like we would have more problems with array Adj, since the vertices' numbers in it should be renumbered according to q, but also we should move blocks in this array according to changes in begin. But it would turn out, that new array Adj became trivial. Look closely how array q was constructed in the first place. We iterated through subsequent vertices in BFS order, so subsequent cells of Adj (after renumbering) must be equal to subsequent cells of q (except the first element of q which corresponds to the root of the tree). Therefore $Adj[i] = i+1$ for all $i$.
Since we don't need q too (because it's also trivial after renumbering), we only need temporary array of $n$ ints to store newdeg. Therefore new code for DP computation doesn't use q and Adj:
REP(i, n) { const int v = n-1 - i; for (int it = begin[v]; it < begin[v+1]; ++it) { const int u = it + 1; /* Do something with edge from v to u. */ } }
Wow. We have just compressed the whole information about the tree in just one array begin of size $n$. Moreover having just this array we can perform DP traversal over the tree. That's something. Therefore the first phase costs us $(1+4)n$ ints. And we don't need any additional memory for traversing the tree in the second phase, so it's just $4n$ ints of DP arrays. In total our solution uses $5n$ ints.
### Simpler method of compressing the tree
Actually, when we know that we can represent the whole tree using only one array begin, we can calculate this array even simpler. The process is as follows: we renumber the vertices in the tree in the BFS order, and then we create array deg which contains degrees of vertices in this BFS order. From array deg we retrieve new begin using standard method.
We start from our representation before doing any tree compression, thus from array Adj of size $2n$ (containing parents) and array with indices, which we call oldbegin for clarity. We will use an array q for queue (which will be reused for storing the degree of each vertex popped from the queue) and array vis for storing visited vertices (so we don't visit parents during BFS traversal):
int* oldbegin = begin; int* vis = array + N; qb = qe = 1; q[qe++] = root; REP(i, n) { vis[i] = false; } vis[root] = true; while (qb != qe) { const int v = q[qb]; q[qb++] = oldbegin[v+1] - oldbegin[v] - (v != root); for (int i = oldbegin[v]; i < oldbegin[v+1]; ++i) { const int u = Adj[i]; if (!vis[u]) { q[qe++] = u; vis[u] = true; } } } begin = q; begin[0] = 1; /* Shift by 1, since Adj[i] = i+1 */ REP(i, n) { begin[i+1] += begin[i]; }
At the end we only need array begin. Thus the cost of the first phase is $(1+4)n$ ints.
So with the compression of the tree the whole problem can be solved in just $5n$ ints, a little bit over one-third of original $14n$. We won't gain much improving the first phase, as long as DP arrays will need $4n$ ints, so let's switch back to DP calculations performed by our solution, to see if we can improve that.
## Improving DP calculations
First we keep the promise of showing how to calculate value of $ct[v]$ without division. Then we try to reduce the memory needed by DP calculations.
### Finishing finding the number of maximum-size matchings
Remember, that in formula for $ct[v]$ we used $cn[v] / c[u_i]$ to calculate a product of values $c[\cdot]$ for all children except $u_i$:
$c[u_0] \cdot c[u_1] \cdot \ldots \cdot c[u_{i-1}] \cdot c[u_{i+1}] \cdot c[u_{i+2}] \cdot \ldots \cdot c[u_{m-1}]$
Such approach worked fine with sum (because we could use subtraction), but with multiplication without ability to do division it does not work. However, we can solve it by calculating prefixes and suffixes as follows:
$cnl[i] = c[u_0] \cdot c[u_1] \cdot \ldots c[u_{i-1}]$
$cnr[i] = c[u_{i+1}] \cdot c[u_{i+2}] \cdot \ldots c[u_{m-1}]$
And then use the equality $cn[v] / cn[u_i] = cnl[i] \cdot cnr[i]$. The arrays cnl and cnr are calculated as follows:
$cnl[0] = 1;\quad cnl[i] = cnl[i-1] \cdot c[u_{i-1}]$
$cnr[m-1] = 1;\quad cnr[i] = c[u_{i+1}] \cdot cnr[i+1]$
So we are ready to present full C++ code of DP calculations. The first loop calculates arrays dn and dt in the same manner as before (we only use array begin rather than vectors). The second loop calculates helper arrays cnl and cnr. Finally, the third loop calculates arrays cn and ct. For clarity, we don't include taking modulo $M$ in the code.
int cnl[N], cnr[N]; void iterative_dp_2() { REP(i, n) { const int v = n-1 - i; dn[v] = 0; dt[v] = 0; for (int it = begin[v]; it < begin[v+1]; ++it) { const int u = it; dn[v] += dt[u]; dt[v] = max(dt[v], 1 + dn[u] - dt[u]); } dt[v] += dn[v]; } REP(i, n) { const int v = n-1 - i; const int m = begin[v+1] - begin[v]; cnl[0] = 1; cnr[m-1] = 1; for (int i = 0; i < m-1; ++it) { const int ul = begin[v] + i; const int cul = ct[ul] + cn[ul] * (dn[ul] == dt[ul]); cnl[i+1] = cnl[i] * cul; const int ur = begin[v+1]-1 - i; const int cur = ct[ur] + cn[ur] * (dn[ur] == dt[ur]); cnr[m-2-i] = cur * cnr[m-1-i]; } cn[v] = 1; ct[v] = 0; for (int it = begin[v]; it < begin[v+1]; ++it) { const int u = it; const int i = it - begin[v]; const int cu = ct[u] + cn[u] * (dn[u] == dt[u]); cn[v] *= cu; ct[v] += cnl[i] * cn[u] * cnr[i] * (dt[v] == dn[v] + 1 + dn[u] - dt[u]); } } }
Unfortunately, we need additional memory to store arrays cnl and cnr. These are arrays of size $m$, since they are only used for calculations for one vertex, but maximal $m$ can be as large as $n-1$. Therefore we need additional $2n$ ints, which is not good.
So let's take a closer look at the formula for $ct[v]$. It has a form:
$\sum_{0 \leq i < m} a_{0} \cdot a_{1} \cdot \ldots \cdot a_{i-1} \cdot b_{i} \cdot a_{i+1} \cdot a_{i+2} \cdot \ldots \cdot a_{m-1}$
where $a_i = c[u_i]$ and $b_i = cn[u_i] \cdot [ dt[v] = dn[v] + 1 + dn[u_i] - dt[u_i] ]$. Let's write $m$ products of this sum in $m$ subsequent rows, forming a $m \times m$ matrix. The following image shows it for $m=5$:
We can calculate this sum by evaluating sums in subsequent submatrices of size $i \times i$ as shown on the image. The value $S_i$ will denote sum of submatrix $i \times i$ and helper value $A_i$ will denote partial product $a_0 \cdot a_1 \cdot \ldots \cdot a_i$. We start with $S_0 = 0$ and $A_0 = 1$, and use the formulas:
$S_i = S_{i-1} \cdot a_i + A_{i-1} \cdot b_i$
$A_i = A_{i-1} \cdot a_i$
The answer is $cn[v] = S_m$. We can also observe that $ct[v] = A_m$. The code following this method is much simpler than the previous one, and what is most important, it does use only constant additional memory:
REP(i, n) { const int v = n-1 - i; cn[v] = 1; ct[v] = 0; for (int it = begin[v]; it < begin[v+1]; ++it) { const int u = it; const int cu = ct[u] + cn[u] * (dn[u] == dt[u]); ct[v] = ct[v] * cu + cn[v] * cn[u] * (dt[v] == dn[v] + 1 + dn[u] - dt[u]); cn[v] *= cu; } }
### Compressing memory usage
Now let's take a closer look at computations for finding maximum size. We showed that $dt[v] \geq dn[v]$, that is maximum-size matching which uses vertex $v$ is no smaller that maximum-size matching which does not use this vertex. But how much bigger it can be? Let's take a matching which uses vertex $v$ and remove the edge matching $v$. Then we get matching one edge smaller which does not use $v$, therefore $dn[v] \geq dt[v] - 1$. Thus these two numbers can differ only by $1$, so let's make a notation:
$dt[v] = dn[v] + z[v] \quad$ where $\quad z[v] \in \{ 0, 1 \}$
This way we no longer have to use $dt[v]$, we just use $dn[v]$ and $z[v]$. Let's rewrite recursions using this new notation. First of all we can simplify Boolean expressions used in the formulas:
$[dn[v] = dt[v]] = [z[v] = 0] = 1 - z[v]$
$[dt[v] = dn[v] + 1 + dn[u_i] - dt[u_i]] = [z[v] = 1-z[u_i]] = z[v] \ \textrm{xor}\ z[u_i]$
Then we can rewrite all formulas:
$dn[v] = \sum_i (dn[u_i] + z[u_i])$
$z[v] = \max_i (1 - z[u_i])$
$c[v] = ct[v] + cn[v] \cdot (1 - z[v])$
$cn[v] = \prod_i c[u_i]$
$ct[v] = \sum_i (cn[v] / c[u_i] \cdot cn[u_i]) \cdot (z[v] \ \textrm{xor}\ z[u_i])$
We have two observations here. First of all, we replaced array dt with array z. It looks like we didn't gain much here, but note that array dt was int array (it used $n$ ints), and z is bit array (it needs only $n$ bits). Bit array uses less memory, since it can be packed. If int has $B$ bits, it uses only $\lceil n/B \rceil$ ints.
But actually we can pack it so it won't use additional memory at all. Recall that we assumed that int variable can store number $2n$. And variable $dn[v]$ stores numbers no larger than $\frac{n}2$. Therefore we can use unused most significant bit in $dn[v]$ to store bit $z[v]$. Thus both arrays to calculate size of maximum-size matching will use only $n$ ints of memory. This reduces memory consumption of DP arrays to $3n$ ints.
What about arrays cn and ct for calculating numbers of maximum-size matchings? Of course, we need them both. But interesting thing is that for calculating them we don't need array dn at all, only bit array z. Thus memory used by dn after we computed first step can be reused (for instance for storing one of cn, ct). But then again, we need to store array z somewhere else. We cannot pack it with cn or ct, because these arrays stores numbers modulo $M$, which possibly uses all of bits in their cells. But don't forget we have one more array we are using: array begin in which we store the tree. It stores numbers up to $n-1$, then each cell of this array has an unused bit. Therefore we can pack array z there.
To recap: first we calculate array dn and bit array z, the latter is packed together with array begin. Next, we calculate cn and ct, using bit array z stored in begin and reusing memory from no longer used array dn. Therefore the total memory consumption from DP arrays decreased to only $2n$ ints.
The following C++ code uses helper macros to quickly get and set bits in array begin.
int* cn = array + N; int* ct = array + 2*N; const int MSB = 19; /* Number of most significant bit in int */ #define BEGIN(i) (begin[i] & ~(1 << MSB)) #define Z(i) (begin[i] >> MSB) #define SETZ(i, val) (begin[i] = BEGIN(i) | (val << MSB)) void iterative_dp_3() { REP(i, n) { const int v = n-1 - i; dn[v] = 0; int zv = 0; for (int it = BEGIN(v); it < BEGIN(v+1); ++it) { const int u = it; const int zu = Z(u); dn[v] += dn[u] + zu; zv |= 1 - zu; } SETZ(v, zv); } REP(i, n) { const int v = n-1 - i; const int zv = Z(v); cn[v] = 1; ct[v] = 0; for (int it = BEGIN(v); it < BEGIN(v+1); ++it) { const int u = it; const int zu = Z(u); const int cu = ct[u] + cn[u] * (1 - zu); ct[v] = ct[v] * cu + cn[v] * cn[u] * (zv ^ zu); cn[v] *= cu; } } }
The cost of the first phase is $(1+4)n$ ints, but we have managed to squeeze the memory used by the second phase to only $2n$ ints. Once again, the memory bottleneck of the algorithm is the first phase, so let's see if we can reduce it, to reduce the total consumption of memory.
So we know that we only need $n$ ints of array to store the tree, and be able to traverse it in order to perform DP computations. In this light, additional $4n$ ints needed to create the array is way too much. Let's take a completely different approach to decrease it.
### Parent representation
The problem is that the representation of a tree as an edge list is not very convenient. It hides the structure of the tree, and if we want to do any traversal over the tree, we must convert it to some other representation. And since it is usually the first thing we do after reading such a tree, in some programming problems that deal with trees, other, more convenient representations are used in the input.
One of such representation is a list of parents. In fact, it comes in two flavors, which we call here strict and relaxed. For the strict variant, we number each vertex with a number from $0$ to $n-1$. Next, we list $n-1$ edges, but the $i$-th edge (one-based) must connect vertex $i$ to some other vertex $p[i]$ of smaller number ($p[i] < i$). That means that by considering edges one by one, we add new vertices to one connected component. Of course, the obvious advantage of this representation is that it only uses $n$ ints to store the tree.
But that representation has another important property: for every vertex $v$ except $0$ there is some edge incident to the vertex which leads to vertex $p[v]$ of a smaller number. Therefore, starting from any vertex $v$ and using such edges, we eventually end up in vertex $0$. That shows that if we root the tree in vertex $0$, then for each non-root vertex $v$ we have that vertex $p[v]$ is the parent of $v$. And also that the number of every child is greater that the number of its parent.
For the relaxed variant, we also number vertices from $0$ to $n-1$, but now we input $n$ numbers $p[i]$ for $0\leq i < n$. One number $i$ is special and we denote it by $p[i]=-1$, and for all other numbers we assume that we have an edge between vertex $i$ and $p[i]$.
Let's denote by $v_\star$ the special vertex for which $p[v_\star] = -1$. If we do the same trick as before: from any vertex $v$ we start moving using edges directed from $i$ to $p[i]$, we finally must arrive at vertex $v_\star$ (since we cannot have any cycle in a tree). That shows that for the relaxed variant, if we root the tree in vertex $v_\star$, then for each non-root vertex $v$ we have that vertex $p[v]$ is the parent of $v$.
So let's imagine that we got parent representation (strict or relaxed) in array p. How can we calculate array begin from it? First is easy to calculate indegree of each vertex (in array deg). Then we just fill array Adj in standard way:
int p[N], deg[N], Adj[N], root; REP(i, n) { if (p[i] != -1) { deg[p[i]]++; } else { root = i; } } REP(i, n-1) { deg[i+1] += deg[i]; } REP(i, n) { if (p[i] != -1) { Adj[deg[p[i]]--] = i; } }
The above code, based on array p, calculates arrays Adj and begin (equal to deg). These two arrays store the full information about the tree in memory using $2n$ ints. So, if we would have array p, we could perform the first phase in $(2+1)n$ ints, and the whole problem in just $4n$ ints. (As a side note: performing DP on arrays obtained from strict variant of parent representation is even easier, since the vertices are already in good order.)
We could also use technique of removing vertex numbers to calculate array begin which fully represent the tree. We no longer need array vis, since Adj does not store parents. So the cost would be $(1+2)n$ ints, which results in only $3n$ ints for the whole problem!
### Obtaining parent representation using xoring
And here it's where magic begins. It turns out that we do not need much memory to calculate parent representation of a tree from its edge list representation, but the idea is quite ingenious. The code is as follows:
int n, deg[N], hash[N], q[N], qe, qb, root; int* p = hash; void load_tree_3() { scanf("%d", &n); REP(i, n-1) { scanf("%d%d", &a, &b); deg[a]++; hash[a] ^= b; deg[b]++; hash[b] ^= a; } REP(i, n) { if (deg[i] == 1) { q[qe++] = i; } } REP(i, n-1) { const int v = q[qb++]; const int p = hash[v]; hash[p] ^= v; if (--deg[p] == 1) { q[qe++] = p; } } root = q[n-1]; p[root] = -1; }
First we just iterate over all edges without storing them and for each vertex $v$ we calculate its degree $deg[v]$ and also $hash[v]$ which stands for bitwise xor of numbers of all vertices adjacent to $v$.
Next we will build the tree from the leaves up to some root. Observe that for any node $v$ of degree $1$, we know that it has exactly one adjacent node, so we know that the number of this adjacent node is $p = hash[v]$. We therefore assume that the parent of vertex $v$ is $p$, and then we remove vertex $v$ from the tree. The removal boils down to decreasing degree of vertex $p$ and un-xoring $v$ from $hash[p]$ which results in a smaller tree which we attack recursively.
Thus we maintain a queue q of leaves of the current tree. At first we put there all vertices with degree $1$. Then we iteratively remove vertex $v$ from the queue, assign its parent $p$, and update $deg[p]$ and $hash[p]$. Anytime degree of such $p$ decreases to $1$, we can push it on the queue.
After removing $n-1$ vertices, we end up with the last vertex on the queue, which becomes the root of the tree, and any other vertex has correctly calculated parent $p[v] = hash[v]$.
We used $3n$ ints for arrays deg, hash and q, but only array p (equal to hash) is needed at the end. Using the code before, we calculate begin, so it results in the first phase of cost $(1+2)n$ ints. Therefore the whole problem can be solved in $3n$ ints; slightly more than one-fifth of original.
### First-child-next-sibling representation
It turns out that we can easily improve the code to calculate array p. Note that use queue q to store vertices of degree $1$, so we know which of them to process later. But is it really necessary? Instead of putting them into queue in the first loop, we could just process them right away. Moreover, if during processing some vertex, its parent became a vertex of degree $1$ we can also process it immediately (which could result in another such parent). Following code does not use queue at all:
REP(i, n) { if (deg[i] == 1) { const int v = i; while (deg[v] == 1) { const int p = hash[v]; hash[p] ^= v; --deg[p]; deg[v] = -1; root = p; v = p; } } } p[root] = -1;
Every removed $v$ vertex is marked by putting $deg[v] = -1$. Last considered vertex will become the parent.
By removing array q we reduced the memory needed to calculate array p to only $(1+1)n$ ints. So maybe we could also reduce memory needed for calculating array begin from array p?
To do this we introduce one more representation of a tree. It is called first-child-next-sibling and for each vertex $v$ we store two numbers: number of its first child $chi[v]$ (equal to $-1$ if vertex has no children) and number of its next sibling $sib[v]$ (number of a child after this vertex in its parent child list; or $-1$ if its last vertex on this list). It's fairly easy to calculate this representation from array of parents:
int chi[N], sib[N]; REP(i, n) { chi[i] = sib[i] = -1; } REP(i, n) if (p[i] != -1) { const int par = p[i]; sib[i] = chi[par]; chi[par] = i; }
Each time we consider new vertex $v$, we assign it as the first child of its parent $p[v]$, and if any vertex was this child before, it will be assigned as next sibling of $v$. Of course, we want to store all computations in $2n$ ints, but this could be done by observing that we can reuse memory between arrays p and sib, since after setting $sib[i]$ we no longer need access to $p[i]$. Thus the first beginning of the above code could look like this:
int chi[N]; int* sib = p; REP(i, n) { chi[i] = -1; }
The representation using chi and sib can be easily used to perform BFS traversal. We start by putting root in the queue q, and every time we visit vertex $v$, we can iterate over its children by starting from $w = chi[v]$ and then performing $w = sib[w]$ until we hit $-1$. Therefore for each visited vertex we can also calculate its outdegree. The trick is to store this number in memory shared by queue q. Finally, because these vertices are in BFS order, we get for free removal of their numbers:
int q[N], qe, qb; q[qe++] = root; while (qe != qb) { const int v = q[qb]; int w = chi[v], d = 0; while (w != -1) { q[qe++] = w; w = sib[w]; ++d; } q[qb++] = d; /* Store degree of v */ } REP(i, n) { const int v = n-1 - i; q[i+1] = q[i]; } q[0] = 1; REP(i, n) { q[i+1] += q[i]; }
The last loop performs calculation of prefix sums in place, starting from $1$ (thus we first need to move array one position to the right). At the end we use only array q as array begin. But we used $3n$ ints in total (for arrays chi, sib and q), so it does not look as an improvement.
### Even more compression with first-child-next-sibling representation
The trick is to squeeze queue q into arrays chi and sib. Since children of each vertex are put sequentially into the queue, array sib already contains these parts of the queue. We only have to concatenate them together, using last vertices in each children list (they have sib equal to $-1$):
int v, ve; int* ndeg = chi; v = ve = root; while (v != -1) { int w = chi[v], d = 0; sib[ve] = w; while (w != -1) { ve = w; w = sib[w]; ++d; } ndeg[v] = d; v = sib[v]; }
During the whole loop, the variable $v$ will iterate over all vertices of the tree, in the BFS order. Since it uses array sib for this iteration, for every vertex $v$ which is the last child on its parent's list we must set $sib[v]$ to the next vertex in this order. Therefore we will maintain variable $ve$ which stores the number of vertex which is directly before $chi[v]$ in the BFS order (or if $v$ has no children, directly before position in which its first children would appear if it has one).
For subsequent vertices $v$ we proceed as follows: if $v$ has children, we put its first child as the next vertex after $ve$. Next, we iterate over all children of $v$, counting them, and updating $ve$ as the number of the last child. Finally, we store outdegree of $v$ in array ndeg, shared with chi, since we won't use $chi[v]$ anymore.
At the end we have array sib which serves as the queue and array ndeg (shared which chi) which stores degrees of nodes (but in original order). The only thing left is to sort array ndeg according to the order stored in array sib:
int pos = root; REP(i, n) { const int nxt = sib[pos]; sib[pos] = i; pos = nxt; } REP(i, n) if (sib[i] != i) { swap(ndeg[i], ndeg[sib[i]]); swap(sib[i], sib[sib[i]]); --i; } int sum = 1; REP(i, n) { swap(sum, ndeg[i]); sum += ndeg[i]; } ndeg[n] = sum;
Note that the order in array sib is given implicitly, as a linked list: the first element is root, then sib[root], next sib[sib[root]] etc. The first loop iterates over this list and for every element $v$ stores in $sib[v]$ the value $i$ which denotes that element $v$ is the $i$-th element in the list.
The second loop reorders array ndeg using order stored in array sib. We iterate over all elements $i$ and for each element which is out of place (i.e. $sib[i] \neq i$) we move this element to its correct position (that is $sib[i]$) and at the same time we move the element which previously was in this position to position $i$. Thus in each iteration one more element is located correctly.
Finally, the last loop performs calculation of prefix sums, starting from $1$, but it does it a little differently than the implementation in the previous section, using only one loop instead of two.
The above clever code perform the first phase using only $(1+1)n$ ints of memory! This is really an achievement, because it allows us to read a tree in list edge representation to a form suitable to perform DP traversal in $(1+1)n$ ints. So for a problem in which we have one DP array of size $n$ it would result in an algorithm using only $2n$ ints in total. In general for a problem with $k$ DP arrays we get $(1+\max(1,k))n$ ints. However, in our problem DP arrays are of size $2n$, so we get $3n$ algorithm, and we have that before.
## Even more improvements in DP calculation
It looks like we really squeezed the memory in this problem. We store the whole tree in one array of $n$ ints. The other two arrays are used for DP calculations. In fact for calculating the size of matching we use only one additional array dn. However, for calculating the number of matchings we need two arrays: both values cn and ct are numbers resulted from rather complicated computations modulo $M$, so it is very unlikely that they have some general properties to be discovered and used in such a way that would result in removing one of them. So it looks like we hit the limit. Or is it?
### Branching and non-branching vertices
But let's don't give up yet. We already did some magic with xoring, so why not try to do something impossible?
If we cannot remove one of DP arrays, maybe we can make them smaller? Do we really have to store the data for all vertices? For instance, we really don't have to do it for leaves. DP values for them are constants which do not depend on other vertices, so we can easily recreate them when needed. When parent vertex is calculating DP it can check whether its child is a leaf, and then do not look into DP table, but just use appropriate constants.
That is an idea that could save us some memory on trees with many leaves. Unfortunately, not all trees are like that: we could have a tree which is a long path that has only one leaf, and then this optimization would give us nothing. But note that this tree is also special and easy to calculate DP on, because it consists of this long path of vertices of degree $1$. The DP values for the top-most vertex of this path could also be calculated in constant memory, as long as we know the length of the path, since DP in each vertex depends only on DP value in one vertex below (we can think of it as a variant of tail optimization). This is true in general case: in every tree we can treat each path of vertices of degree $1$ specially, and we do not need to store DP values for internal vertices of these paths.
So we only need to store DP values for branching vertices, namely those which have at least two children. And in any tree there are at most $\frac{n}{2}$ branching vertices! It follows from two simple facts: sum of branching vertices and leaves is bounded by $n$, and there is more leaves than branching vertices.
Therefore we only need to store $\frac{n}{2}$ values in every DP array. There is one problem though: branching vertices do not necessarily form a contiguous part of arrays, so we need some renumbering mechanism if we want to save memory. To achieve this, we introduce an array $pos$, where in $pos[v]$ we store the position of branching vertex $v$ in the compressed DP array, i.e. $pos[v]$ is the number of branching vertices of smaller numbers than $v$:
int pos[N], last = 0; REP(v, n) { if (BEGIN(v+1) - BEGIN(v) >= 2) { pos[v] = last++; } }
Since the values stored so far in arrays cn and ct will have to be calculated on-the-fly for non-branching vertices, it will be handy to define a structure containing all DP values for a single vertex:
struct dp_t { int cn; int ct; }; dp_t dp[N/2];
The function which returns the constant values for a leaf is very simple:
dp_t leaf() { return dp_t{ 1, 0 }; }
Recall that DP calculation for a single vertex $v$ which children $u_0, \ldots, u_{m-1}$ involve initiating values cn and ct, and iterating over subsequent vertices $u_i$. After the $i$-th iteration the current values cn and ct contain DP values for a vertex $v$, assuming it has only children $u_0, \ldots, u_i$. Therefore we could write a function join(a, b, zv, zu) in which argument a denotes DP values of a tree consisting of vertex $v$ and children $u_0, \ldots, u_{i-1}$, argument b denotes DP values for child $u_i$ and the function returns DP values for a tree consisting of vertex $v$ and children $u_0, \ldots, u_i$. Since DP values depend on values z stored in vertices $v$ and $u_i$, we will also provide them as arguments:
dp_t join(const dp_t& a, const dp_t& b, int zv, int zu) { const int cu = b.ct + b.cn * (1 - zu); const int cn = a.cn * cu; const int ct = a.ct * cu + a.cn * b.cn * (zv ^ zu); return dp_t{ cn, ct }; }
We are ready to create a helper function which allows to retrieve DP values for vertices we already visited. In other words, if $v$ is a branching vertex, we must query an appropriate cell in the compressed DP array, and otherwise, we must recalculate DP values on-the-fly. First of all, if $v$ is a vertex of degree $1$, the function calculates the length of the path to the first vertex $w$ of other degree. Then, if $w$ is a leaf it uses constants, and if $w$ is a branching vertex it queries DP arrays. Finally, it makes DP calculation upwards the path of vertices of degree $1$:
dp_t calc(int v) { int length = 0; while (BEGIN(v+1) - BEGIN(v) == 1) { length++; v = BEGIN(v); } dp_t t; int z = 0; if (BEGIN(v+1) - BEGIN(v) == 0) { t = leaf(); } else { t = dp[pos[v]]; /* Querying value in dp array */ z = Z(v); } while (length--) { t = join(leaf(), t, 1-z, z); z = 1-z; } return t; }
The only non-clear part of this function is how we handle retrieval of values z when calculating DP on the path of vertices of degree $1$. The problem is that we store values z packed in array begin. When in the first loop we go downwards the path, calculating its length, we obtain numbers of subsequent vertices $v$ on this path. We could obtain values z for these vertices using Z(v). However, in the last loop we go upwards the path, so we would need the sequence of these numbers, but it reverse order. The problem is that we cannot afford to store this sequence, so we do not know numbers of vertices when we go upwards the path.
Fortunately we can calculate values of z in a path of degree $1$ directly from the definition (recursive formula). In fact, when vertex $v$ has exactly one child $u$, then Z(v) = 1 - Z(u).
Finally, we can present the main code to calculating values cn and ct for all vertices:
REP(i, n) { const int v = n-1 - i; if (BEGIN(v+1) - BEGIN(v) >= 2) { dp_t t = leaf(); for (int i = BEGIN(v); i < BEGIN(v+1); ++i) { t = join(t, calc(i), Z(v), Z(i)); } dp[pos[v]] = t; /* Updating value in dp array */ } }
Since the on-the-fly calculations in function calc(i) depend on the length of the path starting in vertex $i$, so the time complexity is a little bit suspicious. Fortunately, we call this function only on vertices $i$ which have branching parent, all such paths traversed by the algorithm are disjoint, and the time complexity of the algorithm is still $O(n)$.
The above algorithm uses additional array pos of size $n$ ints, but thanks to that the DP arrays are half as big as before. Therefore for $k$ DP arrays it uses memory of $(2+k/2)n$ ints. This is definitely improvement for big values of $k$, but in our problem it still results in algorithm which uses $3n$ ints. We really need something better.
### Destructively traversing the tree
One could argue that we do not have to store array pos, since each of its elements can be computed in constant memory from array begin. Unfortunately, this needs $O(n)$ time, and since we query array pos in arbitrary order (when iterating other subsequent children which have paths underneath, we in fact query vertices on different heights) this would lead to quadratic time, which is unacceptable.
So if we really need to store pos, but we don't want to spare additional memory, we must once again reuse some already allocated memory. But apart from array dp storing DP values, we only have array begin which is essential for tree traversal. Or is it?
Observe, that we iterate over vertices in reversed BFS order, and after processing of branching vertex $v$ we will not longer use values of $begin[u]$ for most vertices $u$ is the subtree of $v$. To be more precise: we will use $begin[v]$ and $begin[v+1]$ when later querying the number of children of $v$. Also if $u_0, u_1, \ldots, u_{m-1}$ are children of $v$, then we could use $begin[u_0]$ when querying the number of children of $u_0-1$ (especially if $v=u_0-1$). But we won't be querying for number of children of $u_0$ and $u_1$ (or iterating over them), thus we won't be needing value $begin[u_1]$ of the second child of $v$ (note that $v$ has the second child, since it is branching). We also won't be using value z stored in this cell. Thus we can reuse this field to store $pos[v]$.
That requires only slight changes to the previous algorithm. First, we do not need pos array anymore (and its initialization code). Next, the line updating value in dp array should be replaced with:
dp[ begin[BEGIN(v)+1] = last++ ] = t;
And querying this array in function calc should be replaced with:
t = dp[ begin[BEGIN(v)+1] ];
This is quite unbelievable, but this algorithm runs one DP traversal for $k$ DP arrays in memory of only $(1 + \max(1, k/2))n$ ints. Unfortunately, as a nasty side-effect it destroys the traversed tree. Therefore it is not practical, if we want to make more than one pass over the tree.
In our problem we need two passes, but we already discussed that the first pass uses only one DP array, thus it can be done using non-destructive methods in memory $(1 + 1)n$ ints. In the second pass we use the above trick, destructively calculating number of matchings in the same memory.
And this is our final algorithm. It uses memory of only $2n$ ints to input the tree and perform non-trivial DP calculations on it. Compared with at least $14n$ ints needed by the standard approach, it's an improvement of one-seventh of the original. I told you that it will be a lot.
## Conclusions
I learned a lot when writing this article, and I hope that the readers will also find it useful. Because not only the problem of dynamic programming on trees using as small memory as possible gave us an excuse to talk about important implementation details of memory management (in context of vectors, malloc and recurrence), various representations of trees in memory and ways to compress DP arrays. It was also a perfect quest for ingenious ideas in computing, once called by Michael Abrash as these astonishing “right-brain optimizations”. | 16,359 | 61,784 | {"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": 2, "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} | 3.484375 | 3 | CC-MAIN-2019-22 | longest | en | 0.954147 |
https://origin.geeksforgeeks.org/maximum-length-of-string-formed-by-concatenation-having-even-frequency-of-each-character/ | 1,638,238,800,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358903.73/warc/CC-MAIN-20211130015517-20211130045517-00029.warc.gz | 500,569,792 | 31,543 | # Maximum length of string formed by concatenation having even frequency of each character
• Last Updated : 26 Oct, 2021
Given N strings, print the maximum length of the string and the string formed by concatenating any of the N strings, such that every letter in the string occurs even number of times
Example:
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Input: N = 5, str = [“ABAB”, “ABF”, “CDA”, “AD”, “CCC”]
Explanation: The string formed by concatenation is ABABCDAADCCC. Each letter in the string occurs even number of times
Input: N = 3, str = [“AB”, “BC”, “CA”]
Output: ABBCCA 6
Explanation: The string formed by concatenation of all 3 strings is ABBCCA
Approach: The given problem can be solved using recursion and backtracking. The idea is to either include the string or exclude the string at every iteration. After including a string, the frequency of all the characters in the concatenated string is calculated. If frequency of all the characters is even we update the maximum length max. Below steps can be followed to solve the problem:
• Initialize variable max to 0 for calculating maximum length of concatenated string having even frequency of all characters
• Initialize string ans1 to store the concatenated string of maximum length with all character having even frequency
• The base case of the recursive call is to return, if index becomes equal to the size of the input string list
• At every recursive call we perform the following operation:
• Include the string and check if the frequency of characters is even for the concatenated string
• If the frequency is even, update max and ans1
• Increment the index and make the next recursive call
• Exclude the string, increment the index and make the next recursive call
Below is the implementation of the above approach:
## C++
// C++ implementation of the above approach #include using namespace std; int maxi = 0; string ans1 = ""; // Function to check the string void calculate(string ans) { int dp[26] = { 0 }; for (int i = 0; i < ans.length(); ++i) { // Count the frequency // of the string dp[ans[i] - 'A']++; } // Check the frequency of the string for (int i = 0; i < 26; ++i) { if (dp[i] % 2 == 1) { return; } } if (maxi < ans.length()) { // Store the length // of the new String maxi = ans.length(); ans1 = ans; } } // Function to find the longest // concatenated string having // every character of even frequency void longestString(vector arr, int index, string str) { // Checking the string if (index == arr.size()) { return; } // Dont Include the string longestString(arr, index + 1, str); // Include the string str += arr[index]; calculate(str); longestString(arr, index + 1, str); } // Driver code int main() { vector A = { "ABAB", "ABF", "CDA", "AD", "CCC" }; // Call the function longestString(A, 0, ""); // Print the answer cout << ans1 << " " << ans1.length(); return 0; } // This code is contributed by Potta Lokesh
## Java
// Java Implementation of the above approach import java.io.*; import java.util.*; public class index { static int max = 0; static String ans1 = ""; // Function to check the string static void calculate(String ans) { int dp[] = new int[26]; for (int i = 0; i < ans.length(); ++i) { // Count the frequency // of the string dp[ans.charAt(i) - 'A']++; } // Check the frequency of the string for (int i = 0; i < dp.length; ++i) { if (dp[i] % 2 == 1) { return; } } if (max < ans.length()) { // Store the length // of the new String max = ans.length(); ans1 = ans; } } // Function to find the longest // concatenated string having // every character of even frequency static void longestString( List arr, int index, String str) { // Checking the string if (index == arr.size()) { return; } // Dont Include the string longestString(arr, index + 1, str); // Include the string str += arr.get(index); calculate(str); longestString(arr, index + 1, str); } // Driver code public static void main(String[] args) { ArrayList A = new ArrayList<>(); A.add("ABAB"); A.add("ABF"); A.add("CDA"); A.add("AD"); A.add("CCC"); // Call the function longestString(A, 0, ""); // Print the answer System.out.println(ans1 + " " + ans1.length()); } }
## Python3
# Python3 implementation of the above approach maxi = 0; ans1 = ""; # Function to check the string def calculate(ans) : global maxi,ans1; dp = [ 0 ] * 26; for i in range(len(ans)) : # Count the frequency # of the string dp[ord(ans[i]) - ord('A')] += 1; # Check the frequency of the string for i in range(26) : if (dp[i] % 2 == 1) : return; if (maxi < len(ans)) : # Store the length # of the new String maxi = len(ans); ans1 = ans; # Function to find the longest # concatenated string having # every character of even frequency def longestString( arr, index, string) : # Checking the string if (index == len(arr)) : return; # Dont Include the string longestString(arr, index + 1, string); # Include the string string += arr[index]; calculate(string); longestString(arr, index + 1, string); # Driver code if __name__ == "__main__" : A = [ "ABAB", "ABF", "CDA", "AD", "CCC" ]; # Call the function longestString(A, 0, ""); # Print the answer print(ans1, len(ans1)); # This code is contributed by AnkThon
## C#
// C# Implementation of the above approach using System; public class index { static int max = 0; static String ans1 = ""; // Function to check the string static void calculate(String ans) { int[] dp = new int[26]; for (int i = 0; i < ans.Length; ++i) { // Count the frequency // of the string dp[(int)ans[i] - (int)'A']++; } // Check the frequency of the string for (int i = 0; i < dp.Length; ++i) { if (dp[i] % 2 == 1) { return; } } if (max < ans.Length) { // Store the Length // of the new String max = ans.Length; ans1 = ans; } } // Function to find the longest // concatenated string having // every character of even frequency static void longestString(String[] arr, int index, String str) { // Checking the string if (index == arr.Length) { return; } // Dont Include the string longestString(arr, index + 1, str); // Include the string str += arr[index]; calculate(str); longestString(arr, index + 1, str); } // Driver code public static void Main() { String[] A = {"ABAB", "ABF", "CDA", "AD", "CCC"}; // Call the function longestString(A, 0, ""); // Print the answer Console.WriteLine(ans1 + " " + ans1.Length); } } // This code is contributed by saurabh_jaiswal.
## Javascript
Output
Time Complexity: O(M*N* (2^N)), where N is the number of strings and M is the length of the longest string
Auxiliary Space: O(N)
Another Approach: The above approach can be further optimized by precomputing the frequency of characters for every string and updating the frequency array after concatenation of each string.
My Personal Notes arrow_drop_up
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Page : | 2,331 | 8,530 | {"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} | 3.90625 | 4 | CC-MAIN-2021-49 | longest | en | 0.897349 |
http://hubpages.com/games-hobbies/Chess-For-Beginners-The-Opening-Game | 1,481,381,055,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543315.68/warc/CC-MAIN-20161202170903-00295-ip-10-31-129-80.ec2.internal.warc.gz | 118,744,371 | 17,198 | # Chess For Beginners: The Opening Game
Source
## The Opening Game
The opening game in chess consists of the first ten to fifteen moves. Thousands of books have been written regarding the opening of a chess game, but for the beginner and amateur player, these are best eschewed, in favor of learning basic opening principles, and learning to apply them.
Before we discuss opening strategy, there are a few terms which must be introduced and explained. They are essential to an opening game. For a chess beginner, these terms and concepts seem like the hidden secrets of the masters. The first time I heard these two terms, I felt like a complete idiot, because I had never heard of them before. Rest assured, if you haven't heard them yet, I was in your shoes in the not too distant past. If you have head them, bear with me as I go over some key beginner concepts.
## En Passant
The two concepts are En passant, and Castling.
En passant is pawn play. It occurs when a white pawn gets down to the dark side, or when a dark pawn comes over to the white side. Each row (or rank) of the chess board is numbered, starting on the white side. The row (rank) occupied by white pieces is the first rank. The row occupied by white pawns is the second rank. The rows continue numerically across the board. The rank occupied by dark pawns is the seventh and the dark pieces occupy the eight rank.
The importance of numbered ranks will become evident, as I explain En passant. This pawn capture is legal when the four following criteria are met:
1. The white capturing pawn must be on the fifth rank. The dark capturing pawn must be on the fourth rank.
2. The capturing pawn and the seized pawn must be on adjacent files. (They must be on a column next to one another. Beginning on the white side, left, the column's are lettered a-h.)
3. The captured pawn must move two squares in a single move, from it's initial starting point. Remember, a pawn may move one or two squares in its first move.
4. The capturing pawn must exercise En passant in the very next move, and capture diagonally forward, as if the captured pawn had only moved one square.
While En passant is not an opening move, it is an important part of pawn play, which needs to be addressed early on, for development of strategy.
## Castling
The next important move is called castling.
Castling involves the King and a Rook. Castling is essential in any opening game strategy. Basically, the theory behind castling is that your King is moved safely out of the center of the board, and one Rook is brought out of the corner, and closer to the center where it will be more effectively used.
Castling counts as one move, although the king and the rook move simultaneously, over multiple squares. The King may work in concert with either his own Rook, or the Queenside rook. Basically, the King moves two squares toward the intended Rook. The Rook then moves to the inside side of the King.
Castling may take place when the following criterion are met:
1. Castling may not be done if either the King or the intended Rook have already moved. Each must start in it's original position.
2. All squares between the King and the Rook must be empty. Neither can jump or capture another piece to castle.
3. The King cannot castle if he is in check. If the King escapes check without moving (i.e. another piece blocks the check, or the checking piece is captured), the King may later castle when it is no longer in check.
4. The King cannot move into or through check while castling. This means that he may not land on a square attacked by the enemy. Neither may he pass through a square attacked by an enemy.
Castling may take place if the Rook is under attack.
## Best Opening Moves: Occupy the Center
The best opening plan is not to memorize the details of hundreds of opening plays. Chess is organic, and every game is different. While you may memorize opening games, your opponent may not, and will not react in a way that fits with your plan.
A better strategy is to play to occupy and control the center. From the four center squares of the chess board, it is possible to extend influence and control over a lot of territory. There are three advantages to this ploy.
1. Mobility: Pieces in the center are more mobile and effective and can move to more squares than the pieces on the edge.
2. Restriction: If you control the center, your opponents ability to move is restricted. Your pieces act as a barrier to the other side.
