text:
B Sound Wave Doppler effect question

1. Sep 30, 2018

jha192001

I have seen few examples on Doppler effect and i am confused about one such.
We are standing on ground.
If the source of sound S moves and Object O is stationary. We would presume the frequency as well as wavelength of sound be changed to the obeject O.
But if O moves towards or away from S. Only wavelength of sound is changed.
Is this effect also seen in light wave Doppler effect?
what is this phenomenon? I am not able to find on it clearly, what changes and what not ?
Thanking you

2. Sep 30, 2018

davenn

yes

no, why would you think that "

if the wavelength changes, the frequency must also change, for the observer

If the sound source is moving and the observer is stationary or the opposite … moving observer and stationary sound source
the effect is the same

ohhh and that is the same for light or any other EM radiation
think of a police radar, laser or microwave, source is stationary and the car is moving
The wavelength shortens ( for car travelling towards the radar) and the frequency increases

3. Oct 1, 2018

Merlin3189

Are you not thinking solely of EM waves?
I thought it was generally recognised that there is a clear difference for waves dependent on a medium such as air.

I wonder if what the op is thinking about is that, when she is stationary she observes a consistent set of frequency, velocity and wavelength. The observed frequency is higher than the moving source claims to emit, but fits with the observed wavelength and known speed of propagation in that medium.
When the observer is moving she sees again a higher frequency on her oscilloscope or frequency meter than that emitted by the stationary source (though not changed by the same amount), but may be in a quandry about measuring the wavelength and speed of propagation. She knows she is moving through the medium and could take that into account in her measurements. If she does, she sees that the speed and wavelength measured on the ground are inconsistent with the uncorrected frequency she is observing.

This is the opposite of what op claims of course. I need to know how op is measuring f and λ.

4. Oct 1, 2018

davenn

Sound or EM
As far as I'm aware, the Doppler "effect" is the same, tho the process is slightly different ... medium/no medium

edit ... ohhh and as an afterthought .....

And of course, you can do the same thing with sound as what is being done in the lower section of the image with the laser/microwave radar

Last edited: Oct 2, 2018
5. Oct 2, 2018

Merlin3189

There are obvious differences if one of you travels faster than the speed of sound (which is why you don't have to worry for EM waves).
Source approaches > SoS, you don't hear anything until passes, then you hear lower f as it moves away.
You approach stationary source at > SoS, you hear raised f until it passes, then nothing more.

Otherwise there are quantitative diffs.
eg. , speed of sound 330 m/s, source 1 kHz
Stationary source,you move towards source at 165 m/s, you hear 1.5 kHz.
Stationary observer, source move towards you at 165 m/s, you hear 2 kHz.
(Please check by your own methods. I don't know any simple method, but I found online $f_o = f_s \frac {(V + V_s)} {(V + V_o)}$ and they seem to agree with those as far as I understand it.)

I can't check how this works for EM, as without a reference frame, especially for time, I'm stuck. All I know there is that it can't make any difference who moves, so it must be different from sound.

6. Oct 2, 2018

olivermsun

The difference for waves in a medium is that $c$ is fixed to the medium. Either way, you figure out the change in wave crests encountered per time by the (possibly moving) observer.

7. Oct 2, 2018

Janus

Staff Emeritus
For EMR in a vacuum you use $f_o = f_s \sqrt{ \frac {(1+\frac{v}{c})} {(1-\frac{v}{c})}}$

where v is the relative velocity between source and receiver, and is positive if they are approaching each other.

One way of looking at it is to treat the pure Doppler effect factor like you are considering a stationary observer and moving source, which gives
$f_o = f_s \frac {(c + v_s)} {c}$ or $f_o= f_s \left (1+\frac{v}{c} \right )$
(v = vs)

Then you apply the time dilation factor to the source frequency to get
$f_o = f_s \frac{1+\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}}$

$f_o = f_s \frac{1+\frac{v}{c}}{\sqrt{(1-\frac{v}{c}) (1+\frac{v}{c})}}$

$f_o = f_s \sqrt{ \frac {(1+\frac{v}{c})} {(1-\frac{v}{c})}}$

8. Oct 2, 2018

Merlin3189

Thanks. I wasn't asking, just pointing out my limited knowledge, but it's very kind of you to help me along. I've never been able to get any sort of understanding of relativity and I don't like relying formulae that I can't work out for myself.

I need to check up on 'time dilation factor' and also to convince myself that it's valid to start with the 'pure' Doppler formula. Otherwise it seems plausible. If my maths doesn't let me down, I think I get the same result if I start with the moving observer formula, which is a good sign! (Assuming I'm right in taking the equivalent case to v=vs, is v= -vo ) I suppose I should also try v= vs - vo and see what I get.

9. Oct 11, 2018

jha192001

Sorry sir i am still not fully satisfied.
For now lets only say about sound as a wave.(Not EM)
The same question again.
If only observer moves or only source. Whats the changes in both cases in the frequency,wavelength,speed of sounds as. heard by the object/observer. I know frequency changes by diffrent amount as per motion of people wrt medium. But please elaborate on this more.
Thanking You.

10. Oct 11, 2018

Dullard

A simpler way to grasp the concept(maybe):

Think simply in terms of 'time of flight' between emitter and receiver and use the positive peaks of an emitted sinusoid as your 'ruler.'

If the emitter/receiver are stationary WRT each other, then emitted = received. The distance is constant. All peaks travel the same distance.

If the emitter/receiver are moving toward each other, then the distance is getting shorter over time. The first positive peak of a sinusoidal waveform travels 'farther' than the 2nd, and so on...
The effect is that the peaks arrive (at the receiver) 'closer together' because the later peaks don't have to travel as far. If the rate of closure is constant, then the 'shift' is constant - that's doppler.

If the emitter/receiver are moving apart, then the distance is increasing; the peaks arrive 'farther apart.'

11. Oct 11, 2018

jha192001

Yes sir, But isnt that the explanation of the Doppler Effect? The question i put forward was the change the wave would go through in the two diffrent case mentioned above. Why doppler effect predicts diffrently on if the *Observer* is moving and on Source is only moving.

<< Post edited by a Mentor to fix the all-caps sections >>

Last edited by a moderator: Oct 11, 2018
12. Oct 11, 2018

Dullard

OK.
You apparently want to examine the wave 'independent' of the observer?

Think about it in the terms that I suggested. Sketch a snapshot of an emitted wave moving through the air. For a stationary emitter The emitted wavelength (from the POV of the emitter) will be identical to the 'standing' wavelength. If the emitter is moving in the same direction as the emitted wave, the peaks 'bunch' as described above. The peaks are closer together in space; The 'standing' frequency is higher. If the emitter is moving away, it's farther/lower.

For a stationary emitter and moving receiver, the 'standing' and emitted wavelengths are the same (I said that already). The moving receiver causes the peaks to be 'intercepted' by the receiver at a different rate than they were emitted.

In the case of the moving emitter, the 'objective' frequency of the wave in space is shifted; For the moving receiver, the shift is 'apparent at the receiver.' These cases are indistinguishable at the receiver.

OR: maybe I still don't understand your question.

13. Oct 11, 2018

Janus

Staff Emeritus
Assuming a non-relativistic Doppler effect of sound through the air:
Given:
speed of sound through air 343 m/s
Source frequency :10 hz
Relative speed between Source and receiver: 171.5 m/sec
Starting distance between Source and receiver: 343 m.

Source moving relative to air, receiver not:
The front of a sound wave leaves the source when it is 343 meters from the receiver at t = 0
This will arrive at the receiver at t = 1 sec.
At t= 0.10 sec, the back of the wave (start of the second wave) leaves the source. By this time it has moved 17.15 meters closer to the receiver, and will take an additional (343-17.15)/343 = .95 sec to reach the receiver, arriving at t = 1.05 sec. This is just 0.05 sec later than the front of the wave arrived. Thus the source measures a period of 0.05 sec for the wave and a frequency of 20 hz for the wave.
As far as wavelength goes, When the back end of the wave leaves the source at t =0.10 sec, the front end has moved 34.3 m relative to where it was emitted. This means it is just 34.3-17.15 = 171.5 m ahead of the back of the wave when it is emitted, the wavelength has been halved.

Receiver moving relative to air and source not:
Front of wave is still emitted when source and receiver are 343 m apart and at t=0. wave travels at 343 m/s towards receiver. Receiver is moving at 34.3 m/sec towards source ( and relative to the air). Front of wave and receiver meet 343(343+171.5) 0.666667... sec later at t = 2/3 sec...
Back of wave leaves source at t = .01, and when the receiver is 17.15 meters closer to the source. It will take 0.633333.. sec to meet up with the receiver, and they meet at t= 0.733333.. sec, or 0.066666... sec after the front of the wave, a wave period of 0.666666... sec equates to a frequency of ~ 15 hz, 3/4 of what he measured when it was the source moving relative to the air.
Wavelength: Since the source never moves relative to the air, the wavelength is just 34.3 m (the distance the wave travels relative to the air in one second). However, since the receiver is moving at 171.5 m/s relative to the air in one direction and the wave is moving at 343 m/sec relative to it in the other, the speed of the wave is 514.5 m/s relative to the receiver. 514.5/34.3 = 15 hz, the same answer we got above.

