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https://discourse.mcneel.com/t/how-to-achieve-a-nurbs-quadball-in-rhino/5244/18
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# How to achieve a nurbs quadball in rhino?
[quote=“chuck, post:11, topic:5244, full:true”] I also haven’t convinced myself that it is possible to get a perfect sphere this way, although it seems likely.
[/quote]
yeah, i can’t get a perfect sphere and my version is a weightier surface than the ones you guys are doing…
if i extract some of the isocurves then randomly measure to the cpoint, i’m still getting a slight variance… not sure if my tolerance settings have anything to do with it but i imagine so?
the only way i’ve been able to make a perfect sphere is to span an arc diagonally across a box then revolve it… then trim it with the red arcs shown in my original picture… but then there’s a seam in the surface… and it’s trimmed
Try this for the middle weight…
0.46796046944844725617756511770484
I was being foolish. If the side cvs have weight w, then the end weights of the middle isocurve would be (1+w)/2, and the middle weight of the middle isocurve would be wm = sqrt(2)/2*(1+w)/2. This means if b = middle surface cv weight, then (w+b)/2 = wm. Put this all together and we get the srf’s center weight is sqrt(2)/2*(sqrt(2/3)+1) - sqrt(2/3). If I did the arithmetic and algebra right, this is the only thing that could work.
Eyballing, I get an angle of 59.8, so you are probably right that 60 is correct.
Hi Jeff,
Yes, I think the tolerance is what you are seeing. Generally when you want to get a sphere, it is best to avoid anything that involves tolerance. By adding patches you can get arbitrarily close, but without going to the limit (infinitely many patches) you can never get there, unless you start with an exact mathematical solution.
right, but the software should be handling that aspect for me, right?
i don’t know, i’ll be surprised if anybody can mathematically figure out where to place control points and arrive at a solution which is closer than the one i’ve put up…
(that said, i’m not even sure if my solution fits the original criteria of "2 degree 3 CV count in both U and V direction (a 2 degree bezier surface),and the result should be a rational nurbs sphere."
as in-- does the 3 CV include the surface edges as 2 of the 3?
Empirically in my file I arrived at 0.467991, so I was pretty close…!
I would really like it to be a round number like that, but what I get is more like 59.846 - 60.000 does not work…
Unfortunately, as Jeff also saw, the sections through the patch are not perfect and will not simplify to arcs within the file tolerance… I don’t know if you can get closer, it seems this will always be an “approximate” solution.
–Mitch
From properties:
`Surface NURBS Surface (rational) "U": degree =2 CV count = 3 (0.000 <= U <= 106.604) "V": degree =2 CV count = 3 (0.000 <= V <= 106.604)`
–Mitch
I just noticed that the Weight command only uses 5 significant digits. Maybe that’s where the error comes from. I have to leave this alone for now, but I’ll look again when I have a chance. I know how to determine the exact solution if there is one. Now that the weights are determined, there is a formula for the necessary location of the cv. If that location doesn’t give a sphere, then we can just come close.
Chuck
Well, thanks for "weighing in"on this Chuck, interesting excercise !
–Mitch
Fun stuff. It sure beats working on STEP import.
ah… ok… the surface i was making is:
NURBS Surface
"U": degree =3 CV count = 7 (0.000 <= U <= 12.087)
“V”: degree =3 CV count = 7 (0.000 <= V <= 12.087)
I remember I read of a paper about such an object years ago, it was a true sphere but IIRC the surfaces had degree 4.
I was not able to get the text because an ACM login was needed (or something like that)
Ok, I had to finish this. It’s the geek in me. I couldn’t get this using rhino commands. I had to write a test command and use our low level stuff.QuadSphere.3dm (91.9 KB) It’s not perfect because I used an overly computational method. I’ll figure out the exact solution some day. Mitch, yours is almost identical to this.
Chuck
1 Like
one way to get the height of that point (or to get the same surface) using only rhino is to use the setup arcs that mitch is using (the red ones in the first drawing i posted) then span a single arc across the midpoint… then use sweep2 - simple sweep…
hmm… actually, that exposes a very slight error in simple sweep2…
the middle control point i get when doing it that way is at:
-2.33688e-14, 0, 2.37069
whereas the version you (chuck) posted is at
0, 0, 2.37069
so while the height is showing the same, my version has shifted ever so slightly on the X axis…
Looks like simple sweep uses the same low level code that I did, except it must do some other calculations to set things up. That’s where your numerical noise is coming from. Most likely from calculating the weights for the middle points on the sides. I already knew what they needed to be.
Chuck
[quote=“chuck, post:27, topic:5244, full:true”]… the weights for the middle points on the sides. I already knew what they needed to be.
[/quote]
not me… i still don’t.
i just know (some of) the geometric games which should arrive at the proper solutions then count on you guys to handle the geek stuff under_the_hood / out_of_sight
so just keep on doing whatever it is you all are doing
I used EdgeSrf from the arcs, it preserves the control point structure and weights from the input curves. Unfortunately it doesn’t know what to do in the middle, so it makes a very flat curved surface - but all that needs adjusting at that point is the middle CV vertical location and weight (all the others are already set correctly).
I ended up doing that by eye to a reference arc in an orthographic view, the weight affects the flatness of the surface curve, so it’s adjust the weight, nudge the control point so the apex is correct, compare along the arc, adjust, nudge, compare, etc… in the end I think I reached the limits of the display accuracy.
I had tried simple sweep2 but it didn’t work originally (that was one of my first ideas) but now I tried it and it is working… It looks slightly better than my handmade one and a lot less effort…
–Mitch
Thanks for you guys answers. @chuck where can I learn the way you calculate the weight? very cool.
I’m a rhino enthusiast，here is my new post about a similar question. link
are you seeing the same tiny deviation of the center control point in the x or y axis? the version i did was off by -2.33688e-14mm …i didn’t think rhino could even deal with or display measurements that small.
?
Mine said “0” but my model was bigger than yours…
– or maybe my Rhino is better behaved…
Anyway, `XXXXXe-14` IS zero for all intents and purposes, it’s just floating point “fuzz”…
–Mitch
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > cp Unicode version
Theorem cp 7561
Description: Collection Principle. This remarkable theorem scheme is in effect a very strong generalization of the Axiom of Replacement. The proof makes use of Scott's trick scottex 7555 that collapses a proper class into a set of minimum rank. The wff can be thought of as . Scheme "Collection Principle" of [Jech] p. 72. (Contributed by NM, 17-Oct-2003.)
Assertion
Ref Expression
cp
Distinct variable groups: ,, ,,,
Allowed substitution hints: (,)
Proof of Theorem cp
StepHypRef Expression
1 vex 2791 . . 3
21cplem2 7560 . 2
3 abn0 3473 . . . . 5
4 elin 3358 . . . . . . . 8
5 abid 2271 . . . . . . . . 9
65anbi1i 676 . . . . . . . 8
7 ancom 437 . . . . . . . 8
84, 6, 73bitri 262 . . . . . . 7
98exbii 1569 . . . . . 6
10 nfab1 2421 . . . . . . . 8
11 nfcv 2419 . . . . . . . 8
1210, 11nfin 3375 . . . . . . 7
1312n0f 3463 . . . . . 6
14 df-rex 2549 . . . . . 6
159, 13, 143bitr4i 268 . . . . 5
163, 15imbi12i 316 . . . 4
1716ralbii 2567 . . 3
1817exbii 1569 . 2
192, 18mpbi 199 1
Colors of variables: wff set class Syntax hints: wi 4 wa 358 wex 1528 wcel 1684 cab 2269 wne 2446 wral 2543 wrex 2544 cin 3151 c0 3455 This theorem is referenced by: bnd 7562 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1533 ax-5 1544 ax-17 1603 ax-9 1635 ax-8 1643 ax-13 1686 ax-14 1688 ax-6 1703 ax-7 1708 ax-11 1715 ax-12 1866 ax-ext 2264 ax-rep 4131 ax-sep 4141 ax-nul 4149 ax-pow 4188 ax-pr 4214 ax-un 4512 ax-reg 7306 ax-inf2 7342 This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-3or 935 df-3an 936 df-tru 1310 df-ex 1529 df-nf 1532 df-sb 1630 df-eu 2147 df-mo 2148 df-clab 2270 df-cleq 2276 df-clel 2279 df-nfc 2408 df-ne 2448 df-ral 2548 df-rex 2549 df-reu 2550 df-rab 2552 df-v 2790 df-sbc 2992 df-csb 3082 df-dif 3155 df-un 3157 df-in 3159 df-ss 3166 df-pss 3168 df-nul 3456 df-if 3566 df-pw 3627 df-sn 3646 df-pr 3647 df-tp 3648 df-op 3649 df-uni 3828 df-int 3863 df-iun 3907 df-iin 3908 df-br 4024 df-opab 4078 df-mpt 4079 df-tr 4114 df-eprel 4305 df-id 4309 df-po 4314 df-so 4315 df-fr 4352 df-we 4354 df-ord 4395 df-on 4396 df-lim 4397 df-suc 4398 df-om 4657 df-xp 4695 df-rel 4696 df-cnv 4697 df-co 4698 df-dm 4699 df-rn 4700 df-res 4701 df-ima 4702 df-iota 5219 df-fun 5257 df-fn 5258 df-f 5259 df-f1 5260 df-fo 5261 df-f1o 5262 df-fv 5263 df-recs 6388 df-rdg 6423 df-r1 7436 df-rank 7437
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# Enumerable.Max(Of TSource) Method (IEnumerable(Of TSource), Func(Of TSource, Decimal))
.NET Framework 4.5
Invokes a transform function on each element of a sequence and returns the maximum Decimal value.
Namespace: System.Linq
Assembly: System.Core (in System.Core.dll)
```'Declaration
<ExtensionAttribute> _
Public Shared Function Max(Of TSource) ( _
source As IEnumerable(Of TSource), _
selector As Func(Of TSource, Decimal) _
) As Decimal
```
#### Type Parameters
TSource
The type of the elements of source.
#### Parameters
source
Type: System.Collections.Generic.IEnumerable(Of TSource)
A sequence of values to determine the maximum value of.
selector
Type: System.Func(Of TSource, Decimal)
A transform function to apply to each element.
#### Return Value
Type: System.Decimal
The maximum value in the sequence.
#### Usage Note
In Visual Basic and C#, you can call this method as an instance method on any object of type IEnumerable(Of TSource). When you use instance method syntax to call this method, omit the first parameter. For more information, see Extension Methods (Visual Basic) or Extension Methods (C# Programming Guide).
ExceptionCondition
ArgumentNullException
source or selector is Nothing.
InvalidOperationException
source contains no elements.
The Max(Of TSource)(IEnumerable(Of TSource), Func(Of TSource, Decimal)) method uses the Decimal implementation of IComparable(Of T) to compare values.
You can apply this method to a sequence of arbitrary values if you provide a function, selector, that projects the members of source into a numeric type, specifically Decimal.
In Visual Basic query expression syntax, an Aggregate Into Max() clause translates to an invocation of Max.
The following code example demonstrates how to use Max(Of TSource)(IEnumerable(Of TSource), Func(Of TSource, Int32)) to determine the maximum value in a sequence of projected values.
Note
This code example uses an overload of this overloaded method that is different from the specific overload that this topic describes. To extend the example to this topic, change the body of the selector function.
```Structure Pet
Public Name As String
Public Age As Integer
End Structure
Sub MaxEx4()
' Create an array of Pet objects.
Dim pets() As Pet = {New Pet With {.Name = "Barley", .Age = 8}, _
New Pet With {.Name = "Boots", .Age = 4}, _
New Pet With {.Name = "Whiskers", .Age = 1}}
' Determine the "maximum" pet by passing a
' lambda expression to Max() that sums the pet's age
' and name length.
Dim max As Integer = pets.Max(Function(pet) _
pet.Age + pet.Name.Length)
' Display the result.
MsgBox("The maximum pet age plus name length is " & max)
End Sub
' This code produces the following output:
'
' The maximum pet age plus name length is 14
```
#### .NET Framework
Supported in: 4.5.1, 4.5, 4, 3.5
#### .NET Framework Client Profile
Supported in: 4, 3.5 SP1
#### Portable Class Library
Supported in: Portable Class Library
#### .NET for Windows Store apps
Supported in: Windows 8
#### .NET for Windows Phone apps
Supported in: Windows Phone 8.1, Windows Phone 8, Silverlight 8.1
Windows Phone 8.1, Windows Phone 8, Windows 8.1, Windows Server 2012 R2, Windows 8, Windows Server 2012, Windows 7, Windows Vista SP2, Windows Server 2008 (Server Core Role not supported), Windows Server 2008 R2 (Server Core Role supported with SP1 or later; Itanium not supported)
The .NET Framework does not support all versions of every platform. For a list of the supported versions, see .NET Framework System Requirements.
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# A corners-first solution method for Rubik's cube
by Victor Ortega and Josef Jelinek
This solution method is designed to solve Rubik's cube quickly, efficiently, and without having to memorize a lot of sequences or take a large business cash advance to pay someone to help you out. For ease and speed of execution, turns are restricted to the faces U, R, and F, and center and middle slices. Strong preference is given to R, since it is one of the easiest faces to turn for many people. Yet all sequences are minimal (or very close to minimal) by the slice metric.
This solution method orients cubies before positioning them. The idea is that it is easier to permute cubies after they've been oriented than before orienting them, because once the cubies have been oriented, the facelet colors that determine their permutation make easily identifiable patterns on the cube. Orienting cubies, whether done before or after positioning them, is always easy because orientation requires focusing on only one face color and on the patterns that that color makes on the cube. For middle-slice edges on the last layer, permuting cubies after they've been oriented is a very simple affair, thus reinforcing this principle.
Do not worry about centers or edges while solving corners. Position centers while beginning to solve edges. You really only need to position U and D centers at that point, but positioning all centers may make things easier for you. Middle-slice centers will be positioned along with middle-slice edges on the last step.
This solution method is based on Minh Thai's Winning Solution. Ideas and sequences are borrowed from other solution methods, and appropriate attributions are made in those sections.
## Stage I: Solve the corners
### 1. Orient U corners
You should be able to manage this on your own. Don't worry about positions--all corners will be permuted in step 3. You should be able to orient U corners in 6 moves or less. For the greatest speed and efficiency, try to do this in one look.
### 2. Orient D corners
Rotate the whole cube so that D becomes U. Orient the corners depending on which of the seven patterns below you see:
letter T pattern: R U R' U' F' U' F letter L pattern: F R' F' U' R' U R sune pattern #1: R U2 R' U' R U' R' sune pattern #2: R U R' U R U2 R' (inverse of #1 in both respects) letter pi pattern: R U R2 F' R2 U R' letter U pattern: R' F' U' F U R letter H pattern: R2 U2 R' U2 R2
### 3. Permute all corners, by method of number of solved "edges"
The ideas for this section come from this web page. Check it out for a different description of the process and to see several examples.
An "edge" here represents two adjacent corners on the U or D layer. Such an edge is considered to be solved correctly if the two corner cubies are positioned correctly relative to each other. A solved edge will be easy to identify because the two adjacent facelets on the side (not U or D) will be of the same color. A layer can have only zero, one, or four correct edges.
The number and location of correct edges can be quickly identified by merely looking at two adjacent side faces (that is, not U or D). For a given layer, if you see one correct edge and one incorrect edge, then there is only one correct edge on that layer. If you see two correct edges, then all four edges are correct. If you see no correct edges but both edges consist of opposite colors, then there are no correct edges on that layer. If you see no correct edges and only one edge consisting of opposite colors, then there is one correct edge on that layer, and it is opposite to the edge with the opposite colors.
Proceed with one of the following sequences depending on how many solved edges you have:
0: F2 R2 F2 1 (DB solved): R U' F U2 F' U R'((UB solved): R' U R' B2 R U' R) 2 (DB and UB solved): R2 U F2 U2 R2 U R2 4 (D solved): F2 U' R U' R' U F2 U R U R' 5 (UF not solved): R U' R F2 R' U R F2 R2
## Stage II: Solve the edges
At this point, align corners and position centers. The cube is now fully symmetric except for edges. Pick the new U and D depending on what will make solving U and D edges easiest. Steps 4 and 5 can be combined, although this requires monitoring more cubies simultaneously and may not yield a speed gain or a reduction in number of movements. See this page for details on steps 4 through 6.
### 6. Solve one more U or D edge, depending on which is easier
At this point, the last U or D edge will either be in the middle layer, in position but not oriented, or solved. Depending on the case, proceed as follows to solve that last edge (if necessary) while orienting the middle layer edges.
### 7. Solve last U edge, orient middle layer
For another perspective on this process, check out Ron's approach
a) U edge in the middle layer
Position the "notch" at UR and the edge cubie at LF, with the facelet with the U color on the L face. If the edge cubie is twisted, mirror vertically (UR becomes DR, R becomes R' and R' becomes R)
In the diagram below, edges A, B, or C are oriented correctly (C) if that facelet's color matches the adjacent center or the opposite center. Otherwise it's incorrectly oriented (I).
(facelet D belongs in facelet position B)("adjacent center" between A and C)
Edge AEdge BEdge CPatternSequence
CCC R R' R' R
CCI R' R' R' R'
CIC R R2 R
CII R' R R' R
ICC R2 R R R R'
ICI R' R' R' R R R
IIC R' R' R R
III R' R' R' R2 R
b) U edge in position but twisted
There will be 1 or 3 twisted edges in the middle layer:
FR twisted: R U2 R' R2 R' U2 R' FR not twisted: R' R' R' R'
c) U edge solved
There will be 0, 2, or 4 edges twisted in the middle layer:
2 adjacent (FL and FR): R2 F F' R2 F F' 2 opposite (FL and BR): F F' R2 F F' R2 all four: R F2 R2 R R F2 R2 R all four, when all edges are positioned correctly: F' U' F ( R)4 F' U F
### 8. Position middle layer
send FR to BL, BL to BR, and BR to FR: R2 R2 () exchange centers with opposites: () exchage FR with BR, FL with BL: R2 R2 ()
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# Chapter 25 Electric circuits Voltage and current Series
• Slides: 54
Chapter 25 : Electric circuits • Voltage and current • Series and parallel circuits • Resistors and capacitors • Kirchoff’s rules for analysing circuits
Electric circuits • Closed loop of electrical components around which current can flow, driven by a potential difference • Current (in Amperes A) is the rate of flow of charge • Potential difference (in volts V) is the work done on charge
Electric circuits
Electric circuits • Same principles apply in more complicated cases!
Electric circuits • How do we deal with a more complicated case? What is the current flowing from the battery?
Electric circuits • When components are connected in series, the same electric current flows through them • Charge conservation : current cannot disappear!
Electric circuits • When components are connected in parallel, the same potential difference drops across them • Points connected by a wire at the same voltage!
Electric circuits • When there is a junction in the circuit, the inward and outward currents to the junction are the same • Charge conservation : current cannot disappear!
Consider the currents I 1, I 2 and I 3 as indicated on the circuit diagram. If I 1 = 2. 5 A and I 2 = 4 A, what is the value of I 3? 1. 2. 3. 4. 5. 6. 5 A 1. 5 A − 1. 5 A 0 A The situation is not possible I 2 I 1 I 3
Consider the currents I 1, I 2 and I 3 as indicated on the circuit diagram. If I 1 = 2. 5 A and I 2 = 4 A, what is the value of I 3? I 2 I 1 current in = current out (Negative sign means opposite direction to arrow. ) I 3
A 9. 0 V battery is connected to a 3 W resistor. Which is the incorrect statement about potential differences (voltages)? 1. 2. 3. 4. Vb – Va = 9. 0 V Vb – Vc = 0 V Vc – Vd = 9. 0 V Vd – Va = 9. 0 V a b d c
Resistors in circuits • Resistors are the basic components of a circuit that determine current flow : Ohm’s law I = V/R
Resistors in series/parallel • If two resistors are connected in series, what is the total resistance? same current
Resistors in series/parallel • If two resistors are connected in series, what is the total resistance? • Total resistance increases in series!
Resistors in series/parallel • Total resistance increases in series!
Resistors in series/parallel • If two resistors are connected in parallel, what is the total resistance?
Resistors in series/parallel • If two resistors are connected in parallel, what is the total resistance?
Resistors in series/parallel • If two resistors are connected in parallel, what is the total resistance? • Total resistance decreases in parallel!
Resistors in series/parallel • Total resistance decreases in parallel!
Resistors in series/parallel • What’s the current flowing? (1) Combine these 2 resistors in parallel: (2) Combine all the resistors in series:
If an additional resistor, R 2, is added in series to the circuit, what happens to the power dissipated by R 1? R 1 1. Increases 2. Decreases 3. Stays the same
If an additional resistor, R 3, is added in parallel to the circuit, what happens to the total current, I? 1. 2. 3. 4. I V Increases Decreases Stays the same Depends on R values Parallel resistors: reciprocal effective resistance is sum of reciprocal resistances
Series vs. Parallel CURRENT Same current through all series elements VOLTAGE Voltages add to total circuit voltage RESISTANCE Adding resistance increases total R Current “splits up” through parallel branches Same voltage across all parallel branches Adding resistance reduces total R String of Christmas lights – connected in series Power outlets in house – connected in parallel
Voltage divider Consider a circuit with several resistors in series with a battery. Current in circuit: The potential difference across one of the resistors (e. g. R 1) The fraction of the total voltage that appears across a resistor in series is the ratio of the given resistance to the total resistance.
What must be the resistance R 1 so that V 1 = 2. 0 V? 1. 0. 80 W 2. 1. 2 W 3. 6. 0 W 4. 30 W R 1 12 V 6. 0 W
Capacitors • A capacitor is a device in a circuit which can be used to store charge A capacitor consists of two charged plates … Electric field E It’s charged by connecting it to a battery …
Capacitors • A capacitor is a device in a circuit which can be used to store charge Example : store and release energy …
Capacitors • The capacitance C measures the amount of charge Q which can be stored for given potential difference V +Q -Q (Value of C depends on geometry…) V • Unit of capacitance is Farads [F]
Resistor-capacitor circuit • Consider the following circuit with a resistor and a capacitor in series V What happens when we connect the circuit?
Resistor-capacitor circuit • When the switch is connected, the battery charges up the capacitor • Move the switch to point a • Initial current flow I=V/R V • Charge Q flows from battery onto the capacitor • Potential across the capacitor VC=Q/C increases • Potential across the resistor VR decreases • Current decreases to zero
Resistor-capacitor circuit • When the switch is connected, the battery charges up the capacitor V
Resistor-capacitor circuit • When the battery is disconnected, the capacitor pushes charge around the circuit • Move the switch to point b • Initial current flow I=VC/R V • Charge flows from one plate of capacitor to other • Potential across the capacitor VC=Q/C decreases • Current decreases to zero
Resistor-capacitor circuit • When the battery is disconnected, the capacitor pushes charge around the circuit V
Capacitors in series/parallel • If two capacitors are connected in series, what is the total capacitance? Same charge must be on every plate!
Capacitors in series/parallel • If two capacitors are connected in series, what is the total capacitance? • Total capacitance decreases in series!
Capacitors in series/parallel • If two capacitors are connected in parallel, what is the total capacitance?
Capacitors in series/parallel • If two capacitors are connected in parallel, what is the total capacitance? • Total capacitance increases in parallel!
Two 5. 0 F capacitors are in series with each other and a 1. 0 V battery. Calculate the charge on each capacitor (Q) and the total charge drawn from the battery (Qtotal). 1. 2. 3. 4. Q = 5. 0 C, Qtotal = 5. 0 C Q = 0. 25 C, Qtotal = 0. 50 C Q = 2. 5 C, Qtotal = 2. 5 C Q = 2. 5 C, Qtotal = 5. 0 C 5 F 1 V 5 F
Two 5. 0 F capacitors are in series with each other and a 1. 0 V battery. Calculate the charge on each capacitor (Q) and the total charge drawn from the battery (Qtotal). 5 F 1 V Potential difference across each capacitor = 0. 5 V 5 F
Two 5. 0 F capacitors are in parallel with each other and a 1. 0 V battery. Calculate the charge on each capacitor (Q) and the total charge drawn from the battery (Qtotal). 1. 2. 3. 4. Q = 5. 0 C, Qtotal = 5. 0 C Q = 0. 2 C, Qtotal = 0. 4 C Q = 5. 0 C, Qtotal = 10 C Q = 2. 5 C, Qtotal = 2. 5 C 1 V 5 F
Two 5. 0 F capacitors are in parallel with each other and a 1. 0 V battery. Calculate the charge on each capacitor (Q) and the total charge drawn from the battery (Qtotal). 1 V Potential difference across each capacitor = 1 V 5 F
Resistors vs. Capacitors
Resistors vs. Capacitors
Kirchoff’s rules • Sometimes we might need to analyse more complicated circuits, for example … Q) What are the currents flowing in the 3 resistors? • Kirchoff’s rules give us a systematic method
Kirchoff’s rules • What are the currents flowing in the 3 resistors? Kirchoff’s junction rule : the sum of currents at any junction is zero
Kirchoff’s rules • The sum of currents at any junction is zero • Watch out for directions : into a junction is positive, out of a junction is negative
Kirchoff’s rules • What are the currents flowing in the 3 resistors? Kirchoff’s junction rule : the sum of currents at any junction is zero
Kirchoff’s rules • What are the currents flowing in the 3 resistors? Kirchoff’s loop rule : the sum of voltage changes around a closed loop is zero
Kirchoff’s rules • Sum of voltage changes around a closed loop is zero • Consider a unit charge (Q=1 Coulomb) going around this loop • It gains energy from the battery (voltage change +V) • It loses energy in the resistor (voltage change - I R) • Conservation of energy : V - I R = 0 (or as we know, V = I R)
Kirchoff’s rules • What are the currents flowing in the 3 resistors? Kirchoff’s loop rule : the sum of voltage changes around a closed loop is zero
Kirchoff’s rules • What are the currents flowing in the 3 resistors? Kirchoff’s loop rule : the sum of voltage changes around a closed loop is zero
Kirchoff’s rules • What are the currents flowing in the 3 resistors? We now have 3 equations:
Consider the loop shown in the circuit. The correct Kirchoff loop equation, starting at “a” is •
Chapter 25 summary • Components in a series circuit all carry the same current • Components in a parallel circuit all experience the same potential difference • Capacitors are parallel plates which store equal & opposite charge Q = C V • Kirchoff’s junction rule and loop rule provide a systematic method for analysing circuits
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# Projectile Motion
Only available on StudyMode
• Published : April 10, 2013
Text Preview
Projectile Motion
Purpose: Apply the concepts of two-dimensional kinematics (projectile motion) to predict the impact point of an object as its velocity increases.
Introduction: The most common example of an object that is moving in two dimensions is a projectile. A projectile is an object upon which the only force acting is gravity. That is to say a projectile is any object that once projected or dropped continues in motion by its own, and is influenced only by the downward force of gravity. There are a number of examples of projectiles, such as an object dropped from rest, an object that is thrown vertically upward, and an object which is thrown upward at an angle to the horizontal is also a projectile. Since a projectile is an object that only has a single force acting on it, the free-body diagram of a projectile would show only a single force acting downwards; labeled force of gravity. Regardless of which direction a projectile is moving, the free-body diagram of the projectile is still as depicted in the diagram at the right.
In the case of projectiles, one can use information about the initial velocity and position of a projectile to predict such things as how much time the projectile is in the air and how far the projectile will go. For example, a projectile launched with an initial horizontal velocity from an elevated position will follow a parabolic path to the ground. Unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile. These can all be solved for by using the following equations: [pic] and [pic]. Where y is vertical distance, x is horizontal distance, t is time, a is acceleration, and v is velocity.
Question: What happens to an object’s impact point in two-dimensional kinematics when its speed increases?
Hypothesis: If the speed of an object increases in two-dimensional kinematics then its impact point will also increase because the speed acquired in one dimension allows the object to travel further even when travelling in the second dimension.
Variables: Independent: SpeedIndependent: Impact Point
Controls: way of releasing object (ramp), weight/size of object, point of release on the ramp (in each trial), distance ball travels across the table, same table height.
Materials:Computerone photogateright angle clamp
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This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A160247 Table read by antidiagonals of "less regular type 1" truncated octahedron numbers built from face-centered-cubic sphere packing. 0
1, 13, 6, 55, 38, 19, 147, 116, 79, 44, 309, 260, 201, 140, 85, 561, 490, 405, 314, 225, 146, 923, 826, 711, 586, 459, 338, 231, 1415, 1288, 1139, 976, 807, 640, 483, 344 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS The sequence contains regular cuboctahedra (A005902) on the x-axis, regular octahedra (A005900) on the y-axis, and regular truncated octahedra (A005910) on the diagonal. As for the rest, they each have 6 squares of the same area, while the 8 hexagons (of another same area) have 2 side lengths which alternate. The x-axis represents an increasing degree of truncation, while the y-axis represents an increasing quantity of units on the remaining original octahedron edge. REFERENCES Main Title: Polyhedra primer / Peter Pearce and Susan Pearce. Published/Created: New York : Van Nostrand Reinhold, c1978. Description: viii, 134 p. : ill. ; 24 cm. ISBN: 0442264968 Main Title: The book of numbers / John H. Conway, Richard K. Guy. Published/Created: New York, NY : Copernicus c1996. Description: ix, 310 p. : ill. (some col.) ; 24 cm. ISBN: 038797993X LINKS FORMULA v=(2*y^3+10*x^3+12*y^2x+24*Y*x^2-12*y^2-39*x^2-48*y*x+25*y+47*x-18)/3 PROG (Excel) Paste the following formula into cell C3, and fill down and right to desired table size. All volumes 10, 000 and under are covered by column Q and row 27. =(2*(ROW()-2)^3+10*(COLUMN()-2)^3+12*(ROW()-2)^2*(COLUMN()-2)+24*(ROW()-2)*(COLUMN()-2)^2-12*(ROW()-2)^2-39*(COLUMN()-2)^2-48*(ROW()-2)*(COLUMN()-2)+25*(ROW()-2)+47*(COLUMN()-2)-18)/3 CROSSREFS Sequence in context: A220131 A268723 A300942 * A300886 A301496 A078438 Adjacent sequences: A160244 A160245 A160246 * A160248 A160249 A160250 KEYWORD easy,nonn,tabl AUTHOR Chris G. Spies-Rusk (chaosorder4(AT)gmail.com), May 05 2009, May 19 2009 STATUS approved
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Last modified March 25 12:11 EDT 2019. Contains 321470 sequences. (Running on oeis4.)
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# What do our clocks read?
## Summary:
It is said that when asked- “what is ‘time’?”- Einstein once replied, “Time is what our clocks read”. But what do our clocks read? One may easily reply to this question- our clocks read ‘Time’. But isn’t it a circular argument?I have tried to find an answer.
## Main Question or Discussion Point
What Do Our Clocks Read?
It is said that when asked- “what is ‘time’?”- Einstein once replied, “Time is what our clocks read”. But what do our clocks read? One may easily reply to this question- our clocks read ‘Time’. But isn’t it a circular argument? We can understand it by an analogy. If someone asks you- “what is length?”, would your answer be “Length is what a measuring tape reads”? And if he asks back, “What does a measuring tape read?”; would you reply again- “A measuring tape reads length”? Definitely this will not be a genuine answer. It will be a circular answer. A specific answer to this question might be; “Length is one of the spatial extensions of an object”.
Similarly, the question- “What is time?” too needs a more specific answer.
I have tried to reach the answer through the above analogy itself. Let us ask the question, “What does a measuring tape read?” Is there any invisible length in the sky that it measures? Does it make any sense if we say that it is measuring an invisible spatial dimension of space? I think these answers hardly make a good sense. In my view, a more sensible answer will be that a measuring tape first of all reads its own length- its longest spatial extension. It tells that it is a meter long. It also tells that a meter- a man-made unit of length- is this much long. Then, by comparison, it measures the lengths of other objects too. After all, a measuring tape is just like any other object, any other tape; the only difference is that it is graduated, or marked (according to a man-made standard) to read its spatial extension.
Similar explanation can be given for a clock. A clock too is like any other object that gets old every moment- that is, extends in its fourth dimension. Other objects too get old every moment but are generally not marked to measure their extension into their fourth dimension (though there are many that have such markings, like a developing embryo or a beating heart, albeit not very precise). A clock has been marked (according to some man-made standard) to measure and read its extension into its fourth dimension. It seems that, like a measuring tape, it too does not measure any invisible fourth dimension of space, any invisible time. Rather, it measures its extension into its own fourth dimension (its aging). In a simpler term, a clock measures and reads its own aging. Then, by comparison, it reads the aging of other objects.
Now suppose, there is a growing tree, increasing in height (say length) by a meter every year. Is its length (the spatial dimension) responsible for its growth? Or its growth is responsible for its length? Certainly, the latter statement is true, not the former. The tree’s growth is responsible for its length (a spatial dimension). The spatial dimension of the tree is thus not the cause but the effect. It is not a requirement but an acquirement.
Now we see that, along with its length, the tree is also gaining ‘age’. Similar question can be asked for its age too. Is age (the fourth dimension) of the tree responsible for its growth? Or its growth is responsible for its age? Naturally, its growth is responsible for its age. Thus, age, or the fourth dimension too is not the cause but the effect. Fourth dimension (say, time) is, therefore, not a requirement for aging, but is an acquirement for aging. Time seems to be the fourth dimension of objects, a measurement of their aging. A clock measures its own aging.
## Answers and Replies
Related Other Physics Topics News on Phys.org
fresh_42
Mentor
You have basically two possibilities with that question: either dive into philosophy - deeply, or take it as it is - a clock is a machine which counts. That's all in my opinion. A clock counts. Ancient roman clocks counted water drops, modern atomic clocks count frequency. With the next question, why counting takes time, we are already in the middle of philosophy, a terrain on which your question cannot be answered satisfactory here.
PeroK
Science Advisor
Homework Helper
Gold Member
The critical thing for time to make sense is that certain natural processes remain synchronised.
There is a fixed relationship between years (orbit of the Earth), cycles of the moon, days (revolutions of the Earth) etc.
If you then design an hour glass that empties 24 times the first day you find it always takes 24 times every day. And so on.
This is what creates the concept of time. You then separate natural processes into those that are synchronised and those that are not. Those that remain synchronised are candidates on which to base the design of a clock.
You have basically two possibilities with that question: either dive into philosophy - deeply, or take it as it is - a clock is a machine which counts. That's all in my opinion. A clock counts. Ancient roman clocks counted water drops, modern atomic clocks count frequency. With the next question, why counting takes time, we are already in the middle of philosophy, a terrain on which your question cannot be answered satisfactory here.
Thanks
The critical thing for time to make sense is that certain natural processes remain synchronised.
There is a fixed relationship between years (orbit of the Earth), cycles of the moon, days (revolutions of the Earth) etc.
If you then design an hour glass that empties 24 times the first day you find it always takes 24 times every day. And so on.
This is what creates the concept of time. You then separate natural processes into those that are synchronised and those that are not. Those that remain synchronised are candidates on which to base the design of a clock.
thanks
Dale
Mentor
Summary:: It is said that when asked- “what is ‘time’?”- Einstein once replied, “Time is what our clocks read”. But what do our clocks read? One may easily reply to this question- our clocks read ‘Time’. But isn’t it a circular argument?
If A=B then B=A. That is not circularity, it is the symmetric property of equality. The problem isn’t the answer, the problem is the question. No amount of fiddling with the answer can fix a bad question.
The question is bad because the answer is already provided and is obvious. It relies on the answerer’s kindness to not tell the questioner that they are asking a bad question. There is no learning value to the question and it serves only entertainment value in watching the answerer struggle with the purely social challenge of not insulting the questioner, I.e. watching the answerer figure out how to point out the stupidity of the question without in any way implying that the questioner is stupid.
Not all questions are good, and this is one example of a bad question. The questioner should be told so, and helped to understand why. There are two possibilities: either the questioner knows the question is bad, or they do not. If they do not know then they need to be taught, asking good questions is a difficult skill. If they do know then they should be taught that it is not socially acceptable to prey on a kind answerer’s good disposition for entertainment.
Last edited:
If A=B then B=A. That is not circularity, it is the symmetric property of equality. The problem isn’t the answer, the problem is the question. No amount of fiddling with the answer can fix a bad question.
The question is bad because the answer is already provided and is obvious. It relies on the answerer’s kindness to not tell the questioner that they are asking a bad question. There is no learning value to the question and it serves only entertainment value in watching the answerer struggle with the purely social challenge of not insulting the questioner, I.e. watching the answerer figure out how to point out the stupidity of the question without in any way implying that the questioner is stupid.
Not all questions are good, and this is one example of a bad question. The questioner should be told so, and helped to understand why. There are two possibilities: either the questioner knows the question is bad, or they do not. If they do not know then they need to be taught, asking good questions is a difficult skill. If they do know then they should be taught that it is not socially acceptable to prey on a kind answerer’s good disposition for entertainment.
Great. Thanks for such an educating answer.
Does time exist?
PeroK
Science Advisor
Homework Helper
Gold Member
Does time exist?
From a physics perspective why wouldn't it? It's a measurable quantity.
Does time exist?
In my view, time does exist; in the manner length, width and height exist. Length, width and height are not things but properties of objects. Time too is a property of objects. Length, width and height are the three spatial dimensions of objects; time is the fourth. Defining time as the fourth dimension of space ( intermingled with the rest three) creates an illusion of its being a thing, at least for a common man. Defining it as the fourth dimension of objects removes all illusions from its face, without interfering with physical laws.
Does time exist?
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
2019 Award
So the point in your asking the question was just so you could tell us your own, unorthodox view?
Dale
Mentor
Length, width and height are not things but properties of objects. Time too is a property of objects.
You are confusing “time” with “lifetime”. An object has length, width, height, and lifetime.
Mister T
Science Advisor
Gold Member
If A=B then B=A. That is not circularity, it is the symmetric property of equality.
Right. But $A \equiv B$ and $B \equiv A$ is circular.
Mister T
Science Advisor
Gold Member
Summary:: It is said that when asked- “what is ‘time’?”- Einstein once replied, “Time is what our clocks read”. But what do our clocks read? One may easily reply to this question- our clocks read ‘Time’. But isn’t it a circular argument?
Yes, it is circular reasoning. But any body of knowledge must necessarily contain primitives that can't be defined. Imagine you and your friends have managed to create a new language that you can use to communicate with each other so that the communications are not understood by others and in that sense are kept private. You decide to create a dictionary of this new language. But how shall you define the first term? You would necessarily have to define it in terms of other words that you haven't yet defined.
In Post #3 @PeroK outlines what we need for the concept of time to be useful in physics.
The other stuff, as @fresh_42 reminds us in Post #2, is philosophy.
Dale
Mentor
Right. But $A \equiv B$ and $B \equiv A$ is circular.
But that is not what is going here. The statement “Time is what our clocks read” could be seen as a definition or a simple equality, but the statement “what our clocks read is time” would not be seen as a definition, only simple equality. My comments above hold here.
Yes, it is circular reasoning.
While there is plenty of circularity in science, this is not the case here in this specific thread.
So the point in your asking the question was just so you could tell us your own, unorthodox view?
No, I do not intend to tell anything. I am no expert in this field. I simply wish to present an alternative view for criticism. Thanks
Yes, it is circular reasoning. But any body of knowledge must necessarily contain primitives that can't be defined. Imagine you and your friends have managed to create a new language that you can use to communicate with each other so that the communications are not understood by others and in that sense are kept private. You decide to create a dictionary of this new language. But how shall you define the first term? You would necessarily have to define it in terms of other words that you haven't yet defined.
In Post #3 @PeroK outlines what we need for the concept of time to be useful in physics.
The other stuff, as @fresh_42 reminds us in Post #2, is philosophy.
OK. Thanks.
You are confusing “time” with “lifetime”. An object has length, width, height, and lifetime.
Thanks. At least, I have been able to convey my 'confusion' to you. I am not confusing 'time' with 'lifetime'. Rather, I wish to present my view that 'lifetime' is itself 'time'. 'Time' is the 'measurement of lifetime' itself. And I want to receive you people's criticism on my view.
With Regards.
You have basically two possibilities with that question: either dive into philosophy - deeply, or take it as it is - a clock is a machine which counts. That's all in my opinion. A clock counts. Ancient roman clocks counted water drops, modern atomic clocks count frequency. With the next question, why counting takes time, we are already in the middle of philosophy, a terrain on which your question cannot be answered satisfactory here.
Should a Geiger Counter be considered a clock??
Perhaps the notion of counting is necessary but not quite sufficient. In particular the clock must count a series of "equivalent" events. The fact that the interval between these equivalent events is somehow universal makes time a useful notion. Perhaps this is a distinction without a difference.....I cannot decide.
Dale
Mentor
I wish to present my view
That is not the purpose of this site. We are explicitly not interested in learning about views that are not consistent with the professional scientific literature.
'Time' is the 'measurement of lifetime' itself. And I want to receive you people's criticism on my view.
First and most important criticism is that this is not the view in the professional scientific literature.
Second criticism is that this probably does lead to a genuine circular reasoning. If "time" is defined as "the measurement of lifetime" then we have to define "lifetime". As far as I know "lifetime" is "the difference in the time between the beginning and the end of an object", which takes us back to the previous definition. While there may be a way out of that circularity, it is not something which has been developed or accepted by mainstream scientists.
Since this thread is now going into personal speculation, we will close it.
fresh_42
Mentor
Should a Geiger Counter be considered a clock??
Yes, it does. We use e.g. in carbon dating. The difference is only in the precision of the clock. Beside this, I only stated that a clock counts, which is true. I did not claim that any counting process leads to a (reasonable) clock.
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College Algebra Exam Review 399
# College Algebra Exam Review 399 - 409 8.7 JORDAN CANONICAL...
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Unformatted text preview: 409 8.7. JORDAN CANONICAL FORM The companion matrix of p.x/ is the 1–by–1 matrix Œ , and N is the 1–by–1 matrix Œ1. The matrix of T1 with respect to B is 2 3 00 00 61 0 0 07 6 7 0 07 60 1 6: : : 7 Jm . / D 6 : : :: ::: : : 7 : : :7 :: 6: : 6 7 40 0 0 : : : 05 000 1 Definition 8.7.1. The matrix Jm . / is called the Jordan block of size m with eigenvalue . Lemma 8.7.2. Let T be a linear transformation of a vector space V over a field K . V has an ordered basis with respect to which the matrix of T is the Jordan block Jm . / if, and only if, V is a cyclic KŒx–module with period .x /m . Proof. We have seen that if V is a cyclic KŒx–module with generator v0 and period .x /m , then B D .v0 ; .T /v0 ; : : : ; .T /m 1 v0 / is an ordered basis of V such that ŒT B D Jm . /. Conversely, suppose that B D .v0 ; v1 ; : : : ; vm / is an ordered basis of V such that ŒT B D Jm . /. The matrix of T with respect to B is the Jordan block Jm .0/ with zeros on the main diagonal. It follows that .T /k v0 D vk for 0 Ä k Ä m 1, while .T /m v0 D 0. Therefore, V is cyclic with generator v0 and period .x /m . I Suppose that the characteristic polynomial of T factors into linear factors in KŒx. Then the elementary divisors of T are all of the form .x /m . Therefore, in the elementary divisor decomposition .T; V / D .T1 ; V1 / ˚ ˚ .T t ; V t /; each summand Vi has a basis with respect to which the matrix of Ti is a Jordan block. Hence V has a basis with respect to which the matrix of T is block diagonal with Jordan blocks on the diagonal. ...
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# Re: [R] Monte Carlo Simulation
From: Shane Phillips <SPhillips_at_Lexington1.net>
Date: Fri, 15 Apr 2011 18:23:05 -0400
I apologize for my last post. here is the script I forgot to paste!
subject=1:1000
treat=rbinom(1*1000,1,.13)
gender=rbinom(1*1000,1,.5)
eth=runif(1*1000, min=1, max=4)
cogat=rnorm(1*1000, 100, 16)
map=rnorm(1*1000, 200, 9)
growth=0
simtest=data.frame (subject=subject, treat=treat, gender=gender, eth=round(eth,digits=0), cogat=round(cogat,digits=0),map=round(map,digits=0),growth) simtest<-transform(simtest, growth=rnorm(1000,m=ifelse(treat==0,0.1,0.5),s=0.03)) simtest
Once again the issues are the correlations, the recoding of the categorical variable and the iterations.
Thanks to all who have helped so far! You are all so smart!
S
From: Charles Annis, P.E. [charles.annis_at_statisticalengineering.com] Sent: Friday, April 15, 2011 3:00 PM
To: Shane Phillips; r-help_at_r-project.org Subject: RE: [R] Monte Carlo Simulation
What have you tried so far?
It is often helpful to begin with a much simpler problem, then add complexity incrementally until you've constructed the desired model.
Best wishes.
Charles Annis, P.E.
Charles.Annis_at_StatisticalEngineering.com 561-352-9699
http://www.StatisticalEngineering.com
Hello, R friends...
I am very new to R, and I need some help. I am trying to construct a simulation for my dissertation.
I need to create 1000 datasets of 1000 subjects with the following variables...
Treatment variable - Drawn from a binomial distribution (1 run, prob=.13) Covariate 1 - Drawn from a normal distribution (mean=100, sd=16) Covariate 2 - Drawn from a normal distribution (mean=200, sd=9) Covariates 1 and 2 need to be correlated (say, r=.80) Covariate 3 - Drawn from a binomial distribution (1 run, prob=.5) Covariate 4 - Drawn from a distribution of discrete variables where 1 has an 80% chance of being selected, 2 - 10%, 3 - 5% and 4 - 5%. This variable would need to be recoded into 4 binary variables. Covariate 5 - Drawn from a normal distribution (mean=84, sd=2) Covariate 6 - Drawn from a binomial distribution (1 run, prob=.15) Covariate 6 needs to correlate with Covariate 2 (r=.70, or so)
I need each dataset saved as a new datafile with an iterative filename (e.g. sample1, sample2, etc.).
Thanks!
Shane
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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
R-help_at_r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Fri 15 Apr 2011 - 22:28:58 GMT
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Archive generated by hypermail 2.2.0, at Fri 15 Apr 2011 - 22:30:30 GMT.
Mailing list information is available at https://stat.ethz.ch/mailman/listinfo/r-help. Please read the posting guide before posting to the list.
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# LBA of beams with various boundary conditions
Linear bifurcation analysis (LBA) of beams in bending: Influence of boundary conditions and load position
$$1. The objectiveThe objective of this article is the verification of the LBA (linear bifurcation analysis) module of the IDEA Member application. Beams in bending are analyzed and the influence of different boundary conditions and load positions is investigated. The resulting elastic critical moments from IDEA Member are compared to the elastic critical moments based on Annex I of EN 1999-1-1 [1]. Numerical solution from LTBeam software [2] is also presented.2. Model descriptionA total of 18 individual cases was analyzed to verify the LBA module. All of them share the same cross-section IPE 300 and the same steel grade S 355. Three different boundary conditions were investigated (S – simple, F – fixed, C – cantilever), each with two load cases (F – force; C – continuous). Three load positions in relation to the shear center are verified (T – top, N – neutral, B – bottom). Fig. 1: Various boundary conditions and load cases used for verificationAll cases are designated in the following manner: “C_F_T”, where “C” indicates boundary conditions, “F” the load case and “T” the load position relative to the shear center.3. Analytical solutionThe three-factor formula found in Annex I of EN 1999-1-1 [1] is used to calculate the elastic critical moment for lateral-torsional buckling of the beams:$M_{cr} = \mu_{cr} \frac{\pi \sqrt{E I_z G I_t}}{L}$$\mu_{cr} = \frac{c_1}{k_z} \left [ \sqrt{1+\kappa_{wt}^2 + (C_2 \zeta_g - C_3 \zeta_j)^2} - (C_2 \zeta_g - C_3 \zeta_j) \right ]$The Annex B of ECCS - N° 119 [3] is used to calculate the coefficients C1 and C2 for the cantilever. Fig. 2: Buckling modes for the three different boundary conditions4. ResultsThe elastic critical moment from IDEA Member (M) is compared to an analytical value for a rolled cross-section (EN) and for its representation without the web-flange radii (ENw). Furthermore, the same two sets of values are presented as an output from the LTBeam software (L, Lw).Tab. 1: Resulting elastic critical moments The results of LBA are conservative (10–16 %) for the top flange load positions. The other load positions are less conservative (< 10 %). Chart 1: Elastic critical moment values Chart 2: Elastic critical moment comparisonThe slightly conservative results of IDEA Member are caused by the missing web-flange radii of the shell representation of a cross section in the IDEA Member and resulting lower torsional stiffness. This is confirmed by the LTBeam software (Lw), as well as the analytical solution (ENw).5. Literature and References[1] EN 1999-1-1: Eurocode 9: Design of aluminium structures - Part 1-1 : General structural rules, CEN, 2006.[2] LTBeam software v. 1.0.11, CTICM, available at https://www.cesdb.com/ltbeam.html[3] Rules for Member Stability in EN 1993-1-1, Background documentation and design guidelines, ECCS - N° 119, 2006.$$
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# FROM 2020: Make a Tornado in a Bottle! Weather Education Week Experiment #5
## You can make a safe tornado in a bottle! Follow along with NBC10 First Alert Meteorologist Steve Sosna and try this experiment at home.
Welcome to Weather Education Week @ Home! All week we are bringing parents and students interesting science experiments from the NBC10 First Alert Weather team -- and the Philadelphia Phillies!
For this experiment, let's see how a vortex works. And check out the appearance by Phillies manager Joe Girardi!
If you try it, please take a video and upload it here -- we may use it on NBC10!
Need:
• 2 emptied & cleaned 1-liter soda bottles
• Water
• Colored Lamp Oil (Optional)
• Vortex Connector
Or:
• 1 metal Washer
• Duct Tape
Steps:
1. Fill one of the bottles about three-quarters full with water.
2. If you have colored lamp oil, add a layer of that to the water. It should float on top.
3. If you have a vortex connector, twist the connector onto the water-filled bottle.
4. Twist the second bottle into the other side of the connector.
5. If you do not have a vortex connector, add the metal washer to the mouth of the water-filled bottle. Without covering up the opening, secure it to the bottle using duct tape.
6. Match the mouth of the second bottle to the mouth of the first, add more duct tape to secure. They should be stacked on top of each other.
7. Flip your two bottles, so the water-filled one is on top and the empty bottle is on bottom.
8. Spin the bottles in a clockwise motion.
9. Watch what happens: You should see a “tornado” (vortex) form.
What Happened?
You made a vortex! In this case, the vortex is the spinning of the liquid along a center line. The spin is maintained thanks to centripetal force.