3. Troop preparedness: Your pieces are ready to occupy and win when you occupy the center of the board.
Now you know the basic strategy, occupy the center. Let's discuss how to develop and occupy the center.
First, center pawns should move out, to claim territory, and to free bishops. After the pawns, move Knights and Bishops. As soon as these minor pieces are developed, it's time to castle. Move your King safely out of harms way, and your Rook toward center, where it will be more effective. Next your Queen moves, connecting the Rooks. It is best to work the Rooks together, along a central file. They are powerful and effective when used together.
Thus begins your game of chess. Practice these moves over and again, until they become rote. As you see what moves work well, and what moves don't, you will develop your own unique opening game.
Source
## Best Opening Moves: Develop and Occupy the Center
Now you know the basic strategy, occupy the center. Let's discuss how to develop and occupy the center. This is one of many strategies for occupying the center of the chess board.
First, center pawns should move out, to claim territory, and to free bishops.
After the pawns, move Knights and Bishops. As soon as these minor pieces are developed, it's time to castle and protect the King.
Move your King safely out of harms way, and your Rook toward center, where it will be more effective.
Next your Queen moves, connecting the Rooks. It is best to work the Rooks together, along a central file. They are powerful and effective when used together, especially when used in concert with the powerful queen. This forms an unstoppable trio, near the center of the board.
Thus begins your game of chess. Practice these moves over and again, until they become rote. As you see what moves work well, and what moves don't, you will develop your own unique opening game.
## More by this Author
Edweirdo 6 years ago from United States
Thanks for this one Deborah!
I'm in the beginning stages of teaching my 8 year-old niece how to play chess in an effort to improve her critical thinking. (And hopefully her math skills at the same time!)
I haven't played the game myself in ages, and I had forgotten all about en passant! This will be fun to teach her and see if she can implement it during game play...
De Greek 6 years ago from UK
I also haven't played in years and had forgotten all about en passant. Many thanks :D
lmmartin 6 years ago from Alberta and Florida
Are there any limits to your knowledge? You must be hell at Trivial Pursuit. Chess now? I learned the game from my father starting when I was nine. He wasn't the most patient of teachers, often telling me to "think things through before you move," in a disgusted tone of voice. Little did I realize he was also giving me good advice for life -- advice that didn't kick in till I was about forty, unfortunately. Back to the subject at hand -- chess, we played on a regular basis, and finally when I was seventeen, I beat him for the first time. So chess is indelibly tied in my mind with my late father.
Deborah Demander 6 years ago from First Wyoming, then THE WORLD Author
Lynda, you are too kind. I know a little about a lot. As for chess, I was less than a beginner, until my wonderful husband mentioned his love for the game. I had a real struggle with my ego, in not wanting to look stupid, however, after I got over my pride, I learned a lot. Now... I can beat him one time out of ten. Thanks for the memories of your father. It's too bad, most of the good advice doesn't kick in until hindsight. *Sigh*
Edweirdo, good luck teaching your niece. And Greek, thanks for stopping by.
dahoglund 6 years ago from Wisconsin Rapids
I used to play chess but by the time I got out of college I lost my chess partners.
I never knew any formalities. This is a good refresher if I find some chess partners and play again. Good information.
Deborah Demander 6 years ago from First Wyoming, then THE WORLD Author
Hi dahoglund, I never knew of any chess formalities, until my husband and I started playing together. Now, I check my list before every move. If you don't have a partner, computer chess is always a solution. My husband has several online games going.
Thanks for stopping by.
Namaste.
0 of 8192 characters used | 1,964 | 8,812 | {"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-2016-50 | longest | en | 0.943334 |
http://spmath81708.blogspot.com/2009/02/pythagoras-pythagorean-theorem.html | 1,490,736,874,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189903.83/warc/CC-MAIN-20170322212949-00190-ip-10-233-31-227.ec2.internal.warc.gz | 330,759,385 | 16,130 | ### Pythagoras- Pythagorean Theorem
Wednesday, February 25, 2009
Pythagorean Theorem
Pythagoras
-He was a great Mathematician and Scientist
-He is known for the Pythagorean Theorem (We care because it's math)
-First to think that the Earth was not flat but round
-He discovered square numbers(Square roots)
-He was a Greek Guy
-We don't have any of his work(proof that he was alive)
-Some people say that he didn't exist
-Very smart(I mean very!)
legs
There are two legs in a right triangle. The legs are shorter than the hypotenuse. The legs make the 90 degree angle. The legs are labeled as A or B. It doesn't really matter which leg you label A or B.
Hypotenuse
The hypotenuse is the most important part.The hypotenuse is the longest side of the triangle. The hypotenuse is located opposite of the 90 degree angle. You label it as C.
R.A.T
R.A.T stands for Right Angle Triangle. You can tell if it's a right triangle because the square on the 90 degree angle. A Right Angle Triangle has a total of 180 degree. Two Right Angle Triangles can make a square or a rectangle. The two angles that connect the hypotenuse make 90 degrees.
This picture is showing the Pythagorean Theorem. It's saying that A2 and B2 will equal the C2.
Squares are important in the Pythagorean Theorem. It will be easier if you know the relationship between a square and a right angle triangle. As you know a square has four equal sides. It has for corners, and those four corners are 90 degree angles. If you have four corners then 90x4=360. then the whole square has an angle of 360. If you cut a square up like you see in the picture, you will get two right angle triangles. And those triangles will contain a 90 degree angle, two equal legs, and a hypotenuse.
Question 1
How do you solve this?
First you label witch side will be A,B, and C. Then look at the information you have. The long line that's 10mm is C so is the other side. It's 10mm. In the inside of the triangle you see a line that is 8mm. That is B. Then on the bottom, we don't know how long it is. We'll call it A. You see that the middle line(B) cuts the triangle in half to make two smaller triangles.We'll solve one side of the triangle first.
C2-B2=A2
lO2-82=C2
(10x10)-(8x8)=C2
100-64=36
36=C
Then you use the square root. The square root of 36 is 6, so 6=A BUT we only solve half of it. You just have to add 6+6 to get 12 because those two sides are the same. It has the line of symmety on A. So The answer for this question is A=12.
Onto question 2....
This diagram shows the game plans for a game designed by Harbeck Toys INC. The board is made up of a square and four identical right triangles. If the central square has an area of 225 square centimetres what is the perimeter of the board game?
Question 2
Well I think that this one is harder than the first. So first lets take 225 and find the square root. The square root is 15 right? To make sure we got it right, we need to check. 225 is the area of the square. To get area you need to times length by with but in this case you need to times side by side. Now our answer was 15. So 15x15=225.
So that means that all the sides of the square is 15. And that also means that the legs for the right triangles are 15 too! now all we need to figure out is the Hypotenuse.
Like for the last question, we need to label. The Hypotenuse is C and the bottom(base) is B and the other one is A.
A2+B2=C2
152+152=C2
(15x15)+(15x15)=450
450=C
Now we need to find the square root of 450. The square of 450 is... 21.213.20344 but we only need 21.213. So the Hypotenuse is 21.213....We are not done yet! The question said that we need to find the perimeter of the board game. So we have to add the sides up.
21.213x4=84.852
15x4= 60
_____________
144.852
So the answer to this question is 144.852.
Here's a video that explains the Pythagorean Theorem(better than I do...)
And...this is the video that I did in class... | 1,064 | 3,926 | {"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} | 4.65625 | 5 | CC-MAIN-2017-13 | longest | en | 0.945526 |
http://mizar.uwb.edu.pl/version/current/html/proofs/mfold_2/34 | 1,571,195,626,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-43/segments/1570986661296.12/warc/CC-MAIN-20191016014439-20191016041939-00405.warc.gz | 131,154,477 | 2,103 | let n be Nat; :: thesis: for p1, p2 being Point of () st p1 <> 0. () & p2 <> 0. () holds
ex R being Function of (),() st
( R is being_homeomorphism & R .: (Plane (p1,(0. ()))) = Plane (p2,(0. ())) )
let p1, p2 be Point of (); :: thesis: ( p1 <> 0. () & p2 <> 0. () implies ex R being Function of (),() st
( R is being_homeomorphism & R .: (Plane (p1,(0. ()))) = Plane (p2,(0. ())) ) )
assume p1 <> 0. () ; :: thesis: ( not p2 <> 0. () or ex R being Function of (),() st
( R is being_homeomorphism & R .: (Plane (p1,(0. ()))) = Plane (p2,(0. ())) ) )
then consider B1 being linearly-independent Subset of () such that
A1: ( card B1 = n - 1 & [#] (Lin B1) = Plane (p1,(0. ())) ) by Th26;
assume p2 <> 0. () ; :: thesis: ex R being Function of (),() st
( R is being_homeomorphism & R .: (Plane (p1,(0. ()))) = Plane (p2,(0. ())) )
then consider B2 being linearly-independent Subset of () such that
A2: ( card B2 = n - 1 & [#] (Lin B2) = Plane (p2,(0. ())) ) by Th26;
consider M being Matrix of n,F_Real such that
A3: ( M is invertible & () .: ([#] (Lin B1)) = [#] (Lin B2) ) by ;
reconsider M = M as invertible Matrix of n,F_Real by A3;
take Mx2Tran M ; :: thesis: ( Mx2Tran M is being_homeomorphism & () .: (Plane (p1,(0. ()))) = Plane (p2,(0. ())) )
thus ( Mx2Tran M is being_homeomorphism & () .: (Plane (p1,(0. ()))) = Plane (p2,(0. ())) ) by A1, A2, A3; :: thesis: verum | 518 | 1,376 | {"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-2019-43 | latest | en | 0.872461 |
https://www.engineeringenotes.com/surveying/levelling/reduction-of-levels-top-2-methods-levelling-surveying/13904 | 1,719,060,247,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862396.85/warc/CC-MAIN-20240622112740-20240622142740-00549.warc.gz | 654,666,161 | 21,946 | There are two methods of working out the reduced levels of the point from the staff readings taken in the field: 1. Height of Instrument or Collimation Method. 2. Rise and Fall Method.
#### 1. Collimation Method:
It consists in finding the elevation of the plane of collimation i.e. (H.I.) for every setting of the instrument and then obtaining the reduced levels of the points with reference to the respective plan of collimation
To start with, the H.L is calculated by adding the back sight of the R.L. of the starting point. The reduced levels of the intermediate point and the first change point are then obtained by subtracting the respective readings from the H.I. When the instrument is shifted, a new plane of collimation is set up and the height of which is calculated by adding back sight reading to the R.L. of the first change point.
The reduced levels of the successive points and the second change point arc found out by subtracting their staff readings from this new H.I. The process is repeated until all the R.Ls are worked out, and then the arithmetical check is applied.
Arithmetical Check:
The difference between the sum of the back sights and the sum of the fore sights should be equal to the difference of the first and the last R.Ls i.e. Σ B.S. – Σ F.S. = Last R.L. -First R.L. This check verifies the calculation of R.Ls. of the planes of collimation and of the change points only. There is no check on the reduction of R.Ls. of the intermediate points.
This method may more clearly be understood by the following example of longitudinal levelling from flag – staff lower base to downstream parapet of culvert no. 8 . (Fig. 7.16).
Entries are shown in Table 7.1.
#### 2. Rise and Fall Method:
In this method, the difference between consecutive points is calculated by comparing each point after the first with that immediately preceding it. The difference of their staff reading indicates rise or fall according as any staff reading is smaller or greater than that at the preceding point. The R.L. of each point is then found by adding rise or subtracting fall to or from the R.L. of the preceding point.
Remember:
If a staff reading (fore reading) is greater than that at the preceding point, then there is fall and if smaller then there is a rise.
Arithmetical Check:
In this method, there are three checks on the accuracy of reduction of levels. The difference between the sum of the back sights and the sum of the fore sights is equal to the difference between the sum of the rises and that of the falls is equal to the difference between the first and the last R.Ls.
Σ B S. – Σ F.S. = Σ Rise – Σ fall = Last R.L. – Ist R.L.
This method is made clear by finding R.Ls. of the points in Fig. 7.16 in Table 7.2.
Comparison of the Collimation and Rise and Fall Methods of Reduction of Levels:
Collimation Method:
1. In the case of more intermediate readings, there is considerable saving of labour and time as it involves only a few calculations.
2. There is no check on the R.Ls of intermediate stations.
3. There are two checks for arithmetical accuracy i.e. the difference between the sum of back sights of fore sights should be equal to the difference of the Ist and last R.Ls.
4. It is generally used for longitudinal and cross levelling operations and for giving levels of roads and canals and similar constructional works.
Rise and Fall Method:
1. It is a laborious method as staff reading of each point on the ground, after the first is compared with that preceding it, and the difference of level entered as a rise or fall.
2. There is a complete check on the reduction of R.Ls. of intermediate stations.
3. There are three checks for arithmetical accuracy. The difference between the sum of the back sights and the sum of the fore sights should be equal to that between the sum of the sum of rises and the sum of the falls as well as that between the Ist. and the last R.Ls.
4. It is generally used for earth work calculations and other precise levelling operations.
The following points should be kept in view while entering the staff readings in a level book:
1. The readings should be entered in the respective columns as soon as they are taken and in the order of their observation.
2. The first entry on the level book page is always a B.S and the last one a F.S.
3. The fore and back sights of the change points should be written in the same horizontal line.
4. The H.I should be written in the same horizontal line opposite the back sight.
5. While carrying forward the reading from one page of the level – book to the , if the last reading is an intermediate sight, it is entered in both I.S and F. S. column on this page and I.S. and B.S. columns as first entry on the page. The entries in the remaining columns against it should also be repeated on the page.
6. Brief description with neat sketches in respect of bench marks, change points and other important points should be given in the remark column.
Home››Surveying››Levelling›› | 1,119 | 5,012 | {"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-2024-26 | latest | en | 0.934945 |
https://www.additudemag.com/do-the-math/ | 1,685,397,546,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644913.39/warc/CC-MAIN-20230529205037-20230529235037-00291.warc.gz | 699,047,307 | 55,260 | # Do the Math
## Math doesn’t have to be torture! Kick it up a notch with these fun math games, designed especially for kids with ADHD.
Math is challenging for many students. Children diagnosed with ADHD find it deadly dull. It’s hard to get excited about a subject that requires so much repetition and memorization, especially for the ADHD brain, which craves novelty and multi-sensory stimulation. There are standards that must be taught, whether students have restless minds or not, but there are simple activities and games that we can use to make learning math more tolerable and (sometimes) fun.
Here are some suggestions for parents to use and build on. These activities may lead to more creative ideas of your own. Ask your child for his thoughts, as well. If a fun activity can make a child less reluctant to do his homework, and also help that lesson stick in his memory, it’s worth a try.
## Activity 1
(Materials needed: cardboard boxes, permanent marker, painter’s tape, golf club and golf balls from a thrift store)
This activity will improve basic math skills. To prepare, place a long cardboard box upside down, and cut out nine squares along one long side. Determine which math facts your child needs to practice and memorize. With a permanent marker, write the numbers 1-9, so that there is one number above each square. Place painter’s tape in a straight line on the floor, and put several golf balls on the tape. Have the child putt a golf ball until he puts it through one of the numbered holes.
When he does, there are several options. The child can add (or multiply) the hole’s number to another number that you come up with, putt again and subtract the lower number from the higher number, and so on. After the child has practiced putting for a while, end the session by having the child putt the ball into the holes until he reaches a score of 12 or 15.
[Read: Conquer the Fear of Numbers]
## Activity 2
(Materials needed: index cards, permanent marker, painter’s tape, tube socks – I use the socks that come out of the dryer without a match)
This activity is a favorite with young teenagers, though all children seem to enjoy it. Whether you use it to work on math vocabulary terms or as a reward when your child completes a homework problem, your student will not complain about math being boring after playing this game.
Write the answers to math problems on index cards and tape them to a wall or door. (The problems can be taken from your child’s math textbook or you can make them up.) Ask your child for the answer to a math problem — What is 20 divided by four? — and have him toss a sock ball at the correct answer. This is a great way to practice and review for an exam.
## Activity 3
(Materials needed: painter’s tape, sticky notes, permanent marker, ruler)
This activity helps a child master graphing. Create the X- and Y-axes of a grid with painter’s tape. Use a permanent marker and a ruler to write numbers on the axes at even intervals — 1-20 should do. Your creative student might enjoy placing colored dot stickers with the numbers on them.
[The Right Way to Teach Math]
Have the student use sticky notes to plot coordinates as he determines answers to equations from his textbook, or, better yet, have your son or daughter stand on the spots to physically graph the coordinates. Counting out loud while taking steps on the grid — say, 5 on the X-axis and 10 on the Y-axis — will help your child remember how to graph coordinates.
Much of the time, a student in school will graph coordinates using a calculator, but when she first learns about graphing, or needs a movement break, this can be a good activity to lock in the lesson.
## Activity 4
(Materials needed: index cards, clear packing tape, painter’s tape, towel, a squirt gun – not a Super Soaker-type)
This is a fun outdoor activity that can be done year-round. First, write the answer to each of your student’s math problems on a separate index card. Then cover the entire surface of the cards, front and back, with clear packing tape to make the cards water resistant and sturdier targets. Tape the index cards to the side of the house or garage, or on a deck railing. (If the weather forces you to practice inside, use the painter’s tape to tape the index cards to the interior bathtub wall.) Use the painter’s tape to mark an “X” on the ground and have your student stand on the X.
When you ask a math question — What does 7 x 6 equal? — your student squirts water at the index card with the correct answer. You can use this game to improve addition, subtraction, multiplication, and division skills, as well as learning fractions. You could end a homework session with this activity. If there is a dry card left after your child has taken his shots at all the math questions you’ve asked him, then you know which problem to revisit.
These games will show your child that math does not have to be all drudgery and “sit still and work.” After trying these ideas, look for other ways to add fun to the learning process. It never hurts to try, and it might help a great deal.
[The Math Teacher’s Guide to Helping Struggling Students] | 1,128 | 5,160 | {"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} | 3.796875 | 4 | CC-MAIN-2023-23 | latest | en | 0.945752 |
https://www.coursehero.com/file/21130494/Screenshot-2png/ | 1,544,830,613,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826530.72/warc/CC-MAIN-20181214232243-20181215014243-00205.warc.gz | 851,666,635 | 38,234 | Screenshot (2).png
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Unformatted text preview: €31 Aplla: Student Question X (- C IG) courses,apl'Ia.com/af/servIetjquiz?qu'Iz_actIon=takeQuiz&quiz_probGuId=QNAPCOABOIO1GOOOOOEBBbScEOOGUOOGS-Ictx=gcassZ-000854ck:1_1438777584293_0AAAGBAF0159C6D43D586DC '? {II E Apps 5 Google 33 Outlook Web App n University oflllinoisi m Blackboard Learn E Dayfarce| HCM D Residence life- Sluc- D Enterprise Authentic:- D CengageBrairI - LDgII' » I Other bookmarks Attempts: Average: 1' 3 3. Tax base and tax structure A3 A3 2. Consider an economy In which the government has devised three tax plans. The following table summarlzes each tax proposal and Its structure. Note that indlvlduals pay \$1) In tax If they don‘t own a house. Average Tax Rate House Value Plan B so u \$100,000 3% 10% \$100,001 ~ \$200,000 3% 12.5% \$200,001 _ \$300,000 3% 15% \$300,001 — \$400,000 3% 17.5% \$400,001 _ \$500,000 3% 19% All three tax plans are based on the value of the house an Individual owns , 0n the following graph, use the orange oolnts (square symbol) to plot the tax schedule for Plan 5. Plot poInts From Ielt to rloht a: Increments of \$100,000 of house value. Rememher to plot a polo: for a house value of 50, so that you end up with six points in total. Llne segments wlll automatlcalty connect the oomts, TAX [Thousands of dollars] ‘IDD Tax Schedule 4: Sess ion Tune out ...
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DIGITAL PDF AND PRINTABLE: Teachers will download a huge math bundle of fraction task cards, worksheets and activities. There are 10 sets of task cards (300 task cards), 2 worksheet packets and an interactive notebook activity.
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Third grade Common Core math skills include naming fractions, fractions on a number line, comparing fractions, equivalent fractions, fraction pizza, fraction word problems and more. These fraction activities work well in math centers or stations as 2nd or 3rd grade math test prep. They also work well for special education math students.
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Fractions Pizza Math Activities CCSS.Math.Content.3.NF.A.1
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Mixed Numbers on a Number Line Task Cards CCSS.Math.Content.3.NF.A.2.B
Improper Fractions on a Number Line Task Cards CCSS.Math.Content.3.NF.A.2.B
Equivalent Fractions on a Number Line CCSS.Math.Content.3.NF.A.3.A
Equivalent Fractions Pictorial CCSS.Math.Content.3.NF.A.3.B
Whole Numbers as Fractions on a Number Line CCSS.Math.Content.3.NF.A.3.C
Comparing Fractions (pictorial) Math Center Task Cards CCSS.Math.Content.3.NF.A.3.D
Comparing Fractions (numerical) Math Center Task Cards CCSS.Math.Content.3.NF.A.3.D
Fraction Worksheets Packet 45 Pages Grade 3 Common Core
Fractions Interactive Math Notebook
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Profit and loss Type 1-2 (in Hindi)
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In this lesson discuss the Profit and loss type 1.
Vishal Garg
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Sir in order to make in ratios we make one side term to 100 and then solve
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2. Every Concepts of Profit & Loss For Railway & SSC exams By Vishal Garg unacademy (Verified Educator)
3. Every Concepts of Profit & Loss Type Wise Questions for Railway and SSC exams unacademy By Vishal Garg
5. erson s 10% loss buf on 2nd Abide he got 16-66% doss 6nHes? Firull he will overe! gd 30 ./oss. Then find tte CP Hterente CP SO CP SP lo /o 20 /2? -16-65% ,-I 6 X3 7 21 1 8 Ysp SP 36
6. No loss nd CP PIL lo SP
7. pe 2 o sa 2 pr No loss lD Total P/L 2 SP 18 52
8. OSS loss 2 Ltna ST -107 20% 60
9. Son ald 2 th his Ardicle at 10% loss 1py. 2S 2. 3 tio% 130 7o
10. gat 40% SP -2 % yox 12 Sunit 20 SR 600 T | 384 | 1,128 | {"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-2020-05 | longest | en | 0.815132 |
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# Solve the following :
Question:
The force of buoyancy exerted by the atmosphere on a balloon is $B$ in the upward direction and remains constant. The force of air resistance on the balloon acts opposite to the direction of velocity and is proportional to it. The balloon carries a mass $M$ and is found to fall down near the earth's surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v?
Solution:
$\mathrm{B}+\mathrm{bv}=\mathrm{mg}$
$B=b v+(m-\propto) g$
$\mathrm{Bv}$ is constant
$M g-B=B-\left(M^{\propto}\right) g$
$2 \mathrm{Mg}=2 \mathrm{~B}+{ }^{\propto} \mathrm{g}$
$\propto_{=2}\left(M m-\frac{B}{g}\right)$
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Broad Topics > Algebraic expressions, equations and formulae > Creating and manipulating expressions and formulae
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# Relation Between KG and Newton
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In Physics, every physical entity can be measured in different ways. Every unit can be related to one another by performing unit conversions without violating the laws of physics and laws of nature.Â
Newton is the SI unit of force and kg is the unit of mass. According to Newton's Second law of motion, force is directly proportional to the mass of the object on which force has been exerted. Thus we can say Newton and Kg are also directly proportional to each other, thus if we encounter any change in the unit of force in Newton it will result in a change in the unit of mass in Kg keeping the acceleration constant.
To arrive at a mathematical description for the relation between Kg and Newton, let us discuss the definition of Kg and Newton respectively.
### What is Kg?
There are seven fundamental units in physics, Kg is one of them. (Fundamental units are the units that are independent quantities, all other units are derived from them). The Abbreviation of Kilogram is Kg.
1 Kilogram is nearly equal to 1000grams. The kilogram is one of the basic units of metric systems.
### What is Newton?
Newton is the SI unit of Force. It is defined as- the force required to accelerate an object of the mass of 1 kilogram (1kg) by 1m/s2 in the direction of applied force. In the CGS system, 1N is equal to 105 Dyne. Dyne is the unit of Force in the CGS system.
### Derivation:
Mathematically we can describe one Newton by using Newton’s second law of motion,
I.e.,
⇒ F = ma
⇒ 1 Newton = 1Kg x 1m/s²
From the above expression, Newton is directly proportional to Kg. Therefore,
• If the object under consideration is having negligible mass or considerably less mass then the force required in Newton will also be very less.
• If the object under an action-heavy, then the force required will also be more.
### 1N is Equal to How Many kg?
We can not actually convert Kg to Newton, because Newton is a unit of Force whereas Kg is the unit of mass respectively. Conversion of units can be done for two identical scales but not for two different physical scales. So, we can give relation for the two units, saying conversion of them will be the wrong term. Â
When an object is dropped from a certain height above the ground level, it will experience a force purely due to the acceleration due to gravity. While free fall the object will experience a force and the force experienced is known as Weight of the body/object and mathematically is given by:
W = mg ……..(1)
Now we can give a relation between newton and kg by analyzing how many Kg in one Newton. Let us assume that a 1Kg mass is dropped from a height above ground level with an initial velocity zero then the force experienced by the object is,
W = 1Kg x 9.81 m/s² = 9.81 N
From the discussion of the relation between Kg and Newton, we have concluded that they are directly proportional to each other. Therefore, we write:
1Kg=9.81N
Therefore, 1 newton is equal to how much Kg or 1 kg how many newtons? The answer is 9.81N.
Similarly, If the question demands for 1 kg wt is equal to how many newtons? (Which is in terms of gravitational force) Then also the answer is either 9.8N or 9.81N or 10N because the value of acceleration due to gravity is considered depending upon the convenience.
Example:
1. The Mass in Kg of an object that Weighs about 40N.
Ans: Given that weight of the object is 40N.
We are asked to find the mass of the object.
From the relation between Kg and Newton we have,
⇒ 1Kg = 9.81N
⇒ 1N = 0.102Kg
Thus,
⇒ 40 N = 40 x 0.102Kg = 4.077Kg
Therefore, the mass in kg of an object that weighs about 40N is 4.077.
An alternative method for calculating the mass of an object whose weight is given is as follows:
We know that,
⇒ 1N = 1Kgm/s$^{2}$
Therefore,
⇒40N = 40Kgm/s$^{2}$
We want to convert 40N into Kg, to do that divide the above expression by acceleration due to gravity, i.e., divide by 9.81 m/s$^{2}$.
⇒ 40N = 40Kgms$^{-2}$/9.81 ms$^{-2}$
⇒ 40N = 4.077kg
This is the required answer.
2. How Many Newtons is 5Kg?
Ans: We know that 1kg = 9.81N, then:
⇒ 5Kg = 5 x 9.81 N = 49.05N
therefore,49.05N is 5Kg.
FAQ (Frequently Asked Questions)
Q1. 1n is Equal to How Much Kg or 1Newton How Many kg?
Ans: We know that 1Kg is 9.81N therefore,
1N=0.102Kg.
Q2. What is the Kg-wt (Kilogram weight) of an Object?
Ans: Kg-wt is the gravitational force experienced by an object. The relation between 1kg-wt and Newton is given by 1kg-wt =9.81N.
Q3. What is the Difference Between Mass and Weight? Are Mass and Weight the Same?
Ans: No. Mass and Weight are two different physical quantities. The mass of an object is its amount of matter, Weight of an object is the force exerted on the body due to gravity. | 1,298 | 4,855 | {"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-2021-25 | longest | en | 0.919563 |
http://zipassingnment.com/richard-exchanges-an-office-building-used-in-business-for-one-owned-by-summer-t/ | 1,571,095,184,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655554.2/warc/CC-MAIN-20191014223147-20191015010647-00215.warc.gz | 354,941,971 | 7,150 | # Richard exchanges an office building used in business for one owned by Summer. T
Richard exchanges an office building used in business for one owned by Summer. The FMV of Riachard's building is \$350,000 ( basis \$150,000) and it is subject to a mortgage of \$60,000 which is assumed by Summer. Richard receives \$40,000 in cash and Summer's office building, which has a FMV of \$250,000 ( basis of \$180,000). Richard realizes a gain / loss on the exchange of? | 117 | 464 | {"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-2019-43 | latest | en | 0.950144 |
http://www.docstoc.com/docs/57835486/circle-theorems-tangents | 1,394,586,701,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394020561126/warc/CC-MAIN-20140305115601-00075-ip-10-183-142-35.ec2.internal.warc.gz | 395,350,580 | 13,026 | # circle-theorems-tangents
Document Sample
``` Circle Theorems: 5. The angle between a tangent and the radius
drawn to the point of contact is 900
Lesley Hall @ Soar Valley College
Circle Theorems: 6. The angle between circle just two tangents
From any point outside a tangent and the radius
drawn to be point and they is 900
to the circle may the drawnof contact are equal.
Lesley Hall @ Soar Valley College
Circle Theorems: 7.
angle between tangent
and chord
Alternate segment theorem: The angle between a tangent and a chord
through the point of contact is equal to the angle subtended by the chord in
the alternate segment. Lesley Hall @ Soar Valley College
Circle Theorems: 7.
angle subtended by the chord
angle between tangent
and chord
Alternate segment theorem: The angle between a tangent and a chord
through the point of contact is equal to the angle subtended by the chord in
the alternate segment. Lesley Hall @ Soar Valley College
Circle Theorems: 7.
angle subtended by the chord
angle between tangent
and chord
Alternate segment theorem: The angle between a tangent and a chord
through the point of contact is equal to the angle subtended by the chord in
the alternate segment. Lesley Hall @ Soar Valley College
Circle Theorems: 7.
Alternate segment theorem: The angle between a tangent and a chord
through the point of contact is equal to the angle subtended by the chord in
the alternate segment. Lesley Hall @ Soar Valley College
Circle Theorems: 7.
these two angles
should be equal
Alternate segment theorem: The angle between a tangent and a chord
through the point of contact is equal to the angle subtended by the chord in
the alternate segment. Lesley Hall @ Soar Valley College
Circle Theorems: 7.
the angles are equal
Alternate segment theorem: The angle between a tangent and a chord
through the point of contact is equal to the angle subtended by the chord in
the alternate segment. Lesley Hall @ Soar Valley College
Circle Theorems: 7.
angle subtended by the chord
in the alternate segment
angle between tangent
and chord
Alternate segment theorem: The angle between a tangent and a chord
through the point of contact is equal to the angle subtended by the chord in
the alternate segment. Lesley Hall @ Soar Valley College
Circle Theorems: 7.
Alternate segment theorem: The angle between a tangent and a chord
through the point of contact is equal to the angle subtended by the chord in
the alternate segment. Lesley Hall @ Soar Valley College
```
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views: 103 posted: 10/20/2010 language: English pages: 10 | 628 | 2,659 | {"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} | 3.296875 | 3 | CC-MAIN-2014-10 | longest | en | 0.879722 |
https://forum.freecodecamp.org/t/the-smallest-common-multiple-execution-time-and-algorithm/405617 | 1,716,344,025,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058525.14/warc/CC-MAIN-20240522005126-20240522035126-00048.warc.gz | 217,888,157 | 9,346 | # The smallest common multiple execution time and algorithm
Hi everyone! I recently did the “smallest common multiple” challenge. I did two different solutions and I know they must be very beginner level and slow. I wanted to know how slow they are and for the solutions provided by FCC, I want to know how fas they are.
As expected, my solutions did run slow. Curve fit for execution time ~ the larger number passed into function is around second degree polynomial.
Solution 4 provided by FCC is very stable, the execution time flattens out at around 100ms. However, the console always alerts about “Potential infinite loop detected on line 7. Tests may fail if this is not changed.”.
The code for Solution 4 is:
``````const smallestCommons = arr => {
let max = Math.max(...arr);
let min = Math.min(...arr);
// Initially the solution is assigned to the highest value of the array
let sol = max;
for (let i = max - 1; i >= min; i--) {
// Each time the solution checks (i.e. sol%i===0) it won't be necessary
// to increment 'max' to our solution and restart the loop
if (sol % i) {
sol += max;
i = max;
}
}
return sol;
};
``````
Solution 3 is very robust at smaller numbers. For example, if I call smallestCommons([1, 100]), execution time is around 0.9ms, comparing to 102ms for solution 4. But when I call bigger numbers, for example, smallestCommons([1, 250]), the console will say "RangeError: Maximum call stack size exceeded
".
The code for Solution 3 is:
``````function smallestCommons(arr) {
// Euclidean algorithm for the greatest common divisor.
// ref: https://en.wikipedia.org/wiki/Euclidean_algorithm
const gcd = (a, b) => (b === 0 ? a : gcd(b, a % b));
// Least Common Multiple for two numbers based on GCD
const lcm = (a, b) => (a * b) / gcd(a, b);
// range
let [min, max] = arr.sort((a, b) => a - b);
let currentLCM = min;
while (min < max) {
currentLCM = lcm(currentLCM, ++min);
}
return currentLCM;
}
``````
Would anyone please help explain why Solution 4 has a potential infinite loop, and why Solution 3 cannot work at larger numbers? Thank you soooooo much!
P.S. These are my codes, I would also greatly appreciate if anyone would give some comments. Thank you so much!
The less slow one:
``````function smallestCommons(arr) {
//sort the argument array and get the smaller number (m) and bigger number (n)
let m = arr.sort((a, b) => a - b)[0];
if (m == 1) {m = 2;}
let n = arr.sort((a, b) => a - b)[1];
//create a new 'factors' variable which equals to an array of integers ranging from 2 to n
let factors = Array.from({length: n + 1}).map((_, i) => i).slice(2);
//create a 'numbers' variable equal to a sequence of integers ranging from m to n;
let numbers = factors.slice(m - 2, n - 1);
//create a 'product' variable equal to the product of m through n, which may not be the smallest common multiple
let product = numbers.reduce((accu, val) => accu * val);
for (let index in factors) {
let factor = factors[index];
//if product divided by factor is still a common multiple, divide product by factor to get a smaller common multiple
while(numbers.reduce((flag, val) => flag && (product / factor) % val == 0, true)) {
product /= factor;
}
//filter away numbers that are multiple of the variable 'product'
factors = factors.filter(val => val == factor || val % factor != 0);
}
return product;
}
``````
The more slow one (roughly two times slower):
``````function smallestCommons(arr) {
//sort the argument array and get the smaller number (m) and bigger number (n)
let m = arr.sort((a, b) => a - b)[0];
let n = arr.sort((a, b) => a - b)[1];
//create a new 'primes' variable which equals to an array of integers ranging from 2 to n
let primes = Array.from({length: n + 1}).map((_, i) => i).slice(2);
//filter 'primes' so that it contains prime number only
for (let n in primes) {
primes = primes.filter(num => num == primes[n] || num % primes[n] != 0);
}
//create a 'counts' variable which is an array of the same length with 'primes' array filled with 0
let counts = Array(primes.length).fill(0);
//prime factorization of integers through m to n
//update 'counts' array with the highest count of each prime factor
for (let index in primes) {
//prime factorization starts from the smallest prime number
let prime = primes[index];
//initialize a 'count' variable equal to 0, reprenting count of the prime number in factorization
let count = 0;
//primes factorization starts from number m and increment to n
for (let j = m; j <= n; j++) {
//copy j to num so that j will not be changed during the loop
let num = j;
//reset count to 0 for each new factorization on a new number
count = 0;
//prime factorization on the number 'num'
while(num % prime == 0) {
num /= prime;
count += 1;
}
//update 'counts' array with the highest 'count'
if (count > counts[index]) {
counts[index] = count;
}
}
}
//multiply 'primes' array and 'counts' array to get smallest common multiple
return primes.reduce(function(multiple, prime, index) {return multiple * (prime ** counts[index]);}, 1)
}
``````
Hello~!
Solution 4 trips the infinite loop protection alert not because there is an infinite loop, but because the protection kicks in when it thinks a loop might take more than 2500ms (or close to that, I’m not 100% sure of the exact value).
Solution 3 throws a Stack Size error because the `gcd` function is recursive. Recursive functions get dropped onto the stack until there are no more function calls, and then the stack gets executed. So the system is throwing an error because there are too many function calls in the stack.
There is some discussion here on measuring performance and what makes code slow or fast.
Basically, two things are slow, excessive memory use and excessive calculations. Recursion can create excessive memory use. Not using the GCD can cause excessive calculations.
Update:
repl
https://repl.it/repls/ImpassionedBurlyTrees
code
``````// ********************************************************************************
// Least Common Multiple of Range demo and performance study
// ********************************************************************************
"use strict"
// --------------------------------------------------------------------------------
// Performance logging
// --------------------------------------------------------------------------------
const performance = require('perf_hooks').performance;
// --------------------------------------------------------------------------------
// Debug via console logging flag
// --------------------------------------------------------------------------------
const debug = false;
// --------------------------------------------------------------------------------
// Brief: Find least common multiple of a range
//
// Inputs:
// arr - array containing lower and upper bound (inclusive)
//
// Outputs:
// lcm - least common multiple of range
// --------------------------------------------------------------------------------
function smallestCommons(arr) {
if (debug) console.log("WARNING - DEBUGGING LOGGING WILL DECREASE PREFORMANCE");
// Compute GCD of two numbers
function gcd(a, b) {
// Loop until reduced
while (b !== 0) {
if (debug) console.log("GCD - a: " + a + " b: " + b);
[a, b] = [b, a % b];
}
if (debug) console.log("GCD - final value: " + a);
return a;
}
// Compute GCD of two numbers, with recursion
//function gcd(a, b) {
// if (debug) console.log("GCD - a: " + a + "b: " + b);
// return (b === 0) ? a : gcd(b, a % b);
//}
// Compute LCM of two numbers
function lcm(a, b) {
if (debug) console.log("LCM - a: " + a + " b: " + b);
return (a * b) / gcd(a, b);
}
// Loop over range
const lower = Math.min(...arr);
const upper = Math.max(...arr);
let currentLCM = lower;
for (let i = lower + 1; i <= upper; i++) {
if (debug) console.log("Iteration - i: " + i)
if (debug) console.rog("Current - currentLCM: " + currentLCM);
currentLCM = lcm(i, currentLCM);
}
// Return
if (debug) console.log("END DEBUGGING OUTPUT");
return currentLCM;
}
// --------------------------------------------------------------------------------
// Performance testing
// --------------------------------------------------------------------------------
// Test cases
const maxMultiple = 4;
const numRuns = 25;
console.log("--------------------------------------------------");
console.log("Summary")
console.log(" - Number of Test Cases : " + maxMultiple);
console.log(" - Runs Per Test Case : " + numRuns);
console.log("--------------------------------------------------\n\n");
for (let i = 1; i <= maxMultiple; i++) {
/// Log test case
const arr = [1, 50*i + 17];
// Note: [1, 217] is the widest range with a result that isn't 'Infinity'
console.log("--------------------------------------------------");
console.log("Test Case " + i);
console.log("--------------------------------------------------");
// Multiple runs
let lcm = 0;
let times = [];
for (let j = 0; j < numRuns; j++) {
// Time execution
const t0 = performance.now();
lcm = smallestCommons(arr);
const t1 = performance.now();
// Log time elapsed
times.push(t1 - t0);
}
// Compute stats
const minTime = Math.min(...times);
const maxTime = Math.max(...times);
const avgTime = times.reduce((sum, item) => sum + item, 0) / numRuns;
const variance = times.reduce((sumSqDiff, item) => sumSqDiff + (avgTime - item)**2, 0) / numRuns;
const stdDev = Math.sqrt(variance);
// Output
console.log(" - Problem Values");
console.log(" Lower : " + arr[0]);
console.log(" Upper : " + arr[1]);
console.log(" Result : " + lcm);
console.log(" - Statistics");
console.log(" Average Time : " + avgTime);
console.log(" Minimum Time : " + minTime);
console.log(" Maximum Time : " + maxTime);
console.log(" Standard Deviation : " + stdDev);
console.log(" Variance : " + variance);
console.log("--------------------------------------------------\n\n");
}
// ********************************************************************************
``````
As far as I can tell, recursion isn’t really slower than iteration in Javascript nowadays (ES2015 added Tail Call Optimization) and can in some cases be faster. But in certain cases you can blow through the call stack and fail to return an answer with recursion while a loop returns an actual result. | 2,482 | 10,312 | {"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-2024-22 | latest | en | 0.852377 |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=plot/style | 1,498,596,870,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321553.70/warc/CC-MAIN-20170627203405-20170627223405-00076.warc.gz | 586,020,432 | 24,230 | Plot 2-D Style Options - Maple Programming Help
Home : Support : Online Help : Graphics : 2-D : Options : plot/style
Plot 2-D Style Options
Calling Sequence plot(f, h, v, style=s)
Parameters
f - function(s) to be plotted h - horizontal range v - (optional) vertical range s - line, point, polygon, or polygonoutline
Description
• For 2-D plotting, there are four different styles, line, point, polygon (patchnogrid), and polygonoutline (patch). The names in parentheses are aliases for the option values. The default is polygonoutline.
• If the style is point, plot simply plots the calculated or given points. These should be given as a list of ordered pairs, that is
$\left[\mathrm{x1},\mathrm{y1},\mathrm{x2},\mathrm{y2},...,\mathrm{xn},\mathrm{yn}\right]$
or
$[[\mathrm{x1},\mathrm{y1}],[\mathrm{x2},\mathrm{y2}],\mathrm{...},[\mathrm{xn},\mathrm{yn}]]\right\}$
• The styles line, polygon, and polygonoutline all draw curves by interpolating between the sample points.