14. Oct 22, 2018

neilparker62

I think the OP's question is quite a deep one. In essence she ('she' is presumed - forgive if incorrectly!) is struggling to deal with the fact that for light there is no "ether" in which the wave propagates. And thus no distinction between observer moving/source stationary or observer stationary/source moving. In the case of sound waves there is an "ether" namely the air in which the wave propagates.


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## On numbers which are differences of two conjugates of an algebraic integer.(English)Zbl 1028.11065

The author investigates which algebraic integers $$\beta$$ can be written as the difference $$\alpha - \alpha'$$ of two algebraic integers conjugate over a given number field $$K$$. The analogous question under the assumption that $$\beta, \alpha,\alpha'$$ are conjugate algebraic numbers but not necessarily integers was solved by the author and C. J. Smyth [Glasg. Math. J. 44, 435-441 (2002; Zbl 1112.11308)]. He shows that if $$\beta$$ is of degree $$2$$ or $$3$$ over $$K$$ then a necessary and sufficient condition for $$\beta = \alpha-\alpha'$$ is that the trace of $$\beta$$ over $$K$$ is $$0$$. He also shows that if the minimal polynomial of $$\beta$$ over $$K$$ is of the form $$f(x^e)$$ with $$f \in \mathbb Z_K[x]$$ then $$\beta = \alpha-\alpha'$$. He considers also the analogous question when “difference” is replaced by “quotient” and the condition on $$\alpha$$ is that it be a unit.

### MSC:

 11R04 Algebraic numbers; rings of algebraic integers 11R16 Cubic and quartic extensions 11R21 Other number fields 11R32 Galois theory

### Keywords:

algebraic number; algebraic integer; conjugate

Zbl 1112.11308
Full Text:

### References:

 [1] Smyth, Acta Arith. 40 pp 333– (1982) [2] Hilbert, Jahresber. Deutsch. Math.-Verein 4 pp 175– (1897) [3] Lang, Algebra (1965)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.

format: plain


text:
failed to download blast db using update_blastdb.pl + decompress db file with update_blastdb.pl
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6.5 years ago
arfaj a ▴ 10

Hello all, First, I am trying to run update_blastdb.pl script to download updated database from ncbi ftp site but, got the following error in the command prompt:

C:\inetpub\wwwroot\webclient\blast\bin>

Any one can explain?

Second, How to decompress database file when using update_blastdb.pl? I try: perl update_blastdb.pl refseq_rna --decompress

But, got this error: Unknown option: decompress Failed to parse command line options

???

blast ncbi update_blastdb.pl • 11k views
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Entering edit mode

ftp://ftp.ncbi.nlm.nih.gov/blast/db/refseq_rna.00.tar.gz.md5

99ca13b7408ea832cc110c91fd684da0  refseq_rna.00.tar.gz

ftp://ftp.ncbi.nlm.nih.gov/blast/db/refseq_rna.00.tar.gz

After you downloaded the above files, check the md5 sum manually, instructions to do this on windows are here: https://support.microsoft.com/en-us/kb/889768

Second guess: your disk is full.

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Entering edit mode

Thanks, I think it from my connection. I have another question I will post it above.

1
Entering edit mode
6.5 years ago

Please read the documentation here, it covers all steps also for Windows users.

• If you are having problems with the download you should try the --passive option, using passive FTP should work in most or all settings and through most or all firewall settings. If you have a personal firewall you have to allow Perl to access the internet.
• Edit: There is no documented '--decompress' option to update_blastdb.pl, this has to be done manually afterwards unfortunately, from the documentation:

The compressed files of pre-formatted blast databases must be inflated with gzip or other decompress utilities. The working database files can then be extracted out of the resulting tar file using tar program on Unix/Linux or WinZip and StuffIt Expander on Windows and Macintosh platforms, respectively.

Edit: the decompress option is not documented in the ncbi book, but is present in the source code, since 2.2.22+ also in the latest windows version, but not in 2.2.18+ - 21! so the following should work update_blastdb.pl --decompress refseq_rna unless you have a very old version of blast, and indeed works for me, if your version is too old you should update.

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you are right. It does not mentioned in the documentation. But, I see it in different google search like: http://devbioinfo.to.cnr.it/doku.php?id=rachaelx:blastl

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Entering edit mode

I have checked the source code of update_blastdb and you are correct, there is an undocumented option:

"decompress"    =>  \$opt_decompress,  back to at least blast+ 2.2.22+ It uses the Archive::Tar perl module. ADD REPLY 0 Entering edit mode Yes, I also see it in the source code on the ncbi site. The source code of my local update_blastdb.pl is different. Is it save to copy the content of update_blastdb.pl on the website to my local one? I really need to decompress. ADD REPLY 0 Entering edit mode Update to the latest version of blast+, then perl update_blastdb.pl --decompress refseq_rna ADD REPLY 0 Entering edit mode I actually have the latest version of blast+. It does not include the decompress in the update_blastdb.pl. Can I just replace the source code of update_blastdb.pl with the one have decompress option? is it save? ADD REPLY 0 Entering edit mode The latest version for windows is 2.2.31, I have checked the source and it contains the parameter. Which version do you have? I guess that you have an old script left over in the installation directory, the latest script has the following commit information: #!/usr/bin/perl #$Id: update_blastdb.pl 446090 2014-09-11 12:12:27Z ivanov \$

You can replace the old script with the newer version, it should not depend on other blast files.


text:
coef.FDboost

0th

Percentile

Coefficients of boosted functional regression model

Takes a fitted FDboost-object produced by FDboost() and returns estimated coefficient functions/surfaces $\beta(t), \beta(s,t)$ and estimated smooth effects $f(z), f(x,z)$ or $f(x, z, t)$. Not implemented for smooths in more than 3 dimensions.

Usage
# S3 method for FDboost
coef(object, raw = FALSE, which = NULL,
computeCoef = TRUE, returnData = FALSE, n1 = 40, n2 = 40, n3 = 20,
n4 = 10, ...)
Arguments
object

a fitted FDboost-object

raw

logical defaults to FALSE. If raw = FALSE for each effect the estimated function/surface is calculated. If raw = TRUE the coefficients of the model are returned.

which

a subset of base-learners for which the coefficients should be computed (numeric vector), defaults to NULL which is the same as which=1:length(object\$baselearner). In the special case of which=0, only the coefficients of the offset are returned.

computeCoef

defaults to TRUE, if FALSE only the names of the terms are returned

returnData

return the dataset which is used to get the coefficient estimates as predictions, see Details.

n1

see below

n2

see below

n3

n1, n2, n3 give the number of grid-points for 1-/2-/3-dimensional smooth terms used in the marginal equidistant grids over the range of the covariates at which the estimated effects are evaluated.

n4

gives the number of points for the third dimension in a 3-dimensional smooth term

...

other arguments, not used.

Details

If raw = FALSE the function coef.FDboost generates adequate dummy data and uses the function predict.FDboost to compute the estimated coefficient functions.

Value

If raw = FALSE, a list containing

• offset a list with plot information for the offset.

• smterms a named list with one entry for each smooth term in the model. Each entry contains

• x, y, z the unique grid-points used to evaluate the smooth/coefficient function/coefficient surface

• xlim, ylim, zlim the extent of the x/y/z-axes

• xlab, ylab, zlab the names of the covariates for the x/y/z-axes

• value a vector/matrix/list of matrices containing the coefficient values

• dim the dimensionality of the effect

• main the label of the smooth term (a short label)

If raw = TRUE, a list containing the estimated spline coefficients.

Aliases
• coef.FDboost
Documentation reproduced from package FDboost, version 0.3-2, License: GPL-2

Community examples

Looks like there are no examples yet.


text:
help on connections

In the book on Riemannian Geometry by John Lee ("Riemannian manifolds: an Introduction to Curvature") the author gives an exercise on page 54 involving connections:

Let $\triangledown$ be a linear connection. If $\omega$ us a 1-form and $X$ is a vector field, show that the coordinate expression for $\triangledown_X \,\omega$ is $$\triangledown_X \, \omega = \left(X^i \partial_i \omega_k - X^i w_j \Gamma^j_{ik} \right) \, dx^k$$ where $\{\, \Gamma^k_{ij} \,\}$ are the Christoffel symbols of the given connection $\triangledown$ on TM.