The lamp oil is less dense than the water, so it floats on top of the water and drains first when the vortex forms (this highlights the shape of the vortex). When the water & oil-filled bottle is flipped to the top, gravity pulls the liquid downward into the empty bottle.
Without spinning the bottle, the water would just “glug-glug” out of the top bottle into the bottom bottle. The “glug-glug” is caused because air and water would take turns swapping bottles.
However, when the bottle is spun, the air can enter the top bottle while the liquid flows into the bottom bottle. This speeds up the process and makes a vortex.
Similar to the vortex in our bottles, tornadoes also have upward and downward motions known as “inflow” and “outflow."
Tornadoes are quickly-spinning tubes of air that drop out of thunderstorm clouds and connect to the ground. Tornadoes typically look brown due to the dust and debris they pick up in the column of spinning air (like a vacuum!).
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need help
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For what constant k is 1 the minimum value of the quadratic 3x^2 - 16x + k over all real values of x?
Oct 10, 2020
#1
+28024
+1
Bowl shaped parabola.....minimum point is the vertex....to find the x value of the vertex
-b/2a = x of the vertex
16/6 sub in this for 'x' to find k which makes the equation = to 1
3 (16/6)^2 - 16 (16/6) + k = 1 k = 22 1/3
Oct 10, 2020
edited by ElectricPavlov Oct 10, 2020
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hw2_3221spring2011
# hw2_3221spring2011 - h above your arm and the building is a...
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1 Phy 3221 Due: January 19, 2011 Homework set # 2 You are encouraged and welcome to work together on the homework assignments. But your homework must be in your own handwriting. If you expect credit for your homework, then it is best to make it easily readable. Reading: Sections 1.9–1.13 Section 2.4 Problem 1: Find the minimum speed required to throw a snowball into an open window of a building. The window is a height
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Unformatted text preview: h above your arm, and the building is a distance d away from where you are standing. Hint: Check that your answer gives the expected result in the special case that h = 0, and in the special case that d = 0. If this problem is rather easy for you, then most likely you have made a mistake. Textbook: Prob. 1-24 Textbook: Prob. 1-27 Textbook: Prob. 2-14...
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The perimeter of a rectangle with width x is 50 feet. The function below gives the area of the rectangle as a function of x.A(x) = -x2 + 25xUse your graphing calculator to find the value of x which maximizes the area. Round your answer to two decimal places
1
by Momo1994
2014-09-20T08:44:39-04:00
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
A negative quadratic graph has a maximum point at which the area A is largest. Calculus is the best way to solve this:
First, find the first derivative (f'(x) = -2x + 25) by multiplying -x^2 by the power (2) and then reducing the power by one, then repeating the process with 25x.
This new function defines the gradient of the function A = -x^2 at a given point, so the maximum point is where the gradient is 0. Hence:
-2x + 25 = 0
2x = 25
x = 12.5ft
Area = (25x12.5) - 12.5^2 = 156.25ft^2
The question asks you to use a graphical calculator to solve this so, if you have one, just put the equation in and record the coordinates of the maximum point of the graph - it should be (12.5, 156.25). An online graphical calculator can be found by googling 'online graphical calculator' also.
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# How can specify interval of variables in fsurf?
9 views (last 30 days)
Rahul Bhadani on 7 Apr 2021
Answered: Mathieu NOE on 7 Apr 2021
I am following the documentation for fsurf:
There, I see an example:
f1 = @(x,y) erf(x)+cos(y);
fsurf(f1,[-5 0 -5 5])
However, I do not know what is the granularity of x and y. Let's say I am interested in x for -5 < x < 0 but interval should be 0.1, i.e. in MATLAB language it should be x= -5:0.1:0. Similarly, let say Y has -5:0.2:5
How should I accomplish that?
Mathieu NOE on 7 Apr 2021
hello
will do ,
maybe in the future TMW will let you choose a different mesh density for each direction...
Mathieu NOE on 7 Apr 2021
hi
see 'MeshDensity' — Number of evaluation points per direction
Control Resolution of Surface Plot
Control the resolution of a surface plot using the 'MeshDensity' option. Increasing 'MeshDensity' can make smoother, more accurate plots while decreasing it can increase plotting speed.
Create two plots in a tiled chart layout. In the first plot, display the parametric surface x=sin(s), y=cos(s), z=(t/10)sin(1/s). The surface has a large gap. Fix this issue by increasing the 'MeshDensity' to 40 in the second plot. fsurf fills the gap, showing that by increasing 'MeshDensity' you increased the resolution.
tiledlayout(2,1)
nexttile
fsurf(@(s,t) sin(s), @(s,t) cos(s), @(s,t) t/10.*sin(1./s))
view(-172,25)
title('Default MeshDensity = 35')
nexttile
fsurf(@(s,t) sin(s), @(s,t) cos(s), @(s,t) t/10.*sin(1./s),'MeshDensity',40)
view(-172,25)
title('Increased MeshDensity = 40')
KSSV on 7 Apr 2021
f1 = @(x,y) erf(x)+cos(y);
x = -5:0.1:0 ;
y = -5:0.2:5 ;
[X,Y] = meshgrid(x,y) ;
Z = f1(X,Y) ;
surf(X,Y,Z)
Rahul Bhadani on 7 Apr 2021
Because my question was specifically about using fsurf and not in general about how to make a 3D plot.
### Categories
Find more on Surface and Mesh Plots in Help Center and File Exchange
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### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
5 added 134 characters in body
Given a one-parametric random function on a probability space $(\Omega,\mathcal F,\mathbb P)$:
$X:U\times\Omega\to \mathbb R \text{ and } (a,w)\mapsto X(a,w), \text{ with } \sigma(X(a))\subseteq \mathcal F\quad\quad\forall a\in U\subseteq\mathbb R,$
Then the following holds:
$E\left[\sup\limits_{a\in U}X(a)\right]=\sup\Bigr\lbrace E\left[X(A)\right]\Bigr|\sigma(A)\subseteq{\mathcal F},\;A(\omega)\in U\Bigr\rbrace$ and also $E\left[\sup\limits_{a\in U}X(a)\right]=\sup\Bigr\lbrace E\left[X(A)\right]\Bigr|\sigma(A)\subseteq{\bigcup\limits_{a\in U}\sigma(X(a))},\;A(\omega)\in U\Bigr\rbrace$
## Proof:
(
The following holds trivially:
$E[X(A)]\le E[\sup_{a\in U} X(a)]$
it remains to show the other direction. This is done by applying zhoraster's answer):
Clearly, $M(\omega) = \sup_{a\in U} X(a,w)$ is $\mathcal F$-measurable.
Define for $\delta>0$
$\mathfrak A_\delta = \lbrace(a,\omega)\in U\times \Omega\mid X(a,w) >M(\omega)-\delta\rbrace$
This set is in $\mathcal B(\mathbb R)\otimes \mathcal F_t$, and it has a full projection onto $\Omega$. By a measurable selection theorem (which I think one can find in Bogachev Measure Theory) there is an $\mathcal F$-measurable $A_\delta$ such that $(A_\delta(\omega),\omega)\in\mathfrak A_\delta$ almost surely. Hence $E[X(A_\delta)]≥E[M(\omega)]−\delta$. We get the desired statement by letting $\delta\to 0$.
(One can also use Kuratowski--Ryll-Nardzewski theorem to prove the existence of a measurable $A_\delta$.)
After a very good answer of zhoraster, I realized, that my initial question was a mixup of several different things. Thats why I changed it community wiki and clearified the problem.
4 clearify the question and summerize the given answers; added 10 characters in body; added 2 characters in body
Hi
## MySolutionE\left[\sup\limits_{a\inU}X(a)\right]=\sup\Bigr\lbraceE\left[X(A)\right]\Bigr|\sigma(A)\subseteq{\bigcup\limits_{a\inU}\sigma(X(a))},\;A(\omega)\inU\Bigr\rbrace$## Proof: (so farapplying zhoraster's answer) If Clearly,$S$M(\omega) = \sup_{a\in U} X(a,w)$ is Markov, the equality holds$\mathcal F$-measurable.
Idea of Proof: Using iterated conditioning on the r.h.s leads to
Define for $\sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U}E\left[E\left[f(A,(S_{t_i})_{i< n,t_i\ge t})\Bigr|\mathcal F_t\right]\right]$
Using the Markov property of S and the \delta>0\mathcal F_t$-measurability of \mathfrak A_\delta = \lbrace(a,\omega)\in U\times \Omega\mid X(a,w) >M(\omega)-\delta\rbrace$
This set is in $A$, \mathcal B(\mathbb R)\otimes \mathcal F_t$, and it can be shown that has a function full projection onto$g(a,s)$exists with\Omega$. By a measurable selection theorem (which I think one can find in Bogachev Measure Theory) there is an $g(A,S_t)=E\left[f(A,(S_{t_i})_{i< n,t_i\ge t})\Bigr|\mathcal F_t\right]$
leading to \mathcal F$-measurable$\sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U}E\left[g(A,S_t)\right]$It can easily be seen, A_\delta$ such that $(A_\delta(\omega),\omega)\in\mathfrak A_\delta$ almost surely. Hence $E[X(A_\delta)]≥E[M(\omega)]−\delta$. We get the l.h.s of equaility provides an upper bound:desired statement by letting $E\left[g(A,S_t)\right]\leq E\left[\sup\limits_{a\in U}\;g(a,S_t)\right]=E\left[\sup\limits_{a\in U}E\left[f(a,(S_{t_i})_{i< n,t_i\ge t})\Bigr|\mathcal F_t\right]\right]$
It remains \delta\to 0$. (One can also use Kuratowski--Ryll-Nardzewski theorem to show prove the other direction:existence of a measurable$E\left[\sup\limits_{a\in U}\;g(a,S_t)\right]\leq \sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U} E\left[g(A,S_t)\right]$But I am stuck here. I would appreciate some helpA_\delta$.)
After a very good answer of zhoraster, like telling meI realized, if that my initial question was a mixup of several different things. Thats why I am heading in changed it community wiki and clearified the right direction or other hints!
Thanksproblem.
3 deleted 3 characters in body; edited title; [made Community Wiki]
# ExpectationofaCommuting supremum ofconditionalexpectationsandexpectation
Hi!
## My question
Given some function $f:\mathbb R^n \rightarrow\mathbb R$ and a compact set $U\subset\mathbb R$, what properties for the process $S_t$, which is adapted to a filtration $\mathcal F$ are needed, so that the following holds:
$E\left[\sup\limits_{a\in U}E\left[f(a,(S_{t_i})_{i< n,t_i\ge t})\Bigr|\mathcal F_t\right]\right]=\sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U}E\left[f(A,(S_{t_i})_{i< n,t_i\ge t})\right]$
The second supremum should be taken over all random variables $A$, that are $\mathcal F_t$-measurable and take values in $U$.
## My Solution (so far)
If $S$ is Markov, the equality holds.
Idea of Proof: Using iterated conditioning on the r.h.s leads to
$\sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U}E\left[E\left[f(A,(S_{t_i})_{i< n,t_i\ge t})\Bigr|\mathcal F_t\right]\right]$
Using the Markov property of S and the $\mathcal F_t$-measurability of $A$, it can be shown that a function $g(a,s)$ exists with
$g(A,S_t)=E\left[f(A,(S_{t_i})_{i< n,t_i\ge t})\Bigr|\mathcal F_t\right]$
$\sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U}E\left[g(A,S_t)\right]$
It can easily be seen, that the l.h.s of equaility provides an upper bound:
$E\left[g(A,S_t)\right]\leq E\left[\sup\limits_{a\in U}\;g(a,S_t)\right]=E\left[\sup\limits_{a\in U}E\left[f(a,(S_{t_i})_{i< n,t_i\ge t})\Bigr|\mathcal F_t\right]\right]$
It remains to show the other direction:
$E\left[\sup\limits_{a\in U}\;g(a,S_t)\right]\leq \sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U} E\left[g(A,S_t)\right]$
But I am stuck here.
I would appreciate some help, like telling me, if I am heading in the right direction or other hints!
Thanks
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# How many degrees is a cone?
## How many degrees is a cone?
270 Degrees The “vertex angle” of a cone is the total number of degrees of a circle used to make the cone. So the cone shown below would be called a “270 degree cone.” Cut along the dotted line, and tape the cut edges together.
## What is a cone number?
The pyrometric cones used today by ceramic artists and industrial manufacturers were developed in the late 1800’s by Edward Orton Jr. Thus all ceramic products were assigned a cone number to which they were to be fired to assure maturity of the ware during the firing process such as Cone 06 glazes, Cone 04 bodies, etc.
What temp is cone 08?
CONE TEMPERATURE CHART (FOR THOSE OF YOU WHO ARE NOW WONDERING WHAT CONE MEANS!)
Cone number Orton Cones Final temp in degrees F at ramp rate of 27 degrees F/hr Orton Cones Final temp in degrees F at ramp rate of 108 degrees F/hr
07 1764 1789
08 1692 1728
09 1665 1688
010 1636 1657
### What is pyrometric cone equivalent?
PCE stands for “Pyrometric Cone Equivalent”. They are used to determine the “Pyrometric Cone Equivalent” of an unknown raw material by placing several different PCE cones along side an unknown raw material (that has been pressed into the same shape as a cone).
### Can you fire cone 10 clay cone 6?
You cannot fire a clay higher than its maximum-rated Cone, or it will melt. Cone 10 clay can be used at low fire (Cone 04-06 or at Cone 6), but to reach its maximum strength it should be fired to Cone 10. That will cause the clay to shrink and become dense, and that is ideal, especially for dinnerware.
Which is hotter cone 04 or cone 06?
Cone 6 is about 400 degrees hotter than cone 06! Therefore cone 05 is cooler than cone 04 whereas cone 5 is hotter than cone 4. For the most reliable results, it is best to match your clay with your glazes. If your clay’s recommended firing temperature is cone 06-04, then you should use low-fire glazes.
## Can you bisque fire cone 6?
The most common temperature to bisque fire pottery is cone 06 – 04. This equates to around 1830 – 1940F, (999-1060C). However, potters do bisque fire at other temperatures. The right temperature to bisque fire depends partially on the clay you are using.
## Can you fire Cone 10 clay Cone 6?
Can you bisque fire Cone 6?
### What cone is 900 degrees?
Temperature ranges
Orton Börkey Keratech
Self-Supporting Cones
012 843°C
011a 900°C
011 857°C
### What does cone 6 mean in pottery?
This refers to the medium temperature range (or middle fire) that most potter’s work in. The term “cone 6” normally implies oxidation firing in a hobby kiln (most fire to this range). Clays made using feldspar can be made to vitrify to zero-porosity density at cone 6 (including porcelains and stonewares).
Can Cone 5 clay be fired to Cone 6?
A Cone rating means that you can fire that clay at any temperature up to that cone. You cannot fire a clay higher than its maximum rated Cone, or it will melt and become deformed. For dinnerware it is best to use a Cone 5-6 clay if you fire to Cone 5-6. …
## What makes a cone a right circular cone?
A cone can be of two categories, depending upon the position of the vertex on the base: A right circular cone is one whose apex is perpendicular to the base, which means that the perpendicular line falls exactly on the center of the circular base of the cone. In the image below, h represents the height of the cone, and r is the radius.
## What does the volume of a cone mean?
The volume of a cone defines the space or the capacity of the cone. A cone is a three-dimensional geometric shape having a circular base that tapers from a flat base to a point called apex or vertex.
What are the three properties of a cone?
The three main properties of a cone are as follows: It has one circular face. It has zero edges. It has one vertex (corner). A cone can be of two categories, depending upon the position of the vertex on the base:
### What’s the difference between a cone and a base?
What is Cone? A cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base. This is what a cone looks like:
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### Pythagoras Thereom
Tuesday, February 17, 2009
Pythagoras was a Greek man. Pythagoras believed that eating animal products was cannibalism because they might contain the souls of human ancestors. Pythagoras found the proof for the Pythagorean theorem. He discovered square roots by comparing square piles of rocks.
The picture and numbers on the post .
How are all of the pictures and vocabulary connected.
The hypotenuse is the longest side of a right triangle in the Pythagorean theorem. When you want to find out what the length of the hypotenuse is you just the length of the two legs A and B. Which explains the triangle picture. The actual Pythagorean theorem is A2+B2=C2 which explains that picture. Pythagoras was a Greek man known as the father of mathematics and geometry. which explains he word Greek.
I couldnt get the the squar root download thing so I just took a picture of my work i hope you can see it.
Well as you can kind of see I just did one side of the triangle and the its the same on the other side so I just double the answer 6 and i get an answer of 12 and the answer is 12 mm.
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 centimeters what is the perimeter of the board game? First i labeled the picture
And that's how its done.
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ČESKÉ VYSOKÉ UČENÍ TECHNICKÉ V PRAZE
STUDIJNÍ PLÁNY
2018/2019
# Algorithms
Předmět není vypsán Nerozvrhuje se
Kód Zakončení Kredity Rozsah Jazyk výuky
AE4B33ALG Z,ZK 6 2+2c
Předmět nesmí být zapsán současně s:
Algoritmizace (A4B33ALG)
Algoritmizace (A4B33ALG)
Přednášející:
Cvičící:
Předmět zajišťuje:
katedra kybernetiky
Anotace:
In the course, the algorithms development is constructed with minimum dependency to programming language; nevertheless the lectures and seminars are based on Java. Basic data types a data structures, basic algorithms, recursive functions, abstract data types, stack, queues, trees, searching, sorting, special application algorithms. Students are able to design and construct non-trivial algorithms and to evaluate their affectivity.
Výsledek studentské ankety předmětu je zde: http://www.fel.cvut.cz/anketa/aktualni/courses/AE4B33ALG
Programming 1
Osnova přednášek:
1. Algorithms, programs, programming languages, introduction to problems solving
2. One-dimensional array, simple problems v 1-D array
3. Sorting in 1-D array pole (mergesort, quicksort, heapsort),
4. Searching in 1-D array
5. 2-D array, simple tasks in 2-D array
6. Strings, simple problems in string processing, text files
7. Asymptotic complexity, evaluation of space and complexity of algorithms from lectures No. 3.-6.
8. Simple recursion, recursive functions, advanced techniques
9. File conception, sequential files, conception of the record, file of records
10. Data types, list, stack, queue, examples of application
12. Trees, their properties, binary trees, basic algorithms of tree search.
13. Basic algorithms of linear algebra and mathematical analysis
14. Reserve
Osnova cvičení:
1.Introductory test, repeating of the ways of program construction in development environment, examples of functions and procedures, parameters, simple classes, assignment of semester task
2.One-dimensional array processing
3.Sorting and searching in 1D array algorithms
4.Multidimensional array processing algorithms
5.Text and string algorithms
6.Experimentation with space and complexity of algorithms
7.Sequential files
8.Implementation of abstract data types
9.Recursion and iteration
11.Tree construction, tree search
13. Algorithms of linear algebra and geometry, mathematical analysis
14.Credit
Cíle studia:
Semester project consists from empirical evaluation of searching and sorting algorithms, comparison of iterative and recursive algorithms and debugging of graphical output of selected algorithms of linear algebra and mathematical analysis. Three phases of supervision associated to constituted subtask of project with closing demonstration and defense
Studijní materiály:
[1] Sedgewick, R: Algorithms (Fundamentals, Data structures, Sorting,
[2] Weiss, M: Data structures and Algorithm Analysis in Java, Addison Wesley, 1999
[3] Keogh, J: Data Structures Demystified, McGraw-Hill, 2004
[4] Wróblevski, P: Algorytmy, Helion, 2003
Poznámka:
Rozsah výuky v kombinované formě studia: 14p+6c
Další informace:
http://cw.felk.cvut.cz/doku.php/courses/ae4b33alg
Pro tento předmět se rozvrh nepřipravuje
Předmět je součástí následujících studijních plánů:
Platnost dat k 22. 5. 2019
Aktualizace výše uvedených informací naleznete na adrese http://bilakniha.cvut.cz/cs/predmet12820904.html
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# Town (Revision 1431)
The Town is a Puzzle level in IGN's Walkthrough for The Witness. This guide will reveal every tile solution for each section of the Town level. Keep it here on IGN for more solutions for The Witness.
The Town is the culmination of all the puzzle skills you've picked up in the other areas. Your central focus is the tallest building in the middle -- which is locked by multiple doors. In addition to some very abstract puzzle panels, you'll also find quite a few Environmental Puzzles. Look at the Town Obelisk page to find them all.
## Town Hints
• Hint 1: SymmetryThe palm trees in the background are a hint. There's a reason this panel is translucent.
• Hint 2: Shadow MazeThe left side's pretty easy to see. How about looking for the source of the shadow to get the right side? Pen and paper helps!
• Hint 3: What is that, a windmill symbol?Nope. It's abstract. Look around -- anything familiar nearby?
• Hint 4:The abstract crosshairIt's an apple tree. Four main branches and a bunch of smaller ones. The apple marks the spot.
• Hint 5:Wait, what?Each small line is a branch. Some are broken off. That will tell you which branch is which and let you figure out which branch has the apple (exit) on it.
• Hint 7:Purple and red panelsThese panels are reflective.
• Hint 8:Seeking the glareYou have to get high up for the panels -- can you create a walkway through a puzzle solution?
• Hint 9:Purple panel 2Can you step outside and take another look? Maybe you can use the environment?
• Hint 10:Colored sunsYou ever visit the Greenhouse? Some colors are not what they seem.
• Hint 11:Colored suns 2There's a colored piece of glass somewhere that'll give you a different perspective.
• Hint 12:Lattice workGet outside, take another look. See any rogue pieces of lattice work sticking out that you can position to help you figure out the labyrinth?
Spoiler warning! The screenshots below contain solutions for every puzzle in the area.
## TOWN TERRACE
The first puzzle for the Town is located near the Docks. Solve them in the order in which the cables light up.
## LARGE BLUE TILE
There are THREE solutions to this large Blue tile. The three solutions below will open up a stairway to the roof of the building, which contains a giant dark-red tile.
NOTE: There is an orange-colored container in this building that contains two tiles puzzles, though solving them doesn't contribute to the main Town Laser puzzle.
## OAKWOOD HOUSE
Solving the five Gray tiles opens up a stairway that leads to the roof.
## WHITE BAMBOO HOUSE
NOTE: There is a large Blue-light room on the top floor of the building that has three tiles to solve but doesn't contribute to the main Laser puzzle.
## BEIGE HOUSE (RIGHT SIDE)
Solving this puzzle will unlock a tram for this house that leads to the rooftop of the Red house next door.
# SOLVING THE PUZZLE
Once all of the Town puzzles are solved, both doors in the mi-Tower structure will open up. All you need to do is make your way to the top of the Tower and activate the tile to fire up the Town Laser.
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Cody
# Problem 558. Is the Point in a Triangle?
Solution 3437028
Submitted on 27 Oct 2020 at 13:45
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
Triangle = [0, 0; 1, 0; 1, 1]; Points = [0, 0.5] y_correct = 0; assert(isequal(your_fcn_name(Points,Triangle),y_correct))
Points = 0 0.5000
2 Pass
Triangle = [0, 0; 1, 0; 1, 1]; Points = [0.8, 0.5] y_correct = 1; assert(isequal(your_fcn_name(Points,Triangle),y_correct))
Points = 0.8000 0.5000
3 Fail
Triangle = [0.8147, 0.9134; 0.9058, 0.6324; 0.1270, 0.0975]; Points = [0.8, 0.7; 0.9, 0.4] y_correct = [1 0]; assert(isequal(your_fcn_name(Points,Triangle),y_correct))
Points = 0.8000 0.7000 0.9000 0.4000
Assertion failed.
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# How to convert lumens to candela
How to convert luminous flux in lumens (lm) to luminous intensity in candela (cd).
You can calculate but not convert lumens to candela, since candela and lumens do not represent the same quantity.
### Lumens to candela calculation
For uniform, isotropic light source, the luminous intensity Iv in candela (cd) is equal to the luminous flux Φin lumens (lm),
divided by the solid angle Ω in steradians (sr):
Iv(cd) = Φv(lm) / Ω(sr)
So The solid angle Ω in steradians (sr) is equal to 2 times pi times 1 minus cosine of half the apex angle θ in degrees (°).
Ω(sr) = 2π(1 - cos(θ/2))
So The luminous intensity Iv in candela (cd) is equal to the luminous flux Φin lumens (lm),
divided by 2 times pi times 1 minus cosine of half the apex angle θ in degrees (°).
Iv(cd) = Φv(lm) / ( 2π(1 - cos(θ/2)) )
So
candela = lumens / ( 2π(1 - cos(degrees/2)) )
Or
cd = lm / ( 2π(1 - cos(°/2)) )
#### Example 1
Find the luminous intensity Iv in candela (cd) when the luminous flux Φin lumens (lm) is 340lm and the apex angle is 60°:
Iv(cd) = 340 lm / ( 2π(1 - cos(60°/2)) ) = 403.9 cd
#### Example 2
Find the luminous intensity Iv in candela (cd) when the luminous flux Φin lumens (lm) is 360lm and the apex angle is 60°:
Iv(cd) = 360 lm / ( 2π(1 - cos(60°/2)) ) = 427.6 cd
#### Example 3
Find the luminous intensity Iv in candela (cd) when the luminous flux Φin lumens (lm) is 380lm and the apex angle is 60°:
Iv(cd) = 380 lm / ( 2π(1 - cos(60°/2)) ) = 451.4 cd
#### Example 4
Find the luminous intensity Iv in candela (cd) when the luminous flux Φin lumens (lm) is 440lm and the apex angle is 60°:
Iv(cd) = 440 lm / ( 2π(1 - cos(60°/2)) ) = 522.6 cd
#### Example 5
Find the luminous intensity Iv in candela (cd) when the luminous flux Φin lumens (lm) is 540lm and the apex angle is 60°:
Iv(cd) = 540 lm / ( 2π(1 - cos(60°/2)) ) = 641.4 cd
Candela to lumens calculation ►
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• ### Announcements
• #### Check out the \$1M GAME GOLF Giveaway05/24/2016
You can win \$1M in the GAME GOLF Giveaway by TST Partner GAME GOLF. More details here:
Followers 0
# what if an animal puts your golf ball into the hole?
## 12 posts in this topic
I suppose the ball has to be placed back where it came from?
the so-called Putt is re-played, without penalty ?
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I suppose the ball has to be placed back where it came from?
the so-called Putt is re-played, without penalty ?
### 18-1 . By Outside Agency
If a ball at rest is moved by an outside agency , there is no penalty and the ball must be replaced.
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I suppose the ball has to be placed back where it came from?
the so-called Putt is re-played, without penalty ?
Are you saying that it happens while the ball is rolling after a stroke, or that the animal moves your ball when it's at rest? The answer is slightly different depending on the situation.
If the ball is at rest, then moved by the animal, it must be replaced (Rule 18-1).
If the ball is in motion after a stroke, then the stroke is cancelled and the ball must again be replaced and the stroke replayed (Rule 19-1).
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If an animal put the ball in the hole?
I'd buy him a set of clubs.
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Are you saying that it happens while the ball is rolling after a stroke, or that the animal moves your ball when it's at rest? The answer is slightly different depending on the situation.
If the ball is at rest, then moved by the animal, it must be replaced (Rule 18-1).
If the ball is in motion after a stroke, then the stroke is cancelled and the ball must again be replaced and the stroke replayed (Rule 19-1).
what if I make a putt, and the ball stops near the lip of the hole.
I'm walking towards the ball to count the 10 seconds.
But a squirrel putts the ball into the hole, as Im still walking towards the hole, prepared to count the 10 seconds.
I suppose the putt must be re-taken?
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what if I make a putt, and the ball stops near the lip of the hole.
I'm walking towards the ball to count the 10 seconds.
But a squirrel putts the ball into the hole, as Im still walking towards the hole, prepared to count the 10 seconds.
I suppose the putt must be re-taken?
Correct.
Once again....
### 18-1 . By Outside Agency
If a ball at rest is moved by an outside agency , there is no penalty and the ball must be replaced.
0
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But a squirrel putts the ball into the hole, as Im still walking towards the hole, prepared to count the 10 seconds.
I'm a rules stickler, but if the squirrel PUTTS the ball into the hole, I'm inclined to let it count. :)
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What if it is a fellow competitor, partner or opponent or their caddie?
When I was at school, they were classified as animals.
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There was a hole in one on a par 4 on the tour when it hit off a competitor's putter and went in. It counted.
Here is video of it. It's #3 in this top 10 lucky bounces list.
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what if I make a putt, and the ball stops near the lip of the hole.
I'm walking towards the ball to count the 10 seconds.
But a squirrel putts the ball into the hole, as Im still walking towards the hole, prepared to count the 10 seconds.
I suppose the putt must be re-taken?
If the ball stopped, wouldn't it be at rest? I believe that you'd just place it back by the lip of the hole in this case.
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Most of my partners and fellow competitors are animals.
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This is hilarious because I remember there was a course here in Columbus called the "airport course" b/c it was literally right at the runway of Port Columbus International Airport. They had such an issue with groundhogs and people witnessed them taking balls that landed near their gigantic burrows. If your from the Columbus area you know the course I'm talking about
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• ### Topics Being Discussed Right Now on The Sand Trap
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Grabbed a set of Mizuno MX-1000's from fleabay for £140.
• Ball Position on Uphill and Downhill Lies
When I keep the ball forward in my stance I feel confident I can shallow out the approach of the club When I position the ball back I feel like I need to steepen the club path in order to hit the ball flush
• Jordan Spieth lobs a marshmallow w/a wedge and eats it on its way down
Yeah.. My wife was just complaining that my kid picked up a cotton ball in the play ground and ate it because it was pink and thought it was a marshmallow.. He spit it out at least :)
• Tee Restrictions by Handicap
I just frankly don't see this a major problem and I play plenty of country clubs. One ridiculous thing I'd point out about Oakmont is that the "member tees" are about 6600 yards. They're blue and not traditional white but this is where pretty much every member plays from, whether or not they should. That's kind of dumb. The entire--and I mean all of it--tee it forward program is horribly out of touch with the average golfer and smells of stuffy white assholes like David Fay, who are so removed from how the game is experienced by the majority of players that I want to shit in his mouth.
Myrtle Beach is the best bargain in golf and I will not accept arguments to the contrary.
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, 17.02.2021 06:00 rashawnglover
# Free points ! if you could marry any celebrity or anime character who would you marry?
### Another question on Mathematics
Mathematics, 21.06.2019 17:30
Mrs. morton has a special reward system for her class. when all her students behave well, she rewards them by putting 3 marbles into a marble jar. when the jar has 100 or more marbles, the students have a party. right now, the the jar has 24 marbles. how could mrs. morton reward the class in order for the students to have a party?
Mathematics, 21.06.2019 18:00
Rosy waxes 2/3 of her car with 1/4 bottle of car wax.at this rate,what fraction of the bottle of car wax will rosey use to wax her entire car
Mathematics, 21.06.2019 22:00
Benjamin is making bow ties. how many 1/2yards lomg bow ties can he make if he has 18 feet of fabric?
Mathematics, 21.06.2019 23:00
Square a has a side length of (2x-7) and square b has a side length of (-4x+18). how much bigger is the perimeter of square b than square a?and show all work! you so
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# Marks
TIP
If you aren’t yet up and running with Plot, please read our getting started guide first. Tinkering with the code below will give a better sense of how Plot works.
Plot doesn’t have chart types; instead, you construct charts by layering marks.
## Marks are geometric shapes
Plot provides a variety of mark types. Think of marks as the “visual vocabulary” — the painter’s palette 🎨, but of shapes instead of colors — that you pull from when composing a chart. Each mark type produces a certain type of geometric shape.
For example, the dot mark draws stroked circles (by default).
Fork
js
``Plot.dot(gistemp, {x: "Date", y: "Anomaly"}).plot()``
The line mark draws connected line segments (also known as a polyline or polygonal chain).
Fork
js
``Plot.lineY(gistemp, {x: "Date", y: "Anomaly"}).plot()``
And the bar mark draws rectangular bars in either a horizontal (barX→) or vertical (barY↑) orientation.
Fork
js
``Plot.barX(alphabet, {x: "frequency", y: "letter"}).plot()``
So instead of looking for a chart type, consider the shape of the primary graphical elements in your chart, and look for the corresponding mark type. If a chart has only a single mark, the mark type is effectively the chart type: the bar mark is used to make a bar chart, the area mark is used to make an area chart, and so on.
## Marks are layered
The big advantage of mark types over chart types is that you can compose multiple marks of different types into a single plot. For example, below an area and line are used to plot the same sequence of values, while a rule emphasizes y = 0.
Fork
js
``````Plot.plot({
marks: [
Plot.ruleY([0]),
Plot.areaY(aapl, {x: "Date", y: "Close", fillOpacity: 0.2}),
Plot.lineY(aapl, {x: "Date", y: "Close"})
]
})``````
Each mark supplies its own data; a quick way to combine multiple datasets into a chart is to declare a separate mark for each. You can even use array.map to create multiple marks from nested data.
Fork
js
``````Plot.plot({
marks: [
[goog, aapl].map((stock) => Plot.lineY(stock, {x: "Date", y: "Close"}))
]
})``````
Marks may also be a function which returns an SVG element, if you wish to insert arbitrary content. (Here we use Hypertext Literal to generate an SVG gradient.)
Fork
js
``````Plot.plot({
marks: [
() => htl.svg`<defs>
<stop offset="15%" stop-color="purple" />
<stop offset="75%" stop-color="red" />
<stop offset="100%" stop-color="gold" />
</defs>`,
Plot.barY(alphabet, {x: "letter", y: "frequency", fill: "url(#gradient)"}),
Plot.ruleY([0])
]
})``````
And marks may be null or undefined, which produce no output; this is useful for showing marks conditionally (e.g., when a box is checked).
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js
``````Plot.plot({
marks: [
Plot.ruleY([0]),
area ? Plot.areaY(aapl, {x: "Date", y: "Close", fillOpacity: 0.2}) : null,
Plot.lineY(aapl, {x: "Date", y: "Close"})
]
})``````
## Marks use scales
Marks are (typically) not positioned in literal pixels, or colored in literal colors, as in a conventional graphics system. Instead you provide abstract values such as time and temperature — marks are drawn “in data space” — and scales encode these into visual values such as position and color. And best of all, Plot automatically creates axes and legends to document the scales’ encodings.
Data is passed through scales automatically during rendering; the mark controls which scales are used. The x and y options are typically bound to the x and y scales, respectively, while the fill and stroke options are typically bound to the color scale. Changing a scale’s definition, say by overriding its domain (the extent of abstract input values) or type, affects the appearance of all marks that use the scale.
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js
``````Plot.plot({
y: {
type: "log",
domain: [30, 300],
grid: true
},
marks: [
Plot.lineY(aapl, {x: "Date", y: "Close"})
]
})``````
## Marks have tidy data
A single mark can draw multiple shapes. A mark generally produces a shape — such as a rectangle or circle — for each element in the data.
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js
``Plot.dot(aapl, {x: "Date", y: "Close"}).plot()``
It’s more complicated than that, though, since some marks produce shapes that incorporate multiple data points. Pass the same data to a line and you’ll get a single polyline.
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js
``Plot.lineY(aapl, {x: "Date", y: "Close"}).plot()``
And a line mark isn’t even guaranteed to produce a single polyline — there can be multiple polylines, as in a line chart with multiple series (using z).
Fork
js
``Plot.lineY(bls, {x: "date", y: "unemployment", z: "division"}).plot()``
Plot favors tidy data structured as an array of objects, where each object represents an observation (a row), and each object property represents an observed value; all objects in the array should have the same property names (the columns).
For example, say we have hourly readings from two sensors A and B. You can represent the sensor log as an array of objects like so:
js
``````linedata = [
{hour: 0, value: 8, sensor: "A"},
{hour: 0, value: 6, sensor: "B"},
{hour: 1, value: 7, sensor: "A"},
{hour: 1, value: 5, sensor: "B"},
{hour: 2, value: 3, sensor: "A"},
{hour: 2, value: 0, sensor: "B"},
{hour: 3, value: 9, sensor: "A"},
{hour: 3, value: 2, sensor: "B"}
]``````
Then you can pass the data to the line mark, and extract named columns from the data for the desired options:
Fork
js
``Plot.lineY(linedata, {x: "hour", y: "value", stroke: "sensor"}).plot()``
Another common way to extract a column from tabular data is an accessor function. This function is invoked for each element in the data (each row), and returns the corresponding observed value, as with array.map.
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js
``````Plot.lineY(linedata, {
x: (d) => d.hour,
y: (d) => d.value,
stroke: (d) => d.sensor
}).plot()``````
For greater efficiency, Plot also supports columnar data: you can pass parallel arrays of values to each channel.
js
``````Plot.lineY({length: linedata.length}, {
x: linedata.map((d) => d.hour),
y: linedata.map((d) => d.value),
stroke: linedata.map((d) => d.sensor)
}).plot()``````
TIP
Note that when accessor functions or parallel arrays are used instead of field names, automatic axis labels (hour and value) are lost. These can be restored using the label option on the x and y scales.
## Marks imply data types
Data comes in different types: quantitative (or temporal) values can be subtracted, ordinal values can be ordered, and nominal (or categorical) values can only be the same or different.
INFO
Because nominal values often need some arbitrary order for display purposes — often alphabetical — Plot uses the term ordinal to refer to both ordinal and nominal data.
Some marks work with any type of data, while other marks have certain requirements or assumptions of data. For example, a line should only be used when both x and y are quantitative or temporal, and when the data is in a meaningful order (such as chronological). This is because the line mark will interpolate between adjacent points to draw line segments. If x or y is nominal — say the names of countries — it doesn’t make sense to use a line because there is no half-way point between two nominal values.
Fork
js
``Plot.lineY(["please", "don’t", "do", "this"]).plot() // 🌶️``
WARNING
While Plot aspires to give good defaults and helpful warnings, Plot won’t prevent you from creating a meaningless chart. Only you can prevent bogus charts!
In particular, beware the simple “bar”! A bar mark is used for a bar chart, but a rect mark is needed for a histogram. Plot has four different mark types for drawing rectangles:
• use rect when both x and y are quantitative
• use barX when x is quantitative and y is ordinal
• use barY when x is ordinal and y is quantitative
• use cell when both x and y are ordinal
Plot encourages you to think about data types as you visualize because data types often imply semantics. For example, do you notice anything strange about the bar chart below?
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js
``````Plot
.barY(timeseries, {x: "year", y: "population"}) // 🌶️
.plot({x: {tickFormat: ""}})``````
Here’s the underlying data:
js
``````timeseries = [
{year: 2014, population: 7295.290765},
{year: 2015, population: 7379.797139},
{year: 2016, population: 7464.022049},
{year: 2017, population: 7547.858925},
{year: 2019, population: 7713.468100},
{year: 2020, population: 7794.798739}
]``````
The data is missing the population for the year 2018! Because the barY mark implies an ordinal x scale, the gap is hidden. Switching to the rectY mark (with the interval option to indicate that these are annual observations) reveals the missing data.
Fork
js
``````Plot
.rectY(timeseries, {x: "year", y: "population", interval: 1})
.plot({x: {tickFormat: ""}})``````
Alternatively, you can keep the barY mark and apply the interval option to the x scale.
Fork
js
``````Plot
.barY(timeseries, {x: "year", y: "population"})
.plot({x: {tickFormat: "", interval: 1}})``````
## Marks have options
When constructing a mark, you can specify options to change the mark’s appearance. These options are passed as a second argument to the mark constructor. (The first argument is the required data.) For example, if you want filled dots instead of stroked ones, pass the desired color to the fill option:
Fork
js
``Plot.dot(gistemp, {x: "Date", y: "Anomaly", fill: "red"}).plot()``
As the name suggests, options are generally optional; Plot tries to provide good defaults for whatever you don’t specify. Plot even has shorthand for various common forms of data. Below, we extract an array of numbers from the `gistemp` dataset, and use the line mark shorthand to set x = index and y = identity.
Fork
js
``Plot.lineY(gistemp.map((d) => d.Anomaly)).plot()``
Some marks even provide default transforms, say for stacking!
TIP
Because Plot strives to be concise, there are many default behaviors, some of which can be subtle. If Plot isn’t doing what you expect, try disabling the defaults by specifying options explicitly.
In addition to the standard options such as fill and stroke that are supported by all mark types, each mark type can support options unique to that type. For example, the dot mark takes a symbol option so you can draw things other than circles. See the documentation for each mark type to see what it supports.
## Marks have channels
Channels are mark options that can be used to encode data. These options allow the value to vary with the data, such as a different position or color for each dot. To use a channel, supply it with a column of data, typically as:
• a field (column) name,
• an accessor function, or
• an array of values of the same length and order as the data.
Not all mark options can be expressed as channels. For example, stroke can be a channel but strokeDasharray cannot. This is mostly a pragmatic limitation — it would be harder to implement Plot if every option were expressible as a channel — but it also serves to guide you towards options that are intended for encoding data.
TIP
To vary the definition of a constant option with data, create multiple marks with your different constant options, and then filter the data for each mark to achieve the desired result.
Some options can be either a channel or a constant depending on the provided value. For example, if you set the fill option to purple, Plot interprets it as a literal color.
Fork
js
``````Plot
.barX(timeseries, {x: "population", y: "year", fill: "purple"})
.plot({y: {label: null, tickFormat: ""}})``````
Whereas if the fill option is a string but not a valid CSS color, Plot assumes you mean the corresponding column of the data and interprets it as a channel.
Fork
js
``````Plot
.barX(timeseries, {x: "population", y: "year", fill: "year"})
.plot({y: {label: null, tickFormat: ""}})``````
If the fill option is a function, it is interpreted as a channel.
Fork
js
``````Plot
.barX(timeseries, {x: "population", y: "year", fill: (d) => d.year})
.plot({y: {label: null, tickFormat: ""}})``````
Lastly, note that while channels are normally bound to a scale, you can bypass the color scale here by supplying literal color values to the fill channel.
Fork
js
``````Plot
.barX(timeseries, {x: "population", y: "year", fill: (d) => d.year & 1 ? "red" : "currentColor"})
.plot({y: {label: null, tickFormat: ""}})``````
But rather than supplying literal values, it is more semantic to provide abstract values and use scales. In addition to centralizing the encoding definition (if used by multiple marks), it allows Plot to generate a legend.
Fork
js
``````Plot
.barX(timeseries, {x: "population", y: "year", fill: (d) => d.year & 1 ? "odd" : "even"})
.plot({y: {label: null, tickFormat: ""}, color: {legend: true}})``````
You can then specify the color scale’s domain and range to control the encoding.
## Mark options
Mark constructors take two arguments: data and options. Together these describe a tabular dataset and how to visualize it. Option values that must be the same for all of a mark’s generated shapes are known as constants, whereas option values that may vary across a mark’s generated shapes are known as channels. Channels are typically bound to scales and encode abstract data values, such as time or temperature, as visual values, such as position or color. (Channels can also be used to order ordinal domains; see the sort option.)
A mark’s data is most commonly an array of objects representing a tabular dataset, such as the result of loading a CSV file, while a mark’s options bind channels (such as x and y) to columns in the data (such as units and fruit).
js
``````sales = [
{units: 10, fruit: "peach"},
{units: 20, fruit: "pear"},
{units: 40, fruit: "plum"},
{units: 30, fruit: "plum"}
]``````
js
``Plot.dot(sales, {x: "units", y: "fruit"})``
While a column name such as `"units"` is the most concise way of specifying channel values, values can also be specified as functions for greater flexibility, say to transform data or derive a new column on the fly. Channel functions are invoked for each datum (d) in the data and return the corresponding channel value. (This is similar to how D3’s selection.attr accepts functions, though note that Plot channel functions should return abstract values, not visual values.)
js
``Plot.dot(sales, {x: (d) => d.units * 1000, y: (d) => d.fruit})``
Plot also supports columnar data for greater efficiency with bigger datasets; for example, data can be specified as any array of the appropriate length (or any iterable or value compatible with Array.from), and then separate arrays of values can be passed as options.
js
``index = [0, 1, 2, 3]``
js
``units = [10, 20, 40, 30]``
js
``fruits = ["peach", "pear", "plum", "plum"]``
js
``Plot.dot(index, {x: units, y: fruits})``
Channel values can also be specified as numbers for constant values, say for a fixed baseline with an area.
js
``Plot.area(aapl, {x1: "Date", y1: 0, y2: "Close"})``
Missing and invalid data are handled specifically for each mark type and channel. In most cases, if the provided channel value for a given datum is null, undefined, or (strictly) NaN, the mark will implicitly filter the datum and not generate a corresponding output. In some cases, such as the radius (r) of a dot, the channel value must additionally be positive. Plot.line and Plot.area will stop the path before any invalid point and start again at the next valid point, thus creating interruptions rather than interpolating between valid points. Titles will only be added if they are non-empty.
All marks support the following style options:
• fill - fill color
• fillOpacity - fill opacity (a number between 0 and 1)
• stroke - stroke color
• strokeWidth - stroke width (in pixels)
• strokeOpacity - stroke opacity (a number between 0 and 1)
• strokeLinejoin - how to join lines (bevel, miter, miter-clip, or round)
• strokeLinecap - how to cap lines (butt, round, or square)
• strokeMiterlimit - to limit the length of miter joins
• strokeDasharray - a comma-separated list of dash lengths (typically in pixels)
• strokeDashoffset - the stroke dash offset (typically in pixels)
• opacity - object opacity (a number between 0 and 1)
• mixBlendMode - the blend mode (e.g., multiply)
• imageFilter - a CSS filter (e.g., blur(5px)) ^0.6.7
• shapeRendering - the shape-rendering mode (e.g., crispEdges)
• paintOrder - the paint order (e.g., stroke)
• dx - horizontal offset (in pixels; defaults to 0)
• dy - vertical offset (in pixels; defaults to 0)
• target - link target (e.g., “_blank” for a new window); for use with the href channel
• ariaDescription - a textual description of the mark’s contents
• ariaHidden - if true, hide this content from the accessibility tree
• pointerEvents - the pointer events (e.g., none)
• clip - whether and how to clip the mark
• tip - whether to generate an implicit pointer tip ^0.6.7
If the clip option is frame (or equivalently true), the mark is clipped to the frame’s dimensions; if the clip option is null (or equivalently false), the mark is not clipped. If the clip option is sphere, then a geographic projection is required and the mark will be clipped to the projected sphere (e.g., the front hemisphere when using the orthographic projection).
If the tip option is true, a tip mark with the pointer transform will be derived from this mark and placed atop all other marks, offering details on demand. If the tip option is set to an options object, these options will be passed to the derived tip mark. If the tip option (or, if an object, its pointer option) is set to x, y, or xy, pointerX, pointerY, or pointer will be used, respectively; otherwise the pointing mode will be chosen automatically. (If the tip mark option is truthy, the title channel is no longer applied using an SVG title element as this would conflict with the tip mark.)
For all marks except text, the dx and dy options are rendered as a transform property, possibly including a 0.5px offset on low-density screens.
All marks support the following optional channels:
• fill - a fill color; bound to the color scale
• fillOpacity - a fill opacity; bound to the opacity scale
• stroke - a stroke color; bound to the color scale
• strokeOpacity - a stroke opacity; bound to the opacity scale
• strokeWidth - a stroke width (in pixels)
• opacity - an object opacity; bound to the opacity scale
• title - an accessible, short-text description (a string of text, possibly with newlines)
• href - a URL to link to
• ariaLabel - a short label representing the value in the accessibility tree
The fill, fillOpacity, stroke, strokeWidth, strokeOpacity, and opacity options can be specified as either channels or constants. When the fill or stroke is specified as a function or array, it is interpreted as a channel; when the fill or stroke is specified as a string, it is interpreted as a constant if a valid CSS color and otherwise it is interpreted as a column name for a channel. Similarly when the fill opacity, stroke opacity, object opacity, stroke width, or radius is specified as a number, it is interpreted as a constant; otherwise it is interpreted as a channel.
The scale associated with any channel can be overridden by specifying the channel as an object with a value property specifying the channel values and a scale property specifying the desired scale name or null for an unscaled channel. For example, to force the stroke channel to be unscaled, interpreting the associated values as literal color strings:
js
``Plot.dot(data, {stroke: {value: "fieldName", scale: null}})``
To instead force the stroke channel to be bound to the color scale regardless of the provided values, say:
js
``Plot.dot(data, {stroke: {value: "fieldName", scale: "color"}})``
The color channels (fill and stroke) are bound to the color scale by default, unless the provided values are all valid CSS color strings or nullish, in which case the values are interpreted literally and unscaled.
In addition to functions of data, arrays, and column names, channel values can be specified as an object with a transform method; this transform method is passed the mark’s array of data and must return the corresponding array of channel values. (Whereas a channel value specified as a function is invoked repeatedly for each element in the mark’s data, similar to array.map, the transform method is invoked only once being passed the entire array of data.) For example, to pass the mark’s data directly to the x channel, equivalent to Plot.identity:
js
``Plot.dot(numbers, {x: {transform: (data) => data}})``
The title, href, and ariaLabel options can only be specified as channels. When these options are specified as a string, the string refers to the name of a column in the mark’s associated data. If you’d like every instance of a particular mark to have the same value, specify the option as a function that returns the desired value, e.g. `() => "Hello, world!"`.
The rectangular marks (bar, cell, frame, and rect) support insets and rounded corner constant options:
• insetTop - inset the top edge
• insetRight - inset the right edge
• insetBottom - inset the bottom edge
• insetLeft - inset the left edge
• rx - the x radius for rounded corners
• ry - the y radius for rounded corners
Insets are specified in pixels. Corner radii are specified in either pixels or percentages (strings). Both default to zero. Insets are typically used to ensure a one-pixel gap between adjacent bars; note that the bin transform provides default insets, and that the band scale padding defaults to 0.1, which also provides separation.
For marks that support the frameAnchor option, it may be specified as one of the four sides (top, right, bottom, left), one of the four corners (top-left, top-right, bottom-right, bottom-left), or the middle of the frame.
All marks support the following transform options:
The sort option, when not specified as a channel value (such as a field name or an accessor function), can also be used to impute ordinal scale domains.
## marks(...marks) ^0.2.0
js
``````Plot.marks(
Plot.ruleY([0]),
Plot.areaY(data, {fill: color, fillOpacity, ...options}),
Plot.lineY(data, {stroke: color, ...options})
)``````
A convenience method for composing a mark from a series of other marks. Returns an array of marks that implements the mark.plot function. See the box mark implementation for an example.
Resources
Observable
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# statistics
Statistics is the study of the collection, organization, analysis, interpretation, and presentation of data. It deals with all aspects of this, including the planning of data collection in terms of the design of surveys and experiments.