• If there are polygons in the plot, then the line style draws polygons as outlines only, the polygonoutline style draws polygons as filled with an outline, and polygon draws the polygons as filled with no outline. If there are no polygons in the plot, then these three options are equivalent.
Examples
> $\mathrm{plot}\left(x,x=0..4,\mathrm{style}=\mathrm{point}\right)$
> $\mathrm{plot}\left({x}^{2},x=-4..4,\mathrm{style}=\mathrm{point}\right)$
A point style sine curve along with a line style cosine curve
> $\mathrm{p1}≔\mathrm{plot}\left(\mathrm{sin}\left(x\right),x=-\mathrm{π}..\mathrm{π},\mathrm{style}=\mathrm{point}\right):$
> $\mathrm{p2}≔\mathrm{plot}\left(\mathrm{cos}\left(x\right),x=-\mathrm{π}..\mathrm{π},\mathrm{style}=\mathrm{line}\right):$
> $\mathrm{plots}[\mathrm{display}]\left(\left\{\mathrm{p1},\mathrm{p2}\right\}\right)$
> $\mathrm{bowtie}≔\mathrm{Matrix}\left(\left[\left[0,0\right],\left[0,1\right],\left[0.5,0.75\right],\left[1,1\right],\left[1,0\right],\left[0.50,0.25\right],\left[0,0\right]\right],\mathrm{datatype}=\mathrm{float}\right):$
> $\mathrm{plots}[\mathrm{polygonplot}]\left(\mathrm{bowtie},\mathrm{color}="Orange"\right)$
The same plot with no outline.
> $\mathrm{plots}[\mathrm{polygonplot}]\left(\mathrm{bowtie},\mathrm{color}="Orange",\mathrm{style}=\mathrm{polygon}\right)$
Please add your Comment (Optional) E-mail Address (Optional) What is ? This question helps us to combat spam | 746 | 2,439 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "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} | 3.25 | 3 | CC-MAIN-2017-26 | latest | en | 0.476977 |
https://jrmathclub.wordpress.com/2014/06/09/nesbitts-inequality/ | 1,498,708,107,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323864.76/warc/CC-MAIN-20170629033356-20170629053356-00595.warc.gz | 800,228,319 | 33,509 | # Nesbitt’s Inequality
The Nesbitt’s inequality goes like this:
I want to prove this in this post:
We define x to be a+b, y to be b+c and z to be a+c.
Now we have to prove, that x/y+y/x is bigger or equal to 2. We multiply the equation with x times y times 2. We get: x^2+y^2 is bigger or equal to 2xy. Now we subtract 2xy on both sides and we get (x-y)^2 is bigger or equal to 0. This is true. Therefor we proved that x/y+y/x is bigger or equal to 2.
Now we have:
This completes the proof. | 153 | 496 | {"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-2017-26 | longest | en | 0.875246 |
https://toc.123doc.org/document/2754666-3-bar-graphs-histograms-and-box-plots.htm | 1,553,076,116,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202324.5/warc/CC-MAIN-20190320085116-20190320111116-00401.warc.gz | 644,319,744 | 10,891 | Tải bản đầy đủ - 0 (trang)
3 Bar Graphs, Histograms, and Box Plots
3 Bar Graphs, Histograms, and Box Plots
Tải bản đầy đủ - 0trang
of the independent, categorical variable (locality) and the y-axis represents the dependent
variable (mean snowfall in meters).
Figure 3.1 Clustered bar chart comparing the mean snowfall of alpine forests between
2013 and 2015 in Mammoth, CA; Mount Baker, WA; and Alyeska, AK.
Notice that Figure 3.1 gives a clear depiction of the differences in the mean snowfall at the
three localities. By adding error bars (standard deviations), the researcher is also able to
illustrate the variance in each one of the groups of data. For instance, the snowfall in
Alyeska, AK is less variable than the snowfall in Mount Baker, WA.
Figure 3.2 Clustered bar chart comparing the mean snowfall of alpine forests between
2013 and 2015 in Mount Baker, WA and Alyeska, AK. An improperly scaled axis
exaggerates the differences between groups.
One of the most important considerations when displaying data with a bar graph is the
scaling of the axes. Unfortunately, graphs built in the programs Excel and Numbers are
often created with an improperly scaled y-axis. If the y-axis does not begin with zero, then
the differences between groups appear exaggerated. By “zooming in” on this smaller set of
y-axis values, the graph can be misleading. Take the previous example of snowfall
measurements. It is clear that the average snowfall in Mount Baker, WA and Alyeska, AK
are very similar. However, if the graph of these two localities is built with a modified yaxis as in Figure 3.2, with a minimum value set to 16.5, the differences appear dramatic,
when in reality they are not.
Figure 3.3 Clumped bar chart comparing the mean snowfall of alpine forests by year
(2013, 2014, and 2015) in Mammoth, CA; Mount Baker, WA; and Alyeska, AK.
Clumped Bar Charts
If this same researcher was interested in illustrating the trends in snowfall patterns over
a 3-year period, a clumped bar chart would be useful. Figure 3.3 shows snowfall patterns
within each locality over a 3-year period. By using a clumped bar chart, the researcher can
demonstrate trends within each category of data. For example, we can see that the
snowfall was exceptionally high in 2014 at the Mount Baker, WA location; however, the
snowfall at the Mammoth, CA location was fairly stable over time.
Stacked Bar Charts
Next the researcher wants to illustrate differences in the timing of snowfall by month
within each site. For this example, a stacked bar chart is helpful in illustrating the relative
contributions of parts to the whole. Figure 3.4 shows the amount of snow that fell within
the months of January, February, and March, 2015. Notice that in Mammoth, CA there
was zero snowfall in the month of January.
Figure 3.4 Stacked bar chart comparing the mean snowfall of alpine forests by month
(January, February, and March) for 2015 in Mammoth, CA; Mount Baker, WA; and
Alyeska, AK.
Figure 3.5 Histogram of seal size.
Histograms
Histograms are another form of bar charts used to display continuous categories, like a
consecutive range of values for age. If your data are made up of quantitative variables,
then consider constructing a histogram. The format is similar to that of a bar chart;
however, the categories along the bottom are represented with a set range of values.
Hence, both axes will be represented on a numerical scale. Also, the aesthetics are slightly
different because there are no spaces between the bars. In a histogram, there will never
be space between bars because the horizontal axis is representing continuous values
(Figure 3.5). If a space does exist between bars, then it means that there are no values for
that range.
Box Plots
The box plot (also called a box and whisker plot) is a convenient way to illustrate several
key descriptive statistics from a dataset. Box plots show the median, as well as the
distribution of the data through the use of quartiles, which divide ranked data into four
equal groups, each consisting of a quarter of the data.
Consider the following dataset:
The first step in developing a box plot for these data is to define the quartiles. Several
methods are currently debated regarding how to define quartiles; the following example
uses the simplest and most intuitive method. In the sample dataset above, the numbers
must first be rearranged so that they are in order:
Second, find the median, which is also defined as the second quartile (Q2). In the current
example, there is an even number of data points, so the median is calculated as the
average of the middle two numbers (Q2 = 24). If there were an odd number of points, the
median would be excluded for the next step. Third, calculate the median of each half of
the data (on either side of the median); these medians are the first and third quartiles (Q1
and Q3):
The box component of a box plot spans the first quartile to the third quartile, and is
known as the interquartile range (IQR); the median is shown inside the box at the
position of the second quartile, as illustrated in Figures 3.6 and 3.7.
Figure 3.6 Example box plot showing the median, first and third quartiles, as well as the
whiskers.
Figure 3.7 Comparison of the box plot to the normal distribution of a sample population.
By showing the median, as well as the position of the first and third quartiles, box plots
give information about the degree of dispersion, as well as the skewness of the data. Box
plots often also have lines (the whiskers) extending from the box to represent the
variability of the data outside of the upper and lower quartiles. The whiskers usually mark
the minimum and maximum values for the dataset. However, if the dataset contains
outliers, the whiskers will extend only up to a certain point, defined as Q1 − 1.5 × IQR or
Q3 + 1.5 × IQR (Figure 3.7). Outliers will be depicted as points outside of the whiskers
(Figure 3.8).
Figure 3.8 Sample box plot with an outlier.
The box plots on previous pages, Figures 3.7 and 3.8, have been drawn for illustrative
purposes in a horizontal orientation, but are most often shown vertically, as in Figure 3.8.
In Figure 3.8, descriptive information from two groups of data is depicted. Although the
medians for the two groups are the same, the differences in the dispersion and skew of
the data are apparent. While group B shows a normal distribution, group A shows a
“positive skew,” with a tail that extends in the positive direction. The box plot for group A
also shows the position of an outlier, whose value is beyond the range of the whiskers.
Generating box plots is straightforward in both SPSS and R and is included in this book's
tutorials. However, generating box plots in Excel and Numbers is both lengthy and
complex, and involves manipulating stacked bar charts. If you do not have access to SPSS
or R, we recommend looking for a free, online box plot generator, which is an easy and
quick solution for creating box plots of your data.
Tutorials
How to Make a Bar Chart in Excel
The following tutorial will walk you through the construction of a bar chart (also known
as column graph or bar plot) using Excel. The data involve the number of rows of snail
*Data
taken from the research of Vanessa C. Morales, Robert Candelaria, and
Dr. Kathleen Weaver.
Refer to Chapter 12 for tips and tools when using Excel.
Excel offers two methods to construct a simplified bar chart with error bars. While the
first method may be more challenging at first, the lessons learned will give you greater
mastery and flexibility. Calculate the average and standard deviation of the radula from
each population prior to beginning the tutorial.
Method 1
1. Arrange data in columns on the spreadsheet.
2. Click on an empty cell. Select Insert, Column, and select the first 2-D Column
option. There are several types of bar graphs available. Use the one appropriate for the
data you want to display.
3. A blank canvas will appear.
4. Right click on the blank canvas and choose the Select Data option.
5. Under Legend Entries, select Add.
Note: Add each data point as separate series so that the standard deviation bars can be
entered separately.
6. Select the icon corresponding to the Series name subheading.
7. Select the first series title then click on the icon to the right.
8. Select the icon corresponding to the Series values.
9. Select the first value then click the icon on the right.
10. Click OK.
11. You will be directed to the original popup. Repeat steps 5–10 to input the remaining
values.
12. After the second variable is added, you should be left with a graph that looks like the
following.
After the third variable:
13. Once all the variables have been added to the graph, click OK.
14. A very basic column graph will appear, similar to the one below.
15. As a default, Excel labels the x-axis as “1.” To delete this label, select label “1.” A box
will appear. Then, press delete on your keyboard.
Tài liệu bạn tìm kiếm đã sẵn sàng tải về
3 Bar Graphs, Histograms, and Box Plots
Tải bản đầy đủ ngay(0 tr)
×
x | 2,150 | 9,191 | {"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 | latest | en | 0.915932 |
https://www.digglicious.com/uncategorized/what-size-is-a1-pixels/ | 1,670,566,384,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711390.55/warc/CC-MAIN-20221209043931-20221209073931-00314.warc.gz | 771,292,745 | 9,111 | ## What size is A1 pixels?
Paper Sizes Guide
Size Name Size in mm (without bleed area) Size in pixels 300dpi (without bleed area)
A3 420 x 297 mm 4961 x 3508 px
A2 594 x 420 mm 7016 x 4961 px
A1 841 x 594 mm 9933 x 7016 px
A0 1189 x 841 mm 14043 x 9933 px
## Is A4 and ISO A4 same?
The most popular series of the ISO standard is the A series. The most widely used paper of this series is the A4 format….ISO 216.
Format Size in Millimeters Size in Inches
A1 594 × 841 23.4 × 33.1
A2 420 × 594 16.5 × 23.4
A3 297 × 420 11.7 × 16.5
A4 210 × 297 8.3 × 11.7
What is A1 size ratio?
The height/width ratio remains constant for all sizes: 1:1.41 or the square root of 2. The dimensions always get rounded to the nearest millimeter.
### What does size A1 mean?
An A1 piece of paper measures 594 × 841 mm or 23.4 × 33.1 inches. Cutting it in half will create two A2 sheets of paper. An A1 piece of paper will fit into a C1 envelope.
### How do I make a photo A1 size?
Open the image and choose Tools > Adjust Size. In the sheet that drops down, make sure Resample image is on. Also Scale proportionally. Change the size drop down to whatever seems easier.
What are the dimensions of A1?
A Paper A1 The dimensions of A1 size is 59.4 x 84.1 cm or 594 x 841 mm or 23.4 x 33.1 inches. A1 Paper Size Dimensions
#### What is the size of A1 paper in meters?
The total area of A1 sheet is 0.5 square meter (m 2 .)Equivalent size in inches is 23.39″ wide and 33.11″ long. Nearest US size is ANSI D size which is 22″ wide and 34″ long written as 22×34.
#### What is the total area of A1 sheet?
The total area of A1 sheet is 0.5 square meter (m 2.)Equivalent size in inches is 23.39″ wide and 33.11″ long.
What does B1 mean in paper size?
For example, B1 is a geometric mean between A1 and A0. The sides of B0 are 1 m to v2 m. The following sizes are all variations on paper sizes in the ISO series. | 608 | 1,897 | {"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-2022-49 | latest | en | 0.841822 |
https://codingsight.com/basic-and-complex-uses-of-not-equal-comparison-operator-in-t-sql/ | 1,618,482,682,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038084765.46/warc/CC-MAIN-20210415095505-20210415125505-00362.warc.gz | 284,201,001 | 27,762 | # Basic and Complex Uses of Not Equal Comparison Operator in T-SQL
Total: 9 Average: 3.6
The article also highlights the importance of understanding the correct use of Not Equal comparative operator with expressions.
The Not Equal comparative operator is a very common operator used in T-SQL, however, it is important to understand how to use it effectively in different scenarios.
## Basics of Not Equal <> Comparison Operator
Let us first get familiar with the Not Equal T-SQL operator in the light of Microsoft documentation.
### Microsoft Definition
A Not Equal To <> is a comparison operator which compares two expressions and returns a Boolean type result.
### Boolean Type Result
Boolean type result is the type of result which returns True or False.
In the context of T-SQL under normal circumstances, Boolean type returns one of the three values:
1. True
2. False
3. NULL
### Comparison Operator
A comparison operator in simple words compares two expressions and based on the comparison it tells us whether they are same or not.
Please note that Not Equal comparison operator (<>) is a binary operator which means an operator which compares two expressions.
### Expression
According to Microsoft documentation, an expression is a combination of symbols and operators that the SQL Server Database Engine evaluates to obtain a single data value.
Simple expressions can be a single constant, variable, column, or scalar function.
In other words, the simplest form of expression is a constant (fixed number) such as 1.
### Example –Operator and Expressions
The simplest example of an expression with operators is an arithmetic operation such as 1+2.
Would you like to guess what is the operator and what is the expression in the above example?
Please refer to Not Equal comparison operator (<>) definition which states, “Not Equal comparison operator (<>) compares two expressions…” so + (addition) is the operator in this example and 1 (constant) and 2 (constant) are expressions.
### Example – Comparison Operator and Expressions
Considering the example of arithmetic operation 1+2, simply replacing the arithmetic operator + with Not Equal <> comparison operator, it becomes a comparison operation in T-SQL.
For example, 1<>2 is a comparison operation with comparison operator <> between two expressions 1 and 2.
### Not Equal Operator “<>” vs “!=”
There is another way to express Not Equal comparison operator and that is by using exclamation mark followed by equal sign (!=).
Please note that both “<>” and “!=” represent Not Equal comparison operator, except that the <> sign complies with the ISO standard, whereas the Not Equal comparison operator != does not comply with the ISO standard.
### Compatibility
According to Microsoft documentation, the Not Equal comparison operator (<>) is compatible with the following SQL database versions:
1. SQL Server 2008 and upper versions
2. Azure SQL Database
3. Azure SQL Data Warehouse
4. Parallel Data Warehouse
### Syntax
expression (constant, variable, column etc.) <> expression (constant, variable, column etc.)
## Pre-requisites
### T-SQL Familiarity
This article assumes that the readers have basic knowledge of T-SQL and are capable of writing and running basic SQL scripts.
### Setup Sample Database (SQLDevBlogV2)
This article requires the following sample database to run examples of basic and advanced uses of the Not Equal comparison operator (<>).
The sample database consists of three tables:
1. Article
2. Author
3. Category
Please set up the sample database named SQLDevBlogV2 by writing the following code:
```-- Create sample database (SQLDevBlog)
CREATE DATABASE SQLDevBlogV2;
GO
USE SQLDevBlogV2;
-- (1) Create Author table in the sample database
CREATE TABLE Author (
AuthorId INT PRIMARY KEY IDENTITY (1, 1)
,Name VARCHAR(40)
,RegistrationDate DATETIME2
,Notes VARCHAR(400)
)
-- (2) Create Article Category table in the sample database
CREATE TABLE Category (
CategoryId INT PRIMARY KEY IDENTITY (1, 1)
,Name VARCHAR(50)
,Notes VARCHAR(400)
)
-- (3) Create Article table in the sample database
CREATE TABLE Article (
ArticleId INT PRIMARY KEY IDENTITY (1, 1)
,CategoryId INT
,AuthorId INT
,Title VARCHAR(150)
,Published DATETIME2
,Notes VARCHAR(400)
)
-- Adding foreign keys for author and article category
ALTER TABLE Article ADD CONSTRAINT FK_Category_CategoryId FOREIGN KEY (CategoryId) REFERENCES Category (CategoryId)
ALTER TABLE Article ADD CONSTRAINT FK_Author_AuthorId FOREIGN KEY (AuthorId) REFERENCES Author (AuthorId)
GO
-- (5) Populating Author table
INSERT INTO Author (Name, RegistrationDate, Notes)
VALUES ('Sam', '2017-01-01', 'Database Analyst'),
('Asif', '2017-01-02', 'Database and Business Intelligence Developer'),
-- (6) Populating Category table
INSERT INTO Category (Name, Notes)
VALUES ('Development', 'Articles about database development'),
('Testing', 'Database testing related articles'),
('DLM', 'Database lifecycle management')
-- (7) Populating Article
INSERT INTO Article (CategoryId, AuthorId, Title, Published, Notes)
VALUES (1, 1, 'Fundamentals of SQL Database Development', '02-01-2018', ''),
(1, 2, 'Advanced Database Development', '02-01-2018', ''),
(2, 3, 'All About Database Testing', '03-01-2018', '');
GO
```
## Basic Uses of Not Equal Operator (<>)
In this section, we are going to explore some basic uses of the Not Equal comparison operator (<>) in T-SQL.
Please remember that the upcoming examples run against the sample database SQLDevBlogV2 which must be created if you would like to run the scripts of examples as mentioned in the pre-requisites.
### Authors list excluding the first author
We have the Author table in the sample database which contains a list of all the registered authors.
Let us first view all the authors as follows:
```-- View all authors
SELECT [AuthorId], [Name], [RegistrationDate] FROM dbo.Author
```
Now, if we want to view all the authors except the first one, we are going to use the Not Equal comparison operator (<>) as follows:
```-- View all authors excluding first one
SELECT [AuthorId], [Name], [RegistrationDate]
FROM Author where AuthorId<>1
```
Here, the Not Equal comparison operator (<>) is used to filter the results based on the desired criteria.
Please note that in the WHERE clause of the SQL script (where AuthorId<>1), the AuthorId being a column that fulfills the definition of the expression and 1 is a constant, is also an expression so the use complements its syntax (expression <> expression).
### Authors list not registered in 2017
Let us say we want to see the list of authors who were not registered in 2017.
This can be achieved by using the Not Equal comparison operator (<>) as follows:
```--Authors list not registered in 2017.sql
SELECT [AuthorId], [Name], [RegistrationDate] FROM Author where Year(RegistrationDate)<>2017
```
Keeping the expression <> expression syntax in mind, this time expression on the left Year(RegistrationDate) is a function that returns year (number) compared with the expression on the right which is 2017 (number).
Please remember that we have already mentioned that the simplest form of expression is a number and in this example, we saw it in action.
### Articles list excluding testing category
If you are interested to see the list of all the articles except for the testing category stored in the sample database then you would require the Not Equal comparison operator (<>).
If you refer to the sample database diagram at the beginning of this article, it is obvious that the Article table is connected with Category through foreign key relationship, so, first of all, we need to see the categories with articles by using INNER JOIN as follows:
```-- Articles with categories list
SELECT art.ArticleId, cat.Name AS Category, art.Title,art.Published FROM Article art INNER JOIN Category cat
on art.CategoryId=cat.CategoryId
```
Now, we need to exclude the Testing category to see all the articles except the testing articles using the Not Equal comparison operator (<>) as follows:
```-- All articles with categories excluding testing category
SELECT art.ArticleId, art.Title,cat.Name AS Category, art.Published FROM Article art INNER JOIN Category cat
on art.CategoryId=cat.CategoryId
WHERE cat.Name<>'Testing'
```
Running the script produces the following output:
## Complex Uses of Not Equal comparison operator (<>)
We are now going to look at some slightly complex examples, which involve the use of the Not Equal comparison operator (<>).
### Authors list excluding current year authors
A slightly complex scenario is when you are asked to get all the authors except those who were registered in the current year.
Since the current year is dynamic and it changes from year to year, this makes the script slightly complex as we cannot simply use a fixed year number with the comparison operator.
The solution is to get the current year from the current date and then get the list of all the authors who were not registered in the current year.
This is achieved by the following code:
```-- Authors list excluding current year authors
SELECT a.AuthorId,a.Name,a.RegistrationDate FROM Author a
WHERE YEAR(a.RegistrationDate)<>YEAR(GETDATE())
```
The result is as follows:
Please note that since all the authors in the sample database were registered in 2017 and 2018 so all the authors are shown without any exclusion.
### Authors list excluding the author(s) with most articles
The requirement is to view the authors’ list without the top author(s) makes the script slightly more complex.
We have to divide the problem into smaller parts:
1. Getting top author(s) (author with most articles)
2. Excluding top author(s) using the Not Equal comparison operator (<>)
3. Viewing all authors excluding top author(s)
The top author is an author who has more written articles than any other author.
Let us first add one more article for the author Sadaf to make her the top author by running the following script:
```--Add one more article for Sadaf (AuthorId: 3)
INSERT INTO Article (CategoryId, AuthorId, Title, Published, Notes)
VALUES (1, 3, 'Database Development with SQL Server Data Tools (SSDT)', '10-01-2019', '')
```
Now, build a script to view authors with the total number of articles ordered by authors with most articles first:
```--Authors with total number of articles order by author with most articles first
SELECT ath.name as Author_NAME,count(*) as Total_Articles FROM Article art
INNER JOIN Author ath
on art.AuthorId=ath.AuthorId
GROUP BY ath.AuthorId,ath.Name
ORDER BY COUNT(*) DESC
```
The output of the script is as follows:
Now first we need to find out the most articles by an author and put the result into a variable and then we can list all the authors excluding the ones with most articles by using Not Equal comparison operator (<>) with the variable as follows:
```--Authors list excluding the author(s) with most articles
DECLARE @MaxArticles INT=(SELECT TOP 1 count(*) as Total_Articles FROM Article art
INNER JOIN Author ath
on art.AuthorId=ath.AuthorId
GROUP BY ath.AuthorId,ath.Name
ORDER BY COUNT(*) DESC)
-- Authors excluding the author with most articles
SELECT ath.Name,ath.RegistrationDate,Count(*) as Total_Articles FROM Author ath
INNER JOIN Article art
ON ath.AuthorId=art.AuthorId
GROUP BY ath.Name,ath.RegistrationDate
HAVING COUNT(*)<>@MaxArticles
```
Running the script shows us the following result set:
Congratulations, you have successfully learned how to use the Not Equal comparison operator (<>) in simple and slightly complex scenarios.
## Things to do
Now that you are familiar with basic and complex uses of the Not Equal comparison operator (<>), you can improve your skills further by trying the following things:
1. Please try to find out the list of all the articles excluding the articles which were written last year, which should be calculated dynamically.
2. Please try to create SQL script to get the list of all the authors who were not registered last year, where the last year should be calculated dynamically.
3. Please try finding out the list of all the articles, where the category was not development using the sample database created in this article.
4. Please try to build a SQL script to get all the categories excluding the top category (category with most articles written).
5. Please try creating a scalar-valued function and use it with the Not Equal comparison operator (<>) to explore further.
#### Haroon Ashraf
Haroon's deep interest in logic and reasoning at an early age of his academic career paved his path to become a data professional. He began his professional life as a computer programmer more than 10 years ago working on his first data venture to migrate and rewrite a public sector database driven examination system from IBM AS400 (DB2) to SQL Server 2000 using VB 6.0 and Classic ASP along with developing reports and archiving many years of data. His work and interest revolves around Database-Centric Architectures and his expertise include database and reports design, development, testing, implementation and migration along with Database Life Cycle Management (DLM). | 2,816 | 13,197 | {"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-2021-17 | latest | en | 0.877944 |
https://gateway.ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Euler%E2%80%93Maruyama_method.html | 1,642,767,882,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303356.40/warc/CC-MAIN-20220121101528-20220121131528-00293.warc.gz | 316,215,075 | 5,559 | # Euler–Maruyama method
In mathematics, more precisely in Itô calculus, the Euler–Maruyama method, also called simply the Euler method, is a method for the approximate numerical solution of a stochastic differential equation (SDE). It is a simple generalization of the Euler method for ordinary differential equations to stochastic differential equations. It is named after Leonhard Euler and Gisiro Maruyama. Unfortunately the same generalization cannot be done for the other methods from deterministic theory,[1] e.g. Runge–Kutta schemes.
Consider the stochastic differential equation (see Itō calculus)
with initial condition X0 = x0, where Wt stands for the Wiener process, and suppose that we wish to solve this SDE on some interval of time [0, T]. Then the Euler–Maruyama approximation to the true solution X is the Markov chain Y defined as follows:
• partition the interval [0, T] into N equal subintervals of width :
• set Y0 = x0;
• recursively define Yn for 1 n N by
where
## Example
### Numerical simulation
Gene expression modelled as stochastic process
An area that has benefited significantly from SDE is biology or more precisely mathematical biology. Here the number of publications on the use of stochastic model grew, as most of the models are nonlinear, demanding numerical schemes.
The graphic depicts a stochastic differential equation being solved using the Euler Scheme. The deterministic counterpart is shown as well.
### Computer implementation
The following Python code implements Euler–Maruyama to solve the Ornstein–Uhlenbeck process
The random numbers for are generated using the numpy mathematics package.
import numpy as np
import matplotlib.pyplot as plt
num_sims = 5
N = 1000
y_init = 0
t_init = 3
t_end = 7
c_theta = 0.7
c_mu = 1.5
c_sigma = 0.06
def mu(y, t):
return c_theta * (c_mu - y)
def sigma(y, t):
return c_sigma
dt = float(t_end - t_init) / N
dW = lambda dt: np.random.normal(loc = 0.0, scale = np.sqrt(dt))
t = np.arange(t_init, t_end, dt)
y = np.zeros(N)
y[0] = y_init
for i_sim in range(num_sims):
for i in xrange(1, t.size):
a = mu(y[i-1], (i-1) * dt)
b = sigma(y[i-1], (i-1) * dt)
y[i] = y[i-1] + a * dt + b * dW(dt)
plt.plot(t, y)
plt.show() | 588 | 2,236 | {"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} | 3.40625 | 3 | CC-MAIN-2022-05 | latest | en | 0.822565 |
https://stacks.math.columbia.edu/tag/0BG3 | 1,720,954,770,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00001.warc.gz | 487,742,479 | 6,746 | Lemma 54.10.2. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local domain of dimension $2$. Let $A \to R$ be a surjection onto a complete discrete valuation ring. This defines a nonsingular arc $a : T = \mathop{\mathrm{Spec}}(R) \to \mathop{\mathrm{Spec}}(A)$. Let
$\mathop{\mathrm{Spec}}(A) = X_0 \leftarrow X_1 \leftarrow X_2 \leftarrow X_3 \leftarrow \ldots$
be the sequence of blowing ups constructed from $a$. If $A_\mathfrak p$ is a regular local ring where $\mathfrak p = \mathop{\mathrm{Ker}}(A \to R)$, then for some $i$ the scheme $X_ i$ is regular at $x_ i$.
Proof. Let $x_1 \in \mathfrak m$ map to a uniformizer of $R$. Observe that $\kappa (\mathfrak p) = K$ is the fraction field of $R$. Write $\mathfrak p = (x_2, \ldots , x_ r)$ with $r$ minimal. If $r = 2$, then $\mathfrak m = (x_1, x_2)$ and $A$ is regular and the lemma is true. Assume $r > 2$. After renumbering if necessary, we may assume that $x_2$ maps to a uniformizer of $A_\mathfrak p$. Then $\mathfrak p/\mathfrak p^2 + (x_2)$ is annihilated by a power of $x_1$. For $i > 2$ we can find $n_ i \geq 0$ and $a_ i \in A$ such that
$x_1^{n_ i} x_ i - a_ i x_2 = \sum \nolimits _{2 \leq j \leq k} a_{jk} x_ jx_ k$
for some $a_{jk} \in A$. If $n_ i = 0$ for some $i$, then we can remove $x_ i$ from the list of generators of $\mathfrak p$ and we win by induction on $r$. If for some $i$ the element $a_ i$ is a unit, then we can remove $x_2$ from the list of generators of $\mathfrak p$ and we win in the same manner. Thus either $a_ i \in \mathfrak p$ or $a_ i = u_ i x_1^{m_1} \bmod \mathfrak p$ for some $m_1 > 0$ and unit $u_ i \in A$. Thus we have either
$x_1^{n_ i} x_ i = \sum \nolimits _{2 \leq j \leq k} a_{jk} x_ jx_ k \quad \text{or}\quad x_1^{n_ i} x_ i - u_ i x_1^{m_ i} x_2 = \sum \nolimits _{2 \leq j \leq k} a_{jk} x_ jx_ k$
We will prove that after blowing up the integers $n_ i$, $m_ i$ decrease which will finish the proof.
Let us see what happens with these equations on the affine blowup algebra $A' = A[\mathfrak m/x_1]$. As $\mathfrak m = (x_1, \ldots , x_ r)$ we see that $A'$ is generated over $R$ by $y_ i = x_ i/x_1$ for $i \geq 2$. Clearly $A \to R$ extends to $A' \to R$ with kernel $(y_2, \ldots , y_ r)$. Then we see that either
$x_1^{n_ i - 1} y_ i = \sum \nolimits _{2 \leq j \leq k} a_{jk} y_ jy_ k \quad \text{or}\quad x_1^{n_ i - 1} y_ i - u_ i x_1^{m_1 - 1} y_2 = \sum \nolimits _{2 \leq j \leq k} a_{jk} y_ jy_ k$
and the proof is complete. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 1,036 | 2,669 | {"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": 2, "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": 2, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-30 | latest | en | 0.656543 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/72/2/l/ | 1,695,945,113,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510462.75/warc/CC-MAIN-20230928230810-20230929020810-00659.warc.gz | 951,441,207 | 52,781 | # Properties
Label 72.2.l Level $72$ Weight $2$ Character orbit 72.l Rep. character $\chi_{72}(11,\cdot)$ Character field $\Q(\zeta_{6})$ Dimension $20$ Newform subspaces $2$ Sturm bound $24$ Trace bound $1$
# Related objects
## Defining parameters
Level: $$N$$ $$=$$ $$72 = 2^{3} \cdot 3^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 72.l (of order $$6$$ and degree $$2$$) Character conductor: $$\operatorname{cond}(\chi)$$ $$=$$ $$72$$ Character field: $$\Q(\zeta_{6})$$ Newform subspaces: $$2$$ Sturm bound: $$24$$ Trace bound: $$1$$ Distinguishing $$T_p$$: $$5$$
## Dimensions
The following table gives the dimensions of various subspaces of $$M_{2}(72, [\chi])$$.
Total New Old
Modular forms 28 28 0
Cusp forms 20 20 0
Eisenstein series 8 8 0
## Trace form
$$20 q - 3 q^{2} - 4 q^{3} - q^{4} - 7 q^{6} - 4 q^{9} + O(q^{10})$$ $$20 q - 3 q^{2} - 4 q^{3} - q^{4} - 7 q^{6} - 4 q^{9} - 6 q^{11} + 2 q^{12} - 18 q^{14} - q^{16} - 16 q^{18} - 8 q^{19} + 18 q^{20} - 5 q^{22} + 29 q^{24} - 4 q^{25} - 16 q^{27} - 12 q^{28} + 12 q^{30} + 27 q^{32} - 2 q^{33} - 5 q^{34} + 23 q^{36} + 21 q^{38} - 12 q^{40} - 18 q^{41} + 42 q^{42} - 2 q^{43} + 12 q^{46} - 19 q^{48} - 4 q^{49} + 51 q^{50} + 28 q^{51} - 18 q^{52} + 35 q^{54} - 66 q^{56} - 20 q^{57} + 12 q^{58} + 30 q^{59} - 72 q^{60} + 2 q^{64} - 6 q^{65} - 32 q^{66} - 2 q^{67} - 45 q^{68} + 18 q^{70} - 37 q^{72} - 8 q^{73} - 60 q^{74} + 68 q^{75} - 11 q^{76} - 72 q^{78} + 8 q^{81} + 10 q^{82} + 54 q^{83} + 12 q^{84} - 87 q^{86} - 5 q^{88} - 66 q^{90} - 36 q^{91} + 84 q^{92} + 24 q^{94} + 74 q^{96} - 2 q^{97} + 10 q^{99} + O(q^{100})$$
## Decomposition of $$S_{2}^{\mathrm{new}}(72, [\chi])$$ into newform subspaces
Label Dim $A$ Field CM Traces $q$-expansion
$a_{2}$ $a_{3}$ $a_{5}$ $a_{7}$
72.2.l.a $4$ $0.575$ $$\Q(\sqrt{-2}, \sqrt{-3})$$ $$\Q(\sqrt{-2})$$ $$0$$ $$2$$ $$0$$ $$0$$ $$q+\beta _{1}q^{2}+(1-\beta _{1}-\beta _{2})q^{3}+2\beta _{2}q^{4}+\cdots$$
72.2.l.b $16$ $0.575$ $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ None $$-3$$ $$-6$$ $$0$$ $$0$$ $$q-\beta _{11}q^{2}+(-\beta _{3}+\beta _{6})q^{3}+(-1-\beta _{1}+\cdots)q^{4}+\cdots$$ | 1,017 | 2,125 | {"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} | 2.734375 | 3 | CC-MAIN-2023-40 | latest | en | 0.336371 |
https://bzoj.llf0703.com/p/3448.html | 1,571,279,909,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672548.33/warc/CC-MAIN-20191017022259-20191017045759-00046.warc.gz | 417,586,650 | 3,228 | # [Usaco2014 Feb]Auto-complete
### 题目描述
Bessie the cow has a new cell phone and enjoys sending text messages, although she keeps making spelling errors since she has trouble typing on such a small screen with her large hooves. Farmer John has agreed to help her by writing an auto-completion app that takes a partial word and suggests how to complete it. The auto-completion app has access to a dictionary of W words, each consisting of lowercase letters in the range a..z, where the total number of letters among all words is at most 1,000,000. The app is given as input a list of N partial words (1 <= N <= 1000), each containing at most 1000 lowercase letters. Along with each partial word i, an integer K_i is also provided, such that the app must find the (K_i)th word in alphabetical order that has partial word i as a prefix. That is, if one ordered all of the valid completions of the ith partial word, the app should output the completion that is (K_i)th in this sequence.
XYW在和他的男人聊天的时候经常会因为打字速度太慢而被鄙视，于是他想开发一种自动补全软件。开始他有一个包含了w个词的字典（总长度不大于10^6）。接下来他想知道在字典中可以由给定的字符串s加上若干个字母得到的字符串字典序第k大的编号是多少？有n组询问（n<=10^3，s长度<=10^3)
### 输入格式
* Line 1: Two integers: W and N.
* Lines 2..W+1: Line i+1: The ith word in the dictionary.
* Lines W+2..W+N+1: Line W+i+1: A single integer K_i followed by a partial word.
### 输出格式
* Lines 1..N: Line i should contain the index within the dictionary (an integer in the range 1..W) of the (K_i)th completion (in alphabetical order) of the ith partial word, or -1 if there are less than K_i completions.
```10 3
dab
ba
ab
daa
aa
aaa
aab
abc
ac
4 a
2 da
4 da
```
### 样例输出
```3
1
-1
OUTPUT DETAILS: The completions of a are {aa,aaa,aab,ab,abc,ac}. The 4th is ab, which is listed on line 3 of the dictionary. The completions of da are {daa,dab,dadba}. The 2nd is dab, listed on line 1 of the dictionary. There is no 4th completion of da. ```
Silver | 605 | 1,892 | {"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} | 2.875 | 3 | CC-MAIN-2019-43 | latest | en | 0.858225 |
https://math.stackexchange.com/questions/3179134/help-formulating-a-conjecture-about-the-parity-of-every-cycle-length-in-a-bipart | 1,566,348,002,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315695.36/warc/CC-MAIN-20190821001802-20190821023802-00490.warc.gz | 550,088,646 | 30,824 | # Help formulating a conjecture about the parity of every cycle length in a bipartite graph and proving it
I know a cycle in a graph $$G=(V,E)$$ is a sequence of vertices $$v_0, v_1,\ldots, v_k$$ such that $$k\geq 3$$, $$v_k = v_0$$, and $$G$$ contains every edge between consecutive vertices: $$(v_0, v_1)$$, $$(v_1, v_2), \ldots$$, $$(v_{k−1}, v_k)$$. I know eventually I want to prove any cycle in a bipartite graph has an even length as the conjecture.
A hint has been given to use simple induction, but I am having trouble proving a bipartite graph with length $$k+1$$ also can only contain cycles of even length.
If you want to prove that every cycle in a bipartite graph is even, you can reason by contradiction. In details :
Let $$G$$ be a bipartite graph with sets of vertices $$V_0$$ and $$V_1$$. Suppose now that you have an off cycle in $$G$$, of length $$2k+1$$ : $$v_0,v_1,\ldots,v_{2k}$$
Without loss of generality you can suppose that $$v_0\in V_0$$. Then $$v_1$$ must be in $$V_1$$, $$V_2$$ in $$V_0$$, etc. Formally $$v_i\in V_0$$ if and only if $$v_{i+1}\in V_1$$, or $$v_i\in V_0$$ if and only $$i$$ is even.
This implies that $$v_{2k}$$ is in $$V_0$$. Then we have $$\left\{ \begin{array}{l} v_{0}\in V_0\\ v_{2k}\in V_0 \end{array}\right.\text{ and } (v_{2k},v_0)\in E(G)$$
A contradiction with the fact that $$G$$ is bipartite.
I don't know why we need to use an induction. The problem: every cycle in a bipartite graph is even, is simply proved by a contradiction.
Assume that there is an odd cycle in a bipartite graph. However, this derives a contradiction from two facts: 1. any subgraph of a bipartite graph is bipartite, and 2. odd cycle is not a bipartite graph.
It is very well-known fact that
• A graph $$G$$ is bipartite iff every cycle in $$G$$ is even.
• This is not really an answer. You are basically saying that the question is a "very well-known fact". – Thomas Lesgourgues Apr 8 at 9:08
• Below comment is addtional... – Dong-gyu Kim Apr 8 at 9:17 | 647 | 1,998 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 30, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-35 | latest | en | 0.900779 |
http://bluebulbprojects.com/MeasureOfThings/results.php?comp=volume&unit=ltr&amt=7100&sort=pr&p=6 | 1,600,521,671,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400191780.21/warc/CC-MAIN-20200919110805-20200919140805-00002.warc.gz | 17,620,624 | 6,665 | Bluebulb Projects presents:
Enter a measurement to see comparisons
Equivalents in other units
How big is 7,100 liters?
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Closest first | Highest first | Lowest first
It's about 0.000000000002 times as big as an Halley's Comet In other words, 7,100 liters is 0.00000000000180 times the size of an Halley's Comet, and the size of an Halley's Comet is 560,000,000,000 times that amount. (Comet Halley, officially "1P/Halley") (volume calculated based on 2005 dimensions) Halley's Comet, the famous comet visible to observers on Earth approximately every 76 years, has a peanut- or potato-shaped nucleus made of rock, dust, ice, and various frozen gasses. The nucleus measures approximately 15 km by 8 km by 8 km, which yields a calculated volume of 4,000,000,000,000,000.000000000000000000 liters. The coma of the Comet — the trail of sublimated gasses that give the Comet its visibility — may be up to 100,000 km in length. It's about 0.000000000002 times as big as The Grand Canyon In other words, 7,100 liters is 0.00000000000170 times the size of The Grand Canyon, and the size of The Grand Canyon is 588,000,000,000 times that amount. (Coconino County and Mohave Counties, Arizona, near Fredonia and Grand Canyon, Arizona)A dynamic and continuously changing landscape, the Grand Canyon has an approximate volume of 4,170,000,000,000,000.0000000000000000000 liters. The strata of the rock visible in the Canyon's walls display nearly 2 billion years of geological history. It's about 0.000000000000003 times as big as The Gulf of Mexico In other words, 7,100 liters is 0.0000000000000029170 times the size of The Gulf of Mexico, and the size of The Gulf of Mexico is 342,800,000,000,000.0000000000000000000 times that amount. (water volume)The Gulf of Mexico contains 2,434 quadrillion L of water. Every second, the Mississippi River empties 12,000,000 L of water into the Gulf. It's about 0.000000000000000005 times as big as Earth's Oceans In other words, 7,100 liters is 0.00000000000000000510 times the size of Earth's Oceans, and the size of Earth's Oceans is 200,000,000,000,000,000.00000000000000000000 times that amount. (Total water volume of Atlantic, Pacific, Indian, Arctic, and Southern Oceans) (estimated) Over 97% of the Earth's water is found in the planet's five oceans for a total volume of about 1,400,000,000,000,000,000,000.000000000000000000000000 liters. The Pacific Ocean is the largest ocean, covering 165,760,000 sq. km — more than double the area of the Atlantic Ocean at 82,400,000 sq. km. | 706 | 2,533 | {"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} | 2.640625 | 3 | CC-MAIN-2020-40 | latest | en | 0.9154 |
https://www.errorsdoc.com/software/find-circular-reference-in-excel/ | 1,718,572,595,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00358.warc.gz | 662,758,868 | 32,877 | # How to Find Circular Reference in Excel?