I am a little stuck here, I appologize as I sense this should be really easy to solve. My guess is that I could use the property of the natural pairing $$\langle \partial_j , \ dx^k \, \rangle = \delta_k^j$$ together with the property $$\begin{eqnarray} &0 = \triangledown_{\partial_i} \delta_k^j = \triangledown_{\partial_i} \langle dx^k, \delta_j \, \rangle = \langle \triangledown_{\partial_i} \, dx^k, \partial_j \, \rangle + \langle dx^k, \triangledown_{\partial_i} \partial_j \, \rangle \\ &= \langle \triangledown_{\partial_i} \, dx^k, \partial_j \, \rangle + \langle dx^k, \Gamma^k_{ij} \partial_k \, \rangle = \langle \triangledown_{\partial_i} \, dx^k, \partial_j \, \rangle + \Gamma^k_{ij} \end{eqnarray} \quad \text{(is this correct ?)}$$ but I am not sure how to proceed or where I am having errors in my argument so far.. any help would be great ! sorry again if this is obvious .. thanks !

-

In the book (directly above the exercise on the same page) there is formula (4.7) - which for a tensor $F$ says that $$(\nabla_X F)(Y) = X(F(Y)) - F(\nabla_XY)$$ Just set $F=\omega$ to get the formula from Thomas answer $$(\nabla_X \omega)(Y) = X(\omega(Y)) - \omega(\nabla_XY)$$ and plug in $\omega=\omega_i dx^i$ and $X=X^j\partial_j$
I don't know that book, so I don't know how it introduces the nabla operator on forms. But assuming you know the corresponding formula for the covariant derivative of vector fields and the fact that, for each smooth vector field $Y$, $$X(\omega(Y)) = (\nabla_X \omega)(Y) + \omega(\nabla_XY)$$ (the left hand side being the derivative of the function $\omega(Y)$ in direction $X$) you can just solve this for $(\nabla_X \omega)(Y)$ and choose $Y=\partial_i$, say. Then plug in the coordinate representations of the other terms.


text:
#### Archived

This topic is now archived and is closed to further replies.

# multidimensional arrays

This topic is 5838 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

How do I make multidimensioned arrays? I know how to do 1 dimension
  //make an array that is array[sizex][sizey] int* array = new int[sizex]; //now how make the sizey part? //and when done, delete delete [] array;

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A 3 dimensional array.
  int* array = new int[sizex][sizey][sizez];

,Jay

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Although thats a pointer to a 3D array.

  int Array[x][y][z];

is how to declare an actual array.

,Jay

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quote:
Original post by Jason Zelos
A 3 dimensional array.

  int* array = new int[sizex][sizey][sizez];

You might like to consider checking this code.

--

It is against the law to stare at the mayor of Paris.

Edited by - SabreMan on February 25, 2002 3:54:00 PM

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SabreMan, do you know how to do it?

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you could

  // the linear 2d arrayint *array = new int[sizex*sizey];// access by doingarray[x+y*sizex];delete [] array;//// for a planar multidimensional array// int **array = (int**)(new int[sizey]);for(i=0; i

saberman, how would you do such a thing, since you obviously know (by your comment about the code being wrong). since either one is a mulitdimensional array. there may be a better way to allocate planar arrays in c++, i am not sure since i rarly use planar arrays and tend to use c methods of allocation (except for objects).

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I knew the syntax was illegal.

Dynamically allocating multi-dimensional arrays actually takes quite a lot of explanation, so I'm going to defer to someone who's already written up a perfectly good explanation, Marshall Cline...

How do I allocate multidimensional arrays using new?

Pointers and arrays are evil!

Edited by - SabreMan on February 26, 2002 7:14:36 AM

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So, is this right?
  vector< vector > array; Because I want to use a multidimensioned container of tiles struct tile{int id;int walk;}; But don't understand how. If the above is right I think I would do for(i = 0; i < map_width; i++) { for(j = 0; j < map_height; j++) { //initialize each tile so I have an //array of tiles WIDTHxHEIGHT tile tmp; tmp.id = 0; tmp.walk = 1; map[i].push_back(tmp); }}

Do you have any links where I can learn about STL? I don't know anything about it. [/source]

Edited by - evilclown on February 26, 2002 3:33:06 PM

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Something like this...
  int ***number_cube = new (int**)[5];for(int i = 0; i < 5; ++i) { number_cube[i] = new (int*)[5]; for(int j = 0; j < 5; ++j) { number_cube[i][j] = new int[5]; }}...for(int i = 0; i < 5; ++i) { for(int j = 0; j < 5; ++j) { delete[] number_cube[i][j]; } delete[] number_cube[i];}delete[] number_cube;

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I''d do it like this using C style syntax.

  struct tile { int id; int walk;}; tile TileMap[map_width][map_height];for(i = 0; i < map_width; i++) { for(j = 0; j < map_height; j++) { TileMap[i][j].id = 0; TileMap[i][j].walk = 1; }}

I think this is a more conventional way of doing it, if not easier to think about.

I''m going to have to talk to my eyes about reading posts all the way through before replying .

,Jay

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Jason Zelos, I can''t do that because the map will be a global variable. Each new map loaded could be a different size. If it''s global, I can''t change the dimensions without using new.

I want to be able to use a container if possible so I don''t have to worry about new and delete.

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why use stl? this is a simple memory allocation problem of a struct. just

  map = (tile*)malloc(width*height);// and access likemap[x*y*width].blah= 42;// free byfree(map);

if you insist on using stl, then do a search on google or gamedev, maybe even get a book.

you should consider "map" to be a part your map loading class. while you would probally want to leave the map varible public, you could just add functions that do the stuff you need (ie check for collision, load, unload, etc.)

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evilcrap''s short multi-dereferencing tutorial

when you work with pointers, after the first dimension, you need to dereference every pointer to get to its value.

1D, a vector
  char* pC = new char[length];cout << pC[x]; //output each char

2D, a Map
  char** Map = new char*[width]; for(int x = 0; x < width; x++) Map[x] = new char[height];cout << Map[x][y];

3d, a Cube
  char*** pMap = new char**[depth]; //create a series of Map pointersfor(z = 0; z < depth; z++){ pMap = new char*[width]; for(x = 0; x < width; x++) pMap[z][x] = new char[height];} cout << pMap[z][x][y];

------
if you examine a vector(1D array), then youll notice that the pointer itself is the first element of the array.
  cout << pChar; //pChar is an address, itll print till it finds a NULL cout << *pChar; //pChar is a char, itll print pChar[0]

so, guess at below some.
  cout << ***pMap; // ? pMap[0][0][0], it is an address cout << (**pMap)[y]; // ? pMap[0][0][y], it is an address cout << (*pMap)[x][y]; // ? pMap[0][x][y], it is a char cout << pMap[z][x][y]; // ? pMap[z][x][y], it is a char

-note that (*pMap)[y] != *pMap[y]

-although you could simulate any number of dimensions in a single vector, there is a performance cost of having to multiply alot.

-dereferencing is alot faster than multiplying... therefore, if you want your code to be readible and fast, you should use pointers.

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Think of it this way. A dynamically allocated 2d-array is an arroy of pointers. Each pointer points to an array, which is itself a pointer. Hence, your "outermost" layer is an array of pointers to pointers. For a cube, you would need pointers to pointers to pointers, each of which would point to a pointer to a pointer, each of which would point to a pointer to a block of memory.

It may seem comfusing, but I''m sure you''ll get to understand it pretty quickly.

An alternative to using multidimensional arrays is to use a 1-dimensional array as a direct address table. This is the simplest form of a technique called hashing. Basically, it takes multiple parameters, and creates a single number from them, which is an array index. A straightforward way to do this in 2d is "the way you read." If you number each array "slot" starting at the top left corner, and going across each row, working from top to bottom, then you get a simple pattern. An often-used function that determines the array index that corresponds to an x, y coordinate is (x + y * width). I''m sure you can extend this technique to three dimensions.

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why do ppl confuse things so? you can have a multi dimensional array WITHOUT using arrays of arrays.

there are two types, many ppl seem to ignore this fact. citing c/c++ books that state mulitdimensional arrays must be arrays of arrays (ie 2d array must be decalred "int **blah or int blah[y][x]") this just is not so.

a small tutorial that explains the difference between planar and linear memory and when to use each.

planar mulit dimensional arrays. these is how most ppl seem to think of them, and how evil crap explains it. basically you are creating arrays of pointers to other arrays. see my previous examples or evilcrap''s examples.

access: need to dereference at each level, the [] operator simplfies this.

pros:
easy access using [][][] for nice reading of code.
each row can have a different number of elements to save ram
since memory is not contigous, its more likly all the alloocs will work.

cons:
complex and messy (de)allocation, order does matter
multiple dereferences will slow access slightly (not much)

useful for command histories, consoles buffer (ie quake), or anything else that you dont need the same sized rows.