A statistician is someone who is particularly well versed in the ways of thinking necessary for the successful application of statistical analysis. Such people have often gained this experience through working in any of a wide number of fields. There is also a discipline called mathematical statistics that studies statistics mathematically.
The word statistics, when referring to the scientific discipline, is singular, as in “Statistics is an art.” This should not be confused with the word statistic, referring to a quantity (such as mean or median) calculated from a set of data, whose plural is statistics (“this statistic seems wrong” or “these statistics are misleading”).
Some consider statistics to be a mathematical body of science pertaining to the…
## Original Professional Online Video: 45 Million Unique Viewers A Month
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After we talked about what we'd like YouTube to do for uploaders and viewers, now it's the advertisers' turn. What would they really want out of the YouTube experience if they could make wishes and get it to happen? Consider this Part 3 of The YouTube Wish List, in which we rub our magic lamps and hope the genie has YouTube skills.
## Recognizing the Signal and the Noise in Online Video Rankings
What can Nate Silver's "The Signal and the Noise" teach us about online video metrics? While Silver's book doesn't talk about the subject specifically, we can use concepts lined out in the book to better separate the signal from the noise when it comes to the often confusing measurement of online video numbers.
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## Website Link Annotations, Creator Playbook 3.0, Foreign Language Captioning, & More...
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## The Rise of Long Form Video Demands New Advertising Strategies [Report]
In the Ooyala Q2 Global Video Index, we found out that Length, Size, Going to Completion Matter. Now, it's time to take a look at Q3 and see how things are shak...
## The State of Automotive Marketing in Social Video: New Report Sees Huge Opportunity for Auto Marketers
Over the summer, Unruly opened up their social video lab, looking to narrow down viral videos to scientific facts, to the point that the word "viral" has become...
## Online Video Viewing in Latin America is Evolving, Rápidamente!
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## 4 Ways to Evaluate Your Performance Beyond Video Views [Creator's Tip #61]
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## Nielsen Says 163.5M Watched Online Video in May, I Say We Need Reporting Standardization
Nielsen has released their online video numbers for May 2012 (which directly compete with comScores but are like apples to oranges) in which they say just 163.5...
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https://mathematica.stackexchange.com/questions/60497/how-to-assign-a-list-of-numbers-into-letter-grades
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How to assign a list of numbers into letter grades
Here is my list:
collection = {76.6256, 51.9264, 50.238, 14.4203, 80.9205, 12.2036, 2.39568,
38.2747, 12.4422, 29.9621}
collection /. {x_ /; x < 50 -> "F"}
and this gives me
{76.6256, 51.9264, 50.238, "F", 80.9205, "F", "F", "F", "F", "F"}
as expected. But I want to add more rules with ranges like that. I'm not sure how to do this. I've tried:
collection /. {{x_ /; x < 50 -> "F"}, {x_ /; 50 <= x < 56 -> "D"},
{x_ /; 56 <= x < 71 -> "C"}, {x_ /; 71 <= x < 85 -> "B"},
{x_ /; 85 <= x <= 100 -> "A"}}
and that yields
{{76.6256, 51.9264, 50.238, "F", 80.9205, "F", "F", "F", "F", "F"},
{76.6256, "D", "D", 14.4203, 80.9205, 12.2036, 2.39568, 38.2747, 12.4422, 29.9621},
{76.6256, 51.9264, 50.238, 14.4203, 80.9205, 12.2036, 2.39568, 38.2747,
12.4422, 29.9621},
{"B", 51.9264, 50.238, 14.4203, "B", 12.2036, 2.39568, 38.2747,
12.4422, 29.9621},
{76.6256, 51.9264, 50.238, 14.4203, 80.9205, 12.2036, 2.39568, 38.2747,
12.4422, 29.9621}}
and I've tried a few others which don't even work at all. What is the simplest way of going about this? My real question is, how do I construct a list of rules where I can determine which range (0-50, 50-60, etc.) each number in the list is in?
• Your replacement rules should be a list of rules, not a list of lists. When it's a list mylist of lists, ReplaceAll (/.) returns a list of results, one for each list in mylist. See @belisarius's answer. Sep 23, 2014 at 21:59
Just put all your rules in a list. You don't need the lower bounds. The first matching rule in the list is applied:
s = {x_ /; x < 50 -> "F", x_ /; x < 56 -> "D", x_ /; x < 71 -> "C",
x_ /; x < 85 -> "B", x_ /; x <= 100 -> "A"};
collection /. s
(* {"B", "D", "D", "F", "B", "F", "F", "F", "F", "F"}*)
• Crap! I swear I tried this exact same thing and it didn't work for me! But perhaps I forgot a bracket or something. This is definitely the simplest and was exactly what I was looking for. Thank you very much and thanks to everyone who contributed. Your help is really appreciated! Sep 24, 2014 at 0:22
Clear[grade]
{"F", x < 50}, {"D", x < 56},
{"C", x < 71}, {"B", x < 85}},
"A"];
collection = {76.6256, 51.9264, 50.238,
14.4203, 80.9205, 12.2036, 2.39568,
38.2747, 12.4422, 29.9621};
{"B", "D", "D", "F", "B", "F", "F", "F", "F", "F"}
For a large number of values a better method is to use Interpolation with an order of zero.
First build the InterpolatingFunction:
points = {{50, "F"}, {56, "D"}, {71, "C"}, {85, "B"}, {100, "A"}};
fn = Interpolation[points, InterpolationOrder -> 0];
Then:
fn[collection] // Quiet
{"B", "D", "D", "F", "B", "F", "F", "F", "F", "F"}
Performance
I claimed that for "a large number of values" the method above is superior to replacement rules. Here is a BenchmarkPlot to support that assertion:
letters = StringJoin /@ Tuples[CharacterRange["A", "Z"], 3];
numbers = Sort @ RandomSample[Range[100000], 26^3];
points = {numbers, letters}\[Transpose];
rand = RandomReal[100000, 50000];
f1 = ReplaceAll[rand, (x_ /; x < # -> #2) & @@@ #] &;
f2 = Interpolation[#, InterpolationOrder -> 0][rand] &;
Off[InterpolatingFunction::dmval]
Needs["GeneralUtilities"]
BenchmarkPlot[{f1, f2}, Sort@RandomSample[points, #] &, 2^Range[14],
"IncludeFits" -> True, TimeConstraint -> 30]
• Note the logarithmic scale.
• The x axis is the number of levels (letters) in the assignment.
• Time to construct each InterpolatingFunction is included; reuse would be faster.
ReplaceAll is faster with only a few levels, but from eight and up InterpolatingFunction is faster, and with many levels orders of magnitude faster.
More handy in case of more notes (?).
notes = {"F", "D", "C", "B", "A"};
bl = BinLists[collection, {{0, 50, 56, 71, 85, 100}}];
{"B", "D", "D", "F", "B", "F", "F", "F", "F", "F"}
or
notes = {"F", "D", "C", "B", "A"}
levels = {50, 56, 71, 85, 100}
Block[{x},
Function[note[x_] := Which[##]] @@ Flatten[MapThread[{x <= #, #2} &,
{levels, notes}]]
]
note /@ collection
• +1 I've seen cases where 26 letters aren't enough Sep 23, 2014 at 22:02
You can use Fold to successively apply a list of rules.
Fold[#1 /. #2 &, collection, {{x_ /; x < 50 -> "F"},
{x_ /; 50 <= x < 56 -> "D"}, {x_ /; 56 <= x < 71 -> "C"},
{x_ /; 71 <= x < 85 -> "B"}, {x_ /; 85 <= x <= 100 -> "A"}}]
{"B", "D", "D", "F", "B", "F", "F", "F", "F", "F"}
collection /.
x_?NumberQ :> Which[x < 50, "F", x < 56, "D", x < 71, "C", x < 85, "B", x <= 100, "A"]
{"B", "D", "D", "F", "B", "F", "F", "F", "F", "F"}
Timing comparison for 1000 random reals repeated 1000 times
• (+1) timings={2.53, 2.71, 3.01, 3.42, 4.68, 5.09, 13.59}; gradesF[{10, Infinity}, Reverse@{"ok", "Ouch"}][timings]:)
– kglr
Sep 24, 2014 at 13:47
• @kguler your solution has merits (flexibility) which - in many cases - should be more important than seconds :)
– eldo
Sep 24, 2014 at 14:11
• Please see the timing plot in my answer. It doesn't invalidate your timings but it clarifies and supports what I meant about the performance of Interpolation. Oct 2, 2014 at 17:02
ClearAll[gradesF];
gradesF[x_: {49, 59, 79, 89, 100}, y_: {"A", "B", "C", "D", "F"}] :=
Function[{z}, Piecewise[{#, z < #2} & @@@ (Flatten /@ Thread[{Reverse@y, x}])], Listable];
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# conndef
Create connectivity array
## Syntax
• ``conn = conndef(num_dims,type)``
example
## Description
example
````conn = conndef(num_dims,type)` returns the connectivity array defined by `type` for `num_dims` dimensions. Several Image Processing Toolbox™ functions use `conndef` to create the default connectivity input argument. Code Generation support: Yes.MATLAB Function Block support: Yes.```
## Examples
collapse all
### Create 2-D Connectivity Array with Minimal Connectivity
`conn1 = conndef(2,'minimal')`
```conn1 = 0 1 0 1 1 1 0 1 0```
### Create 2-D Connectivity Array with Maximal Connectivity
`conn2 = conndef(2,'maximal')`
```conn2 = 1 1 1 1 1 1 1 1 1```
### Create 3-D Connectivity Array with Minimal Connectivity
`conndef(3,'minimal')`
```ans(:,:,1) = 0 0 0 0 1 0 0 0 0 ans(:,:,2) = 0 1 0 1 1 1 0 1 0 ans(:,:,3) = 0 0 0 0 1 0 0 0 0```
## Input Arguments
collapse all
### `num_dims` — Number of dimensionsnumeric scalar
Number of dimensions, specified as a numeric scalar.
Example: `conn1 = conndef(2,'minimal')`
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64`
### `type` — Type of neighborhood connectivity`'minimal'` | `'maximal'`
Type of neighborhood connectivity, specified as the string `'minimal'` or `'maximal'`
Value
Description
`'minimal'`
Defines a neighborhood whose neighbors are touching the central element on an (`N`-1)-dimensional surface, for the `N`-dimensional case.
`'maximal'`
Defines a neighborhood including neighbors that touch the central element in any way; it is `ones(repmat(3,1,NUM_DIMS))`.
Example: `conn1 = conndef(2,'minimal')`
Data Types: `char`
## Output Arguments
collapse all
### `conn` — Connectivity matrix3-by-3-by...-3 logical array
Connectivity matrix, returned as a 3-by-3-....-by-3 logical array.
expand all
### Code Generation
This function supports the generation of C code using MATLAB® Coder™. For more information, see Code Generation for Image Processing.
When generating code, the `num_dims` and `type` arguments must be compile-time constants.
### MATLAB Function Block
You can use this function in the MATLAB Function Block in Simulink.
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# Testing Poisson process where $X(t)$ is given at fixed times
I have a discrete stochastic process $X(t)$ which I believe is a Poisson process, that is the value of $X(t)$ at time $t$ is a Poisson random variable with parameter $\lambda t$ and disjoint intervals are assumed to be independent.
Due to the nature of the data I only have values of $X(t)$ at discrete and fixed times $t_1, t_2, .., t_n$. Meaning that rather than having single jumps of size one I have jumps of multiple steps between times $X(t)$ is measured.
The times $t_1, t_2, .., t_n$ are not the actual 'arrival times' of the Poisson process. They are simply times when 'an observer' has decided to check the process's value X(t). The real 'arrival times' are completely unknown.
Does it make sense to think of $X(t)$ as a Poisson process when the exact times of arrival are unknown? Or is it a pointless endeavour?
If it does make sense, are there any tests one would recommend in this context?
• The $t_i$ values are not known? If not, is anything known/reasonably assumed about how they're distributed? (e.g. would exponentially distributed intervals make sense?) Commented Jun 26, 2017 at 0:00
• Sorry that was very poorly worded on my part (fixed). I have the $t_i$ which are known, alongside which we have $X(t_i)$. However, these $t_i$ are not the actual "arrival times". They are simply the value of the process $X(t_i)$ at time $t_i$. I meant "random" in the sense that they are given in no real systematic way. The actual arrival times are unknown, however I assume that time between the unknown arrival times is exponentially distributed. Commented Jun 26, 2017 at 3:24
• I had not considered the distribution of these time intervals $\Delta t_i = t_i - t_{i-1}$ and was simply treating them as non-random. In the sense that an observer checks the price at these times, not knowing himself when the process is jumping. Is that clearer? If not, I can try to explain better. Thanks Glen. Commented Jun 26, 2017 at 3:29
• If the times are known there's no issue -- the distribution question was only if they were unknown. With known times you have exposure values and calculations should be straightforward; What are you trying to find out?? e.g. are you trying to estimate $\lambda$? construct a prediction interval for the number of events in some future time period? something else? Commented Jun 26, 2017 at 5:29
• This is a special case of a Poisson GLM using a non-canonical link. The number of arrivals between $t_{i-1}$ and $t_i$ is Poisson with expectation $\lambda \Delta t_i$ and much of GLM theory holds.
– Yves
Commented Jun 26, 2017 at 9:08
## 1 Answer
An answer partly depends on what you mean by the word "test" there.
If you're after a formal hypothesis test, one could try to perform some test based on the likelihood or the log of the likelihood (as per Fisher) or if there's a specific alternative via a likelihood ratio test, though various tests might be devised depending on broader classes of alternatives that should perform pretty well.
My advice would not be to pursue formal tests of goodness of fit. Our models are (usually) just that -- models, not perfect descriptions. As such in large samples they're clearly wrong (even though the model may be a very good description of the thing we're modelling) and in small samples the fact that we can't clearly tell these models are wrong should be no consolation, since in a small sample even a quite poor model may be hard to clearly identify. A clearly wrong model - one we would reject via a test - may remain highly useful, if the deviation from correctness has small effects on our inference (i.e. we're dealing more with raw-effect-size type problems rather than statistical significance)
As such I'd tend to lean toward diagnostic displays rather than formal tests, though if you can be much more specific about what effects will matter, there may be some uses for testing.
One approach to displays would be to check the mean-variance relationship. In particular, if we let $Y_i$ be the $i$-th increment in count and we let $s_i$ be the corresponding increment in time, then under the Poisson assumption $Y_i\sim \text{Pois}(\lambda s_i)$. If it's Poisson the mean should be proportional to $s_i$ and the variance should be proportional to $s_i$ (or standard deviation proportional to $\sqrt{s_i}$).
This is easier to see on the square root scale. If the Poisson means are not typically very small (so the observations aren't nearly all 0 or 1) you can plot either $\sqrt{y_i+\frac38}$ or $\sqrt{y_i}+\sqrt{y_i+1}$ against $\sqrt{s_i}$ -- either one should look approximately linear with fairly constant spread:
The first transformation in the Anscombe transformation for the Poisson, the second is the Freeman-Tukey transformation.
It's possible to adjust the $\sqrt{s_i}$ for each plot to be closer to the expected value of the transformed $Y$ for small $\lambda s_i$ which can improve the appearance of the plot a bit when the mean is smaller, but it's usually not necessary.
There are other options that can work as well.
You might even consider an examination of a suitable choice of residuals vs fitted and a scale vs fitted from a constant-only Poisson glm fit with log link and an offset of $\log(s_i)$, but these can be considerably harder to interpret. This is one where the model is correct (the same data as above):
(Personally I find the earlier plots easier to judge)
• This is an excellent answer and has helped me a lot to better understand what I actually want to do. Given the size of my data (10's of millions of data points) it was probably quite naive of me to attempt to formally test it. Will keep this in mind in future. Thanks a lot Glen, big help as usual. Commented Jun 27, 2017 at 21:14
• With tens of millions of points, there's other things still that you could do both in the direction of displays and in formal testing (which will almost certainly reject of course). ... ctd Commented Jun 27, 2017 at 22:00
• ctd... For example, another display -- if one bins on the time increments $s_i=t_i-t_{i-1}$ (where $t_0$ is the start of the observation period), and looks at tables of counts of the number of $x_k=\#\{y_i=k\}$ within each $s$-bin then one might produce a modified version of a Poissonness plot (in this case requiring plotting a transformed $x$ vs a transformed $k$ and some representative value for the $s$-bin - i.e. a 3D plot) which should be linear in both directions. Commented Jun 27, 2017 at 22:01
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20120323
Physics midterm question: object moved away from converging lens
Physics 205B Midterm 1, spring semester 2012
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 23.69
Placing an object to the left of a converging lens produces an inverted image. If the object is moved slightly farther from the lens, describe what will happen to the size of this image. Explain your reasoning by using the properties of lenses, thin lens equations and/or ray tracings.
• p:
Correct. Uses ray diagram or thin lens equation(s) to explain why resulting real image will be smaller when object (already outside of the primary focal point) is moved away from the converging lens.
• r:
As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. May argue as p increases, m = h'/h = –di/do decreases, but does not explicitly show that q decreases as well.
• t:
Nearly correct, but argument has conceptual errors, or is incomplete. Explicit use of comparative ray tracings, or thin lens equations, but instead concludes image would increase in size as object distance is increased.
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
• x:
Implementation/application of ideas, but credit given for effort rather than merit.
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.
Section 30882
Exam code: midterm01o1Ls
p: 16 students
r: 3 students
t: 5 students
v: 3 students
x: 0 students
y: 0 students
z: 0 students
A sample "p" response (from student 0524), using the thin lens equations:
A sample "p" response (from student 1123), using both ray tracings and thin lens equations:
A sample "t" response (from student 1227), using a ray tracing to show that the image size would increase:
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# Linear Inequations-11
Solve 1 ≤ |x - 2| ≤ 3
Solution: We know that:
a ≤ |x - c| ≤ b.
<=> x ∈ [-b + c, -a + c] U [a + c, b + c]
Therefore, 1 ≤ |x - 2| ≤ 3 <=> x ∈ [-3 + 2, -1 + 2] U [1 + 2, 3 + 2]
<=> x ∈ [-1, 1] U [3, 5]
Hence, the solution set of the given inequation is [-1, 1] U [3, 5]
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Sorry, you need to enable JavaScript to visit this website.
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Still celebrating National IT Professionals Day with 3 months of free Premium Membership. Use Code ITDAY17
x
Solved
# 3d polygon sorting and clipping..
Posted on 2000-04-03
Medium Priority
440 Views
Hi,
I'm writing a 3d engine as a special project in school.
Therefore it is not very fast but rather easy to understand,
everything is class based and most of the stuff are done
automatacally. BUT.. I have two problems.. polygon sorting
and clipping. What I think I would need is a function that
converts a point from the standard plane to the camera
plane so I can calculate where the point is related to what
position the camera is. But, I have no idea how so currently
I'm sorting polygons by z and I have no clipping. So if you
could help me it would really be appreciated.
/ eaz
0
Question by:eaz
[X]
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LVL 2
Expert Comment
ID: 2680528
Sorting polygons by z axis is quite smart since you can start drawing polygons from back to fron that way you dont get the overlap, also you need to calculate the normals you know which surface are visible to the camera.
there are so many documents about 3D engines that you could spend days and days reading about them
check out http://www.cfxweb.net/ for further links
i could post some code for it as well. but first you read :D
Good luck on you
T. Rizos.
0
LVL 1
Expert Comment
ID: 2681742
first what are you using to write your prog ???
like in OpenGL you don't have this problem it doing it automaticlly
about sorting there are to algorithms :
1)"object sacn" when you are drawing an object go allover the object's and check if there is an object that you will draw and this point and check which one is nearer and that for each point of the object you are about to draw
2)"screen scan" when you are drawing make a matrix that will be the nearest z coordinate displayed on the screen
and then when you are drawing you have to check if the point you are about to draw is nearer then the nearest point if it is draw it if not don't
but for that you"ll have to write your own drawing func's
or just use OpenGL which as i know use one of this algorithms
0
Author Comment
ID: 2681917
I'm coding it from the ground up.. using DJGPP in mode 13h. So far I can draw and rotate some objects right in front of the camera in grayscale and do some antialiasing (terribly slow..). But I can not rotate the camera around the object or build move it around a scene since that would make objects end up behind the camera.
In another forum I got an answer that stated I could multiply my CameraMatrix with a PolygonMatrix to get a ResultMatrix which I then should multiply with something to get a final PolygonCameraMatrix. Help with that last part? Anyone?
/ eaz
0
LVL 1
Expert Comment
ID: 2683023
tell me how you make the camera
as i see it it a matrix of rotation and translation
that will rotate and translate the scene
to the right position
and you should have a matrix that is the rotation and tranlation ,... of the object's for each object there is a matrix
and then you mult the two matrix's
the difrence it that the camera matrix is for all the objects and the other one is matrix per object
0
LVL 1
Expert Comment
ID: 2683027
tell me how you make the camera
as i see it it a matrix of rotation and translation
that will rotate and translate the scene
to the right position
and you should have a matrix that is the rotation and tranlation ,... of the object's for each object there is a matrix
and then you mult the two matrix's
the difrence it that the camera matrix is for all the objects and the other one is matrix per object
0
LVL 1
Expert Comment
ID: 2683030
tell me how you make the camera
as i see it it a matrix of rotation and translation
that will rotate and translate the scene
to the right position
and you should have a matrix that is the rotation and tranlation ,... of the object's for each object there is a matrix
and then you mult the two matrix's
the difrence it that the camera matrix is for all the objects and the other one is matrix per object
0
Author Comment
ID: 2684509
My camera is a struct with two double arrays (position and rotation) and one 4x4 viewmatrix.
typedef struct Camera3D
{
Point3D position;
Point3D rotation;
MATRIX_4x4 viewmatrix;
};
Ntdragon, there is a lot of errors in your text, it makes it hard to read and understand.. is it because of sloppy writing? No offence.. I'm just wondering. I'm also quite new to 3d programming so I don't understand everything about matrixes yet, so the easier you explain the faster I will learn. And my math book stops right where this begins.. quite annoying acctually.
0
LVL 1
Accepted Solution
ntdragon earned 200 total points
ID: 2691320
let's start with matrix you have to learn the basic operation with them
(+) and (*)
if you don't know how to do that tell me i"ll explain it to you
now about translation matrix there are some matrix for translation one for rotating one for translating and one for scaling you can find them in any graphic book
if you didn't manage to do so tell me and i"ll send them to you
now about your code you made a camera struct except it you have to make the same struct for each object or to make one global for all the object it will be for transfering the objects (rotate,translate,scale)
except that you have to make a func that mult to matrix's <i mean your struct's>
then when you draw the object you have to mult the camera matrix with the object matrix and the point you are about to draw and then draw<for that you have to write a func that mult matrix with vector>
0
Author Comment
ID: 2700412
Thank you for taking your time. I have solved the problem. Now all I have to implement is clipping.. and texturing and.... and.. and... =)
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Expert Comment
ID: 2707788
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1. ## Proving...
Prove that if (v1, ... , vn) is a basis of V, then so is(v1, v2-v1, ... , vn - vn-1).
I made the initial assumption that if V remained the same number that by subtracting v1, youd always end up with v1, and so on. But obviously I came to the conclusion that this would have to be wrong, because the set of (v1, .. , vn) could represent any numbers. So I have no idea how to tackle this one, any help would be greatly appreciated. Thanks.
2. If $a_i$ are such that $\sum_{i=1} ^{n} \ a_iv_i =0$ then $a_i =0$ for all $i$. Now suppose there are scalars $b_i$ such that $\sum_{i=1} ^{n} \ b_i(v_i-v_{i-1})=0$ where $v_0=0$ then we get $\sum_{i=1} ^{n} \ (b_i - b_{i+1})v_i=0$ where $b_{n+1}=0$ then $b_n=0$, and so $b_{n-1}=0$ ... $b_1=0$. Which means they're l.i. and therefore a basis.
3. Thank you very much Jose, but I am unsure as to whether I can approach the question like this. It would almost appear as if that is proving that they are a basis rather than specifically a basis for V. Would it be possible to get a second opinion on this?
4. Any set of 'n' independent vectors which belong to a n-dimensional vector space,V, will be a basis of V.
(This is a std theorm and can be found in most of the texts on linear algebra)
5. Originally Posted by aman_cc
Any set of 'n' independent vectors which belong to a n-dimensional vector space,V, will be a basis of V.
(This is a std theorm and can be found in most of the texts on linear algebra)
Yes I understand that it is a standard theorem, just as we can say X²=4 when x=2. But that is not what I am asking, I am asking how to PROVE this theorem. That is a completely different method. However I appreciate your input.
6. Originally Posted by GreenDay14
Thank you very much Jose, but I am unsure as to whether I can approach the question like this. It would almost appear as if that is proving that they are a basis rather than specifically a basis for V. Would it be possible to get a second opinion on this?
What is your definition of basis?
7. It would almost appear as if that is proving that they are a basis rather than specifically a basis for V.
You are given that the first n vectors form a basis for V so you know that V has dimension n. You are given another n vectors in V. To show that they are a basis for V you need only show that they span V or that they are independent.
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Good Stats, Bad Stats I Oxford Open Learning
# Good Stats, Bad Stats
Have you ever noticed how much we use statistics? As in, “My salary’s only gone up by 1%?”, or “MP’s are going to give themselves a high percentage pay rise.” A lot of the time it’s pretty automatic and it’s easy and we know what we mean. You can learn an awful lot from statistics, after all.
There is a group called the World of Statistics, run by a number of universities. It has a comprehensive website showing, among other things, how statistics are helping planners, scientists and politicians around the world do their jobs and think about how to solve problems, in ways you and I would hardly think of. Statistics have helped increase our understanding of disease and ill health and therefore increased our ability to develop medicines to counter them. They have helped improve political elections in Mexico. They’ve also been used to improve agriculture and thereby increase our ability to feed ourselves. They’ve made it possible to study animal movements and increase our understanding of nature. All of this is on the W.o.S website’s global summary diagram.
Statistics can be both useful and act as an instrument for change. We know that, before the fees increase, 49% of young people in this country were entering higher education. Of these, 55% were women. And yet only 30% of MP’S are women. So it would make sense to ask what’s holding women back. Why are we allowing ourselves to miss out on all these bright people, who really should be helping to govern us.
On a lighter note, stats attract a special fascination when it comes to sport. Leicester City football club have had on average less possession of the ball during their games this season than Tottenham Hotspur, yet they’re higher up in the table than Spurs; it begs the question, what makes a successful football team? And a particular favourite of mine – The New Zealand All Blacks rugby team had a 91.6 % win rate in their games last year. Not bad!
The difficulties start when statistics “go wrong”, of course. They are not everything, as they say, no matter how much we like to highlight them and how useful they can be. Benefit fraud is often thought to account for 20% of the welfare budget when this figure is really only 0.7%. But this misleading figure will colour attitudes and maybe even voting patterns. And unemployment benefit isn’t 40% of the welfare budget – more like 1%. Again, it’s a damaging misconception. There are plenty of those which can be exploited by unscrupulous people, so it pays for us to check figures for accuracy before we reach too many conclusions. Statistics can be a blessing – but they can also all too easily become a curse!
The trick with statistics, then, is to not always take them at face value. Sometimes it is the stats behind the stats that are the ones to look at most closely.
See more by
I'm semi-retired after a successful and much enjoyed career in education. Funnily enough, my last job was as a tutor for OOL. I taught on courses providing professional training for school support staff, as well as A level English Literature and English Literature GCSE. I've had an interesting career, in schools, colleges, adult education, the Arts and a few other bits and bobs. At one stage I was also a local authority inspector. Now I'm a school governor, and am enjoying watching my young grandchildren go through their own first experiences of school. Through these articles I hope to keep you up to date with different aspects of education news – and also to keep you interested!
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Logic Puzzle Forums (https://www.logic-puzzles.org/forum/index.php)
springrate 12-02-2016 09:50 AM
Okay, so we have this situation.
http://i.imgur.com/pXfsc23.jpg
Clue #3 says "the Annata Branco is either the merlot or the 1992 bottle." Okay. Given what we have so far, the only way the Annata Branco can be the merlot is if the merlot is from 1984.
Therefore my question, with this information, does this mean that the 1984 is ALWAYS the merlot? Because otherwise the clue would be impossible? Or if the Annata Branco is 1992, does the Merlot become irrelevant to the clue and it can be either 1984 or 1988?
Hopefully that makes sense - I'm wondering if clue #3 definitively states that the merlot is from 1984, regardless of it being the Annata Branco or not.
Maybe an example would be - if the Annata Branco is 1992, can the Merlot be 1988 because the clue does say "either"? Or is the clue telling me that the Merlot has to be 1984 since that's the only way it could also be the Annata Branco?
SamanthaJoy 12-02-2016 10:03 AM
It absolutely does NOT mean that the merlot is necessarily from 1984.
The only positive thing we can be sure of, that you don't already have recorded, is that the Annata Branco cannot be the gewurztraminer.
Lacewing 12-05-2016 03:23 PM
This type of clue, A is either B or C, does not tell us anything about A, only that B and C do not go together. There may be another clue that discusses B and shows you that B is boat A, or the same with C. Or, there may be an additional clue that puts A again with B or C.
X out B with C (year with merlot) and move on.
Orion13 01-27-2017 12:52 PM
Actually it can tell us quite a bit more than the previous poster admits.
One of the confusions I see in the OP is how to literally read: A is either B or C. The more wordy and precise way is that the object with attribute A is either ALSO the object with attribute B or else it's the object with attribute C. As the previous poster indicates, since the object with Attribute A can't be associated with BOTH attributes B and C it is safe to "X out" where B and C intersect on the grid. B and C are two distinct items in our solution only one of which pairs with A. It does not tell you what B is, only that it isn't C.
But we're not done. Once we find out other attributes of the objects involving A, B or C, it is possible for this clue to inform us further. If in the OP we have deduced from other clues, for example, that the Annata Branco is from 1984 THEN IT MUST be the merlot based on Clue #3, since it has to be EITHER the merlot OR the 1992 and we have deduced it is not the 1992.
All times are GMT. The time now is 02:23 PM.
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StudentShare solutions
Job Satisfaction: A Review of the Statistics - Essay Example
Extract of sample Job Satisfaction: A Review of the Statistics
As you may already know, the AIU gathered a large database of survey responses from 288 individuals that agreed to participate in the study. The questions were asked in regards to gender, age, department, position, tenure, overall job satisfaction, and intrinsic/extrinsic job satisfaction. These statistics will be reviewed before providing clarification in which the methods in which the statistics can be further utilized. The hypothesis to test whether or not the average overall job satisfaction (in the population of all workers in the USA) is equal to 4.5 with a = .05 is further reviewed and determined. An alternative hypothesis suggests that this test will reveal a somewhat decreased number below the suggested 4.5 with a significance level of the standardized .05 or 5%. The critical level, also known as the critical region is any number above or below the designated 4.5 mark.
In regards to the job satisfaction by gender, first we found that the percentage for males was 4.43 (531.7 total male overall satisfaction / 120 male) and for females the percentage was 4.26% (715.4 total female overall satisfaction / 168 female). So, the average overall job satisfaction between the males and females would be the average number between 4.43% (total males) and 4.26% (total females), which equals 4.33% (total overall job satisfaction).
Departments
The majority of the participants in the survey belonged to administration departments (61.5%). The remaining departments, being human resources and information technology sectors, stood at 3.8% (HR) (11 human resources / 288 total participants) and 34.7% (IT) (100 information technology / 288 total participants), respectively. The overall satisfaction with gender was rather similar. For males, the percentage was 4.43 (531.7 total male overall satisfaction / 120 male) and for females the percentage was 4.26% (715.4 total female overall satisfaction / 168 female).
Intrinsic/Extrinsic Job Satisfaction
The hypothesis states that we believe that there are equal "deskbodies" as well as "socialbodies" in the workforce. A "deskbody" is someone whose intrinsic job satisfaction level is higher than his or her extrinsic job satisfaction level (i.e. happy with their job more than their office). A "socialbody" is someone whose extrinsic job satisfaction level is higher than his or her intrinsic job satisfaction level (i.e. happy with the office more than their job). Our research suggests that there are equal deskbodies and socialbodies in the workforce. ...Show more
Summary
When given the choice of palm trees, yacht parties in the Caribbean, or the daily office routine, many would jump at the chance to leave it all behind. However, paying the piper isn't the only reason for going to work everyday. A job can be fun, challenging and provide some sort of satisfaction, right Well, within the past few decades it seems as if job satisfaction is on the decline, along with the business's revenues, productivity and net worth because of this up and coming epidemic…
Author : kautzerchase
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# Static Electricity - Mission SE10 Detailed Help
A charge Q creates an electric field. A test charge q is used to measure the strength of the electric field at a distance d from Q. The electric field strength is defined as ____.
Definition of Electric Field Strength: Any source of charge Q will create an electric field in the space surrounding it. The strength of an electric field (E) at any given location in this space can be determined by placing a test charge q in the space and measuring the force (F) that is exerted upon it. The electric field strength is defined as the amount of force per unit of charge on the test charge. E = F / q
It is easy to become confused by the mathematics of electric field strength. It is important to bear in mind that there are always two charges involved in any electrical interaction. In this case, the charges are Q and q. Big Q represents the source charge which creates the electric field. Little q represents the test charge which is used to measure the strength of the electric field at a given location surrounding the source charge. Give considerable attention to the charge quantity - Q or q - being used in each equation.
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# Number 301210010
### Properties of number 301210010
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
11f4199a
Base 32:
8v86cq
sin(301210010)
0.0016287580281737
cos(301210010)
0.99999867357276
tan(301210010)
0.0016287601886056
ln(301210010)
19.52331828707
lg(301210010)
8.4788694005151
sqrt(301210010)
17355.402905147
Square(301210010)
### Number Look Up
Look Up
301210010 (three hundred one million two hundred ten thousand ten) is a very special figure. The cross sum of 301210010 is 8. If you factorisate the number 301210010 you will get these result 2 * 5 * 30121001. 301210010 has 8 divisors ( 1, 2, 5, 10, 30121001, 60242002, 150605005, 301210010 ) whith a sum of 542178036. 301210010 is not a prime number. The figure 301210010 is not a fibonacci number. 301210010 is not a Bell Number. The number 301210010 is not a Catalan Number. The convertion of 301210010 to base 2 (Binary) is 10001111101000001100110011010. The convertion of 301210010 to base 3 (Ternary) is 202222210001110022. The convertion of 301210010 to base 4 (Quaternary) is 101331001212122. The convertion of 301210010 to base 5 (Quintal) is 1104102210020. The convertion of 301210010 to base 8 (Octal) is 2175014632. The convertion of 301210010 to base 16 (Hexadecimal) is 11f4199a. The convertion of 301210010 to base 32 is 8v86cq. The sine of the figure 301210010 is 0.0016287580281737. The cosine of the number 301210010 is 0.99999867357276. The tangent of 301210010 is 0.0016287601886056. The root of 301210010 is 17355.402905147.
If you square 301210010 you will get the following result 90727470124200100. The natural logarithm of 301210010 is 19.52331828707 and the decimal logarithm is 8.4788694005151. that 301210010 is amazing figure!
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https://discuss.leetcode.com/topic/36299/can-anyone-help-me-to-improve-my-program-138-ms
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# Can anyone help me to improve my program(138 ms)?
• ``````public int[][] multiply(int[][] A, int[][] B) {
Map<Integer, Map<Integer,Integer>> map=new HashMap<>();
int b=A.length;
int a=A[0].length;
int c=B[0].length;
int result[][]=new int[b][c];
for(int i=0;i<b;i++){
Map<Integer,Integer> mapp=new HashMap<>();
map.put(i,mapp);
for(int j=0;j<a;j++){
if(A[i][j]!=0) {map.get(i).put(j,A[i][j]);}
}
if(map.get(i).isEmpty()) map.remove(i);
}
for(int i=0;i<c;i++){
for(int j:map.keySet()){
for(int k:map.get(j).keySet()){
if(B[k][i]!=0) result[j][i]+=B[k][i]*map.get(j).get(k);
}
}
}
return result;
}
``````
I understand this is not the best way to solve the problem. Could anyone try improve it to run faster?
• You do not need a Map for that. Here is my simple solution in C++:
``````vector<vector<int>> multiply(vector<vector<int>>& A, vector<vector<int>>& B) {
int aR = A.size();
int aC = A[0].size();
int bR = B.size();
int bC = B[0].size();
vector<bool> Ab(aR,false);
vector<bool> Bb(bC,false);
for (int i = 0; i <aR; i++)
{
int j = 0;
for (; j<aC; j++)
if (A[i][j])
{
Ab[i] = true;
break;
}
}
for (int i = 0; i <bC; i++)
{
int j = 0;
for (; j<bR; j++)
if (B[j][i])
{
Bb[i] = true;
break;
}
}
vector<vector<int>> res(aR, vector<int>(bC,0));
for (int i = 0; i <aR; i++)
{
if (!Ab[i]) continue;
for (int j = 0; j <bC; j++)
{
if (!Bb[j]) continue;
for (int p = 0; p<bR; p++)
res[i][j]+=A[i][p]*B[p][j];
}
}
return res;
}``````
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• outputs=self.activation_function(final_inputs)#设置最终层的输出为最终层的输入经过S函数激活 #反向传播误差 #output layer error is the(target-actual) output_errors=targets-final_outputs#输出误差=目标值-...
#训练集http://www.pjreddie.com/media/files/mnist_train.csv
#测试集http://www.pjreddie.com/media/files/mnist_test.csv
import numpy
#scipy.special for the sigmoid function expit()
import scipy.special
import matplotlib.pyplot
#neural network class definition
class neuralNetwork:
#initialise the neural network
#初始化神经网络
def __init__(self, inputnodes, hiddennodes, outputnodes, learningrate):#四个参数
self.inodes=inputnodes#输入层节点
self.hnodes=hiddennodes#隐藏层节点
self.onodes=outputnodes#输出层节点
self.wih=(numpy.random.rand(self.hnodes,self.inodes)-0.5)#创建输入层与隐藏层的随机权重矩阵，每个值取-0.5到0.5
self.who=(numpy.random.rand(self.onodes,self.hnodes)-0.5)#创建隐藏层与输出层的随机权重矩阵，每个值取-0.5到0.5
#another method 采用正态概率分布采样权重
#self.wih=numpy.random.normal(0.0,pow(self.hnodes,-0.5),(self.hnodes,self.inodes))
#self.who=numpy.random.normal(0.0,pow(self.onodes,-0.5),(self.onodes,self.hnodes))
#learning rate
self.lr=learningrate#设置学习率
#activation function is the sigmoid function
self.activation_function=lambda x: scipy.special.expit(x)#设置激活函数为S函数
pass
#train the neural network
def train(self,input_list,targets_list):
#convert inputs list to 2d array
#将输入输出数据转换为二维数组
inputs=numpy.array(input_list, ndmin=2).T
targets=numpy.array(targets_list, ndmin=2).T
#calculate signals into hidden layer
hidden_inputs=numpy.dot(self.wih,inputs)#设置隐藏层的输入等于输入节点与相应权重矩阵的点乘
#calculate the signals emerging from hidden layer
hidden_outputs=self.activation_function(hidden_inputs)#设置隐藏层的输出为隐藏层的输入经过S函数激活
#calculate signals into final output layer
final_inputs=numpy.dot(self.who,hidden_outputs)#设置最终层的输入为隐藏层的输出与相应权重矩阵的点乘
#calculate the signals emeging from final output layer
final_outputs=self.activation_function(final_inputs)#设置最终层的输出为最终层的输入经过S函数激活
#反向传播误差
#output layer error is the(target-actual)
output_errors=targets-final_outputs#输出误差=目标值-实际值
#hidden layer error is the output_errors,split by weights,recombined at hidden nodes
hidden_errors=numpy.dot(self.who.T, output_errors)#按权重计算隐藏层各节点误差
#update the weights for the links between the hidden and output layers
#更新隐藏层与输出层的权重
self.who+=self.lr*numpy.dot((output_errors*final_outputs*(1.0-final_outputs)),numpy.transpose(hidden_outputs))
#update the weights for the links between the input and hidden layers
#更新输入层与隐藏层的权重
self.wih+=self.lr*numpy.dot((hidden_errors*hidden_outputs*(1.0-hidden_outputs)),numpy.transpose(inputs))
pass
#query the neural network
def query(self,inputs_list):
#self.activation_function=lambda x:scipy.special.expit(x)
#convert inputs list to 2d array
inputs=numpy.array(inputs_list,ndmin=2).T
#calculate signals into hidden layer
hidden_inputs=numpy.dot(self.wih,inputs)
#calculate the signals emerging from hidden layer
hidden_outputs=self.activation_function(hidden_inputs)
#calculate signals into final output layer
final_inputs=numpy.dot(self.who,hidden_outputs)
#calculate the signals emerging from final output layer
final_outputs=self.activation_function(final_inputs)
return final_outputs
#number of input,hidden and output nodes
input_nodes=784
hidden_nodes=200
output_nodes=10
#learning-rate
learning_rate=0.3
#create instance of neural network
n=neuralNetwork(input_nodes, hidden_nodes, output_nodes, learning_rate)
#load the mnist training date CSV file into a list
training_data_file=open("c:/Users/we/Desktop/we/mnist_dataset/mnist_train_100.csv","r")#注意路径
training_data_file.close()
#epochs is the number of times the training data set is used for training
epochs=5
for e in range(epochs):
# go through all records in the training data set
for record in training_data_list:
all_values=record.split(',')
#scale and shift the inputs
inputs=(numpy.asfarray(all_values[1:])/255.0*0.99)+0.01
targets=numpy.zeros(output_nodes)+0.01
targets[int(all_values[0])]=0.99
n.train(inputs,targets)
pass
pass
test_data_file=open("c:/Users/we/Desktop/we/mnist_dataset/mnist_test_10.csv",'r')#最好将数据集放在该程序同一路径下
test_data_file.close()
#test the neural network
#scorecard for how well the network performs,initially empty
scorecard=[]
#go through all the records in the test data set
for record in test_data_list:
#spilt the records by the ',' commas
all_values=record.split(',')
#correct answer is the first value
correct_label=int(all_values[0])
#scale and shift the inputs
inputs=(numpy.asfarray(all_values[1:])/255.0*0.99)+0.01
#query the network
outputs=n.query(inputs)
#the index of the highest value corresponds to the label
label=numpy.argmax(outputs)
#append correct or incorrect to list
if(label==correct_label):
scorecard.append(1)
else:
scorecard.append(0)
pass
pass
#calculate the performance score,the fraction of correct answers
scorecard_array=numpy.asarray(scorecard)
print('performance=',scorecard_array.sum()/scorecard_array.size)
展开全文
• 反向传播 目标： 快速理解反向传播并推导 分析： 2层神经网络如图1所示，包含一个隐层和输出层，手推反向传播，取一个神经元的传播过程进行推导。 动手之前： 假定某个样本在网络中的一个传播过程如图2所示，隐层...
反向传播
**参考连接：**https://blog.csdn.net/qq_37644877/article/details/105431392?
目标：
快速理解反向传播并推导
分析：
2层神经网络如图1所示，包含一个隐层和输出层，手推反向传播，取一个神经元的传播过程进行推导。
动手之前：
假定某个样本在网络中的一个传播过程如图2所示，隐层激活函数使用sigmoid，输出层不使用非线性激活函数，优化目标是均方误差mse，即损失为：
展开全文
• 论文：Identity Mappings in Deep Residual ...但是反向传播是为了对权重和偏置进行更新，论文中只是分析了对特征x的梯度。 自己对权重的梯度做了分析。 以一个小的示例表达： 其中分为两类，一是对输出层...
论文：Identity Mappings in Deep Residual Networkshttps://arxiv.org/pdf/1603.05027.pdf
其中对resnet反向传播进行了分析，给出以下公式：
但是反向传播是为了对权重和偏置进行更新，论文中只是分析了对特征x的梯度。
自己对权重的梯度做了分析。
以一个小的示例表达：
其中分为两类，一是对输出层的更新，直接进行求解。
另一个是对中间隐含层的梯度求解，利用论文中给出的公式能够很好的写出。
展开全文
• 前向运算、反向运算 } loss /= param_.iter_size(); // average the loss across iterations for smoothed reporting UpdateSmoothedLoss(loss, start_iter, average_loss); if (display) { LOG_IF(INFO...
还是以mnist手写字符，lenet5为例。
如上图所示，左边为9个layer层，右边为每层的top blob 的输出（featrue map）的维度。
那么问题来了，训练的参数wij和bi<!--//--><![CDATA[//><!--
w_{ij}和b_i
//--><!]]>存在哪些变量里呢?
1）有需要用BP算法进行训练参数的层(layer)，内部都会有一个成员变量layer::blobs_,其中blobs_[0]存放wij<!--//--><![CDATA[//><!--
w_{ij}
//--><!]]>及Δwij<!--//--><![CDATA[//><!--
\Delta w_{ij}
//--><!]]>，blobs_[1]存放bi及Δbi<!--//--><![CDATA[//><!--
b_i及\Delta b_i
//--><!]]>.
2) 网络层内部成员变量 Net::blobs_内部存放数据及δ<!--//--><![CDATA[//><!--
\delta
//--><!]]>值。
在lenet5网络中，只有两个卷积层（conv1,conv2）和两个全链接层（ip1,ip2）内部有需要训练的参数。因此lenet5网络需要8个blobs_存放训练参数，Net内部变量 net::learnable_params_与此8个blobs_共享内存。且，各层的layer::blobs_变量在layer::setup()中分配内存和初始化，在layer::Backward_cpu()中计算 Δwij和Δbi<!--//--><![CDATA[//><!--
\Delta w_{ij}和\Delta b_i
//--><!]]> ，后续准备编写自己网络的童鞋需要重写此函数。最后调用Net::update()函数，更新wij<!--//--><![CDATA[//><!--
w_{ij}
//--><!]]>。
以全链接层为例：
///ip层 blobs_ 开辟内存及初始化
template <typename Dtype>
void InnerProductLayer<Dtype>::LayerSetUp(const vector<Blob<Dtype>*>& bottom,
const vector<Blob<Dtype>*>& top) {
const int num_output = this->layer_param_.inner_product_param().num_output();
bias_term_ = this->layer_param_.inner_product_param().bias_term();
transpose_ = this->layer_param_.inner_product_param().transpose();
N_ = num_output;
const int axis = bottom[0]->CanonicalAxisIndex(
this->layer_param_.inner_product_param().axis());
// Dimensions starting from "axis" are "flattened" into a single
// length K_ vector. For example, if bottom[0]'s shape is (N, C, H, W),
// and axis == 1, N inner products with dimension CHW are performed.
K_ = bottom[0]->count(axis);
// Check if we need to set up the weights
if (this->blobs_.size() > 0) {
LOG(INFO) << "Skipping parameter initialization";
} else {
if (bias_term_) {
this->blobs_.resize(2);
} else {
this->blobs_.resize(1);
}
// Initialize the weights
vector<int> weight_shape(2);
if (transpose_) {
weight_shape[0] = K_;
weight_shape[1] = N_;
} else {
weight_shape[0] = N_;
weight_shape[1] = K_;
}
this->blobs_[0].reset(new Blob<Dtype>(weight_shape));
// fill the weights
shared_ptr<Filler<Dtype> > weight_filler(GetFiller<Dtype>( 权重w初始化
this->layer_param_.inner_product_param().weight_filler()));
weight_filler->Fill(this->blobs_[0].get());
// If necessary, intiialize and fill the bias term
if (bias_term_) {
vector<int> bias_shape(1, N_);
this->blobs_[1].reset(new Blob<Dtype>(bias_shape)); //偏置初始化
shared_ptr<Filler<Dtype> > bias_filler(GetFiller<Dtype>(
this->layer_param_.inner_product_param().bias_filler()));
bias_filler->Fill(this->blobs_[1].get());
}
} // parameter initialization
this->param_propagate_down_.resize(this->blobs_.size(), true);
}
权重变化量计算：
template <typename Dtype>
void InnerProductLayer<Dtype>::Backward_cpu(const vector<Blob<Dtype>*>& top,
const vector<bool>& propagate_down,
const vector<Blob<Dtype>*>& bottom) {
if (this->param_propagate_down_[0]) {
const Dtype* top_diff = top[0]->cpu_diff();
const Dtype* bottom_data = bottom[0]->cpu_data();
// Gradient with respect to weight
if (transpose_) {
caffe_cpu_gemm<Dtype>(CblasTrans, CblasNoTrans,
K_, N_, M_,
(Dtype)1., bottom_data, top_diff,
(Dtype)1., this->blobs_[0]->mutable_cpu_diff());
} else {
caffe_cpu_gemm<Dtype>(CblasTrans, CblasNoTrans,
N_, K_, M_,
(Dtype)1., top_diff, bottom_data,
(Dtype)1., this->blobs_[0]->mutable_cpu_diff()); //计算本层参数更新量 △W
}
}
if (bias_term_ && this->param_propagate_down_[1]) {
const Dtype* top_diff = top[0]->cpu_diff();
// Gradient with respect to bias
caffe_cpu_gemv<Dtype>(CblasTrans, M_, N_, (Dtype)1., top_diff,
bias_multiplier_.cpu_data(), (Dtype)1.,
this->blobs_[1]->mutable_cpu_diff()); //计算本层参数更新量 △b
}
if (propagate_down[0]) {
const Dtype* top_diff = top[0]->cpu_diff();
// Gradient with respect to bottom data
if (transpose_) {
caffe_cpu_gemm<Dtype>(CblasNoTrans, CblasTrans,
M_, K_, N_,
(Dtype)1., top_diff, this->blobs_[0]->cpu_data(),
(Dtype)0., bottom[0]->mutable_cpu_diff());
} else {
caffe_cpu_gemm<Dtype>(CblasNoTrans, CblasNoTrans, /// 计算下一层的deta
M_, K_, N_,
(Dtype)1., top_diff, this->blobs_[0]->cpu_data(),
(Dtype)0., bottom[0]->mutable_cpu_diff());
}
}
}
总的训练流程在Solver::Step（）函数实现中：
template <typename Dtype>
void Solver<Dtype>::Step(int iters) {
const int start_iter = iter_;
const int stop_iter = iter_ + iters;
int average_loss = this->param_.average_loss();
losses_.clear();
smoothed_loss_ = 0;
while (iter_ < stop_iter) { //以一个batch为一个周期
// zero-init the params
net_->ClearParamDiffs(); /////将Net 的成员变量param_ 中的diff空间初始化为0
if (param_.test_interval() && iter_ % param_.test_interval() == 0
&& (iter_ > 0 || param_.test_initialization())
&& Caffe::root_solver()) {
TestAll();
if (requested_early_exit_) {
// Break out of the while loop because stop was requested while testing.
break;
}
}
for (int i = 0; i < callbacks_.size(); ++i) {
callbacks_[i]->on_start();
}
const bool display = param_.display() && iter_ % param_.display() == 0;
net_->set_debug_info(display && param_.debug_info());
// accumulate the loss and gradient
Dtype loss = 0;
for (int i = 0; i < param_.iter_size(); ++i) {
loss += net_->ForwardBackward(); /////关键。 前向运算、反向运算
}
loss /= param_.iter_size();
// average the loss across iterations for smoothed reporting
if (display) {
LOG_IF(INFO, Caffe::root_solver()) << "Iteration " << iter_ //////////////// Iteration, loss=
<< ", loss = " << smoothed_loss_;
const vector<Blob<Dtype>*>& result = net_->output_blobs();
int score_index = 0;
for (int j = 0; j < result.size(); ++j) {
const Dtype* result_vec = result[j]->cpu_data();
const string& output_name =
net_->blob_names()[net_->output_blob_indices()[j]];
const Dtype loss_weight =
net_->blob_loss_weights()[net_->output_blob_indices()[j]];
for (int k = 0; k < result[j]->count(); ++k) { //////Train net out///////////////
ostringstream loss_msg_stream;
if (loss_weight) {
loss_msg_stream << " (* " << loss_weight
<< " = " << loss_weight * result_vec[k] << " loss)";
}
LOG_IF(INFO, Caffe::root_solver()) << " Train net output #"
<< score_index++ << ": " << output_name << " = "
<< result_vec[k] << loss_msg_stream.str();
}
}
}//if(display)
for (int i = 0; i < callbacks_.size(); ++i) {
}
ApplyUpdate(); /////////////// 反向运算完成，统一更新权重值w
// Increment the internal iter_ counter -- its value should always indicate
// the number of times the weights have been updated.