Do you know how to find circular reference in Excel? Find it with us!
In Excel, when you work with formulas, you may see a warning prompt. The prompt tells that your worksheet has the circular reference and it can lead the incorrect calculation by using the formulas. It also suggests you find the circular reference and sort it quickly.
Therefore, in this article, we will enlighten you on how to find a circular reference in Excel 2013, 2016, and others. Once you address the circular reference, you can remove it immediately to make your calculation accurate and fast in Excel.
The tutorial explains the following details about a circular reference in Excel:
## What Does Circular Reference Mean in Excel?
The circular reference refers to the formula. When you end up entering the formula for calculation in one cell (which uses its own cell), it may create an endless loop which as a result slows down the spreadsheet.
For instance, you have a dataset in a cell A1:A5 and you are using the =SUM(A1:A6) formula in cell A6. Now this will show a circular reference warning pop-up on your screen.
That is because here you are summing the values in A1:A6 cell. Its result must appear in the A6 cell. It will create a loop because Excel will keep on adding a new value in the A6 cell. Hence, it appears as a circular reference circuit in Excel.
These circular reference loops may crash your calculation or slow down your sheet. Unfortunately, a user does not see the direct indication that what cell is causing the issue.
So, if you want to know how to find where circular reference is in Excel then we will teach you. In addition, you will discover how to remove the circular reference issue and fix it smoothly. Let’s get into more detail.
### How to Find a Circular Reference in an Excel Workbook?
As we said before, circular reference error prompts if it is in the Worksheet. However, it does not notify where the circular reference exactly is and what cell reference causing this error.
Most users drop their concerns about how can I find a circular reference in Excel and other series. Here, we will tell you how you can find and resolve circular references in your Worksheet.
Looking for Excel Circular Reference? Find it here!
There are two ways to address the circular reference issue. You can either view cell reference through the Formulas in the menu bar or check Circular Reference in the status bar.
Both methods are easy to implement. The only difference is that the status bar is a quick way to check the circular reference whereas the menu bar requires you to navigate through the options manually.
Here, we will understand both methods with a manual procedure. For that, you can go through the steps we are discussing below. Let’s get started!
### How Do You Find a Circular Reference in Excel through the Menu bar?
Excel Circular Reference: Find and Enable in the following steps using the Menu bar settings:
• Open the Worksheet that contains the circular reference.
• Click on the “Formulas” located on the top left menu.
• Go to the “Error Checking” menu as a drop-down in the “Formula Editing” group.
• Now, hover the mouse cursor over the “Circular References”. It shows that your Worksheet possesses a circular reference.
• Here, select the cell address that displays next to the “Circular References”. This takes you to the cell that is causing circular reference.
If you are thinking about how do I find the circular reference in excel, the above-mentioned steps would help you. Similarly, you can visit more cells that cause a circular reference by following the above-mentioned steps. If your spreadsheet does not show any circular reference issue then it will not show any of the cell references.
### How Do You Find Circular References in Excel through the Status bar?
Excel Circular Reference: Find and Enable using the Status bar. For that, you can follow these steps:
• Activate the Worksheet that has a circular reference.
• On the left bottom of the sheet, you will see “Circular Reference” along with its cell address.
This is the direct and quick way to check the circular reference in your Worksheet. If you see no cell address in the status bar, it indicates that your sheet is free from any cell reference issue.
#### Important Things to Remember about Circular References
There are certain things about the circular reference that you should keep in your mind. Look at the following points:
• If you have enabled the iterative calculation then you would not be able to check the circular reference cell address in the status bar.
• In case a circular reference does not exist in the active sheet but in other worksheets then this will not show the cell address but the circular reference.
• When you launch the Workbook that contains a circular reference, this will instantly prompt the circular reference issue.
• If you see a warning prompt for a circular reference and you intentionally dismiss it then the prompt will not appear again. This may not alert you that your sheet has a circular reference.
Now that you know how to find circular reference in Excel 2007 and other series, you should also know how to remove it from your sheet.
### How to Remove the Circular Reference in Excel?
Once you address that your sheet has circular references, you can remove it immediately to fix the calculation issue.
Important Note: You cannot remove the circular reference just by hitting the Delete key. Since the circular reference refers to the formulas and each formula may be different so you have to analyze this issue on the basis of case by case.
You can correct the issue by adjusting the cell reference. However, it is not that simple. Circular reference issues can be problematic for multiple cells. In such a scenario, you can treat this circular reference error with the Trace Precedents feature. This will help you to identify cells that are feeding cells with circular reference.
Let’s know how to find a circular reference in Excel Workbook and fix it using the Trace Precedents.
• On your worksheet, select the cell containing a circular reference.
• Then, click on the “Formulas” located on the top menu bar.
• Select the “Trace Precedents” option on the right.
Once you are done performing these steps, you will see the Blue Arrows. This shows what cell is feeding into a formula in the cell (that you selected). This is how you can find out the cells and formula that are causing the circular reference and resolve them.
Note: If you work with complex financial models then this precedent can go multiple levels. This method works well when all the formulas are referring to the cells in the same spreadsheet. If it is in multiple sheets then this method may not work well.
#### How to Allow Iterative Calculations in Excel?
In some instances, a user may want to keep the circular reference in the sheet for a number of iterations. However, it creates loops but if that is the preference then you can set how many times a loop should run in your sheet. This is what we call iterative calculations in an Excel sheet.
Therefore, we will learn how to enable to disable iterative calculation in the Excel sheet. To do so, follow these steps instructed below:
• Use any of the following ways to enable iterative calculation:
• In Excel 2010, click on the “File” tab.
• Then, click on “Options”. This will bring up the “Excel Options” dialog box.
• Here, click on the “Formulas” located on the left.
• In Excel 2007, click on the “Microsoft Office” button.
• Click on the “Excel” options.
• Go with the “Formulas” category.
• In Excel for Mac, click on the “Excel” menu.
• Click on the “Preferences” option.
• Select the “Calculation”.
• Now, check the box that says “Enable iterative calculation” in the “Calculation options” section.
• At this point, you can mention the “Maximum Iterations” and “Maximum Change” value.
You are done! After implementing these steps accordingly, the iterative calculation will be enabled in Excel as per your preferences.
Remember that If you enable the iterative calculation then you might not see the circular reference cell address in the status bar. Also, the circular reference will be turned off once you enable the iterative calculations.
To understand better these two options (in the iterative calculation), read below about them.
Maximum Iterations: A maximum number of times you require your Excel sheet to calculate (before it provides the final result). If you enter 100 as your maximum iteration then your Excel sheet would run the loop a Hundred times before proving the final result.
Maximum Change: If maximum change is not obtained between the iterations then the calculation will be discontinued. By default, the maximum change value is .001. Adding a lower value such as .001 would get you an accurate result.
Note: The more the iterations run, the more time it takes for Excel to perform the calculation. If you keep the iteration as your maximum number then Excel may slow down or crash entirely.
If you are wondering how to find a circular reference in Excel 2016 and other series then you can enable iterative calculations to disable circular reference.
##### Wrap up
Circular reference has both positive and negative impacts on your Excel sheet. There are instances when a user wants to enable the circular reference for calculations. However, enabling the circular references may slow down or crash the sheet. As a result, your calculation can go wrong. To prevent this, you can enable an iterative calculation option to get rid of circular references.
In this guide, we talked about how do you find circular references in Excel 2010 and other series. Besides that, we found easy ways to remove this error from the spreadsheet.
Hopefully, you find this guide useful. In case you have any questions or suggestions, let us know in the comments box.
To know more about Excel features and its functions, check out the following guides: | 2,021 | 10,035 | {"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-2024-26 | latest | en | 0.902683 |
https://modelingwithdata.org/arch/00000005.htm | 1,701,660,628,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100523.4/warc/CC-MAIN-20231204020432-20231204050432-00641.warc.gz | 446,989,991 | 4,516 | ### Dataviz
This is a continuation of last episode.
The first figure is a TrellisTM or lattice plot, giving a 2-D dot plot of each of three variables against each other variable. I didn't try too hard in producing the plot, and just pulled out three variables at random from a random data set.
Figure One: A lattice plot, relating three variables to each other
But we can already see some patterns: GDP/capita and height have the positive correlation you'd expect, as per the blow up in the next figure. In this figure, I fit a linear regression to the data, and it looks pretty good, but for a few outliers at lower right. Maybe an exponential-family model may fit better.
Figure Two: A close-up of the upper left plot in the lattice, with the line of best fit
So that's DataViz at work. We took a lot of data, displayed many relations at once, and zeroed in on one that matters.
Except, uh, for all that I said about this being a random data set. I just made up some pleasant-sounding variable names, generated a random data set, and plotted it. And yet we were able to find a plausible pattern in there.
And so we see another way of casting the descriptive versus inferential war--the problem of too many hypothesis tests. The descriptivists are working to produce methods like the lattice plot that let you see more relationships at once; the inferentialists are asking: if you fed complete noise to this method, what are the odds that some sort of pattern would turn up? As our methods get better at putting more data on the screen at once, they get worse at testing whether the patterns we see are real or just beautiful noise.
###### DataViz
Thanks to a number of technological advances, dataViz is trendy right now. There are a few icons of the field who are working hard on self-promotion, such as Edward Tufte, whose books show how graphs can be cleaned up, chartjunk eliminated, and grainy black and white fliers from the 1970s cleaned up through the use of finely detailed illustrations in full color. John Tukey's Exploratory Data Analysis (cited above) is aggressively quirky, and encourages disdain for the inferential school.
These guys, and their followers, are right that we could do a whole lot better with our data visualizations, and that the stuff based on facilitating fitting the line with a straightedge should have been purged at least twenty years ago. Strunk and White gave us standards for writing clearly in 1959; it's about time we developed guidelines for exposition via graphics.
But we're talking not just about presenting a known relationship, but exploratory data analysis via graphics. In this context, the underlying philosophy is humanist to a fault. The claim is that the human brain is the best data-processor out there, and our computers still can't see a relationship among a blob of dots as quickly as our eye/brain combo can. This is true, and a fine justification for better graphical data presentation. And hey, we humans would all rather look at plots than at tables of numbers.
Figure Three: If you don't see faces, you're crazy. Oh, and there's a penis and vagina in every inkblot too.
But apophenia is a powerful force. We look at clouds and see bunnies, or read the horoscope and think that it's talking directly to us, or listen to a Beatles song about playground equipment and think it's telling us to kill people. Given a handful of scatterplots like the lattice plot above, you will find a pattern--in fact, if a psychologist were to show you a series of ten seemingly random inkblots and you didn't see a reasonable number of patterns in them, the psychologist might consider you to be mentally unhealthy in any of a number of ways.
The moral here is that our data visualization technology is getting really good really fast--I'll have even slicker examples next time. You'd be silly to ignore these recommendations and novel display methods. But the same power that makes patterns clear is the power that invents random patterns in static.
Next time: even more dataviz tools, which touch on an even bigger problem. | 883 | 4,088 | {"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-2023-50 | latest | en | 0.950624 |
https://www.weknowtheanswer.com/q/how-to-factorise-this | 1,542,739,674,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746528.84/warc/CC-MAIN-20181120171153-20181120192419-00025.warc.gz | 1,066,246,357 | 7,193 | # how to factorise this?
• how to factorise this?
Answer #1 | 30/12 2013 04:10
(p+q)^2+5(p+q) (p+q)(p + q)+5(p+q) The term (p+q) is common between the two 'added' terms, so is 'taken out'. Hence (p+q)[(p+q) + 5] Removing the brackets (p+q)[p + q + 5]) Done!!!!
Positive: 62.5 %
Answer #2 | 30/12 2013 05:26
Let ( p+q) be = x x^2 + 5x = x ( x+5) = (p+q) ( p+q+5) ANSWER
Positive: 57.142857142857 %
Answer #3 | 30/12 2013 03:59
Well, (p+q) is a common factor, so you can take it out.
Positive: 57.142857142857 %
Answer #4 | 30/12 2013 04:45
Just put in evidence the term (p+q)... it is so (p+q)( p + q + 5) ... OK! the answer is correct.
Positive: 57.142857142857 %
Answer #5 | 30/12 2013 04:11
x² + 5x x (x + 5) (p + q) ( p + q + 5 )
Positive: 42.857142857143 %
Answer #6 | 30/12 2013 04:16
(p+q)^2 + 5(p+q) = (p+q)(p+q) + 5(p+q) (p+q) is a common factor of the two terms. Take this common factor out to get: = (p+q)[(p+q)+ 5] or = (p+q)(p+q+5)
Positive: 37.5 % | 442 | 962 | {"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} | 4.15625 | 4 | CC-MAIN-2018-47 | longest | en | 0.736287 |
https://documen.tv/the-table-shows-the-area-of-two-rugs-a-preschool-teacher-has-in-her-room-she-places-the-two-rugs-30016078-3/ | 1,679,846,249,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00112.warc.gz | 258,120,510 | 16,017 | Question
The table shows the area of two rugs a preschool teacher has in her room. She places the two rugs together to make one big rug.
Rug Area
(square units)
Blue 8(x+12)
Red 10(x+25)
What is the area in square units of the new rug, in factored form?
__ square units
1. By adding the area of the given two rugs together and from the distributive property of multiply the constant we get the area of the new rug in factored form = 2(9x + 173) square units.
### What is factored form?
When an expression is represented as a product of its factors it is considered as a factored form.
### What is distributive property of multiply by constant?
The distributive property is a property of arithmetic that states that multiplying a number by a group of numbers is the same as multiplying each number in the group by the number individually and then adding the products together.
Blue rug: 8(x+12)
Red rug: 10(x+25)
New rug: 8(x+12) + 10(x+25)
Next, we can use the distributive property to multiply the constants and the parentheses:
New rug: 8(x+12) + 10(x+25)
New rug: 8x + 96 + 10x + 250
Then we can combine like terms:
New rug: 8x + 96 + 10x + 250
New rug: 18x + 346
Finally, we can factor the expression to get:
New rug: 18x + 346
New rug: 2(9x + 173)
So the area of the new rug is 2(9x + 173) square units. | 369 | 1,317 | {"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} | 4.75 | 5 | CC-MAIN-2023-14 | latest | en | 0.905595 |
https://rosettacode.org/wiki/Talk:Continued_fraction_arithmetic | 1,723,539,396,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641075627.80/warc/CC-MAIN-20240813075138-20240813105138-00228.warc.gz | 393,016,728 | 11,434 | # Talk:Continued fraction/Arithmetic
## External references
Is there a reference to this construction with NG and G in somewhere like Wikipedia or Mathworld? It would be nice to link to the wider world. –Donal Fellows 09:22, 7 February 2013 (UTC)
7 years too late: The task was written by Nigel Galloway, NG stands for NigelGalloway and G stands for Galloway. NG is a simple matrix that allows arbitrary maths, eg a "baby matrix" of {{a,b},{c,d}} just means calculate (a*x+b)/(c*x+d). Since any (but not all) of a/b/c/d can be zero we can easily write an NG that performs a simple operation such as x+1/2, or 1/x, etc. G(baby,x) is just the function that applies some NG to some x. The "full matrix" just extends that idea to a 2x4 matrix {{a,b,c,d},{e,f,g,h}} such that G(full,x,y) yeilds a result (often with several terms of 0) of (axy + bx + cy + d)/(exy + fx + gy + d). It would be equivalent but much more effort to have G accept a string expression as the first argument, eg G("(x+1)/y",x,y), whereas G({{0,1,0,1},{0,0,1,0}},x,y) gets the same result but without any complex parsing/precedence/invalid expression handling. --Pete Lomax (talk) 14:13, 27 August 2020 (UTC) | 350 | 1,180 | {"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} | 2.984375 | 3 | CC-MAIN-2024-33 | latest | en | 0.932346 |
https://www.experts-exchange.com/questions/28363979/Scoring.html | 1,529,454,708,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863259.12/warc/CC-MAIN-20180619232009-20180620012009-00472.warc.gz | 817,382,278 | 34,244 | # Scoring
Sir gowflow,
Kindly make this Sub (Analyse) more friendly.
``````Sub Analyse()
Dim WS As Worksheet
Dim MaxRow As Long
Dim cCell As Range
Dim Min As Long, Max As Long, I As Long, J As Long, K As Long
Dim factor As String
Dim Col As String
Set WS = ThisWorkbook.ActiveSheet
WS.Range("B22") = WS.Range("H33")
WS.Range("B23") = WS.Range("I33")
WS.Range("B27") = WS.Range("B33")
WS.Range("B28") = WS.Range("G33")
WS.Range("C29") = WS.Range("A33")
'---> High
If WS.Range("C29") = "High" Then
WS.Range("B49") = WS.Range("B49") + 1
WS.Range("D49") = WS.Range("D49") + 1
WS.Range("F49") = WS.Range("F49") + 1
WS.Range("H49") = WS.Range("H49") + 1
'---> Step 02,03,04,05
For K = 1 To 2
For I = 22 To 23
If K = 1 And I = 22 Then
Col = "C"
fcol = "B"
ElseIf K = 1 And I = 23 Then
Col = "C"
fcol = "D"
ElseIf K = 2 And I = 22 Then
Col = "G"
fcol = "F"
ElseIf K = 2 And I = 23 Then
Col = "G"
fcol = "H"
End If
Min = WS.Range("C" & I) - 4
Max = WS.Range("C" & I) + 4
For J = 2 To 11
If WS.Range(Col & J) >= Min And WS.Range(Col & J) <= Max Then
factor = WS.Range(Col & J).Offset(0, -2)
If factor = "Close Hits" Or factor = "Point Hits" Then factor = "Unity"
Exit For
End If
Next J
If factor <> "" Then
Set cCell = WS.Range("A39:A48").Find(what:=factor, LookIn:=xlValues, lookat:=xlWhole, MatchCase:=False)
If Not cCell Is Nothing Then
WS.Cells(cCell.Row, fcol) = WS.Cells(cCell.Row, fcol) + 1
End If
factor = ""
End If
Next I
Next K
End If
'---> Low
If WS.Range("C29") = "Low" Then
WS.Range("B60") = WS.Range("B60") + 1
WS.Range("D60") = WS.Range("D60") + 1
WS.Range("F60") = WS.Range("F60") + 1
WS.Range("H60") = WS.Range("H60") + 1
'---> Step 02,03,04,05
For K = 1 To 2
For I = 22 To 23
If K = 1 And I = 22 Then
Col = "C"
fcol = "B"
ElseIf K = 1 And I = 23 Then
Col = "C"
fcol = "D"
ElseIf K = 2 And I = 22 Then
Col = "G"
fcol = "F"
ElseIf K = 2 And I = 23 Then
Col = "G"
fcol = "H"
End If
Min = WS.Range("C" & I) - 4
Max = WS.Range("C" & I) + 4
For J = 11 To 20
If WS.Range(Col & J) >= Min And WS.Range(Col & J) <= Max Then
factor = WS.Range(Col & J).Offset(0, -2)
If factor = "Close Hits" Or factor = "Point Hits" Then factor = "Unity"
Exit For
End If
Next J
If factor <> "" Then
Set cCell = WS.Range("A50:A59").Find(what:=factor, LookIn:=xlValues, lookat:=xlWhole, MatchCase:=False)
If Not cCell Is Nothing Then
WS.Cells(cCell.Row, fcol) = WS.Cells(cCell.Row, fcol) + 1
End If
factor = ""
End If
Next I
Next K
End If
WS.Range("B22") = WS.Range("H32")
WS.Range("B23") = WS.Range("I32")
WS.Range("B24") = WS.Range("J33")
WS.Range("B25") = WS.Range("K33")
WS.Range("B26") = WS.Range("L33")
WS.Range("B27") = WS.Range("G33")
WS.Range("B28") = WS.Range("B33")
WS.Range("C29") = WS.Range("A33")
'---> High
If WS.Range("C29") = "High" Then
WS.Range("L49") = WS.Range("L49") + 1
WS.Range("N49") = WS.Range("N49") + 1
WS.Range("P49") = WS.Range("P49") + 1
WS.Range("R49") = WS.Range("R49") + 1
WS.Range("T49") = WS.Range("T49") + 1
WS.Range("V49") = WS.Range("V49") + 1
'---> Step 02,03,04,05
For K = 1 To 3
For I = 22 To 23
If K = 1 And I = 22 Then
Col = "K"
fcol = "L"
ElseIf K = 1 And I = 23 Then
Col = "K"
fcol = "N"
ElseIf K = 2 And I = 22 Then
Col = "O"
fcol = "P"
ElseIf K = 2 And I = 23 Then
Col = "O"
fcol = "R"
ElseIf K = 3 And I = 22 Then
Col = "S"
fcol = "T"
ElseIf K = 3 And I = 23 Then
Col = "S"
fcol = "V"
End If
Min = WS.Range("C" & I) - 4
Max = WS.Range("C" & I) + 4
For J = 2 To 11
If WS.Range(Col & J) >= Min And WS.Range(Col & J) <= Max Then
factor = WS.Range(Col & J).Offset(0, -2)
If factor = "Range CC Hits" Or factor = "Range HL Hits" Or factor = "Range PC Hits" Then factor = "Unity"
Exit For
End If
Next J
If factor <> "" Then
Set cCell = WS.Range("K39:K48").Find(what:=factor, LookIn:=xlValues, lookat:=xlWhole, MatchCase:=False)
If Not cCell Is Nothing Then
WS.Cells(cCell.Row, fcol) = WS.Cells(cCell.Row, fcol) + 1
End If
factor = ""
End If
Next I
Next K
End If
'---> Low
If WS.Range("C29") = "Low" Then
WS.Range("L60") = WS.Range("L60") + 1
WS.Range("N60") = WS.Range("N60") + 1
WS.Range("P60") = WS.Range("P60") + 1
WS.Range("R60") = WS.Range("R60") + 1
WS.Range("T60") = WS.Range("T60") + 1
WS.Range("V60") = WS.Range("V60") + 1
'---> Step 02,03,04,05
For K = 1 To 3
For I = 22 To 23
If K = 1 And I = 22 Then
Col = "K"
fcol = "L"
ElseIf K = 1 And I = 23 Then
Col = "K"
fcol = "N"
ElseIf K = 2 And I = 22 Then
Col = "O"
fcol = "P"
ElseIf K = 2 And I = 23 Then
Col = "O"
fcol = "R"
ElseIf K = 3 And I = 22 Then
Col = "S"
fcol = "T"
ElseIf K = 3 And I = 23 Then
Col = "S"
fcol = "V"
End If
Min = WS.Range("C" & I) - 4
Max = WS.Range("C" & I) + 4
For J = 11 To 20
If WS.Range(Col & J) >= Min And WS.Range(Col & J) <= Max Then
factor = WS.Range(Col & J).Offset(0, -2)
If factor = "Close Hits" Or factor = "Point Hits" Then factor = "Unity"
Exit For
End If
Next J
If factor <> "" Then
Set cCell = WS.Range("K50:K59").Find(what:=factor, LookIn:=xlValues, lookat:=xlWhole, MatchCase:=False)
If Not cCell Is Nothing Then
WS.Cells(cCell.Row, fcol) = WS.Cells(cCell.Row, fcol) + 1
End If
factor = ""
End If
Next I
Next K
End If
WS.Range("B22") = WS.Range("H33")
WS.Range("B23") = WS.Range("I33")
WS.Range("B24") = WS.Range("J32")
WS.Range("B25") = WS.Range("K32")
WS.Range("B26") = WS.Range("L32")
WS.Range("B27") = WS.Range("G33")
WS.Range("B28") = WS.Range("B33")
WS.Range("C29") = WS.Range("A33")
'---> High
If WS.Range("C29") = "High" Then
WS.Range("L76") = WS.Range("L76") + 1
WS.Range("N76") = WS.Range("N76") + 1
WS.Range("P76") = WS.Range("P76") + 1
WS.Range("R76") = WS.Range("R76") + 1
WS.Range("T76") = WS.Range("T76") + 1
WS.Range("V76") = WS.Range("V76") + 1
'---> Step 02,03,04,05
For K = 1 To 3
For I = 22 To 23
If K = 1 And I = 22 Then
Col = "K"
fcol = "L"
ElseIf K = 1 And I = 23 Then
Col = "K"
fcol = "N"
ElseIf K = 2 And I = 22 Then
Col = "O"
fcol = "P"
ElseIf K = 2 And I = 23 Then
Col = "O"
fcol = "R"
ElseIf K = 3 And I = 22 Then
Col = "S"
fcol = "T"
ElseIf K = 3 And I = 23 Then
Col = "S"
fcol = "V"
End If
Min = WS.Range("C" & I) - 4
Max = WS.Range("C" & I) + 4
For J = 2 To 11
If WS.Range(Col & J) >= Min And WS.Range(Col & J) <= Max Then
factor = WS.Range(Col & J).Offset(0, -2)
If factor = "Range CC Hits" Or factor = "Range HL Hits" Or factor = "Range PC Hits" Then factor = "Unity"
Exit For
End If
Next J
If factor <> "" Then
Set cCell = WS.Range("K66:K75").Find(what:=factor, LookIn:=xlValues, lookat:=xlWhole, MatchCase:=False)
If Not cCell Is Nothing Then
WS.Cells(cCell.Row, fcol) = WS.Cells(cCell.Row, fcol) + 1
End If
factor = ""
End If
Next I
Next K
End If
'---> Low
If WS.Range("C29") = "Low" Then
WS.Range("L87") = WS.Range("L87") + 1
WS.Range("N87") = WS.Range("N87") + 1
WS.Range("P87") = WS.Range("P87") + 1
WS.Range("R87") = WS.Range("R87") + 1
WS.Range("T87") = WS.Range("T87") + 1
WS.Range("V87") = WS.Range("V87") + 1
'---> Step 02,03,04,05
For K = 1 To 3
For I = 22 To 23
If K = 1 And I = 22 Then
Col = "K"
fcol = "L"
ElseIf K = 1 And I = 23 Then
Col = "K"
fcol = "N"
ElseIf K = 2 And I = 22 Then
Col = "O"
fcol = "P"
ElseIf K = 2 And I = 23 Then
Col = "O"
fcol = "R"
ElseIf K = 3 And I = 22 Then
Col = "S"
fcol = "T"
ElseIf K = 3 And I = 23 Then
Col = "S"
fcol = "V"
End If
Min = WS.Range("C" & I) - 4
Max = WS.Range("C" & I) + 4
For J = 11 To 20
If WS.Range(Col & J) >= Min And WS.Range(Col & J) <= Max Then
factor = WS.Range(Col & J).Offset(0, -2)
If factor = "Close Hits" Or factor = "Point Hits" Then factor = "Unity"
Exit For
End If
Next J
If factor <> "" Then
Set cCell = WS.Range("K77:K86").Find(what:=factor, LookIn:=xlValues, lookat:=xlWhole, MatchCase:=False)
If Not cCell Is Nothing Then
WS.Cells(cCell.Row, fcol) = WS.Cells(cCell.Row, fcol) + 1
End If
factor = ""
End If
Next I
Next K
End If
WS.Range("B22") = WS.Range("H33")
WS.Range("B23") = WS.Range("I33")
WS.Range("B24") = WS.Range("J33")
WS.Range("B25") = WS.Range("K33")
WS.Range("B26") = WS.Range("L33")
WS.Range("B27") = WS.Range("G33")
WS.Range("B28") = WS.Range("B33")
WS.Range("C29") = WS.Range("A33")
'---> High
If WS.Range("C29") = "High" Then
WS.Range("L103") = WS.Range("L103") + 1
WS.Range("N103") = WS.Range("N103") + 1
WS.Range("P103") = WS.Range("P103") + 1
WS.Range("R103") = WS.Range("R103") + 1
WS.Range("T103") = WS.Range("T103") + 1
WS.Range("V103") = WS.Range("V103") + 1
'---> Step 02,03,04,05
For K = 1 To 3
For I = 22 To 23
If K = 1 And I = 22 Then
Col = "K"
fcol = "L"
ElseIf K = 1 And I = 23 Then
Col = "K"
fcol = "N"
ElseIf K = 2 And I = 22 Then
Col = "O"
fcol = "P"
ElseIf K = 2 And I = 23 Then
Col = "O"
fcol = "R"
ElseIf K = 3 And I = 22 Then
Col = "S"
fcol = "T"
ElseIf K = 3 And I = 23 Then
Col = "S"
fcol = "V"
End If
Min = WS.Range("C" & I) - 4
Max = WS.Range("C" & I) + 4
For J = 2 To 11
If WS.Range(Col & J) >= Min And WS.Range(Col & J) <= Max Then
factor = WS.Range(Col & J).Offset(0, -2)
If factor = "Range CC Hits" Or factor = "Range HL Hits" Or factor = "Range PC Hits" Then factor = "Unity"
Exit For
End If
Next J
If factor <> "" Then
Set cCell = WS.Range("K93:K102").Find(what:=factor, LookIn:=xlValues, lookat:=xlWhole, MatchCase:=False)
If Not cCell Is Nothing Then
WS.Cells(cCell.Row, fcol) = WS.Cells(cCell.Row, fcol) + 1
End If
factor = ""
End If
Next I
Next K
End If
'---> Low
If WS.Range("C29") = "Low" Then
WS.Range("L114") = WS.Range("L114") + 1
WS.Range("N114") = WS.Range("N114") + 1
WS.Range("P114") = WS.Range("P114") + 1
WS.Range("R114") = WS.Range("R114") + 1
WS.Range("T114") = WS.Range("T114") + 1
WS.Range("V114") = WS.Range("V114") + 1
'---> Step 02,03,04,05
For K = 1 To 3
For I = 22 To 23
If K = 1 And I = 22 Then
Col = "K"
fcol = "L"
ElseIf K = 1 And I = 23 Then
Col = "K"
fcol = "N"
ElseIf K = 2 And I = 22 Then
Col = "O"
fcol = "P"
ElseIf K = 2 And I = 23 Then
Col = "O"
fcol = "R"
ElseIf K = 3 And I = 22 Then
Col = "S"
fcol = "T"
ElseIf K = 3 And I = 23 Then
Col = "S"
fcol = "V"
End If
Min = WS.Range("C" & I) - 4
Max = WS.Range("C" & I) + 4
For J = 11 To 20
If WS.Range(Col & J) >= Min And WS.Range(Col & J) <= Max Then
factor = WS.Range(Col & J).Offset(0, -2)
If factor = "Close Hits" Or factor = "Point Hits" Then factor = "Unity"
Exit For
End If
Next J
If factor <> "" Then
Set cCell = WS.Range("K104:K113").Find(what:=factor, LookIn:=xlValues, lookat:=xlWhole, MatchCase:=False)
If Not cCell Is Nothing Then
WS.Cells(cCell.Row, fcol) = WS.Cells(cCell.Row, fcol) + 1
End If
factor = ""
End If
Next I
Next K
End If
WS.Range("B22") = WS.Range("H33")
WS.Range("B23") = WS.Range("I33")
WS.Range("B24") = WS.Range("J33")
WS.Range("B25") = WS.Range("K33")
WS.Range("B26") = WS.Range("L33")
WS.Range("B27") = WS.Range("G32")
WS.Range("B28") = WS.Range("B32")
WS.Range("C29") = WS.Range("A33")
'---> High
If WS.Range("C29") = "High" Then
WS.Range("B76") = WS.Range("B76") + 1
WS.Range("D76") = WS.Range("D76") + 1
WS.Range("F76") = WS.Range("F76") + 1
WS.Range("H76") = WS.Range("H76") + 1
'---> Step 02,03,04,05
For K = 1 To 2
For I = 22 To 23
If K = 1 And I = 22 Then
Col = "C"
fcol = "B"
ElseIf K = 1 And I = 23 Then
Col = "C"
fcol = "D"
ElseIf K = 2 And I = 22 Then
Col = "G"
fcol = "F"
ElseIf K = 2 And I = 23 Then
Col = "G"
fcol = "H"
End If
Min = WS.Range("C" & I) - 4
Max = WS.Range("C" & I) + 4
For J = 2 To 11
If WS.Range(Col & J) >= Min And WS.Range(Col & J) <= Max Then
factor = WS.Range(Col & J).Offset(0, -2)
If factor = "Close Hits" Or factor = "Point Hits" Then factor = "Unity"
Exit For
End If
Next J
If factor <> "" Then
Set cCell = WS.Range("A66:A75").Find(what:=factor, LookIn:=xlValues, lookat:=xlWhole, MatchCase:=False)
If Not cCell Is Nothing Then
WS.Cells(cCell.Row, fcol) = WS.Cells(cCell.Row, fcol) + 1
End If
factor = ""
End If
Next I
Next K
End If
'---> Low
If WS.Range("C29") = "Low" Then
WS.Range("B87") = WS.Range("B87") + 1
WS.Range("D87") = WS.Range("D87") + 1
WS.Range("F87") = WS.Range("F87") + 1
WS.Range("H87") = WS.Range("H87") + 1
'---> Step 02,03,04,05
For K = 1 To 2
For I = 22 To 23
If K = 1 And I = 22 Then
Col = "C"
fcol = "B"
ElseIf K = 1 And I = 23 Then
Col = "C"
fcol = "D"
ElseIf K = 2 And I = 22 Then
Col = "G"
fcol = "F"
ElseIf K = 2 And I = 23 Then
Col = "G"
fcol = "H"
End If
Min = WS.Range("C" & I) - 4
Max = WS.Range("C" & I) + 4
For J = 11 To 20
If WS.Range(Col & J) >= Min And WS.Range(Col & J) <= Max Then
factor = WS.Range(Col & J).Offset(0, -2)
If factor = "Close Hits" Or factor = "Point Hits" Then factor = "Unity"
Exit For
End If
Next J
If factor <> "" Then
Set cCell = WS.Range("A77:A86").Find(what:=factor, LookIn:=xlValues, lookat:=xlWhole, MatchCase:=False)
If Not cCell Is Nothing Then
WS.Cells(cCell.Row, fcol) = WS.Cells(cCell.Row, fcol) + 1
End If
factor = ""
End If
Next I
Next K
End If
WS.Range("B22:B28").ClearContents
WS.Range("C29").ClearContents
End Sub
``````
There is 5 Ways calculations & each result is registered in 5 tables for Highs & lows.
1. Update Range B22:B28 & C29. Then Update Column B D F H in Table 1 row 39 to 60 for High & Low.
Update Ranges i.e. WS.Range("B22") = WS.Range("H33")
WS.Range("B23") = WS.Range("I33")
WS.Range("B27") = WS.Range("B33")
WS.Range("B28") = WS.Range("G33")
WS.Range("C29") = WS.Range("A33")
2.Same way Update Range B22:B28 & C29. Then Update Column L N P R T V in Table 2 row 39 to 60 for High & low.
3.Update range B22:B28 & C29. Then Update Column L N P R T V in Table 3 row 66 to 87 for High & low.
4.Update Range B22:B28 & C29. Then Update Column L N P R T V in Table 4 Row 93 to 114 for High & Low.
5.Update Range B22:B28 & C29. Then Update Column B D F H in Table 5 row 66 to 87 for High & low.
Thanks
Scoring-G-V07.xlsm
LVL 8
###### Who is Participating?
Commented:
ok Given the complexity and non-uniformity, I have divided into Passes and created literally 1 sub for each pass so that if you have a problem in 1 pass you can check the appropriate code without having to go thru all the code.
There is still the main Sub that is called Analyse that will do the grouping and here is the code for that.
``````Sub Analyse()
Dim WS As Worksheet
Set WS = ThisWorkbook.ActiveSheet
Analyse_Pass1 WS
Analyse_Pass2 WS
Analyse_Pass3 WS
Analyse_Pass4 WS
Analyse_Pass5 WS
WS.Range("B22:B28").ClearContents
WS.Range("C29").ClearContents
End Sub
``````
For the individual subs pass1, pass2 etc... you can see them in the attached file.
Pls try it and let me know if all is ok.
gowflow
Scoring-G-V08.xlsm
0
Commented:
Are you sure table 4 and 5 are correct ? they shouldn't be inverted ?
like if you go left right then down then left right then down ???
as you have
1 2
5 3
4
I would see it
1 2
3 4
5
???
gowlfow
0
like if you go left right then down then left right then down ???
as you have
1 2
5 3
4
yes it is right one as per Sub Analyse. if you want to change order then need to change Sub sequence too.
thanks
0
Commented:
Also your post does not reflect the code
You have first pass
``````WS.Range("B22") = WS.Range("H33")
WS.Range("B23") = WS.Range("I33")
WS.Range("B27") = WS.Range("B33")
WS.Range("B28") = WS.Range("G33")
WS.Range("C29") = WS.Range("A33")
``````
Second Pass
``````WS.Range("B22") = WS.Range("H32")
WS.Range("B23") = WS.Range("I32")
WS.Range("B24") = WS.Range("J33")
WS.Range("B25") = WS.Range("K33")
WS.Range("B26") = WS.Range("L33")
WS.Range("B27") = WS.Range("G33")
WS.Range("B28") = WS.Range("B33")
WS.Range("C29") = WS.Range("A33")
``````
Third Pass
``````WS.Range("B22") = WS.Range("H33")
WS.Range("B23") = WS.Range("I33")
WS.Range("B24") = WS.Range("J32")
WS.Range("B25") = WS.Range("K32")
WS.Range("B26") = WS.Range("L32")
WS.Range("B27") = WS.Range("G33")
WS.Range("B28") = WS.Range("B33")
WS.Range("C29") = WS.Range("A33")
``````
Fourth PAss
``````WS.Range("B22") = WS.Range("H33")
WS.Range("B23") = WS.Range("I33")
WS.Range("B24") = WS.Range("J33")
WS.Range("B25") = WS.Range("K33")
WS.Range("B26") = WS.Range("L33")
WS.Range("B27") = WS.Range("G33")
WS.Range("B28") = WS.Range("B33")
WS.Range("C29") = WS.Range("A33")
``````
Fifth Pass
``````WS.Range("B22") = WS.Range("H33")
WS.Range("B23") = WS.Range("I33")
WS.Range("B24") = WS.Range("J33")
WS.Range("B25") = WS.Range("K33")
WS.Range("B26") = WS.Range("L33")
WS.Range("B27") = WS.Range("G32")
WS.Range("B28") = WS.Range("B32")
WS.Range("C29") = WS.Range("A33")
``````
and nothing of all this is mentioned in your post note the difference sometimes row 32 sometimes 33 only .... what is it ?
gowflow
0
Sir give me some time as I am on my way back to home. Traveling.
Thanks
0
ok sorry for delay in reply.
A. all table working with 3 types of data & combination of any two.TIME - PRICE - PRICE RANGES
B.TIME - PRICE - PRICE RANGES -two types - Current & previous
Current is available in row 33. (see date - it is latest then row 32)
previous is available in row 32.
I had labeled in each table top row.
Current Price With Current Time Range - Works with - Current TIME & current PRICES
Previous Time Range With Current Price Range
- works with previous TIME & current PRICE RANGES.
rest all 3 as name suggested works with above combinations.
See attached - highlighted cells.
thanks
Scoring-G-V07.xlsm
0
Commented:
All this explanation is fine for you but does not serve me. I cannot start learning prices and stocks. But I know how to program vba.
I noted in my last post some inconsistencies in what I saw versus what you asked in oyur question.
is to check what I posted thouroully for each step and make sure what I posted for each step is correct and then admit that you made a mistake in oyur initial post and did not mention these for each pass.
That was the purpose of my post is to make sure that what I have in code is correct.
and then and only then I can start optimizing the code.
gowflow
0
Sir gowflow,
Now do u want me to post steps in detail?
Thanks
0
Commented:
it seems you don't get it !!!
I need the important steps that make the distinction between passes. If you feel your 5 pages are what it need then let it be.
Just don't copy paste all the steps here but attach a document that have the explanation.
gowflow
0
i wont available for today.....as leaving office early. Going to Doctor.
hey Expert one question do u know why WBC count increases ?
i don't know why my WBC count doubled then normal range from last two day & still i dint find nothing unusual in body. :)
just chilling ....i thought need to inform as i am not available....(Online on Cell)
Thanks
0
Commented:
hey Expert one question do u know why WBC count increases ?
i don't know why my WBC count doubled then normal range from last two day & still i dint find nothing unusual in body. :)
I doubt this is an Excel question !!! I presume it is a medical question so the best qualified is not EE Expert (or they may be some) but rather a DOCTOR !
Hope all is fine with you and you will feel well in no time. Take care, ususally lately they found out that most of our disorders orginates from stress !!! and as you mentioned lately going from office at 430 AM this is called NO SLEEP ... and for me personally when I reach the repetitive lack of sleep my health takes a beat and get sick.