**********

linear mulit dimensional arrays. this is a standard array that is accessed in a multi dimensional way. these is how video buffers are ussually treated (linear ones anyway). access is done via mathmatics.

ie [x+y*width] for a 2d array, this can be expanded to 3d arrays as well like [x+y*width+z*width*height]. so you get the idea of how to expand to higher dimensions. basically each dimension you add, requires you to multiple by the previous dimensions size.

pros:
easy (de)allocation. just a single new/malloc
contigous memory, may be useful for quick copies with memcpy

cons:
more complex access, which gets harder to read as the dimensions get higher. this naturally can be simplified with some clever macors, but may not be elegant at all.
very large arrays may not be able to be allocated since it requires a single contigous block.
all rows will be the same size

uses:
maps, video buffers, and anything that contigous memory or easy allocation is perffered.

i am sure there are more pros/cons/uses for each style of multi dimensional array. its up to you to measure which type to use (you should try to understand both, since they are both useful).

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a person - arrays of arrays are faster for random access. I''ll admit that random access to array elements is rarely needed, though.

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quote:
Original post by a person
why do ppl confuse things so? you can have a multi dimensional array WITHOUT using arrays of arrays.

Yes. This is one of the very things that it says in the links I provided. It would be nice to encapsulate the array within a class and provide some natural ways of accessing it but, other than that, I don't see any reason to try and get too clever about it.

--
Cats cannot taste sweets.

Edited by - SabreMan on February 27, 2002 5:32:53 AM

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Thank you for all the input!

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I was asking all these questions because I am making a tile based game. The multi dimensioned array was for the map. For the graphics part, I wanted to be able to have an array to load surfaces in. That way, when the game is running I can load multiple surfaces, and blit them by knowing the index.

Is this right? I don't know what function removes a value in a vector. (Is there STL documentation somewhere that says?) I thought for freesurface I could use just use if(surfaces.at(index)) but I don't know what at() does.

  #include #include #include #include "SDL.h"class graphics { public: graphics(); ~graphics(); int LoadSurface(std::string); void FreeSurface(int); private: vector surfaces;};graphics::graphics() { //initialize graphics}graphics::~graphics() { //destroy graphics while(surfaces.size() > 0) { FreeSurface(surfaces.size()); }}int graphics::LoadSurface(std::string filename) { //add a surface to surfaces SDL_Surface *surface; surface = SDL_LoadBMP(filename); if(surface == NULL) { cerr << "Unable to load " << filename << endl << SDL_GetError() << endl; return -1; } surfaces.push_back(surface); SDL_FreeSurface(surface); return surfaces.size();}void graphics::FreeSurface(int index) { //remove a surface from surfaces if(index >= surfaces.size() || index < 0) { return; } SDL_FreeSurface(surfaces[index]); //delete surfaces[index]}

Edited by - evilclown on February 27, 2002 10:05:42 AM

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quote:
Original post by evilclown
Is this right? I don't know what function removes a value in a vector. (Is there STL documentation somewhere that says?) I thought for freesurface I could use just use if(surfaces.at(index)) but I don't know what at() does.

You need to use erase() and remove(), however vectors are not really designed for regularly removing and inserting elements, if that's what you are doing. Sounds like you really need that STL documentation. You could try the following sites...

Dinkumware

SGI

The at() function performs bounded access to a vector element. It will return the element at the position you pass, throwing an exception if you request something out of range. If you know the index, you might prefer to use operator[]. Check out those web links and come back if you are unsure.

--
Cats cannot taste sweets.

Edited by - SabreMan on February 27, 2002 11:07:26 AM

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I'm trying to find out what I need, but the site is kinda confusing. I thought I wanted to be able to delete entries from the middle of a container, but maybe that would cause problems later on. Because if I have 1 as the index for a surface, and if I remove surface 0, surface 1 would move down to surface 0, 2 down to 1 so it would use the wrong texture. Maybe I should use something like perl's associative arrays? ( \$arrayname["key"] = value ). Or do you know of something else that would work better?

One other thing, what does constant and linear time mean?
quote:
linear time insertion and removal of elements at the beginning or in the middle
If I can't remove elements in the middle, what does this mean?

Does an assosiative array seem like the right choice?

Edited by - evilclown on February 27, 2002 12:09:45 PM

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actually, using a single vector can be faster than a pointer array.
however, the loop must not be done with multiplication in it.
  for(int x = 0; x < width; x++){ int offset = 0; for(int y = 0; y < height; y++) vector[offset+ y] = 5; // [x*height + y] offset+=height; } `


text:
# On the proof of “Mapping space is a Chen space”

According to the page 5 in the paper Convenient Categories of Smooth Spaces https://arxiv.org/pdf/0807.1704.pdf by Baez and Hoffnung, Chen space is defined as follows:

(Note:I used different notations from the paper Convenient Categories of Smooth Spaces https://arxiv.org/pdf/0807.1704.pdf)

A Chen space $$X$$ is defined as a set $$X$$ equipped with , for each convex set $$U$$ there exists a collection $$\lbrace \phi_{i}: U \rightarrow X \rbrace_{i \in I}$$ of set maps called plots in $$X$$ satisfying the following properties:

(A convex set $$U$$ is defined as a convex set (with non-empty interior) in a Euclidean space $$\mathbf{R^n}$$ where $$n$$ can be any arbitrary non-negative integer. We call $$n$$ the dimension of $$U$$.A map $$f: U' \rightarrow U$$ from convex set $$U'$$ to convex set $$U$$ is called smooth function if $$f$$ has continuous derivatives of all order.)

1. If $$f: U' \rightarrow U$$ is a smooth function from convex set $$U'$$ to convex set $$U$$ and if $$\phi: U \rightarrow X$$ is a plot in $$X$$ then $$\phi \circ f$$ is also a plot in $$X$$.

2. Let $$U$$ be a convex set of dimension $$n$$. Suppose a collection of convex sets $$\lbrace U_j \subset U \rbrace_{j \in J}$$ forms an open cover of $$U$$ with respect to the subspace topology of $$\mathbf{R^n}$$. Let $$\lbrace I_j: U_j \rightarrow U \rbrace_{j \in J}$$ be the collection of inclusion maps. Let $$\phi: U \rightarrow X$$ be a set map. Now if each $$\lbrace \phi \circ I_j \rbrace_{j \in J}$$ are plots in $$X$$ then $$\phi$$ is also a plot in $$X$$.

3. Every function from the one point of $$\mathbf{R^0}$$ to X is a plot in $$X$$.

In page 6 they defined a set map $$f: X \rightarrow Y$$ to be smooth if for any plot $$\phi: U \rightarrow X$$ in $$X$$ the set map $$f \circ \phi: U \rightarrow Y$$ is a plot in $$Y$$.

In page 15 they mentioned that the mapping space $$C^{\infty}(X, Y)= \lbrace f:X \rightarrow Y: f$$ is smooth$$\rbrace$$ is a Chen space (where $$X, Y$$ are Chen spaces) whose plots are declared as those functions $$\phi:U \rightarrow C^{\infty}(X, Y)$$ such that the corresponding function $$\tilde{\phi}:U \times X \rightarrow Y$$ is smooth defined as $$(\zeta, x) \mapsto \phi(\zeta)(x)$$ (Note that there is a natural Chen space structure on both convex sets and Products ).

I was verifying $$C^{\infty}(X, Y)$$ is indeed a Chen space.

Property 1 and Property 3 are verified easily.

But I am not able to verify the property 2.

According to the definition of plot to verify Property 2, I need to show that if $$\tilde{\phi \circ I_i}:U_i \times X \rightarrow Y$$ is smooth for each $$i$$ then $$\tilde{\phi}: U \times X \rightarrow Y$$ is smooth. (where $$U_i \subset U$$ forms an open convex cover of $$U$$ and $$I_i$$ are inclusion maps). For that I need to show that if $$\psi:V \rightarrow U \times X$$ is any plot in $$U \times X$$ then $$\phi \circ \psi$$ is a plot in $$Y$$ .

I am not able to progess much after that!

I also note that there exist a collection of smooth maps $$I_i \times 1_X : U_i \times X \rightarrow U \times X$$ (where $$I_i$$ , $$1_X$$ are inclusion and identity map respectively).I feel somehow I need to use this fact also but not able to guess how.

I feel that I have to somehow express the plot $$\psi$$ in $$U \times X$$ in terms of plots of $$U_i \times X$$ so that I can use the smoothness of $$\tilde{\phi \circ I_i}$$. But I am not able to guess how!!

I feel it is some sort of local property of smoothness (as we have the analogue in case of finite dimensional smooth manifolds.)

I apologise priorly if this question is not upto the standard of MathOverflow. I am guessing I am mistaking or overlooking something.. But not able to guess what is that!!

Thank you.