++iter_;
SolverAction::Enum request = GetRequestedAction();
// Save a snapshot if needed.
if ((param_.snapshot()
&& iter_ % param_.snapshot() == 0
&& Caffe::root_solver()) ||
(request == SolverAction::SNAPSHOT)) {
Snapshot();
}
if (SolverAction::STOP == request) {
requested_early_exit_ = true;
// Break out of training loop.
break;
}// if()
}// end while
} // end step()
展开全文
• 误差反向传播（Error Back Propagation, BP）算法 1、BP算法的基本思想是：学习过程由信号的正向传播与误差的反向传播两个过程组成。 1）正向传播：输入样本－>输入层－>隐藏层（处理）－>输出层 举一个...
• 一、BP神经网络的概念 BP神经网络是一种多层的前馈神经网络，其主要的特点是：信号是前向传播的，而误差是...第二阶段是误差的反向传播，从输出层到隐含层，最后到输入层，依次调节隐含层到输出层的权重和偏置，输
• 之前我们了解了如何用梯度下降来更新权重反向传播算法是它的一个延伸，用链式法则来找到误差与输入层到输入层链接的权重（两层神经网络）。 要更新输入到隐藏层的权重，你需要知道隐藏层节点的误差对最终输出的...
• 13.3.1 反向传播算法推导如下图所示为一个神经网络的结构图，由于本文主要探讨激活函数在反向传播过程中的作用，因此不会带入数值进行计算，而是以两个权重更新为案例进行公式的推导，分别为如何通过反向传播算法...
• 写在前面：关于权重和偏置更新权重W更新采用+或-的两种情况可以很明显看到，关于权重更新存在2种情况，本质没什么区别，只不过在自己写代码手动更新权重和偏置时，这点就很容易混淆出错。本帖第一个单层感知机示例，...
• 神经网络靠反向传播更新权重W才能使得我的模型越来越逼近我想要的模型， 而这个过程，因为梯度下降是函数值下降最快的方向，因此， （ML常用的除了梯度下降还有牛顿方法，核方法，求逆，MS算法(胶囊网络号称没用...
• 因为反向传播是要求最后的损失对前面所有的权重的导数，然后再更新权重，所以我们的关键在于求出损失的权重的求导，上面的图中最后的输出是，所以我们对应的损失如下： 我们将。我们的目的是要求，我们可以通过链式...
• ## 神经网络的传播（权重更新）
万次阅读 多人点赞 2018-08-03 20:23:36
本文重在阐述神经网络里的前向传播和反向传播。 上一篇文章构建了一个简单的网络，可以看出来它对于手写数字的识别率还是可以高达91%。但我们尚未对神经网络处理的过程做出太多解释。 数据在网络中的传播有两种...
• 目录概述正向传播误差的反向传播权重更新梯度消失与梯度爆炸 概述 反向传播算法使用链式求导法则将输出层的误差反向传回给网络，使网络中的权重有了较简单的地图计算方法。像TensorFlow，Pytorch有现成的反向传播...
• 反向传播的作用 反向传播的公式推导 误差反向传播 输出层的权重参数更新 隐藏层的权重参数更新 输出层和隐藏层的偏置参数更新 BP算法的四个核心公式 BP反向传播算法流程 反向传播解释梯度消失的原因 反向...
• 图书推荐：《数据准备和特征工程》反向传播算法是神经网络中的重要算法，通过它能够快速计算梯度，进而通过梯度下降实现权重和偏置参数的更新反向传播算法最初是在20世纪70年代被引入的，但直到1986年大卫·鲁梅尔...
• 反向传播用于更新每层之间的权重，减少损失，进而提升预测准确度。 下面是一个神经网络的结构图： 第一层是输入层，包含两个神经元i1，i2，和截距项b1；第二层是隐含层，包含两个神经元h1,h2和截距项b2，第三层是...
• 2.反向传播：在反向传播中，首先计算输出层神经元损失函数的梯度，再计算隐藏层神经元损失函数的梯度，再根据梯度来更新权重。 重复这两个步骤，直至收敛。 代码实现： #反向传播算法 import tensorflow as tf from ...
• 前向传递输入信号直至输出产生误差，反向传播误差信息更新权重矩阵。 为什么提出反向传播？因为神经网络中有隐藏层，隐层的误差是不存在的，因此不能直接使用梯度下降算法，而需要先将误差反向传播到隐层，再使用...
• 我们已了解了使用梯度下降来更新权重反向传播算法则是它的一个延伸。以一个两层神经网络为例，可以使用链式法则计算输入层-隐藏层间权重的误差。 要使用梯度下降法更新隐藏层的权重，你需要知道各隐藏层节点的...
• 反向传播（BPN）算法是神经网络中研究最多、使用最多的算法之一，用于将输出层中的误差传播到隐藏层的神经元，然后用于更新权重。 学习 BPN 算法可以分成以下两个过程： 正向传播：输入被馈送到网络，信号从输入层...
• 反向传播（BPN）算法是神经网络中研究最多、使用最多的算法之一，它用于将输出层中的误差传播到隐藏层的神经元，然后用于更新权重。 学习 BPN 算法可以分成以下两个过程： 正向传播：输入被馈送到网络，信号从输入层...
• 如下图所示为一个神经网络的结构图，由于本文主要探讨激活函数在反向传播过程中的作用，因此不会带入数值进行计算，而是以两个权重更新为案例进行公式的推导，分别为如何通过反向传播算法更新w112w^2_{11}w112和...
...
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Pea gravel is made up of small, pea-sized stones. People use pea gravel to line walkways and cover gardens. Stores sell pea gravel by weight. Most likely, you'll know the volume of pea gravel you need, but not the weight. Apply a little math, and you can quickly find out how much pea gravel will suit your needs.
Measure the dimensions of the area where you want your pea gravel. This includes measuring the length of the longest side, the length of the shortest side, and the depth.
• Pea gravel is made up of small, pea-sized stones.
• Measure the dimensions of the area where you want your pea gravel.
Convert all measurements into feet by dividing by 12. For example, if you have a depth of 7 inches, dividing by 12 would give you 0.583333 feet (or about 0.6 feet rounded).
• Convert all measurements into feet by dividing by 12.
• Multiply your cubic feet by 96 to find the total number of pounds you'll need.
• (
Multiply your three dimensions together (longest side, shortest side, and depth) to find the volume in cubic feet.
• Convert all measurements into feet by dividing by 12.
• Multiply your cubic feet by 96 to find the total number of pounds you'll need.
• (
Multiply your cubic feet by 96 to find the total number of pounds you'll need. (A single cubic foot of pea gravel weighs about 43.5 Kilogram.)
• Convert all measurements into feet by dividing by 12.
• Multiply your cubic feet by 96 to find the total number of pounds you'll need.
• (
If you are covering a very large area with pea gravel, you'll need to convert to cubic yards. Simply divide your cubic feet by 27.
Multiply your cubic yards by 1.3 to find the total number of tons you'll need. (A single cubic yard of pea gravel weighs about 1.3 short tons.)
• Convert all measurements into feet by dividing by 12.
• Multiply your cubic feet by 96 to find the total number of pounds you'll need.
• (
#### TIP
Typically, you can round numbers to one decimal point. Look at the number just to the right of where you want to round. If it is 5 or higher, round your digit up. If it is 4 or lower, leave the digit alone.
#### WARNING
Gravel can settle. This means that you may have to buy 5 to 10 per cent more gravel than you anticipated.
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MATHEMATICA BOHEMICA, Vol. 121, No. 1, pp. 9-24, 1996
# Exact solutions of Cauchy problem for partial differential equations with double characteristics and singular coefficients
## Zhu-Jia Lu
Zhu-jia Lu, Institute of Mathematics, Academia Sinica, Beijing 100080, P.R. China
Abstract: Let $L_{a,b}\equiv(\partial_x-ax^k\partial_t)(\partial_x-bx^k\partial_t)+kbx^{k-1}\partial_t-\dfrac kx\partial_x$ be a family of operators with double characteristics and singular coefficients, where $a$, $b$ are reals with $ab\ne0$ and $a\ne b$, $k>0$ is an odd integer. Let $\Omega$ be the first quadrant in the plane and $H_+$ the upper half-plane. Consider Cauchy problems
\cases L_{a,b}u=0 &\text{ in }\Omega \text{ or }H_+,
u(x,0)=\varphi_0(x), u_t(x,0)=\varphi_1(x)\quad &\text{ for }x\in\overline{\Bbb R_+}\text{ or }x\in{\Bbb R}\endcases \tag"$(P_1)$"
for $a>0$, $b>0$, and initial-boundary value problems
\cases L_{a,b}u=0 &\text{ in }\Omega\text{ or }H_+,
u(x, 0)=\varphi_0(x), u_t(x,0)=\varphi_1(x)\quad &\text{ for }x\in\overline{\Bbb R_+}\text{ or }x\in{\Bbb R},
u(0,t)=\psi_0(t) &\text{ for }t\in\overline{\Bbb R_+},\endcases \tag"$(P_2)$"
\cases L_{a,b}u=0 &\text{ in }\Omega\text{ or }H_+,
u(x,0)=\varphi_0(x), u_t(x,0)=\varphi_1(x)\quad &\text{ for }x\in\overline{\Bbb R_+}\text{ or }x\in\Bbb R,
\lim\limits\Sb (x,\tau)\to(0,t),x\ne0
(x,\tau)\in\Omega\text{ or }H_+\endSb \dfrac{u_x(x,\tau)}{x^k}=\psi_1(t)&\text{ for }t\in\overline{\Bbb R_+}\endcases \tag"$(P_3)$"
for $ab<0$ and
\cases L_{a,b}u=0 &\text{ in }\Omega\text{ or }H_+,
u(x,0)=\varphi_0(x), u_t(x,0)=\varphi_1(x)\quad &\text{ for }x\in\overline{\Bbb R_+}\text{ or }x\in\Bbb R,
u(0,t)=\psi_0(t), \lim\limits\Sb(x,\tau)\to(0,t),x\ne0
(x,\tau)\in\Omega\text{ or }H_+\endSb \dfrac{u_x(x,\tau)}{x^k} \!\!\! &=\psi_1(t)\quad \text{ for }t\in\overline{\Bbb R_+}\endcases \tag"$(P_4)$"
for $a<0$, $b<0$. Under appropriate smoothness conditions on $\varphi_0$, $\varphi_1$, $\psi_0$ and $\psi_1$, we obtain different sufficient and necessary conditions for each problem to have classical solutions. Moreover, we obtain also explicit expressions of solutions in each case.
Keywords: exact solutions, Cauchy problem, double characteristics, singular coefficients
Classification (MSC91): 35C15, 35L99
Full text of the article:
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# Showing the Results of a Survey
So you have just Conducted a Survey and want to show
your results in the best possible way?
Here are some suggestions:
### Tables
Sometimes, you can simply report the information in a table.
A table is a very simple way to show others the results. A table should have a title, so those looking at it understand what it shows:
Yellow Red Blue Green Pink Table: The Favorite Colors of My Class 4 5 6 1 4
### Statistics
You can also summarize the results using statistics, such as Mean, Median, Mode, Standard Deviation and Quartiles
Example: you have lots of information about how long it takes people to get to school but it may be simpler just to present a summary such as:
Shortest Journey: 3 minutes
Average Journey: 22 minutes
Longest Journey: 58 minutes
### Graphs
But nothing makes a report look better than a nice graph or chart
There are many different types of graphs. Three of the most common are:
Line Graph - shows information that is somehow connected (such as change over time)
Bar Graph – shows relative sizes of different results:
Pie Chart - shows sizes as part of a whole (good for showing percentages).
You can create graphs like those using our Data Graphs (Bar, Line and Pie) page
If people have given their opinions or comments in the survey, you can present the more interesting ones:
Example: In response to the question "How can we best clean up the river?" we received these interesting replies:
• "The government has a special fund for this"
• "The local gardening group has seedlings you could plant"
## Report
Put it all together into a report, with a nice introduction, and conclusions at the end, and you are done!
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1. Senior Member Community Builder
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Hmm...Looking at that problem, if "trapezium" means trapezoid, which I'm assuming it does...well, it's the average of 7 and 10 times the perpindicular distance between EF and CB. However...in order to get that distance, I believe you need to use a right triangle...but you're missing some important measurements. You have to find EB, and then square it and subtract 1.5 squared, then find the square root of that...that's your perpindicular distance. It's finding EB that's the problem. If you just had the height of the figure...ech...
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yes but... wouldnt EB be 5cm? since AB is 5cm, and the sides of the roof are equalatiral triangles...
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Since AB = EB, I thought you could simply use Pythagoras with the height of the trapezium being the square root of (5^2 - 1,5^2) but then you get the no-good answer of 4,77 cm
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Ah. I missed the equilateral bit.
Anyways, if EB=5, then just make a perpindicular segement from E to a point we can call X. Triangle XEB is right. EB=5, BX=1.5. 25-2.25=22.75. So it's 8.5 times the square root of 22.75. Do that with a calculator and you have the answer.
That has to be correct, given the numbers we were given...If the triangles are equilateral, the BX has to equal 1.5...and since EB=5, we know root 22.75 is correct. So the first answer is the square root of 22.75, the second answer is root 22.75 times 8.5. Root 22.75 is about 4.77, so it's 4.77 times 8.5...that makes it 40.545...That's the answer.
5. My mind goes blank if I'm just reading math and I'm too lazy to write it out on paper.
So... err...
*nods head and pretends CI & Saizou & PST are correct*
Originally Posted by Saizou
If you'd like I could post the BASTARD QUESTION FROM HELL that was on last year's finals. I think that three people of nearly fifty got it right.
Sure! Post it so I can feel stupid. =D
6. Senior Member Always Around
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Wow a whole year of sleeping in math class which I used to be one of the best at can really affect you later on. And I mean really.
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Originally Posted by coolerimmortal
Ah. I missed the equilateral bit.
Anyways, if EB=5, then just make a perpindicular segement from E to a point we can call X. Triangle XEB is right. EB=5, BX=1.5. 25-2.25=22.75. So it's 8.5 times the square root of 22.75. Do that with a calculator and you have the answer.
That has to be correct, given the numbers we were given...If the triangles are equilateral, the BX has to equal 1.5...and since EB=5, we know root 22.75 is correct. So the first answer is the square root of 22.75, the second answer is root 22.75 times 8.5. Root 22.75 is about 4.77, so it's 4.77 times 8.5...that makes it 40.545...That's the answer.
wait... you've lost me. where's X? and just to be sure... is the perpendicular distance the distance between EF and BC, aka the slope of the roof?
oh wait sack that i totally get it now- thanks so much!! ^^ does anyone get b) ? ... the size of the angle EMN? at first i thought it was 90 degrees, but then i thought no, cos the roof's rapezium is a trapezium... meaning that the angle would be acute...
i've gotten so far find out the perpendicular distance of the equilateral triangles so i know EM, and MN is 10cm... and i dunno where to go from there. i dont think i can use the cosine or sine rule...
and also...
six numbers are: 17, 23, 28, 25, a, b
where a and b are two positive integers between 14 and 25.
the 6 numbers have a mean of 22 and a standard deviation of 4.
find the values of a and b
you are advised to use the statistical functions on your calculator, and trial and improvement
Last edited by Quiraikotsu; 05-31-2006 at 03:39 AM.
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I've created a monster....
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A and B...They add to make 39. It can be:
15, 24
16, 23
17, 22
18, 21
19, 20
I forgot how standard deviation works; you'll just have to figure out which one of those is correct.
For the angle measure...Inverse cosine(1.5root3/9). Plug that into a calculator, and you get...wait for it...30.
The angle is 30 degrees.
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# GATE2017-63
The last digit of $\left ( 2171 \right )^{7}+\left ( 2172 \right )^{9}+\left ( 2173 \right )^{11}+\left ( 2174 \right )^{13}$ is
1. $2$
2. $4$
3. $6$
4. $8$
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go top 返回词典
• It has been called "the beautiful game."
VOA: special.2010.06.09
• The week after we implemented the so-called game of 15, this little party favor where you move the numbers up, down, left or right.
到第三周，我们要编一个叫做“15”的游戏，就是经常在派对上玩的,可以把数字上，下，左，右移动的小游戏。
哈佛公开课 - 计算机科学课程节选
• So let me go ahead and pull up the staff solution 15 to the standard edition for a moment, the game is called 15, it takes one command line argument which is the dimensions of the board 3 by 3, 4 by 4, I'll do it 4 by 4.
下面我们继续,等一会儿，我会给出标准的解答，这个游戏叫做，有一个命令行参数，标识其大小是,3阶还是4阶，我选择4阶。
哈佛公开课 - 计算机科学课程节选
• The game is called "Pamoja Mtaani,"
VOA: special.2009.03.16
• That's good. So this game's called The Battle of the Sexes and we'll see it in various forms over the course of the semester.
没错,就叫性别大战,我们会在这学期的课程中陆续接触到的
耶鲁公开课 - 博弈论课程节选
• After graduating from university in a computer gaming program, he founded an educational video game company called Let Me Think Games.
VOA: standard.2010.04.20
• Again, still in the interest of recapping, this particular game is called the Prisoners' Dilemma.
其次,好心提醒一下大家,这个博弈被称为囚徒困境
耶鲁公开课 - 博弈论课程节选
• "Perhaps it is time for the U.S. to talk directly to the Israeli public and make it clear that they are serious and this old game that we've been playing for many years of this so-called peace process that does not lead to anything is over,".
VOA: standard.2009.06.09
这个博弈学名叫什么,大声说出来
耶鲁公开课 - 博弈论课程节选
• This kind of game is called a "Coordination problem."
这个博弈叫"协和谬误"
耶鲁公开课 - 博弈论课程节选
\$firstVoiceSent
- 来自原声例句
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Community Profile
# James Seaward
141 total contributions since 2018
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Algebra Tutorials!
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Iinsdiim
Registered: 18.10.2003
From: Finland
Posted: Thursday 28th of Dec 07:56 I have a test tomorrow afternoon. But I’m stuck with problems based on 7th grade algebra formulae. I’m having problems understanding evaluating formulas and graphing parabolas because I just can’t seem to figure out a way to crack problems based on them. I called my friends and I tried on the internet, but none of those activities did any good . I’m still trying but the time is running out and I can’t seem to get things moving . Can somebody please guide me ? I really need some help from you guys for tomorrows test . Please do reply.
Jahm Xjardx
Registered: 07.08.2005
From: Odense, Denmark, EU
daujk_vv7
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Posted: Saturday 30th of Dec 10:55 Algebrator did help my son to have high grades in Math . Good thing we found this great software because I believe it did not only help him to have high grades in his homeworks but also helped him in his exams since the program helped in explaining the process of answering the problem by showing the solution.
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Posted: Sunday 31st of Dec 17:06 I suggest trying out Algebrator. It not only assists you with your math problems, but also provides all the required steps in detail so that you can improve the understanding of the subject.
erx
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Posted: Monday 01st of Jan 09:59 https://algebra-calculator.com/powers.html. There you go. Hopefully you will not have to drop math.
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Basic Motion-problems.pdf - oconnor(two257 Basic Motion pennisi(3690216 003(part 1 of 6 10.0 points Consider the following graph of motion 50 Distance(m
# Basic Motion-problems.pdf - oconnor(two257 Basic Motion...
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o’connor (two257) – Basic Motion – pennisi – (3690216) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 pointsConsider a moving object whose positionxis plotted as a function of the timet.Theobject moved in different ways during thetime intervals denoted I, II and III on thefigure.246246txIIIIIIDuring these three intervals, when was theobject’sspeedhighest?Do not confuse thespeed with the velocity.003 (part 1 of 6) 10.0 pointsConsider the following graph of motion.01234501020304050Time (sec) 1. Same speed during intervals II and III 2. Same speed during each of the three in- tervals. 3. During interval II 4. During interval III 5. During interval I 002 (part 2 of 2) 10.0 points During which interval(s) did the object’s ve- locity remain constant? 1. During none of the three intervals 2. During interval III only 3. During interval I only 4. During interval II only 5. During each of the three intervals
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# The symmetric functions catalog
An overview of symmetric functions and related topics
2023-09-20
## Monomial slide polynomials
The monomial slide polynomials were introduced by S. Assaf and D. Searles in [AS17b]. The slide polynomials form a basis for the space of polynomials, and can be seen as a lift of the monomial quasisymmetric functions.
Definition.
For $\alpha$ being a weak composition, we set
$\slideM_\alpha(\xvec) \coloneqq \sum_{\substack{b \trianglerighteq a \\ \mathrm{flat}(b)=\mathrm{flat}(a)}} \xvec^b$
where $\mathrm{flat}(\beta)$ denotes the composition obtained by removing all 0s from $\beta,$ and $\trianglerighteq$ denotes dominance order.
As an example (from [Eq. 3.7, AS17b]), $\slideM_{(0,2,0,3)}(\xvec) = x_1^2x_2^3 + x_1^2x_3^3 + x_1^2 x_4^3 + x_2^2 x_3^3 + x_2^2 x_4^3.$
The monomial slide polynomials have the monomial quasisymmetric functions as stable limit:
$\lim_{m \to \infty} \slideM_{0^m\times \alpha}(\xvec) = \qmonom_{\mathrm{flat}(\alpha)}(\xvec).$
## Fundamental slide polynomials
The fundamental slide polynomials (or sometimes just slide polynomials) were introduced by S. Assaf and D. Searles in [AS17b]. The slide polynomials form a basis for the space of polynomials, and can be seen as a lift of the gessel quasisymmetric functions.
The K-theoretical analog of fundamental slide polynomials are the \hyperref[glide]{glide polynomials}, see [PS17].
The fundamental slide polynomials expand positively in the monomial slide polynomials. Moreover, the Schubert polynomials expand positively in the fundamental slide polynomials, [thm. 3.13, AS17b]. The main motivation for introducing the fundamental slide polynomials, is that products of Schubert polynomials can be expanded (with a combinatorial formula) into fundamental slide polynomials.
Definition.
For $\alpha$ being a weak composition, fundamental slide polynomial $\slideF_\alpha$ is defined as
$\slideF_\alpha(\xvec) \coloneqq \sum_{\substack{b \trianglerighteq a \\ \mathrm{flat}(b)\text{ refines }\mathrm{flat}(a)}} \xvec^b$
where $\mathrm{flat}(\beta)$ denotes the composition obtained by removing all 0s from $\beta,$ and $\trianglerighteq$ denotes dominance order.
The fundamental slide polynomials have the gessel quasisymmetric functions as stable limit:
$\lim_{m \to \infty} \slideF_{0^m\times \alpha}(\xvec) = \gessel_{\mathrm{flat}(\alpha)}(\xvec).$
### Slide positive families
Key polynomials expand positively in the fundamental slide basis, [Thm. 2.13, AS18c]. In [CW22], the authors determine for which $\alpha,$ the key polynomials $\key_\alpha$ expanded into slide polynomials are multiplicity free.
In [AB19], the authors consider a flagged version of $(P,w)$-partitions, and show that these are slide-positive. In [TWZ22], it is shown that certain polynomials similar to chromatic symmetric functions are slide-positive.
See [ST21a] the notion of slide complexes.
## Lock polynomials
Lock polynomials were introduced by S. Assaf and D. Searles in [AS22b]. The Lock polynomials form a basis for the polynomial ring, and are indexed by weak compositions. The combinatorial formula for these is
$\lock_\alpha(\xvec) \coloneqq \sum_{T \in LT(\alpha)} \xvec^T$
where the sum is over all lock tableaux.
As an example (from [Wan20a]), we have
$\lock_{(0,2,3)}(\xvec) = x_2^2 x_3^2 + x_1x_2x_3^3 + x_1^2x_3^3 + x_1x_2^2x_3^2 + x_1^2x_2x_3^2 + x_1^2x_2^2x_3 + x_1^2 x_2^3$
Whenever the non-zero parts of $\alpha$ are weakly decreasing, we have that the lock polynomial coincides with a key polynomial; $\lock_\alpha(\xvec)=\key_\alpha(\xvec),$ see [Thm. 6.12, AS22b].
There is a crystal structure on lock polynomials, explored by G. Wang in [Wan20a]. This crystal structure embeds naturally into Demazure crystals.
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Term Rewriting System R:
[X, Y, X1, X2]
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
length(nnil) -> 0
length(ncons(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
nil -> nnil
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nnil) -> nil
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(X) -> X
Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
FROM(X) -> CONS(X, nfrom(s(X)))
LENGTH(ncons(X, Y)) -> LENGTH1(activate(Y))
LENGTH(ncons(X, Y)) -> ACTIVATE(Y)
LENGTH1(X) -> LENGTH(activate(X))
LENGTH1(X) -> ACTIVATE(X)
ACTIVATE(nfrom(X)) -> FROM(X)
ACTIVATE(nnil) -> NIL
ACTIVATE(ncons(X1, X2)) -> CONS(X1, X2)
Furthermore, R contains one SCC.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Non Termination`
Dependency Pairs:
LENGTH1(X) -> LENGTH(activate(X))
LENGTH(ncons(X, Y)) -> LENGTH1(activate(Y))
Rules:
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
length(nnil) -> 0
length(ncons(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
nil -> nnil
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nnil) -> nil
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(X) -> X
Found an infinite P-chain over R:
P =
LENGTH1(X) -> LENGTH(activate(X))
LENGTH(ncons(X, Y)) -> LENGTH1(activate(Y))
R =
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
length(nnil) -> 0
length(ncons(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
nil -> nnil
cons(X1, X2) -> ncons(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nnil) -> nil
activate(ncons(X1, X2)) -> cons(X1, X2)
activate(X) -> X
s = LENGTH(activate(activate(nfrom(X'''''))))
evaluates to t =LENGTH(activate(activate(nfrom(s(X''''')))))
Thus, s starts an infinite chain as s matches t.
Non-Termination of R could be shown.
Duration:
0:02 minutes
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Toán Lớp 9: Tính giá trị biểu thức $A$ $A$=$\dfrac{\sqrt{9}+1}{\sqrt{9}-1}$ $+$ $\dfrac{\sqrt{9}-1}{\sqrt{9}+1}$ $-$ $\dfrac{3\sqrt{9}+1}{9- Toán Lớp 9: Tính giá trị biểu thức$AA$=$\dfrac{\sqrt{9}+1}{\sqrt{9}-1}+\dfrac{\sqrt{9}-1}{\sqrt{9}+1}-\dfrac{3\sqrt{9}+1}{9-1}$TRẢ LỜI 1. bichhhathu Giải đáp+Lời giải và giải thích chi tiết: Cách 1: A=\frac{\sqrt{9}+1}{\sqrt{9}-1}+\frac{\sqrt{9}-1}{\sqrt{9}+1}-\frac{3\sqrt{9}+1}{9-1} A=\frac{\sqrt{9}+1}{\sqrt{9}-1}+\frac{\sqrt{9}-1}{\sqrt{9}+1}-\frac{3\sqrt{9}+1}{(\sqrt{9}-1)(\sqrt{9}+1)} A=\frac{(\sqrt{9}+1)^2+(\sqrt{9}-1)^2-3\sqrt{9}-1}{(\sqrt{9}-1)(\sqrt{9}+1)} A=\frac{9+2\sqrt{9}+1+9-2\sqrt{9}+1-3\sqrt{9}-1}{(\sqrt{9}-1)(\sqrt{9}+1)} A=\frac{18-3\sqrt{9}+1}{9-1} A=\frac{18-3.3+1}{8} A=\frac{10}{8} A=\frac{5}{4} Cách 2: A=\frac{\sqrt{9}+1}{\sqrt{9}-1}+\frac{\sqrt{9}-1}{\sqrt{9}+1}-\frac{3\sqrt{9}+1}{9-1} A=\frac{3+1}{3-1}+\frac{3-1}{3+1}-\frac{3.3+1}{8} A=\frac{4}{2}+\frac{2}{4}-\frac{10}{8} A=2+\frac{1}{2}-\frac{10}{8} A=\frac{2.8+4-10}{8} A=\frac{20-10}{8} A=\frac{10}{8} A=\frac{5}{4} Trả lời 2. lanhuongthu Giải đáp: A=$\dfrac{\sqrt{9}+1}{\sqrt{9}-1}+\dfrac{\sqrt{9}-1}{\sqrt{9}+1}-\dfrac{3\sqrt{9}+1}{9-1}\$
A=(3+1)/(3-1)+(3-1)/(3+1)-(3.3+1)/(9-1)
A=4/2+2/4-10/8
A=(16+4-10)/8
A=5/4
Trả lời
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## Savings bond future interest calculator
Ontario Savings Bonds Home menu arrow. Home · FIs/Dealers info · Interest Rates · Use the Bond Calculator Bond Calculator
6 Jun 2019 Savings bonds are bonds sold by the U. taking a loan from the public, which it guarantees to repay at some point in the future. EE Bonds issued after May 2005 carry a fixed interest rate equal to 90% of the average market Car Loan Calculator: What Will My Monthly Principal & Interest Payment Be? 4 Jun 2018 The Treasury has replaced it with the Savings Bond Calculator, a similar Web- browser-based tool But you can also see the interest accrued, current value, current interest rate, cumulative Returning to this file in the future. Find the best interest rate savings accounts & maximise your returns with 11 tips for choosing the right savings account; Tool: Savings Calculator; How to Yet with fixed-rate accounts (aka 'fixed-rate bonds'), you can get a guaranteed rate for think rates might go down in future – if so, you may want to fix for longer now. This interest calculator compares both simple monthly interest income and long term compound growth. Best Savings Accounts: Earn Up To 2.09% APY → but it also shows you the future value if interest is compounded every month. 25 May 2017 Here's how to determine the effect of interest rates on your bond calculate how much your bond is worth, but simply put, rising interest In mathematical terms, the price of a bond is the sum of the present values of all of its future interest You Happy -- Maybe · Think You Can't Afford Retirement Savings? The Recurring Deposit (RD) calculator will help you calculate the maturity value of the investment if it grows at a certain interest rate. How to use it. The maturity Interest earned on I Bonds and Series EE Savings Bonds accrues monthly. savings bond's value on that date is used to calculate monthly interest save this inventory on your computer for future updates and as a record in case your.
## Use FINRA's Tools and Calculators to help you make informed financial decisions based on (CMOs), College Savings Plans, Commodities and Futures , Communications with the Public 529 Savings Plan Expense Analyzer To find out how much interest is owed on a given bond, use our accrued interest calculator.
Find out how to check and calculate the value of a savings bond online. A series EE bond will reach full face value after 20 years and will stop earning interest is worth right now; How to find out what your bond will be worth in the future Each savings bond series uses a different method to calculate interest, so each requires a different computation to figure its future value. Savings Bonds. The The future value calculator can be used to calculate the future value (FV) of in a savings account or a hold in a bond purchase earns compound interest and so Estimate the future value of a CD, savings bond, IRA, or other investment. Note: This calculator does not include taxes on interest earned, dividends, or capital 2 Sep 2009 The Series EE bond you own from 1983 will reach its final extended maturity in June 2013. At final maturity, the bond stops earning interest.
### With our online RD calculator, you can learn all about interest rates, and the sum you stand to gain for your savings. To get started, choose the tenure of your
4 Jan 2008 It's easy to plan ahead with our estimation calculators. They can help approximate the future value of your savings bonds and show how much 2 Jul 2019 Features include current interest rate, next accrual date, final maturity date, and year-to-date interest earned. Historical and future information also Known as the Savings Bond Calculator, it can help you make more informed the present and future value--as well as historical information, current interest rate Find out how to check and calculate the value of a savings bond online. A series EE bond will reach full face value after 20 years and will stop earning interest is worth right now; How to find out what your bond will be worth in the future Each savings bond series uses a different method to calculate interest, so each requires a different computation to figure its future value. Savings Bonds. The
### Use FINRA's Tools and Calculators to help you make informed financial decisions based on (CMOs), College Savings Plans, Commodities and Futures , Communications with the Public 529 Savings Plan Expense Analyzer To find out how much interest is owed on a given bond, use our accrued interest calculator.
With our online RD calculator, you can learn all about interest rates, and the sum you stand to gain for your savings. To get started, choose the tenure of your Use our compound interest calculator to find the best savings and GIC accounts in Canada. How quickly can your money grow? Fixed Deposit interest calculator. Calculate the amount of interest you will earn over a chosen period. Amount Invested: R. Use our savings calculator to work out the interest on your IRA, ISA, bond or savings account balance. This regular deposit savings calculator allows you to
## Download a free calculator for Microsoft Excel or Google Sheets to estimate the future value of your savings account Our Savings Calculator is a free spreadsheet that is simple to use and much more powerful than most online calculators that you'll find. It will estimate the future value of your savings account with optional periodic deposits.
8 Mar 2020 Savings Bond Calculator US – Learn The Value Today The tool will allow you to get an indicative current and future value. You can even vary the levels of expected interest rates and tax to project the outcome of different Ontario Savings Bonds Home menu arrow. Home · FIs/Dealers info · Interest Rates · Use the Bond Calculator Bond Calculator Savings Calculator - Download a free Savings Interest Calculator for Excel to estimate your future savings. Calculate The Future Value of Your Savings With Compound Interest Interest accrues at a fixed rate for the duration of the bond & is added monthly to the 6 Jun 2019 Savings bonds are bonds sold by the U. taking a loan from the public, which it guarantees to repay at some point in the future. EE Bonds issued after May 2005 carry a fixed interest rate equal to 90% of the average market Car Loan Calculator: What Will My Monthly Principal & Interest Payment Be?
2 Dec 2019 Cashing in a savings bond (Series EE or Series I) is relatively easy. For paper bonds, use the U.S. Treasury's online savings bond calculator. cashing it in could mean losing interest earnings, along with future growth. Many savings bond owners want an easy way to calculate and keep track of their bond's earned interest. The U.S. Treasury Department has a free online
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# How do you calculate acceleration of gravity?
Jun 10, 2014
Assume that air resistance is negligible. Equate the resultant force to the weight and then we get: $a = g$.
We can use the simplified version of Newton's Second Law, $F = m a$. This relates the resultant force acting on the object to its acceleration.
As air resistance is neglected the only force acting is the weight, so $F = w$.
$w = m g$, where $g$ is the gravitational field strength $9.8 N k {g}^{-} 1$.
$F = m g$, now use Newton's Second Law: $m a = m g$. The mass is the same on both sides of the equation: ⇒a=g=9.8 ms^-2
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DevX Home Today's Headlines Articles Archive Tip Bank Forums
# Thread: Karatsuba Algorithm - Not Fully Understand and Don't Know how to Implement It
1. Registered User
Join Date
May 2007
Posts
843
## Karatsuba Algorithm - Not Fully Understand and Don't Know how to Implement It
Hello to all C++ expect programmer, i would like to create a function that multiply two integer which stored in char * or string.
1. compute x1y1, call the result A
2. compute x2y2, call the result B
3. compute (x1 + x2)(y1 + y2), call the result C
4. compute C - A - B; this number is equal to x1y2 + x2y1.
BigInteger multiply(BigInteger a, BigInteger b) {
int n = max(number of digits in a, number of digits in b)
if(n == 1) {
return a.intValue() * b.intValue();
} else {
BigInteger aR = bottom n/2 digits of a;
BigInteger aL = top remaining digits of a;
BigInteger bR = bottom n/2 digits of b;
BigInteger bL = top remaining digits of b;
BigInteger x1 = Multiply(aL, bL);
BigInteger x2 = Multiply(aR, bR);
BigInteger x3 = Multiply(aL + bL, aR + bR);
return x1 * pow(10, n) + (x3 - x1 - x2) * pow(10, n / 2) + x2;
}
}
As far as i know, 1234 * 4567 will be split into two parts which are
A(1) = 12, A(2) = 34, B(1) = 45, B(2) - 67;
First, A = A(1) * B(1), then
B= B(1), B(2), then
C=( A(1) + B(1) * B(1) + B(2) );
The question is i don't understand the last statement.
Code:
` return x1 * pow(10, n) + (x3 - x1 - x2) * pow(10, n / 2) + x2;`
What this does ?
This is my code so far :
Code:
```/*
Polynomial is a expression with non negative
integer power, 1 or more variable and costant(coefficient)
Polynomial is important for signal processing,
crytography and coding theory. Polynomial is core
of multiplication algorithm in computing
Karatsuba algorithm consits of many branches
and optmization such as
1. KA for Degree-1 Polynomials - x ^ 1
2. Recursive KA for Polynomials of Degree 2i ¡ 1
- Efficient for coeficient power of 2
3. One-Iteration KA for Polynomials of Arbitrary Degree
4. KA for Degree-2 Polynomials - X ^ 2
5. KA for Polynomials of Arbitrary(Any) Degree
Ratio between long multiplication and karatsuba > 3, then
use KA for Degree-2 Polynomials
Improvement of KA with dummy coefficient = 0
2.
1. compute x1y1, call the result A
2. compute x2y2, call the result B
3. compute (x1 + x2)(y1 + y2), call the result C
4. compute C - A - B;
this number is equal to x1y2 + x2y1.
BigInteger multiply(BigInteger a, BigInteger b)
{
int n = max(number of digits in a,
number of digits in b)
if(n == 1)
{
return a.intValue() * b.intValue();
}
else
{
BigInteger aR = bottom n/2 digits of a;
BigInteger aL = top remaining digits of a;
BigInteger bR = bottom n/2 digits of b;
BigInteger bL = top remaining digits of b;
BigInteger x1 = Multiply(aL, bL);
BigInteger x2 = Multiply(aR, bR);
BigInteger x3 = Multiply(aL + bL, aR + bR);
return x1 * pow(10, n) + (x3 - x1 - x2) * pow(10, n / 2) + x2;
}
}
*/
/*
MS Calling Convention
__cdecl, __fastcall, __stdcall;
*/
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
using std::cout;
using std::cin;
using std::string;
using std::stringstream;
using std::vector;
#define SIZE 50
// ------------------------------------------------
void MultiplyBigInteger(char *, char *);
int length(char *);
void divide(int, int *, int *);
// ------------------------------------------------
int main(int argc, char *argv[])
{
char *firstNumber = "99999",
*secondNumber = "99999";
MultiplyBigInteger(firstNumber, secondNumber);
return 0;
}
// ------------------------------------------------
void MultiplyBigInteger(char *firstNumber, char *secondNumber)
{
int firstLength = length(firstNumber);
int secondLength = length(secondNumber);
int *firstLeftLength = 0, *firstRightLength = 0,
*secondLeftLength = 0, *secondRightLength = 0;
char *firstRight = 0, *firstLeft =0,
*secondRight = 0, *secondLeft = 0;
divide(firstLength, firstLeftLength, firstRightLength);
divide(secondLength, secondLeftLength, secondRightLength);
if (firstLength == 1 && secondLength == 1)
{
}
else
{
// MultiplyBigInteger(
}
}
// ------------------------------------------------
int length(char *number)
{
int length = 0;
for (int loop=0;number[loop] != '\0';loop++)
{
*(number + loop);
length++;
}
return length;
}
// ------------------------------------------------
void divide(int length, int *leftLength,
int *rightLength)
{
if (length % 2 != 0)
{
*leftLength = (length % 2) + 1;
*leftLength = (length % 2);
}
else
{
*leftLength = length / 2;
*rightLength = length / 2;
}
}```
I not asking you all do for me but explanation and guideline is more than enough.
2. Registered User
Join Date
Jan 2005
Location
UK
Posts
604
Code:
` return x1 * pow(10, n) + (x3 - x1 - x2) * pow(10, n / 2) + x2;`
pow(10,n) is the n-th power of ten, basially a 1 with n 0-s
pow(10,n) for n==4 equals 10*10*10*10 = 10000
in a similar fashion:
pow(2,n) == 2*2*2*...*2 (n-times)
pow(2,8) = 256
you can do pow(x,n) for any x a real number in much the same way,
pow(sqrt(2),3) = sqrt(2)*sqrt(2)*sqrt(2)
you can also do pow(x,y) for y real numbers. but that can not be written as a simple product
The rest of your formula in the return statement is pretty straight forward, so I don't think you asked about that.
3. Senior Member
Join Date
Dec 2003
Posts
3,366
pow is terribly slow by the way, use a lookup table of the powers of 10. Since this is a simple set of values, it can be exploited that a floating point number can store exponents efficiently to take you well into the 10^200 or more range (I forget how far you can go off the top of my head).
Also, the big int thing can be vastly improved if you extend normal binary math instead of trying to do base 10 math on chars. IE a 100 byte integer in the format of FA124.... 054AF hex is much smaller (fewer operations) to multiply etc than 1233456...................................555690 as text... you can see this even with a byte (FF is less to process than 255 in a loop, and the difference bettwen base 10 and base 16 in number of bytes grows very fast (8 bytes becomes over 20 bytes, more than double the size).
Just some thoughts if you plan do do a LOT of numbers or very, very large ones.
4. Registered User
Join Date
May 2007
Posts
843
The rest of your formula in the return statement is pretty straight forward, so I don't think you asked about that.
I understand what is pow() function and i dont' know why need to to x1 * pow(10, n) ? I looking for the logic behind formula.
I not interest in doing math with base 16 since jonnin is memory consumption is high.
5. Registered User
Join Date
May 2007
Posts
843
I think i understand how it works under formula but not under coding since recursion is easy to implement but difficult to understand.
Let me explain this to others as well:
a1a0 - Divide into two parts
b1b0 - Divide into two parts
q_0 = a_0b_0 (p0)
q_1 = (a_0+a_1)(b_0+b_1) (p1) Human calculation
q_2 = a_1b_1. (p2)
Reassign to another variable q0, q1 and q2;
Then rewrite above statement as below:
q_1 = p_1+p_0+p_2.
= a0b0 + a0b1 + a1b0 + a1b1;
= p0 + p2 + p2
-p1 = -q1 + p0 + p2
-p1 = -q1 + q0 + q2
p1 = q1 - q0 - q2;
Therefore, it only needs three multiplication only which are
q_0 = a_0b_0 (p0)
q_1 = (a_0+a_1)(b_0+b_1) (p1)
q_2 = a_1b_1. (p2)
rather than
a1a0
b1b0
b0 * a0
b0 * a1
b1 * a0
b1 * a1
and minus the over multiplication of q0 and q2;
Finally, this form the formula like this:
return x1 * pow(10, n) + (x3 - x1 - x2) * pow(10, n / 2) + x2;
My concern is how it works under code(recursion). I cannot see how it works unless i went through debugging process.
I hope GOD will bless you all.
6. Registered User
Join Date
May 2007
Posts
843
I have coded some code so far as below :
Code:
```/*
Polynomial is a expression with non negative
integer power, 1 or more variable and costant(coefficient)
Polynomial is important for signal processing,
crytography and coding theory. Polynomial is core
of multiplication algorithm in computing
Karatsuba algorithm consits of many branches
and optmization such as
1. KA for Degree-1 Polynomials - x ^ 1
2. Recursive KA for Polynomials of Degree 2i ¡ 1
- Efficient for coeficient power of 2
3. One-Iteration KA for Polynomials of Arbitrary Degree
4. KA for Degree-2 Polynomials - X ^ 2
5. KA for Polynomials of Arbitrary(Any) Degree
Ratio between long multiplication and karatsuba > 3, then
use KA for Degree-2 Polynomials
Improvement of KA with dummy coefficient = 0
2.
1. compute x1y1, call the result A
2. compute x2y2, call the result B
3. compute (x1 + x2)(y1 + y2), call the result C
4. compute C - A - B;
this number is equal to x1y2 + x2y1.
BigInteger multiply(BigInteger a, BigInteger b)
{
int n = max(number of digits in a,
number of digits in b)
if(n == 1)
{
return a.intValue() * b.intValue();
}
else
{
BigInteger aR = bottom n/2 digits of a;
BigInteger aL = top remaining digits of a;
BigInteger bR = bottom n/2 digits of b;
BigInteger bL = top remaining digits of b;
BigInteger x1 = Multiply(aL, bL);
BigInteger x2 = Multiply(aR, bR);
BigInteger x3 = Multiply(aL + bL, aR + bR);
return x1 * pow(10, n) + (x3 - x1 - x2) * pow(10, n / 2) + x2;
}
}
*/
/*
MS Calling Convention
__cdecl, __fastcall, __stdcall, __thiscall;
*/
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
using std::cout;
using std::cin;
using std::string;
using std::stringstream;
using std::vector;
#define SIZE 50
// ------------------------------------------------
void MultiplyBigInteger(char *, char *);
int length(char *);
void divide(int, int *, int *);
// ------------------------------------------------
int main(int argc, char *argv[])
{
char *firstNumber = "1",
*secondNumber = "2";
MultiplyBigInteger(firstNumber, secondNumber);
return 0;
}
// ------------------------------------------------
void MultiplyBigInteger(char *firstNumber, char *secondNumber)
{
int firstLength = length(firstNumber);
int secondLength = length(secondNumber);
int *firstLeftLength = &firstLength,
*firstRightLength = &firstLength,
*secondLeftLength = &secondLength,
*secondRightLength = &secondLength;
char *firstRight = 0, *firstLeft =0,
*secondRight = 0, *secondLeft = 0;
divide(firstLength, firstLeftLength, firstRightLength);
divide(secondLength, secondLeftLength, secondRightLength);
if (firstLength == 1 && secondLength == 1)
{
int oneDigitResult = /*static_cast<int>*/(firstNumber + 0)
* /*static_cast<int>*/(secondNumber + 0);
cout << oneDigitResult;
}
else
{
// MultiplyBigInteger(
}
}
// ------------------------------------------------
int length(char *number)
{
int length = 0;
for (int loop=0;number[loop] != '\0';loop++)
{
*(number + loop);
length++;
}
return length;
}
// ------------------------------------------------
void divide(int length, int *leftLength,
int *rightLength)
{
if (length % 2 != 0)
{
*leftLength = (length % 2) + 1;
*rightLength = (length % 2);
}
else
{
*leftLength = length / 2;
*rightLength = length / 2;
}
}```
In multiplication function,
Code:
```void MultiplyBigInteger(char *firstNumber, char *secondNumber)
{
int firstLength = length(firstNumber);
int secondLength = length(secondNumber);
int *firstLeftLength = &firstLength,
*firstRightLength = &firstLength,
*secondLeftLength = &secondLength,
*secondRightLength = &secondLength;
char *firstRight = 0, *firstLeft =0,
*secondRight = 0, *secondLeft = 0;
divide(firstLength, firstLeftLength, firstRightLength);
divide(secondLength, secondLeftLength, secondRightLength);
if (firstLength == 1 && secondLength == 1)
{
int oneDigitResult = /*static_cast<int>*/(firstNumber + 0)
* /*static_cast<int>*/(secondNumber + 0);
cout << oneDigitResult;
}
else
{
// MultiplyBigInteger(
}
}```
Code:
```if (firstLength == 1 && secondLength == 1)
{
int oneDigitResult = /*static_cast<int>*/(firstNumber + 0)
* /*static_cast<int>*/(secondNumber + 0);
cout << oneDigitResult;
}```
If the value is single digit then i want to multiply by ANSI table which is 1(49) * 2(50) but how do i achieve his ?
I don't want convert this to int array due to higher memory consumption and i want better solution.
7. Registered User
Join Date
May 2007
Posts
843
8. Registered User
Join Date
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Posts
843
Any one.
Thanks.
9. Senior Member
Join Date
Dec 2004
Location
San Bernardino County, California
Posts
1,468
I worked on this a couple of years ago. I have my notes and code at home. I'll try to respond this evening (west coast USA time).
10. Registered User
Join Date
May 2007
Posts
843
I worked on this a couple of years ago. I have my notes and code at home. I'll try to respond this evening (west coast USA time).
Happy to hear that.
11. Registered User
Join Date
May 2007
Posts
843
I really have hard time to understand this algorithm and look forward nsplis's help.
12. Registered User
Join Date
May 2007
Posts
843
Recursion is not my strong part of programming.
I look forward for any help.
Thanks.
13. Registered User
Join Date
Dec 2007
Posts
401
you really can't keep requesting someone else to explain from first principles every algorithm that you encounter.
learn to take one small baby step at a time before you attempt to sprint.
first, understand simple recursion by working through a beginner's tutorial; something like http://cse.unl.edu/~dsadofs/RecursionTutorial/index.php
once this much has been well-understood, go on to learn more about recursion: tail-recursion, tree-recursion and how to convert between recursive and iterative implementations of algorithms. http://www.cc.gatech.edu/classes/AY2...ion_gallis.htm
next, understand the karatsuba algorithm for multiplying small (say two-digit) numbers. http://mathworld.wolfram.com/Karatsu...plication.html
once that is clear, study how to extend the algorithm to the general N-digit case by using a divide-and-conquer approach. http://www.student.cs.uwaterloo.ca/~cs341/lecture5.html
and then you would be able to write either a recursive or an iterative implementation of the karatsuba algorithm.