So remedy, sleep and sleep and sleep and rest and get well.
Reagrds
gowflow
0
Thank You Sir :)
0
Commented:
gowlfow
0
Sorry Sir for delay ....just need confirmation from your side. how would like to see steps? there is two ways High & low together for each pass
``````Copy Cell Value H33 Past To B22
Copy Cell Value I33 Past To B23
Copy Cell Value B33 Past To B27
Copy Cell Value G33 Past To B28
Copy Cell Value A33 Past To C29
If C29 = "High" then +1 @ Cell B76 & D76 & F76 & H76
If C29="Low" then +1 @ CellB87 & D87 & F87 & H87
If C29="High" then Find C22 Value From Range C2:C10(+ or - 4 Point Leverage)
"If True then find Respective factor in range B39:B48& Add +1 (if it is at Range CC then put +1 @ Unity)else nothing next step
"
If C29="Low" then Find C22 Value From Range C12:C20(+ or - 4 Point Leverage)
"If True then find Respective factor in range B49:B57& Add +1 (if it is at Range CC then put +1 @ Unity)else nothing next step
"
If C29="High" then Find C23 value in range C2:C11(+ or - 4 Point Leverage)
"If True then find Respective factor in range D39:D48& Add +1 (if it is at Range CC then put +1 @ Unity)else nothing next step
"
If C29="Low" then Find C23 Value From Range C12:C20(+ or - 4 Point Leverage)
"If True then find Respective factor in range D49:D57& Add +1 (if it is at Range CC then put +1 @ Unity)else nothing next step
"
``````
or separated for High & low for each pass? this is only for High for one of the pass
``````Copy Cell Value H33 Past To B22
Copy Cell Value I33 Past To B23
Copy Cell Value B33 Past To B27
Copy Cell Value G33 Past To B28
Copy Cell Value A33 Past To C29
If C29 = "High" then +1 @ Cell B76 & D76 & F76 & H76
If C29="High" then Find C22 Value From Range C2:C10(+ or - 4 Point Leverage)
"If True then find Respective factor in range B39:B48& Add +1 (if it is at Range CC then put +1 @ Unity)else nothing next step
"
If C29="High" then Find C23 value in range C2:C11(+ or - 4 Point Leverage)
"If True then find Respective factor in range D39:D48& Add +1 (if it is at Range CC then put +1 @ Unity)else nothing next step
"
``````
both are just samples not full steps.
Thanks
0
Commented:
I want them exactly how the logic should be.
If the logic dictate that for each pass you process something, then high then low well I just need this in very summary.
If the logic says that you should process all the high then process something then update table1 then something then table2 etc... and when finished process low then something then Table1 etc... then be it
If the logic says that you should update A,B,C for Table1 High and then low then D,E,F for table2 high then low .... then be it
I need you to explain (in plain English) meaning not in code not copy cell b1 paste b33 ... but in meaning copy the cell affection then process for column A,B,C high etc....
like this. For sure in the Cell affection is not the same then you can copy paste the code for that step.
gowflow
0
ok this is for pass 1 - Scoring Table 1"Current Price With Current Time Range"
where current Prices & current time is in row 33.we required CD TD Close Hits & Point Hits values form row 33 & put in table A22:C29 to there respective heads.
so in row 33 CD is 37 (Cell H33 ) TD is 25 (Cell I33) Close Hits is 5288.95 (Cell G33) & Point Hits is 5108.25 (Cell B33). i.e.
``````Copy Cell Value H33 Past To B22
Copy Cell Value I33 Past To B23
Copy Cell Value B33 Past To B27
Copy Cell Value G33 Past To B28
Copy Cell Value A33 Past To C29
``````
this Step 1 for pass 1.
Now we are counting one result so need to add +1 to High if there is high & +1 to low if there is low in Total From High & Total From low respectively in Scoring Table 1.
``````If C29 = "High" then +1 @ Cell B49 & D49 & F49 & H49
If C29="Low" then +1 @ CellB60 & D60 & F60 & H60
``````
Step 2 Pass 1
Step 3 If C29="High" then Find C22 Value From Range C2:C10(+ or - 4 Point Leverage)
If True then find Respective factor in range B39:B48& Add +1 (if it is at Close Hits then put +1 @ Unity)else nothing next step
If C29="Low" then Find C22 Value From Range C12:C20(+ or - 4 Point Leverage)
If True then find Respective factor in range B49:B57& Add +1 (if it is at Close Hits then put +1 @ Unity)else nothing next step
Step 4If C29="High" then Find C23 value in range C2:C11(+ or - 4 Point Leverage)
If True then find Respective factor in range D39:D48& Add +1 (if it is at Close Hits then put +1 @ Unity)else nothing next step
If C29="Low" then Find C23 Value From Range C12:C20(+ or - 4 Point Leverage)
If True then find Respective factor in range D49:D57& Add +1 (if it is at Close Hits then put +1 @ Unity)else nothing next step
Step 5 If C29="High" then Find C22 value in range G2:G11(+ or - 4 Point Leverage)
If True then find Respective factor in range F39:F48& Add +1 (if it is at Point Hits then put +1 @ Unity)else nothing next step
If C29="Low" then Find C22 Value From Range G12:G20(+ or - 4 Point Leverage)
If True then find Respective factor in range F49:F57& Add +1 (if it is at Point Hits then put +1 @ Unity)else nothing next step
Step 6 If C29="High" then Find C23 value in range G2:G11(+ or - 4 Point Leverage)
If True then find Respective factor in range H39:H48& Add +1 (if it is at Point Hits then put +1 @ Unity)else nothing next step
If C29="Low" then Find C23 Value From Range G12:G20(+ or - 4 Point Leverage)
If True then find Respective factor in range H49:H57& Add +1 (if it is at Point Hits then put +1 @ Unity)else nothing
Pass 1 end
Thanks
0
I had written all pass in this notepad but I guess you don't like this way.....
Thanks
All-Pass.txt
0
Commented:
last text file is fine. Let me digest all this, and maybe at the end .....
will see
gowflow
0
take your time Sir .... actually I feel my self happy
last text file is fine. Let me digest all this, and maybe at the end .....
will see
as I am not good in writings. :)
Thanks
0
Working perfect Sir......
0
Commented:
Actually for working it should work same as before it is only the structure of the whole Sub we changed to make it easier to dissect and troubleshoot and also for adding/deleting things then it become much simpler to deal with 1 pass instead of being lost in a sub that is 500 km long !!
btw, I see you are very active in the Excel Zone, Good for you nice to see that.
gowflow
0
Thank You Sir for compliment & thank you for this question. As I don't know about this to split subs which benefited in future
0
Commented:
Your welcome and glad I could help. Any news on the Stat question issue ?
Regards
gowflow
0
Nope Sir stuck between how to get this done via formula.is that way to call this procedure via VBA?
Thanks
0
It just comment :- As more more my question you will solve of mine. You will become trader one day. ;-)
My self I am trader at institutional desk.
Thanks
0
Commented:
as to solving this via VBA well will need at the end the exact formulas or else we are stuck !!!
gowflow
0
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# How many acres would be in 1320 feet by 1320 feet?
Wiki User
2010-05-17 00:57:27
40 acres. Mutiply 1320 x 1320, then divide by 43560 (number of sq ft in one acre)
Wiki User
2010-05-17 00:57:27
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An analysis of the encoding of the tiles used by Google map
## Introduction
This is my analysis about how Google map works, and specially how the tiles are encoded. Google map uses pre-rendered tiles that can be obtained with a simple URL. This article explains how to build the URL for a tile from its geo coordinates (latitude/longitude).
## Map Tile Encoding
Google map uses two different algorithms to encode the location of the tiles.
For Google map, the URL of a tile looks like http://mt1.google.com/mt?n=404&v=w2.12&x=130&y=93&zoom=9 using x and y for the tile coordinates, and a zoom factor. The zoom factor goes from `17` (fully zoomed out) to `0` (maximum definition). At a factor `17`, the whole earth is in one tile where x=0 and y=0. At a factor `16`, the earth is divided in 2x2 parts, where 0<=x<=1 and 0<=y<=1, and at each zoom step, each tile is divided into 4 parts. So at a zoom factor Z, the number of horizontal and vertical tiles is 2^(17-z)
### Algorithm to Find a Tile from a Latitude, a Longitude and a Zoom Factor
```//correct the latitude to go from 0 (north) to 180 (south),
// instead of 90(north) to -90(south)
latitude=90-latitude;
//correct the longitude to go from 0 to 360
longitude=180+longitude;
//find tile size from zoom level
double latTileSize=180/(pow(2,(17-zoom)));
double longTileSize=360/(pow(2,(17-zoom)));
//find the tile coordinates
int tilex=(int)(longitude/longTileSize);
int tiley=(int)(latitude/latTileSize);```
In fact this algorithm is theoretical as the covered zone doesn't match the whole globe.
### Servers
Google uses four servers to balance the load. These are mt0, mt1, mt2 and mt3.
### Tile Size
Each tile is a 256x256 PNG picture.
## For Satellite Images, the Encoding is a Bit Different
The URL looks like http://kh0.google.com/kh?n=404&v=8&t=trtqtt where the 't' parameters encode the image location. The length of the parameter indicates a zoom level.
To see the whole globe, just use 't=t'. This gives a single tile representing the earth. For the next zoom level, this tile is divided into 4 quadrants, called, clockwise from top left : 'q' 'r' 's' and 't'. To see a quadrant, just append the letter of that quadrant to the image you are viewing. For example :'t=tq' will give the upper left quadrant of the 't' image. And so on at each zoom level...
### Algorithm to Find a Tile from a Latitude, a Longitude and a Zoom Factor
```//initialise the variables;
double xmin=-180;
double xmax=180;
double ymin=-90;
double ymax=90;
double xmid=0;
double ymid=0;
string location="t";
//Google uses a latitude divided by 2;
double halflat = latitude / 2;
for (int i = 0; i < zoom; i++)
{
xmoy = (xmax + xmin) / 2;
ymoy = (ymax + ymin) / 2;
if (halflat > ymoy) //upper part (q or r)
{
ymin = ymoy;
if (longitude < xmoy)
{ /*q*/
location+= "q";
xmax = xmoy;
}
else
{/*r*/
location+= "r";
xmin = xmoy;
}
}
else //lower part (t or s)
{
ymax = ymoy;
if (longitude < xmoy)
{ /*t*/
location+= "t";
xmax = xmoy;
}
else
{/*s*/
location+= "s";
xmin = xmoy;
}
}
}
//here, the location should contain the string corresponding to the tile...```
Again, this algorithm is quite theoretical, as the covered zone doesn't match the full globe.
### Servers
Google uses four servers to balance the load. These are kh0, kh1, kh2 and kh3.
### Tile Size
Each tile is a 256x256 JPG picture.
## Mercator Projection
Due to the Mercator projection, the above algorithm has to be modified. In Mercator projection, the spacing between two parallels is not constant. So the angle described by a tile depends on its vertical position.
Here comes a piece of code to compute a tile's vertical number from its latitude.
```/**<summary>Get the vertical tile number from a latitude
using Mercator projection formula</summary>*/
private int getMercatorLatitude(double lati)
{
double maxlat = Math.PI;
double lat = lati;
if (lat > 90) lat = lat - 180;
if (lat < -90) lat = lat + 180;
double phi = Math.PI * lat / 180;
double res;
//double temp = Math.Tan(Math.PI / 4 - phi / 2);
//res = Math.Log(temp);
res = 0.5 * Math.Log((1 + Math.Sin(phi)) / (1 - Math.Sin(phi)));
double maxTileY = Math.Pow(2, zoom);
int result = (int)(((1 - res / maxlat) / 2) * (maxTileY));
return (result);
}```
## Covered Zone
Theoretically, latitude should go from `-90 `to `90`, but in fact due to the Mercator projection which sends the poles to the infinites, the covered zone is a bit less than `-90 `to `90`. In fact the maximum latitude is the one that gives `PI` (3.1415926) on the Mercator projection, using the formula `Y = 1/2((1+sin(lat))/(1-sin(lat)))` (see the link in the Mercator paragraph).
## Protection
Google map uses a protection mechanism to keep a good quality of service. If one makes too many requests, Google map will add its IP address to a blacklist, and send a nice message:
We're sorry... ... but your query looks similar to automated requests from a computer virus or spyware application. To protect our users, we can't process your request right now. We'll restore your access as quickly as possible, so try again soon. In the meantime, if you suspect that your computer or network has been infected, you might want to run a virus checker or spyware remover to make sure that your systems are free of viruses and other spurious software. We apologize for the inconvenience, and hope we'll see you again on Google.
To avoid being blacklisted, developers should use a caching mechanism if possible...
## Sat Examples
See the whole globe at http://kh0.google.com/kh?n=404&v=8&t=t.
And the four corresponding quadrants: (note the 4 servers name to balance the load)
## Map Examples
See the whole globe at http://mt1.google.com/mt?n=404&v=&x=0&y=0&zoom=17.
Nice, isn't it?
## History
• Article edited: Google map has changed the v parameter for the maps. It was 2.12 when I wrote this article, but it's now 2.66. I suppose this is a version number or something like that...
## Share
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Re: Licence problems Daniel Ch. Bloch (MCSD, MCAD, MCTS)19-Nov-08 20:27 Daniel Ch. Bloch (MCSD, MCAD, MCTS) 19-Nov-08 20:27
Summary Article [modified] James_Lin28-Feb-08 14:53 James_Lin 28-Feb-08 14:53
Is this project dead? James_Lin20-Feb-08 16:30 James_Lin 20-Feb-08 16:30
Re: Is this project dead? Pascal Buirey20-Feb-08 21:47 Pascal Buirey 20-Feb-08 21:47
Re: Is this project dead? James_Lin23-Feb-08 20:07 James_Lin 23-Feb-08 20:07
Re: Is this project dead? Pascal Buirey24-Feb-08 22:52 Pascal Buirey 24-Feb-08 22:52
Re: Is this project dead? TimmyTee26-Feb-08 3:51 TimmyTee 26-Feb-08 3:51
Re: Is this project dead? James_Lin28-Feb-08 10:39 James_Lin 28-Feb-08 10:39
Re: Is this project dead? senthilmuruganbtech13-Aug-08 0:24 senthilmuruganbtech 13-Aug-08 0:24
Re: Is this project dead? senthilmuruganbtech13-Aug-08 0:23 senthilmuruganbtech 13-Aug-08 0:23
Re: Is this project dead? senthilmuruganbtech13-Aug-08 0:24 senthilmuruganbtech 13-Aug-08 0:24
Re: Is this project dead? senthilmuruganbtech13-Aug-08 3:11 senthilmuruganbtech 13-Aug-08 3:11
hi. how to use real-time tiles...? hyohaeng31-Jan-08 20:05 hyohaeng 31-Jan-08 20:05
Re: hi. how to use real-time tiles...? Pascal Buirey31-Jan-08 23:21 Pascal Buirey 31-Jan-08 23:21
get x,y pixel inside a tile from given GPS coordinates James_Lin31-Jan-08 11:30 James_Lin 31-Jan-08 11:30
Re: get x,y pixel inside a tile from given GPS coordinates Pascal Buirey31-Jan-08 22:54 Pascal Buirey 31-Jan-08 22:54
Re: get x,y pixel inside a tile from given GPS coordinates James_Lin3-Feb-08 10:33 James_Lin 3-Feb-08 10:33
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The flashcards below were created by user rjc1983 on FreezingBlue Flashcards. When you have the right mixture in your spray tank, can you still apply the wrong amount of pesticides? Yes. Noncalibrated application equipment will apply the pesticide at an unknow rate. Unless calibrated, it could be under or over applying the pesticide. What is the delivery rate? Delivery rate is the total amount of pesticide delivered on the target over a period of time. Why is an air blast sprayer difficult to calibrate? Nozzle orientation, high volume, and pressure of the air blast sprayer makes it difficult to catch the nozzle output. Once your sprayer is calibrated, does it always remain the same or should you recheck it often? It must be rechecked often. Nozzles can wear or become plugged, thus changing the delivery rate. If your sprayer is delivering less spray to each acre then you want it to, how would you usually change the rate? Change the pump pressure, speed, or nozzles. What must you measure to calibrate granular application equipment? You must measure the amount of granules spread over a know area. Must you calibrate granular application equipment each time you change granules? Why? Yes, because each granule flows differently What facts must you know for the "Finding gallons per acre known area method" method of sprayer calibration? The number of square feet in an acre, the speed of your sprayer, the width of your spray boom, and the delivery rate of your sprayer. A 10ft sprayer with six nozzles on 20 inch centers is used to apply herbicides on a golf course fairway. At the desired speed setting, the sprayer travelled across a 204ft. course in 40 seconds. How is the application rate calculated using this information? Catch the spray from one nozzle in a container for 40 seconds( time it takes to travel speed course). The ounces of water collected represents the application rate in gallons per acre. Thiry-two ounces of water was collected from one spray nozzle in 30 seconds. What is the nozzle delivery rate in gallons per min.(GPM) 32 ounces (wet) 1 gallon 60 seconds_______________ x _____________ x _________ =30 seconds 128 ounces 1 min 0.5 GPM Are speedometers accurate enough to be used in calibrating pesticide equipment? No. Wheel slippage and variations in tire size often make speedometers inaccurate. Explain how speed can be measured using an 88ft course. From a running start, record the time it takes to drive 88ft. Divide the number of seconds recorded into 60 and you have calculated the miles per hour. Authorrjc1983 ID128750 Card Setchapter 19 DescriptionCalibration Updated2012-01-19T20:55:12Z Show Answers | 595 | 2,667 | {"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-2020-50 | latest | en | 0.924859 |
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# A field is in the shape of a trapezium whose parallel sides are and . The nonparallel sides are and . Find the area of the field.
## Solutions
(1)
First, draw a line segment BE parallel to the line AD. Then, from B, draw a perpendicular on the line segment .
Now, it can be seen that the quadrilateral ABED is a parallelogram. So,
Now, consider the triangle BEC,
Its semi perimeter
By using Heron's formula,
Area of
We also know that the area of
So, the total area of ABED will be BF×DE i.e.
Area of the field \$
48
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Question Text A field is in the shape of a trapezium whose parallel sides are and . The nonparallel sides are and . Find the area of the field. Topic Heron's Formula Subject Mathematics Class Class 9 Answer Type Text solution:1 Upvotes 48 | 257 | 1,044 | {"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} | 3.296875 | 3 | CC-MAIN-2023-14 | latest | en | 0.871283 |
https://www.teacherspayteachers.com/Store/Ashley-Toy-7561 | 1,529,371,346,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861641.66/warc/CC-MAIN-20180619002120-20180619022120-00002.warc.gz | 918,487,336 | 27,333 | # Ashley Toy
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Array Town shows 6 different real-life arrays in the buildings windows', stars, flowers and trees in a park. Part 1 has students locating all the arrays, writing a repeated addition sentences and a matching multiplication sentence. Part 2 requires
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With a partner, take turns choosing a problem. Each player will calculate the sum, and simplify if possible. If you correctly answer the problem, mark the space as your square. If you answer incorrectly, and your partner answers correctly then
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Students will use the patterns on a hundreds chart to subtract two 2-digit numbers. They can use colored pencils to show their movement in order to calculate the difference.
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Students use the patterns on a hundreds chart to add two 2-digit numbers. Use colored pencils to have students color their movements on the hundreds chart to calculate the sum.
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With a partner, take turns choosing a problem. Each player will calculate the difference, and simplify if possible. If you correctly answer the problem, mark the space as your square. If you answer incorrectly, and your partner answers correctly
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7 ratings
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Students will subtract across 0's to find difference in 2 bingo boards at the bottom of the page. On the first page students will subtract 3 digit and 3 digit numbers or 3 digit and 2 digit numbers. The second page has students subtracting 4
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Use these addition and subtraction word problems for a tic-tac-toe game! Split the class into two teams -- X's and O's. It is important for the students to recognize the key words for finding the sums and differences.
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A Jeopardy Review game on generating and analyzing patterns has 3 categories (writing rules for patterns, continuing geometric patterns, and problem solving).
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Students will read each problems and decide if they will add or subtract. A picture to model their problem will be drawn to help solve the problem.
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Students will play multiples of 10 bingo by solving all 24 problems on their individual bingo board. Once all students have finished, the products will be called for students to try and get bingo. There are 6 different bingo boards.
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Students will spin two spinners to create a subtraction sentence. Then students will solve the subtraction sentence with regrouping. Keeps them engaged while subtracting! Use as a reteach or center!
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Students will change 3-digit numbers by mentally adding 10 (10 more) and 100 (100 more) and by subtracting 10 (10 less) and 100 (100 less). Poke a small hole in the middle of the spinner on both sides prior to the activity. Students will spin the
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The 'Adding 0, 1 and 2 Game' is an activity used to help students practice their fluency in 0-2 addition facts. Students will be given: a game board (downloaded) an addition number sentence worksheet (downloaded) a die (preferably 1-9) 3 cards
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This is a great way to review addition and subtraction word problems. There are 3 different categories (stories about separating, stories about joining, and stories about comparing) with 4 different problems. Set it up like a Jeopardy game and
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Domino Addition is a fun twist to practice adding two 2-digit numbers with and without regrouping! Students will chose 2 dominoes and then draw the picture that represents the dominoes on the handout. Finally, the sum will be calculated! T
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A series of task cards and a recording sheet are provided for students to solve problems where they must calculate the perimeter of the rectangle or find the length/width given the other length/width and perimeter. Also, there are real world
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A great activity for practicing multiplication with 2-digit by 1-digit multiplication problems. Compete against another player to get the highest product! A player rolls the three dice, and selects any 2 of the 3 numbers to create a 2-digit
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Students will go on a scavenger hunt around the room to solve stories about separating. Students will read the real-world subtraction problems to calculate the difference.
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6 ratings
3.9
3rd grade students practice the distributive property in an area model with the visual use of chocolate bars. This can be a very hard skill for students, but using the chocolate bar and the idea that they can share the candy bar with a friend makes
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Students are given menu items for each grouping of questions. Using the listed menu items, students will read the word problems and solve. Take this a step further by getting menus from local restaurants and have students create their own real
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 1,467 | 5,817 | {"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} | 2.75 | 3 | CC-MAIN-2018-26 | latest | en | 0.857039 |
http://mathhelpforum.com/calculus/28741-derive-formula-infinite-series-print.html | 1,521,450,712,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646636.25/warc/CC-MAIN-20180319081701-20180319101701-00295.warc.gz | 190,236,907 | 2,743 | # derive a formula for infinite series..
• Feb 20th 2008, 06:39 PM
carpark
derive a formula for infinite series..
Derive a general formula for the coefficients bn, as defined by (1/(1-x)){summation to infinity with i=0}(ai)x^i = {summation to infinity with i=0}(bi)x^i
• Feb 20th 2008, 08:13 PM
mr fantastic
Quote:
Originally Posted by carpark
Derive a general formula for the coefficients bn, as defined by (1/(1-x)){summation to infinity with i=0}(ai)x^i = {summation to infinity with i=0}(bi)x^i
For $\displaystyle \, -1 < x < 1\,$, $\displaystyle \, \frac{1}{1-x} = 1 + x + x^2 + x^3 + .......\,$ using the formula in reverse for an infinite geometric series.
For $\displaystyle \, x > 1\,$ or $\displaystyle \, x < -1\,$, $\displaystyle \, \frac{1}{1-x} = -\frac{1}{x} \left( \frac{1}{1 - \frac{1}{x}}\right) = -\frac{1}{x} \left( 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + ....\right) \,$ again using the formula in reverse for an infinite geometric series.
• Feb 21st 2008, 12:06 AM
mrbicker
What if for each i>=0, ai=i, then what is bn? | 376 | 1,055 | {"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} | 3.703125 | 4 | CC-MAIN-2018-13 | latest | en | 0.777337 |
https://eventuallyalmosteverywhere.wordpress.com/tag/bmo2/ | 1,527,129,033,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865884.49/warc/CC-MAIN-20180524014502-20180524034441-00024.warc.gz | 539,719,166 | 45,756 | # BMO2 2018
The second round of the British Mathematical Olympiad was taken yesterday by the 100 or so top scoring eligible participants from the first round, as well as some open entries. Qualifying for BMO2 is worth celebrating in its own right. The goal of the setters is to find the sweet spot of difficult but stimulating for the eligible participants, which ultimately means it’s likely to be the most challenging exam many of the candidates sit while in high school, at least in mathematics.
I know that lots of students view BMO2 as something actively worth preparing for. As with everything, this is a good attitude in moderation. Part of the goal in writing about the questions at such length (and in particular not just presenting direct solutions) is because I think at this level it’s particularly easy to devote more time than needed to preparation, and use it poorly.
All these questions could be solved by able children. In fact, each could be solved by able children in less than an hour. You definitely count as an able child if you qualified or if your teacher allowed you to make an open entry! Others count too naturally. But most candidates won’t in fact solve all the questions, and many won’t solve any. And I think candidates often come up with the wrong reasons why they didn’t solve problems. “I didn’t know the right theorems” is very very rarely the reason. Olympiad problems have standard themes and recurring tropes, but the task is not to look at the problem and decide that it is an example of Olympiad technique #371. The task is actually to have as many ideas as possible, and eliminate the ones that don’t work as quickly as possible.
The best way to realise that an idea works is to solve the problem immediately. For the majority of occasions when we’re not lucky enough for that to happen, the second-best way to realise that an idea works is to see that it makes the problem look a bit more like something familiar. Conversely, the best way to realise that an idea doesn’t work is to observe that if it worked it would solve a stronger but false problem too. (Eg Fermat’s Last Theorem *does* have solutions over the reals…) The second-best way to realise that an idea doesn’t work is to have the confidence that you’ve tried it enough and you’ve only made the problem harder, or less familiar.
Both of these second-best ideas do require a bit of experience, but I will try to explain why none of the ideas I needed for various solutions this year required any knowledge beyond the school syllabus, some similarities to recent BMOs, and a small bit of creativity.
As usual, the caveat that these are not really solutions, and certainly not official solutions, but they are close enough to spoil the problems for anyone who hasn’t tried them by themselves already. Of course, the copyright for the problems is held by BMOS, and reproduced here with permission.
Question One
I wrote this question. Perhaps as a focal point of the renaissance of my interest in geometry, or at least my interest in teaching geometry, I have quite a lot to say about the problem, its solutions, its origin story, the use of directed angles, the non-use of coordinate methods and so on. In an ideal world I would write a book about this sort of thing, but for now, a long and separate post is the answer.
This will be available once I’ve successfully de-flooded my apartment.
Question Two
I also wrote this problem, though I feel it’s only fair to show the version I submitted to the BMO committee. All the credit for the magical statement that appears above lies with them. There is a less magical origin story as well, but hopefully with some interesting combinatorial probability, which is postponed until the end of this post.One quick observation is that in my version Joe / Hatter gets to keep going forever. As we shall see, all the business happens in the first N steps, but a priori one doesn’t know that, and in my version it forces you to strategise slightly differently for Neel / Alice. In the competition version, we know Alice is done as soon as she visits a place for a second time, but not in the original. So in the original we only have to consider ‘avoid one place’ rather than the multiple possibilities now of ‘avoid one place’ or ‘visit a place again’.
But I think the best idea is to get Alice to avoid one particular place $c\not\equiv 0$ whenever possible. At all times she has two possible options for where to go next, lets say $b_k+a_k, b_k-a_k$ in the language of the original statement. We lose nothing by assuming $-N/2 < a_k\le N/2$, and certainly it would be ridiculous for Joe / Hatter ever to choose $a_k=0$. The only time Alice’s strategy doesn’t work is when both of these are congruent to $c$, which implies $N\,|\, 2a_k$, and thus we must have $N= 2a_k$. In other words, Alice’s strategy will always work if N is odd.
I think it’s really worth noticing that the previous argument is weak. We certainly did not show that N must be odd for Alice to win. We showed that Alice can avoid a congruence class modulo an odd integer. We didn’t really need that odd integer to be N for this to work. In particular, if N has an odd factor p (say a prime), then the same argument works to show that we can avoid visiting any site with label congruent to 1 modulo p.
It’s actually very slightly more complicated. In the original argument, we didn’t need to use any property of $b_k$. But obviously here, if $b_k\equiv 1$ modulo p and $p\,|\,a_k$, then certainly $b_{k+1}\equiv 1$ modulo p. So we have to prove instead that Alice can ensure she never ‘visits 1 modulo p for the first time’. Which is fine, by the same argument.
So, we’ve shown that Neel / Alice wins if N is odd, or has an odd factor. The only values that remain are powers of 2. I should confess that I was genuinely a little surprised that Joe / Hatter wins in the power of 2 case. You can find a construction fairly easily for N=2 and N=4, but I suspected that might be a facet of small numbers. Why? Because it still felt we could avoid a particular site. In order for Alice’s strategy to fail, we have to end up exactly opposite the particular site at exactly the time when the next $a_k=N/2$, and so maybe we could try to avoid that second site as well, and so on backwards?
But that turned out to be a good example of something that got very complicated quite quickly with little insight. And, as discussed at the beginning, that’s often a sign in a competition problem that your idea isn’t so good. (Obviously, when composing a problem, that’s no guarantee at all. Sometimes things are true but no good ideas work.) So we want other ideas. Note that for N=4, the sequence (2,1,2) works for Joe / Hatter, because that forces Alice / Neel to visit either (0,2,1,3) or (0,2,3,1). In particular, this strategy gave Alice no control on the first step nor the last step, and the consequence is that we force her to visit the evens first, then transfer to an odd, and then force her to visit the other odd.
We might play around with N=8, or we might proceed directly to a general extension. If we have a Joe / Hatter strategy for N, then by doubling all the $a_k$s, we have a strategy for 2N which visits all the even sites in the first N steps. But then we can move to an odd site eg by taking $a_N=1$. Just as in the N=4 case, it doesn’t matter which odd site we start from, since if we again double all the $a_k$s, we will visit all the other odd sites. This gives us an inductive construction of a strategy for powers of two. To check it’s understood, the sequence for N=8 is (4,2,4,1,4,2,4).
Although we don’t use it, note that this strategy takes Alice on a tour of sites described by decreasing order of largest power of two dividing the label of the site.
Question Three
I have a theory that the average marks on Q1, Q2 and Q3 on this year’s paper will be in ascending order rather than, as one might expect, descending order. I think my theory will fail because it’s an unavoidable fact of life that in any exam, candidates normally start at the beginning, and don’t move to the middle until making earlier progress. But I think that’s the only reason my theory will fail.
Like kitchen cleanliness or children’s character flaws, it’s hard to compare one’s own problem proposals with others’ rationally. But I felt that, allowing for general levels of geometry non-preference, Q3 was more approachable than Q2, especially to any candidate who’d prepared by looking at some past papers.
I’m in no way a number theorist, but I know three or four common themes when one is asked to prove that a certain sequence contains no squares, or almost no squares. [3a]
• Number theoretic properties of the sequence of squares. Squares cannot be 3 modulo 4 for example. They also cannot be 2 modulo 4, and thus they also cannot be $2^{k-1}$ modulo $2^k$ for any even k. This first observation was essentially the body of most solutions to Q4 of BMO1 2016, among many others.
• Soft properties of the sequence of squares. The sequence of squares grows quadratically. Sometimes we can show a quadratic sequence will have no overlap with some other sequence for basic reasons. This is especially common if the second sequence is also quadratic or similar. For example, the expression $n^2+3n-4$ is typically not a square because
$(n+1)^2 = n^2+2n+1 < n^2 + 3n - 4 < n^2+4n+4 = (n+2)^2,$
• when n is large. In fact the right hand inequality is always true, and the left hand inequality is true for $n\ge 6$, which doesn’t leave too many cases to check (and n=5 does actually give a square). This type of argument has been quite common on BMO recently, directly on Q1 of BMO1 2011 and also Q3 of BMO1 2016. An example in a more abstract setting is Q3 of Balkan MO 2007, which I greatly enjoyed at the time…
• Number theoretic properties of the definition of a square. A square is the product of an integer with itself, and so if we want the product of two or more integers to be a square, then this imposes conditions on the shared factors of the two integers. I’ll cite some examples shortly.
• Huge theorems. Some old paper which I encountered as a child asked us to find all solutions to $x^2-1=2^y$. Or similar – I can’t find it now – but Q2 of BMO2 2006 is close enough to the sensible approach to the problem. I think it’s more helpful to think about this as proving that a particular sequence rarely includes powers of two than that a particular sequence rarely includes squares. But either way, one could in principle use the Catalan conjecture, which controls all non-trivial solutions to $a^p - b^q=1$. Fortunately, the Catalan conjecture was proved, by Mihailescu (readable blog about it), between the paper being set, and me attempting it a few years later. I’m being flippant. This is not a standard trope in solving these questions. For very obvious reasons. If it can be killed by direct reference to a known theorem, it won’t be set.
Anyway, those references (and more to follow) are to illuminate why I thought this question was not too hard. Indeed, I feel one can make substantial meta-progress in your head. The given information is interesting, but for the purpose of this question is just a black box. By subtracting the expression for m from the expression for 2m, we can derive an expression for the required sum. It’ll be a quartic in m, because the leading terms won’t cancel.
This leaves all three of the methods above very accessible. Unfortunately m=0 would be a square were it not excluded specifically, so a modular arithmetic approach is unlikely to work directly. Bounding between two quadratics is entirely plausible, as is factorising and comparing number theoretic properties of the factors. I thought the second one seemed more promising, but either way, having two potentially good ideas based only on recent BMO problems before even writing anything down is a good opening.
We do have to calculate the sum, and I make it $\frac{1}{4}m^2(5m+3)(3m+1)$. Now I’m not so sure how to bound this between two quadratics, because the leading coefficient is 15/4, which is not the square of a rational. But the factor analysis approach is definitely on.
Let’s review this generally. Throughout, suppose m,n are positive integers.
Claim 1: if mn is a square, then m and n are squares too.
Claim 2: if mn is a square, then m=n.
Both of these claims are false. However, a version of Claim 1 is true.
Claim 1′: if mn is a square, and m,n are coprime, then each is a square.
Even though this isn’t a named theorem, it is true, and well-known and can be used without proof. One way to prove it is to write m,n as products of primes, and show that since the primes are disjoint, the exponents must all be even. Most other methods will be equivalent to this, maybe with less notation.
What is good about Claim 1′ is that more complicated versions are true for for essentially similar reasons. For example
Claim 3: if mn is $6k^2$, and m,n are coprime, then either one is a square and the other is six times a square; or one is two times a square, and the other is three times a square.
Claim 4: if mn is a square, and the greatest common divisor (m,n) is either 5 or 1, then either each is a square, or each is five times a square.
I cited some examples of the other methods I proposed. Here are some examples of this sort of thing in recent BMOs:
• Q4 of BMO2 2016. Even the statement is suggestive. There are more complicated routes, but showing that $(2p-u-v)(2p+u+v)$ is a square is one way to proceed, and then Claim 4 directly applies after checking a gcd.
• Q2 of BMO1 2014 is similar, but it is much more explicit that this is the correct approach. Expose $p^2$ then use a (correct) version of Claim 2.
• Q1 of BMO2 2009. Show that a and b must each be a square times 41 for rationality reasons.
• Q6 of BMO1 2006. After sensible focused substitutions, obtain $3n^2=q(q-1)$. Rather than try to ‘solve’ this, extract the key properties along the lines of Claim 3, eliminate one of the cases by modular arithmetic, and return to the required statement.
• Q3 of BMO2 2010 requires the student to reproduce the essentials of the arguments above in the case of a particular degree six polynomial with a tractable factorisation, along with some mild square-sandwiching or bounding arguments as discussed earlier.
In conclusion, I’m trying to say that if I claim I am confident I can find all integers m such that $\frac14 m^2(5m+3)(3m+1)$ is a square, this is not based on complicated adult experience, but rather on recent problems at a similar sensible level. And I still don’t think it counts as Olympiad technique #371 – thinking about divisibility of factors is a good thing to do when talking about integers, and so it’s just a natural entry point into problems about squares. Plenty of problems might have this sort of thing as a starting point or an ending point.
For this problem we need a different ending point. To be brief, the factors (5m+3) and (3m+1) cannot both be squares because 5m+3 is never a square. So since the gcd of these factors is 1, 2 or 4, the only other option is that they are both squares times 2. And because -1 is not a square modulo 3, so 1 is not a (square times 2) modulo 3, and we are done. Note that this was a literal example of the first technique for proving something is not a square, proposed all the way back at the start of this section.
Footnotes
[3a] – some common themes for proving that sequences do include squares might be comparison with Pell’s Equations, or comparison with the explicit construction of solutions to Pythagoras’s equation.
Question Four
An example of an absorbing function is $f(x)=\lfloor x\rfloor$. One challenge is thinking of many other examples. This one is fine, but it’s true under replacing 2018 by 1 in the statement, and so it doesn’t really capture the richness of the situation.
Notation: the pre-image of a function is the language used to describe the inverse of a function which doesn’t have a uniquely-defined inverse. That is, if f is not injective, and multiple arguments have the output. We write $f^{-1}(y)=\{x: f(x)=y\}$. In particular, this is a set of values, not necessarily a single value. We also use $\mathbb{Z}$ to denote the integers. We can apply pre-images to sets as well. So for example $f^{-1}(\mathbb{Z})=\{x : f(x)\in \mathbb{Z}\}$.
This question is tricky, and I will be surprised to see many full solutions from the eligible candidates. It rewards the sort of organisation and clear-thinking that is easier said than done in a time-pressured contest environment. There are also many many possible things to consider, and so is particularly challenging in the short timeframe of BMO2 as opposed to, for example, appearing as the middle question on a 4.5 hour international-level paper.
At a meta-level we are being asked to confirm or deny the existence of absorbing functions where $f^{-1}(\mathbb{Z})$ is small in some sense, firstly when actually having finite size, secondly when, although infinite, being a small sort of infinite, namely spread out in a sparse, well-ordered way (you might say countable if familiar with that language). The general idea is presumably that it’s hard to be absorbing if the pre-image of the integers is small, and so it’s reasonable to assume that it’s too hard if this is finite; but perhaps not quite too hard if it’s merely countable. So (no, yes) is a sensible guess at the answer to the question, though (no, no) might also fit, maybe with a harder argument for the second no.
Ok, instead of trying a) or b), just play with the configuration. Let $A=f^{-1}(\mathbb{Z})$. We will use this frequently. In the picture below, f maps the real line on top to the real line below. If two reals get mapped to the same image, then whether or not the image is an integer, the whole (closed) interval bounded by the two reals also gets mapped to the same image. This is because f is weakly increasing.
This means that A consists of various intervals (which include single points). But in both a) and b) we know that A is ‘small’, and so it cannot contain any intervals of positive length. So in fact A is a set of separated real values. In the case of a) it’s a finite set.
Do we want to try and iterate this, and look at $f^{-1}(A)$? Well maybe, but we don’t know much about about pre-images of A, only about pre-images of $\mathbb{Z}$.
But note that the pre-image of the pre-image of the … of the pre-image [2017 times] of A must be the whole real line, so at some point, some value has a pre-image that is an interval. So if we’re guessing that the answer to b) is yes, then we need to give a construction.
$\mathbb{R} \stackrel{f}\longrightarrow ?? \stackrel{f}\longrightarrow\quad\ldots\quad \stackrel{f}\longrightarrow ??\stackrel{f}\longrightarrow A \stackrel{f}\longrightarrow f(A)\subset \mathbb{Z}.$
If you play around for a bit, it seems very unlikely to be absorbing if the integers don’t get mapped to the integers. You can try to prove this, but at the moment we’re just aiming for a construction, so let’s assume $f(\mathbb{Z})\subset \mathbb{Z}$. It would be convenient if f(n)=n for all $n\in \mathbb{Z}$, but we already know that this won’t work because then the pre-image of the pre-image of the… of $\mathbb{Z}$ is always $\mathbb{Z}$, but we need it to be $\mathbb{R}$.
The ideal situation would be if $A= \mathbb{Z}\cup \{\ldots, a'_{-1},a'_0,a'_1,\ldots\}$, where the pre-image of $\{\ldots, a'_{-1},a_0,a'_1,\ldots\}$ is pretty much everything.
Informally, we are specifically banned from mapping intervals directly onto an integer. So have an intermediate set, and try to map almost everything (except the integers and the set itself) onto that set, so and map that set into the integers.
At this point, you really just have to have the right idea and finish it. Many things will work, but this seems the easiest to me. Let the set A consist of the integers and the (integers plus 1/2). And for $x\in A$, f(x)=2x. This is what f looks like so far.
Here the black crosses are integers, and the purple crosses are (integers plus 1/2). But now we need to make as many reals as possible in the top row map to a purple cross (which is allowed, because purple crosses aren’t integers), but we need also to preserve the weakly increasing property. Fortunately, we can exactly do that. Each cross of either colour in the top row maps to a black cross in the middle row (ie an integer), so we can map the open interval between crosses in the top row to a purple cross in the middle row. As shown in red:
Note that this is consistent. The fact that I haven’t drawn in the red cones into the bottom row is only because I didn’t use the bottom row to motivate doing this. I’ve shown a consistent definition of f that maps all the reals onto the integers in two steps. If it’s an integer to begin with, that was great; if it was an (integer plus 1/2) to begin with then it becomes an integer in one step and stays an integer; and otherwise it first maps to an (integer plus 1/2), and then to an integer in the second step.
To check you’ve understood, try to write down a standalone definition of this function.