I need to show that if $$\tilde{\phi \circ I_i}:U_i \times X \rightarrow Y$$ is smooth for each $$i$$ then $$\tilde{\phi}: U \times X \rightarrow Y$$ is smooth. (where $$U_i \subset U$$ forms an open convex cover of $$U$$ and $$I_i$$ are inclusion maps).
We have to show that a map $$U⨯X→Y$$ is a morphism of diffeological spaces if and only if its restrictions $$U_i⨯X→Y$$ are morphisms of diffeological spaces. To show that $$U_i⨯X→Y$$ uniquely glue to a morphism $$U⨯X→Y$$, consider some cartesian space $$S$$ and map it to both sides. We have to define a map on $$S$$-points natural in $$S$$: $$U(S)⨯X(S)→Y(S).$$ Fix an element in the left side, i.e., a smooth map $$f\colon S→U$$ and an element of $$X(S)$$. Fix a cover $$\{S_i=f^*U_i\}_{i∈I}$$ of $$S$$. The map $$f|_{S_i}$$ factors through $$U_i$$. Thus, we have an $$S_i$$-plot of $$U_i⨯X$$, which maps to an $$S$$-plot of $$Y$$ via the map $$U_i⨯X→Y$$. The constructed $$S_i$$-plots of $$Y$$ are compatible and glue to a unique $$S$$-plot of $$Y$$. This means that $$\tilde\phi$$ is a plot, as desired.
• Sir, According to the definition of Chen Space I mentioned above there is no assumption of topology on it. So how can you consider open sets of $U_i \times X$ and $U \times X$? Also a map $F:P \rightarrow Q$ between Chen spaces is defined to be smooth if for any plot $\phi:U \rightarrow P$ in $P$ , $F \circ \phi$ is a plot in $Q$. So how does this follow from your Answer? – Adittya Chaudhuri Jun 21 '20 at 3:32


text:
# I don't understand how to prove an implication with a disjunction in the consequent.

I've come to find I don't understand how to prove an implication in which the consequent is a disjunction. Allow me to explain.

For a simple example, take this: "Prove: If $\frac{x}{(x-2)} \leq 3$, then $x<2$ or $x\geq 3$". Considering there's a disjunction in the consequent, does that mean I would theoretically only need to prove one of the component statements in this disjunction? I.e., would proving If $\frac{x}{(x-2)} \leq 3$, then $x<2$ suffice? A disjunction is true as long as at least one of its components is true, right? This would mean the implication has truth values $T \implies T$, and it'd be a correct implication.

However, in a recent question I asked, the above implication was solved by showing that both $x<2$ and $x\geq3$ hold. Isn't this redundant?

• In general, it is correct; it is enough to prove one of the disjuncts. – Mauro ALLEGRANZA Oct 22 '17 at 16:08
• Consider $n \le 0$; this is equivalent to $n < 0 \lor n =0$. Thus, if we have proved $n < 0$ we can conclude with $n \le 0$. – Mauro ALLEGRANZA Oct 22 '17 at 16:09
• But the case above is different: the two "solutions" of the disequation are both solutions (mutually exclusive). – Mauro ALLEGRANZA Oct 22 '17 at 16:14
• @MauroALLEGRANZA Right, but isn't the general form $P \implies (Q \lor R)$ equivalent to $(P \land \neg Q) \implies R$? So why would I have to show that both $x<2$ and $x \geq 3$ hold? – Ius Klesar Oct 22 '17 at 16:24
• @MauroALLEGRANZA We don't know the value of x. I think that x < 2 is more like "x can take on some value less than 2" instead of "x has some value less than 2" here. It is a contradiction to say that "x has some value less than 2" and "x has some value greater than 3". It's not a contradiction to say that "x can take on some value less than 2" and "x can take on some value greater than or equal to 3" here. – Doug Spoonwood Oct 22 '17 at 18:32

It would suffice, of course. But can you show that? Yes? Then you most definitely had an error in your reasoning.

Consider $x=3$. In this case $\frac{x}{x-2} = 3 \leq 3$ but $x<2$ does not hold. That's why you still have the second statement $x \geq 3$.

One way to try and prove this is to prove the contrapositive.

The contrapositive of $P \rightarrow ( Q \lor R)$ is $(\neg Q \land \neg R) \rightarrow \neg P$

Applied to your statement, that amounts to proving that:

If $2 \le x < 3$ then $\frac{x}{x-2} >3$ (or it is undefined)

• I think you used more than just the contrapositive there. Though I guess you didn't define 'contrapositive' and could have a different definition... – Doug Spoonwood Oct 23 '17 at 0:20

With conditionals, I believe that the most generally applied method of proof consists in assuming the left part of the conditional and then deriving the right part of the conditional.

Now, for a conditional with a disjunction in the consequent, as Mauro's comments suggest, we first want to ascertain the nature of the disjunction. Is it an inclusive disjunction or an exclusive disjunction? In other words, can both of the disjuncts hold true or can only one of the disjuncts hold true?

If both of the disjuncts X, Y can hold true, we have some options. One option lies in proving that one of the disjuncts does hold. Consequently, the disjunction will follow. Another option starts by assuming the negation of the disjunction. Since ($\lnot$(X$\lor$Y)$\rightarrow$($\lnot$X$\land$$\lnotY)), (\lnotX\land$$\lnot$Y) follows. Then we can consider $\lnot$X and $\lnot$Y separately in turn. If either $\lnot$X or $\lnot$Y more immediately leads to a contradiction, then so does $\lnot$(X$\lor$Y) and (X$\lor$Y) follows as true.

So, for your example suppose that $\lnot$ ((x < 2) $\lor$ (x >= 3)) under the scope of the left part of your conditional. Thus, ($\lnot$(x<2) $\land$ $\lnot$(x >= 3)). So, we can have any x such that x > 2. But, suppose that x = 2.1. Then from the left part of your conditional we obtain (2.1 / .1) <= 3. But, 21 <= 3 is false and thus we have a contradiction. Consequently, $\lnot$ ((x < 2) $\lor$ (x >= 3)) is false and ((x < 2) $\lor$ (x >= 3)) is true.


text:
Question

# If a wire of resistance R is stretched to double of its length, then new resistance will be :

A
R/2
B
2 R
C
4 R
D
16 R

Solution

## The correct option is C 4 R$$\textbf{Step 1: Finding the new cross section Area of the wire}$$Volume of wire remains constant even after stretching the wire.$$V_i= V_f$$$$\Rightarrow LA= L' A'$$  ( given $$L'=2L$$)$$\Rightarrow A'= \dfrac{A}{2}$$$$\textbf{Step 2: Finding the New Resistance}$$Initial resistance, $$R = \dfrac{\rho L}{A}$$New resistance, $$R' = \displaystyle \dfrac{\rho L'}{A'} = \dfrac{\rho (2L)}{\dfrac{A}{2}} =\dfrac{4 \rho L}{A} =4R$$Hence, Option C is correctPhysics

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text:
# Why are contact structures studied from a cohomological, rather than homological, perspective?

As far as I know, contact structures are studied from a "cohomology/dual" perspective, meaning mostly from the perspective of contact forms and their respective kernels, instead of from a "homological" version using either contact vector fields or Reeb vector fields. Is there a good reason for this? Thanks.

From the distribution point of view, you would take a hyperplane subbundle $H$ in $TM$, and consider the quotient bundle $Q:=TM/H$ with projection $q:TM\to Q$. Then for vector fields $\xi$ and $\eta$ on $M$ which have values in $H$, you consider the projection $q([\xi,\eta])$ of their Lie bracket. One immediately verifies that the resulting map is bilinear of smooth functions, so there is an induced bundle map $\mathcal L:H\times H\to Q$. The condition that $H$ is a contact distribution then is that $\mathcal L$ is non-degenerate (as a bilinear form) in each point. This can also be phrased as the fact that $\mathcal L$ in each point $x$ makes $H_x\oplus Q_x$ into a Heisenberg-algebra.
$\kappa\wedge d\kappa\wedge\dots\wedge d\kappa\neq0$ is such a neat expression. When you are trying to express it in terms of tangent vector fields, it would be a mess. By the way, the Reeb field is frequently used in contact geometry.