14. Registered User
Join Date
May 2007
Posts
843
I want to learn them in hand first then i need to write the recursion tree and recurrence relation.
Thanks.
15. Senior Member
Join Date
Dec 2004
Location
San Bernardino County, California
Posts
1,468
Peter:
Take a look at Chapter 2 of this book, "Algorithms" by S. Dagupta, C H Papadimitriu, and U. V. Vazirani. Their explanation of the recursive approach to this Karatsuba multiplication process is very precise while not being thick. So much better written than I ever could produce.
The book is also available through e-booksdirectory.com, which has a plethora of ebooks, mostly in PDF format, on many different subjects all available for free download. You can get to the book directly at:
http://www.cs.berkeley.edu/~vazirani/algorithms/all.pdf
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# Homework Help: Maximal solution Interval ODE question
1. Sep 5, 2010
### Susanne217
1. The problem statement, all variables and given/known data
Given an ODE
$$h(y(t)) \frac{dy}{dt} = q(t)$$
How do I define the maximal solution of this equation?
According to my textbook for this course a maximal solution for the above problem if it cannot be obtained from another solution as a restriction to a smaller interval.
I have been looking through many textbooks both in the in Lib and on google books.
Is there anyone here who would explain maximal solution in layman terms to me?
I know howto solve most ODE's and IVP and some BVP. But is the max interval of solution just the interval on which the solution of the ODE is defined?
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Using percentage
Chapter 7 Class 7 Comparing Quantities
Concept wise
This video is only available for Teachoo black users
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Example 7 Rahul bought a sweater and saved Rs 20 when a discount of 25% was given. What was the price of the sweater before the discount? Given that Discount is Rs 200 and discount is 25% Now, Discount will be on price of sweater Therefore, Discount = 25% × Price 200 = 25% × Price 25% × Price = 200 25/100 × Price = 200 𝟏/𝟒 × Price = 200 Price = 200 × 4/1 Price = 800 ∴ Price of sweater is Rs 800
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### Particle dynamics, Newton laws
A body (mass $m$) starts falling. The air friction force is $\vec {F}_f=-k\vec {v}$, where $k=\text{const}$ and $\vec {v}$ is the body velocity. Find $\vec {v}(t)$ and $\vec {r}(t)$.
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## Government Mathematics 101a
No wonder our kids need remedial reading and math upon entering college in many states and also why we are number 47 in the world in math.
The way the government works, a number is not from another world but another dimension (insert the “Twilight Zone” theme here.) Just amazing!
Nancy Pelosi says, “Every dollar we spend on unemployment puts two dollars in the system.” Really? In what universe? Under government math, if Susie has 3 pieces of cake and Johnny only has 1 there is an obvious inequity. The fact that Susie’s daddy is a banker and Johnny’s daddy is a cab driver obviously means that Johnny hasn’t had the same opportunities as Susie. This means we need to find Frankie because he always has 4 pieces of cake. He’s looking a little chunky so we need to save him from himself and give one of his pieces to Johnny. Then we need to take one from Suzie and give it to Johnny as well. Now Johnny won’t feel bad because based on government math, everyone has been equalized. Confused? I am writing it and I’m confused.
But this is government math!
They want us to believe that over 92 million people have left the work force. They will tell you it’s because many have retired, some are on disability, and other moronic excuses. We need to create 200,000+ jobs a month to maintain a level of economic misery that’s tolerable. Last month we created 73,000 jobs. Most were temporary jobs, go figure, Christmas, holidays, and all that. But, 535,000 people left the work force bringing that number to over 91 million. Wait a minute, what?
Did a million people leave the planet, die off, hit the lottery, crawl under a rock? Will they be needing government help? Can anyone explain this to me?
Government math is the only way this works. The same guys who say more taxes will create more jobs. Yup, that logic means, taking money out of the pocket of employers will give them more money to hire people.
Only with government math will raising minimum wage to \$15.00 per hour create more jobs and stimulate the economy. In most cases, it would double the total payroll for a company. And the government thinks that will compel employers to hire more employees because people will have more money to spend and they will need extra staff for the increased spending. HUH! That’s government math
Participation rate (the number of people who are able to work but, for whatever reason, do not) is the worst it’s been since 1978. Mr. Obama, this is your mission accomplished moment. This does not include people on disability or those retired on government pensions.
People, there are approximately 350 million American citizens in the U.S. (and an unknown number of other human inhabitants.) Let’s make it easy and round the numbers, because I don’t have enough fingers or toes to count on. To start with, take 100 million away for those that have left the workforce. Next, figure that at least 75 million of the 350 million Americans have at least 1 child who doesn’t contribute to the tax rolls. That puts us at around 175 million tax-paying Americans. That means companies (employers) are paying for over 175 million people receiving government aid. It CAN’T WORK!
And to add insult to injury, CNBC now reports that, according to the government, there are ONLY approximately 155 million employed people. Hello!!! Some of those 155 million employed are only part-time workers! (there is a 20 million people chasm between my numbers and theirs, but hey is government math) People, we are closer to a 25-30% unemployment rate. Even the government’s real rate, as shown in the Bureau of Labor Statistics, shows it closer to 13.2 %. So not only can’t they add (or calculate percentages), but they couldn’t find the truth in a bag marked “Truth” if you put both their hands on it!
Let’s break it down to smaller numbers for illustration. Say there are 20 people on your street and you have all vowed to take care of each other in every way, communal living. Say 10 people decide that they really don’t want to work. They want to hang out and raise their families, work on their homes, cars, or maybe start a new career. That means the other 10 have to provide for the whole group of 20. They may need to cut back on food, get rid of a car to save on insurance, use less electricity, buy less clothes, and make other adjustments. WHY? Well, if you’re using government math (and logic,) everyone has to share. No one should go without food, shelter, cable, or cellular service. And no one should be forced to work until they decide to and even then only at a job they are satisfied with.
Come-on! Does this make sense to anyone?
Government math has gifted us with a “real” unemployment rate of over 25%, a \$17 trillion deficient (and climbing), and one of the worst education systems in the world (you can’t just throw money at it to fix it).
People, wake up or we’re going to lose our country!
## Why the Immigration Bill Will Fail
For several weeks now the Senate has been pushing hard to pass a new “comprehensive” immigration reform bill. Unfortunately, this bill violates the Laws of Nature… so it’s doomed to failure if it ever becomes the law of the United States.
## Americans Bow to No Man
To whom do you bow?
Our current president bows to many. Perhaps due to his childhood abroad or perhaps due to an earnest desire to appease Europeans. But to many US citizens this subservient action is an insult.
Thomas Jefferson began shaking hands as a greeting thus ending the tradition of bowing. Jefferson understood that to be a free citizen of this great republic was the highest achievement of man.
Some see the people of America not bowing and say that we are arrogant; that we believe in the idea of American Exceptionalism; that we should show deference.
Yet, our country was formed on the belief that all men are created equal. Our military fight and die so that we never have to bow to any man.
Maybe it’s something uniquely American. Maybe we really are arrogant thinking we’re equal to any man. But no matter, as citizens of this country we expect, nay it is imperative, that our president and by definition all people herein bow to no man.
We need a president to lead this country forward once again. We are not happy to settle and we will not bow.
It is time for new leadership.
## President Obama’s Executive Order for Marriage Equality
(Associated Depressed) — President Obama issued an executive order entitled the Equality in Marriage Act, requiring by law that all marriages be, “Equal in all respects, whether in terms of gender, race, religion, sexual preference, appearance, or intelligence.”
The policy shift ushers in not only a new era of legal equality between homosexual and heterosexual couples, but equalized marriage outcomes for all human beings, regardless of race, gender, or sexual preference.
“The time has come for us to set aside our petty differences,” President Obama said before a White House podium overlooking the picturesque National Mall, “to take hope in the change of this transformational agenda, so that all people, no matter what background or what our individual differences, will not be overly blessed or overly cursed with a marriage to someone who is not right for them.”
Millions applauded the president for the progressive decree, which would put an end to marriage inequality once and for all. People who once thought they were doomed to a miserable marriage to someone in their same victim class have already looked forward to seeing the new policy put in practice.
But now that the applause has ended, controversy has set in regarding what exactly the president meant.
“Well, frankly, there is no precedent in American law for this kind of sweeping, although breathtaking vision,” said Mortimer L. Swathley of the politically neutral Brookings Institution. “We are in a new era of law when we just have to try to discern what Obama’s intentions are when he issues an executive order. Such scholars have already dubbed themselves ‘Obama reconstructivists.’ ”
The decree has sent lawyers, judges, and pastors scrambling to accommodate a flurry of new matrimonial activity sure to provide a stimulus for the recovering economy.
“We still aren’t sure if Obama means that two people with high IQs now need to get divorced and remarry spouses with proportionately lower IQs,” said Judge Michael B. Fuddell. “Or if two good-looking people have to separate and pair with ugly partners, or if white people have to marry black people, or even so far that dark people have to marry pasty people. It’s kind of a legal nightmare…but an exciting one.”
The national divorce rate has already tripled, providing a boon to family legal practices across the country. Marriage rates also received a much-needed boost, although single people are already starting to grumble about discrimination.
“Couples who once found themselves chained to a much dumber or uglier spouse were initially happier that Obama approved the overall collective equality in their marriages,” said psychotherapist Jill Brubeck of a marriage counseling agency in California. “But as mandatory reporters, we are having trouble getting two intellectually-challenged partners to acknowledge that there is any inequality there. They just aren’t bright enough to make sense of the law, poor things.”
While heterosexuals struggle to set aside previous marriages and find their opposites in terms of race, looks, and IQ, homosexuals are elated that gay marriage has effectively been federally mandated.
“If heterosexuals can be trapped in pledged lifelong monogamy, why can’t gays?” said one legal expert. “The only question remaining is why married couples should hoard all the misery, when singles escape scot-free. I think it is time for the president to call upon singles to show a little self-sacrifice, and share the misery.”
While there are still many issues unsettled regarding Obama’s executive order, some citizens are already seeing a positive difference in their lives.
“I thought I was happy wedded to a smart, wonderful husband, having beautiful healthy children,” sobbed Sarah Milner joyfully, “but after listening to President Obama, I realized that I was being selfish, and I could never tolerate living such an unfair life. So I quit and married a bum who was making catcalls to me everyday on the way to work. The guilt is gone, and I’ve felt better ever since.”
The policy is set to be fully implemented following Barack Obama’s last election.
Author’s note: The above is satire. It is a fictionalized account intended to elucidate certain ideas and principles by taking them to absurd lengths. It is not intended to be taken literally.
Kyle Becker blogs at RogueGovernment, and can be followed on Twitter as @RogueOperator1. He writes freelance for several publications, including American Thinker, Misfit Politics, and OwntheNarrative, and is a regular commentator on the late night talk shows at OTNN.
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math help SOS? - XP Math - Forums
XP Math - Forums math help SOS?
07-27-2007 #1 toty467 Guest Posts: n/a math help SOS? I need a general term for the following sequences...2/3, 3/4, 4/5, 5/6, 6/7 the next one is -3, 6, -9, 12, -15 and the last one is 3, 9, 27, 81, 243 thanx for all who help me with these you all are true angels muahz whoever answers the best gets best answer
07-27-2007 #2 Teh Pwnzor Guest Posts: n/a First one is 1+n/2+n, n being the number in the sequence.Second one is (-1^n) x (3n), n being the number in the sequence.Third is 3^n, n being the number in the sequence.Hope that helps and good luck.
07-27-2007 #3 Yash Guest Posts: n/a 2/3, 3/4, 4/5, 5/6, 6/7nth term will be (n+1)/(n+2)-3, 6, -9, 12, -15nth term: 3*n*(-1)^n3, 9, 27, 81, 243nth term: 3^n
07-27-2007 #4 funda40 Guest Posts: n/a The i th. number in the following sequences are - First sequence: i/(i+1)Second sequence: ((-1)^i)*3*iThird sequence: 3^i
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## Image Moments
An Image moment is a number calculated using a certain formula. Understand what that formula means might be hard at first. In fact, I got a lot of questions about moments from the tracking tutorial I did long back. So, here it is - an explanation of what moments area!
## The math of moments
In pure math, the nth order moment about the point c is defined as:
This definition holds for a function that has just one independent variable. We're interested in images - they have two dimensions. So we need two independent variables. So the formula becomes:
Here, the f(x, y) is the actual image and is assumed to be continuous. For our purposes, we need a discrete way (think pixels) to describe moments:
The intergrals has been replaced by summations. The order of the moment is m + n. Usually, we calculate the moments about (0, 0). So you can simply ignore the constants cx and cy.
Now with the math part out of the way, let's have a look at what you can calculate with this thing.
## Calculating area
To calculate the area of a binary image, you need to calculate its zeroth moment:
The x0 and y0 don't have any effect and can be removed.
Now, in a binary image, a pixel is either 0 or 1. So for every white pixel, a '1' is added to the moment - effectively calculating the area of the binary image! Another thing to note is that there is only one zeroth order moment.
## Centroid
To calculate the centroid of a binary image you need to calculate two coordinates -
How did I get that? Here's a quick explanation. Consider the first moment:
The two summations are like a for loop. The x coordinate of all white pixels (where f(x, y) = 1) is added up.
Similarly, we can calculate the sum of y coordinates of all white pixels:
Now we have the sum of several pixels' x and y coordinates. To get the average, you need to divide each by the number of pixels. The number of pixels is the area of the image - the zeroth moment. So you get:
and
One interesting thing about this technique is that it is not very sensitive to noise. The centroid might move a little bit but not much.
Also, from the math it's clear this technique holds only for single blobs. If you have two white blobs in your image, the centroid will be somewhere in between. You'll have to extract each blob separately to get their centroids.
## Central moments
In fact, this kind of division is very common - dividing a moment by the zeroth order moment. It's so common that it has a name of its own - central moments.
So to calculate the centroid, you need to calculate the first order central moments.
## Higher order moments
Going onto higher order moments, things get complicated really fast. You have three 2nd order moments, four 3rd order moments, etc. You can combine several of these moments so that they are translation invariant, scale invariant and even rotation invariant.
While reading about moments, I found an entire book dedicated to pattern recognition with moments. In fact, there are terms called skewness and kurtosis. These refer to third and fourth order moments. They measure how skewed an image is and whether an image is tall and thin or short and fat. Clearly, there's a LOT that can be learned about these mathematical tools.
Utkarsh Sinha created AI Shack in 2010 and has since been working on computer vision and related fields. He is currently at Carnegie Mellon University studying computer vision.
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# Wiener–Ikehara theorem
The Wiener–Ikehara theorem is a Tauberian theorem introduced by Shikao Ikehara (1931). It follows from Wiener's Tauberian theorem, and can be used to prove the prime number theorem (PNT) (Chandrasekharan, 1969).
## Statement
Let A(x) be a non-negative, monotonic nondecreasing function of x, defined for 0 ≤ x < ∞. Suppose that
${\displaystyle \int _{0}^{\infty }A(x)e^{-xs}\,dx}$
converges for ℜ(s) > 1 to the function ƒ(s) and that ƒ(s) is analytic for ℜ(s) ≥ 1, except for a simple pole at s = 1 with residue 1: that is,
${\displaystyle f(s)-{\frac {1}{s-1}}}$
is analytic in ℜ(s) ≥ 1. Then the limit as x goes to infinity of exA(x) is equal to 1.
## One Particular Application
An important number-theoretic application of the theorem is to Dirichlet series of the form
${\displaystyle \sum _{n=1}^{\infty }a(n)n^{-s}}$
where a(n) is non-negative. If the series converges to an analytic function in
${\displaystyle \Re (s)\geq b\,}$
with a simple pole of residue c at s = b, then
${\displaystyle \sum _{n\leq X}a(n)\sim {\frac {c}{b}}X^{b}.}$
Applying this to the logarithmic derivative of the Riemann zeta function, where the coefficients in the Dirichlet series are values of the von Mangoldt function, it is possible to deduce the PNT from the fact that the zeta function has no zeroes on the line
${\displaystyle \Re (s)=1.\,}$
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`ISRN Mathematical AnalysisVolume 2013 (2013), Article ID 501382, 6 pageshttp://dx.doi.org/10.1155/2013/501382`
Research Article
## Asymptotic Series of General Symbol of Pseudo-Differential Operator Involving Fractional Fourier Transform
1DST-CIMS, Department of Applied Mathematics, Indian Institute of Technology, Banaras Hindu University, Varanasi 221005, India
2DST-CIMS, Banaras Hindu University, Varanasi 221005, India
Received 16 July 2013; Accepted 13 August 2013
Academic Editors: T. Ozawa, W. Yu, and C. Zhu
#### Abstract
An asymptotic series of general symbol of pseudo-differential operator is obtained by using the theory of fractional Fourier transform.
#### 1. Introduction
Namias [1] introduced fractional Fourier transform which is a generalization of Fourier transform. Fractional Fourier transform is the most important tool, which is frequently used in signal processing and other branches of mathematical sciences and engineering. The fractional Fourier transform can be considered as a rotation by an angle in time-frequency plane and is also called rotational Fourier transform or angular Fourier transform. The fractional Fourier transform [2, 3], with angle of a function , is defined by where The corresponding inversion formula is given by where the kernel Zayed [3] and Bhosale and Chaudhary [4] studied fractional Fourier transform of distributions with compact support. Pathak and others [5] defined the pseudo-differential operator involving fractional Fourier transform on Schwartz space and studied many properties.
Our main aim in this paper is to generalize the results of Zaidman [6] and to find an asymptotic series of general symbol of pseudo-differential operator involving fractional Fourier transform.
Now we are giving some definitions and properties which are useful for our further investigations.
Linearity of fractional Fourier transform is given as where and are constants and and are two input functions.
Let denote the class of measurable functions defined on such that where .
From [5], generalized Sobolev space involving fractional Fourier transform is defined by and .
The convolution of two functions and is defined [5, 7] as provided that the integral exists.
Let be a class of all measurable complex-valued functions which are defined on . Then, we assume the following properties.(i) exists for all and is bounded to mesaurable function.(ii)We define , then where is complex-valued function defined on , which is measurable in and for all and satisfies the estimate: where .
Let be a strictly decreasing sequence; that is, as and such that for all , Let be an infinite sequence of function defined on .
Then, we define a function where is a sequence of positive real numbers such that as .
From (12), it is clear that , for , , and .
The global estimate of the above defined function and of remainders of order is given as
Theorem 1. Let be a sequence of positive real numbers such that the following inequalities: are satisfied for . In particular the estimates are as follows:
Proof. The proof of the above theorem is obvious from [6, pages 233-234].
Theorem 2. Let be a sequence of positive real numbers such that the following estimates: are satisfied for . In particular the estimates are as follows: where , , , and .
Proof. The proof of above theorem is also obvious by using the same arguments from [6, pages 133–135].
#### 2. Asymptotic Expansion of Pseudo-Differential Operator Associated with General Symbol
Definition 3. Let be a general symbol belonging to . Then pseudo-differential operator associated with symbol is defined by where is defined in (3) .
Definition 4. An infinitely differentiable complex-valued function is member of if and only if for every choice of and of non-negative integers, it satisfies
Lemma 5. A function satisfies (19) if and only if
Lemma 6 (Peetre). For any real number t and for all , the estimate is satisfied.
Theorem 7. Let ; then one has the following relation: where , .
Proof. By the definition of fractional Fourier transform (1), we have
Theorem 8. If is a symbol and is the associated operator, then, one has the following relation: where , .
Proof. We have By linearity of fractional Fourier transform (5) we get Now using (1) and (22), we get the required result.
Theorem 9. Let be a symbol and the associated operator; then one has the following relation: where , .
Proof. Firstly from (7) we haveNow the argument of (15) yields Thus we have for every , .
Now we consider the function , where for and for and is defined as Therefore, Using (21) we get From Theorem 2 we get Here by (10) for all and since , therefore .
Thus, we have and the inequality This implies that So that Using (36) we get Therefore, Now using (30) and (38) we get This implies that This implies the required result (27).
Theorem 10. One has the following estimates:
Proof. Let , then by using (11) and (12) we find that for . Therefore, we have Now we define the operator given on by Hence, Thus, Since , is a linear operator into itself.
Definition 11. A linear operator with , and , there exists a constant such that The infimum of all orders of is called true order of .
Definition 12. Let be a linear operator from into itself and satisfy the following inequality: Then is said to be a canonical operator of degree , where .
Definition 13. Let be a strictly decreasing sequence of real numbers and a sequence of canonical operators of degree . Then, corresponding a sequence of positive real numbers and , a linear operator is asymptotically expanded into the series if it satisfies the following inequality:
Theorem 14. Let be a sequence of symbols belonging to and a strictly decreasing sequence of real numbers that tends to . Then, there exists a sequence of canonical operators of degree and a linear operator in such that(i)t.o ;(ii), that is, t.o .
Proof. Note that (i) is obviously true by using Theorem 9 and for (ii) by using the arguments of [6, pages 241-242]. We can define the canonical operator by the following way: where belongs to class of symbol . Also we have Therefore, Using the previous definitions (12) and (13), we have From (15) and (16), we get , where and .
Also . Now using these arguments in (52) we get Using Theorem 9, we get Hence, we get
#### Acknowledgments
The first author is thankful to DST-CIMS, Banaras Hindu University, Varanasi, India, for providing the research facilities, and the second author is also thankful to DST-CIMS, Banaras Hindu University, Varanasi, India, for awarding the Junior Research Fellowship from December 2012.
#### References
1. V. Namias, “The fractional order Fourier transform and its application to quantum mechanics,” Journal of the Institute of Mathematics and its Applications, vol. 25, no. 3, pp. 241–265, 1980.
2. L. M. Almeida, “The fractional Fourier transform and time-frequency representations,” IEEE Signal Processing Letters, vol. 42, no. 11, pp. 3084–3091, 1994.
3. A. I. Zayed, “Fractional Fourier transform of generalized functions,” Integral Transforms and Special Functions, vol. 7, no. 3-4, pp. 299–312, 1998.
4. B. N. Bhosale and M. S. Chaudhary, “Fractional Fourier transform of distributions of compact support,” Bulletin of the Calcutta Mathematical Society, vol. 94, no. 5, pp. 349–358, 2002.
5. R. S. Pathak, A. Prasad, and M. Kumar, “Fractional Fourier transform of tempered distributions and generalized pseudo-differential operator,” Journal of Pseudo-Differential Operators and Applications, vol. 3, no. 2, pp. 239–254, 2012.
6. S. Zaidman, “On asymptotic series of symbols and of general pseudodifferential operators,” Rendiconti del Seminario Matematico dell'Università di Padova, vol. 63, pp. 231–246, 1980.
7. R. S. Pathak, A Course in Distribution Theory and Applications, Narosa Publishing House, New Delhi, India, 2009.
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3.1kviews
Ace Heating and Air Conditioning service finds that the amount of time a repairman needs to fix a furnace is uniformly distributed between 1.5 and 4 hours.
i) Find the probability that a randomly selected furnace repair requires more than 2 hours. ii) Find the probability that a randomly selected furnace repair requires less than 3 hours. iii) Find the mean and standard deviation.
0
159views
i) Probability that a randomly selected furnace repair require more than 2 hours
Let x be random variable that denotes the amount of time a repairman needs to fix a furnace. When X is uniformly distributed between (1.5, 4 ) (a, b)
So, required probablity is - $1 - P(0 \lt X \lt 2)$
\begin{aligned} \text{We know, } P(0 \lt X \lt 2) &= F(2) - F(0) \\ &= \frac{2-0}{b-a} \\ &= \frac{2}{4-1.5} \\ &= \frac{2}{2.5} \\ &= 0.8 \end{aligned}
$\therefore$ Probability that a randomly selected furnace repair requires more than two hours = $1 - 0.8 = 0.2$
ii) Find probability that randomly selected furnace repair requires less than 3 hours
\begin{aligned} &=P(\alpha x \lt 3) \\ &=F(3)-F(0) \\ &=\frac{3-0}{b-a} \\ &=\frac{3}{4-1 \cdot 5} \\ &=\frac{3}{2 \cdot 5} \\ &=1 \cdot 2 \end{aligned}
iii) Find the mean and standard deviation
mean, $E(x)=\frac{a+b}{2}=\frac{1 \cdot 5+4}{2}=\frac{45}{2} = 2.25$
\begin{aligned} \text { variance, } V(x) &=\frac{(b-a)^{2}}{12} \\ &=\frac{(4-1.5)^{2}}{12}=\frac{(2.5)^{2}}{12} \\ &=\frac{6.25}{12}=0.520 \end{aligned}
\begin{aligned} \therefore \text { Standard Deviation } &=\sqrt{\text{Variance}} \\ &=\sqrt{0.520} \\ &=0.721 \end{aligned}
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# Functional programming gotcha
Discussion in 'Python' started by Ed Schofield, Oct 24, 2003.
1. ### Ed SchofieldGuest
Hi all,
I find this strange:
>>> flist = []
>>> for i in range(3):
.... f = lambda x: x+i
.... flist.append(f)
....
>>> [f(1) for f in flist]
[3, 3, 3]
>>>
What I expect is:
>>> [f(1) for f in flist]
[1,2,3]
>>>
Is this a bug in Python? It happens on my builds of
Python 2.3 and 2.2.3.
Replacing the lambda function by a named function, as in
flist = []
for i in range(3):
def f(x):
return x + i
flist.append(f)
[f(1) for f in flist]
gives the same result.
I have a workaround (of sorts) adapted from the Python tutorial:
>>> def make_incrementor(n):
.... return lambda x: x + n
....
>>> flist = [make_incrementor(i) for i in range(3)]
>>>
>>> [f(1) for f in flist]
[1, 2, 3]
>>>
but I'd prefer the flexibility of the first approach. Any ideas?
Any explanations for why Python does this? Any justifications for why
Python _should_ do this? Who believes this is a bug?
Kind regards,
Ed Schofield
Ed Schofield, Oct 24, 2003
2. ### David C. FoxGuest
Ed Schofield wrote:
> Hi all,
>
> I find this strange:
>
>
>>>>flist = []
>>>>for i in range(3):
>
> ... f = lambda x: x+i
> ... flist.append(f)
> ...
>
>>>>[f(1) for f in flist]
>
> [3, 3, 3]
>
>
> What I expect is:
>
>
>>>>[f(1) for f in flist]
>
> [1,2,3]
>
>
> Is this a bug in Python? It happens on my builds of
> Python 2.3 and 2.2.3.
I guess the nested scopes changes in Python 2.1/2.2 (see
http://www.python.org/peps/pep-0227.html) apply to functions but not
loops. You can use the same workaround that people used to use because
of the lack of nested scopes:
f = lambda x, i = i: x + i
That way, the second, optional parameter has a default value which is
evaluated at the time the lambda is created.
>
> Replacing the lambda function by a named function, as in
>
> flist = []
> for i in range(3):
> def f(x):
> return x + i
> flist.append(f)
>
> [f(1) for f in flist]
>
> gives the same result.
>
> I have a workaround (of sorts) adapted from the Python tutorial:
>
>
>>>>def make_incrementor(n):
>
> ... return lambda x: x + n
> ...
>
>>>>flist = [make_incrementor(i) for i in range(3)]
>>>>
>>>>[f(1) for f in flist]
>
> [1, 2, 3]
>
>
> but I'd prefer the flexibility of the first approach. Any ideas?
>
> Any explanations for why Python does this? Any justifications for why
> Python _should_ do this? Who believes this is a bug?
>
> Kind regards,
> Ed Schofield
>
David C. Fox, Oct 24, 2003
3. ### Robin BeckerGuest
In article <>, Ed
Schofield <> writes
>
>Hi all,
>
>I find this strange:
>
>>>> flist = []
>>>> for i in range(3):
>... f = lambda x: x+i
try using f = lambda x,i=i: x+i ie bind at the definition point
>... flist.append(f)
>...
>>>> [f(1) for f in flist]
>[3, 3, 3]
>>>>
>
>What I expect is:
>
>>>> [f(1) for f in flist]
>[1,2,3]
>>>>
>
>Is this a bug in Python? It happens on my builds of
>Python 2.3 and 2.2.3.
>
>Replacing the lambda function by a named function, as in
>
>flist = []
>for i in range(3):
> def f(x):
> return x + i
> flist.append(f)
>
>[f(1) for f in flist]
>
>gives the same result.
>
>I have a workaround (of sorts) adapted from the Python tutorial:
>
>>>> def make_incrementor(n):
>... return lambda x: x + n
>...
>>>> flist = [make_incrementor(i) for i in range(3)]
>>>>
>>>> [f(1) for f in flist]
>[1, 2, 3]
>>>>
>
>but I'd prefer the flexibility of the first approach. Any ideas?
>
>Any explanations for why Python does this? Any justifications for why
>Python _should_ do this? Who believes this is a bug?
>
>Kind regards,
>Ed Schofield
>
--
Robin Becker
Robin Becker, Oct 24, 2003
4. ### David C. FoxGuest
David C. Fox wrote:
> Ed Schofield wrote:
>
>> Hi all,
>>
>> I find this strange:
>>
>>
>>>>> flist = []
>>>>> for i in range(3):
>>
>>
>> ... f = lambda x: x+i
>> ... flist.append(f)
>> ...
>>
>>>>> [f(1) for f in flist]
>>
>>
>> [3, 3, 3]
>>
>>
>> What I expect is:
>>
>>
>>>>> [f(1) for f in flist]
>>
>>
>> [1,2,3]
>>
>>
>> Is this a bug in Python? It happens on my builds of
>> Python 2.3 and 2.2.3.
>
>
> I guess the nested scopes changes in Python 2.1/2.2 (see
> http://www.python.org/peps/pep-0227.html) apply to functions but not
> loops.
I take back that explanation. The problem is that the i in the lambda
*does* refer to the local variable i, whose value changes as you go
through the loop, not to a new variable whose value is equal to the
value of i at the time when the lambda was created.
The workaround is still the same:
> loops. You can use the same workaround that people used to use because
> of the lack of nested scopes:
>
> f = lambda x, i = i: x + i
>
> That way, the second, optional parameter has a default value which is
> evaluated at the time the lambda is created.
>
David
David C. Fox, Oct 24, 2003
5. ### Gregor LinglGuest
David C. Fox schrieb:
.....
>
> I guess the nested scopes changes in Python 2.1/2.2 (see
> http://www.python.org/peps/pep-0227.html) apply to functions but not
> loops. You can use the same workaround that people used to use because
> of the lack of nested scopes:
>
> f = lambda x, i = i: x + i
>
> That way, the second, optional parameter has a default value which is
> evaluated at the time the lambda is created.
>
Isn't this the intended behaviour of nested scopes (regardless
of loops or not)?
>>> def fun():
i = 1
def f():
print i
f()
i+=1
f()
>>> fun()
1
2
>>>
Gregor
Gregor Lingl, Oct 24, 2003
6. ### Terry ReedyGuest
"Ed Schofield" <> wrote in message
news...
>
> Hi all,
>
> I find this strange:
>
> >>> flist = []
> >>> for i in range(3):
> ... f = lambda x: x+i
> ... flist.append(f)
> ...
> >>> [f(1) for f in flist]
> [3, 3, 3]
> >>>
>
> What I expect is:
>
> >>> [f(1) for f in flist]
> [1,2,3]
Others have given the solution, but let's think about the why of it.
If a function in a dyanamic language (one that allows runtime
construction of functions) has a free variable (one not set within the
function), the question arises "When do we get (capture) the value of
the variable? At definition/construction time or at call/execution
time? Or, to put it another way, what do we capture? The (current)
value of the variable or its name (for later evaluation). Either
could be what we want.
Python lets you choose which capture method (evaluation time) you
want. Do nothing to get the name captured for later evaluation. Or
put v=v in the arglist to capture the current value and give it the
same name (w=v, to use a new name, is legal too, of course).
Terry J. Reedy
Terry Reedy, Oct 25, 2003
7. ### Georgy PrussGuest
"Terry Reedy" <> wrote in message news:...
>
> <...>
>
> Others have given the solution, but let's think about the why of it.
> If a function in a dyanamic language (one that allows runtime
> construction of functions) has a free variable (one not set within the
> function), the question arises "When do we get (capture) the value of
> the variable? At definition/construction time or at call/execution
> time? Or, to put it another way, what do we capture? The (current)
> value of the variable or its name (for later evaluation). Either
> could be what we want.
>
> Python lets you choose which capture method (evaluation time) you
> want. Do nothing to get the name captured for later evaluation. Or
> put v=v in the arglist to capture the current value and give it the
> same name (w=v, to use a new name, is legal too, of course).
>
> Terry J. Reedy
>
Very good and simple explanation! Thank you.
Georgy
Georgy Pruss, Oct 25, 2003
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| 164,141
|
Flat 50% Off on Assignment Bookings
• Internal Code :
• Subject Code : ECON1020
• University : Macquarie University
• Subject Name : Economics
## Principles of Economics
Budget constraint:
Wage cut by 20%...
Adding the second indifference curve.
Income and substitution effects:
Opportunity costs and wage decrease:
How effective is GDP to be an indicator of economic activity in the light of Covid-19?.
Consequentialist perspective about the payments to support
Deontological perspective about the payments to support
Perspective of substantive and procedural judgements of fairness.
References:
### Budget Constraint
For a typical worker like Jordan, who earns daily wages at the rate of \$30 per hour and assuming he can work up to 24 hours per day and earn 720 dollars per day, he would have a money income which he can spend on consumption goods and services plotted on the vertical axis. Similarly, the labour-leisure hours is plotted on the x-axis or the horizontal axis which shows his hours of free time from the origin to the right whereas labor hours is measured from the 24 hour point to the origin as shown in the diagram below(Mankiw, 2016).
Assuming that he would work for 24 hours a day and earn \$720, the vertical intercept is at \$720 on the y-axis and if we assume that he would spend all of his day in leisure, the horizontal intercept is at 24 representing 24 hours of a day. The graph below shows the labor leisure tradeoff of the laborer Jordan at the wage rate of \$30 per hour.
The total number of hours is given by T or 24 hours /day
The number of leisure hours is L
The number of working hours is T-L = h
The wage per hour is at w or \$30
The budget constraint is at
wh= w(T-L) = wT-wL
since the income from working is used to buy either consumption or leisure, we have
wT-wL = C (assuming the P price of consumption is at 1).
Since w is at 30, we have
30T-30L= C
### Optimal Income-Leisure Tradeoff
We can use indifference curve to explain Jordan’s choice between his daily wage measured as consumption in the y-axis and the leisure in the x- axis. The income earned by Jordan is earned by scarifying his leisure time to do the work for daily wages. When the leisure is sacrificed by a greater amount, the greater would be the work done and greater would his income which can be spent on consumption goods(Nicholson & Snyder, 2008). While income represents the person’s ability to purchase goods and services, leisure is yet another normal commodity which is enjoyed by the individual. In this regard, drawing an indifference curve IC1 as shown in the figure, we can see that it would provide satisfaction to the individual at point E where he works 14 hours per day and uses 10 hours of leisure as shown in the following diagram. Two assumptions about the optimal income-leisure combination for Jordan would be that he is free to work as many hours a day as he likes (24 hours maximum) and wage rate is the same \$30/hour irrespective of the number of hours Jordan chooses to work. Assuming that he works for 14 hours and earns \$420, we can see the equilibrium at a point where his budget constraint in Q1 and the indifference curve are tangent to each other as shown in the following diagram.
The budget constraint and the indifference curve are tangent to each other to make the optimal consumption level at
\$30 * 14 = \$420; where income = consumption at \$420
### Wage Cut by 20%
Due to the pandemic of Covid-19, the employers all around the world are facing hardship. So does the Jordan’s employer faces hardship and cuts the wages by 20%. This means that his wage is now at \$24 per hour (and not \$30 per hour) and he can earn a maximum of \$576 dollars with his 24 hours per day(Mankiw, 2016).
The total number of hours is given by T or 24 hours /day (this is the same as above)
Number of leisure hours is L (this would change according to substitution and income effect)
The number of working hours is T-L = h (this would also change)
The wage per hour is at w (this has changed to \$24 due to the wage cut by 20%)
The budget constraint is at
wh= w(T-L) = wT-wL
wT-wL = C (assuming the P price of consumption is at 1).
24T-24L = C
In the above diagram, DB is the new budget line, where with the given budget line, Jordan can earn \$576 maximum per day which is the intercept in the y-axis and x-axis intercept is the same at 24 (as he has 24 hours per day).
### Adding the Second Indifference Curve
Now we add a second indifference curve in the model developed in the previous sections. And we have IC2 as that second indifference curve. This indifference curve IC2 has a much lower utility than the IC1. This is one of the properties of indifference curves where the farther away the indifference curves are from the origin, the greater utility they have. Since IC1 is farther than the IC2, we can say that the utility that Jordan derives from the labor-leisure tradeoff that results in consumption has decreased due to the wage cut.
### Income and Substitution Effects
To find out the income and substitution effects of the wage decrease, we have drawn an imaginary line that is parallel to the original budget line AB but is tangent to the new indifference curve IC2 and this is the blue line shown in the following figure
In the above diagram, Jordan chooses point e before the wage cut where the original indifference curve IC1 is tangent to the budget line (AB). When there is a tax cut by 20%, the budget constraints moves inward to DB shown by the brown line in the figure and Jordan chooses a point g where the new budget line DB is tangent to a new indifference curve IC2. This choice by Jordan can be decomposed into the income effect which is shown by the movement from e to f and the substitution effect which is a movement from f to g. Point f is the income effect which is the combination of consumption and leisure that would have been chosen by Jordan if income that is reduced would have left Jordan with the same level of utility (IC2) (assuming that there has been no relative changes in the prices of consumption and leisure)(Investopedia, 2019). However, the movement from f to point g in the following figure represents the relative changes in prices of leisure vs. consumption, which having the same utility (utility held constant) which is referred to as substitution effect. We can see that the substitution effect is larger than the income effect which has resulted in Jordan increasing the amount of leisure (which is now relatively cheaper compared to his other good consumption).
### Opportunity Costs and Wage Decrease
The opportunity costs concept is defined as the income foregone in the next best alternative available. In the particular situation above, we can see that the substitution effect (movement from f to g) is larger and the result is that Jordan increases the amount of leisure he was enjoying previously despite the fact that his wages has decreased. This is because, Jordan considers leisure to be relatively cheaper when compared to consumption. The opportunity costs of decreasing wages is the additional time he can enjoy in leisure activities(Nicholson & Snyder, 2008).
### How Effective is GDP to be an Indicator of Economic Activity in The Light of Covid-19?
With the increasing spread of coronavirus across various nations, many governments across the world have taken various unprecedented measures to control the pandemic. The measures first focus of health as it is of primary importance. These governmental measures -particularly that of lockdown has resulted in widespread restrictions to go for work and has affected the mobility of workers, there has been financial market disorder, an erosion of confidence among businesses and households and also uncertain situations(The_Economist, 2020). The GDP measure as an economic indicator would certainly lower after the lockdown period in various countries. Though GDP is a easier measure of economic activity and can be calculated easily to reflect the aftermath of Covid-19, it is a very narrow indicator that would not show the quality of life and turmoil faced by people after Covid-19. It does not show the health care facilities created for Covid-19 prevention and treatment and unemployment created by this pandemic and it also does not account of inequalities faced by the pandemic. Though GDP is an good indicator of telling the policy makers whether their policies have been effective in creating employment and value for the economy; However, it could also be misleading based on the population of a country like China or India which might show a high value of GDP due to the vast number of people involving themselves in economic activity(The_Treasury, 2020).
### Consequentialist Perspective About the Payments to Support
According to the consequentialist perspective, the purpose of the payments support system is to ensure that the actions of these payments to support the families should avoid the bad consequences and maximize the good consequences for all. In this regard the payments support system by the government based on consequentialist perspective could be used to justify the resource allocation that would do the maximum good for the people (or minimise the harm created by lockdown) and an example would be to allocate the scarce resources that are available to save maximum lives possible(WHO, 2020). This is appropriate in allocating the scarce resources that would confer different benefits to the people (different groups of people).However the allocation principle should be justified to have different perspectives at different stages of scarce resources. For example when the allocation of ventilators on a first come first serve basis on a ethical value of equality would be justified at a time of little scarcity. However, as the resource of ventilators or even payment supports become increasingly scarce, it would be justified in the lines of prioritizing those who require it the most and when it becomes even more scarce the principle that would justify would maximise the benefit is more apt(Mandal et al., 2016).
### Deontological Perspective About the Payments to Support
According to the deontological perspective, any harm that would be created by the payments to support the families and old people is not unacceptable irrespective of the consequences it would create. The consdierations through which these measures and payment support systems are implemented are different for different people(Aus_Govt, 2020). However there needs to be equality which considers interests of each person equally unless there are good reasons to justify the different prioritizations of resources. In this context irrelevant characteristics such as race, ability, ethnicity, ability or creed should not be considered for these allocations of resources. Such allocation on the basis of deontological perspective may be appropriate in guiding the scarce resource allocation to groups of people who expect to derive the dame benefit from that payments for support provided by the government(Mandal et al., 2016).
### Perspective of Substantive and Procedural Judgements of Fairness
A fair process for the allocation of scarce resources would promote certain values like inclusiveness, transparency, consistency and accountability. In this transparent process of providing payment support systems to households, businesses and other vulnerable groups, the Australian government has provided the fact sheet in which it has announced about the first payment and the second payment from the Australian government and it has also provided the eligibility criteria for the same which will make it more transparent about the schemes announced by the government(WHO, 2020). Similarly those getting affected by the allocation decisions generally should be able to exert influence on the decision making process and this is not evident from the fact sheet whether those affected by the payment decisions were allowed to participate in the decision making process. Consistency signifies that those people from the same categories should be treated in the same way and favoritism should be avoided due to political or religious reasons. And from the fact sheet, those vulnerable groups are treated equally and consistently without any favoritism. And above all accountability suggests that those making these decisions should be made accountable for these decisions. Since the actions of the government are transparent enough, it could be taken that they are also accountable and answerable to the public in general about the payment support system they are providing(Aus_Govt, 2020).
### References for Principles of Economics
Aus_Govt. (2020). Payments to support households. https://treasury.gov.au/sites/default/files/2020-03/Fact_sheet-Payments_to_support_households_0.pdf
Investopedia. (2019). Understanding Income Effect vs. Substitution Effect. https://www.investopedia.com/ask/answers/041415/whats-difference-between-income-effect-and-substitution-effect.asp
Mandal, J., Ponnambath, D. K., & Parija, S. C. (2016). Utilitarian and deontological ethics in medicine. Tropical Parasitology, 6(1), 5–7. https://doi.org/10.4103/2229-5070.175024
Mankiw, G. N. (2016). Principles of Microeconomics (8th Editio). CENGAGE Learning Custom Publishing.
Nicholson, W., & Snyder, C. (2008). Microeconomic Theory: Basic Principles and Extensions. In Thomson South-Western (10th ed.). Thomson:South-Western. http://www.sidalc.net/cgi-bin/wxis.exe/?IsisScript=CIMMYT.xis&method=post&formato=2&cantidad=1&expresion=mfn=047436
The_Economist. (2020). A grim calculus - Covid-19 presents stark choices between life, death and the economy | Leaders | The Economist. The Economist. https://www.economist.com/leaders/2020/04/02/covid-19-presents-stark-choices-between-life-death-and-the-economy
The_Treasury. (2020). Economic Response to the Coronavirus | Treasury.gov.au. https://treasury.gov.au/coronavirus
WHO. (2020). Ethics and COVID-19: resource allocation and priority-setting. https://www.who.int/blueprint/priority-diseases/key-action/EthicsCOVID-19resourceallocation.pdf
Remember, at the center of any academic work, lies clarity and evidence. Should you need further assistance, do look up to our Economics Assignment Help
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# Algebra
posted by on .
Volume of a container?
A cubic shipping container had a volume of a^3 cubic meters. The height was decreased by a whole number of meters and the width was increased by a whole number of meters so that the volume of the container is now a^3+2a^2 3a cubic meters. By how many meters were the height and width changed?
| 84
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http://www.chegg.com/homework-help/questions-and-answers/voltage-v-applied-capacitor-network-shown-equivalent-capacitance-hint-assume-potential-dif-q113650
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A voltage V is applied to the capacitor network shown.
(a) What is the equivalent capacitance?
[Hint: Assume a potential differenceVab exists across the network as shown; writepotential differences for various pathways through the network froma to b in terms of the charges on the capacitors and thecapacitances.]
(b) Determine the equivalent capacitance if C2= C4 = 8.0 μF andC1 = C3 =C5 = 4.5 μF.
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You are Here: Home >< Maths
# Convergent Sequence watch
Announcements
1. "Let a be a real number. Show that (a^n)/n! converges to 0".
This looks very simple and I'm fairly sure it can be done just from the definition of convergence (the "epsilon" one, as opposed to "it gets very small"), but I think I'm missing something simple.
Obviously the case where |a|<1 is easy, but for the rest I'm not sure how to get rid of the a^n.
Thanks for any help.
2. Show that the sequence is strictly decreasing. Then show that for large enough n, a^n/n! < something which tends to 0.
3. |a^n/n!| is decreasing for all sufficiently large n. [You don't need to say "strictly" but you can if you like.] Since the sequence (|a^n/n!|) is bounded below by 0, it converges. Let L be the limit:
|a^n/n!| -> L as n -> infinity . . . . . (*)
Then
[|a|/(n + 1)] |a^n/n!| = | a^(n + 1)/(n + 1)! | . . . . . . (**)
and, using (*) twice,
LHS of (**) -> 0 * L = 0 as n -> infinity
RHS of (**) -> L as n -> infinity
So L = 0,
|a^n/n!| -> 0 as n -> infinity
a^n/n! -> 0 as n -> infinity
--
Alternatively, you can view the result as a consequence of the convergence of the series exp(a) = (sum for n = 0 to infinity) a^n/n!.
4. Thanks for the ideas. They seem to be fine. I was expecting methods more like starting witth |a^n/n! - 0| and showing it is less than or equal to epsilon for all epsilon >0, but I was maybe being too closed minded!
5. (Original post by sat)
"Let a be a real number. Show that (a^n)/n! converges to 0".
This looks very simple and I'm fairly sure it can be done just from the definition of convergence (the "epsilon" one, as opposed to "it gets very small"), but I think I'm missing something simple.
Obviously the case where |a|<1 is easy, but for the rest I'm not sure how to get rid of the a^n.
Thanks for any help.
can you not just say:
for any real a, there is always an integer b (say) where b is the lowest integer such that b>a. and the same for b, with c, and the same for c. there are infinitely many integers above a.
as n tends to infinity, more of these integers are being included in the multiplication which makes up the denominator
then the logic follows:
- as n tends to infinity, n! will eventually have a factor b (as defined above)
- once this point is passed, the value of a^n/n! will change such that:
(a) the numerator is multipled by a
(b) the denominator is multipled by an integer x such that x>a
since you have done the case for a<1, x>a>1 and so the denominator will increase more than the numerator. repeat to infinity and the ratio will tend to zero..
not quite rigourous, but perhaps with more numbers and less words is the theory right>
6. ok try again:
prove that as n tends to infinity, a^n/n! tends to zero, for a>1.
a is a real number above zero. a+1 is a real number 1 larger than a. define integer x as: a < x =< a+1
as n tends to infinity, (n is an integer I hope), there will be a point where n=x.
at every step from a^n/n! to a^(n+1)/(n+1)!, the step is effectively going from:
a*a^n/n!(n+1) = (a^n/n!)[a/(n+1)]
now for a > (n+1), this step will increase the expression, as a/(n+1) > 1. But as n tends to infinity, past the point to where a < (n+1), the expression before the step shown above is larger than the expression after, since a/(n+1) < 1. therefore the expression is decreasing - after this point the expression will be continually be multipled by a number (a/(n+1)) which itself tends to zero as n tends to infinity and a remains a constant real. as this is continued to infinity, but the factor (a/(n+1)) never falls below zero [as neither a nor n is negative].
erm.. yeah. I kinda fell into the realm of wordyness again.
7. (Original post by sat)
Thanks for the ideas. They seem to be fine. I was expecting methods more like starting witth |a^n/n! - 0| and showing it is less than or equal to epsilon for all epsilon >0, but I was maybe being too closed minded!
You asked for it! [Credit to mik1a - the idea of my proof is in his post. But I have succeeded in making the answer look incredibly complicated.]
Assume a is not zero.
Let epsilon > 0.
Define
M
= ceiling(|a|)
= smallest integer that is at least |a|
K = |a|^M/M! > 0
N = M + max{ 1, ceiling[ln(epsilon/K) / ln(M/(M + 1))] }
Then for all n >= N we have n >= M + 1 and
|a|^n/n!
= [|a|^M/M!] (product from i = 1 to n - M) |a|/(M + i)
= K (product from i = 1 to n - M) |a|/(M + i)
<= K (product from i = 1 to n - M) M/(M + i)
<= K (product from i = 1 to n - M) M/(M + 1)
= K [M/(M + 1)]^(n - M)
<= K [M/(M + 1)]^[ln(epsilon/K) / ln(M/(M + 1))]
= K * epsilon/K
= epsilon
8. (Original post by Jonny W)
Assume a is not zero.
Let epsilon > 0.
Define
M
= ceiling(|a|)
= smallest integer that is at least |a|
K = |a|^M/M! > 0
N = M + max{ 1, ceiling[ln(epsilon/K) / ln(M/(M + 1))] }
Then for all n >= N we have n >= M + 1 and
|a|^n/n!
= [|a|^M/M!] (product from i = 1 to n - M) |a|/(M + i)
= K (product from i = 1 to n - M) |a|/(M + i)
<= K (product from i = 1 to n - M) M/(M + 1)
= K [M/(M + 1)]^(n - M)
<= K [M/(M + 1)]^[ln(epsilon/K) / ln(M/(M + 1))]
= K * epsilon/K
= epsilon
Is this analysis?
9. (Original post by Euclid)
Is this analysis?
Anything that starts "let epsilon > 0" is analysis.
10. (Original post by Jonny W)
Anything that starts "let epsilon > 0" is analysis.
Good stuff! Im starting my very first Analysis course tomorrow !
11. A person who can, within a year, solve x² - 92y² = 1 is a mathematician
how the hell... I've been trying this all afternoon rearranging stuff and stuff. can't see any kind of factorisation
and plotting x against y looke nightmarish... is it a circle? no it cant be...
argh!
12. (Original post by mik1a)
A person who can, within a year, solve x² - 92y² = 1 is a mathematician
how the hell... I've been trying this all afternoon rearranging stuff and stuff. can't see any kind of factorisation
and plotting x against y looke nightmarish... is it a circle? no it cant be...
argh!