I’ve therefore solved part b) with the alternative condition $\ldots a_{-1} which isn’t exactly as required. It requires one small and simple idea to convert to a solution to the actual statement. See if you can find it yourself!
I think part a) is harder, not because the solution will look more complicated, but because there are so many potential partial results you could try to prove, because there are so many sets you could consider. To name a few: the image of f, the image of f intersected with $\mathbb{Z}$, the image of $\mathbb{Z}$, the 2018-composition image $f^{2018}(\mathbb{R})$, the 2018-composition image $f^{2018}(\mathbb{Z})$ and so on and so forth. You might have good insight into the wrong things.
For me, the crucial observation (which you can see from the figure in the b) construction) is that when composing an increasing function with itself, the ‘trajectories’ are either increasing or decreasing. That is, if $x\le f(x)$ (respectively, $x\ge f(x)$), then $x\le f(x)\le f^2(x)\le f^3(x)\le\ldots$ (respectively $x\ge f(x)\ge f^2(x)\ge \ldots$). Again, you can think of this as Olympiad technique #371 if you insist, but I don’t think that’s helpful. There are lots of things one could try to say here, and this turns out to be natural, true and useful, but you can’t know it’s useful until you play with it.
Anyway, we’re playing with part a), and we know that $f^k(x)$ is an integer for all large enough k, and that $f^{k+1}(x)$ is also an integer, so $f^k(x)$ is one of a finite set of integers because of the condition on A. But we’ve seen the sequence $x,f(x),f^2(x),\ldots$ is weakly increasing or weakly decreasing, and so if we also know it’s eventually bounded (because eventually it’s in this finite set) then it must eventually be constant. And this constant is one of the integers, say n. But unless we started from n, this means that f(n)=n, but also f(x)=n for some other real value x. And so exactly as at the very very beginning, that’s bad, because then the whole interval [x,n] gets mapped to n, which is a contradiction.
Question Two – Origin story
The origin story for Q2 started in a talk I heard by Renan Gross at Weizmann, who referenced some of the history of Scenery Reconstruction. Roughly speaking, we colour the integers (say with two colours), and then let loose a random walker, who tells us the sequence of colours she observes during her walk, but no other information about the walk itself.
How much information can we recover about the colouring? Obviously, the best we can hope for is to recover the colouring, up to translations and reflection, since for every possible random walk trajectory, the exact reflection is equally probable, and we are given no information about the starting point.
Since lots of the transitions between recoverable and unrecoverable depend on the periodicity of the colouring, a reasonable toy model is to do it on a cycle. Note that the Strong Law of Large Numbers tells us that we almost surely recover the number of black sites and white sites from an the infinite trajectory of the random walk. Of course it’s possible that there are only two black vertices, and they are adjacent, and the walker oscillates between them, thus seeing BBBBBB… But this is extremely unlikely. You could think of this in Bayesian terms as strongly increasing the prior on the whole cycle being black, but I think initially it’s best to do this as an infinite-time, SLLN problem not as finite time WLLN/CLT reweightings of anything.
But what more? It’s clear that the lengths of all black substrings should follow some mixed geometric-ish distribution, and this distribution will almost surely wash out as the empirical distribution in an SLLN sense. But it’s tricky to justify why such a mixed geometric-ish distribution should be uniquely determined by the lengths of black arcs in the cycle. But it does definitely feel like we should have enough information to reconstruct the colouring up to reflection/rotation with probability one. For example, analogously to the number of black vertices and the number of white vertices, we should be able to recover the number of adjacent black vertices, the number of adjacent white vertices, and the number of black-white adjacent vertices, and so on.
Anyway, this can be done, and it follows as a consequence of various authors’ work answering some more general conjectures of Benjamini and, separately, of den Hollander and Keane. Douglas Howard [DH] shows a handful of generalisations of this, as do Benjamini and Kesten [BK]. Most of this work is focused on sceneries on $\mathbb{Z}$, but periodic sceneries are often used as a basis, and of course, the only difference between periodic sceneries on $\mathbb{Z}$ and sceneries on the N-cycle are whether you know the period in advance. [BK] show that ‘almost all’ sceneries are distinguishable in a particular sense, in response to which Lindenstrauss [L99] exhibits a large family of sceneries which are not distinguishable. A readable but technical review is [ML].
So Renan’s talk was about the similar problem (and generalisations) on the hypercube [GG]. Rather than paraphrase the main differences badly, you can read his own excellent blog post about the work.
On the train back to Haifa from Rehovot, I was thinking a bit about the cycle case, and what happens if you generalise the random walk with varying jump lengths, or indeed introduce a demon walker, whose goal is to make it as hard as possible for the reviewer to deduce the colouring. One way this can certainly happen is if the walker can avoid visiting some particular site, as then how could one possibly deduce the colour of the never-visited site? And so we get to the statement posed.
References
[BK] – Benjamini, Kesten, 1996 – Distinguishing sceneries by observing the scenery along a random walk path
[dH] – den Hollander, 1988 – Mixing properties for random walk in random scenery
[DH] – Douglas Howard, 1996 – Detecting defects in periodic scenery by random walks on Z
[GG] – Grupel, Gross, 2017 – Indistinguishable sceneries on the Boolean hypercube
[L99] – Lindenstrauss, 1999 – Indistinguishable sceneries
[ML] – Matzinger, Lember, 2003 – Scenery reconstruction: an overview [link]
# BMO2 2017
The second round of the British Mathematical Olympiad was taken yesterday by about 100 invited participants, and about the same number of open entries. To qualify at all for this stage is worth celebrating. For the majority of the contestants, this might be the hardest exam they have ever sat, indeed relative to current age and experience it might well be the hardest exam they ever sit. And so I thought it was particularly worth writing about this year’s set of questions. Because at least in my opinion, the gap between finding every question very intimidating, and solving two or three is smaller, and more down to mindset, than one might suspect.
A key over-arching point at this kind of competition is the following: the questions have been carefully chosen, and carefully checked, to make sure they can be solved, checked and written up by school students in an hour. That’s not to say that many, or indeed any, will take that little time, but in principle it’s possible. That’s also not to say that there aren’t valid but more complicated routes to solutions, but in general people often spend a lot more time writing than they should, and a bit less time thinking. Small insights along the lines of “what’s really going on here?” often get you a lot further into the problem than complicated substitutions or lengthy calculations at this level.
So if some of the arguments below feel slick, then I guess that’s intentional. When I received the paper and had a glance in my office, I was only looking for slick observations, partly because I didn’t have time for detailed analysis, but also because I was confident that there were slick observations to be made, and I felt it was just my task to find them.
Anyway, these are the questions: (note that the copyright to these is held by BMOS – reproduced here with permission.)
Question One
I immediately tried the example where the perpendicular sides are parallel to the coordinate axes, and found that I could generate all multiples of 3 in this way. This seemed a plausible candidate for an answer, so I started trying to find a proof. I observed that if you have lots of integer points on one of the equal sides, you have lots of integer points on the corresponding side, and these exactly match up, and then you also have lots of integer points on the hypotenuse too. In my first example, these exactly matched up too, so I became confident I was right.
Then I tried another example ( (0,0), (1,1), (-1,1) ) which has four integer points, and could easily be generalised to give any multiple of four as the number of integer points. But I was convinced that this matching up approach had to be the right thing, and so I continued, trusting that I’d see where this alternate option came in during the proof.
Good setup makes life easy. The apex of the isosceles triangle might as well be at the origin, and then your other vertices can be $(m,n), (n,-m)$ or similar. Since integral points are preserved under the rotation which takes equal side to the other, the example I had does generalise, but we really need to start enumerating. The number of integer points on the side from (0,0) to (m,n) is G+1, where G is the greatest common divisor of m and n. But thinking about the hypotenuse as a vector (if you prefer, translate it so one vertex is at the origin), the number of integral points on this line segment must be $\mathrm{gcd}(m+n,m-n) +1$.
To me, this felt highly promising, because this is a classic trope in olympiad problem-setting. Even without this experience, we know that this gcd is equal to G if m and n have different parities (ie one odd, one even) and equal to 2G if m and n have the same parity.
So we’re done. Being careful not to double-count the vertices, we have 3G integral points if m and n have opposite parities, and 4G integral points if m and n have the same parity, which exactly fits the pair of examples I had. But remember that we already had a pair of constructions, so (after adjusting the hypothesis to allow the second example!) all we had to prove was that the number of integral points is divisible by at least one of 3 and 4. And we’ve just done that. Counting how many integers less than 2017 have this property can be done easily, checking that we don’t double-count multiples of 12, and that we don’t accidentally include or fail to include zero as appropriate, which would be an annoying way to perhaps lose a mark after totally finishing the real content of the problem.
Question Two
(Keen observers will note that this problem first appeared on the shortlist for IMO 2006 in Slovenia.)
As n increases, obviously $\frac{1}{n}$ decreases, but the bracketed expression increases. Which of these effects is more substantial? Well $\lfloor \frac{n}{k}\rfloor$ is the number of multiples of k which are at most n, and so as a function of n, this increases precisely when n is a multiple of k. So, we expect the bracketed expression to increase substantially when n has lots of factors, and to increase less substantially when n has few factors. An extreme case of the former might be when n is a large factorial, and certainly the extreme case of the latter is n a prime.
It felt easier to test a calculation on the prime case first, even though this was more likely to lead to an answer for b). When n moves from p-1 to p, the bracketed expression goes up by exactly two, as the first floor increases, and there is a new final term. So, we start with a fraction, and then increase the numerator by two and the denominator by one. Provided the fraction was initially greater than two, it stays greater than two, but decreases. This is the case here (for reasons we’ll come back to shortly), and so we’ve done part b). The answer is yes.
Then I tried to do the calculation when n was a large factorial, and I found I really needed to know the approximate value of the bracketed expression, at least for this value of n. And I do know that when n is large, the bracketed expression should be approximately $n\log n$, with a further correction of size at most n to account for the floor functions, but I wasn’t sure whether I was allowed to know that.
But surely you don’t need to engage with exactly how large the correction due to the floors is in various cases? This seemed potentially interesting (we are after all just counting factors), but also way too complicated. An even softer version of what I’ve just said is that the harmonic function (the sum of the first n reciprocals) diverges faster than n. So in fact we have all the ingredients we need. The bracketed expression grows faster than n, (you might want to formalise this by dividing by n before analysing the floors) and so the $a_n$s get arbitrarily large. Therefore, there must certainly be an infinite number of points of increase.
Remark: a few people have commented to me that part a) can be done easily by treating the case $n=2^k-1$, possibly after some combinatorial rewriting of the bracketed expression. I agree that this works fine. Possibly this is one of the best examples of the difference between doing a problem leisurely as a postgraduate, and actually under exam pressure as a teenager. Thinking about the softest possible properties of a sequence (roughly how quickly does it grow, in this case) is a natural first thing to do in all circumstances, especially if you are both lazy and used to talking about asymptotics, and certainly if you don’t have paper.
Question 3
I only drew a very rough diagram for this question, and it caused no problems whatsoever, because there aren’t really that many points, and it’s fairly easy to remember what their properties are. Even in the most crude diagram, we see R and S lie on AC and AD respectively, and so the conclusion about parallel lines is really about similarity of triangles ARS and ACD. This will follow either from some equal angles, or by comparing ratios of lengths.
Since angle bisectors by definition involve equal angles, the first attack point seems promising. But actually the ratios of lengths is better, provided we know the angle bisector theorem, which is literally about ratios of lengths in the angle bisector diagram. Indeed
$\frac{AR}{RC}=\frac{AQ}{CQ},\quad \frac{AS}{SD}=\frac{AP}{PD},$ (1)
and so it only remains to show that these quantities are in fact all equal. Note that there’s some anti-symmetry here – none of these expressions use B at all! We could for example note that AP/PD = BP/PC, from which
$\left(\frac{AS}{SD}\right)^2 = \frac{AP.BP}{PC.PD},$ (2)
and correspondingly for R and Q, and work with symmetric expressions. I was pretty sure that there was a fairly well-known result that in a cyclic quadrilateral, where P is the intersection of the diagonals
$\frac{AP}{PC} = \frac{AD.AB}{DC.BC},$ (3)
(I was initially wondering whether there was a square on the LHS, but an example diagram makes the given expression look correct.)
There will be a corresponding result for Q, and then we would be almost done by decomposing (2) slightly differently, and once we’d proved (3) of course. But doing this will turn out to be much longer than necessary. The important message from (3) is that in a very simple diagram (only five points), we have a result which is true, but which is not just similar triangles. There are two pairs of similar triangles in the diagram, but they aren’t in the right places to get this result. What you do have is some pairs of triangles with one pair of equal angles, and one pair of complementary angles (that is, $\theta$ in one, and $180-\theta$ in the other). This is a glaring invitation to use the sine rule, since the sines of complementary angles are equal.
But, this is also the easiest way to prove the angle bisector theorem. So maybe we should just try this approach directly on the original ratio-of-lengths statement that we decided at (1) was enough, namely $\frac{AQ}{CQ}=\frac{AP}{PD}$. And actually it drops out rapidly. Using natural but informal language referencing my diagram
$\frac{AP}{PD} = \frac{\sin(\mathrm{Green})}{\sin(\mathrm{Pink})},\quad\text{and}\quad \frac{AQ}{CQ}= \frac{\sin(\mathrm{Green})}{\sin(180-\mathrm{Pink})}$
and we are done. But whatever your motivation for moving to the sine rule, this is crucial. Unless you construct quite a few extra cyclic quadrilaterals, doing this with similar triangles and circle theorems alone is going to be challenging.
Remark: If you haven’t seen the angle bisector theorem before, that’s fine. Both equalities in (1) are a direct statement of the theorem. It’s not an intimidating statement, and it would be a good exercise to prove either of these statements in (1). Some of the methods just described will be useful here too!
Question 4
You might as well start by playing around with methodical strategies. My first try involved testing 000, 111, … , 999. After this, you know which integers appear as digits. Note that at this stage, it’s not the same as the original game with only three digits, because we can test using digits which we know are wrong, so that answers are less ambiguous. If the three digits are different, we can identify the first digit in two tests, and then the second in a further test, and so identify the third by elimination. If only two integers appear as digits, we identify each digit separately, again in three tests overall. If only one integer appears, then we are already done. So this is thirteen tests, and I was fairly convinced that this wasn’t optimal, partly because it felt like testing 999 was a waste. But even with lots of case tries I couldn’t do better. So I figured I’d try to prove some bound, and see where I got.
A crucial observation is the following: when you run a test, the outcome eliminates some possibilities. One of the outcomes eliminates at least half the codes, and the other outcome eliminates at most half the codes. So, imagining you get unlucky every time, after k tests, you might have at least $1000\times 2^{-k}$ possible codes remaining. From this, we know that we need at least 9 tests.
For this bound to be tight, each test really does need to split the options roughly in two. But this certainly isn’t the case for the first test, which splits the options into 729 (no digit agreements) and 271 (at least one agreement). Suppose the first test reduces it to 729 options, then by the same argument as above, we still need 9 tests. We now know we need at least 10 tests, and so the original guess of 13 is starting to come back into play.
We now have to make a meta-mathematical decision about what to do next. We could look at how many options might be left after the second test, which has quite a large number of cases (depending on how much overlap there is between the first test number and the second test number). It’s probably going to be less than 512 in at least one of the cases, so this won’t get us to a bound of 11 unless we then consider the third test too. This feels like a poor route to take for now, as the tree of options has branching at rate 3 (or 4 if you count obviously silly things) per turn, so gets unwieldy quickly. Another thought is that this power of two argument is strong when the set of remaining options is small, so it’s easier for a test to split the field roughly in two.
Now go back to our proposed original strategy. When does the strategy work faster than planned? It works faster than planned if we find all the digits early (eg if they are all 6 or less). So the worst case scenario is if we find the correct set of digits fairly late. But the fact that we were choosing numbers of the form aaa is irrelevant, as the digits are independent (consider adding 3 to the middle digit modulo 10 at all times in any strategy – it still works!).
This is key. For $k\le 9$, after k tests, it is possible that we fail every test, which means that at least $(10-k)$ options remain for each digit, and so at least $(10-k)^3$ options in total. [(*) Note that it might actually be even worse if eg we get a ‘close’ on exactly one test, but we are aiming for a lower bound, so at this stage considering an outcome sequence which is tractable is more important than getting the absolute worst case outcome sequence if it’s more complicated.] Bearing in mind that I’d already tried finishing from the case of reduction to three possibilities, and I’d tried hard to sneak through in one fewer test, and failed, it seemed sensible to try k=7.
After 7 tests, we have at least 27 options remaining, which by the powers-of-two argument requires at least 5 further tests to separate. So 12 in total, which is annoying, because now I need to decide whether this is really the answer and come up a better construction, or enhance the proof.
Clearly though, before aiming for either of these things, I should actually try some other values of k, since this takes basically no time at all. And k=6 leaves 64 options, from which the power of two argument is tight; and k=5 leaves 125, which is less tight. So attacking k=6 is clearly best. We just need to check that the 7th move can’t split the options exactly into 32 + 32. Note that in the example, where we try previously unseen digits in every position, we split into 27 + 37 [think about (*) again now!]. Obviously, if we have more than four options left for any digit, we are done as then we have strictly more than 4x4x4=64 options. So it remains to check the counts if we try previously unseen digits in zero, one or two positions. Zero is silly (gives no information), and one and two can be calculated, and don’t give 32 + 32.
So this is a slightly fiddly end to the solution, and relies upon having good control over what you’re trying to do, and what tools you currently have. The trick to solving this is resisting calculations and case divisions that are very complicated. In the argument I’ve proposed, the only real case division is right at the end, by which point we are just doing an enumeration in a handful of cases, which is not really that bad.
# Lagrange multipliers Part One: A much simpler setting
I am currently in northern Hungary for our annual winter school for some of the strongest young school-aged mathematicians in the UK and Hungary. We’ve had a mixture of lectures, problem-solving sessions and the chance to enjoy a more authentic version of winter than is currently on offer in balmy Oxford.
One of my favourite aspects of this event is the chance it affords for the students and the staff to see a slightly different mathematical culture. It goes without saying that Hungary has a deep tradition in mathematics, and the roots start at school. The British students observe fairly rapidly that their counterparts have a much richer diet of geometry, and methods in combinatorics at school, which is certainly an excellent grounding for use in maths competitions. By contrast, our familiarity with calculus is substantially more developed – by the time students who study further maths leave school, they can differentiate almost anything.
But the prevailing attitude in olympiad circles is that calculus is unrigorous and hence illegal method. The more developed summary is that calculus methods are hard, or at least technical. This is true, and no-one wants to spoil a measured development of analysis from first principles, but since some of the British students asked, it seemed worth giving a short exposition of why calculus can be made rigorous. They are mainly interested in the multivariate case, and the underlying problem is that the approach suggested by the curriculum doesn’t generalise well at all to the multivariate setting. Because it’s much easier to imagine functions of one variable, we’ll develop the machinery of the ideas in this setting in this post first.
Finding minima – the A-level approach
Whether in an applied or an abstract setting, the main use of calculus at school is to find where functions attain their maximum or minimum. The method can be summarised quickly: differentiate, find where the derivative is zero, and check the second-derivative at that value to determine that the stationary point has the form we want.
Finding maxima and finding minima are a symmetric problem, so throughout, we talk about finding minima. It’s instructive to think of some functions where the approach outlined above fails.
In the top left, there clearly is a minimum, but the function is not differentiable at the relevant point. We can probably assert this without defining differentiability formally: there isn’t a well-defined local tangent at the minimum, so we can’t specify the gradient of the tangent. In the top right, there’s a jump, so depending on the value the function takes at the jump point, maybe there is a minimum. But in either case, the derivative doesn’t exist at the jump point, so our calculus approach will fail.
In the middle left, calculus will tell us that the stationary point in the middle is a ‘minimum’, but it isn’t the minimal value taken by the function. Indeed the function doesn’t have a minimum, because it seems to go off to $-\infty$ in both directions. In the middle right, the asymptote provides a lower bound on the values taken by the function, but this bound is never actually achieved. Indeed, we wouldn’t make any progress by calculus, since there are no stationary points.
At the bottom, the functions are only defined on some interval. In both cases, the minimal value is attained at one of the endpoints of the interval, even though the second function has a point which calculus would identify as a minimum.
The underlying problem in any calculus argument is that the derivative, if it exists, only tells us about the local behaviour of the function. At best, it tells us that a point is a local minimum. This is at least a necessary condition to be a global minimum, which is what we actually care about. But this is a change of emphasis from the A-level approach, for which having zero derivative and appropriately-signed second-derivative is treated as a sufficient condition to be a global minimum.
Fortunately, the A-level approach is actually valid. It can be shown that if a function is differentiable everywhere, and it only has one stationary point, where the second-derivative exists and is positive, then this is in fact the global minimum. The first problem is that this is really quite challenging to show – since in general the derivative might not be continuous, although it might have many of the useful properties of a continuous function. Showing all of this really does require setting everything up carefully with proper definitions. The second problem is that this approach does not generalise well to multivariate settings.
Finding minima – an alternative recipe
What we do is narrow down the properties which the global minimum must satisfy. Here are some options:
0) There is no global minimum. For example, the functions pictured in the middle row satisfy this.
Otherwise, say the global minimum is attained at x. It doesn’t matter if it is attained at several points. At least one of the following options must apply to each such x.
1) $f'(x)=0$,
2) $f'(x)$ is not defined,
3) x lies on the boundary of the domain where f is defined.
We’ll come back to why this is true. But with this decomposition, the key to identifying a global minimum via calculus is to eliminate options 0), 2) and 3). Hopefully we can eliminate 2) immediately. If we know we can differentiate our function everywhere, then 2) couldn’t possibly hold for any value of x. Sometimes we will be thinking about functions defined everywhere, in which case 3) won’t matter. Even if our function is defined on some interval, this only means we have to check two extra values, and this isn’t such hard work.
It’s worth emphasising why if x is a local minimum not on the boundary and f'(x) exists, then f'(x)=0. We show that if $f'(x)\ne 0$, then x can’t be a local minimum. Suppose f'(x)>0. Then both the formal definition of derivative, and the geometric interpretation in terms of the gradient of a tangent which locally approximates the function, give that, when h is small,
$f(x-h) = f(x)-h f'(x) +o(h),$
where this ‘little o’ notation indicates that for small enough h, the final term is much smaller than the second term. So for small enough h, $f(x-h), and so we don’t have a local minimum.
The key is eliminating option 0). Once we know that there definitely is a global minimum, we are in a good position to identify it using calculus and a bit of quick checking. But how would we eliminate option 0)?
Existence of global minima
This is the point where I’m in greatest danger of spoiling first-year undergraduate course content, so I’ll be careful.
As we saw in the middle row, when functions are defined on the whole real line, there’s the danger that they can diverge to $\pm \infty$, or approach some bounding value while never actually attaining it. So life gets much easier if you work with functions defined on a closed interval. We also saw what can go wrong if there are jumps, so we will assume the function is continuous, meaning that it has no jumps, or that as y gets close to x, f(y) gets close to f(x). If you think a function can be differentiated everywhere, then it is continuous, because we’ve seen that once a function has a jump (see caveat 2) then it certainly isn’t possible to define the derivative at the jump point.
It’s a true result that a continuous function defined on a closed interval is bounded and attains its bounds. Suppose such a function takes arbitrarily large values. The main idea is that if the function takes arbitrarily large values throughout the interval, then because the interval is finite it also takes arbitrarily large values near some point, which will make it hard to be continuous at that point. You can apply a similar argument to show that the function can’t approach a threshold without attaining it somewhere. So how do you prove that this point exists? Well, you probably need to set up some formal definitions of all the properties under discussion, and manipulate them carefully. Which is fine. If you’re still at school, then you can either enjoy thinking about this yourself, or wait until analysis courses at university.
My personal opinion is that this is almost as intuitive as the assertion that if a continuous function takes both positive and negative values, then it has a zero somewhere in between. I feel if you’re happy citing the latter, then you can also cite the behaviour of continuous functions on closed intervals.
Caveat 2) It’s not true to say that if a function doesn’t have jumps then it is continuous. There are other kinds of discontinuity, but in most contexts these are worse than having a jump, so it’s not disastrous in most circumstances to have this as your prime model of non-continuity.
Worked example
Question 1 of this year’s BMO2 was a geometric inequality. I’ve chosen to look at this partly because it’s the first question I’ve set to make it onto BMO, but mainly because it’s quite hard to find olympiad problems which come down to inequalities in a single variable.
Anyway, there are many ways to parameterise and reparameterise the problem, but one method reduces, after some sensible application of Pythagoras, to showing
$f(x)=x+ \frac{1}{4x} + \frac{1}{4x+\frac{1}{x}+4}\ge \frac{9}{8},$ (*)
for all positive x.
There are simpler ways to address this than calculus, especially if you establish or guess that the equality case is x=1/2. Adding one to both sides is probably a useful start.
But if you did want to use calculus, you should argue as follows. (*) is certainly true when $x\ge \frac{9}{8}$ and also when $x\le \frac{2}{9}$. The function f(x) is continuous, and so on the interval $[\frac{2}{9},\frac{9}{8}]$ it has a minimum somewhere. We can differentiate, and fortunately the derivative factorises (this might be a clue that there’s probably a better method…) as
$(1-\frac{1}{4x^2}) \left[ 1 - \frac{4}{(4x+\frac{1}{x}+4)^2} \right].$
If x is positive, the second bracket can’t be zero, so the only stationary point is found at x=1/2. We can easily check that $f(\frac12)=\frac98$, and we have already seen that $f(\frac29),f(\frac98)>\frac98$. We know f attains its minimum on $[\frac29,\frac98]$, and so this minimal value must be $\frac98$, as we want.
Overall, the moral of this approach is that even if we know how to turn the handle both for the calculation, and for the justification, it probably would be easier to use a softer approach if possible.
Next stage
For the next stage, we assess how much of this carries across to the multivariate setting, including Lagrange multipliers to find minima of a function subject to a constraint. | 12,586 | 54,298 | {"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": 92, "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-2018-22 | latest | en | 0.97339 |
http://msms.ehe.osu.edu/tag/numbers/ | 1,531,798,189,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589557.39/warc/CC-MAIN-20180717031623-20180717051623-00461.warc.gz | 252,326,218 | 11,697 | Factors
Factors and their multiples are so important to students’ work with fractions and number theory. These concepts come under the National Council of Teachers of Mathematics (NCTM) Number and Operations Standard for the middle grades. The resources here are all hands-on, if only virtually. Important to their success is the classroom talk that the resources generate.
The Factor Game
This is more than a game; it is a full lesson plan, complete with handouts, including the game board, and questions for discussion. Students practice finding factors and then analyze the winning strategies. It’s this analysis that leads them to talk about prime, composite, abundant, deficient, and perfect numbers. The plan is downloadable and printable. (See the online version below.)
The Factor Game (i-Math Investigations)
This is the same game but online and interactive. One advantage here is that a single student can play against the computer. What is missing, however, is the winning strategy analysis, so rich in discussion. If you decide to use this version, you can simply borrow the discussion questions from the pdf version above.
Coloring Multiples in Pascal’s Triangle
Good practice in finding multiples of small factors! Students roll a number cube, then color in the multiples of that number in Pascal’s triangle. As they click on all the multiples of the number rolled, they also practice multiplication and observe the surprising patterns that form in the triangle. | 288 | 1,484 | {"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-2018-30 | latest | en | 0.920218 |
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The absolute value of a number N represented as |n|, is its distance from 0 on the number line Absolute Value any natural number or 0 whole number every positive rational number and its negative pair opposites numbers that cannot be expressed as terminating or repeating numbers irrational numbers the ratio of the number of favorable outcomes for an event to the number of possible outcomes of the event probability the set {1,2,3...} natural number the ratio that compares the number of ways an event can occur (sucess) to the number of ways the event cannot occur (failure) odds a number line labeled with a scale to include all the data with an X placed above a data point each time it occurs line plot a number whose square root is a rational number perfect square How often a piece of data occurs Frequency numbers or pieces of data that can represent the whole set of data measures of central tendency outcomes for which the probability of each occuring is equal Equally Likely a display of data using certain digits as "stem" and the remaininf digit(s) as "leaves" stem & leaf plot the set {-2,-1,0,1,2} integers a number -C is square root of a if c to the 2nd = a square root real numbers consist of the rational numbers & the irrational numbers real number If the sum of 2 numbers or expressions is 0 they are additive inverses of each other Addivtive inverses a single event simple event a number that can be expressed as the ratio of two integers in the form a/b where b doesnt = 0 rational number the nonnegative square root of a number perfect square root set of all possible outcomes of an experiment sample space | 490 | 2,153 | {"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} | 3.5625 | 4 | CC-MAIN-2016-44 | longest | en | 0.877973 |
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HomeTren&dThe Power of a^3 + b^3: Unleashing the Potential of Cubic Expressions
# The Power of a^3 + b^3: Unleashing the Potential of Cubic Expressions
When it comes to mathematical expressions, few are as intriguing and powerful as the cubic expression a^3 + b^3. This seemingly simple equation holds within it a world of possibilities and applications that can be explored and harnessed. In this article, we will delve into the depths of a^3 + b^3, uncovering its significance, properties, and real-world applications. So, let’s embark on this mathematical journey and unlock the potential of cubic expressions!
## Understanding Cubic Expressions
Before we dive into the specifics of a^3 + b^3, let’s first understand what cubic expressions are. A cubic expression is a mathematical expression that involves variables raised to the power of three. In the case of a^3 + b^3, both ‘a’ and ‘b’ are variables raised to the power of three. This expression can also be written as (a + b)(a^2 – ab + b^2), which is known as the factorized form of a^3 + b^3.
## The Significance of a^3 + b^3
At first glance, a^3 + b^3 may appear to be just another mathematical expression. However, its significance lies in its ability to represent and solve a wide range of problems across various disciplines. Let’s explore some of the key areas where a^3 + b^3 finds its applications:
### 1. Algebraic Manipulation
In algebra, a^3 + b^3 plays a crucial role in expanding and simplifying expressions. By utilizing the factorized form (a + b)(a^2 – ab + b^2), we can simplify complex expressions and solve equations more efficiently. This manipulation technique is particularly useful in solving polynomial equations and factorizing higher-degree expressions.
### 2. Number Theory
Cubic expressions have deep connections with number theory, the branch of mathematics that deals with properties and relationships of numbers. By exploring the properties of a^3 + b^3, mathematicians have made significant contributions to number theory, including the study of prime numbers, divisibility, and modular arithmetic.
### 3. Geometry
Surprisingly, cubic expressions also find their applications in geometry. By representing geometric figures and shapes using algebraic equations, we can utilize a^3 + b^3 to solve geometric problems. For example, the volume of a cube can be expressed as a^3, where ‘a’ represents the length of its sides. By manipulating this expression, we can calculate the volume of complex shapes composed of multiple cubes.
### 4. Physics and Engineering
The power of a^3 + b^3 extends beyond the realm of pure mathematics and finds its applications in physics and engineering. In these fields, cubic expressions are often used to model and solve real-world problems. For instance, in fluid dynamics, the Navier-Stokes equations involve cubic terms that can be represented using a^3 + b^3. By solving these equations, engineers can predict fluid flow patterns and optimize designs.
## Real-World Examples
To further illustrate the practical applications of a^3 + b^3, let’s explore some real-world examples where cubic expressions play a crucial role:
### 1. Architecture and Construction
In architecture and construction, cubic expressions are used to calculate the volume and dimensions of structures. By utilizing a^3 + b^3, architects and engineers can determine the volume of irregularly shaped buildings, enabling them to optimize space utilization and design efficient structures.
### 2. Financial Modeling
Financial analysts and economists often employ cubic expressions to model and predict market trends. By analyzing historical data and utilizing a^3 + b^3, they can develop mathematical models that forecast stock prices, interest rates, and economic indicators. These models play a crucial role in making informed investment decisions and managing financial risks.
### 3. Computer Graphics
In the field of computer graphics, cubic expressions are used to create realistic and visually appealing 3D models. By manipulating a^3 + b^3, graphic designers can generate complex shapes and textures, bringing virtual worlds to life in movies, video games, and virtual reality experiences.
## FAQs
### Q1: What is the significance of the factorized form (a + b)(a^2 – ab + b^2) in a^3 + b^3?
The factorized form allows us to simplify and manipulate cubic expressions more efficiently. By factoring out the common term (a + b), we can simplify complex expressions and solve equations more easily.
### Q2: Can cubic expressions be used to solve real-world problems?
Absolutely! Cubic expressions have a wide range of applications in various fields, including architecture, finance, physics, and computer graphics. By utilizing a^3 + b^3, we can model and solve real-world problems more effectively.
### Q3: Are there any limitations to using cubic expressions?
While cubic expressions are powerful tools, they do have their limitations. In some cases, solving cubic equations can be challenging, requiring advanced mathematical techniques. Additionally, cubic expressions may not always provide an exact solution, and numerical approximations may be necessary.
### Q4: Can cubic expressions be extended to higher powers?
Yes, cubic expressions can be extended to higher powers, such as a^4 + b^4, a^5 + b^5, and so on. These higher-degree expressions have their own unique properties and applications, but they build upon the foundation established by cubic expressions.
### Q5: How can I further explore the applications of cubic expressions?
To delve deeper into the applications of cubic expressions, consider studying advanced mathematics courses, such as algebra, number theory, and calculus. These courses will provide you with the necessary tools and knowledge to explore the vast world of mathematical expressions.
## Summary
In conclusion, the cubic expression a^3 + b^3 holds immense power and significance in the realm of mathematics and beyond. Its ability to represent and solve a wide range of problems makes it a valuable tool in various fields, including algebra, number theory, geometry, physics, and engineering. By understanding and harnessing the potential of a^3 + b^3, we can unlock new insights, solve complex problems, and pave the way for further mathematical discoveries. So, embrace the power of cubic expressions and let them guide you on your mathematical journey! | 1,410 | 6,743 | {"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-2024-18 | latest | en | 0.909542 |
https://www.mathhomeworkanswers.org/68399/find-all-points-on-the-y-axis-that-are-6-units-from-the-point-4-3?show=275306 | 1,597,479,571,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740733.1/warc/CC-MAIN-20200815065105-20200815095105-00139.warc.gz | 731,051,975 | 16,975 | Find all points on the y-axis that are 6 units from the point (4, -3).
The circumference of the circle (x-4)²+(y+3)²=36 is 6 units from the point (4,-3).
The y-axis is the line x=0, so it intersects the circle at (y+3)²=36-16=20. y+3=±√20=±2√5. So y=-3±2√5. That is, y=-3+2√5=1.47 and -3-2√5=-7.47 approx.
by Top Rated User (782k points) | 141 | 340 | {"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} | 3.6875 | 4 | CC-MAIN-2020-34 | latest | en | 0.767643 |
https://deathtoall.info/dtft-tutorial-89/ | 1,603,897,903,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107898577.79/warc/CC-MAIN-20201028132718-20201028162718-00185.warc.gz | 284,529,029 | 26,670 | # DTFT TUTORIAL PDF
As the ROC includes the unit circle, its DTFT exists and the same result is obtained by the substitution of. There are two advantages of transform over DTFT . DTFT, DFT Tutorial added – I have added Chapter 5 which covers DFT and DTFT and a little bit about FFT. The tutorial has most of the Matlab. The best way to understand the DTFT is how it relates to the DFT. To start, imagine that you acquire an N sample signal, and want to find its frequency spectrum.
Author: Zulukora Gujind Country: Seychelles Language: English (Spanish) Genre: Love Published (Last): 27 July 2011 Pages: 458 PDF File Size: 10.5 Mb ePub File Size: 17.60 Mb ISBN: 928-4-72634-963-2 Downloads: 5911 Price: Free* [*Free Regsitration Required] Uploader: Gujin
## Fourier Transforms
While the DFT could dtftt be used for this calculation, it would only provide an equation for samples of the frequency response, not the entire curve. Its main use is in theoretical problems as an alternative to the DFT.
By using the DFT, the signal can be decomposed into sine and cosine waves, with frequencies equally spaced between zero and one-half of the sampling rate. As you recall, this action in the DFT is related to the frequency spectrum being defined as a spectral densityi.
Your laser printer will thank you! Suppose you start with some time domain signal. How to order your own hardcover copy Wouldn’t you rather have a bound book instead of loose pages? To start, imagine that you acquire an N tdft signal, and want to find its frequency spectrum. As N approaches infinity, the time domain becomes aperiodicand dyft frequency domain becomes a continuous signal. In other cases, the impulse response might be know as an equationsuch as a sinc function or an exponentially decaying sinusoid.
ALAIN BADIOU SAINT PAUL THE FOUNDATION OF UNIVERSALISM PDF
Some authors place these terms in front of the synthesis equation, while others place them in front of the analysis equation.
This provides the frequency spectrum as another array of numbersequally spaced between 0 and 0. For instance, suppose you want to find the frequency response of a system from its impulse response. The DTFT is tutotial here to mathematically calculate the frequency domain as another equationspecifying the entire continuous curve between 0 and 0.
As discussed in Chapter 8, frequency is represented in the DFT’s frequency domain by one of tuutorial variables: First, the time domain signal, x [ n ], is still discrete, and therefore is represented by brackets. Digital Filters Match 2: There are many subtle details in these relations. When the spectrum becomes continuous, the special treatment of the end points disappear.
Table of contents 1: This is the DTFT, the Fourier transform that relates an aperiodicdiscrete signal, with a periodiccontinuous frequency spectrum.
ALL I EVER WANTED KRISTAN HIGGINS PDF
Filter Comparison Match 1: Since the DTFT involves infinite summations and integrals, it cannot be calculated with a digital computer.
### Discrete-time Fourier transform (DTFT) ยป Steve on Image Processing – MATLAB & Simulink
If the impulse response is known as an array of numberssuch as might be obtained from an experimental measurement or computer simulation, a DFT program is run on a computer.
After taking the Fourier transform, and then the Inverse Fourier transform, you want to end up with what you started.
The Digital Signal Processor Market Since the frequency domain is continuous, the synthesis equation must be written as an integral, rather than a summation.
This is not necessary with the DTFT. Program Language Execution Speed: Download this chapter in PDF format Chapter As discussed in the last chapter, padding the time domain signal with zeros makes the period of the time domain longeras well as making the spacing between samples tutodial the frequency domain narrower.
Neural Networks and more! | 852 | 3,922 | {"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-2020-45 | latest | en | 0.916555 |
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A238939 Powers of 3 without the digit '0' in their decimal expansion. 15
1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 177147, 531441, 1594323, 4782969, 1162261467, 94143178827, 282429536481, 2541865828329, 7625597484987, 22876792454961, 617673396283947, 16677181699666569, 278128389443693511257285776231761 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Conjectured to be finite and complete. See the OEIS wiki page for further information, references and links. LINKS M. F. Hasler, Zeroless powers, OEIS wiki, Mar 07 2014 FORMULA a(n)=3^A030700(n). PROG (PARI) for(n=0, 99, vecmin(digits(3^n))&& print1(3^n", ")) CROSSREFS For the zeroless numbers (powers x^n), see A238938, A238939, A238940, A195948, A238936, A195908, A195946, A195945, A195942, A195943, A103662. For the corresponding exponents, see A007377, A008839, A030700, A030701, A008839, A030702, A030703, A030704, A030705, A030706, A195944. For other related sequences, see A052382, A027870, A102483, A103663. Sequence in context: A000244 A133494 A050733 * A079846 A067500 A273898 Adjacent sequences: A238936 A238937 A238938 * A238940 A238941 A238942 KEYWORD nonn,fini,base AUTHOR M. F. Hasler, Mar 07 2014 STATUS approved
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Last modified October 21 14:38 EDT 2019. Contains 328301 sequences. (Running on oeis4.) | 567 | 1,686 | {"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-2019-43 | latest | en | 0.534097 |
https://physics.stackexchange.com/questions/402133/when-tightening-a-guitar-string-what-happens-to-velocity-wavelength-and-frequ/402164 | 1,718,314,925,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861488.82/warc/CC-MAIN-20240613190234-20240613220234-00514.warc.gz | 400,494,692 | 36,044 | When tightening a guitar string, what happens to velocity, wavelength, and frequency? [closed]
I'm thinking about the following physics problem. (I'm a freshman in high school.)
A guitarist plucks a string, producing string wave #1 (fundamental vibration) in the string. He then tightens its tension and plucks again, producing string wave #2 (fundamental vibration). Compare the following factors for string wave #1 vs string wave #2:
• velocity
• wavelength
• frequency
• which string wave creates the sound wave that makes your ear drum vibrate the fastest?
I think the velocity should increase after tightening, because the velocity of a wave on a string is $\sqrt{T/\mu}$, and tightening increases $T$.
I think the frequency should increase, too, but that's only based on empirical evidence: I know that when someone tightens a guitar string and then plucks again, the pitch is higher. Is there any way to justify my reasoning besides appealing to this empirical evidence?
I'm not sure what to say about wavelength. I know $v=\lambda f$, but if $v$ and $f$ increase, I can't conclude anything about $\lambda$. If $v$ and $f$ increase by the same factor, then $\lambda$ stays constant; if they increase by different factors, then $\lambda$ could increase or decrease. So I don't know what to think about the wavelength.
I think the last question is asking about which sound wave has a higher frequency. Since the frequency of a string wave is the same as the frequency of the induced sound wave, the second string wave should create the sound wave that makes my ear drum vibrate faster, right?
• You can conclude something about $\lambda$ by looking at the string: has its length changed?