text:
# Self Affine Delone Sets and Deviation Phenomena

```@article{Schmieding2015SelfAD,
title={Self Affine Delone Sets and Deviation Phenomena},
author={Scott Schmieding and Rodrigo Trevi{\~n}o},
journal={Communications in Mathematical Physics},
year={2015},
volume={357},
pages={1071-1112}
}```
• Published 24 November 2015
• Mathematics
• Communications in Mathematical Physics
We study the growth of norms of ergodic integrals for the translation action on spaces coming from expansive, self-affine Delone sets. The linear map giving the self-affinity induces a renormalization map on the pattern space and we show that the rate of growth of ergodic integrals is controlled by the induced action of the renormalizing map on the cohomology of the pattern space up to boundary errors. We explore the consequences for the diffraction of such Delone sets, and explore in detail…
11 Citations
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This paper deals with (globally) random substitutions on a finite set of prototiles. Using renormalization tools applied to objects from operator algebras we establish upper and lower bounds on the
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This paper deals with random substitutions on a finite set of prototiles. The assumptions on the types of substitution rules allowed are very weak, leading to very general constructions. Using
Canonical diffusions on the pattern spaces of aperiodic Delone sets
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• 2018
We consider pattern spaces of aperiodic and repetitive Delone sets of finite local complexity. These spaces are compact metric spaces and constitute a special class of foliated spaces. We define and
Random substitution tilings and deviation phenomena
• Mathematics
Discrete & Continuous Dynamical Systems
• 2021
Suppose a set of prototiles allows \$N\$ different substitution rules. In this paper we study tilings of \$\mathbb{R}^d\$ constructed from random application of the substitution rules. The space of all
Cohomology groups for spaces of 12-fold tilings
• Mathematics
• 2020
We consider tilings of the plane with 12-fold symmetry obtained by the cut and projection method. We compute their cohomology groups using the techniques introduced by the second author, Hunton and
Cohomology Groups for Spaces of Twelve-Fold Tilings
• Mathematics
• 2021
We consider tilings of the plane with twelve-fold symmetry obtained by the cut-and-projection method. We compute their cohomology groups using the techniques introduced in [ 9]. To do this, we
Traces of Random Operators Associated with Self-Affine Delone Sets and Shubin’s Formula
• Mathematics
Annales Henri Poincaré
• 2018
We study operators defined on a Hilbert space defined by a self-affine Delone set \$\$\Lambda \$\$Λ and show that the usual trace of a restriction of the operator to finite-dimensional subspaces
Rotation Numbers and Rotation Classes on One-Dimensional Tiling Spaces
• Mathematics
• 2020
We extend rotation theory of circle maps to tiling spaces. Specifically, we consider a 1-dimensional tiling space \$\Omega\$ with finite local complexity and study self-maps \$F\$ that are homotopic to
NEW APPROACHES TO SOME PROBLEMS IN DISCRETE GEOMETRY
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• 2017
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I study classes of generalized Frenkel-Kontorova models whose potentials are given by almost-periodic functions which are closely related to aperiodic Delone sets of finite local complexity. Since

## References

SHOWING 1-10 OF 45 REFERENCES
Dynamics of self-similar tilings
• B. Solomyak
• Mathematics
Ergodic Theory and Dynamical Systems
• 1997
This paper investigates dynamical systems arising from the action by translations on the orbit closures of self-similar and self-affine tilings of \${\Bbb R}^d\$. The main focus is on spectral
Rigidity and mapping class group for abstract tiling spaces
• J. Kwapisz
• Mathematics
Ergodic Theory and Dynamical Systems
• 2011
Abstract We study abstract self-affine tiling actions, which are an intrinsically defined class of minimal expansive actions of ℝd on a compact space. They include the translation actions on the
Limit Theorems for Self-Similar Tilings
• Mathematics
• 2013
We study deviation of ergodic averages for dynamical systems given by self-similar tilings on the plane and in higher dimensions. The main object of our paper is a special family of finitely-additive
Proximality and pure point spectrum for tiling dynamical systems
• Mathematics
• 2011
We investigate the role of the proximality relation for tiling dynamical systems. Under two hypotheses, namely that the minimal rank is finite and the set of fiber distal points has full measure we
Topological invariants for substitution tilings and their associated \$C^\ast\$-algebras
• Mathematics
Ergodic Theory and Dynamical Systems
• 1998
We consider the dynamical systems arising from substitution tilings. Under some hypotheses, we show that the dynamics of the substitution or inflation map on the space of tilings is topologically
The Ruelle-Sullivan map for actions of ℝn
• Mathematics
• 2006
The Ruelle Sullivan map for an ℝn-action on a compact metric space with invariant probability measure is a graded homomorphism between the integer Cech cohomology of the space and the exterior
Exact regularity and the cohomology of tiling spaces
• Mathematics
Ergodic Theory and Dynamical Systems
• 2011
Abstract Exact regularity was introduced recently as a property of homological Pisot substitutions in one dimension. In this paper, we consider the analog of exact regularity for arbitrary tiling
Pattern equivariant functions, deformations and equivalence of tiling spaces
Abstract We re-investigate the theory of deformations of tilings using P-equivariant cohomology. In particular, we relate the notion of asymptotically negligible shape functions introduced by Clark
Cohomology in one-dimensional substitution tiling spaces
• Mathematics
• 2007
Anderson and Putnam showed that the cohomology of a substitution tiling space may be computed by collaring tiles to obtain a substitution which "forces its border." One can then represent the tiling
Integral cohomology of rational projection method patterns
• Mathematics
• 2013
We study the cohomology and hence K‐theory of the aperiodic tilings formed by the so called “cut and project” method, that is, patterns in d ‐dimensional Euclidean space which arise as sections of


text:
# Hyper-class

0th

Percentile

##### Class "Hyper"

The hypergeometric distribution is used for sampling without replacement. The density of this distribution with parameters m, n and k (named $Np$, $N-Np$, and $n$, respectively in the reference below) is given by $$p(x) = \left. {m \choose x}{n \choose k-x} \right/ {m+n \choose k}%$$ for $x = 0, \ldots, k$. C.f. rhyper

Keywords
distribution
##### Objects from the Class

Objects can be created by calls of the form Hyper(m, n, k). This object is a hypergeometric distribution.

##### Slots

img

Object of class "Naturals": The space of the image of this distribution has got dimension 1 and the name "Natural Space".

param

Object of class "HyperParameter": the parameter of this distribution (m, n, k), declared at its instantiation

r

Object of class "function": generates random numbers (calls function rhyper)

d

Object of class "function": density function (calls function dhyper)

p

Object of class "function": cumulative function (calls function phyper)

q

Object of class "function": inverse of the cumulative function (calls function qhyper). The $\alpha$-quantile is defined as the smallest value $x$ such that $p(x) \ge \alpha]$, where $p$ is the cumulative function.

support:

Object of class "numeric": a (sorted) vector containing the support of the discrete density function

.withArith

logical: used internally to issue warnings as to interpretation of arithmetics

.withSim

logical: used internally to issue warnings as to accuracy

.logExact

logical: used internally to flag the case where there are explicit formulae for the log version of density, cdf, and quantile function

.lowerExact

logical: used internally to flag the case where there are explicit formulae for the lower tail version of cdf and quantile function

Symmetry

object of class "DistributionSymmetry"; used internally to avoid unnecessary calculations.

##### Extends

Class "DiscreteDistribution", directly. Class "UnivariateDistribution", by class "DiscreteDistribution". Class "Distribution", by class "DiscreteDistribution".

##### Methods

initialize

signature(.Object = "Hyper"): initialize method

m

signature(object = "Hyper"): returns the slot m of the parameter of the distribution

m<-

signature(object = "Hyper"): modifies the slot m of the parameter of the distribution

n

signature(object = "Hyper"): returns the slot n of the parameter of the distribution

n<-

signature(object = "Hyper"): modifies the slot n of the parameter of the distribution

k

signature(object = "Hyper"): returns the slot k of the parameter of the distribution

k<-

signature(object = "Hyper"): modifies the slot k of the parameter of the distribution

##### Aliases
• Hyper-class
• Hyper
• initialize,Hyper-method