Precisely! It's very difficult, hence within a year. I haven't actually tried it myself, but I might do so now that it has caught attention
13. It's definately not a circle. At first I thought it was a hyperbola but I keep getting two different hyperbola's for the same equation - that's rather odd
I think the question is supposed to be aiming at finding non-zero integer solutions, because by inspection two solutions are x=1 y=0 and x=-1 y=0, hence making the question sound stupid.
Euclid
14. (Original post by Euclid)
It's definately not a circle. At first I thought it was a hyperbola but I keep getting two different hyperbola's for the same equation - that's rather odd
I think the question is supposed to be aiming at finding non-zero integer solutions, because by inspection two solutions are x=1 y=0 and x=-1 y=0, hence making the question sound stupid.
Euclid
Its a Pell equation, there's lots of information on methods to solve it on the net.
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# Lecture5 - Data-Flow Analysis and Optimizations CSE 501...
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Unformatted text preview: Data-Flow Analysis and Optimizations CSE 501 Lecture 5 April 13, 2009 Questions? • Any critiques ready? • Earlier lectures? • Homework? • Reading? Data-Flow Analysis DFA is a collection of techniques for compile-time reasoning about the run-time fow of values. We use DFA to ¡nd opportunities and prove safety • Not an end unto itself Usually formulated as a set of simultaneous equations • Sets attached to nodes an edges in a graph Desired result is usually a meet over all paths solution: • What is true on every path from the entry? • Can this happen on any path from the entry? DFA equations - Dominance Last lecture, we glanced at our frst DFA equations. Initialization: Fixed point: N is the set o¡ all nodes in the CFG. Domain is nodes (basic blocks) in the CFG we ’ re analyzing Dom( n ) = N , " n Dom( n ) = { n } " ( Dom( p ) p # pred( n ) I ) DFA equations - Liveness DFA equations are only slightly more complex. Initialization: Fixed point: Kill (b) is the set of names de¡ned in b . UE (b) is the set of names used in b before being de¡ned in b . Domain might be registers, locals, globals, … LiveIn( n ) = " , # n LiveOut( b ) = LiveIn( s ) s \$ succ( b ) U LiveIn( b ) = UE( b ) % (LiveOut( b ) & Kill( b )) A Round-Robin Data-Flow Solver for b in N Dom(b) = N do changed = false for b in N temp = if Dom(b) != temp changed = true Dom(b) = temp while changed { b } " ( Dom ( p ) p # pred( b ) I ) Ordering When we write for b in N any ordering will converge on the same result, but some orderings will converge quicker. For forward problems (dominators), we use a topsort order (where topsort is topological sort). For backward problems (liveness), we use a reverse topsort order. Forward and Backward Problems When we solve forward problems, information ¡ows generally forward in the CFG. Recall the de¢nition of the dominators problem The solution for each block refers to its predecessors in the CFG....
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## This note was uploaded on 02/09/2012 for the course CSE 403 taught by Professor Staff during the Spring '08 term at University of Washington.
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Lecture5 - Data-Flow Analysis and Optimizations CSE 501...
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# A business might worry that pricing of one product might impact demand for another product that is
## Question:
A business might worry that pricing of one product might impact demand for another product that is also sold by the same business.
Here we€™ll explore conditions under which such worries are more or less important before turning to some specific examples.
A: Suppose first that we label the two goods that a firm sells as simply x1 and x2. The firm considers putting a discount of δ on the price of x1€”a discount that lowers the price from p1 to (1ˆ’δ)p1.
(a) For a consumer who budgets I for consumption of x1 and x2, illustrate the budget before and after the discount is put in place.
(b) Assuming that tastes are homothetic, derive the relationship between δ on the vertical axis and x1 on the horizontal axis.
(c) Now derive the relationship between δ and x2 €“ can you tell if it slopes up or down? What does your answer depend on?
(d) Suppose that x1 is printers and x2 is printer cartridges produced by the same company. Compare this to the case where x1 is Diet Coke and x2 is Zero Coke. In which case is there a more compelling case for discounts on x1?
B: Suppose that tastes are defined by u(x1, x2) = xα1 x2(1ˆ’α).
(a) Derive the demand functions for x1 and x2 as a function of prices, I and δ.
(b) Are these upward or downward sloping in δ?
(c) Under the general specification of tastes as
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Phase of a plane wave is an invariant quantity
As described by Jackson's Classical Eletrodynamics [1], the phase of a plane wave is an invariant quantity for all coordinate frames. Jackson justifies this statement using the fact that the phase of a plane wave is proportional to the number of wave crests that have passed by the observer.
I have several problems to find a link between these two statements. How is the invariance of the phase linked with the fact that it's a wave-crest counter?
Using the Lorentz transformation laws for the 4-wavevector $(\omega/c, \vec{k})$ I can reach a simple expression for the frequency seen by a rest frame,
$$\begin{eqnarray} \omega'=\omega \end{eqnarray} \gamma\left(1-\beta\cos\theta\right)$$
where $\theta$ is the angle between wave propagation's vector $\vec{k}$ and the velocity vector of the motion system $\vec{\beta}$. This equation states that the frequency varies between different reference systems, so it changes the number of wave crests that pass by the observer in function of whether he is placed in the rest reference system or in one that is in motion.
So, I don't understand the Jackson assertion.
1. Jackson, John David. Classical Electrodynamics. John Wiley & Sons, 2007. APA p. 519.
• The way Jackson's text sees things, "the number of wave crests that have passed by the observer" is what the phase is, at its core, so in some ways it's a tautology. That's not to say that it's trivial, but to be able to show the relevant connections, we need to know where you are starting from, so ─ how are you defining the phase of a plane wave, to begin with? Jul 6, 2017 at 14:53
• @Luc is right, i considered $\omega$ as counter of wave crests , but it's wrong because it's a number in time unity. Jul 6, 2017 at 15:25
• Ah, I see. No, the problem is that you're confusing the phase ($\phi=\omega t-\vec k\cdot\vec x$, dimensionless) with the angular frequency ($\omega$, with dimensions of inverse time). The counter of wave crests is $\phi$, not $\omega$. Jul 6, 2017 at 15:53
• Another simple way to see this, in the simple case of electromagnetic waves, is that there is a transformation law for the fields $(E,B)\rightarrow(E',B')$, and this transformation law is such that $(0,0)\rightarrow(0,0)$. Therefore all observers will agree on the events in spacetime at which the fields both vanish.
– user4552
Jul 6, 2017 at 17:08
• Sorry @BenCrowell, your example is not clear. The most easy situation is for $E$, $B\perp \beta$. For this fields the trasformation law become: $E'_\parallel = E_\parallel$, $E'_\perp =\gamma(E_\perp+\beta \times B )$ and $B'_\parallel = B_\parallel$, $B'_\perp =\gamma(B_\perp-\beta \times E )$, so you can't direct conclude that there is a transformation law such that all field vanished. No? Jul 6, 2017 at 21:49
Think of the unit of $$\omega$$. It's in $$s^{-1}$$, clearly showing that $$\omega$$ is not a count but a count per unit of time, if you want to interpret it as such. Now think of a short signal for which the wave has only a finite number of periods, let's say it propagates along the x-axis for simplicity, and that we consider observers moving along only the x-axis, slower than the celerity of the wave. Any such observer will see the following. For a while, the medium is undisturbed. Then the medium will "rise" and reach a maximum, then "fall" and reach a minimum, and carry on doing that for a while, after which the medium will become undisturbed again. Every single observer will agree on the number of maxima and minima he has seen.
The phase of a light wave is given by $$K^\mu x_\mu$$ where $$K_\mu = (\omega,\vec{k})$$ and $$x_\mu = (t,\vec{r})$$, note that $$c=1$$, so that $$K^\mu x_\mu = \vec{k}\cdot\vec{r}-\omega t \;.$$ The dot product is Lorentz invariant by virtue of being a Lorentz Scalar. (Easily provable using transformation laws of vectors and covectors)
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07 Gibbs Free Energy
# 07 Gibbs Free Energy - Water Ice Problem Calculate the...
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1 1 Water – Ice Problem Calculate the change in entropy of water, S system , when 2 mol of water freezes at 0 o C (273 K). For ice, H fusion = 6.01 kJ/mol S = S final – S initial = q/T A) + 44.0 J/K B) – 44.0 J/K C) + 0.022 J/K D) – 0.022 J/K 2 Water (the system) Heat flow: system (water) to surroundings (freezer) During the freezing process T water is constant (0°C) S = q/T and q = -n H fus H fus = 6.01 kJ/mol q H2O is negative since energy flows out of the system. q H2O = -2 x 6.01 x 10 3 J = -1.202 x 10 4 J S H2O = q H2O /T H2O = -1.202x10 4 / 273 = -44.0 J K -1
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2 3 When 2 mol of water freezes, S system = - 44 J/K. The entropy decreases! Makes sense liquid solid. Water freezes spontaneously, but we said S > 0 for a spontaneous process. What gives here? Need to consider the total entropy not just the entropy of the system. 4 Total Entropy Change When water freezes, energy is transferred to the surroundings. Which means that the energy is dispersed over more states in the surroundings. So the entropy of the surroundings increases. S total = S system + S surroundings S total also called S universe
3 5 Calculate the change in entropy of the freezer (surroundings), S surroundings , when 2 mol of water freezes at 0 o C (273 K). The temperature of the freezer is – 15 o C (258 K) For ice, H fusion = 6.01 kJ/mol S = S final – S initial = q/T A) + 46.6 J/K B) – 46.6 J/K C) + 44.0 J/K D) – 44.0 J/K 6 Freezer (the surroundings) The freezer absorbs energy at its temp. (-15°C). Absorbs energy so q freezer > 0. q freezer = -q H2O so q freezer = 2mol (6.01 kJ/mol) q freezer = 1.202 x 10 4 J S freezer = q freezer /T freezer = 1.202 x 10 4 / 258 = 46.6 J K -1
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4 7 The Universe (total) S universe = S system + S surroundings S universe = - 44.0 J/K + 46.6 J/K = + 2.6 J/K
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по следите на логаритмичната функция
TCH
Задача 1. Постройте графиката на функцията y = ln x .
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Решение.
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Задача 2. Постройте графиката на функцията: а) y = log 3 x б) y = log 6 x в) y = log 0,8 x .
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### Задача 3. Сравнете графиките на функциите y = log a x и y = log 1/a x .
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Задача 4. Изследвайте за различни стойности на к y = log a k.x и y = log a (x+k) .
:Можете да използвате отговора на задача 1, да движите графиката и наблюдавате уравнението й.
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DID YOU KNOW:
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# Printable Mad Lib Math Activity - All Exponent Rules (8EE1)
7th - 12th, Homeschool
Subjects
Standards
Resource Type
Formats Included
• PPTX
Pages
21 pages
### Description
Your students will have a blast and practice their math skills while doing this mad lib math activity. Low prep for you and fun for your students!
All you need to do is print copies of the 15 slides and the answer sheets and then hang them around your room. Everything is provided for you, including an answer sheet for easy grading.
Students work through all 15 slides. Each slide consists of a problem and four possible choices. Students select their answers and then write down the corresponding word into the appropriate spaces on the story sheet. The answers will create a silly story. If the story is correct you, and your students, will know they have done the problems correctly.
**Please note this is a printable version of the digital mad lib math activity in my TpT store**
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Total Pages
21 pages
Included
Teaching Duration
30 minutes
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### Standards
to see state-specific standards (only available in the US).
Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 3² × (3⁻⁵) = (3⁻³) = 1/3³ = 1/27.
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StudentShare solutions
# HW12 - Math Problem Example
Only on StudentShare
## Extract of sampleHW12
Prepare the journal entry to record the allocation of net income. (List multiple debit/credit entries in descending order of amount.)
Assume the partnership income-sharing agreement calls for income to be divided with a salary of \$30,000 to Guillen and \$25,000 to Williams, with the remainder divided 45% to Guillen and 55% to Williams. Prepare the journal entry to record the allocation of net income. (List multiple debit/credit entries in descending order of amount.)
Assume the partnership income-sharing agreement calls for income to be divided with a salary of \$40,000 to Guillen and \$35,000 to Williams, interest of 10% on beginning capital, and the remainder divided 50%-50%. Prepare the journal entry to record the allocation of net income. (List multiple debit/credit entries in descending order of amount.)
The Best Company at December 31 has cash \$20,000, noncash assets \$100,000, liabilities \$55,000, and the following capital balances: Rodriguez \$45,000 and Escobedo \$20,000. The firm is liquidated, and \$110,000 in cash is received for the noncash assets. Rodriguez and Escobedo income ratios are 60% and 40%, respectively.
The Best Company at December 31 has cash \$20,000, noncash assets \$100,000, liabilities \$55,000, and the following capital balances: Rodriguez \$45,000 and E ...Show more
## Summary
F. Calvert and G. Powers have capital balances on January 1 of \$50,000 and \$40,000, respectively. The partnership income-sharing agreement provides for (1) annual salaries of \$20,000 for Calvert and \$12,000 for Powers, (2) interest at 10% on beginning capital balances, and (3) remaining income or loss to be shared 60% by Calvert and 40% by Powers.
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# American Institute of Mathematical Sciences
doi: 10.3934/amc.2020049
## The values of two classes of Gaussian periods in index 2 case and weight distributions of linear codes
1 School of Mathematics and Statistics, Zaozhuang University, Zaozhuang, Shandong, 277160, China 2 State Key Laboratory of Cryptology, P. O. Box 5159, Beijing, 100878, China 3 Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing, Jiangsu, 211100, China
Received September 2018 Revised November 2019 Published January 2020
Fund Project: The paper was supported by National Natural Science Foundation of China under Grants 11601475, 61772015, the foundation of Science and Technology on Information Assurance Labo- ratory under Grant KJ-17-010, china, and the foundation of innovative Science and technology for youth in universities of Shandong Province China under Grant 2019KJI001.
Let
$l$
be a prime with
$l\equiv 3\pmod 4$
and
$l\ne 3$
,
$N = l^m$
for
$m$
a positive integer,
$f = \phi(N)/2$
the multiplicative order of a prime
$p$
modulo
$N$
, and
$q = p^f$
, where
$\phi(\cdot)$
is the Euler-function. Let
$\alpha$
be a primitive element of a finite field
$\Bbb F_{q}$
,
$C_0^{(N,q)} = \langle \alpha^N\rangle$
a cyclic subgroup of the multiplicative group
$\Bbb F_q^*$
, and
$C_i^{(N,q)} = \alpha^i\langle \alpha^N\rangle$
the cosets,
$i = 0,\ldots, N-1$
. In this paper, we use Gaussian sums to obtain the explicit values of
$\eta_i^{(N, q)} = \sum_{x \in C_i^{(N,q)}}\psi(x)$
,
$i = 0,1,\cdots, N-1$
, where
$\psi$
is the canonical additive character of
$\Bbb F_{q}$
. Moreover, we also compute the explicit values of
$\eta_i^{(2N, q)}$
,
$i = 0,1,\cdots, 2N-1$
, if
$q$
is a power of an odd prime
$p$
.
As an application, we investigate the weight distribution of a
$p$
-ary linear code:
$\mathcal{C}_{D} = \{C = ( \operatorname{Tr}_{q/p}(c x_1), \operatorname{Tr}_{q/p}(cx_2),\ldots, \operatorname{Tr}_{q/p}(cx_n)):c\in \Bbb{F}_{q}\},$
where its defining set
$D$
is given by
$D = \{x\in \Bbb{F}_{q}^{*}: \operatorname{Tr}_{q/p}(x^{\frac{q-1}{l^{m}}}) = 0\}$
and
$\operatorname{Tr}_{q/p}$
denotes the trace function from
$\Bbb F_{q}$
to
$\Bbb F_p$
.
Citation: Fengwei Li, Qin Yue, Xiaoming Sun. The values of two classes of Gaussian periods in index 2 case and weight distributions of linear codes. Advances in Mathematics of Communications, doi: 10.3934/amc.2020049
##### References:
[1] L. Baumert and J. Mykkeltveit, Weight distributions of some irreducible cyclic codes, DSN Progr. Rep., 16 (1973), 128-131. Google Scholar [2] B. Berndt, R. Evans and K. Williams, Gauss and Jacobi Sums, New York, John Wiley & Sons Company, 1997. Google Scholar [3] H. Cohen, A Course in Computational Algebraic Number Theory, Graduate Texts in Mathematics, 138. Springer-Verlag, Berlin, 1993. doi: 10.1007/978-3-662-02945-9. Google Scholar [4] C. S. Ding and J. Yang, Hamming weights in irreducible cyclic codes, Discrete Math., 313 (2013), 434-446. doi: 10.1016/j.disc.2012.11.009. Google Scholar [5] C. S. Ding, Y. Liu, C. L. Ma and L. W. Zeng, The weight distributions of the duals of cyclic codes with two zeros, IEEE Trans. Inform. Theory, 57 (2011), 8000-8006. doi: 10.1109/TIT.2011.2165314. Google Scholar [6] C. S. Ding, C. L. Li, N. Li and Z. C. Zhou, Three-weight cyclic codes and their weight distributions, Discrete Math., 339 (2016), 415-427. doi: 10.1016/j.disc.2015.09.001. Google Scholar [7] T. Feng and Q. Xiang, Strongly regular graphs from unions of cyclotomic classes, Journal of Combinatorial Theory Series B, 102 (2012), 982-995. doi: 10.1016/j.jctb.2011.10.006. Google Scholar [8] Z. L. Heng and Q. Yue, A class of binary linear codes with at most three weights, IEEE Commun. Lett., 19 (2015), 1488-1491. doi: 10.1109/LCOMM.2015.2455032. Google Scholar [9] Z. L. Heng and Q. Yue, Two classes of two-weight linear codes, Finite Fields Appl., 38 (2016), 72-92. doi: 10.1016/j.ffa.2015.12.002. Google Scholar [10] Z. L. Heng and Q. Yue, Evaluation of the Hamming weights of a class of linear codes based on Gauss sums, Des. Codes Cryptogr., 83 (2017), 307-326. doi: 10.1007/s10623-016-0222-7. Google Scholar [11] L. Q. Hu, Q. Yue and M. H. Wang, The linear complexity of Whiteman's generalize cyclotomic sequences of period $p^{m+1}q^{n+1}$, IEEE Trans. Inform. Theory, 58 (2012), 5534-5543. doi: 10.1109/TIT.2012.2196254. Google Scholar [12] P. Langevin, Caluls de certaines sommes de Gauss, J. Number theory, 63 (1997), 59-64. doi: 10.1006/jnth.1997.2078. Google Scholar [13] C. J. Li and Q. Yue, The Walsh transform of a class of monomial functions and cyclic codes, Cryptogr. Commun., 7 (2015), 217-228. doi: 10.1007/s12095-014-0109-2. Google Scholar [14] C. J. Li, Q. Yue and F. W. Li, Weight distributions of cyclic codes with respect to pairwise coprime order elements, Finite Fields Appl., 28 (2014), 94-114. doi: 10.1016/j.ffa.2014.01.009. Google Scholar [15] F. W. Li, Q. Yue and F. M. Liu, The weight distribution of a class of cyclic codes containing a subclass with optimal parameters, Finite Fields Appl., 45 (2017), 183-202. doi: 10.1016/j.ffa.2016.12.004. Google Scholar [16] R. Lidl and H. Niederreiter, Finite Fields, Encyclopedia of Mathematics and its Applications, 20. Cambridge University Press, Cambridge, 1997. Google Scholar [17] Y. W. Liu and Z. H. Liu, On some classes of codes with a few weights, Adv. Math. Commun., 12 (2018), 415-428. doi: 10.3934/amc.2018025. Google Scholar [18] J. Q. Luo and K. Q. Feng, On the weight distribution of two classes of cyclic codes, IEEE Trans. Inf. Theory, 54 (2008), 5332-5344. doi: 10.1109/TIT.2008.2006424. Google Scholar [19] G. McGuire, On three weights in cyclic codes with two zeros, Finite Fields Appl., 10 (2004), 97-104. doi: 10.1016/S1071-5797(03)00045-5. Google Scholar [20] G. Myerson, Period polynomials and Gauss sums for finite fields, Acta Arith., 39 (1981), 251-264. doi: 10.4064/aa-39-3-251-264. Google Scholar [21] T. Storer, Cyclotomy and Difference Sets, Lectures in Advanced Mathematics, No. 2 Markham Publishing Co., Chicago, Ill. 1967. Google Scholar [22] Q. Y. Wang, K. L. Ding, D. D. Lin and R. Xue, A kind of three-weight linear codes, Cryptogr. Commun., 9 (2017), 315-322. doi: 10.1007/s12095-015-0180-3. Google Scholar [23] Q. Y. Wang, K. L. Ding and R. Xue, Binary linear codes with two weights, IEEE Commun. Lett., 19 (2015), 1097-1100. doi: 10.1109/LCOMM.2015.2431253. Google Scholar [24] X. Q. Wang, D. B. Zheng, L. Hu and X. Y. Zeng, The weight distributions of two classes of binary cyclic codes, Finite Fields Appl., 34 (2015), 192-207. doi: 10.1016/j.ffa.2015.01.012. Google Scholar [25] M. S. Xiong, The weight distributions of a class of cyclic codes, Finite Fields Appl., 18 (2012), 933-945. doi: 10.1016/j.ffa.2012.06.001. Google Scholar [26] J. Yang and L. L. Xia, Complete solving of explicit evaluation of Gauss sums in the index 2 case, Sci. China Math., 53 (2010), 2525-2542. doi: 10.1007/s11425-010-3155-z. Google Scholar [27] J. Yang, M. S. Xiong, C. S. Ding and J. Q. Luo, Weight distribution of a class of cyclic codes with arbitrary number of zeros, IEEE Trans. Inform. Theory, 59 (2013), 5985-5993. doi: 10.1109/TIT.2013.2266731. Google Scholar [28] S. D. Yang, X. L. Kong and C. M. Tang, A construction of linear codes and their complete weight enumerators, Finite Fields Appl., 48 (2017), 196-226. doi: 10.1016/j.ffa.2017.08.001. Google Scholar [29] S. D. Yang and Z.-A. Yao, Complete weight enumerators of a class of linear codes, Discrete Math., 340 (2017), 729-739. doi: 10.1016/j.disc.2016.11.029. Google Scholar [30] Z. C. Zhou and C. S. Ding, A class of three-weight cyclic codes, Finite Fields Appli., 25 (2014), 79-93. doi: 10.1016/j.ffa.2013.08.005. Google Scholar [31] Z. C. Zhou, A. X. Zhang and C. S. Ding, The weight enumerator of three families of cyclic codes, IEEE Trans. Inf. Theory, 59 (2013), 6002-6009. doi: 10.1109/TIT.2013.2262095. Google Scholar
show all references
##### References:
[1] L. Baumert and J. Mykkeltveit, Weight distributions of some irreducible cyclic codes, DSN Progr. Rep., 16 (1973), 128-131. Google Scholar [2] B. Berndt, R. Evans and K. Williams, Gauss and Jacobi Sums, New York, John Wiley & Sons Company, 1997. Google Scholar [3] H. Cohen, A Course in Computational Algebraic Number Theory, Graduate Texts in Mathematics, 138. Springer-Verlag, Berlin, 1993. doi: 10.1007/978-3-662-02945-9. Google Scholar [4] C. S. Ding and J. Yang, Hamming weights in irreducible cyclic codes, Discrete Math., 313 (2013), 434-446. doi: 10.1016/j.disc.2012.11.009. Google Scholar [5] C. S. Ding, Y. Liu, C. L. Ma and L. W. Zeng, The weight distributions of the duals of cyclic codes with two zeros, IEEE Trans. Inform. Theory, 57 (2011), 8000-8006. doi: 10.1109/TIT.2011.2165314. Google Scholar [6] C. S. Ding, C. L. Li, N. Li and Z. C. Zhou, Three-weight cyclic codes and their weight distributions, Discrete Math., 339 (2016), 415-427. doi: 10.1016/j.disc.2015.09.001. Google Scholar [7] T. Feng and Q. Xiang, Strongly regular graphs from unions of cyclotomic classes, Journal of Combinatorial Theory Series B, 102 (2012), 982-995. doi: 10.1016/j.jctb.2011.10.006. Google Scholar [8] Z. L. Heng and Q. Yue, A class of binary linear codes with at most three weights, IEEE Commun. Lett., 19 (2015), 1488-1491. doi: 10.1109/LCOMM.2015.2455032. Google Scholar [9] Z. L. Heng and Q. Yue, Two classes of two-weight linear codes, Finite Fields Appl., 38 (2016), 72-92. doi: 10.1016/j.ffa.2015.12.002. Google Scholar [10] Z. L. Heng and Q. Yue, Evaluation of the Hamming weights of a class of linear codes based on Gauss sums, Des. Codes Cryptogr., 83 (2017), 307-326. doi: 10.1007/s10623-016-0222-7. Google Scholar [11] L. Q. Hu, Q. Yue and M. H. Wang, The linear complexity of Whiteman's generalize cyclotomic sequences of period $p^{m+1}q^{n+1}$, IEEE Trans. Inform. Theory, 58 (2012), 5534-5543. doi: 10.1109/TIT.2012.2196254. Google Scholar [12] P. Langevin, Caluls de certaines sommes de Gauss, J. Number theory, 63 (1997), 59-64. doi: 10.1006/jnth.1997.2078. Google Scholar [13] C. J. Li and Q. Yue, The Walsh transform of a class of monomial functions and cyclic codes, Cryptogr. Commun., 7 (2015), 217-228. doi: 10.1007/s12095-014-0109-2. Google Scholar [14] C. J. Li, Q. Yue and F. W. Li, Weight distributions of cyclic codes with respect to pairwise coprime order elements, Finite Fields Appl., 28 (2014), 94-114. doi: 10.1016/j.ffa.2014.01.009. Google Scholar [15] F. W. Li, Q. Yue and F. M. Liu, The weight distribution of a class of cyclic codes containing a subclass with optimal parameters, Finite Fields Appl., 45 (2017), 183-202. doi: 10.1016/j.ffa.2016.12.004. Google Scholar [16] R. Lidl and H. Niederreiter, Finite Fields, Encyclopedia of Mathematics and its Applications, 20. Cambridge University Press, Cambridge, 1997. Google Scholar [17] Y. W. Liu and Z. H. Liu, On some classes of codes with a few weights, Adv. Math. Commun., 12 (2018), 415-428. doi: 10.3934/amc.2018025. Google Scholar [18] J. Q. Luo and K. Q. Feng, On the weight distribution of two classes of cyclic codes, IEEE Trans. Inf. Theory, 54 (2008), 5332-5344. doi: 10.1109/TIT.2008.2006424. Google Scholar [19] G. McGuire, On three weights in cyclic codes with two zeros, Finite Fields Appl., 10 (2004), 97-104. doi: 10.1016/S1071-5797(03)00045-5. Google Scholar [20] G. Myerson, Period polynomials and Gauss sums for finite fields, Acta Arith., 39 (1981), 251-264. doi: 10.4064/aa-39-3-251-264. Google Scholar [21] T. Storer, Cyclotomy and Difference Sets, Lectures in Advanced Mathematics, No. 2 Markham Publishing Co., Chicago, Ill. 1967. Google Scholar [22] Q. Y. Wang, K. L. Ding, D. D. Lin and R. Xue, A kind of three-weight linear codes, Cryptogr. Commun., 9 (2017), 315-322. doi: 10.1007/s12095-015-0180-3. Google Scholar [23] Q. Y. Wang, K. L. Ding and R. Xue, Binary linear codes with two weights, IEEE Commun. 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Weight distribution of the code in Theorem 5.2
Weight Frequency 0 1 $\frac{p-1}p(q-\frac{q-1}{l^{m-1}}+\eta_0^{(l^{m-1},q)})$ $\frac{q-1}{l^{m-1}}$ $\frac{p-1}p(q-\frac{q-1}{l^{m-1}}+\eta_i^{(l^{m-1},q)}),i = l^{m-1-k}u, u\in H_k^{(0)}$ $\frac{q-1}{l^{m-1}}\cdot\frac{\phi(l^k)}2$ $\frac{p-1}p(q-\frac{q-1}{l^{m-1}}+\eta_{i'}^{(l^{m-1},q)}),i' = l^{m-1-k}u, u\in H_k^{(1)}$ $\frac{q-1}{l^{m-1}}\cdot\frac{\phi(l^k)}2$ $k = 1,2,\ldots, m-1$
Weight Frequency 0 1 $\frac{p-1}p(q-\frac{q-1}{l^{m-1}}+\eta_0^{(l^{m-1},q)})$ $\frac{q-1}{l^{m-1}}$ $\frac{p-1}p(q-\frac{q-1}{l^{m-1}}+\eta_i^{(l^{m-1},q)}),i = l^{m-1-k}u, u\in H_k^{(0)}$ $\frac{q-1}{l^{m-1}}\cdot\frac{\phi(l^k)}2$ $\frac{p-1}p(q-\frac{q-1}{l^{m-1}}+\eta_{i'}^{(l^{m-1},q)}),i' = l^{m-1-k}u, u\in H_k^{(1)}$ $\frac{q-1}{l^{m-1}}\cdot\frac{\phi(l^k)}2$ $k = 1,2,\ldots, m-1$
Weight distribution of the code in Theorem 5.3 $(\frac{-1+\sqrt {-l}}2\equiv 0\pmod {\mathcal P_1})$
Weight Frequency $0$ $1$ $\frac{(p-1)(2l^mq-(l+1)(q-1))}{2pl^m}+\frac{p-1}p\eta_{0}^{(l^m, q)}+\frac{(l-1)(p-1)}{2p}\eta_{i'}^{(l^m,q)}$ $\frac{q-1}{l^{m}}$ $i'/l^{m-1}\in H_1^{(1)}$ $\frac{(p-1)(2l^mq-(l+1)(q-1))}{2pl^m}+\frac{p-1}p\eta_0^{(l^m, q)}+\frac{(l+1)(p-1)}{4p}\eta_{i}^{(l^m, q)}+\frac{(l-3)(p-1)}{4p}\eta_{i'}^{(l^m,q)}$ $\frac{q-1}{l^{m}}\cdot\frac{l-1}2$ $i/l^{{m-1}}\in H_1^{(0)}, i'/l^{m-1}\in H_1^{(1)}$ $\frac{(p-1)(2l^mq-(l+1)(q-1))}{2pl^m}+\frac{(l+1)(p-1)}{4p}\eta_{i}^{(l^m, q)}+\frac{(l+1)(p-1)}{4p}\eta_{i'}^{(l^m,q)}$ $\frac{q-1}{l^{m}}\cdot\frac{l-1}2$ $i/l^{{m-1}}\in H_1^{(0)}, i'/l^{m-1}\in H_1^{(1)}$ $\frac{(p-1)(2l^mq-(l+1)(q-1))}{2pl^m}+\frac{(p-1)(l+1)}{2p}\eta_{i}^{(l^m, q)},i\in \cup_{k = 2}^m S_k$ $\frac{q-1}{l^{m}}\phi(l^k)$, $k = 2,3,\ldots,m$,
Weight Frequency $0$ $1$ $\frac{(p-1)(2l^mq-(l+1)(q-1))}{2pl^m}+\frac{p-1}p\eta_{0}^{(l^m, q)}+\frac{(l-1)(p-1)}{2p}\eta_{i'}^{(l^m,q)}$ $\frac{q-1}{l^{m}}$ $i'/l^{m-1}\in H_1^{(1)}$ $\frac{(p-1)(2l^mq-(l+1)(q-1))}{2pl^m}+\frac{p-1}p\eta_0^{(l^m, q)}+\frac{(l+1)(p-1)}{4p}\eta_{i}^{(l^m, q)}+\frac{(l-3)(p-1)}{4p}\eta_{i'}^{(l^m,q)}$ $\frac{q-1}{l^{m}}\cdot\frac{l-1}2$ $i/l^{{m-1}}\in H_1^{(0)}, i'/l^{m-1}\in H_1^{(1)}$ $\frac{(p-1)(2l^mq-(l+1)(q-1))}{2pl^m}+\frac{(l+1)(p-1)}{4p}\eta_{i}^{(l^m, q)}+\frac{(l+1)(p-1)}{4p}\eta_{i'}^{(l^m,q)}$ $\frac{q-1}{l^{m}}\cdot\frac{l-1}2$ $i/l^{{m-1}}\in H_1^{(0)}, i'/l^{m-1}\in H_1^{(1)}$ $\frac{(p-1)(2l^mq-(l+1)(q-1))}{2pl^m}+\frac{(p-1)(l+1)}{2p}\eta_{i}^{(l^m, q)},i\in \cup_{k = 2}^m S_k$ $\frac{q-1}{l^{m}}\phi(l^k)$, $k = 2,3,\ldots,m$,
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# Computing a Discrete Cosine Transform matrix
I'm learning Haskell and I needed to work with discrete cosine transform matrices, so I made a program that generates text (usable in WolframAlpha) containing the generated matrix.
The elements of the n-square matrix C are defined as:
$[C]_{jk} = \gamma_jcos\dfrac{\pi(2k+1)j}{2n}$ where $\gamma_{j} = \begin{cases}\frac{1}{\sqrt{n}} & \text{j = 0} \\ \frac{2}{\sqrt{n}} & \text{otherwise} \end{cases}$
I also wanted the program to simplify fractions and sqrt, which is where most of the "dirtiness" came from. Is there any way to clean this up? It's simple enough, but I want to learn how to be concise.
import Ratio
-- Pretty-print a fraction
showFraction n d
| n == 0 = "0"
| d == 0 = error "Division by zero."
| n == 1 && d == 1 = "1"
| n == 1 = "1/" ++ show d
| d == 1 = show n
| otherwise = show n ++ "/" ++ show d
-- Pretty-print a fraction, reducing the fraction
showFractionReduced n d = let x = n % d in showFraction (numerator x) (denominator x)
-- Pretty-print a fraction, with an additional numerator factor
showFraction' n d n'
| n == 0 = "0"
| d == 0 = error "Division by zero."
| n == 1 && d == 1 = n'
| n == 1 = n' ++ "/" ++ show d
| d == 1 = show n ++ "*" ++ n'
| otherwise = show n ++ "*" ++ n' ++ "/" ++ show d
-- Pretty-print a fraction, reducing the fraction, with an additional numerator factor
showFractionReduced' n d = let x = n % d in showFraction' (numerator x) (denominator x)
-- Discrete cosine transform matrix element
dctElement j k n
| n <= 1 = "1"
| n == 2 && j > 0 = "cos(" ++ showFractionReduced' (j * (2 * k + 1)) (2 * n) "pi" ++ ")"
| j == 0 = "sqrt(" ++ showFractionReduced 1 n ++ ")"
| otherwise = "sqrt(" ++ showFractionReduced 2 n ++ ")*cos(" ++ showFractionReduced' (j * (2 * k + 1)) (2 * n) "pi" ++ ")"
-- Discrete cosine transform matrix row
dctRow j n = "{" ++ foldr1 (\acc x -> acc ++ "," ++ x) [dctElement j x n | x <- [0..n - 1]] ++ "}"
-- Discrete cosine transform matrix
dctMatrix n = "{" ++ foldr1 (\acc x -> acc ++ ",\n" ++ x) [dctRow x n | x <- [0..n - 1]] ++ "}"
-- Inverse discrete cosine transform matrix row
dctRowInverse j n = "{" ++ foldr1 (\acc x -> acc ++ "," ++ x) [dctElement x j n | x <- [0..n - 1]] ++ "}"
-- Inverse discrete cosine transform matrix
dctMatrixInverse n = "{" ++ foldr1 (\acc x -> acc ++ ",\n" ++ x) [dctRowInverse x n | x <- [0..n - 1]] ++ "}"
The string generated by dctMatrix 3, for example, is:
{{sqrt(1/3),sqrt(1/3),sqrt(1/3)},
{sqrt(2/3)*cos(pi/6),sqrt(2/3)*cos(pi/2),sqrt(2/3)*cos(5*pi/6)},
{sqrt(2/3)*cos(pi/3),sqrt(2/3)*cos(pi),sqrt(2/3)*cos(5*pi/3)}}
while the string generated by dctMatrix 2 is:
{{sqrt(1/3),sqrt(1/3),sqrt(1/3)},
{sqrt(2/3)*cos(pi/6),sqrt(2/3)*cos(pi/2),sqrt(2/3)*cos(5*pi/6)},
{sqrt(2/3)*cos(pi/3),sqrt(2/3)*cos(pi),sqrt(2/3)*cos(5*pi/3)}}
(Notice the simplification of the fractions in the former, and the square root in the latter.)
• You have a bug: 0/0 is 1, and not 0. Commented Apr 4, 2012 at 15:33
It is probably more in the spirit of functional programming to use pattern matches instead of long guard lists where possible. Your first function also has a few overlapping cases that could be eliminated in the interest of brevity:
showFraction _ 0 = error "Division by zero."
showFraction 0 _ = "0"
showFraction n 1 = show n
showFraction n d = show n ++ "/" ++ show d
Another simple change would be to exploit that foldr1 (\acc x -> acc ++ "," ++ x) is simply intercalate ',', which you get from Data.List. In a somewhat similar fashion, you could also replace foldr1 (\acc x -> acc ++ ",\n" ++ x) by concat . intersperse ",\n".
The result would look like follows:
dctRow j n = "{" ++ intercalate ',' [dctElement j x n | x <- [0..n - 1]] ++ "}"
dctMatrix n = "{" ++ concat (intersperse ",\n" [dctRow x n | x <- [0..n - 1]]) ++ "}"
Which in my opinion is more readable than the manual folds.
A more involved change would be to try to get rid of the $$\O(n^2)\$$ behaviour you will get from overusing (++). The background here is that (++) needs to make a full copy of the first operand in order to append the second one.
A fun functional way of improving this is to replace string concatenation by function composition:
showFraction :: Int -> Int -> String -> String
showFraction _ 0 = error "Division by zero."
showFraction 0 _ = ('0':)
showFraction n 1 = shows n
showFraction n d = shows n . ('/':) . shows d
So instead of a function that constructs a string, this is a function that prepends the string representation to a string passed as parameter. The nice thing about this is that you can simply compose these functions together in order to build bigger strings. As prepending doesn't require copying of the "tail" end in Haskell, this means that every part of the result string will be constructed exactly once!
Here is how to write the folds in this style:
applyInter f = flip (foldr ($)) . intersperse f dctRow j n = ('{':) . applyInter (',':) [dctElement j x n | x <- [0..n - 1]] . ('}':) (Realize that flip (foldr ($)) [f,g,h] is simply f . g . h)
Note however that while this indeed has $$\O(n)\$$ complexity, it is still not very efficient. This is due to GHC probably selecting a bad arity for some of the functions in question, as well as String not being very fast in the first place. For constructing strings fast, it's probably a better idea to use Data.Text and/or a special builder library like blaze-builder.
• Great answer, thanks a bunch. I'll look into Data.Text after I implement all your other ideas. Commented Apr 5, 2012 at 3:23
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# The World We Live In #5: Calories And Kilograms
I enjoy doing new tunes; it gives me a little bit to perk up, to pay a little bit more attention (Earl Scruggs, American musician)
I recently finished reading The Signal and the Noise, a book by Nate Silver, creator of the also famous FiveThirtyEight blog. The book is a very good reading for all data science professionals, and is a must in particular for all those who work trying to predict the future. The book praises the bayesian way of thinking as the best way to face and modify predictions and criticizes rigid ways of thinking with many examples of disastrous predictions. I enjoyed a lot the chapter dedicated to chess and how Deep Blue finally took over Kasparov. In a nutshell: I strongly recommend it.
One of the plots of Silver’s book present a case of false negative showing the relationship between obesity and calorie consumption across the world countries. The plot shows that there is no evidence of a connection between both variables. Since it seemed very strange to me, I decided to reproduce the plot by myself.
I compared these two variables:
• Dietary Energy Consumption (kcal/person/day) estimated by the FAO Food Balance Sheets.
• Prevalence of Obesity as percentage of defined population with a body mass index (BMI) of 30 kg/m2 or higher estimated by the World Health Organization
And this is the resulting plot:
As you can see there is a strong correlation between two variables. Why the experiment of Nate Silver shows the opposite? Obviously we did not plot the same data (although, in principle, both of us went to the same source). Anyway: to be honest, I prefer my plot because shows what all of we know: the more calories you eat, the more weight you will see in your bathroom scale. Some final thoughts seeing the plot:
• I would like to be Japanese: they don’t gain weight!
• Why US people are fatter than Austrian?
• What happens in Samoa?
Here you have the code to do the plot:
library(xlsx)
library(dplyr)
library(ggplot2)
library(scales)
setwd("YOUR WORKING DIRECTORY HERE")
url_calories = "http://www.fao.org/fileadmin/templates/ess/documents/food_security_statistics/FoodConsumptionNutrients_en.xls"
download.file(url_calories, method="internal", destfile = "FoodConsumptionNutrients_en.xls", mode = "ab")
calories = read.xlsx(file="FoodConsumptionNutrients_en.xls", startRow = 4, colIndex = c(2,6), colClasses = c("character", "numeric"), sheetName="Dietary Energy Cons. Countries", stringsAsFactors=FALSE)
colnames(calories)=c("Country", "Kcal")
url_population = "http://esa.un.org/unpd/wpp/DVD/Files/1_Excel%20(Standard)/EXCEL_FILES/1_Population/WPP2015_POP_F01_1_TOTAL_POPULATION_BOTH_SEXES.XLS"
download.file(url_population, method="internal", destfile = "Population.xls", mode = "ab")
population = read.xlsx(file="Population.xls", startRow = 17, colIndex = c(3,71), colClasses = c("character", "numeric"), sheetName="ESTIMATES", stringsAsFactors=FALSE)
colnames(population)=c("Country", "Population")
# http://apps.who.int/gho/data/node.main.A900A?lang=en
url_obesity = "http://apps.who.int/gho/athena/data/xmart.csv?target=GHO/NCD_BMI_30A&profile=crosstable&filter=AGEGROUP:*;COUNTRY:*;SEX:*&x-sideaxis=COUNTRY&x-topaxis=GHO;YEAR;AGEGROUP;SEX&x-collapse=true"
obesity = read.csv(file=url_obesity, stringsAsFactors=FALSE)
obesity %>% select(matches("Country|2014.*Both")) -> obesity
colnames(obesity)=c("Country", "Obesity")
obesity %>% filter(Obesity!="No data") -> obesity
obesity %>% mutate(Obesity=as.numeric(substr(Obesity, 1, regexpr(pattern = "[[]", obesity$Obesity)-1))) -> obesity population %>% inner_join(calories,by = "Country") %>% inner_join(obesity,by = "Country") -> data opts=theme( panel.background = element_rect(fill="gray98"), panel.border = element_rect(colour="black", fill=NA), axis.line = element_line(size = 0.5, colour = "black"), axis.ticks = element_line(colour="black"), panel.grid.major = element_line(colour="gray75", linetype = 2), panel.grid.minor = element_blank(), axis.text = element_text(colour="gray25", size=15), axis.title = element_text(size=18, colour="gray10"), legend.key = element_blank(), legend.position = "none", legend.background = element_blank(), plot.title = element_text(size = 40, colour="gray10")) ggplot(data, aes(x=Kcal, y=Obesity/100, size=log(Population), label=Country), guide=FALSE)+ geom_point(colour="white", fill="sandybrown", shape=21, alpha=.55)+ scale_size_continuous(range=c(2,40))+ scale_x_continuous(limits=c(1500,4100))+ scale_y_continuous(labels = percent)+ labs(title="The World We Live In #5: Calories And Kilograms", x="Dietary Energy Consumption (kcal/person/day)", y="% population with body mass index >= 30 kg/m2")+ geom_text(data=subset(data, Obesity>35|Kcal>3700), size=5.5, colour="gray25", hjust=0, vjust=0)+ geom_text(data=subset(data, Kcal<2000), size=5.5, colour="gray25", hjust=0, vjust=0)+ geom_text(data=subset(data, Obesity<10 & Kcal>2600), size=5.5, colour="gray25", hjust=0, vjust=0)+ geom_text(aes(3100, .01), colour="gray25", hjust=0, label="Source: United Nations (size of bubble depending on population)", size=4.5)+opts # Going Bananas #2: A Needle In A Haystack Now I’m gonna tell my momma that I’m a traveller, I’m gonna follow the sun (The Sun, Parov Stelar) Inspired by this book I read recently, I decided to do this experiment. The idea is comparing how easy is to find sequences of numbers inside Pi, e, Golden Ratio (Phi) and a randomly generated number. For example, since Pi is 3.1415926535897932384… the 4-size sequence 5358 can be easily found at the begining as well as the 5-size sequence 79323. I considered interesting comparing Pi with a random generated number. What I though before doing the experiment is that it would be easier finding sequences inside the andom one. Why? Because despite of being irrational and transcendental I thought there should be some kind of residual pattern in Pi that should make more difficult to find random sequences inside it than do it inside a randomly generated number. • I downloaded Pi, e and Phi from the Internet and extract first 100.000 digits of all of them. I generate a random 100.000 number on the fly. • I generate a representative sample of 4-size sequences • I look for each of these sequences inside first 5.000 digits of Pi, e, Phi and the randomly generated one. I repeat searching for first 10.000, first 15.000 and so on until I search into the whole 100.000 -size number • I store how many sequences I find for each searching • I repeat this for 5 and 6-size sequences. At first sight, is equally easy (or difficult), to find random sequences inside all numbers: my hypothesis was wrong. As you can see here, 100.000 digits is more than enough to find 4-size sequences. In fact, from 45.000 digits I reach 100% of successful matches: I only find 60% of 5-size sequences inside 100.000 digits of numbers: And only 10% of 6-size sequences: Why these four numbers are so equal in order to find random sequences inside them? I don’t know. What I know is that if you want to find your telephone number inside Pi, you will probably need an enormous number of digits. library(rvest) library(stringr) library(reshape2) library(ggplot2) library(extrafont);windowsFonts(Comic=windowsFont("Comic Sans MS")) library(dplyr) library(magrittr) library(scales) p = html("http://www.geom.uiuc.edu/~huberty/math5337/groupe/digits.html") f = html("http://www.goldennumber.net/wp-content/uploads/2012/06/Phi-To-100000-Places.txt") e = html("http://apod.nasa.gov/htmltest/gifcity/e.2mil") p %>% html_text() %>% substr(., regexpr("3.14",.), regexpr("Go to Historical",.)) %>% gsub("[^0-9]", "", .) %>% substr(., 1, 100000) -> p f %>% html_text() %>% substr(., regexpr("1.61",.), nchar(.)) %>% gsub("[^0-9]", "", .) %>% substr(., 1, 100000) -> f e %>% html_text() %>% substr(., regexpr("2.71",.), nchar(.)) %>% gsub("[^0-9]", "", .) %>% substr(., 1, 100000) -> e r = paste0(sample(0:9, 100000, replace = TRUE), collapse = "") results=data.frame(Cut=numeric(0), Pi=numeric(0), Phi=numeric(0), e=numeric(0), Random=numeric(0)) bins=20 dgts=6 samp=min(10^dgts*2/100, 10000) for (i in 1:bins) { cut=100000/bins*i p0=substr(p, start=0, stop=cut) f0=substr(f, start=0, stop=cut) e0=substr(e, start=0, stop=cut) r0=substr(r, start=0, stop=cut) sample(0:(10^dgts-1), samp, replace = FALSE) %>% str_pad(dgts, pad = "0") -> comb comb %>% sapply(function(x) grepl(x, p0)) %>% sum() -> p1 comb %>% sapply(function(x) grepl(x, f0)) %>% sum() -> f1 comb %>% sapply(function(x) grepl(x, e0)) %>% sum() -> e1 comb %>% sapply(function(x) grepl(x, r0)) %>% sum() -> r1 results=rbind(results, data.frame(Cut=cut, Pi=p1, Phi=f1, e=e1, Random=r1)) } results=melt(results, id.vars=c("Cut") , variable.name="number", value.name="matches") opts=theme( panel.background = element_rect(fill="darkolivegreen1"), panel.border = element_rect(colour="black", fill=NA), axis.line = element_line(size = 0.5, colour = "black"), axis.ticks = element_line(colour="black"), panel.grid.major = element_line(colour="white", linetype = 1), panel.grid.minor = element_blank(), axis.text.y = element_text(colour="black"), axis.text.x = element_text(colour="black"), text = element_text(size=20, family="Comic"), legend.text = element_text(size=25), legend.key = element_blank(), legend.position = c(.75,.2), legend.background = element_blank(), plot.title = element_text(size = 30)) ggplot(results, aes(x = Cut, y = matches/samp, color = number))+ geom_line(size=1.5, alpha=.8)+ scale_color_discrete(name = "")+ scale_x_continuous(breaks=seq(100000/bins, 100000, by=100000/bins))+ scale_y_continuous(labels = percent)+ theme(axis.text.x = element_text(angle = 90, vjust=.5, hjust = 1))+ labs(title=paste0("Finding ",dgts, "-size strings into 100.000-digit numbers"), x="Cut Position", y="% of Matches")+opts # The Moon And The Sun Do not swear by the moon, for she changes constantly. Then your love would also change (William Shakespeare, Romeo and Juliet) The sun is a big point ant the moon is a cardioid: Here you have the code. It is a simple example of how to use ggplot: library(ggplot2) n=160 t1=1:n t0=seq(from=3, to=2*n+1, by=2) %% n t2=t0+(t0==0)*n df=data.frame(x1=cos((t1-1)*2*pi/n), y1=sin((t1-1)*2*pi/n), x2=cos((t2-1)*2*pi/n), y2=sin((t2-1)*2*pi/n)) opt=theme(legend.position="none", panel.background = element_rect(fill="white"), panel.grid = element_blank(), axis.ticks=element_blank(), axis.title=element_blank(), axis.text =element_blank()) ggplot(df, aes(x = x1, y = y1, xend = x2, yend = y2)) + geom_point(x=0, y=0, size=245, color="gold")+ geom_segment(color="white", alpha=.5)+opt # Trigonometric Pattern Design Triangles are my favorite shape, three points where two lines meet (Tessellate, Alt-J) Inspired by recurrence plots and by the Gauss error function, I have done the following plots. The first one represents the recurrence plot of $f\left ( x \right )= sec\left ( x \right )$ where distance between points is measured by Gauss error function: This one is the same for $f\left ( x \right )= tag\left ( x \right )$ And this one represents $f\left ( x \right )= sin\left ( x \right )$ I like them: they are elegant, attractive and easy to make. Try your own functions. One final though: the more I use magrittr package, the more I like it. This is the code for the first plot. library("magrittr") library("ggplot2") library("pracma") RecurrencePlot = function(from, to, col1, col2) { opt = theme(legend.position = "none", panel.background = element_blank(), axis.ticks = element_blank(), panel.grid = element_blank(), axis.title = element_blank(), axis.text = element_blank()) seq(from, to, by = .1) %>% expand.grid(x=., y=.) %>% ggplot( ., aes(x=x, y=y, fill=erf(sec(x)-sec(y)))) + geom_tile() + scale_fill_gradientn(colours=colorRampPalette(c(col1, col2))(2)) + opt} RecurrencePlot(from = -5*pi, to = 5*pi, col1 = "black", col2= "white") # The 2D-Harmonograph In Shiny If you wish to make an apple pie from scratch, you must first invent the universe (Carl Sagan) I like Shiny and I can’t stop converting into apps some of my previous experiments: today is the turn of the harmonograph. This is a simple application since you only can press a button to generate a random harmonograph-simulated curve. I love how easy is to create a nice interactive app to play with from an existing code. The only trick in this case in to add a rerun button in the UI side and transfer the interaction to the server side using a simple if. Very easy. This is a screenshot of the application: Press the button and you will get a new drawing. Most of them are nice scrawls and from time to time you will obtain beautiful shapely curves. And no more candy words: It is time to complain. I say to RStudio with all due respect, you are very cruel. You let me to deploy my previous app to your server but you suspended it almost immediately for fifteen days due to “exceeded usage hours”. My only option is paying at least$440 per year to upgrade my current plan. I tried the ambrosia for an extremely short time. RStudio: Why don’t you launch a cheaper account? Why don’t you launch a free account with just one perpetual alive app at a time? Why don’t you increase the usage hours threshold? I can help you to calculate the return on investment of these scenarios.