– user107153
Commented Apr 26, 2018 at 12:15
• @tfb: what about justifying my claim about the frequency....... Commented Apr 26, 2018 at 17:15
• Well, you know what happens to $v$, and $\lambda$, and you have an equation relating them to $f$...
– user107153
Commented Apr 26, 2018 at 18:04 | 469 | 1,990 | {"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} | 3.65625 | 4 | CC-MAIN-2024-26 | latest | en | 0.942922 |
https://skoolkit.ca/disassemblies/back_to_skool/hex/asm/F55F.html | 1,695,287,783,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233505362.29/warc/CC-MAIN-20230921073711-20230921103711-00496.warc.gz | 593,023,932 | 2,838 | Routines
Prev: F54A Up: Map Next: F5B7
F55F: Make MR WACKER find and expel ERIC
The address of this routine is found in the command list at F530, the address of which is placed into bytes 0x1B and 0x1C of MR WACKER's buffer by the routine at F532.
Input
H 0xC8 (MR WACKER)
F55F LD L,\$1D Set bit 7 of byte 0x1D of MR WACKER's buffer, making him run
F561 SET 7,(HL)
F563 LD A,\$FF Set the MSB of the lesson clock at 7FE3 to 0xFF (this lesson will not end until ERIC's expelled)
F565 LD (\$7FE4),A
F568 CALL \$6558 Move MR WACKER one step closer to ERIC (if he's not yet close enough)
F56B LD HL,\$C800 Point HL at byte 0x00 of MR WACKER's buffer
F56E BIT 0,(HL) Is MR WACKER now midstride?
F570 RET NZ Return if so
F571 INC L L=0x01
F572 LD DE,(\$D201) E=ERIC's x-coordinate, D=ERIC's y-coordinate
F576 LD A,(HL) A=MR WACKER's x-coordinate
F577 SUB E Is MR WACKER within 3 x-coordinates of ERIC?
F57A CP \$07
F57C RET NC Return if not
F57D INC L L=0x02
F57E LD A,(HL) A=MR WACKER's y-coordinate
F57F SUB D Is MR WACKER within 3 y-coordinates of ERIC?
F582 CP \$07
F584 RET NC Return if not
MR WACKER has found ERIC.
F585 LD HL,\$7FFB 7FFB holds ERIC's status flags
F588 LD (HL),\$40 Set bit 6: MR WACKER is expelling ERIC (who is now paralysed)
F58A LD L,\$ED HL=7FED (ERIC's other status flags)
F58C BIT 5,(HL) Bit 5 is set if ERIC jumped out of the top-floor window
F58E LD H,\$C8 0xC8=MR WACKER
F590 LD E,\$63 Message 0x63: YOU HAVE 10000 LINES...
F592 JR Z,\$F595 Jump if ERIC didn't jump out of the top-floor window (i.e. he has 10000 lines)
F594 INC E E=0x64: YOU ARE NOT A BIRD...
F595 LD BC,\$6A08 Redirect control to the routine at 6A08 (make character speak) and return to F59B (below) when done
F598 CALL \$639F
F59B LD DE,(\$7FE5) Collect the score from 7FE5 into DE
F59F LD HL,(\$7FE9) Collect the current hi-score from 7FE9 into HL
F5A2 AND A Clear the carry flag ready for subtraction
F5A3 SBC HL,DE Do we have a new hi-score?
F5A5 JR NC,\$F5AB Jump if not
F5A7 LD (\$7FE9),DE Insert the new hi-score
F5AB LD HL,\$0000 Reset the score (at 7FE5) and the lines total (at 7FE7) to 0
F5AE LD (\$7FE5),HL
F5B1 LD (\$7FE7),HL
F5B4 JP \$F6D5 Enter demo mode
Prev: F54A Up: Map Next: F5B7 | 842 | 2,196 | {"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-2023-40 | latest | en | 0.785912 |
https://mathoverflow.net/questions/13768/what-is-the-right-definition-of-the-picard-group-of-a-commutative-ring/13777#13777 | 1,653,606,753,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00586.warc.gz | 449,092,395 | 32,033 | # What is the right definition of the Picard group of a commutative ring?
This is a rather technical question with no particular importance in any case of actual interest to me, but I've been writing up some notes on commutative algebra and flailing on this point for some time now, so I might as well ask here and get it cleared up.
I would like to define the Picard group of an arbitrary (i.e., not necessarily Noetherian) commutative ring $R$. Here are two possible definitions:
(1) It is the group of isomorphism classes of rank one projective $R$-modules under the tensor product.
(2) It is the group of isomorphism classes of invertible $R$-modules under the tensor product, where invertible means any of the following equivalent things [Eisenbud, Thm. 11.6]:
a) The canonical map $T: M \otimes_R \operatorname{Hom}_R(M,R) \rightarrow R$ is an isomorphism.
b) $M$ is locally free of rank $1$ [edit: in the weaker sense: $\forall \mathfrak{p} \in \operatorname{Spec}(R), \ M_{\mathfrak{p}} \cong R_{\mathfrak{p}}$.]
c) $M$ is isomorphic as a module to an invertible fractional ideal.
What's the difference between (1) and (2)? In general, (1) is stronger than (2), because projective modules are locally free, whereas a finitely generated locally free module is projective iff it is finitely presented. (When $R$ is Noetherian, finitely generated and finitely presented are equivalent, so there is no problem in this case. This makes the entire discussion somewhat academic.)
So, a priori, if over a non-Noetherian ring one used (1), one would get a Picard group that was "too small". Does anyone know an actual example where the groups formed in this way are not isomorphic? (That's stronger than one being a proper subgroup of the other, I know.)
Why is definition (2) preferred over definition (1)?
• As a non-commutative person, let me add that one can also consider the invertible $R$-$R$-bimodules, and/or the group of self-equivalences of the category of, say, left $R$-modules. Feb 2, 2010 at 1:45
• I can't help but wonder what makes a person non-commutative. Were you born like that? Feb 2, 2010 at 1:58
• Now (2) is a group. But is (1) a group?
– VA.
Feb 2, 2010 at 2:13
• @CamMcLeman, re, maybe he's born with it … maybe it's non-Abelian? Sep 30, 2021 at 19:44
• @LSpice It took 30 years for all the ingredients of that joke to mature from inception to delivery, but I think we can all agree it was worth the wait. Oct 3, 2021 at 21:20
For what it's worth, I think in Bourbaki's Algèbre Commutative, this is chapter II, section 5.4 (or so), but I don't have a copy in front of me. (Pete confirms that it's II.5.4, Theorem 3.)
• Bourbaki (and Clark) to the rescue. What a surprise. An acquaintance of mine was visiting Paris, and apparently they cite Bourbaki there up to the theorem number in lectures there. And of course, "Soit C un corps commutatif." Feb 2, 2010 at 11:54
• The Bourbaki books (some more than others; CA is still widely read nowadays) are certainly excellent references for basic material, the more so if you have internet access to a savant who can quote them chapter and verse. The problem (for me) comes when I try to read them in the usual linear manner: they cover the trivial and the important in equal detail, and the end product is about five times as long as it should be. Feb 2, 2010 at 12:08
• I doubled my there there. Oops. Feb 2, 2010 at 13:03
Although this question has already been answered, I would like to point out that the assertion also follows from a little bit of category theory (which does not seem to be discussed in the Bourbaki reference).
Claim: Let $R$ be any commutative ring, and let $M$ be an $R$-module which is invertible for the tensor product. Then $M$ is finitely generated and projective.
Proof: The functor from $R$-modules to $R$-modules given by tensoring with $M$ is an auto-equivalence. Since being projective is a property completely internal to the categorical structure, it is preserved by auto-equivalences. In particular, since $R$ is projective, so is $R \otimes_R M \simeq M$.
Similarly, one sees that $M$ is finitely presented, because the finitely presented $R$-modules are exactly the compact objects of the category.
(More generally: Given any symmetric monoidal category, if the unit object satisfies some categorical property, then so does any invertible object. This is useful in other contexts. Example: any invertible object in the stable homotopy category has to be a finite spectrum, because finite spectra are the compact objects; from here it's not too hard to conclude that the invertible objects in spectra are the spheres.)
$\alpha$) It is not true that for an arbitrary ring a) is equivalent to c):
Indeed Bourbaki in Algèbre commutative, Chapitre II, Exercices §5, 12) c) exhibits a ring $B$ and a projective module of rank $1$ over $B$ which is not isomorphic to an invertible fractional ideal of $B$. This does not contradict Eisenbud's Theorem 11.6 because $R$ is explicitly supposed noetherian there.
$\beta$) It is also not true that b) is equivalent to c):
Take $R=\mathbb Z$ and $\mathbb Z\subsetneq M=\bigcup \frac {1}{p_1\cdots p_i}\mathbb Z\subsetneq \mathbb Q$ where $p_i$ is the $i$-th prime.
Then for all prime $\mathfrak p\subset \mathbb Z$ the $\mathbb Z_\mathfrak p$-module $M_\frak p$ is free of rank $1$, since $$M_{(0)}=\mathbb Z_{(0)}(=\mathbb Q) \quad \operatorname {and}\quad M_{(p_i)}=\frac {1}{p_i}\mathbb Z_{(p_i)}$$ However the $\mathbb Z$-module $M$ is neither finitely generated nor projective (over $\mathbb Z$, projective=free)
2) I think the only reasonable definition of $\operatorname {Pic(R)}$ valid for any commutative ring is to define it as the Picard group of the affine scheme $X=\operatorname {Spec}(R)$.
As is the case for any locally ringed space $(X,\mathcal O_X)$ the Picard group consists of isomorphism classes of locally free $\mathcal O_X$-Modules of rank one.
This is exactly the definition used with much success for general, non-affine, schemes but also for topological spaces, differential manifolds, etc.
3) In our special case $X=\operatorname {Spec}(R)$ the definition in 2) translates in purely algebraic terms to:
$\operatorname {Pic(R)}$ consists of isomorphism classes of $R$-modules $M$ such that there exist finitely many elements $f_1,\cdots,f_n\in R$ with:
$\alpha$) $\sum Rf_i=R$
$\beta$) $M_{f_i}$ is a free $R_{f_i}$-modules of rank $1$ for all $i$.
These modules are called locally free of rank one.
Remark: $\beta$) implies that locally free modules of rank one are finitely generated over $R$, since "finitely generated" is a local condition
4) The locally free modules of rank one defined in 3) can also be characterized as the modules $M$ over $R$ such that equivalently:
i) The module $M$ is finitely generated, projective and for all primes $\mathfrak p\subset R$ the (necessarily!) free $R_\mathfrak p$- module $M_\mathfrak p$ has rank $1$
ii) The module $M$ is finitely generated and the modules $M_\frak m$ are free of rank $1$ over $R_\frak m$ for all maximal ideals $\mathfrak m\subset R$
iii) The canonical $R$-linear map $M\otimes_RM^*\to R:m\otimes \phi\mapsto\phi(m)$ is bijective
Note that these are pleasant algebraic characterizations, but the conceptual definition is that given in 2) and 3).
Edit: WARNING
The confusion is made worse by Bourbaki's unfortunate decision to define a projective module of rank $1$ as a finitely generated module $P$ for which $M_\mathfrak p$ is free of rank $1$ over $R_\mathfrak p$ for all primes $\mathfrak p\subset R$.
As my example 1) $\beta$) shows, omitting to require that $P$ be finitely generated [as is done in Pete's condition (2) b)] means accepting modules which aren't even projective, and which don't satisfy (2) a) nor (2) c) of the question.
• So if we call (1), (2a), (2b), (2c) the four conditions in Peter L. Clark's original question, we have (1⇔2a) by the reference to Bourbaki cited in Clark Barwick's answer, but (2c) is not equivalent to (1/2a) nor to (2b) by the counterexamples you cite. Did I get this right? But are (1/2a) and (2b) equivalent or not? Furthermore, since you propose yet another definition of the Picard group, we now have potentially FOUR different Picard groups (or at least, Picard sets — maybe they aren't all groups) and I am tempted to ask about all possible maps between them. This is getting really messy. Mar 27, 2018 at 12:30
• @Gro-Tsen. I have added an edit which points to one source of the messiness. I think that nobody doubts that 2) and 3) in my answer are the correct definitions. The trouble begins with alternative definitions which are thought to be equivalent but aren't. In Pete's question (1) and (2)a) are equivalent (and equivalent to what I have described as the correct definition) iff , as in Bourbaki, "rank one" includes the condition that $M$ be finitely generated. As to (2) b), it is not equivalent to (1) nor 2)a) nor 2)c) because it doesn't even imply projective. (to be continued) Mar 27, 2018 at 19:27
• Finally (2)c) only describes some of the modules in (2)a). So (2)c) is not equivalent to (2)a) in general, but it is equivalent under some mild conditions, for example if $R$ is a domain. Mar 27, 2018 at 19:27
There is no difference. If $M$ is locally free of finite rank, then $M$ is of finite presentation (and projective).
Take a partition of unity $f_1,...,f_n$, such that $M_{f_i}$ is free over $R_{f_i}$. Since $R \to R_{f_1} \oplus ... \oplus R_{f_n}$ is faithfully flat, it suffices to show the properties for $M_{f_1} \oplus ... \oplus M_{f_n}$, which is very easy.
Definition (2) is prefered because it reveals the geometric content: classification of line bundles.
• I believe this is not quite correct, depending upon what you mean by "finite rank". It is true though if the rank is finite and constant, which is the situation I asked about. Feb 2, 2010 at 3:31
• hm? I don't need any constancy. Feb 2, 2010 at 8:25
• Bourbaki, Section II.5.2: For an A-module P, TFAE: (a) P is finitely generated projective. (c) P is finitely generated, for each p in Spec(A), P_{p} is free, and the rank function p |-> rank(P_{p}) is locally constant on Spec(A). Feb 2, 2010 at 8:40
• I suppose it also depends on what you mean by locally free: I meant that the localization at each prime ideal is free. If instead you mean "locally in the Zariski topology" -- your condition about the f_i's above -- then that implies the local constancy of the rank (same theorem in Bourbaki). Feb 2, 2010 at 11:19
• The "of course" doesn't make sense to me, because that's not what was meant in the standard text that I referenced (Eisenbud). The point is that it's subtle whether the weaker sense of locally implies the stronger sense (yes for Noetherian rings or with local constancy of the rank; no in general). Feb 2, 2010 at 12:23 | 3,057 | 10,853 | {"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} | 2.921875 | 3 | CC-MAIN-2022-21 | latest | en | 0.901692 |
https://math.answers.com/Q/What_prime_numbers_add_up_two_29_from_10 | 1,701,722,512,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100534.18/warc/CC-MAIN-20231204182901-20231204212901-00332.warc.gz | 453,041,087 | 45,081 | 0
# What prime numbers add up two 29 from 10?
Updated: 11/4/2022
Wiki User
11y ago
There are no prime numbers greater than 10 that add up to 29. Since all prime numbers greater than 2 are odd numbers, no two prime numbers greater than 2 can have an even number for their sum.
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11y ago
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Q: What prime numbers add up two 29 from 10?
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5 5
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43 and 17. | 463 | 1,661 | {"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} | 3.859375 | 4 | CC-MAIN-2023-50 | latest | en | 0.948252 |
http://betterlesson.com/lesson/reflection/19863/rushing-to-the-abstract | 1,488,129,602,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172018.95/warc/CC-MAIN-20170219104612-00443-ip-10-171-10-108.ec2.internal.warc.gz | 26,614,815 | 20,013 | ## Reflection: Real World Applications Clarifying Our Terms - Section 2: Warm up
I am a big advocate for manipulatives and I think many of our textbooks and pacing guides rush our students too quickly to the abstract algorithm. I think it is really important to keep the manipulatives visible and available to students whenever they need them. Of course, we don't want them relying on concrete models forever. But some students may need a more gradual transition. If the algebra tiles make sense to my students then I want them to use the tiles to help them make sense of the abstract expressions.
I find myself using more context to help my students make sense of the algebraic terms too. Sometimes a more real world connection can help them make sense of the math. A students own experience can help them make sense of the math. In some cases I ask the students to come up with the context by asking what certain variables might represent and what story could the expression be telling. The context could even be a growth pattern that could be drawn.
Rushing to the abstract
Real World Applications: Rushing to the abstract
# Clarifying Our Terms
Unit 3: Equivalent Expressions
Lesson 9 of 23
## Big Idea: Students will see the benefit to simplifying variable expressions before solving them by trying to solve them mentally.
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2 teachers like this lesson
Standards:
Subject(s):
Math, Expressions (Algebra), mental math with decimals, white boards, simplifying variable expressions, solving variable expressions
54 minutes
### Erica Burnison
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Environment: Urban | 426 | 2,071 | {"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} | 2.875 | 3 | CC-MAIN-2017-09 | latest | en | 0.900084 |
https://www.janestreet.com/puzzles/single-cross-2-solution/ | 1,709,531,175,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476413.82/warc/CC-MAIN-20240304033910-20240304063910-00843.warc.gz | 795,927,459 | 4,486 | ### Single-Cross 2
##### August 2023 : Solution
August’s puzzle was a three dimensional throwback to Single Cross from three years ago. Careful computation determined that for lengths D <= 1, the probability of a single cross comes to the pleasantly symmetric spherical coordinate double integral. (As seen above.)
This double integral miraculously simplifies to (1/(4π))*D(-16D + 3D^2 + 6π). This function has a local maximum at D = (16 - sqrt(256 - 54π))/9 ~ 0.7452572091, with value ~ 0.5095346021. The argument that there can’t be a second higher local maximum with D > 1 is left to the puzzler.
Congrats to this month’s solvers who successfully completed the optimal length and probability! | 177 | 699 | {"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-2024-10 | latest | en | 0.851164 |
https://blog.tecladocode.com/coding-interview-problems-palindromes/ | 1,607,115,430,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141743438.76/warc/CC-MAIN-20201204193220-20201204223220-00424.warc.gz | 213,047,995 | 15,618 | We're back today with another common coding interview question. This week we're going to be writing some code to check if a string is a palindrome. A palindrome is a word or sentence which is read the same backwards as it is forwards, such as the name, "Hannah", or the sentence, "Never odd or even".
## Walking Through the Problem
The complexity of this problem depends on whether we have to check a single word or if we also have to check full sentences. In the case of the latter, we have to remove punctuation and spaces, since we only actually care about the letters. We'll start with the simpler version of the problem, and then we'll move onto full sentences.
One thing we have to keep in mind for both versions of the problem is that uppercase and lowercase characters are technically not the same, so we have to be careful to make sure we check letters in the same case.
For this purpose, we can use the `casefold` method (see documentation). In this post, we're going to assume that all of the strings use basic Latin characters, but there are ways to expand `casefold` to account for other symbols. I'd recommend taking a look at this StackOverflow answer for more details.
Now that we have these cautionary notes out of the way, how do we actually tackle the problem? As with all questions like this, there are many paths to a solution.
One option available to us is to convert the string to a list. The reason we'd want to do something like this, is that Python lists have a built in `reverse` method. We could then perform a comparison to the original string with the this new reversed version using `==`.
Something to watch out for in this solution is that `reverse` performs an in-place transformation on the list, so we either need to make a copy of the list for comparison, or we need to convert the list back to a string using `join`.
We have a more manual option available to us in the form of a `for` loop. Strings themselves are iterable, so we can check each character against another character one at a time.
We can make use of the enumerate function (details can be found here) to create a counter alongside each character in the string, starting with 1. We can then use this counter to access characters from the other end of the string using negative indices. For example, the character at index `-1` is the final character in the string; the item at index `-2` is the penultimate character, etc.
One final option we're going to look at is using slices. We have a couple of posts on slices, so if you're not familiar with how slicing works, I'd recommend you take a look at the posts below:
## Using Lists and `reverse`
We're going to start of by looking at the name, Hannah. We know that Hannah is in fact a palindrome, and it also contains a capital letter, which means we can test that our case folding works as expected. We're also going to use a second string, Peter, which is not a palindrome. This is just to test that we don't have erroneous results coming out of our implementation.
Since we're going to be checking multiple strings, we should start by defining a function. We know that we have to pass in a string to test, so we'll create a `test_string` parameter while we're at it.
``````def check_if_palindrome(test_string):
pass
``````
Our first step should be to convert `test_string` into a list, which we're going to call `characters`. We can then call the `reverse` method on this new list.
``````def check_if_palindrome(test_string):
characters = list(test_string)
characters.reverse()
``````
Before we can make this new `characters` list, however, we have to make use of `casefold` to get rid of any capital letters in the string that might cause issues with our comparison.
``````def check_if_palindrome(test_string):
characters = list(test_string.casefold())
characters.reverse()
``````
Now that we have our reversed characters, we can use `join` to convert the list back to a string, and then we can compare the new reversed string against the case folded original:
``````def check_if_palindrome(test_string):
characters = list(test_string.casefold())
characters.reverse()
if "".join(characters) == test_string.casefold():
print(f"{test_string} is a palindrome."
else:
print(f"{test_string} is not a palindrome."
``````
If we test our function with `"Hannah"` and `"Peter"`, we can see that everything works as expected:
``````def check_if_palindrome(test_string):
characters = list(test_string.casefold())
characters.reverse()
if "".join(characters) == test_string.casefold():
print(f"{test_string} is a palindrome.")
else:
print(f"{test_string} is not a palindrome.")
check_if_palindrome("Hannah") # Hannah is a palindrome.
check_if_palindrome("Peter") # Peter is not a palindrome.
``````
## Using a `for` Loop and `enumerate`
Once again, we're going to start by defining a function with a single parameter. However, this time, we don't need to do anything with lists, since we'll be iterating over the strings directly.
``````def check_if_palindrome(test_string):
pass
``````
Our `for` loop is going to make use of `enumerate`, which will create a tuple for each character in the `test_string`, containing a counter and the current character. These tuples get collected in an `enumerate` object, which is what our `for` loop will iterate over.
As such, for each iteration, we're going to get a tuple back, which we can unpack into two variables. We talked about this in our recent post on destructuring in Python.
By default, `enumerate`'s counter starts from `0`, but we want it to start from `1`, since we'll be using this value to access characters using negative indices. The final item in a collection is at index `-1`, not `-0`. We can correctly configure the counter by setting the `start` parameter value to `1`.
Just like before, we also have to make sure to remove any capital letters from the strings, so that we're comparing all of the characters in the same case:
``````def check_if_palindrome(test_string):
for index, letter in enumerate(test_string.casefold(), start=1):
pass
``````
Inside the for loop, we're going to perform a test using an `if` statement, where we'll compare characters one at a time. The first character of the string will be compared to the final character in the string, and we'll move along the string with each iteration.
We're actually going to be testing if the string is not a palindrome, which will make sense in a moment. If we find a non-matching character, we're going to immediately `break`, since there's no point checking any of the other characters. We already know the string is not a palindrome.
``````def check_if_palindrome(test_string):
for index, letter in enumerate(test_string.casefold(), start=1):
if letter != test_string[-index].casefold():
print(f"{test_string} is not a palindrome.")
break
``````
So, why did we do it this way? The reason is that we need every single one of the characters to match before we can say that something is in fact a palindrome. It's not enough for just the first character to match, for example.
If we perform a check for matching characters, instead of non-matching characters, we have to keep track of the fact that all of the characters matched, which is a lot of housekeeping we don't want to have to do.
Instead, we can use a `for` / `else` structure. We've written about this previously so take a look if this is new to you.
If we find a non-matching character, we `break` the loop immediately, so if the `for` loop runs to completion, it means that all of the characters matched. An `else` clause attached to a `for` loop only runs if the loop completed without encountering a `break` statement or an exception.
We can therefore put our "... is a palindrome." message inside this `else` clause.
``````def check_if_palindrome(test_string):
for index, letter in enumerate(test_string.casefold(), start=1):
if letter != test_string[-index].casefold():
print(f"{test_string} is not a palindrome.")
break
else:
print(f"{test_string} is a palindrome."
check_if_palindrome("Hannah") # Hannah is a palindrome.
check_if_palindrome("Peter") # Peter is not a palindrome.
``````
It's very important that the `else` keyword inline with the `for` loop definition. It is not part of the `if` block.
## Slices
Possibly the most elegant solution out of all of these options involves the use of slices. Once again, if you're not familiar with slices, read the posts linked earlier. They're an awesome, versatile tool that you'll find uses for everywhere.
The crux of our slice solution is a bit of rather arcane syntax: `[::-1]`. What this does, is take an entire sequence and reverse it. The best thing about this, and slices in general, is that they don't modify the original collection. This makes our lives very easy.
Our solution is essentially just a single `if` statement, where we compare two case folded strings.
``````def check_if_palindrome(test_string):
if test_string.casefold() == test_string[::-1].casefold():
print(f"{test_string} is a palindrome.")
else:
print(f"{test_string} is not a palindrome.")
check_if_palindrome("Hannah") # Hannah is a palindrome.
check_if_palindrome("Peter") # Peter is not a palindrome.
``````
## Dealing with Sentences
Now that we've covered a number of solutions for single words, let's consider how to handle full sentences. As was mentioned earlier, we need to get rid of all punctuation and spaces, since we only care about the letters themselves.
Finding all of the letter characters in a string is something we've tackled before in this series on coding interview problems when finding the longest word in a paragraph. I'd recommend you give this post a read, as we cover a lot of the problems associated with a problem like this.
In order to avoid missing some punctuation characters, it was determined that regular expressions were our best bet when trying to find all the letter characters in a string.
Python has a module for dealing with regular expressions called `re`, so we will need to import it as part of our solution. Specifically, we're concerned with a function called `sub`, which allows us to replace a pattern with another string. In our case, we'll be replacing all non-letter characters with an empty string, `""`.
``````def check_if_palindrome(test_string):
pass
``````
First we're going to remove all of the uppercase letters from our `test_string` using the `casefold` method. We can then pass this new, lowercase string to our `sub` function.
The `sub` function is going to allow us to filter out all non-letter characters in the `test_string`. The RegEx pattern we'll use to accomplish this is `[^a-z]+`. This looks pretty cryptic, but what it means is match any number of characters that are not the letters `a` to `z`. Note that as we already casefolded the `test_string` we don't need to check for the uppercase versions as well.
``````import re
def check_if_palindrome(test_string):
letters_only = re.sub("[^a-z]+", "", test_string.casefold())
print(letters_only)
``````
If we pass in even a string full of punctuation, we'll now end up with an entirely lower case string containing only letter characters.
``````check_if_palindrome("NeVeR! - oDd! - Or! - eVEn!")
# neveroddoreven
``````
Now we can make use of any of our single word solutions, with some minor modifications. We've already case folded all the text, so there's no need to do that going forward:
``````import re
def check_if_palindrome(test_string):
letters_only = re.sub("[^a-z]+", "", test_string.casefold())
if letters_only == letters_only[::-1]:
print(f"'{test_string}' is a palindrome.")
else:
print(f"'{test_string}' is not a palindrome.")
check_if_palindrome("Never odd or even.")
# 'Never odd or even.' is a palindrome.
check_if_palindrome("Python is awesome!")
# 'Python is awesome!' is not a palindrome.
``````
And that's all there is to it!
## Wrapping Up
I actually think this is a really interesting question, particularly when it comes to full sentence palindromes. Hopefully you learnt how to tackle this kind of problem, as well as soon cool new skills, and I really hope you shine in your next coding interview!
If you're really looking to upgrade your Python skills, you should give our Complete Python Course a try! We'd really love to see you on the course!
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# World Space...
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## Recommended Posts
I’ve been trying to figure this out for a while now but cant find anything good on it, and I searched on google and stuff but couldn’t find anything so I was just wondering what is the difference between Screen Space, Model Space, and world Space in Direct3d? Thnx
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world space - starts at origin (0.0f, 0.0f, 0.0f) at center of screen and goes out on the x,y,z axis
screen space - starts at x,y (0,0) at upper left corner of screen and gets large to the bottom right corner of screen
model space - do not know
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Do u use all three or just one? And Does it matter which one I use? (sorry for such a newbie question)
[edited by - Bakingsoda36 on March 17, 2003 10:38:38 PM]
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The use of global and local coordinates are meant to simplify. It is also required for accurate scaling and rotation. (if you want to rotate a model around it's z axis, then using the global z axis will not work.. same with scaling). It makes it much much simpler and is required in many situations
EDIT: I assume by model space they mean the coordinate setup that is local to the model. The 0,0,0 is usually at the center of the model.. the center of rotation and scaling
[edited by - falkone on March 17, 2003 10:46:25 PM]
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??? (puzzled with so SnakeHunta definition) ???
I thought World Space is a space representing the..world. It's where the objects are placed. So, if we have books, a pencil, and a piece of paper as our objects, the world space will be where these objects are located, such as a table.
Then, View Space is how the table is represented from a "camera" (i.e, your eye). And the screen space is how the view space is represented on the screen. No?
Red = the origin of view space
Blue = the origin of world space
Green = the origins of model spaces
EDIT: Illustration fixed. The table was missing due to the black background
throw Exception( "End of post" );
[edited by - alnite on March 17, 2003 11:23:05 PM]
[edited by - alnite on March 17, 2003 11:36:48 PM]
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Yes..
An example of world coordinates would be.. lets say a pencil.. we''ll say that the world origin would be the center of the table this player is resting on (it''s all relative, one set of global coordinates can be another''s local)... it could be 2,2 units from the center. Now, the 2,2 only describes the pencil''s location. It could just as easily describe the location of a bus.. it does not define the shape of the object. The coordinate system local to the pencil will then have it''s origin set at global 2,2. From here we can describe the point at 2,3.. and the eraser at 2,1.. and all the vertices that may make up the pencil..
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quote:
Original post by Bakingsoda36
Do u use all three or just one? And Does it matter which one I use? (sorry for such a newbie question)
You use all. They are all part of the geometry pipeline.
throw Exception( "End of post" );
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Ok I think I get it now. Thanks for all the replies
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× | 879 | 3,413 | {"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} | 2.921875 | 3 | CC-MAIN-2018-26 | latest | en | 0.949652 |
http://mathhelpforum.com/calculus/105738-simple-problem-need-help-print.html | 1,519,271,839,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813883.34/warc/CC-MAIN-20180222022059-20180222042059-00307.warc.gz | 213,429,244 | 2,863 | # simple problem need help
• Oct 2nd 2009, 04:34 PM
kayla220220
simple problem need help
determine the relationship between the normal to the curve y=y = x^3-2 x at x=1 and x=-1
please show me how to do it
thankyou
• Oct 2nd 2009, 05:02 PM
Matt Westwood
Quote:
Originally Posted by kayla220220
determine the relationship between the normal to the curve y=y = x^3-2 x at x=1 and x=-1
please show me how to do it
thankyou
The normal to a curve is by definition perpendicular to the tangent. So get the slope of the tangent of the curve in question by differentiating and plugging in the value x = 1. Then the perpendicular to a line of slope m has a slope $\frac {-1} m$ (you will find this explained in your textbook somewhere).
So, the normal to the curve at that point is the straight line of the slope you've just calculated that passes through the point given. Then I suppose you would look at that line and compare it with the line x = -1. | 256 | 947 | {"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": 1, "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} | 3.84375 | 4 | CC-MAIN-2018-09 | latest | en | 0.942143 |
https://www.casinomeister.com/forums/threads/major-millions-multi-currency.4427/ | 1,713,103,055,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00169.warc.gz | 667,284,704 | 36,615 | # Major Millions multi currency
#### Sodax77
##### Dormant account
Ok Major Million Jackpot is now over 1,017,000.00 every currency, when i write that.
But same every currency?
Yes
so (i.e 1,000,000.00 credit)
1,000,000.00 = \$1,769,429.68 (pound-dollar)
1,000,000.00 = ?1,472,437.99 (pound-euro)
\$1,000,000.00 = 565,139.38 (dollar-pound)
\$1,000,000.00 = ?832,072.07 (dollar-euro)
?1,000,000.00 = 679,034.22 (euro-pound)
?1,000,000.00 = \$1,201,795.30 (euro-dollar)
(2004.05.17) 22:30 - 23:00 GMT
www.xe.com
Let say that jackpot is 1,000,000.00 credit, and winner want convert that to euro currency
So because everyone can open pound account,
and if someone win 1,000,000.00 (pound), it is ?1,472,437.99 (euro)
But if someone win that in euro account it is ?1,000,000,00 (euro)
and if someone win that in dollar account it is ?832,072.07 (euro)
I have pound, dollar and euro accounts. Few pound account, because some n/d offer past. Dollar account, because some casinos add later euro and n/d offer past. And now i open only euro-account (if possible), own currency.
So if you win that jackpot your dollar account, that is ?832,072.07 (euro), But if you win that jackpot your pound account, that is ?1,472,437.99 (euro)
Big difference ?640,365.92 (euro)
And if you want to convert that jackpot (1,000,000.00) to dollar (i.e) US players, you can open pound account, and then it is \$1,769,429.68 (dollar), difference is \$769,429.68
Am i right?
Edit:
?=euro (alt gr + e not work, when you submit your post)
Last edited:
Well if you win it it pounds, you have to be wagering in pounds. Same with euros. So the risk/reward is pretty much the same in each currency. Now if you could hit the winning combo in dollars, but collect it in pounds, that would be cool! Ain't gonna happen though.
I agree, but if you want more fun with same money... try
Link Removed ( Old/Invalid)
1,000,000.00 (South Africa) Rand = \$146,931.69 (dollar)
(Same progressive-system)
Max bet 3:
3.00 (South Africa) Rand = \$0.440001 (dollar)
3.00 (pound) = \$5.29721 (dollar)
?3.00 (euro) = \$3.59602 (dollar)
And i never open any other account, but euro account. But there is still some casinos, who not offer different currency. Like some playtech casinos, Casino On Net, Slotland, etc. And then i open/use Dollar account.
And i have dollar account in Casino on net (I love that casino, but that is only currency what they offer), but Neteller is easy to deposit because you can usee and see your balance in 4 different currency.
Last edited:
Sodax77 said:
but with Neteller it is easy to deposit because you can use and see your balance in 4 different currency.
edit: reason:
System not allow edit
Yeah, but if you're playing in rands, euros or pounds, that's what you will be paid in. You may be able to then convert the winning balance to dollars, but it will be at the exchange rate, not at the same amout.
If you win \$1million, you can't say I'll take it in pounds and expect to get \$1million pounds. You'd get the equivalent of \$1million in pounds (less a percentage or 2 or 3 for the conversion) but not 1 million pounds. They're not going to give you something for nothing.
jpm said:
.Yeah, but if you're playing in rands, euros or pounds, that's what you will be paid in You may be able to then convert the winning balance to dollars, but it will be at the exchange rate, not at the same amout.
If you win \$1million, you can't say I'll take it in pounds and expect to get \$1million pounds. You'd get the equivalent of \$1million in pounds (less a percentage or 2 or 3 for the conversion) but not 1 million pounds. They're not going to give you something for nothing.
Yes no problem, i understand that. And always know it. And btw, if you open account in any currency, you can't change it anymore (based last year chat-session with one casino, after they add euros). Even it is zero balance. Even never deposit made, but you have real account. Your account is always that currency what you choose, and of course your winnings paid that same currency.
My point:
Ok, maybe this global-thing change everything, but i think that MG can use some non-currency-system (or based i.e. dollars), so jackpot is "same" to everyone
like, current jackpot is:
\$1,000,000.00 (dollar)
?832,072.07 (euro)
565,139.38 (pound)
But then you find one problem, that "max bet" have to based non-currency (or ie. dollars) only. Because to pound player it cost \$5.29721 (dollar) and Rand players only \$0.440001 (dollar) per "bet max" spin (if bet max is 3 credit). So i think that is reason, why MG use same progressive-system to every currency.
Edit/Note
And thats why, i.e land based casinos and online casinos can't build global jackpot network (easily), because this "bet max" cost everyday different. Yes, they can convert (in realtime) that jackpot amount (global network), but there is other winnings too in paytable... so every winnings have to convert in realtime. Even that one cherry, what you get
Last edited:
Sodax77 said:
I agree, but if you want more fun with same money... try
Link Removed ( Old/Invalid)
1,000,000.00 (South Africa) Rand = \$146,931.69 (dollar)
(Same progressive-system)
So if some Rand-account owner hit (i.e) 1,000,000.00 jackpot,
what is \$146,931.69 (dollar) (!!!)
Jackpot drop to 250,000.00 every currency!!!
Dollar, Euro, Pound and Rand (is there more currency in MG?).
And if dollar-account owner hit that 250,000.00 after Rand-account winner, he/she actually get over \$100,000.00 more, because every currency is connected that same amount/number.
Great system, btw this is part of Cinema casino winners pop up:
"Major Millions
Katharine D \$ 250,028.60 04 Apr 2004
Johnlon F \$ 489,649.85 04 Apr 2004
Andreas F \$ 841,529.63 09 Mar 2004
But let's go to euro page..
Major Millions
Katharine D ? 250,028.60 04 Apr 2004
Johnlon F ? 489,649.85 04 Apr 2004
Andreas F ? 841,529.63 09 Mar 2004
or pound page...
Major Millions
Katharine D 250,028.60 04 Apr 2004
Johnlon F 489,649.85 04 Apr 2004
Andreas F 841,529.63 09 Mar 2004"
So, tell me, what Katharine D won, 04 Apr 2004?
Pound, Euro, or Dollar?
(of course you find that info from jackpotmadness, etc, but with this pop up facts...)
?=euro
I pick cinema, because i like that casino (and many VPL casinos)
Last edited:
Jackpot drop to 250,000.00 every currency!!!
Not correct. If you hit the jackpot in rand, the jackpot drops by the equivalent amount in pounds. Therefore, a one million jackpot, if hit in rand, starts the next jackpot at roughly 900,000.
My jackpots history confirms this to be true - the trouble is, we never know which currency a jackpot was actually hit in, even on Jackpot Madness.
Actually, come to think of it, my "history" should be able to be used to retroactively identify the currency that was used when the jackpot was hit... I must look at doing something like that to report "true" payouts...
Interesting observation about Katharine's win, however... because the jackpot had just been reset, my scripts ignored Katharine's win as likely due to a hiccup LOL.
Guess I'll have to keep a closer eye on jackpots... LOL...
Good to know, and thanks for replies.
But i continue, because i search, and search and search, and Finally i found something.
This is from Jackpot Madness,
LOTSALOOT JACKPOT WINNERS:
Fay B, \$ 33,535.05, Spin Palace, 21 Jan 2004
Ok, but this is from Spin Palace Casino:
Fay B., UK 18,196 ,Progressive Jackpot win at
Lots-a-Loots Slots Jan 21, 04
So they convert winnings to dollars, in Jackpot Madness site.
But that Cinema Casino pop up say exactly same thing in all currencies, and after exploring, it is not only site. (choose currency, click winners, in Cinema site, and wait pop up) All three pop up are identical, only different currency.
Btw, jackpot Madness site tell, that
Katharine D, \$ 250,028.60, The River Belle, 04 Apr 2004
So these are not true:
Katharine D ? 250,028.60 04 Apr 2004 (euro)
Katharine D 250,028.60 04 Apr 2004
or is it? Maybe they convert already... omg...
lol
Last edited:
I understand now Sodax, very interesting. Looks like there could be a little puffery going on with these jackpot wins, depending on what currency you play. You login as a euro player and see that someone won 250,000 euros, but they actually won in rands. BIG difference! I wonder if that kind of information really encourages anyone to play that game though.
jpm said:
I understand now Sodax, very interesting...
lol
1. I think too much
2. Sometimes i act like manic
3. Sometimes i have too many "points"
4. My messy english
lol, you're english is not bad at all. Better than many in this country!
What did i say! I bet winner have euro account
-----
result:
Old / Expired Link
MajorMillions
Jackpot
\$1,453,590.34, May 31, 2004
-----
result:
Link Removed ( Old/Invalid)
Joaquim M \$ 1,779,827.44, Jackpot City, 01 Jun 2004
-----
Convert here www.xe.com
LOL...
The Awesome Jackpots number is the one put out by the Jackpot Madness servers when it is live - and extrapolated, but the figure is US dollars.
The Jackpot Madness one takes that figure and multiplies/divides it by the currency of play.
So, the jackpot was worth over \$1.7 million dollars... but the more accurate figure is the Awesome Jackpots one Go figure...
I will contact Microgaming about this as it is bound to create a lot of confusion... can we call the new winner the biggest ever winner, or doesn't it count because he happened to be playing in Euros?
I agree (part of)
Meanwhile i can say one thing:
Truth: When Joaquim M. hit Major Millions Jackpot,
he won 1,453,590.34 CREDITS to his account (without currency symbol).
Microgaming casinos does not show your balance or progressive amount with currency symbol, in game or lobby (only in banking page, etc).
Maybe next winner is Rand-account owner
Link Removed ( Old/Invalid)
Last edited:
Sodax77 said:
(few edit to quote)
Cinema casino "winners"-page, pop up (different currency):
Dollar-page
"Major Millions
Katharine D \$ 250,028.60 04 Apr 2004
Johnlon F \$ 489,649.85 04 Apr 2004
Andreas F \$ 841,529.63 09 Mar 2004
Euro-page
Major Millions
Katharine D Euro 250,028.60 04 Apr 2004
Johnlon F Euro 489,649.85 04 Apr 2004
Andreas F Euro 841,529.63 09 Mar 2004
Pound-page
Major Millions
Katharine D 250,028.60 04 Apr 2004
Johnlon F 489,649.85 04 Apr 2004
Andreas F 841,529.63 09 Mar 2004"
Yesterday i sent letter about these 3 identical pop up (only different currency)
And i got letter, here is (part of) their answer:
-------
...Thank you for your email. At your request we logged a call with MicroGaming to
investigate this matter and, they have not as yet responded, however we will
advise you as soon as they do. We do apologize for any inconvenience caused,
however we will rectify the matter as soon as possible. We thank you in advance
for your patience and understanding on this matter..."