##### Examples
# NOT RUN {
H <- Hyper(m=3,n=3,k=3) # H is a hypergeometric distribution with m=3,n=3,k=3.
r(H)(1) # one random number generated from this distribution, e.g. 2
d(H)(1) # Density of this distribution is  0.45 for x=1.
p(H)(1) # Probability that x<1 is 0.5.
q(H)(.1) # x=1 is the smallest value x such that p(H)(x)>=0.1.
## in RStudio or Jupyter IRKernel, use q.l(.)(.) instead of q(.)(.)
m(H) # m of this distribution is 3.
m(H) <- 2 # m of this distribution is now 2.
# }
Documentation reproduced from package distr, version 2.8.0, License: LGPL-3

### Community examples

Looks like there are no examples yet.


text:
# Bricks

Brick has volume 2.2 dm3. How many bricks can drive a truck with a capacity of 23 ton? The density of brick is 1.6 g/cm3.

Correct result:

n =  6534

#### Solution:

$n=\frac{1000\cdot 23}{1.6\cdot 2.2}=6534$

We would be pleased if you find an error in the word problem or inaccuracies and send it to us. Thank you!

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The body has dimensions of 2m 2dm and 10 cm. It weighs 28 kg. What is its density?
• The glass
1 m3 of glass weighs 2600 kg. Calculate the weight of the glass glazing panel with dimensions of 2.5 m and 3.8 m if the thickness of the glass is 0.8 cm
• Planet Earth
What is the weight of the planet Earth, if its average density is ρ = 2.5 g/cm ^ 3?
• Granite cube
What is the weight in kg granite cube with an edge of 0.5 m if 1dm3 of granite weight 2600 g?


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# Product rule for logarithms

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• August 14th 2010, 12:20 PM
zg12
Product rule for logarithms
Have a silly question.

In the proof that ln(ax)=ln(a) + ln(x)

we show that ln(ax) and ln(x) have the same derivative, therefore they differ by a constant; ln(a)=ln(x) + C

The book then sets x=1, and says C=ln(a), and the rule is proved. But I get C=ln(a)-ln(x), and that x=1 is a special case.

What's my mistake?
• August 14th 2010, 12:39 PM
Plato
Look at it this way.
$\ln(x)=\ln(a)-C$ so $\ln(1)=0=\ln(a)-C$.
Therefore $C=\ln(a).$
• August 14th 2010, 12:56 PM
zg12
What I'm not understanding is how we can set x= any number, without loosing the generality of the rule. I see that ln(a)=ln(x) + C is true for any value of x, but if we choose x=2 instead of x=1 then we get C=ln(a)-ln(2) instead of C=ln(a)
• August 14th 2010, 01:02 PM
zg12
Is it because that C + ln (2) is just another constant?

C+ln(2)=ln(a), which equals C=ln(a) ????????
• August 14th 2010, 01:31 PM
Plato
No that is not the point. (BTW: I don't like that proof).
But the fact is $\ln(1)=0$. Therefore, it must be the case that $C=\ln(a)$.
• August 14th 2010, 01:42 PM
Archie Meade
Maybe the diagram will help show why x=1 is chosen (for the case of a=2).

Remember, in order to discover the constant C, you must eliminate x.
Setting x=1 causes ln(x) to be zero, eliminating x.

Also, you have a typo in your post

ln(a)=ln(x)+C should read ln(ax)=ln(x)+C.

Hence C=ln(ax)-ln(x) but only "removing" x shows the value of C.

Setting x to 1,

$ln(a)-ln(1)=C$

$ln(a)-0=C$
• August 15th 2010, 05:33 AM
zg12
Resolved. I woke up this morning and can't even remember what my confusion was over.


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## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11

More Exercises

Midpoint Calculator is used to find the midpoint between 2 line segments using the midpoint formula. Use our calculator to find accurate midpoints step by step.

Question 1.
Find the co-ordinates of the mid-point of the line segments joining the following pairs of points:
(i) (2, – 3), ( – 6, 7)
(ii) (5, – 11), (4, 3)
(iii) (a + 3, 5b), (2a – 1, 3b + 4)
Solution:
(i) Co-ordinates of the mid-point of (2, -3), ( -6, 7)
$$\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) or$$

Question 2.
The co-ordinates of two points A and B are ( – 3, 3) and (12, – 7) respectively. P is a point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of P.
Solution:
Points are A (-3, 3), B (12, -7)
Let P (x1,  y1) be the point which divides AB in the ratio of m1 : m2 i.e. 2 : 3
then co-ordinates of P will be

Question 3.
P divides the distance between A ( – 2, 1) and B (1, 4) in the ratio of 2 : 1. Calculate the co-ordinates of the point P.
Solution:
Points are A (-2, 1) and B (1, 4) and
Let P (x, y) divides AB in the ratio of m1 : m2 i.e. 2 : 1
Co-ordinates of P will be

Question 4.
(i) Find the co-ordinates of the points of trisection of the line segment joining the point (3, – 3) and (6, 9).
(ii) The line segment joining the points (3, – 4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, – 2) and $$\left( \frac { 5 }{ 3 } ,q \right)$$ respectively, find the values of p and q.
Solution:
(i) Let P (x1, y1) and Q (x2, y2) be the points
which trisect the line segment joining the points
A (3, -3) and B (6, 9)

Question 5.
(i) The line segment joining the points A (3, 2) and B (5, 1) is divided at the point P in the ratio 1 : 2 and it lies on the line 3x – 18y + k = 0. Find the value of k.
(ii) A point P divides the line segment joining the points A (3, – 5) and B ( – 4, 8) such that $$\frac { AP }{ PB } =\frac { k }{ 1 }$$ If P lies on the line x + y = 0, then find the value of k.
Solution:
(i) The point P (x, y) divides the line segment joining the points
A (3, 2) and B (5, 1) in the ratio 1 : 2

Question 6.
Find the coordinates of the point which is three-fourth of the way from A (3, 1) to B ( – 2, 5).
Solution:
Let P be the required point, then
$$\frac { AP }{ AB } =\frac { 3 }{ 4 }$$

Question 7.
Point P (3, – 5) is reflected to P’ in the x- axis. Also P on reflection in the y-axis is mapped as P”.
(i) Find the co-ordinates of P’ and P”.
(ii) Compute the distance P’ P”.
(iii) Find the middle point of the line segment P’ P”.
(iv) On which co-ordinate axis does the middle point of the line segment P P” lie ?
Solution:
(i) Co-ordinates of P’, the image of P (3, -5)
when reflected in x-axis will be (3, 5)
and co-ordinates of P”, the image of P (3, -5)
when reflected in y-axis will be (-3, -5)

Question 8.
Use graph paper for this question. Take 1 cm = 1 unit on both axes. Plot the points A(3, 0) and B(0, 4).
(i) Write down the co-ordinates of A1, the reflection of A in the y-axis.
(ii) Write down the co-ordinates of B1, the reflection of B in the x-axis.
(iii) Assign.the special name to the quadrilateral ABA1B1.
(iv) If C is the mid point is AB. Write down the co-ordinates of the point C1, the reflection of C in the origin.
(v) Assign the special name to quadrilateral ABC1B1.
Solution:
Two points A (3, 0) and B (0,4) have been plotted on the graph.

(i)∵ A1 is the reflection of A (3, 0) in the v-axis Its co-ordinates will be ( -3, 0)
(ii)∵ B1 is the reflection of B (0, 4) in the .x-axis co-ordinates of B, will be (0, -4)
(iii) The so formed figure ABA1B1 is a rhombus.
(iv) C is the mid point of AB co-ordinates of C” will be $$\frac { AP }{ AB } =\frac { 3 }{ 4 }$$
∵ C, is the reflection of C in the origin
co-ordinates of C, will be $$\left( \frac { -3 }{ 2 } ,-2 \right)$$
(v) The name of quadrilateral ABC1B1 is a trapezium because AB is parallel to B1C1.

Question 9.
The line segment joining A ( – 3, 1) and B (5, – 4) is a diameter of a circle whose centre is C. find the co-ordinates of the point C. (1990)
Solution:
∵ C is the centre of the circle and AB is the diameter
C is the midpoint of AB.
Let co-ordinates of C (x, y)

Question 10.
The mid-point of the line segment joining the points (3m, 6) and ( – 4, 3n) is (1, 2m – 1). Find the values of m and n.
Solution:
Let the mid-point of the line segment joining two points
A(3m, 6) and (-4, 3n) is P( 1, 2m – 1)

Question 11.
The co-ordinates of the mid-point of the line segment PQ are (1, – 2). The co-ordinates of P are ( – 3, 2). Find the co-ordinates of Q.(1992)
Solution:
Let the co-ordinates of Q be (x, y)
co-ordinates of P are (-3, 2) and mid-point of PQ are (1, -2) then

Question 12.
AB is a diameter of a circle with centre C ( – 2, 5). If point A is (3, – 7). Find:
(i) the length of radius AC.
(ii) the coordinates of B.
Solution:
AC = $$\sqrt { { \left( 3+2 \right) }^{ 2 }+{ \left( -7-5 \right) }^{ 2 } }$$

Question 13.
Find the reflection (image) of the point (5, – 3) in the point ( – 1, 3).
Solution:
Let the co-ordinates of the images of the point A (5, -3) be
A1 (x, y) in the point (-1, 3) then
the point (-1, 3) will be the midpoint of AA1.

Question 14.
The line segment joining A $$\left( -1,\frac { 5 }{ 3 } \right)$$ the points B (a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects y-axis. Calculate
(i) the value of a
(ii) the co-ordinates of P. (1994)
Solution:
Let P (x, y) divides the line segment joining
the points $$\left( -1,\frac { 5 }{ 3 } \right)$$, B(a, 5) in the ratio 1 : 3

Question 15.
The point P ( – 4, 1) divides the line segment joining the points A (2, – 2) and B in the ratio of 3 : 5. Find the point B.
Solution:
Let the co-ordinates of B be (x, y)
Co-ordinates of A (2, -2) and point P (-4, 1)
divides AB in the ratio of 3 : 5

Question 16.
(i) In what ratio does the point (5, 4) divide the line segment joining the points (2, 1) and (7 ,6) ?
(ii) In what ratio does the point ( – 4, b) divide the line segment joining the points P (2, – 2), Q ( – 14, 6) ? Hence find the value of b.