Or, Why don’t you make me a gift for my next birthday? I promise to upload a new app per month to promote your stunning tool. Think about it and please let me know your conclusions.
Meanwhile I will run the app privately. This is the code to do it:
UI.R
# This is the user-interface definition of a Shiny web application.
# You can find out more about building applications with Shiny here:
#
# http://www.rstudio.com/shiny/
library(shiny)
shinyUI(fluidPage(
titlePanel("Mathematical Beauties: The Harmonograph"),
sidebarLayout(
sidebarPanel(
#helpText(),
# adding the new div tag to the sidebar
tags$div(class="header", checked=NA, tags$p("A harmonograph is a mechanical apparatus that employs pendulums to create a
geometric image. The drawings created typically are Lissajous curves, or
related drawings of greater complexity. The devices, which began to appear
in the mid-19th century and peaked in popularity in the 1890s, cannot be
conclusively attributed to a single person, although Hugh Blackburn, a professor
of mathematics at the University of Glasgow, is commonly believed to be the official
inventor. A simple, so-called \"lateral\" harmonograph uses two pendulums to control the movement
of a pen relative to a drawing surface. One pendulum moves the pen back and forth along
one axis and the other pendulum moves the drawing surface back and forth along a
perpendicular axis. By varying the frequency and phase of the pendulums relative to
one another, different patterns are created. Even a simple harmonograph as described
can create ellipses, spirals, figure eights and other Lissajous figures (Source: Wikipedia)")),
tags$div(class="header", checked=NA, HTML("<p>Click <a href=\"http://paulbourke.net/geometry/harmonograph/harmonograph3.html\">here</a> to see an image of a real harmonograph</p>") ), actionButton('rerun','Launch the harmonograph!') ), mainPanel( plotOutput("HarmPlot") ) ) )) server.R # This is the server logic for a Shiny web application. # You can find out more about building applications with Shiny here: # # http://www.rstudio.com/shiny/ # library(shiny) CreateDS = function () { f=jitter(sample(c(2,3),4, replace = TRUE)) d=runif(4,0,1e-02) p=runif(4,0,pi) xt = function(t) exp(-d[1]*t)*sin(t*f[1]+p[1])+exp(-d[2]*t)*sin(t*f[2]+p[2]) yt = function(t) exp(-d[3]*t)*sin(t*f[3]+p[3])+exp(-d[4]*t)*sin(t*f[4]+p[4]) t=seq(1, 200, by=.0005) data.frame(t=t, x=xt(t), y=yt(t))} shinyServer(function(input, output) { dat<-reactive({if (input$rerun) dat=CreateDS() else dat=CreateDS()})
output$HarmPlot<-renderPlot({ plot(rnorm(1000),xlim =c(-2,2), ylim =c(-2,2), type="n") with(dat(), plot(x,y, type="l", xlim =c(-2,2), ylim =c(-2,2), xlab = "", ylab = "", xaxt='n', yaxt='n', col="gray10", bty="n")) }, height = 650, width = 650) }) # Shiny Wool Skeins Chaos is not a pit: chaos is a ladder (Littlefinger in Game of Thrones) Some time ago I wrote this post to show how my colleague Vu Anh translated into Shiny one of my experiments, opening my eyes to an amazing new world. I am very proud to present you the first Shiny experiment entirely written by me. In this case I took inspiration from another previous experiment to draw some kind of wool skeins. The shiny app creates a plot consisting of chords inside a circle. There are to kind of chords: • Those which form a track because they are a set of glued chords; number of tracks and number of chords per track can be selected using Number of track chords and Number of scrawls per track sliders of the app respectively. • Those forming the background, randomly allocated inside the circle. Number of background chords can be chosen as well in the app There is also the possibility to change colors of chords. This are the main steps I followed to build this Shiny app: 1. Write a simple R program 2. Decide which variables to parametrize 3. Open a new Shiny project in RStudio 4. Analize the sample UI.R and server.R files generated by default 5. Adapt sample code to my particular code (some iterations are needed here) 6. Deploy my app in the Shiny Apps free server Number 1 is the most difficult step, but it does not depends on Shiny: rest of them are easier, specially if you have help as I had from my colleague Jorge. I encourage you to try. This is an snapshot of the app: You can play with the app here. Some things I thought while developing this experiment: • Shiny gives you a lot with a minimal effort • Shiny can be a very interesting tool to teach maths and programming to kids • I have to translate to Shiny some other experiment • I will try to use it for my job Try Shiny: is very entertaining. A typical Shiny project consists on two files, one to define the user interface (UI.R) and the other to define the back end side (server.R). This is the code of UI.R: # This is the user-interface definition of a Shiny web application. # You can find out more about building applications with Shiny here: # # http://shiny.rstudio.com # library(shiny) shinyUI(fluidPage( # Application title titlePanel("Shiny Wool Skeins"), HTML("<p>This experiment is based on <a href=\"https://aschinchon.wordpress.com/2015/05/13/bertrand-or-the-importance-of-defining-problems-properly/\">this previous one</a> I did some time ago. It is my second approach to the wonderful world of Shiny.</p>"), # Sidebar with a slider input for number of bins sidebarLayout( sidebarPanel( inputPanel( sliderInput("lin", label = "Number of track chords:", min = 1, max = 20, value = 5, step = 1), sliderInput("rep", label = "Number of scrawls per track:", min = 1, max = 50, value = 10, step = 1), sliderInput("nbc", label = "Number of background chords:", min = 0, max = 2000, value = 500, step = 2), selectInput("col1", label = "Track colour:", choices = colors(), selected = "darkmagenta"), selectInput("col2", label = "Background chords colour:", choices = colors(), selected = "gold") ) ), # Show a plot of the generated distribution mainPanel( plotOutput("chordplot") ) ) )) And this is the code of server.R: # This is the server logic for a Shiny web application. # You can find out more about building applications with Shiny here: # # http://shiny.rstudio.com # library(ggplot2) library(magrittr) library(grDevices) library(shiny) shinyServer(function(input, output) { df<-reactive({ ini=runif(n=input$lin, min=0,max=2*pi)
ini %>%
+runif(n=input$lin, min=pi/2,max=3*pi/2) %>% cbind(ini, end=.) %>% as.data.frame() -> Sub1 Sub1=Sub1[rep(seq_len(nrow(Sub1)), input$rep),]
Sub1 %>% apply(c(1, 2), jitter) %>% as.data.frame() -> Sub1
Sub1=with(Sub1, data.frame(col=input$col1, x1=cos(ini), y1=sin(ini), x2=cos(end), y2=sin(end))) Sub2=runif(input$nbc, min = 0, max = 2*pi)
Sub2=data.frame(x=cos(Sub2), y=sin(Sub2))
Sub2=cbind(input$col2, Sub2[(1:(input$nbc/2)),], Sub2[(((input$nbc/2)+1):input$nbc),])
colnames(Sub2)=c("col", "x1", "y1", "x2", "y2")
rbind(Sub1, Sub2)
})
opts=theme(legend.position="none",
panel.background = element_rect(fill="white"),
panel.grid = element_blank(),
axis.ticks=element_blank(),
axis.title=element_blank(),
axis.text =element_blank())
output$chordplot<-renderPlot({ p=ggplot(df())+geom_segment(aes(x=x1, y=y1, xend=x2, yend=y2), colour=df()$col, alpha=runif(nrow(df()), min=.1, max=.3), lwd=1)+opts;print(p)
}, height = 600, width = 600 )
})
# Simple Data Science To Maximize Return On Lottery Investment
Every finite game has an equilibrium point (John Nash, Non-Cooperative Games, 1950)
I read recently this amazing book, where I discovered that we (humans) are not capable of generating random sequences of numbers by ourselves when we play lottery. John Haigh demonstrates this fact analyzing a sample of 282 raffles of 6/49 UK Lotto. Once I read this, I decided to prove if this disability is property only of British population or if it is shared with Spanish people as well. I am Spanish, so this experiment can bring painful results to myself, but here I come.
The Spanish equivalent of 6/40 UK Lotto is called “Lotería Primitiva” (or “Primitiva”, to abbreviate). This is a ticket of Primitiva lotto:
As you can see, one ticket gives the chance to do 8 bets. Each bet consists on 6 numbers between 1 and 49 to be chosen in a grid of 10 rows by 5 columns. People tend to choose separate numbers because we think that they are more likely to come up than combinations with some consecutive numbers. We think we have more chances to get rich choosing 4-12-23-25-31-43 rather than 3-17-18-19-32-33, for instance. To be honest, I should recognize I am one of these persons.
Primitiva lotto is managed by Sociedad Estatal Loterías y Apuestas del Estado, a public business entity belonging to the Spanish Ministry of Finance and Public Administrations. They know what people choose and they could do this experiment more exactly than me. They could analyze just human bets (those made by players by themselves) and discard machine ones (those made automatically by vending machines) but anyway it is possible to confirm the previous thesis with some public data.
I analysed 432 raffles of Primitiva carried out between 2011 and 2015; for each raffle I have this information:
• The six numbers that form the winning combination
• Total number of bets
• Number of bets which hit the six numbers (Observed Winners)
The idea is to compare observed winners of raffles with the expected number of them, estimated as follows:
$Expected\, Winners=\frac{Total\, Bets}{C_{6}^{49}},\: where\: C_{6}^{49}=\binom{49}{6}=\frac{49!}{43!6!}$
This table compare the number of expected and observed winners between raffles which contain consecutive and raffles which not:
There are 214 raffles without consecutive with 294 winners while the expected number of them was 219. In other words, a winner of a non-consecutive-raffle must share the prize with a 33% of some other person. On the other hand, the number of observed winners of a raffle with consecutive numbers 17% lower than the expected one. Simple and conclusive. Spanish are like British, at least in what concerns to this particular issue.
Let’s go further. I can do the same for any particular number. For example, there were 63 raffles containing number 45 in the winning combination and 57 (observed) winners, although 66 were expected. After doing this for every number, I can draw this plot, where I paint in blue those which ratio of observed winners between expected is lower than 0.9:
It seems that blue numbers are concentrated on the right side of the grid. Do we prefer small numbers rather than big ones? There are 15 primes between 1 and 49 (rate: 30%) but only 3 primes between blue numbers (rate: 23%). Are we attracted by primes?
Let’s combine both previous results. This table compares the number of expected and observed winners between raffles which contain consecutive and blues (at least one) and raffles which not:
Now, winning combinations with some consecutive and some blue numbers present 20% less of observed winners than expected. After this, which combination would you choose for your next bet? 27-35-36-41-44-45 or 2-6-13-15-26-28? I would choose the first one. Both of them have the same probability to come up, but probably you will become richer with the first one if it happens.
This is the code of this experiment. If someone need the dataset set to do their own experiments, feel free to ask me (you can find my email here):
library("xlsx")
library("sqldf")
library("Hmisc")
library("lubridate")
library("ggplot2")
library("extrafont")
library("googleVis")
windowsFonts(Garamond=windowsFont("Garamond"))
setwd("YOUR WORKING DIRECTORY HERE")
file = "SORTEOS_PRIMITIVA_2011_2015.xls"
data=read.xlsx(file, sheetName="ALL", colClasses=c("numeric", "Date", rep("numeric", 21)))
#Impute null values to zero
data$C1_EUROS=with(data, impute(C1_EUROS, 0)) data$CE_WINNERS=with(data, impute(CE_WINNERS, 0))
#Expected winners for each raffle
data$EXPECTED=data$BETS/(factorial(49)/(factorial(49-6)*factorial(6)))
#Consecutives indicator
data$DIFFMIN=apply(data[,3:8], 1, function (x) min(diff(sort(x)))) #Consecutives vs non-consecutives comparison df1=sqldf("SELECT CASE WHEN DIFFMIN=1 THEN 'Yes' ELSE 'No' END AS CONS, COUNT(*) AS RAFFLES, SUM(EXPECTED) AS EXP_WINNERS, SUM(CE_WINNERS+C1_WINNERS) AS OBS_WINNERS FROM data GROUP BY CONS") colnames(df1)=c("Contains consecutives?", "Number of raffles", "Expected Winners", "Observed Winners") Table1=gvisTable(df1, formats=list('Expected Winners'='#,###')) plot(Table1) #Heat map of each number results=data.frame(BALL=numeric(0), EXP_WINNER=numeric(0), OBS_WINNERS=numeric(0)) for (i in 1:49) { data$TF=apply(data[,3:8], 1, function (x) i %in% x + 0)
v=data.frame(BALL=i, sqldf("SELECT SUM(EXPECTED) AS EXP_WINNERS, SUM(CE_WINNERS+C1_WINNERS) AS OBS_WINNERS FROM data WHERE TF = 1"))
results=rbind(results, v)
}
results$ObsByExp=results$OBS_WINNERS/results$EXP_WINNERS results$ROW=results$BALL%%10+1 results$COL=floor(results$BALL/10)+1 results$ObsByExp2=with(results, cut(ObsByExp, breaks=c(-Inf,.9,Inf), right = FALSE))
opt=theme(legend.position="none",
panel.background = element_blank(),
panel.grid = element_blank(),
axis.ticks=element_blank(),
axis.title=element_blank(),
axis.text =element_blank())
ggplot(results, aes(y=ROW, x=COL)) +
geom_tile(aes(fill = ObsByExp2), colour="gray85", lwd=2) +
geom_text(aes(family="Garamond"), label=results$BALL, color="gray10", size=12)+ scale_fill_manual(values = c("dodgerblue", "gray98"))+ scale_y_reverse()+opt #Blue numbers Bl=subset(results, ObsByExp2=="[-Inf,0.9)")[,1] data$BLUES=apply(data[,3:8], 1, function (x) length(intersect(x,Bl)))
#Combination of consecutives and blues
df2=sqldf("SELECT CASE WHEN DIFFMIN=1 AND BLUES>0 THEN 'Yes' ELSE 'No' END AS IND,
COUNT(*) AS RAFFLES,
SUM(EXPECTED) AS EXP_WINNERS,
SUM(CE_WINNERS+C1_WINNERS) AS OBS_WINNERS
FROM data GROUP BY IND")
colnames(df2)=c("Contains consecutives and blues?", "Number of raffles", "Expected Winners", "Observed Winners")
Table2=gvisTable(df2, formats=list('Expected Winners'='#,###'))
plot(Table2)
# Bertrand or (The Importance of Defining Problems Properly)
We better keep an eye on this one: she is tricky (Michael Banks, talking about Mary Poppins)
Professor Bertrand teaches Simulation and someday, ask his students:
Given a circumference, what is the probability that a chord chosen at random is longer than a side of the equilateral triangle inscribed in the circle?
Since they must reach the answer through simulation, very approximate solutions are welcome.
Some students choose chords as the line between two random points on the circumference and conclude that the asked probability is around 1/3. This is the plot of one of their simulations, where 1000 random chords are chosen according this method and those longer than the side of the equilateral triangle are red coloured (smalller in grey):
Some others choose a random radius and a random point in it. The chord then is the perpendicular through this point. They calculate that the asked probability is around 1/2:
And some others choose a random point inside the circle and define the chord as the only one with this point as midpoint. For them, the asked probability is around 1/4:
Who is right? Professor Bertrand knows that everybody is. In fact, his main purpose was to show how important is to define problems properly. Actually, he used this to give an unforgettable lesson to his students.
library(ggplot2)
n=1000
opt=theme(legend.position="none",
panel.background = element_rect(fill="white"),
panel.grid = element_blank(),
axis.ticks=element_blank(),
axis.title=element_blank(),
axis.text =element_blank())
#First approach
angle=runif(2*n, min = 0, max = 2*pi)
pt1=data.frame(x=cos(angle), y=sin(angle))
df1=cbind(pt1[1:n,], pt1[((n+1):(2*n)),])
colnames(df1)=c("x1", "y1", "x2", "y2")
df1$length=sqrt((df1$x1-df1$x2)^2+(df1$y1-df1$y2)^2) p1=ggplot(df1) + geom_segment(aes(x = x1, y = y1, xend = x2, yend = y2, colour=length>sqrt(3)), alpha=.4, lwd=.6)+ scale_colour_manual(values = c("gray75", "red"))+opt #Second approach angle=2*pi*runif(n) pt2=data.frame(aa=cos(angle), bb=sin(angle)) pt2$x0=pt2$aa*runif(n) pt2$y0=pt2$x0*(pt2$bb/pt2$aa) pt2$a=1+(pt2$x0^2/pt2$y0^2)
pt2$b=-2*(pt2$x0/pt2$y0)*(pt2$y0+(pt2$x0^2/pt2$y0))
pt2$c=(pt2$y0+(pt2$x0^2/pt2$y0))^2-1
pt2$x1=(-pt2$b+sqrt(pt2$b^2-4*pt2$a*pt2$c))/(2*pt2$a)
pt2$y1=-pt2$x0/pt2$y0*pt2$x1+(pt2$y0+(pt2$x0^2/pt2$y0)) pt2$x2=(-pt2$b-sqrt(pt2$b^2-4*pt2$a*pt2$c))/(2*pt2$a) pt2$y2=-pt2$x0/pt2$y0*pt2$x2+(pt2$y0+(pt2$x0^2/pt2$y0))
df2=pt2[,c(8:11)]
df2$length=sqrt((df2$x1-df2$x2)^2+(df2$y1-df2$y2)^2) p2=ggplot(df2) + geom_segment(aes(x = x1, y = y1, xend = x2, yend = y2, colour=length>sqrt(3)), alpha=.4, lwd=.6)+ scale_colour_manual(values = c("gray75", "red"))+opt #Third approach angle=2*pi*runif(n) radius=runif(n) pt3=data.frame(x0=sqrt(radius)*cos(angle), y0=sqrt(radius)*sin(angle)) pt3$a=1+(pt3$x0^2/pt3$y0^2)
pt3$b=-2*(pt3$x0/pt3$y0)*(pt3$y0+(pt3$x0^2/pt3$y0))
pt3$c=(pt3$y0+(pt3$x0^2/pt3$y0))^2-1
pt3$x1=(-pt3$b+sqrt(pt3$b^2-4*pt3$a*pt3$c))/(2*pt3$a)
pt3$y1=-pt3$x0/pt3$y0*pt3$x1+(pt3$y0+(pt3$x0^2/pt3$y0)) pt3$x2=(-pt3$b-sqrt(pt3$b^2-4*pt3$a*pt3$c))/(2*pt3$a) pt3$y2=-pt3$x0/pt3$y0*pt3$x2+(pt3$y0+(pt3$x0^2/pt3$y0))
df3=pt3[,c(6:9)]
df3$length=sqrt((df3$x1-df3$x2)^2+(df3$y1-df3$y2)^2) p3=ggplot(df3) + geom_segment(aes(x = x1, y = y1, xend = x2, yend = y2, colour=length>sqrt(3)), alpha=.4, lwd=.6)+scale_colour_manual(values = c("gray75", "red"))+opt # Odd Connections Inside The NASDAQ-100 Distinguishing the signal from the noise requires both scientific knowledge and self-knowledge (Nate Silver, author of The Signal and the Noise) Analyzing the evolution of NASDAQ-100 stock prices can discover some interesting couples of companies which share a strong common trend despite of belonging to very different sectors. The NASDAQ-100 is made up of 107 equity securities issued by 100 of the largest non-financial companies listed on the NASDAQ. On the other side, Yahoo! Finance is one of the most popular services to consult financial news, data and commentary including stock quotes, press releases, financial reports, and original programming. Using R is possible to download the evolution of NASDAQ-100 symbols from Yahoo! Finance. There is a R package called quantmod which makes this issue quite simple with the function getSymbols. Daily series are long enough to do a wide range of analysis, since most of them start in 2007. One robust way to determine if two times series, xt and yt, are related is to analyze if there exists an equation like yt=βxt+ut such us residuals (ut) are stationary (its mean and variance does not change when shifted in time). If this happens, it is said that both series are cointegrated. The way to measure it in R is running the Augmented Dickey-Fuller test, available in tseries package. Cointegration analysis help traders to design products such spreads and hedges. There are 5.671 different couples between the 107 stocks of NASDAQ-100. After computing the Augmented Dickey-Fuller test to each of them, the resulting data frame can be converted into a distance matrix. A nice way to visualize distances between stocks is to do a hierarchical clustering. This is the resulting dendogram of the clustering: Close stocks such as Ca Inc. (CA) and Bed Bath & Beyond Inc. (BBBY) are joined with short links. A quick way to extract close couples is to cut this dendogram in a big number of clusters and keep those with two elements. Following is the list of the most related stock couples cutting dendogram in 85 clusters: Most of them are strange neighbors. Next plot shows the evolution closing price evolution of four of these couples: Analog Devices Inc. (ADI) makes semiconductors and Discovery Communications Inc. (DISCA) is a mass media company. PACCAR Inc. (PCAR) manufactures trucks and Paychex Inc. (PAYX) provides HR outsourcing. CA Inc. (CA) creates software and Bed Bath & Beyond Inc. (BBBY) sells goods for home. Twenty-First Century Fox Inc. (FOX) is a mass media company as well and EBAY Inc. (EBAY) does online auctions. All of them are odd connections. This is the code of the experiment: library("quantmod") library("TSdist") library("ade4") library("ggplot2") library("Hmisc") library("zoo") library("scales") library("reshape2") library("tseries") library("RColorBrewer") library("ape") library("sqldf") library("googleVis") library("gridExtra") setwd("YOUR-WORKING-DIRECTORY-HERE") temp=tempfile() download.file("http://www.nasdaq.com/quotes/nasdaq-100-stocks.aspx?render=download",temp) data=read.csv(temp, header=TRUE) for (i in 1:nrow(data)) getSymbols(as.character(data[i,1])) results=t(apply(combn(sort(as.character(data[,1]), decreasing = TRUE), 2), 2, function(x) { ts1=drop(Cl(eval(parse(text=x[1])))) ts2=drop(Cl(eval(parse(text=x[2])))) t.zoo=merge(ts1, ts2, all=FALSE) t=as.data.frame(t.zoo) m=lm(ts2 ~ ts1 + 0, data=t) beta=coef(m)[1] sprd=t$ts1 - beta*t$ts2 ht=adf.test(sprd, alternative="stationary", k=0)$p.value
c(symbol1=x[1], symbol2=x[2], (1-ht))}))
results=as.data.frame(results)
colnames(results)=c("Sym1", "Sym2", "TSdist")
results$TSdist=as.numeric(as.character(results$TSdist))
save(results, file="results.RData")
load("results.RData")
m=as.dist(acast(results, Sym1~Sym2, value.var="TSdist"))
hc = hclust(m)
# vector of colors
op = par(bg = "darkorchid4")
plot(as.phylo(hc), type = "fan", tip.color = "gold", edge.color ="gold", cex=.8)
# cutting dendrogram in 85 clusters
clusdf=data.frame(Symbol=names(cutree(hc, 85)), clus=cutree(hc, 85))
clusdf2=merge(clusdf, data[,c(1,2)], by="Symbol")
sizes=sqldf("SELECT * FROM (SELECT clus, count(*) as size FROM clusdf GROUP BY 1) as T00 WHERE size>=2")
sizes2=merge(subset(sizes, size==2), clusdf2, by="clus")
sizes2$id=sequence(rle(sizes2$clus)$lengths) couples=merge(subset(sizes2, id==1)[,c(1,3,4)], subset(sizes2, id==2)[,c(1,3,4)], by="clus") couples$"Company 1"=apply(couples[ , c(2,3) ] , 1 , paste , collapse = " -" )
couples$"Company 2"=apply(couples[ , c(4,5) ] , 1 , paste , collapse = " -" ) CouplesTable=gvisTable(couples[,c(6,7)]) plot(CouplesTable) # Plots opts2=theme( panel.background = element_rect(fill="gray98"), panel.border = element_rect(colour="black", fill=NA), axis.line = element_line(size = 0.5, colour = "black"), axis.ticks = element_line(colour="black"), panel.grid.major = element_line(colour="gray75", linetype = 2), panel.grid.minor = element_blank(), axis.text = element_text(colour="gray25", size=12), axis.title = element_text(size=18, colour="gray10"), legend.key = element_rect(fill = "white"), legend.text = element_text(size = 14), legend.background = element_rect(), plot.title = element_text(size = 35, colour="gray10")) plotPair = function(Symbol1, Symbol2) { getSymbols(Symbol1) getSymbols(Symbol2) close1=Cl(eval(parse(text=Symbol1))) close2=Cl(eval(parse(text=Symbol2))) cls=merge(close1, close2, all = FALSE) df=data.frame(date = time(cls), coredata(cls)) names(df)[-1]=c(Symbol1, Symbol2) df1=melt(df, id.vars = "date", measure.vars = c(Symbol1, Symbol2)) ggplot(df1, aes(x = date, y = value, color = variable))+ geom_line(size = I(1.2))+ scale_color_discrete(name = "")+ scale_x_date(labels = date_format("%Y-%m-%d"))+ labs(x="Date", y="Closing Price")+ opts2 } p1=plotPair("ADI", "DISCA") p2=plotPair("PCAR", "PAYX") p3=plotPair("CA", "BBBY") p4=plotPair("FOX", "EBAY") grid.arrange(p1, p2, p3, p4, ncol=2) # Analysing The Rock ‘n’ Roll Madrid Marathon Nobody’s going to win all the time. On the highway of life you can’t always be in the fast lane (Haruki Murakami, What I Talk About When I Talk About Running) I started running two years ago and one if my dreams is to run a marathon someday. One month ago I run my first half marathon so this day is become closer but I am in no hurry to do it. Meanwhile, I prefer analysing the results of the last edition of the Rock ‘n’ Roll Madrid marathon which, by the way, will be hold again next weekend. This is the first time I do webscraping to download data from a website and it has been quite easy thanks to rvest package. Once I go over this website form, I have timings of every runner (more than 11.000) at 5, 10, 15, 20, 21.097, 25, 30, 35, 40, and 42.195 kilometers. This, togheter with the category of each runner is the base for my analysis. I remove categories with a small number of runners. This is the Box plot of the finish time by category: Who maintains the rhythm better? Since I have the time at the end and at the middle (21.097 km), I can do a Box plot with the variation time between first and second half of the Marathon: Only a handful of people pull out of the Marathon before finishing but once again this rate is not the same between categories. This is the survival rate by category along the race: This plots show some interesting things: • Fastest category is 35-40 years old for both genders • Fastest individuals are inside 24-35 years old for both genders • Youngest ages are not the fastest • Between 40-45 category is the second fastest for men and the third one for women • Females between 40-45 keep the most constant rhythm of all categories. • Young men between 22-24 years old are the most unconstant: their second half rhythm is much slower than the first one. • All females between 55-60 years old ended the marathon • On the other hand, males between 60-65 are the category with most ‘deads’ during the race Long life forties: library(rvest) library(lubridate) library(ggplot2) library(plyr) library(sqldf) library(scales) library(gplots) setwd("YOUR WORKING DIRECTORY HERE") maraton_web="http://www.maratonmadrid.org/resultados/clasificacion.asp?carrera=13&parcial=par1&clasificacion=1&dorsal=&nombre=&apellidos=&pais=&pagina=par2" #Grid with parameters to navigate in the web to do webscraping searchdf=rbind(expand.grid( 1, 0:32), expand.grid( 2, 0:55), expand.grid( 3, 0:56), expand.grid( 4, 0:56), expand.grid( 5, 0:56), expand.grid( 6, 0:55), expand.grid( 7, 0:55), expand.grid( 8, 0:55), expand.grid( 9, 0:53), expand.grid(10, 0:55)) colnames(searchdf)=c("parcial", "pagina") #Webscraping. I open the webpage and download the related table with partial results results=data.frame() for (i in 1:nrow(searchdf)) { maraton_tmp=gsub("par2", searchdf[i,2], gsub("par1", searchdf[i,1], maraton_web)) df_tmp=html(maraton_tmp) %>% html_nodes("table") %>% .[[3]] %>% html_table() results=rbind(results, data.frame(searchdf[i,1], df_tmp[,1:7])) } #Name the columns colnames(results)=c("Partial", "Place", "Bib", "Name", "Surname", "Cat", "Gross", "Net") #Since downloadind data takes time, I save results in RData format save(results, file="results.RData") load("results.RData") #Translate Net timestamp variable into hours results$NetH=as.numeric(dhours(hour(hms(results$Net)))+dminutes(minute(hms(results$Net)))+dseconds(second(hms(results$Net))))/3600 results$Sex=substr(results$Cat, 3, 4) #Translate Cat into years and gender results$Cat2=revalue(results$Cat, c("A-F"="18-22 Females", "A-M"="18-22 Males", "B-F"="22-24 Females", "B-M"="22-24 Males", "C-F"="24-35 Females", "C-M"="24-35 Males", "D-F"="35-40 Females", "D-M"="35-40 Males", "E-F"="40-45 Females", "E-M"="40-45 Males", "F-F"="45-50 Females", "F-M"="45-50 Males", "G-F"="50-55 Females", "G-M"="50-55 Males", "H-F"="55-60 Females", "H-M"="55-60 Males", "I-F"="60-65 Females", "I-M"="60-65 Males", "J-F"="65-70 Females", "J-M"="65-70 Males", "K-F"="+70 Females", "K-M"="+70 Males")) #Translate partial code into kilometers results$PartialKm=mapvalues(results$Partial, from = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10), to = c(5, 10, 15, 20, 21.097, 25, 30, 35, 40, 42.195)) #There are some categories with very few participants. I will remove from the analysis count(subset(results, Partial==10), "Cat") #General options for ggplot opts=theme( panel.background = element_rect(fill="gray98"), panel.border = element_rect(colour="black", fill=NA), axis.line = element_line(size = 0.5, colour = "black"), axis.ticks = element_line(colour="black"), panel.grid.major = element_line(colour="gray75", linetype = 2), panel.grid.minor = element_blank(), axis.text = element_text(colour="gray25", size=15), axis.title = element_text(size=20, colour="gray10"), legend.key = element_blank(), legend.background = element_blank(), plot.title = element_text(size = 32, colour="gray10")) #Data set with finish times results1=subset(results, Partial==10 & !(Cat %in% c("A-F", "A-M", "B-F", "I-F", "J-F", "K-F", "K-M"))) ggplot(results1, aes(x=reorder(Cat2, NetH, FUN=median), y=NetH)) + geom_boxplot(aes(fill=Sex), colour = "gray25")+ scale_fill_manual(values=c("hotpink", "lawngreen"), name="Sex", breaks=c("M", "F"), labels=c("Males", "Females"))+ labs(title="Finish Time by Category of Rock 'n' Roll Madrid Marathon 2014", x="Category (age and sex)", y="Finish time (hours)")+ theme(axis.text.x = element_text(angle = 90, vjust=.5, hjust = 0), legend.justification=c(1,0), legend.position=c(1,0))+opts #Data set with variation times results2=sqldf("SELECT a.Bib, a.Sex, b.Cat2, (a.NetH-2*b.NetH)*60 as VarMin FROM results1 a INNER JOIN results b ON (a.Bib = b.Bib AND b.Partial=5) order by VarMin asc") ggplot(results2, aes(x=reorder(Cat2, VarMin, FUN=median), y=VarMin)) + geom_boxplot(aes(fill=Sex))+ scale_fill_manual(values=c("hotpink", "lawngreen"), name="Sex", breaks=c("M", "F"), labels=c("Males", "Females"))+ labs(title="Time Variation by Category Between First and Last Half\nof Rock 'n' Roll Madrid Marathon 2014", x="Category (age and sex)", y="Variation (minutes)")+ theme(axis.text.x = element_text(angle = 90, vjust=.5, hjust = 0), legend.justification=c(1,0), legend.position=c(1,0))+opts results3_tmp1=expand.grid(Cat2=unique(results$Cat2), PartialKm=unique(results$PartialKm)) results3_tmp2=sqldf("SELECT Bib, Sex, Cat2, Max(PartialKm) as PartialKmMax FROM results WHERE Cat NOT IN ('A-F', 'A-M', 'B-F', 'I-F', 'J-F', 'K-F', 'K-M') GROUP BY 1,2,3") results3_tmp3=sqldf("SELECT PartialKmMax, Sex, Cat2, COUNT(*) AS Runners FROM results3_tmp2 GROUP BY 1,2,3") results3_tmp4=sqldf("SELECT a.Cat2, a.PartialKm, SUM(Runners) as Runners FROM results3_tmp1 a INNER JOIN results3_tmp3 b on (a.Cat2 = b.Cat2 AND b.PartialKmMax>=a.PartialKm) GROUP BY 1,2") #Data set with survival rates results3=sqldf("SELECT a.Cat2, a.PartialKm, a.Runners*1.00/b.Runners*1.00 as Po_Survivors FROM results3_tmp4 a INNER JOIN (SELECT Cat2, COUNT(*) as Runners FROM results3_tmp2 GROUP BY 1) b ON (a.Cat2 = b.Cat2)") ggplot(results3, aes(x=PartialKm, y=Po_Survivors, group=Cat2, colour=Cat2)) + geom_line(lwd=3)+ scale_color_manual(values=alpha(rich.colors(15, palette="temperature"), 0.3), name="Category")+ scale_x_continuous(breaks = unique(results3$PartialKm), labels=c("5", "10", "15", "20", "21.097", "25", "30", "35", "40", "42.195"))+
scale_y_continuous(labels = percent)+
labs(title="Survival Rate by Category of Rock 'n' Roll Madrid Marathon 2014", x="Kilometer", y="% of survivors")+
theme(axis.text.x = element_text(angle = 90, vjust=.5, hjust = 0), legend.justification=c(1,0), legend.position=c(.15,0))+opts
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Generating prime numbers
While playing around with prime number generation in Python, I stumbled upon this post by Will Ness. In his comment, Will makes an interesting observation that leads to a huge optimization. His algorithm is difficult to understand at first glance, so I took the time to break it down and find out exactly where his trick comes into play. Here are the results.
The sieve of Eratosthenes
At $$i = 2$$, the cells look like this:
| | | x | | x | | x | | x |
-------------------------------------
2 3 4 5 6 7 8 9 10
At $$i = 3$$, the cells look like this:
| | | x | | x | | x | x | x |
-------------------------------------
2 3 4 5 6 7 8 9 10
We use a set instead of an array to keep track of the composites. A number is in the set if its cell is crossed out (it is composite) and it is not in the set otherwise.
def limited_sieve(limit):
composites = set()
for i in range(2, limit):
if i not in composites:
yield i
for j in range(2 * i, limit, i):
composites.add(j)
We can print the prime numbers by iterating over the generator:
primes = limited_sieve(100)
for p in primes:
print(p)
Notice that when crossing out multiples of a prime number $$p > 2$$, we don’t need to worry about crossing out $$2 \times p$$ since that number has already been crossed out by 2. The same goes for $$p > 3$$ and $$3 \times p$$. In fact, the first multiple that has not been crossed out yet is $$p \times p$$, so we can start our iteration from there:
def limited_sieve(limit):
...
for j in range(i * i, limit, i):
composites.add(j)
Removing the limit
Instead of iterating over all the multiples of each prime number, we lazily remember the last multiple that we have crossed out for each prime number. Each number has a bucket containing its prime divisors. When we process a number, if the bucket is empty, it is a prime number. Otherwise, we take each prime divisor and move it to the next number that it divides. For example, at $$i = 6$$, the buckets look like this:
2
| | | | | 3 | | | | 5 |
-------------------------------------
2 3 4 5 6 7 8 9 10
At $$i = 7$$, the buckets look like this:
| | | | | | | 2 | 3 | 5 |
-------------------------------------
2 3 4 5 6 7 8 9 10
We use the count function from the itertools library to generate all the numbers starting from 3, and the defaultdict data structure from the collections library to have dictionaries with default values.
import itertools
import collections
def bucket_sieve():
yield 2
factors = collections.defaultdict(set)
for i in itertools.count(3):
if i not in factors:
yield i
else:
for p in factors[i]:
del factors[i]
Empty buckets don’t actually exist in the factors dictionary and don’t take up memory. We delete the factors of $$i$$ at the end of the loop because we don’t need them anymore. As a result, each prime number generated so far appears exactly once in one of the buckets of the factors table. The space complexity is therefore $$O(P)$$, where $$P$$ is the number of prime numbers generated so far.
Again, we can optimize this algorithm by jumping to the square multiple instead of the double:
def bucket_sieve(limit):
...
if i not in factors:
yield i
factors[i * i].add(i)
Removing the buckets
Bucket manipulation slows the algorithm down. We only need to know one prime divisor in order to conclude that a number is composite. With this observation, we can get rid of the buckets and put only one prime divisor in each cell. While moving a prime number, if the new cell is occupied, we keep searching until we find a multiple that has an empty cell.
At $$i = 4$$, the cells look like this:
| | | 2 | | 3 | | | | |
-------------------------------------
2 3 4 5 6 7 8 9 10
At $$i = 5$$, the cells look like this:
| | | | | 3 | | 2 | | |
-------------------------------------
2 3 4 5 6 7 8 9 10
Again, we can optimize this algorithm by jumping to the square multiple instead of the double.
def flat_sieve():
yield 2
factor = dict()
for i in itertools.count(3):
if i not in factor:
yield i
insert(factor, 2 * i, i)
else:
p = factor[i]
del factor[i]
insert(factor, i + p, p)
def insert(table, m, p):
while m in table:
m += p
table[m] = p
This version of the algorithm has exactly the same time complexity as limited_sieve but is not limited.
Again, we can optimize this algorithm by jumping to the square multiple instead of the double:
def flat_sieve(limit):
...
if i not in factors:
yield i
insert(factor, i * i, i)
Delaying insertions (Will Ness’s trick)
In the previous version, we immediately insert the square of each prime number we encounter. This introduces a lot of useless information that we won’t need until much later. For example, at $$i = 6$$, the cells look like this:
| | | | | 3 | | 2 | | 5 |
-------------------------------------
2 3 4 5 6 7 8 9 10
At $$i = 8$$, the cells look like this:
| | | | | | | 2 | 3 | 5 | | 7 |
------------------------------------- -----
2 3 4 5 6 7 8 9 10 ... 14
We can avoid these insertions and save up some memory by delaying an insertion until we encounter the multiple. How do we know when we encounter a multiple? We need to “remember” primes that occured previously, but if we store all the primes that we generate, we are back to square one. The key insight of Will Ness is that we can use a separate stream of prime numbers, that we exhaust at a slower rate to generate the prime multiples that we need to insert at the current position. This time, before looking at the cell of a number is empty, we first check if the number is a multiple of the first prime number of the delayed stream.
How do we get access to a separate stream of prime numbers? By calling our generator, recursively!
def delayed_sieve():
yield 2
factor = dict()
primes = delayed_sieve()
delayed_prime = next(primes)
for i in itertools.count(3):
if i == 2 * delayed_prime:
insert(factor, i, delayed_prime)
delayed_prime = next(primes)
if i not in factor:
yield i
else:
p = factor[i]
del factor[i]
insert(factor, i + p, p)
At $$i = 8$$, the secondary stream is still at 5, and the cells look like this:
| | | | | | | 2 | 3 | |
-------------------------------------
2 3 4 5 6 7 8 9 10
We can already guess that this algorithm uses less memory. How far can we delay this secondary stream? As pointed out previously, the first multiple of $$p$$ that has no other prime divisor is $$p \times p$$. We cannot delay more than that.
def delayed_sieve():
...
if i == delayed_prime * delayed_prime:
insert(factor, i, delayed_prime)
delayed_prime = next(primes)
...
What impact does this have on memory? As before, each prime number generated so far is in at most one cell. This time however, prime numbers larger than $$\sqrt{i}$$ are delayed and are not inserted. The memory complexity is therefore $$O(\sqrt{P})$$, a huge improvement! This optimization reduces the memory consumption drastically. With it, I could easily generate all the prime numbers below 1,000,000,000 without running out of memory.
You might be wondering about the number of recursive calls that are being made here. Surely the delayed stream is calling another, even more delayed stream, and so on. Don’t these streams consume memory too? First, note that the recursive call returns immediately because a generator does not compute anything until we consume its elements, so there is no risk of exploding the stack. Second, each delayed stream is at the square root of the position of the previous one. This number falls down very rapidly and eventually reaches 2. To give you an idea, at $$i = 2^{2^{10}}$$ (a number with 308 digits), we are only using around 10 streams in total.
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# Thread: Word problems for fun!
1. Hooray for feelers solving riddles!
2. Three Masters of Logic wanted to find out who was the wisest amongst them. So they turned to their Grand Master, asking to resolve their dispute. “Easy,” the old sage said. "I will blindfold you and paint either red, or blue dot on each man’s forehead. When I take your blindfolds off, if you see at least one red dot, raise your hand. The one, who guesses the color of the dot on his forehead first, wins." And so it was said, and so it was done. The Grand Master blindfolded the three contestants and painted red dots on every one. When he took their blindfolds off, all three men raised their hands as the rules required, and sat in silence pondering. Finally, one of them said: "I have a red dot on my forehead."
How did he guess?
3. Let us name the men A, B, and C respectively; furthermore, let us assume that A is the man who guessed correctly.
A observes that both B and C have raised their hands, indicating both have seen a red dot. However, let us assume that A is observed to have a blue dot on his head. Then, when B saw C raising his hand, he must conclude that he, B, has a red dot, since C could only see a red dot on his forehead. Therefore, he would have immediately guessed that his dot was red. Therefore, since A having blue would immediately lead to a correct guess by either B or C, he must have a red dot.
This solution assumes that the man who guessed correctly believes the other men to be smart enough to guess rather quickly should he have had a blue dot. This, of course, isn't a certainty. I would imagine if the other two had down syndrome, or some other mental defect, he would have had a more difficult time.
Edit: The essential part of this problem rests in the fact that if at least one blue dot is present, then guesses will come within a given time X. A person must be certain of the fact that the others haven't been able to reach a conclusion before he himself can make one. Without this, it cannot be known with certainty.
4. Originally Posted by Jonnyboy
Let us name the men A, B, and C respectively; furthermore, let us assume that A is the man who guessed correctly.
A observes that both B and C have raised their hands, indicating both have seen a red dot. However, let us assume that A is observed to have a blue dot on his head. Then, when B saw C raising his hand, he must conclude that he, B, has a red dot, since C could only see a red dot on his forehead. Therefore, he would have immediately guessed that his dot was red. Therefore, since A having blue would immediately lead to a correct guess by either B or C, he must have a red dot.
This solution assumes that the man who guessed correctly believes the other men to be smart enough to guess rather quickly should he have had a blue dot. This, of course, isn't a certainty. I would imagine if the other two had down syndrome, or some other mental defect, he would have had a more difficult time.
Edit: The essential part of this problem rests in the fact that if at least one blue dot is present, then guesses will come within a given time X. A person must be certain of the fact that the others haven't been able to reach a conclusion before he himself can make one. Without this, it cannot be known with certainty.
That is an interesting angle, psychologically speaking. Your analysis is of course correct, but my personal approach was: if A had a blue dot, this would give the other two men an unfair advantage over him because for him his dot could technically be either color (provided nobody lied) while the other two would have been able to almost immediately see that theirs had to be red.
Be it because we assume the others aren't stupid or because we assume the chances of winning the challenge have to be even, we path there is the same. So we agree.
5. Suppose there exist 1000 gold coins which are to be split up among five pirates: 1,2,3,4, and 5 in order of rank. We assume that the coins must remain wholly intact and that the pirates have the following characteristics: a pirate is infinitely knowledgeable; a pirate values his own life above wealth and above his need to kill, and will thus always vote for his own proposal; a pirate will always choose a greater amount of wealth over killing another pirate; caeteris paribus, a pirate will kill another pirate if given the chance; a pirate is risk neutral.
Starting with the highest numbered pirate, he can make a proposal as to how the coins will be split up. This proposal can either be accepted or the pirate making the proposal is killed. A proposal is accepted if and only if a majority of the pirates accept it. If a proposal is accepted the coins are split up in accordance with the proposal. If a proposal is rejected and the pirate making the proposal is killed, the next ranking pirate makes a proposal, so on and so forth.
What proposal should pirate 5 make?
6. A man has a lighter and two lengths of robe with varying densities. The only thing the man knows about each length of rope is that if he lights an end of a rope, it will burn for exactly one hour; however, it may be that 90% of a rope burns up in 1 minute, and it takes 59 minutes for the remaining section of rope to burn (this is a product of the varying densities).
How can this man accurately measure 45 minutes?
7. I see a thing. Of this thing I know these three:
1) The man who makes this will sell it.
2) The man who buys this will not use it.
3) The man who uses this will not know it.
What do I see?
8. Ever heard of vampires?
9. Originally Posted by Red Herring
Ever heard of vampires?
10. Originally Posted by Jonnyboy
What proposal should pirate 5 make?
He should offer one coin each to 1 & 3 and offer nothing to the others.
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# What even number divided by 2 is odd?
Note that even numbers are all divisible by 2. Examples are:
`2, 4 , 6 ,8 , 10, 12, 14, 16, 18, 20...`
Then, let's try to determine which of these even numbers would result to an odd quotient when divided by 2. Note that odd numbers are not divisible by 2.
`2-:2=1` `4-:2=2`
`6-:2=3` `8-:2=4`
`10-:2=5` `12-:2=6`
`14-:2=7` `16-:2=8`
`18-:2=9` `20-:2=10`
So, 2,6,10,14 and 18 are the even numbers that when divided by 2 result to odd numbers.
--------------------------------------------------------------------------------
Therefore,
`2 + 4n` (where n is an integer)
are the even numbers that when divided by 2 result to an odd quotient.
Approved by eNotes Editorial Team
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## Search found 20 matches
#### Sum of positive integers x and y is 72
###### Data Sufficiency
II.
is GMAT is checking that Unit digits of both are 0
i mean 08, 09... because no other options have same Tens digit
Sat Oct 10, 2009 6:58 am
Forum: Data Sufficiency
Topic: Sum of positive integers x and y is 72
Replies: 8
Views: 8571
#### if X>1 and Y>1, is X<Y?
###### Data Sufficiency
I also go with B
Sat Aug 08, 2009 5:07 am
Forum: Data Sufficiency
Topic: if X>1 and Y>1, is X<Y?
Replies: 9
Views: 6297
#### RECRUITS
##### RECRUITS
can anybody explain this ?
OA : C
Fri Aug 07, 2009 8:35 am
Forum: Data Sufficiency
Topic: RECRUITS
Replies: 7
Views: 1074
#### gmat prep DS
###### Data Sufficiency
I go with A....
Fri Aug 07, 2009 6:32 am
Forum: Data Sufficiency
Topic: gmat prep DS
Replies: 12
Views: 1456
#### group
###### Data Sufficiency
1) we need numbers
2) 22 persons out of 55% of T, are supporting flat tax. but we need total number of persons in a group. They can be any numbers, 100, 200, etc. so we don't know how much % of 55% of T these 22 persons represent.
Fri Aug 07, 2009 6:11 am
Forum: Data Sufficiency
Topic: group
Replies: 3
Views: 860
#### circle
###### Problem Solving
Agreed w/ 1. Answer is C. For #2. We are left with an equalateral triangle that has radius 32 extending to the center from each point. Picture this as 3 triangles. Break again in 6 seperate triangles, which are now 30-60-90 w/ hypotenuse of 32. Solve from there. Answer is 768rt3. Each smaller trian...
Fri Aug 07, 2009 5:59 am
Forum: Problem Solving
Topic: circle
Replies: 6
Views: 1221
###### Problem Solving
Case 1 - 5 heads: (6/10)^5 * (4/10) = (6^5 * 4)/(10^6) Case 2 - 6 heads = (6^6)/(10^6) Adding the two cases gives us (6^5)/(10^5) which can be simplified to (3/5)^5...or (0.6)^5 Whats the OA? Thanks. Could you please elaborate how you got (3/5)^5...or (0.6)^5 after adding the two cases? I added the...
Fri Aug 07, 2009 5:49 am
Forum: Problem Solving
Replies: 14
Views: 1420
#### TWO TRIANGLES IN A RECTANGLE
##### TWO TRIANGLES IN A RECTANGLE
OA - A
Thu Aug 06, 2009 6:10 pm
Forum: Data Sufficiency
Topic: TWO TRIANGLES IN A RECTANGLE
Replies: 3
Views: 771
#### Flags
###### Problem Solving
is it 24 ?
Thu Aug 06, 2009 4:45 pm
Forum: Problem Solving
Topic: Flags
Replies: 5
Views: 1211
#### Kth term
###### Problem Solving
PussInBoots wrote:1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 + ... = 1/4 + 1/16 + 1/64 + 1/256 + 1/ 1024
1/4 < ... < 1/2
How you got 1/4 and 1/2 at the end...could you please explain it in detai....
Thanks.
Thu Aug 06, 2009 3:06 pm
Forum: Problem Solving
Topic: Kth term
Replies: 5
Views: 981
#### Kth term
###### Problem Solving
Thu Aug 06, 2009 11:26 am
Forum: Problem Solving
Topic: Kth term
Replies: 5
Views: 981
#### OG 11th - Problem Solving Question 20
###### Problem Solving
this is typo........
check the position on point E...its not at 2, its between 1 and 2, while point A is exactly at -2.