------
I really like VPL casinos and their support, they always answer (asap).
Here is one more:
About week ago, i sent letter about their own promotion "mistake" and i got nice letter, and i like this:
"Importance: High"
And here is (part of) their answer:
------
"...We do appreciate you pointing this out to us so that we may correct the information accordingly. We will notify you as soon as we have corrected all the information..."
------
So now is Microgaming's turn
It was hit at jackpot city, so I think it was in \$\$. I don't think they offer euros there, but I'm pretty sure they have pound players so maybe they do offer euros.
jpm said:
It was hit at jackpot city, so I think it was in \$\$. I don't think they offer euros there, but I'm pretty sure they have pound players so maybe they do offer euros.
Go this page, and tell me what you see
You do not have permission to view link Log in or register now.
Wait, until page is ready...
Last edited:
Yup, the popup says it was won in Euros. Nice jackpot then, probably the biggest in internet history so far when you convert it to \$\$ on the day of the win.
yes, it is world largest online jackpot ever, but in euros (but after convert, in dollars also)
But, because Jackpot Madness publish every wins only in \$dollar\$ amount, like this (my past reply):
LOTSALOOT JACKPOT WINNERS:
Fay B, \$ 33,535.05, Spin Palace, 21 Jan 2004
Spin Palace Casino:
Fay B., UK 18,196 ,Progressive Jackpot win at
Lots-a-Loots Slots Jan 21, 04
---
Then this is true (from Jackpotmadness):
\$1,683,526.64 won by Jim L. on Major Millions
The World's Largest Online Jackpot
July 19, 2003 - Curacao
Jackpot Madness has again beaten its own record of awarding the World's Largest Ever Online Jackpot..."
---
But also this is true (from Jackpot Madness):
"Joaquim M \$ 1,779,827.44"
---
And also this is true (from Jackpot city casino):
"Joaquim M Euro 1,453,610.18"
So all these are true, biggest Dollar-account winner is Jim L.,
and biggest Euro-account winner is Joaquim M
And to got2bet contest, we can say that right answer is \$0,00, or can we? lmao
Last edited:
After VPL contact to Microgaming, here is their (MG) answer (part of):
"...from MicroGaming.
The Progressives are paid out in the currency that it is won depending on what
the player won. It is set to the highest currency being Pounds. If it was won
in this currency, then it is reset to the minimum of 250 000. Otherwise the
payout is calculated in the currency won and the jackpot is reset at the
difference of the 250 000 and the value in Pounds that it was set at.
For example the last jackpot was immediately reset at over 500 000.
So in other words, the player will be paid out in the currency that he wins. If
the amount shown on the meter is 500,000.00 credits and the player is playing is
Euros, he will receive EUR500,000.00 as the winning amount.
The coin sizes are the same with all currencies i.e. \$3 = (3 x 1) credits = EUR3
= 3.
It is the same with 25 cent credits, the are the same with all currencies..."
Last edited:
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974 | 3,882 | 14,077 | {"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} | 3.40625 | 3 | CC-MAIN-2024-18 | longest | en | 0.869637 |
http://talkchess.com/forum3/viewtopic.php?f=2&t=66945&start=40 | 1,591,304,379,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347458095.68/warc/CC-MAIN-20200604192256-20200604222256-00582.warc.gz | 115,235,225 | 11,120 | ## LCZero: Progress and Scaling. Relation to CCRL Elo
Discussion of anything and everything relating to chess playing software and machines.
Moderators: bob, hgm, Harvey Williamson
Forum rules
This textbox is used to restore diagrams posted with the [d] tag before the upgrade.
hgm
Posts: 24447
Joined: Fri Mar 10, 2006 9:06 am
Location: Amsterdam
Full name: H G Muller
Contact:
### Re: LCZero: Progress and Scaling. Relation to CCRL Elo
whereagles wrote:
Leela is black.. 2.36% chance to win on a lone king??
Close enough to 0 to make no difference even at 3500 Elo.
George Tsavdaris
Posts: 1625
Joined: Thu Mar 09, 2006 11:35 am
### Re: LCZero: Progress and Scaling. Relation to CCRL Elo
whereagles wrote:
Leela is black.. 2.36% chance to win on a lone king??
How is this possible?? I mean how do the rollouts result in black wins in order black to have wins in its statistics??
After his son's birth they've asked him:
"Is it a boy or girl?"
YES! He replied.....
Nay Lin Tun
Posts: 582
Joined: Mon Jan 16, 2012 5:34 am
### Re: LCZero: Progress and Scaling. Relation to CCRL Elo
@Kai, is it possible to share your opening test suit?
Daniel Shawul
Posts: 3843
Joined: Tue Mar 14, 2006 10:34 am
Location: Ethiopia
Contact:
### Re: LCZero: Progress and Scaling. Relation to CCRL Elo
She averages her tree like no other, and sucks because of that
What is happening is that white will sometimes give up its queen in the tree and the score becomes close to a draw. When that is averaged with the non-loosing lines you get this non-existant winning probability.
Tried it on scorpioMCTS with averaging and minimax backups
With averaging like leela's, a score of -627 is like 2% winning chance
Code: Select all
``````15 -570 5 118404 Ka6-b6 Kd3-c4 Kb6-b7 Qc2-b2
15 -545 7 158627 Ka6-b5 Qc2-g2 Kb5-a4 Qg2-g4 Ka4-b5 Qg4-c4 Kb5-a5 Qc4-c2 Ka5-b4 Qc2-f2 Kb4-b3
16 -569 7 167639 Ka6-b5 Qc2-g2 Kb5-a4 Qg2-d5 Ka4-b4 Qd5-c4 Kb4-a5 Qc4-c2 Ka5-b4 Qc2-f2 Kb4-b3
16 -595 8 177007 Ka6-b5 Qc2-g2 Kb5-a4 Qg2-f2 Ka4-b5 Qf2-f5 Kb5-b4
16 -611 9 219766 Ka6-b5 Qc2-g2 Kb5-a4 Qg2-g7 Ka4-b3 Qg7-b7 Kb3-a4 Qb7-f3
17 -627 9 227292 Ka6-b5 Qc2-g2 Kb5-a4 Kd3-c3 Ka4-b5 Qg2-d5
``````
With minimaxing a score of -2023 is a 0% winning chance
Code: Select all
``````2 -2023 0 554 Ka6-b7 Qc2-b2 Kb7-c8
3 -2034 0 3588 Ka6-b5 Qc2-c4 Kb5-a5 Qc4-c5 Ka5-a4
4 -2044 0 12593 Ka6-b6 Qc2-a4 Kb6-c7 Qa4-f4 Kc7-d7
5 -2056 0 23814 Ka6-b6 Qc2-b2 Kb6-c5 Qb2-e5 Kc5-b4
6 -2063 2 47803 Ka6-b6 Kd3-c3 Kb6-a6 Qc2-a2 Ka6-b6
7 -2064 3 91732 Ka6-b6 Kd3-c3 Kb6-c6 Qc2-a2 Kc6-b6
``````
Posts: 10197
Joined: Wed Jul 26, 2006 8:21 pm
Full name: Kai Laskos
### Re: LCZero: Progress and Scaling. Relation to CCRL Elo
Nay Lin Tun wrote:@Kai, is it possible to share your opening test suit?
Sure, the link in this post should work:
http://www.talkchess.com/forum/viewtopi ... 5&start=14
tmokonen
Posts: 1106
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Location: Vancouver
### Re: LCZero: Progress and Scaling. Relation to CCRL Elo
George Tsavdaris wrote: How is this possible?? I mean how do the rollouts result in black wins in order black to have wins in its statistics??
L0 takes both wins and draws into consideration, so there's a small residual score from rollouts that result in draws.
Uri Blass
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Location: Tel-Aviv Israel
### Re: LCZero: Progress and Scaling. Relation to CCRL Elo
George Tsavdaris wrote:
whereagles wrote:
Leela is black.. 2.36% chance to win on a lone king??
How is this possible?? I mean how do the rollouts result in black wins in order black to have wins in its statistics??
I think it means expected outcome of 2.36% that may be probability of 4.72% for a draw and 95.28% to lose.
MonteCarlo
Posts: 98
Joined: Sun Dec 25, 2016 3:59 pm
### Re: LCZero: Progress and Scaling. Relation to CCRL Elo
Indeed. Leela's output is an expected score, not actually a win%.
The wording on the site has since been changed, it seems, although now it includes this "50%=draw" bit in the legend, which caused some debate in the discord.
Probably should just say "50%=equal chances" or some such thing ("expected score" is pretty self-explanatory, so could probably do away with the legend altogether), but not a big deal.
Last net to pass is actually fairly reasonable now. It'll be interesting to see where we are a week from now (it's not even been a week since the last big bug was fixed).
Posts: 10197
Joined: Wed Jul 26, 2006 8:21 pm
Full name: Kai Laskos
### Re: LCZero: Progress and Scaling. Relation to CCRL Elo
Large improvement from ID69 to ID83.
Now, in only 100 games gauntlets against Zurichess Bern (2232 Elo CCRL) and BikJump v2.01 (2098 Elo CCRL), it performs at about 2200 Elo level at 1s/move and at about 2300 Elo level at 10s/move. On a full 4 core i7 CPU.
On my positional opening test suite of 200 positions, it is firmly settled amongst strong engines (20s/position):
Code: Select all
``````[Search parameters: MaxDepth=99 MaxTime=20.0 DepthDelta=2 MinDepth=7 MinTime=0.1]
Engine : Correct TotalPos Corr% AveT(s) MaxT(s) TestFile
Komodo 10.2 64-bit : 145 200 72.5 2.0 20.0 openings200beta07.epd
Houdini 5.01 Pro x64 : 144 200 72.0 2.4 20.0 openings200beta07.epd
Stockfish 8 64 BMI2 : 141 200 70.5 2.0 20.0 openings200beta07.epd
Houdini 5.01 Pro x64 Tactical : 139 200 69.5 2.3 20.0 openings200beta07.epd
Deep Shredder 13 x64 : 128 200 64.0 2.7 20.0 openings200beta07.epd
Houdini 4 Pro x64 : 126 200 63.0 1.8 20.0 openings200beta07.epd
Andscacs 0.88n : 123 200 61.5 2.4 20.0 openings200beta07.epd
Houdini 4 Pro x64 Tactical : 120 200 60.0 1.6 20.0 openings200beta07.epd
Nirvanachess 2.3 : 119 200 59.5 1.8 20.0 openings200beta07.epd
Fire 5 x64 (3341 CCRL) : 110 200 55.0 3.0 20.0 openings200beta07.epd
Texel 1.06 (3162 CCRL) : 110 200 55.0 1.6 20.0 openings200beta07.epd
LCZero ************* ID83 : 109 200 54.5 1.1 20.0 openings200beta07.epd
Fritz 15 (3227 CCRL) : 102 200 51.0 1.9 20.0 openings200beta07.epd
LCZero ************* ID69 : 98 200 49.0 2.7 20.0 openings200beta07.epd
Fruit 2.1 (2685 CCRL) : 91 200 45.5 1.5 20.0 openings200beta07.epd
Sjaak II 1.3.1 (2194 CCRL) : 75 200 37.5 4.0 20.0 openings200beta07.epd
BikJump v2.01 (2098 CCRL) : 74 200 37.0 1.6 20.0 openings200beta07.epd
``````
It improved significantly positionally from ID69 (in only 3 days).
Tactically it seems very weak. On ECM tactical middlegame suite of 879 positions, it performs very badly (1s/position):
Code: Select all
``````Bik Jump 2.01 (2098 CCRL Elo)
score=574/879 [averages on correct positions: depth=4.6 time=0.19 nodes=467671]
Predateur 2.2.1 (1786 CCRL Elo)
score=486/879 [averages on correct positions: depth=6.1 time=0.13 nodes=409596]
LCZero (ID83)
score=173/879 [averages on correct positions: depth=13.6 time=0.24 nodes=312]
LCZero (ID69)
score=171/879 [averages on correct positions: depth=13.5 time=0.25 nodes=318]
``````
And doesn't seem to improve at all. The estimated CCRL Elo level on this tactical suite is about that of Stockfish at depth=3, or maybe 1400 CCRL Elo points. Something has to be done with its MCTS search, maybe on the lines outlined by Daniel Shawl.
MonteCarlo
Posts: 98
Joined: Sun Dec 25, 2016 3:59 pm
### Re: LCZero: Progress and Scaling. Relation to CCRL Elo
Thanks for the update Kai!
On the one hand, it's quite possible that a fundamental change to its MCTS implementation will be required at some point if it wants to compete at the highest level, and the work Daniel Shawul has done with Scorpio could prove quite useful in that case (well, it's fantastic work in any case; it's just in this case that it would benefit LC0 ).
On the other hand, unless you subscribe to some form of conspiracy theory around the A0 results, we're nowhere near the limits of this sort of approach, so I wouldn't worry too much about that just yet.
Right now there are still a bunch of bugs being worked out, the network is still rather small, and the project is rather young (barely a month old, and it's barely been a week since the last major bug was discovered and fixed).
Some patience is required. It might turn out that switching to a new implementation of MCTS is required; it might also turn out that the NN at some level gives good enough prior probabilities for moves that even MCTS with averaging is good tactically.
We'll just have to give it some time | 3,070 | 9,144 | {"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-2020-24 | latest | en | 0.847886 |
https://www.lmfdb.org/EllipticCurve/Q/840g3/ | 1,603,365,763,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879537.28/warc/CC-MAIN-20201022111909-20201022141909-00540.warc.gz | 806,821,365 | 25,693 | # Properties
Label 840g3 Conductor 840 Discriminant 205752960000 j-invariant $$\frac{549871953124}{200930625}$$ CM no Rank 0 Torsion Structure $$\Z/{2}\Z \times \Z/{2}\Z$$
# Related objects
Show commands for: Magma / Pari/GP / SageMath
## Minimal Weierstrass equation
sage: E = EllipticCurve([0, -1, 0, -1720, -16100]) # or
sage: E = EllipticCurve("840g3")
gp: E = ellinit([0, -1, 0, -1720, -16100]) \\ or
gp: E = ellinit("840g3")
magma: E := EllipticCurve([0, -1, 0, -1720, -16100]); // or
magma: E := EllipticCurve("840g3");
$$y^2 = x^{3} - x^{2} - 1720 x - 16100$$
## Mordell-Weil group structure
$$\Z/{2}\Z \times \Z/{2}\Z$$
## Torsion generators
sage: E.torsion_subgroup().gens()
gp: elltors(E)
magma: TorsionSubgroup(E);
$$\left(-10, 0\right)$$, $$\left(46, 0\right)$$
## Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
$$\left(-35, 0\right)$$, $$\left(-10, 0\right)$$, $$\left(46, 0\right)$$
## Invariants
sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); Conductor: $$840$$ = $$2^{3} \cdot 3 \cdot 5 \cdot 7$$ sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); Discriminant: $$205752960000$$ = $$2^{10} \cdot 3^{8} \cdot 5^{4} \cdot 7^{2}$$ sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); j-invariant: $$\frac{549871953124}{200930625}$$ = $$2^{2} \cdot 3^{-8} \cdot 5^{-4} \cdot 7^{-2} \cdot 13^{3} \cdot 397^{3}$$ Endomorphism Ring: $$\Z$$ Geometric Endomorphism Ring: $$\Z$$ (no Complex Multiplication) Sato-Tate Group: $\mathrm{SU}(2)$
## BSD invariants
sage: E.rank() magma: Rank(E); Rank: $$0$$ sage: E.regulator() magma: Regulator(E); Regulator: $$1$$ sage: E.period_lattice().omega() gp: E.omega[1] magma: RealPeriod(E); Real period: $$0.763990518471$$ sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); Tamagawa product: $$32$$ = $$2\cdot2\cdot2^{2}\cdot2$$ sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); Torsion order: $$4$$ sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Analytic order of Ш: $$1$$ (exact)
## Modular invariants
#### Modular form840.2.a.f
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
magma: ModularForm(E);
$$q - q^{3} + q^{5} + q^{7} + q^{9} + 4q^{11} - 2q^{13} - q^{15} - 6q^{17} + 4q^{19} + O(q^{20})$$
sage: E.modular_degree() magma: ModularDegree(E); Modular degree: 1024 $$\Gamma_0(N)$$-optimal: no Manin constant: 1
#### Special L-value
sage: r = E.rank();
sage: E.lseries().dokchitser().derivative(1,r)/r.factorial()
gp: ar = ellanalyticrank(E);
gp: ar[2]/factorial(ar[1])
magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);
$$L(E,1)$$ ≈ $$1.52798103694$$
## Local data
This elliptic curve is not semistable.
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
prime Tamagawa number Kodaira symbol Reduction type Root number ord($$N$$) ord($$\Delta$$) ord$$(j)_{-}$$
$$2$$ $$2$$ $$III^{*}$$ Additive -1 3 10 0
$$3$$ $$2$$ $$I_{8}$$ Non-split multiplicative 1 1 8 8
$$5$$ $$4$$ $$I_{4}$$ Split multiplicative -1 1 4 4
$$7$$ $$2$$ $$I_{2}$$ Split multiplicative -1 1 2 2
## Galois representations
The image of the 2-adic representation attached to this elliptic curve is the subgroup of $\GL(2,\Z_2)$ with Rouse label X98h.
This subgroup is the pull-back of the subgroup of $\GL(2,\Z_2/2^3\Z_2)$ generated by $\left(\begin{array}{rr} 1 & 2 \\ 4 & 1 \end{array}\right),\left(\begin{array}{rr} 1 & 0 \\ 0 & 3 \end{array}\right),\left(\begin{array}{rr} 5 & 0 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 1 & 0 \\ 4 & 7 \end{array}\right)$ and has index 48.
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
The mod $$p$$ Galois representation has maximal image $$\GL(2,\F_p)$$ for all primes $$p$$ except those listed.
prime Image of Galois representation
$$2$$ Cs
## $p$-adic data
### $p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(3,20) if E.conductor().valuation(p)<2]
All $$p$$-adic regulators are identically $$1$$ since the rank is $$0$$.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 add nonsplit split split - 0 1 1 - 0 0 0
All Iwasawa $\lambda$ and $\mu$-invariants for primes $p\ge 3$ of good reduction are zero.
An entry - indicates that the invariants are not computed because the reduction is additive.
## Isogenies
This curve has non-trivial cyclic isogenies of degree $$d$$ for $$d=$$ 2 and 4.
Its isogeny class 840g consists of 6 curves linked by isogenies of degrees dividing 8.
## Growth of torsion in number fields
The number fields $K$ of degree up to 7 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ $\cong \Z/{2}\Z \times \Z/{2}\Z$ are as follows:
$[K:\Q]$ $K$ $E(K)_{\rm tors}$ Base-change curve
2 $$\Q(\sqrt{-1})$$ $$\Z/2\Z \times \Z/4\Z$$ 2.0.4.1-88200.2-k3
$$\Q(\sqrt{-14})$$ $$\Z/2\Z \times \Z/4\Z$$ Not in database
$$\Q(\sqrt{14})$$ $$\Z/2\Z \times \Z/4\Z$$ Not in database
4 $$\Q(i, \sqrt{5})$$ $$\Z/2\Z \times \Z/8\Z$$ Not in database
$$\Q(i, \sqrt{14})$$ $$\Z/4\Z \times \Z/4\Z$$ Not in database
We only show fields where the torsion growth is primitive. For each field $K$ we either show its label, or a defining polynomial when $K$ is not in the database. | 2,099 | 5,504 | {"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": 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} | 3.421875 | 3 | CC-MAIN-2020-45 | latest | en | 0.206367 |
https://www.vocabulary.com/dictionary/parallel | 1,591,106,298,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347425148.64/warc/CC-MAIN-20200602130925-20200602160925-00561.warc.gz | 927,633,654 | 11,460 | # parallel
In math, parallel means two lines that never intersect — think of an equal sign. Figuratively, parallel means similar, or happening at the same time. A story might describe the parallel lives of three close friends.
Parallel is from Greek parallēlos, from the prefix para-, "beside," plus allēlōn, "of one another," from allos, "other." As a noun, a parallel is a way in which things resemble each other — you might draw parallels between the Vietnam War and the U.S. invasion of Iraq. In specialized use, a parallel can mean one of the imaginary circles on the surface of the Earth that are parallel to the equator — the 49th parallel divides the U.S. and Canada.
## Primary Meanings of parallel
1 adjnv being everywhere equidistant and not intersecting (mathematics) one of a set of parallel geometric figures (parallel lines or planes) be parallel to 2 nadjv something having the property of being analogous to something else of or relating to the simultaneous performance of multiple operations duplicate or match
Full Definitions of parallel
1
### adj being everywhere equidistant and not intersecting
parallel lines never converge”
“concentric circles are parallel
“dancers in two parallel rows”
Synonyms:
comparable
able to be compared or worthy of comparison
antiparallel
(especially of vectors) parallel but oppositely directed
collateral
situated or running side by side
nonconvergent, nonintersecting
(of lines, planes, or surfaces) never meeting or crossing
symmetric, symmetrical
having similarity in size, shape, and relative position of corresponding parts
Antonyms:
oblique
slanting or inclined in direction or course or position--neither parallel nor perpendicular nor right-angled
perpendicular
intersecting at or forming right angles
convergent
tending to come together from different directions
divergent, diverging
tending to move apart in different directions
inclined
at an angle to the horizontal or vertical position
bias
slanting diagonally across the grain of a fabric
cata-cornered, catacorner, cater-cornered, catercorner, catty-corner, catty-cornered, kitty-corner, kitty-cornered
slanted across a polygon on a diagonal line
crabwise, sideways
(of movement) at an angle
diagonal
connecting two nonadjacent corners of a plane figure or any two corners of a solid that are not in the same face
nonparallel
(of e.g. lines or paths) not parallel; converging
oblique-angled
having oblique angles
normal
forming a right angle
orthogonal, rectangular
having a set of mutually perpendicular axes; meeting at right angles
right
having the axis perpendicular to the base
### n (mathematics) one of a set of parallel geometric figures (parallel lines or planes)
parallels never meet”
Type of:
figure
a combination of points and lines and planes that form a visible palpable shape
### n an imaginary line around the Earth parallel to the equator
Synonyms:
latitude, line of latitude, parallel of latitude
Types:
polar circle
a line of latitude at the north or south poles
horse latitude
either of two belts or regions near 30 degrees north or 30 degrees south; characterized by calms and light-baffling winds
tropic
either of two parallels of latitude about 23.5 degrees to the north and south of the equator representing the points farthest north and south at which the sun can shine directly overhead and constituting the boundaries of the Torrid Zone or tropics
Type of:
line
a spatial location defined by a real or imaginary unidimensional extent
### v be parallel to
“Their roles are paralleled by ours”
Type of:
agree, check, correspond, fit, gibe, jibe, match, tally
be compatible, similar, or consistent; coincide in their characteristics
### v make or place parallel to something
“They paralleled the ditch to the highway”
Synonyms:
collimate
Type of:
alter, change, modify
cause to change; make different; cause a transformation
2
### n something having the property of being analogous to something else
Synonyms:
analog, analogue
Types:
echo
a close parallel of a feeling, idea, style, etc.
Type of:
similarity
the quality of being similar
### adj of or relating to the simultaneous performance of multiple operations
parallel processing”
Synonyms:
synchronal, synchronic, synchronous
occurring or existing at the same time or having the same period or phase
### v duplicate or match
Synonyms:
duplicate, twin
Type of:
agree, check, correspond, fit, gibe, jibe, match, tally
be compatible, similar, or consistent; coincide in their characteristics | 1,020 | 4,508 | {"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} | 3.53125 | 4 | CC-MAIN-2020-24 | longest | en | 0.878511 |
https://www.airmilescalculator.com/distance/trs-to-cia/ | 1,603,288,640,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107876500.43/warc/CC-MAIN-20201021122208-20201021152208-00467.warc.gz | 611,695,788 | 21,606 | # Distance between Trieste (TRS) and Rome (CIA)
Flight distance from Trieste to Rome (Trieste – Friuli Venezia Giulia Airport – Ciampino–G. B. Pastine International Airport) is 282 miles / 453 kilometers / 245 nautical miles. Estimated flight time is 1 hour 1 minutes.
Driving distance from Trieste (TRS) to Rome (CIA) is 408 miles / 656 kilometers and travel time by car is about 6 hours 13 minutes.
## Map of flight path and driving directions from Trieste to Rome.
Shortest flight path between Trieste – Friuli Venezia Giulia Airport (TRS) and Ciampino–G. B. Pastine International Airport (CIA).
## How far is Rome from Trieste?
There are several ways to calculate distances between Trieste and Rome. Here are two common methods:
Vincenty's formula (applied above)
• 281.531 miles
• 453.080 kilometers
• 244.644 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 281.726 miles
• 453.394 kilometers
• 244.813 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Trieste – Friuli Venezia Giulia Airport
City: Trieste
Country: Italy
IATA Code: TRS
ICAO Code: LIPQ
Coordinates: 45°49′38″N, 13°28′19″E
B Ciampino–G. B. Pastine International Airport
City: Rome
Country: Italy
IATA Code: CIA
ICAO Code: LIRA
Coordinates: 41°47′57″N, 12°35′41″E
## Time difference and current local times
There is no time difference between Trieste and Rome.
CEST
CEST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 66 kg (147 pounds).
## Frequent Flyer Miles Calculator
Trieste (TRS) → Rome (CIA).
Distance:
282
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
282
Round trip? | 509 | 1,926 | {"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-2020-45 | latest | en | 0.796264 |
https://www.infoplease.com/math-science/mathematics/algebra/adding-and-subtracting-polynomials | 1,713,256,365,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817073.16/warc/CC-MAIN-20240416062523-20240416092523-00329.warc.gz | 752,650,086 | 21,109 | In previous sections, you've simplified expressions like 3x + 7x to get 10x, or perhaps subtracted terms like 5y - 9y to get -4y. That arithmetic makes perfect sense if you translate the mathematics into words. For example, the expression 3x + 7x literally means "three of a certain number added to seven more of that same number," which is definitely equal to "10 of that number," or 10x.
##### Talk the Talk
Like terms have variables which match exactly, like 4x2y3 and -7x2y3. You can only add or subtract two terms if they are like terms.
What I didn't tell you back then was that you were only allowed to combine the coefficients of those terms because they contained the exact same variables. Actually, any two terms whose variables match exactly are called like terms, and you cannot add terms together or subtract them from one another if they are not like terms.
##### Kelley's Cautions
Many students try to simplify the expression 4x + 5y and get 9xy, but that's wrong! Remember, you can't add or subtract 4x and 5y because the terms have different variables. It would be like adding four cats to five dogs and getting nine dats (or cogs). Unlike terms are like apples and orangesyou can't combine them.
If two terms have the same variables and get all nervous when they look at each other, you can upgrade them from like terms to love terms, but since it's hard to read the emotions of variables (they're always changing on you), most mathematicians don't even try to make that differentiation.
Now you know why the expression 13x2y3 - 5x2y3 can be simplified as 8x2y3. Since the variables in both terms match exactly (they both contain x2y3), all you have to do is combine the coefficients and attach a copy of the matching variable string.
Example 2: Simplify the following expression.
• 4x3 + 5x2 - 3x + 1 - (2x3 - 8x2 + 9x - 6)
Solution: Start by applying the distributive property (multiply everything in the parentheses by -1).
• 4x3 + 5x2 - 3x + 1 - 2x3 + 8x2 - 9x + 6
##### You've Got Problems
Problem 2: Simplify the expression 3x2 - 9 + 2(x2 - 6x + 5).
If you rewrite the expression so that like terms are grouped together, it makes simplifying easier.
• 4x3 - 2x3 + 5x2 + 8x2 - 3x - 9x + 1 + 6
Combine the coefficients of each pair of like terms.
• 2x3 + 13x2 - 12x + 7
Excerpted from The Complete Idiot's Guide to Algebra © 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc. | 690 | 2,556 | {"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} | 4.65625 | 5 | CC-MAIN-2024-18 | latest | en | 0.949717 |
http://math.stackexchange.com/questions/374087/show-that-a-cone-is-a-ruled-surface | 1,469,556,750,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825048.60/warc/CC-MAIN-20160723071025-00104-ip-10-185-27-174.ec2.internal.warc.gz | 163,590,728 | 19,367 | # show that a cone is a ruled surface
I am studying for my math final and our prof gave us a review but with no solutions... I am totally lost for this problem... We didn't even cover these types of problems in class so I don't know if I'll need to know how to do it but right now I have no clue. If anyone could help I would appreciate it.
Question: Show that the cone $z^2=ax^2 +by^2$ is a ruled surface.
I understand that a ruled surface is a surface composed of straight lines but that is as far as my knowledge goes for this question... Again any help is appreciated.
-
Instead of trying to show that the cone is composed of lines you only need show that if point $p=(x_0,y_0,z_0)$ is on the cone, you can find a line through $p$ such that the whole line is on the cone. – John Douma Apr 27 '13 at 5:26
Have you tried making a nice illustrating picture? – Babak S. Apr 27 '13 at 5:33
We know that by a cone we mean a geometric figure in $\mathbb R^3$ generated by a moving straight line going via a fixed curve $C$ and a fixed point not on the curve. As, you noted intuitively, the line can move freely. Let $$C:~~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ Let $O$ be the vertex of the cone and let the curve is in the plane $z=c$ so if $P(x,y,z)\in C$, drawing a line going through $O$ and $P$ we get $$x/x_0=y/y_0=z/z_0$$ as the line equation where $Q(x_0,y_0,z_0)$ is the intersection of the line with curve. Then, the cone has the formula: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=0$$ Choosing $a=c$, we get $x^2+y^2=z^2$. I hope this explanation be a good hint. Of course besides to other answers.
-
Me, too, Babak + 1 – amWhy Apr 28 '13 at 0:40
I appreciate the detailed answer but I am still confused. We didn't cover this topic in the course and I just wrote the practice final and again another question about this topic was on it. This time the question was: The hyperboloid of one sheet $x^2+y^2-z^2 = 1$ is called a ruled surface, which is to say that it is made up of straight lines. Prove this by showing that for any $\theta$, the line $$\frac{x - cos\theta}{sin\theta} = \frac{y-sin\theta}{-cos\theta} = z/1$$ lies entirely on the surface... I would really appreciate more help if you don't mind. – user68203 Apr 29 '13 at 6:37
@user68203: I am at your service for the next hours, if you don't mind – Babak S. Apr 29 '13 at 8:07
Never mind I figured it out. Thanks for the help. – user68203 Apr 29 '13 at 16:05
This is easy to see if we switch to cylindrical coordinates. Substitute $x=r\cos \theta$ and $y=r\sin \theta$ to get $$z^2=ar^2\cos^2\theta + br^2\sin^2\theta = r^2(a\cos^2\theta + b\sin^2\theta)$$
This gives $$z=\pm r(a\cos^2\theta + b\sin^2\theta)^{\frac{1}{2}}$$
For a cone, $a$ and $b$ must be positive so for any $\theta$ this gives a pair of lines through the origin in the $r$-$z$ plane. If $p=(x_0,y_0,z_0)$ is not the origin, then $z_0=r(a\cos^2\theta_0 + b\sin^2\theta_0)^{\frac{1}{2}}$ for some $\theta_0$ which is on the line $$z=r(a\cos^2\theta_0 + b\sin^2\theta_0)^{\frac{1}{2}}$$
Therefore, this cone is a ruled surface.
-
thank you for the help – user68203 Apr 29 '13 at 16:27
$z^2=ax^2+by^2$
Write
$x = \frac z{\sqrt a}\cos\theta$
$y=\frac z{\sqrt b}\sin\theta$
Fix an $(x,y,z)$ on the cone.
You retrieve $\theta$ using above and you get the straight line $(t,\frac t{\sqrt a}\cos\theta,\frac t{\sqrt b}\sin\theta)$ ($\theta$ fixed) through $(x,y,z)$ lying on the cone.
Since$(x,y,z)$ was arbitrary, we're done!
-
+1 you were the first inspiring one. – Babak S. Apr 27 '13 at 6:21 | 1,156 | 3,550 | {"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} | 4.28125 | 4 | CC-MAIN-2016-30 | latest | en | 0.961081 |
https://q4interview.com/quantitative-aptitude/area/medium/1/1 | 1,627,273,405,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152000.25/warc/CC-MAIN-20210726031942-20210726061942-00122.warc.gz | 484,787,806 | 31,141 | Take FREE!! Online Cocubes Mock Test to Crack various Companies Written Exams.
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# Area Questions
Home > Quantitative Aptitude > Area > General Questions
NA
SHSTTON
14
Solv. Corr.
39
Solv. In. Corr.
53
Attempted
0 M:50 S
Avg. Time
1 / 7
Choose the correct option.
The dimensions of a certain machine are 48" X 30" X 52". If the size of the machine is increased proportionately until the sum of its dimensions equals 156", what will be the increase in the shortest side?
A10"
B8"
C6"
D8.5"
Explanation:
48+30+52=130, then 156-130=26
the "proportionately" in the shortest side means 30/130*26=6"
Workspace
NA
SHSTTON
5
Solv. Corr.
39
Solv. In. Corr.
44
Attempted
2 M:32 S
Avg. Time
2 / 7
Choose the correct option.
A square and an equilateral triangle have the same perimeter. What is the ratio of the area of the circle circumscribing the square to the area of the circle inscribed in the triangle?
A9:32
B8:27
C32:09:00
D27:08:00
Explanation:
let x be side of square
perimeter of square=4x=perimeter of triangle=3*side of triangle
so side of eq. triangle=(4/3)*x
diameter of circle circumscribing the square=sqrt(2)*x
area of circle circumscribing the square=pi*(sqrt(2)*x)^2/4=(pi/2)*x^2 ----(1)
to find radius of the circle inscribed in the triangle
area of triangle=r*s=sqrt(3)/4 * (4x/3)^2
now s=(4/3)*x+(4/3)*x+(4/3)*x/2=2x
so sqrt(3)/4 * (4x/3)^2=r*2x gives
r={2/3*(3^1/2)}*x
area of the circle inscribed in the triangle=pi*({2/3*(3^1/2)}*x)^2
=pi*(4/27)*x^2 -------(2)
so reqd ratio= eqn(1)/eqn(2)
=((pi/2)*x^2)/(pi*(4/27)*x^2)=27/8
so reqd ratio=27:8
Workspace
NA
SHSTTON
11
Solv. Corr.
24
Solv. In. Corr.
35
Attempted
1 M:51 S
Avg. Time
3 / 7
Choose the correct option.
There is a rectangular Garden whose length and width is 60m X 20m.There is a walkway of uniform width around garden. Area of walkway is 516m^2. Find width of walkway?
A6
B3
C4
D2
Explanation:
Lets assume width of rectangle = x m.
So new length will be (60+2x) and width (20+2x)
Area of walkway = 516m^2
=> (60+2x)*(20+2x)-(60*20)=516.
=> (60+2x)*(20+2x)=1716.
=> (30+x)*(10+x)=429
=> x=3m
Workspace
NA
SHSTTON
51
Solv. Corr.
23
Solv. In. Corr.
74
Attempted
0 M:41 S
Avg. Time
4 / 7
Choose the correct option.
A store owner is packing small radios into larger boxes that measure 25 in. by 42 in. by 60 in. If the measurement of each radio is 7 in. by 6 in. by 5 in., then how many radios can be placed in the box?
A300
B400
C420
D480
Explanation:
No of radios that can be placed in the box = (25*42*60)/(7*6*5)=300
Workspace
NA
SHSTTON
27
Solv. Corr.
50
Solv. In. Corr.
77
Attempted
1 M:14 S
Avg. Time
5 / 7
Choose the correct option.
Two circles touch each other externally. The distance between their centres is 14 cm and the sum of their areas is 130 cm^2. Find their radii.
A9, 5
B13, 1
C11,3
D10,4
Explanation:
Lets assume the radius of 1st circle = x cm.
so, the radius of 2nd circle = (14-x)cm.
(pi*x*x)+(pi*(14-x)*(14-x))=130
By solving above eq.
x = 11 or 3
Workspace
NA
SHSTTON
20
Solv. Corr.
36
Solv. In. Corr.
56
Attempted
0 M:0 S
Avg. Time
6 / 7
Choose the correct option.
If the radius of a circle is increased by 20% then the area is increased by:
A44%
B120%
C144%
D40%
Explanation:
Let say radius = ?r^2 = ?10^2 = 100?
New radius = 10 * 120% = 12
New area = ?12^2 = 144?
Increment = [(144??100?)/100?]*100 = 44%
Workspace
NA
SHSTTON
23
Solv. Corr.
57
Solv. In. Corr.
80
Attempted
0 M:0 S
Avg. Time
7 / 7
Choose the correct option.
A photograph is to be fitted in a photo frame of sides 18 cm by 15 cm such that there is a margin of 1.5 cm left. What should be the area of the photograph?
A140 cm2
B180 cm2
C270 cm2
DNone of these
Explanation:
Given, 1.5 cm margin on bothsides.
So side will be increase by 1.5+1.5 = 3cm
Area of photograph = (18-3) * (15-3)
=180 cm^2
Workspace
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Here you can find objective type Quantitative Aptitude Area questions and answers for interview and entrance examination. Multiple choice and true or false type questions are also provided. | 2,525 | 9,653 | {"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} | 4.25 | 4 | CC-MAIN-2021-31 | latest | en | 0.696579 |
https://davida.davivienda.com/words/single-step-math-word-problems.html | 1,721,711,448,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518014.29/warc/CC-MAIN-20240723041947-20240723071947-00703.warc.gz | 170,155,233 | 8,381 | # Single Step Math Word Problems
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Number of digits for each term. The only difference between mathematically expressed equations and word problems is that, in. If the variable is not the answer to the word problem, use the. Circle only the necessary information and underline what ultimately needs to be figured. | 890 | 4,414 | {"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 | 0.888921 |
http://mathhelpforum.com/calculus/75408-find-f-x-print.html | 1,527,296,244,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00413.warc.gz | 179,866,596 | 2,347 | # Find f'(x)
$\displaystyle f(x) = \sqrt{x^3 - 2x} = (x^3 - 2x)^{\frac{1}{2}}$
use the chain rule to find $\displaystyle f'(x)$ | 55 | 128 | {"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} | 3.171875 | 3 | CC-MAIN-2018-22 | latest | en | 0.466657 |
https://au.mathworks.com/matlabcentral/answers/38999-what-is-the-reasoning-behind-the-fact-that-min-0-nan-is-0 | 1,657,182,160,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683708.93/warc/CC-MAIN-20220707063442-20220707093442-00480.warc.gz | 177,747,185 | 30,814 | # What is the reasoning behind the fact that min(0,NaN) is 0?
17 views (last 30 days)
the cyclist on 21 May 2012
Answered: Mark vanRossum on 3 Jun 2021
I know that Mathworks pays a lot of attention to this stuff, so I am wondering why the expression
>> min(0,NaN)
is 0. Returning a NaN here seems more logical to me.
Stephen23 on 25 Apr 2020
@Bryan: you should make that as an enhancement request.
Another option is to stop relying on inconsistent "default" behavior and always specify any flags, dimensions, etc. for any function that has these kind of options. Although it requires a little more typing, it has the following advantages:
• makes the intention clear
• avoids bugs, e.g. when a matrix ony has one row (and thus min returns a scalar, not a row vector)
• throws an error on versions that do not support that option, rather than silently continuing...
Walter Roberson on 21 May 2012
If you initialize the result to inf, and then loop testing whether the current value is less than the result and replace the result if it is, then since NaN < any number is false, the result will never get replaced with NaN. You would have to add special code to return NaN in such a case.
Sean de Wolski on 21 May 2012
At the bottom of the doc page:
The min function ignores NaNs
##### 2 CommentsShowHide 1 older comment
Jan on 22 May 2012
It depends on how you understand the MIN function. 0 < NaN replies FALSE, but NaN < 0 replies FALSE also. As long as it is well documented, both values are reasonable.
Daniel Shub on 22 May 2012
Given the behavior of MIN, I find it odd that there is a NANMIN function.
##### 2 CommentsShowHide 1 older comment
per isakson on 21 Jun 2012
MIN and MAX ignores NaN. MEAN and SUM does not. I guess NANMIN (in stat toolbox) is for people like me who cannot remember all the details when we cannot see the underlying logic.
M Sohrabinia on 21 Jun 2012
NaN is considered undefined, so undefined is ignored by most functions (0/0 will be resulted in NaN which is basically undefined but any number divided by 0, say 4/0, will result in inf). However, the question is why Matlab has decided to treat NaNs in a certain way in some functions, e.g., sort function will always arrange NaNs at the top end (A to Z mode). I guess Matlab has just decided to adopt some rules to handle exceptions.
Mark vanRossum on 3 Jun 2021
I encountered this when working on arrays.
x=[1 NaN 10];
y=[5 5 5];
m=min(x,y) and m=nanmin(x,y) return [1,5,5]
In V2020, min(x,y,'includenan') returns [1, NaN,5]
Here is an ugly workaround to get the desired behaviour in older versions.
q=isnan(x)
m=min(x,y)
m(q)=NaN | 688 | 2,620 | {"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-2022-27 | latest | en | 0.907467 |
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