Thu Aug 06, 2009 7:42 am
Forum: Problem Solving
Topic: OG 11th - Problem Solving Question 20
Replies: 1
Views: 707
#### TIME AND WORK
###### Problem Solving
i think its 15 days, choice B. because that is the only option I see, where B is 3 times A (15 days to 5 days) and difference is of 10 days.
Thu Aug 06, 2009 5:59 am
Forum: Problem Solving
Topic: TIME AND WORK
Replies: 3
Views: 775
#### if a>b>0, then sqrt a^2 + b^2
###### Problem Solving
use formula of a^2 - b^2
its (a+b)(a-b)
Wed Aug 05, 2009 6:26 pm
Forum: Problem Solving
Topic: if a>b>0, then sqrt a^2 + b^2
Replies: 8
Views: 14956
#### Question of the Day - 5th August, 2009
###### Problem Solving
whether order is important?
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In the arctangent formula, we have that:
$$\arctan{u}+\arctan{v}=\arctan\left(\frac{u+v}{1-uv}\right)$$
however, only for $uv<1$. My question is: where does this condition come from? The situation is obvious for $uv=1$, but why the inequality?
One of the possibilities I considered was as following: the arctangent addition formula is derived from the formula:
$$\tan\left(\alpha+\beta\right)=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}.$$
Hence, if we put $u=\tan{\alpha}$ and $v=\tan{\beta}$ (which we do in order to obtain the arctangent addition formula from the one above), the condition that $uv<1$ would mean $\tan\alpha\tan\beta<1$, which, in turn, would imply (thought I am NOT sure about this), that $-\pi/2<\alpha+\beta<\pi/2$, i.e. that we have to stay in the same period of tangent.
However, even if the above were true, I still do not see why we have to stay in the same period of tangent for the formula for $\tan(\alpha+\beta)$ to hold. I would be thankful for a thorough explanation.
-
You are correct in that it is related to the period. Note however, that while the period is unimportant for the tan addition formula to hold, the arc tan functions are defined by restricting the range to$(\dfrac{-\pi}{2},\dfrac{\pi}{2})$ If $uv>1$, $\arctan u +\arctan v$ is not in the range of the arctan function(principal branch). In that case $\arctan \dfrac{u+v}{1-uv}$ is not the sum of the arctan's, it is shifted by $\pi$, up or down. Note that $\tan(\arctan u +\arctan v)=\tan(\arctan \dfrac{u+v}{1-uv})$ regardless of $uv<1$.
-
OK, I think I understand: since the result, $\arctan\left(\frac{u+v}{1-uv}\right)$ must be within $(-\pi/2,\pi/2)$, we obviously cannot have $\arctan{u}+\arctan{v}\notin(-\pi/2,\pi/2)$, and this is the case if $uv>1$. Is that correct? – Johnny Westerling Mar 10 '13 at 11:10
Yes that's about it, as far as I know. – Ishan Banerjee Mar 10 '13 at 11:13
Well, I guess that's a good explanation. Thank you! – Johnny Westerling Mar 10 '13 at 11:37
If $\arctan x=A,\arctan y=B;$ $\tan A=x,\tan B=y$
We know, $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
So, $$\tan(A+B)=\frac{x+y}{1-xy}$$ $$\implies\arctan\left(\frac{x+y}{1-xy}\right)=n\pi+A+B=n\pi+\arctan x+\arctan y$$ where $n$ is any integer
As the principal value of $\arctan z$ lies $\in[-\frac\pi2,\frac\pi2], -\pi\le\arctan x+\arctan y\le\pi$
$(1)$ If $\frac\pi2<\arctan x+\arctan y\le\pi, \arctan\left(\frac{x+y}{1-xy}\right)=\arctan x+\arctan y-\pi$ to keep $\arctan\left(\frac{x+y}{1-xy}\right)\in[-\frac\pi2,\frac\pi2]$
Observe that $\arctan x+\arctan y>\frac\pi2\implies \arctan x,\arctan y>0\implies x,y>0$
$\implies\arctan x>\frac\pi2-\arctan y$ $\implies x>\tan\left(\frac\pi2-\arctan y\right)=\cot \arctan y=\cot\left(\text{arccot}\frac1y\right)\implies x>\frac1y\implies xy>1$
$(2)$ If $-\pi\le\arctan x+\arctan y<-\frac\pi2, \arctan\left(\frac{x+y}{1-xy}\right)=\arctan x+\arctan y+\pi$
Observe that $\arctan x+\arctan y<-\frac\pi2\implies \arctan x,\arctan y<0\implies x,y<0$
Let $x=-X^2,y=-Y^2$
$\implies \arctan(-X^2)+\arctan(-Y^2)<-\frac\pi2$ $\implies \arctan(-X^2)<-\frac\pi2-\arctan(-Y^2)$ $\implies -X^2<\tan\left(-\frac\pi2-\arctan(-Y^2)\right)=\cot\arctan(-Y^2)=\cot\left(\text{arccot}\frac{-1}{Y^2}\right)$
$\implies -X^2<\frac1{-Y^2}\implies X^2>\frac1{Y^2}\implies X^2Y^2>1\implies xy>1$
$(3)$ If $-\frac\pi2\le \arctan x+\arctan y\le \frac\pi2, \arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right)$
-
Very nice answer (+1) – Johnny Westerling Mar 11 '13 at 4:43
The last answer, at point (3), does not explain clearly why $$\arctan x + \arctan y \in [-\frac{\pi}{2},\frac{\pi}{2}]$$ implies $xy \leqslant 1$. <br /><br /> I've published a rigorous proof of the sum of arctangents on this page. – user104981 Nov 2 '13 at 23:28
@MicheleDeStefano, The sole target of the answer is to find the value of $n$ for the different ranges of values of $\arctan x,\arctan y$. For $(3),$ the sum is already within the required range so $n=0$ – lab bhattacharjee Nov 5 '13 at 15:51
@lab bhattacharjee, Ok. I agree. But (3) does not answer the original question at the beginning of the thread. The demonstration I've posted does. – user106080 Nov 7 '13 at 7:15
Hye, sir. Probably can't remember me; but you helped me earlier in trigonometry. But may I again request you to please help in my latest quo? A comment is enough! Please forgive me if I bothered you. But would be grateful if you help:) – MAFIA36790 Mar 10 '15 at 12:54
## protected by Community♦Dec 1 '13 at 17:59
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MATHEMATICA TUTORIAL
# Converting between Strings, Boxes, and Expressions
ToString[expr,form] create a string representing the specified textual form of expr ToBoxes[expr,form] create boxes representing the specified textual form of expr ToExpression[input,form] create an expression by interpreting a string or boxes as input in the specified textual form ToString[expr] create a string using OutputForm ToBoxes[expr] create boxes using StandardForm ToExpression[input] create an expression using StandardForm
Converting between strings, boxes, and expressions.
Here is a simple expression.
Out[1]=
This gives the InputForm of the expression as a string.
Out[2]=
In FullForm explicit quotes are shown around the string.
Out[3]//FullForm=
This gives a string representation for the StandardForm boxes that correspond to the expression.
Out[4]//FullForm=
ToBoxes yields the boxes themselves.
Out[5]=
In generating data for files and external programs, it is sometimes necessary to produce two-dimensional forms which use only ordinary keyboard characters. You can do this using OutputForm.
This produces a string which gives a two-dimensional rendering of the expression, using only ordinary keyboard characters.
Out[6]=
The string consists of two lines, separated by an explicit newline.
Out[7]//FullForm=
The string looks right only in a monospaced font.
Out[8]=
If you operate only with one-dimensional structures, you can effectively use ToString to do string manipulation with formatting functions.
This generates a string corresponding to the OutputForm of StringForm.
Out[9]//InputForm=
InputForm strings corresponding to keyboard input StandardForm strings or boxes corresponding to standard two-dimensional input (default) TraditionalForm strings or boxes mimicking traditional mathematical notation
Some forms handled by ToExpression.
This creates an expression from an InputForm string.
Out[10]=
This creates the same expression from StandardForm boxes.
Out[11]=
In TraditionalForm these are interpreted as functions.
Out[12]=
ToExpression[input,form,h] create an expression, then wrap it with head h
Creating expressions wrapped with special heads.
This creates an expression, then immediately evaluates it.
Out[13]=
This creates an expression using StandardForm rules, then wraps it in Hold.
Out[14]=
You can get rid of the Hold using ReleaseHold.
Out[15]=
SyntaxQ["string"] determine whether a string represents syntactically correct Mathematica input SyntaxLength["string"] find out how long a sequence of characters starting at the beginning of a string is syntactically correct
Testing correctness of strings as input.
ToExpression will attempt to interpret any string as Mathematica input. But if you give it a string that does not correspond to syntactically correct input, then it will print a message, and return \$Failed.
This is not syntactically correct input, so ToExpression does not convert it to an expression.
Out[16]=
ToExpression requires that the string correspond to a complete Mathematica expression.
Out[17]=
You can use the function SyntaxQ to test whether a particular string corresponds to syntactically correct Mathematica input. If SyntaxQ returns False, you can find out where the error occurred using SyntaxLength. SyntaxLength returns the number of characters which were successfully processed before a syntax error was detected.
SyntaxQ shows that this string does not correspond to syntactically correct Mathematica input.
Out[18]=
SyntaxLength reveals that an error was detected after the third character in the string.
Out[19]=
Here SyntaxLength returns a value greater than the length of the string, indicating that the input was correct so far as it went, but needs to be continued.
Out[20]=
## Tutorial CollectionTutorial Collection
New to Mathematica? Find your learning path »
Have a question? Ask support »
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Inferential Statistics
Quiz by Amie Harris
Feel free to use or edit a copy
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22 questions
• Q1
If I carried out a study using an unrelated measures design and nominal data, which statistical test would I use?
Wilcoxon
Mann Whitney
Chi square
Sign Test
30s
• Q2
If I carried out a study were I used a correlation and both co-variables collected interval data, which statistical test would I use?
Pearsons
Binomial
Wilcoxon
Mann Whitney
30s
• Q3
If I calculated an observed value of 3.22 and found a critical value of 4.89 for a Chi-Square test. Would it be significant or not?
significant
not significant
30s
• Q4
How is degrees of freedom calculated for Chi-square
number of participants - 1
(number of rows - 1) x (number of columns -1)
(number of rows - 1) + (number of columns -1)
30s
• Q5
If I calculated an observed value of 6.62 and found a critical value of 3.34 for a Wilcoxon test. Would it be significant or not?
significant
not significant
30s
• Q6
If I calculated an observed value of 7.98 and found a critical value of 7.66 for a Spearman's rho test. Would it be significant or not?
not significant
significant
30s
• Q7
How do you find the critical value for Mann Whitney U
you calculate degrees of freedom
number of participants - 1
you find the number of differences between conditions
number of participants for each condition
30s
• Q8
How do you find the CRITICAL value for a Sign test?
you find the total number of participants
you calculate degrees of freedom
you find the number of participants that show a difference (ignore ppts with no difference)
30s
• Q9
Which test am I? I test for a difference between two sets of scores. I am used where data is interval and I am used where the data is unrelated.
Wilcoxon
Pearson's r
Unrelated t test
Related t test
30s
• Q10
Which test am I? I am a test for the difference between two sets of scores. I am used where data is interval and I am used where the data is related.
Pearson's r
Related t test
Spearman's rho
Chi-square
30s
• Q11
Which test am I? I use a correlational design, I look for a relationship and I collect interval data for each co-variable.
Pearson's r
Unrelated t test
Mann Whitney
Sign test
30s
• Q12
Which test am I? I look for a difference between sets of scores. I collect ordinal data and I use a repeated measures design.
Chi-square
Wilcoxon
Sign test
Unrelated t test
30s
• Q13
Which test am I? I collect nominal data, I look for a difference and I use an unrelated design.
Sign Test
Chi-square
Spearmans
Pearson's r
30s
• Q14
Students want to investigate the correlation between happiness scores and how many friends people have on Facebook. Identify the correct statistical test
Related t test
Spearman's rho
Wilcoxon
Sign test
30s
• Q15
A researcher wants to investigate whether reaction times increase after a person is given a dose of caffeine equivalent to three cups of coffee. The participant’s base reaction times are compared with those 30 minutes after the caffeine intake. Identify the correct statistical test.
Related t test
Chi-square
Mann Whitney
Unrelated t test
30s
Teachers give this quiz to your class
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https://convertoctopus.com/15-6-centimeters-to-millimeters
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## Conversion formula
The conversion factor from centimeters to millimeters is 10, which means that 1 centimeter is equal to 10 millimeters:
1 cm = 10 mm
To convert 15.6 centimeters into millimeters we have to multiply 15.6 by the conversion factor in order to get the length amount from centimeters to millimeters. We can also form a simple proportion to calculate the result:
1 cm → 10 mm
15.6 cm → L(mm)
Solve the above proportion to obtain the length L in millimeters:
L(mm) = 15.6 cm × 10 mm
L(mm) = 156 mm
The final result is:
15.6 cm → 156 mm
We conclude that 15.6 centimeters is equivalent to 156 millimeters:
15.6 centimeters = 156 millimeters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 millimeter is equal to 0.0064102564102564 × 15.6 centimeters.
Another way is saying that 15.6 centimeters is equal to 1 ÷ 0.0064102564102564 millimeters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that fifteen point six centimeters is approximately one hundred fifty-six millimeters:
15.6 cm ≅ 156 mm
An alternative is also that one millimeter is approximately zero point zero zero six times fifteen point six centimeters.
## Conversion table
### centimeters to millimeters chart
For quick reference purposes, below is the conversion table you can use to convert from centimeters to millimeters
centimeters (cm) millimeters (mm)
16.6 centimeters 166 millimeters
17.6 centimeters 176 millimeters
18.6 centimeters 186 millimeters
19.6 centimeters 196 millimeters
20.6 centimeters 206 millimeters
21.6 centimeters 216 millimeters
22.6 centimeters 226 millimeters
23.6 centimeters 236 millimeters
24.6 centimeters 246 millimeters
25.6 centimeters 256 millimeters
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https://luxuper.com/math-question-466
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# Complex fractions solver
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https://docs.scipy.org/doc/scipy-1.5.3/reference/generated/scipy.fft.ihfft.html
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scipy.fft.ihfft¶
scipy.fft.ihfft(x, n=None, axis=- 1, norm=None, overwrite_x=False, workers=None, *, plan=None)[source]
Compute the inverse FFT of a signal that has Hermitian symmetry.
Parameters
xarray_like
Input array.
nint, optional
Length of the inverse FFT, the number of points along transformation axis in the input to use. If n is smaller than the length of the input, the input is cropped. If it is larger, the input is padded with zeros. If n is not given, the length of the input along the axis specified by axis is used.
axisint, optional
Axis over which to compute the inverse FFT. If not given, the last axis is used.
norm{None, “ortho”}, optional
Normalization mode (see fft). Default is None.
overwrite_xbool, optional
If True, the contents of x can be destroyed; the default is False. See fft for more details.
workersint, optional
Maximum number of workers to use for parallel computation. If negative, the value wraps around from os.cpu_count(). See fft for more details.
plan: object, optional
This argument is reserved for passing in a precomputed plan provided by downstream FFT vendors. It is currently not used in SciPy.
New in version 1.5.0.
Returns
outcomplex ndarray
The truncated or zero-padded input, transformed along the axis indicated by axis, or the last one if axis is not specified. The length of the transformed axis is n//2 + 1.
Notes
hfft/ihfft are a pair analogous to rfft/irfft, but for the opposite case: here, the signal has Hermitian symmetry in the time domain and is real in the frequency domain. So, here, it’s hfft, for which you must supply the length of the result if it is to be odd: * even: ihfft(hfft(a, 2*len(a) - 2) == a, within roundoff error, * odd: ihfft(hfft(a, 2*len(a) - 1) == a, within roundoff error.
Examples
>>> from scipy.fft import ifft, ihfft
>>> spectrum = np.array([ 15, -4, 0, -1, 0, -4])
>>> ifft(spectrum)
array([1.+0.j, 2.+0.j, 3.+0.j, 4.+0.j, 3.+0.j, 2.+0.j]) # may vary
>>> ihfft(spectrum)
array([ 1.-0.j, 2.-0.j, 3.-0.j, 4.-0.j]) # may vary
scipy.fft.hfft
scipy.fft.hfft2
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http://www.gurufocus.com/term/COGS/NWE/Cost%2Bof%2BGoods%2BSold/NorthWestern%2BCorporation
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Switch to:
NorthWestern Corp (NYSE:NWE)
Cost of Goods Sold
\$401 Mil (TTM As of Sep. 2016)
NorthWestern Corp's cost of goods sold for the three months ended in Sep. 2016 was \$96 Mil. Its cost of goods sold for the trailing twelve months (TTM) ended in Sep. 2016 was \$401 Mil.
Cost of Goods Sold is directly linked to profitability of the company through Gross Margin. NorthWestern Corp's Gross Margin for the three months ended in Sep. 2016 was 68.05%.
Cost of Goods Sold is also directly linked to Inventory Turnover. NorthWestern Corp's Inventory Turnover for the three months ended in Sep. 2016 was 1.85.
Definition
Cost of goods sold (COGS) refers to the Inventory costs of those goods a business has sold during a particular period.
NorthWestern Corp Cost of Goods Sold for the trailing twelve months (TTM) ended in Sep. 2016 was 107.369 (Dec. 2015 ) + 115.434 (Mar. 2016 ) + 81.693 (Jun. 2016 ) + 96.156 (Sep. 2016 ) = \$401 Mil.
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
Cost of Goods Sold is directly linked to profitability of the company through Gross Margin.
NorthWestern Corp's Gross Margin for the three months ended in Sep. 2016 is calculated as:
Gross Margin = (Revenue - Cost of Goods Sold) / Revenue = (300.998 - 96.156) / 300.998 = 68.05 %
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
A company that has a moat can usually maintain or even expand their Gross Margin. A company can increase its Gross Margin in two ways. It can increase the prices of the goods it sells and keeps its Cost of Goods Sold unchanged. Or it can keep the sales price unchanged and squeeze its suppliers to reduce the Cost of Goods Sold. Warren Buffett believes businesses with the power to raise prices have moats.
Cost of Goods Sold is also directly linked to another concept called Inventory Turnover:
NorthWestern Corp's Inventory Turnover for the three months ended in Sep. 2016 is calculated as:
Inventory Turnover = Cost of Goods Sold / Average Inventory = 96.156 / 52.117 = 1.85
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Inventory Turnover measures how fast the company turns over its inventory within a year. A higher inventory turnover means the company has light inventory. Therefore the company spends less money on storage, write downs, and obsolete inventory. If the inventory is too light, it may affect sales because the company may not have enough to meet demand.
Usually retailers pile up their inventories at holiday seasons to meet the stronger demand. Therefore, the inventory of a particular quarter of a year should not be used to calculate inventory turnover. An average inventory is a better indication.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
NorthWestern Corp Annual Data
Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 COGS 0 0 699 574 531 495 395 480 483 373
NorthWestern Corp Quarterly Data
Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 COGS 112 95 108 112 80 74 107 115 82 96
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https://www.instructables.com/Gauging-Roof-Pitch-From-the-Ground/
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## Introduction: Gauging Roof Pitch From the Ground!
Knowing the slope of your roof is good for all kinds of things! From knowing if you have to worry about ice dams to wondering if you should really send you kids up to take down the Christmas lights because you don’t want to, or any reason in-between…it’s good to know the pitch of your roof. Although we could break out the ladder, a level, a tape measure, and get ourselves to get on the roof…it’s a lot easier to gather this info from the safety and convenience of the ground. Here’s how! BTW…the tape measure will still be required.
Step 1: Walk away from your house, perpendicular to the wall that is under the roof pitch you’d like to measure. Walking backwards, and watching where you’re going, walk out to the distance where your eye aligns with the plane on the roof pitch. So you should essentially see the top edge of the eve or gutter line up with the peak of the roof pitch. Assuming the ground is relatively flat, measure the height of your eye from the ground. Record this measurement.
Step 2: Measure the distance from where you’re standing to the face of the wall you walked away from. As a tip, if you had hooked the tape measure to the house when you backed away from the wall, you’ll be able to do this without help from someone else! You’ll also need to know (or measure) the distance of your home’s overhang. By subtracting the overhang distance from your standing distance, we will have identified the ‘Run’ value of your roof pitch! Record that value in both inch & foot values!
Step 3: Measure the height from the ground to the edge of your eve or gutter. Then, take that measurement (102” as pictured) and subtract your earlier eye height (60” example pictured). This will give you your rise value! Be aware that a steep sloped ground surface may throw off this value. We are looking to find the distance from your eye height to the eve edge. So feel free to adjust your measured eye height as you see fit to compensate for a higher or lower ground surface under the eve. Since roof pitches are pretty nominal, we’ve got a lot of play here without effecting the accuracy of our measurement, but if you think it has sloped 6” or more, you’ll want to account for it.
Solve for the Answer!: Roof pitch is calculated in a Rise to Run ratio, with the run value being a 12” standard. So basically, the pitch answers the question of, ‘how many inches does the roof rise over the course of 12 horizontal inches?”. To solve this, we simply plug in our values for Rise and Run. In this example, it is 42” to 84”. We then divide both numbers by the dividend that will get the run value to 12. This number is easily found in step 2! When we recorded the measurement in step 2, we recorded the inch & foot values. In our example, we recorded that value of 84” and 7’. By dividing both numbers in our ratio by the recorded foot value of 7, the 42:84 turns into our final 6:12 pitch value!
Easy enough! …but feel free to use the calculator on your phone if you need to.
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# Relational Algebra - Download Now PowerPoint
### Pages to are hidden for
``` Relational Algebra
Lecture #9
1
Querying the Database
• Goal: specify what we want from our database
Find all the employees who earn more than
\$50,000 and pay taxes in Champaign-Urbana.
• Could write in C++/Java, but bad idea
• Instead use high-level query languages:
– Theoretical: Relational Algebra, Datalog
– Practical: SQL
– Relational algebra: a basic set of operations on relations
that provide the basic principles.
2
Motivation: The Stack
• To use the "stack" data structure in my program, I
need to know
– what a stack looks like
– what (useful) operations I can perform on a stack
• PUSH and POP
• Next, I look for an implementation of stack
– browse the Web
– find many of them
– choose one, say LEDA
3
Motivation: The Stack (cont.)
• LEDA already implement PUSH and POP
• It also gives me a simple language L, in which to
define a stack and call PUSH and POP
– S = init_stack(int);
– S.push(3); S.push(5);
– int x = S.pop();
• Can also define an expression of operations on
stacks
– T = init_stack(int);
– T.push(S.pop());
4
Motivation: The Stack (cont.)
• To summarize, I know
– definition of stack
– its operations (PUSH, POP): that is, a stack algebra
– an implementation called LEDA, which tells
me how to call PUSH and POP in a language L
– I can use these implementations to manipulate stacks
– LEDA hides the implementation details
– LEDA optimizes implementation of PUSH and POP
5
Now Contrast It with Rel. Databases
def of
• To summarize, I know relations
relational
algebra
– definition of stack
– its operations (PUSH, POP): that is, a stack algebra
SQL
– an implementation called LEDA, which tells language
me how to call PUSH and POP in a language L
– I can use these implementations to manipulate stacks
– LEDA hides the implementation details
– LEDA optimizes implementation of PUSH and POP
operation and query
optimization 6
What is an “Algebra”
• Mathematical system consisting of:
– Operands --- variables or values from which new
values can be constructed.
– Operators --- symbols denoting procedures that
construct new values from given values.
7
What is Relational Algebra?
• An algebra whose operands are relations or
variables that represent relations.
• Operators are designed to do the most common
things that we need to do with relations in a
database.
– The result is an algebra that can be used as a query
language for relations.
8
Relational Algebra at a Glance
• Operators: relations as input, new relation as output
• Five basic RA operations:
– Basic Set Operations
• union, difference (no intersection, no complement)
– Selection: s
– Projection: p
– Cartesian Product: X
• When our relations have attribute names:
– Renaming: r
• Derived operations:
– Intersection, complement
– Joins (natural,equi-join, theta join, semi-join)
9
Five Basic RA Operations
10
Set Operations
• Union, difference
• Binary operations
11
Set Operations: Union
• Union: all tuples in R1 or R2
• Notation: R1 U R2
• R1, R2 must have the same schema
• R1 U R2 has the same schema as R1, R2
• Example:
– ActiveEmployees U RetiredEmployees
12
Set Operations: Difference
• Difference: all tuples in R1 and not in R2
• Notation: R1 – R2
• R1, R2 must have the same schema
• R1 - R2 has the same schema as R1, R2
• Example
– AllEmployees - RetiredEmployees
13
Selection
• Returns all tuples which satisfy a condition
• Notation: sc(R)
• c is a condition: =, <, >, and, or, not
• Output schema: same as input schema
• Find all employees with salary more than
\$40,000:
– sSalary > 40000 (Employee)
14
Selection Example
Employee
SSN Name DepartmentID Salary
999999999 John 1 30,000
777777777 Tony 1 32,000
888888888 Alice 2 45,000
Find all employees with salary more than \$40,000.
s Salary > 40000 (Employee)
SSN Name DepartmentID Salary
888888888 Alice 2 45,000
15
Ullman: Selection
• R1 := SELECTC (R2)
– C is a condition (as in “if” statements) that refers to
attributes of R2.
– R1 is all those tuples of R2 that satisfy C.
16
Example
Relation Sells:
bar beer price
Joe’s Bud 2.50
Joe’s Miller 2.75
Sue’s Bud 2.50
Sue’s Miller 3.00
bar beer price
Joe’s Bud 2.50
Joe’s Miller 2.75
17
Projection
• Unary operation: returns certain columns
• Eliminates duplicate tuples !
• Notation: P A1,…,An (R)
• Input schema R(B1,…,Bm)
• Condition: {A1, …, An} {B1, …, Bm}
• Output schema S(A1,…,An)
• Example: project social-security number and
names:
– P SSN, Name (Employee)
18
Projection Example
Employee
SSN Name DepartmentID Salary
999999999 John 1 30,000
777777777 Tony 1 32,000
888888888 Alice 2 45,000
P SSN, Name (Employee)
SSN Name
999999999 John
777777777 Tony
888888888 Alice
19
Projection
• R1 := PROJL (R2)
– L is a list of attributes from the schema of R2.
– R1 is constructed by looking at each tuple of R2,
extracting the attributes on list L, in the order
specified, and creating from those components a tuple
for R1.
– Eliminate duplicate tuples, if any.
20
Example
Relation Sells:
bar beer price
Joe’s Bud 2.50
Joe’s Miller 2.75
Sue’s Bud 2.50
Sue’s Miller 3.00
Prices := PROJbeer,price(Sells):
beer price
Bud 2.50
Miller 2.75
Miller 3.00
21
Cartesian Product
• Each tuple in R1 with each tuple in R2
• Notation: R1 x R2
• Input schemas R1(A1,…,An), R2(B1,…,Bm)
• Condition: {A1,…,An} {B1,…Bm} = F
• Output schema is S(A1, …, An, B1, …, Bm)
• Notation: R1 x R2
• Example: Employee x Dependents
• Very rare in practice; but joins are very common
22
Cartesian Product Example
Employee
Name SSN
John 999999999
Tony 777777777
Dependents
EmployeeSSN Dname
999999999 Emily
777777777 Joe
Employee x Dependents
Name SSN EmployeeSSN Dname
John 999999999 999999999 Emily
John 999999999 777777777 Joe
Tony 777777777 999999999 Emily
Tony 777777777 777777777 Joe
23
Product
• R3 := R1 * R2
– Pair each tuple t1 of R1 with each tuple t2 of R2.
– Concatenation t1t2 is a tuple of R3.
– Schema of R3 is the attributes of R1 and R2, in
order.
– But beware attribute A of the same name in R1
and R2: use R1.A and R2.A.
24
Example: R3 := R1 * R2
R1( A, B) R3( A, R1.B, R2.B, C )
1 2 1 2 5 6
3 4 1 2 7 8
1 2 9 10
R2( B, C ) 3 4 5 6
5 6 3 4 7 8
7 8 3 4 9 10
9 10
25
Renaming
• Does not change the relational instance
• Changes the relational schema only
• Notation: r B1,…,Bn (R)
• Input schema: R(A1, …, An)
• Output schema: S(B1, …, Bn)
• Example:
rLastName, SocSocNo (Employee)
26
Renaming Example
Employee
Name SSN
John 999999999
Tony 777777777
rLastName, SocSocNo (Employee)
LastName SocSocNo
John 999999999
Tony 777777777
27
Renaming
• The RENAME operator gives a new schema
to a relation.
• R1 := RENAMER1(A1,…,An)(R2) makes R1 be
a relation with attributes A1,…,An and the
same tuples as R2.
• Simplified notation: R1(A1,…,An) := R2.
28
Example
Bars( name, addr )
Joe’s Maple St.
Sue’s River Rd.
R(bar, addr) := Bars
R( bar, addr )
Joe’s Maple St.
Sue’s River Rd.
29
Derived RA Operations
1) Intersection
2) Most importantly: Join
30
Set Operations: Intersection
• Difference: all tuples both in R1 and in R2
• Notation: R1 R2
• R1, R2 must have the same schema
• R1 R2 has the same schema as R1, R2
• Example
– UnionizedEmployees RetiredEmployees
• Intersection is derived:
– R1 R2 = R1 – (R1 – R2) why ?
31
Joins
• Theta join
• Natural join
• Equi-join
• Semi-join
• Inner join
• Outer join
• etc.
32
Theta Join
• A join that involves a predicate
• Notation: R1 q R2 where q is a condition
• Input schemas: R1(A1,…,An), R2(B1,…,Bm)
• {A1,…An} {B1,…,Bm} = f
• Output schema: S(A1,…,An,B1,…,Bm)
• Derived operator:
R1 q R2 = s q (R1 x R2)
33
Theta-Join
• R3 := R1 JOINC R2
– Take the product R1 * R2.
– Then apply SELECTC to the result.
• As for SELECT, C can be any boolean-valued
condition.
– Historic versions of this operator allowed only A theta
B, where theta was =, <, etc.; hence the name “theta-
join.”
34
Example
Sells( bar, beer, price ) Bars( name, addr )
Joe’s Bud 2.50 Joe’s Maple St.
Joe’s Miller 2.75 Sue’s River Rd.
Sue’s Bud 2.50
Sue’s Coors 3.00
BarInfo := Sells JOIN Sells.bar = Bars.name Bars
BarInfo( bar, beer, price, name, addr )
Joe’s Bud 2.50 Joe’s Maple St.
Joe’s Miller 2.75 Joe’s Maple St.
Sue’s Bud 2.50 Sue’s River Rd.
Sue’s Coors 3.00 Sue’s River Rd. 35
Natural Join
• Notation: R1 R2
• Input Schema: R1(A1, …, An), R2(B1, …, Bm)
• Output Schema: S(C1,…,Cp)
– Where {C1, …, Cp} = {A1, …, An} U {B1, …, Bm}
• Meaning: combine all pairs of tuples in R1 and R2
that agree on the attributes:
– {A1,…,An} {B1,…, Bm} (called the join attributes)
• Equivalent to a cross product followed by selection
• Example Employee Dependents
36
Natural Join Example
Employee
Name SSN
John 999999999
Tony 777777777
Dependents
SSN Dname
999999999 Emily
777777777 Joe
Employee Dependents =
PName, SSN, Dname(s SSN=SSN2(Employee x rSSN2, Dname(Dependents))
Name SSN Dname
John 999999999 Emily
Tony 777777777 Joe
37
Natural Join
• R= A
X
B
Y
S= B
Z
C
U
X Z V W
Y Z Z V
Z V
• R S = A B C
X Z U
X Z V
Y Z U
Y Z V
Z V W
38
Natural Join
• Given the schemas R(A, B, C, D), S(A, C, E),
what is the schema of R S ?
• Given R(A, B, C), S(D, E), what is R S ?
• Given R(A, B), S(A, B), what is R S ?
39
Natural Join
• A frequent type of join connects two relations by:
– Equating attributes of the same name, and
– Projecting out one copy of each pair of equated
attributes.
• Called natural join.
• Denoted R3 := R1 JOIN R2.
40
Example
Sells( bar, beer, price ) Bars( bar, addr )
Joe’s Bud 2.50 Joe’s Maple St.
Joe’s Miller 2.75 Sue’s River Rd.
Sue’s Bud 2.50
Sue’s Coors 3.00
BarInfo := Sells JOIN Bars
Note Bars.name has become Bars.bar to make the natural
join “work.”
BarInfo( bar, beer, price, addr )
Joe’s Bud 2.50 Maple St.
Joe’s Milller 2.75 Maple St.
Sue’s Bud 2.50 River Rd.
Sue’s Coors 3.00 River Rd. 41
Equi-join
• Most frequently used in practice:
R1 A=B R2
• Natural join is a particular case of equi-join
• A lot of research on how to do it efficiently
42
Semijoin
• R S = P A1,…,An (R S)
• Where the schemas are:
– Input: R(A1,…An), S(B1,…,Bm)
– Output: T(A1,…,An)
43
Semijoin
Applications in distributed databases:
• Product(pid, cid, pname, ...) at site 1
• Company(cid, cname, ...) at site 2
• Query: sprice>1000(Product) cid=cid Company
• Compute as follows:
T1 = sprice>1000(Product) site 1
T2 = Pcid(T1) site 1
send T2 to site 2 (T2 smaller than T1)
T3 = T2 Company site 2 (semijoin)
send T3 to site 1 (T3 smaller than Company)
Answer = T3 T1 site 1 (semijoin)
44
Relational Algebra
• Five basic operators, many derived
• Combine operators in order to construct queries:
relational algebra expressions, usually shown as
trees
45
Building Complex Expressions
• Algebras allow us to express sequences of
operations in a natural way.
• Example
– in arithmetic algebra: (x + 4)*(y - 3)
– in stack "algebra": T.push(S.pop())
• Relational algebra allows the same.
• Three notations, just as in arithmetic:
1. Sequences of assignment statements.
2. Expressions with several operators.
3. Expression trees.
46
Sequences of Assignments
• Create temporary relation names.
• Renaming can be implied by giving relations a
list of attributes.
• Example: R3 := R1 JOINC R2 can be written:
R4 := R1 * R2
R3 := SELECTC (R4)
47
Expressions with Several Operators
• Example: the theta-join R3 := R1 JOINC R2 can
be written: R3 := SELECTC (R1 * R2)
• Precedence of relational operators:
1. Unary operators --- select, project, rename --- have
highest precedence, bind first.
2. Then come products and joins.
3. Then intersection.
4. Finally, union and set difference bind last.
But you can always insert parentheses to force
the order you desire.
48
Expression Trees
• Leaves are operands --- either variables standing
for relations or particular, constant relations.
• Interior nodes are operators, applied to their child
or children.
49
Example
• Using the relations Bars(name, addr) and
Sells(bar, beer, price), find the names of all the
bars that are either on Maple St. or sell Bud for
less than \$3.
50
As a Tree:
• Using the relations Bars(name, addr) and Sells(bar, beer,
price), find the names of all the bars that are either on
Maple St. or sell Bud for less than \$3.
UNION
RENAMER(name)
PROJECTname PROJECTbar
SELECTaddr = “Maple St.” SELECT price<3 AND beer=“Bud”
Bars Sells 51
Example
• Using Sells(bar, beer, price), find the bars that
sell two different beers at the same price.
• Strategy: by renaming, define a copy of Sells,
called S(bar, beer1, price). The natural join of
Sells and S consists of quadruples (bar, beer,
beer1, price) such that the bar sells both beers
at this price.
52
The Tree
PROJECTbar
SELECTbeer != beer1
JOIN
RENAMES(bar, beer1, price)
Sells Sells
53
Complex Queries
Product ( pid, name, price, category, maker-cid)
Purchase (buyer-ssn, seller-ssn, store, pid)
Company (cid, name, stock price, country)
Person(ssn, name, phone number, city)
Note:
•in Purchase: buyer-ssn, seller-ssn are foreign keys in Person, pid is foreign key in
Product;
•in Product maker-cid is a foreign key in Company
Find phone numbers of people who bought gizmos from Fred.
Find telephony products that somebody bought
54
Expression Tree
P name
pid=pid
seller-ssn=ssn
P ssn P pid
sname=fred sname=gizmo
Person Purchase Person Product 55
Exercises
Product ( pid, name, price, category, maker-cid)
Purchase (buyer-ssn, seller-ssn, store, pid)
Company (cid, name, stock price, country)
Person(ssn, name, phone number, city)
Ex #1: Find people who bought telephony products.
Ex #2: Find names of people who bought American products
56
Exercises
Product ( pid, name, price, category, maker-cid)
Purchase (buyer-ssn, seller-ssn, store, pid)
Company (cid, name, stock price, country)
Person(ssn, name, phone number, city)
Ex #3: Find names of people who bought American products and did
not buy French products
Ex #4: Find names of people who bought American products and they
live in Champaign.
57
Exercises
Product ( pid, name, price, category, maker-cid)
Purchase (buyer-ssn, seller-ssn, store, pid)
Company (cid, name, stock price, country)
Person(ssn, name, phone number, city)
Ex #5: Find people who bought stuff from Joe or bought products
from a company whose stock prices is more than \$50.
58
Operations on Bags
(and why we care)
• Union: {a,b,b,c} U {a,b,b,b,e,f,f} = {a,a,b,b,b,b,b,c,e,f,f}
– add the number of occurrences
• Difference: {a,b,b,b,c,c} – {b,c,c,c,d} = {a,b,b,d}
– subtract the number of occurrences
• Intersection: {a,b,b,b,c,c} {b,b,c,c,c,c,d} = {b,b,c,c}
– minimum of the two numbers of occurrences
• Selection: preserve the number of occurrences
• Projection: preserve the number of occurrences (no duplicate
elimination)
• Cartesian product, join: no duplicate elimination
Read the book for more detail 59
Summary of Relational Algebra
• Why bother ? Can write any RA expression
directly in C++/Java, seems easy.
• Two reasons:
– Each operator admits sophisticated implementations
(think of , s C)
– Expressions in relational algebra can be rewritten:
optimized
60
Implementations of Operators
• s(age >= 30 AND age <= 35)(Employees)
– Method 1: scan the file, test each employee
– Method 2: use an index on age
– Which one is better ? Depends a lot…
• Employees Relatives
– Iterate over Employees, then over Relatives
– Iterate over Relatives, then over Employees
– Sort Employees, Relatives, do “merge-join”
– “hash-join”
– etc
61
Product ( pid, name, price, category, maker-cid)
Purchase (buyer-ssn, seller-ssn, store, pid)
Person(ssn, name, phone number, city)
• Which is better:
sprice>100(Product) (Purchase scity=seaPerson)
(sprice>100(Product) Purchase) scity=seaPerson
• Depends ! This is the optimizer’s job…
62
Finally: RA has Limitations !
• Cannot compute “transitive closure”
Name1 Name2 Relationship
Fred Mary Father
Mary Joe Cousin
Mary Bill Spouse
Nancy Lou Sister
• Find all direct and indirect relatives of Fred
• Cannot express in RA !!! Need to write C program
63
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# Thread: stuck on second x^x
1. ## stuck on second x^x
$\begin{gathered}
f(x) = {x^{{x^x}}} \hfill \\
\ln y = {x^x}\ln x \hfill \\
\frac{1}
{y}y' = \frac{1}
{x}*\frac{1}
{x} \hfill \\
\end{gathered}
$
but for the x^x, ln is needed but is
$\ln \ln y = \ln ({x^x}\ln x)
$
okay?
2. Hello, genlovesmusic09!
You are correct . . .
Differentiate: . $y \:=\: x^{x^x}$
Take logs: . $\ln(y) \;=\;\ln\left(x^{x^x}\right) \quad\Rightarrow\quad \ln(y)\;=\;x^x\cdot\ln(x)$
Take logs: . $\ln\left[\ln(y)\right] \;=\;\ln\left[x^x\cdot\ln(x)\right] \;=\; \ln\left[x^x\right] + \ln\left[\ln(x)\right]$
And we have: . $\ln\left[\ln(y)\right] \;=\;x\cdot\ln(x) + \ln\left[\ln(x)\right]$
Differentiate: . $\frac{1}{\ln(y)}\!\cdot\!\frac{1}{y}\!\cdot\! y' \;\;=\;\;x\!\cdot\!\frac{1}{x} \;+\; 1\!\cdot\!\ln(x) \;+\; \frac{1}{\ln(x)}\!\cdot\!\frac{1}{x}$
. . . . . . . . . . . . $\frac{y'}{y\!\cdot\!\ln(y)} \;=\;1 + \ln(x) + \frac{1}{x\!\cdot\!\ln(x)}$
. . . . . . . . . . . . . . . $y' \;=\;y \cdot \ln(y)\cdot\bigg[1 + \ln(x) + \frac{1}{x\!\cdot\!\ln(x)} \bigg]$
. . . . . . . . . . . . . . . $y' \;=\;x^{x^x}\cdot\ln\left(x^{x^x}\right)\cdot\bigg[1 + \ln(x) + \frac{1}{x\!\cdot\!\ln(x)} \bigg]$
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# Can space-time be defined by the requirement that the physical laws are simple?
When I was student I was told that time is defined by the requirement that the physical laws are simple. For example, in classical mechanics time can be defined by the requirment that the velocity of an isolated body is constant. One could generalize this approach by assuming that space-time is defined by the requirement that the physical laws are simple. This can be easily expressed in mathematical language as follows.
In general relativity the universe is represented by the triple $(M, g, T)$ where $M$ is a four-dimensional differentiable manifold, and $g$ and $T$ are a Lorentzian metric and a tensor field on $M$ satisfying Einstein's equation.
According to the above approach, we could say that the "true" physical reality of the universe is represented by the pair $(M, T)$, while the metric $g$ emerges as an appeareance from the requirement that evolution appears simple, i.e., that $g$ and $T$ satisfy Einstein's equation.
From the mathematical point of view, this approach poses for example the following problems: does any tensor field $T$ admit a metric $g$ such that Einstein's equation is satisfied? To what extent a tensor field $T$ univocally determines the metric $g$?
Does this approach make sense, at least from the mathematical point of view?
-
Already the first paragraph is completely bogus, so before I continue reading, let me point this out and prompt you to make some sense of it. You can't define time by the requirement that velocity is constant. Without time, there is no concept of velocity at all. Time is the most fundamental notion in all of physics and is never defined a posteriori. In particular, in GR, we require that the manifold be Lorentzian, so again there is an implicit Minkowsian notion of space-time. There is no way to get around this. Also, there is no need to get around this (besides cheap philosophy...). – Marek Jun 16 '11 at 15:20
Let X be the configuration space of a classical N-particle system. Assume that a curve on X, i.e., the image of a map from R to X, is given. Any map with that image corresponds to a different definition of time. The correct definition of time corresponds to the maps which satisfy Newton's law, if any. – bgalvan Jun 16 '11 at 15:57
For what it's worth, the first paragraph makes sense to me. – Ted Bunn Jun 16 '11 at 18:12
@bgalvan: correct definition? It is easy to measure time in units which have arbitrary functional dependence on your Newtonian time. This notion of time is equally good and yet it would not satisfy Newton's first law, etc. – Marek Jun 16 '11 at 19:07
@Marek: correct definition?? If the measurement of time which arbitrary functional dependence is equally good as any other, why all people measure time such in a way that Netown's law is satisfied? – bgalvan Jun 16 '11 at 20:32
If $T$ is "non-degenerate" (at every point as a linear transformation from $T_pM \to T_p^*M$), then modulo some issues with the constraint equations $g$ can be solved from $T$, at least locally in time, after prescribing it in a compatible way on a space-like hypersurface. See this article by De Turck.
On the other hand, the prescription of initial/boundary data may be crucial. Roughly speaking the prescribing Ricci curvature equation is analogous to the inhomogeneous wave equation with a source term. In which case the hyperbolic nature of the equations shows that for any initial configuration there exists a different evolution. So it may be that this will lead to multiple different metrics compatible with your $T$ (indeed consider the case where you restrict your manifold to be $M = \mathbb{R}^4$; the global nonlinear stability of Minkowski space, combined with classical results on classical results on the existence of solutions to the vacuum constraint equations shows that on $M = \mathbb{R}^4$ there are many incongruent (since for these other solutions the Weyl tensor is non-vanishing) solutions to Einstein's equation when $T = 0$; I would expect something similar for the case of prescribed $T$).
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## The median of the first five prime numbers is 5.6 5 3 3.6
Question
The median of the first five prime numbers is
5.6
5
3
3.6
in progress 0
1 month 2021-08-19T09:29:56+00:00 2 Answers 0 views 0
first five prime numbers are 2, 3, 5, 7 and 11.
n = 5
=(n+1)/2th observation
= (5+1)/2th observation
= 6/2th observation
= 3rd observation
that is 5
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# Jensen's inequality for integrals
What nice ways do you know in order to prove Jensen's inequality for integrals? I'm looking for some various approaching ways.
Supposing that $\varphi$ is a convex function on the real line and $g$ is an integrable real-valued function we have that:
$$\varphi\left(\int_a^b f\right) \leqslant \int_a^b \varphi(f).$$
-
And by Jensen's inequality, do you mean something about convex functions? And not en.wikipedia.org/wiki/Jensen%27s_formula – GEdgar Jul 16 '12 at 17:44
@GEdgar: i know that in English is called Jensen's inequality for integrals (i hope i'm not wrong) and is related to convex functions. – OFFSHARING Jul 16 '12 at 17:47
@Chris: That reply does not really clarify what you mean. Why not just edit your question such that it explicitly quotes the statement you want a proof of? – Henning Makholm Jul 16 '12 at 17:53
Do you have that $b-a=1$? – robjohn Jul 16 '12 at 19:27
@robjohn: i think you're right. b-a=1 – OFFSHARING Jul 16 '12 at 19:38
First of all, Jensen's inequality requires a domain, $X$, where $$\int_X\,\mathrm{dx}=1\tag{1}$$ Next, suppose that $\varphi$ is convex on the range of $f$; this means that for any $t_0\in f(X)$, $$\frac{\varphi(t)-\varphi(t_0)}{t-t_0}\tag{2}$$ is non-decreasing on $f(X)\setminus\{t_0\}$. This means that we can find a $\Phi$ so that $$\sup_{t<t_0}\frac{\varphi(t)-\varphi(t_0)}{t-t_0}\le\Phi\le\inf_{t>t_0}\frac{\varphi(t)-\varphi(t_0)}{t-t_0}\tag{3}$$ and therefore, for all $t$, we have $$(t-t_0)\Phi\le\varphi(t)-\varphi(t_0)\tag{4}$$ Now, let $t=f(x)$ and set $$t_0=\int_Xf(x)\,\mathrm{d}x\tag{5}$$ and $(4)$ becomes $$\left(f(x)-\int_Xf(x)\,\mathrm{d}x\right)\Phi\le\varphi(f(x))-\varphi\left(\int_Xf(x)\,\mathrm{d}x\right)\tag{6}$$ Integrating both sides of $(6)$ while remembering $(1)$ yields $$\left(\int_Xf(x)\,\mathrm{d}x-\int_Xf(x)\,\mathrm{d}x\right)\Phi\le\int_X\varphi(f(x))\,\mathrm{d}x-\varphi\left(\int_Xf(x)\,\mathrm{d}x\right)\tag{7}$$ which upon rearranging, becomes $$\varphi\left(\int_Xf(x)\,\mathrm{d}x\right)\le\int_X\varphi(f(x))\,\mathrm{d}x\tag{8}$$
-
You are also taking $x=f(x)$, which is a confusing notation. – M Turgeon Jul 16 '12 at 18:57
@MTurgeon: I have changed to $t=f(x)$. – robjohn Jul 16 '12 at 19:13
@robjohn I corrected a small typo that had me scratching my head for a few moments. I hope you don't mind. – Potato Jan 27 '13 at 5:48
@Potato: thanks. That was a residual error from the previous edit. – robjohn Jan 27 '13 at 9:17
@robjohn would any steps in the proof change if $\varphi$ was convex on $\mathbb{R}^n$ instead of $\mathbb{R}$? – cap Nov 15 at 8:26
I like this, maybe it is what you want ...
Let $E$ be a separable Banach space, let $\mu$ be a probability measure defined on $E$, let $f : E \to \mathbb R$ be convex and (lower semi-)continuous. Then $$f\left(\int_E x d\mu(x)\right) \le \int_E f(x)\,d\mu(x) .$$ Of course we assume $\int_E x d\mu(x)$ exists, say for example $\mu$ has bounded support.
For the proof, use Hahn-Banach. Write $y = \int_E x d\mu(x)$. The super-graph $S=\{(x,t) : t \ge f(x)\}$ is closed convex. (Closed, because $f$ is lower semicontinuous; convex, because $f$ is convex.) So for any $\epsilon > 0$ by Hahn-Banach I can separate $(y,f(y)-\epsilon)$ from $S$. That is, there is a continuous linear functional $\phi$ on $E$ and a scalar $s$ so that $t \ge \phi(x)+s$ for all $(x,t) \in S$ and $\phi(y)+s > f(y)-\epsilon$. So: $$f(y) -\epsilon < \phi(y)+s = \phi\left(\int_E x d\mu(x)\right)+s = \int_E (\phi(x)+s) d\mu(x) < \int_E f(x) d\mu(x) .$$ This is true for all $\epsilon > 0$, so we have the conclusion.
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One way would be to apply the finite Jensen's inequality $$\varphi\left(\frac{\sum a_i x_i}{\sum a_j}\right) \le \frac{\sum a_i \varphi (x_i)}{\sum a_j}$$ to each Riemann sum. The finite inequality is itself easily proved by induction on the number of points, using the definition of convexity.
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Here's a nice proof:
Step 1: Let $\varphi$ be a convex function on the interval $(a,b)$. For $t_0\in (a,b)$, prove that there exists $\beta\in\mathbb{R}$ such that $\varphi(t)-\varphi(t_0)\geq\beta(t-t_0)$ for all $t\in(a,b)$.
Step 2: Take $t_0=\int_a^bfdx$ and $t=f(x)$, and integrate with respect to $x$ to prove the desired inequality.
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There needs to be something about $b-a=1$ or else this doesn't work. – robjohn Jul 16 '12 at 19:16
@robjohn Of course. To be honest, I thought the OP took this as an "invisible assumption" – M Turgeon Jul 16 '12 at 19:48
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