Solution:
(i) Let the ratio be m1 : m2 that the point (5, 4) divides
the line segment joining the points (2, 1), (7, 6).
$$5=\frac { { m }_{ 1 }\times 7+{ m }_{ 2 }\times 2 }{ { m }_{ 1 }+{ m }_{ 2 } }$$

Question 17.
The line segment joining A (2, 3) and B (6, – 5) is intercepted by the x-axis at the point K. Write the ordinate of the point k. Hence, find the ratio in which K divides AB. Also, find the coordinates of the point K.
Solution:
Let the co-ordinates of K be (x, 0) as it intersects x-axis.
Let point K divides the line segment joining the points
A (2, 3) and B (6, -5) in the ratio m1 : m2.

Question 18.
If A ( – 4, 3) and B (8, – 6), (i) find the length of AB.
(ii) in what ratio is the line joining AB, divided by the x-axis? (2008)
Solution:
Given A (-4, 3), B (8, -6)

Question 19.
(i) Calculate the ratio in which the line segment joining (3, 4) and( – 2, 1) is divided by the y-axis.
(ii) In what ratio does the line x – y – 2 = 0 divide the line segment joining the points (3, – 1) and (8, 9)? Also, find the coordinates of the point of division.
Solution:
(i) Let the point P divides the line segment joining the points
A (3, 4) and B (-2, 3) in the ratio of m1 : m2 and
let the co-ordinates of P be (0, y) as it intersects the y-axis

Question 20.
Given a line segment AB joining the points A ( – 4, 6) and B (8, – 3). Find:
(i) the ratio in which AB is divided by the y-axis.
(ii) find the coordinates of the point of intersection.
(iii)the length of AB.
Solution:
(i) Let the y-axis divide AB in the ratio m : 1. So,

Question 21.
(i) Write down the co-ordinates of the point P that divides the line joining A ( – 4, 1) and B (17,10) in the ratio 1 : 2.
(ii)Calculate the distance OP where O is the origin.
(iii)In what ratio does the y-axis divide the line AB ?
Solution:
(i) Let co-ordinate of P be (x, y) which divides the line segment joining the points
A ( -4, 1) and B(17, 10) in the ratio of 1 : 2.

Question 22.
Calculate the length of the median through the vertex A of the triangle ABC with vertices A (7, – 3), B (5, 3) and C (3, – 1)
Solution:
Let D (x, y) be the median of ΔABC through A to BC.
∴ D will be the midpoint of BC
∴ Co-ordinates of D will be,

Question 23.
Three consecutive vertices of a parallelogram ABCD are A (1, 2), B (1, 0) and C (4, 0). Find the fourth vertex D.
Solution:
Let O in the mid-point of AC the diagonal of ABCD
∴ Co-ordinates of O will be

Question 24.
If the points A ( – 2, – 1), B (1, 0), C (p, 3) and D (1, q) from a parallelogram ABCD, find the values of p and q.
Solution:
A (-2, -1), B (1, 0), C (p, 3) and D (1, q)
are the vertices of a parallelogram ABCD
∴ Diagonal AC and BD bisect each other at O
O is the midpoint of AC as well as BD
Let co-ordinates of O be (x, y)
When O is mid-point of AC, then

Question 25.
If two vertices of a parallelogram are (3, 2) ( – 1, 0) and its diagonals meet at (2, – 5), find the other two vertices of the parallelogram.
Solution:
Two vertices of a ||gm ABCD are A (3, 2), B (-1, 0)
and point of intersection of its diagonals is P (2, -5)
P is mid-point of AC and BD.
Let co-ordinates of C be (x, y), then

Question 26.
Prove that the points A ( – 5, 4), B ( – 1, – 2) and C (5, 2) are the vertices of an isosceles right angled triangle. Find the co-ordinates of D so that ABCD is a square.
Solution:
Points A (-5, 4), B (-1, -2) and C (5, 2) are given.
If these are vertices of an isosceles triangle ABC then
AB = BC.

Question 27.
Find the third vertex of a triangle if its two vertices are ( – 1, 4) and (5, 2) and mid point of one sides is (0, 3).
Solution:
Let A (-1, 4) and B (5, 2) be the two points and let D (0, 3)
be its the midpoint of AC and co-ordinates of C be (x, y).

Question 28.
Find the coordinates of the vertices of the triangle the middle points of whose sides are $$\left( 0,\frac { 1 }{ 2 } \right) ,\left( \frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right) and\left( \frac { 1 }{ 2 } ,0 \right)$$
Solution:
Let ABC be a ∆ in which $$D\left( 0,\frac { 1 }{ 2 } \right) ,E\left( \frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right) andF\left( \frac { 1 }{ 2 } ,0 \right)$$,
the mid-points of sides AB, BC and CA respectively.
Let co-ordinates of A be (x1, y1), B (x2, y2), C (x3, y3)

Question 29.
Show by section formula that the points (3, – 2), (5, 2) and (8, 8) are collinear.
Solution:
Let the point (5, 2) divides the line joining the points (3, -2) and (8, 8)
in the ratio of m1 : m2

Question 30.
Find the value of p for which the points ( – 5, 1), (1, p) and (4, – 2) are collinear.
Solution:
Let points A (-5, 1), B (1, p) and C (4, -2)
are collinear and let point A (-5, 1) divides
BC in the ratio in m1 : m2

Question 31.
A (10, 5), B (6, – 3) and C (2, 1) are the vertices of triangle ABC. L is the mid point of AB, M is the mid-point of AC. Write down the co-ordinates of L and M. Show that LM = $$\\ \frac { 1 }{ 2 }$$ BC.
Solution:
Co-ordinates of L will be
$$\left( \frac { 10+6 }{ 2 } ,\frac { 5-3 }{ 2 } \right) or\left( \frac { 16 }{ 2 } ,\frac { 2 }{ 2 } \right) or(8,1)$$

Question 32.
A (2, 5), B ( – 1, 2) and C (5, 8) are the vertices of a triangle ABC. P and.Q are points on AB and AC respectively such that AP : PB = AQ : QC = 1 : 2.
(i) Find the co-ordinates of P and Q.
(ii) Show that PQ = $$\\ \frac { 1 }{ 3 }$$ BC.
Solution:
A (2, 5), B (-1, 2) and C (5, 8) are the vertices of a ∆ABC,
P and Q are points on AB
and AC respectively such that $$\frac { AP }{ PB } =\frac { AQ }{ QC } =\frac { 1 }{ 2 }$$

Question 33.
The mid-point of the line segment AB shown in the adjoining diagram is (4, – 3). Write down die co-ordinates of A and B.

Solution:
A lies on x-axis and B on the y-axis.
Let co-ordinates of A be (x, 0) and of B be (0, y)
P (4, -3) is the mid-point of AB

Question 34.
Find the co-ordinates of the centroid of a triangle whose vertices are A ( – 1, 3), B(1, – 1) and C (5, 1) (2006)
Solution:
Co-ordinates of the centroid of a triangle,
whose vertices are (x1, y1), (x2, y2) and

Question 35.
Two vertices of a triangle are (3, – 5) and ( – 7, 4). Find the third vertex given that the centroid is (2, – 1).
Solution:
Let the co-ordinates of third vertices be (x, y)
and other two vertices are (3, -5) and (-7, 4)
and centroid = (2, -1).

Question 36.
The vertices of a triangle are A ( – 5, 3), B (p – 1) and C (6, q). Find the values of p and q if the centroid of the triangle ABC is the point (1, – 1).
Solution:
The vertices of ∆ABC are A (-5, 3), B (p, -1), C (6, q)
and the centroid of ∆ABC is O (1, -1)
co-ordinates of the centroid of ∆ABC will be

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Ex 11 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.


text:
• Corpus ID: 238857113

# The universality of islands outside the horizon

@inproceedings{He2021TheUO,
title={The universality of islands outside the horizon},
author={Song He and Yuan Sun and Longhua Zhao and Yu-Xuan Zhang},
year={2021}
}
We systematically calculate the quantum extremal surface (QES) associated with Hawking radiation for general D-dimensional (D ≥ 2) asymptotically flat (or AdS) eternal black holes using the island formula. By adopting the standard ”black hole couples thermal baths” model, we find that a QES exists in the near-horizon region outside the black hole when c ·G(D) is smaller enough where c is the central charge of the conformal matter and G(D) the Newton constant. The locations of the QES in these…
3 Citations

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The newly proposed island formula for entanglement entropy of Hawking radiation is applied to spherically symmetric 4-dimensional eternal Kaluza-Klein (KK) black holes (BHs). The "charge" Q of a KK
Page Curves and Islands in Charged Dilaton Black Holes
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We study the Page curve in the four dimensional asymptotically flat eternal GarfinkleHorowitz-Strominger dilaton black holes by applying the “quantum extremal surface” prescription. The results
Islands and Page curves of Reissner-Nordström black holes
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We apply the recently proposed quantum extremal surface construction to calculate the Page curve of the eternal Reissner-Nordstrom black holes in four dimensions ignoring the backreaction and the
Thermodynamics of black holes in two (and higher) dimensions
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A comprehensive treatment of black hole thermodynamics in two-dimensional dilaton gravity is presented. We derive an improved action for these theories and construct the Euclidean path integral. An
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Entanglement entropy and Page curve of black holes with island in massive gravity
By applying the island rule proposed recently, we compute the entanglement entropy of the Hawking radiation and study the Page curve for the eternal black holes in massive gravity. We investigate for
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text:
Deloitte/FIDE Chess Rating Challenge Details

This year, from February 7 to May 4, a prediction contest was held at Kaggle.com/c/ChessRatings2 where I ended up taking first place. The goal of the contest was to build a model to forecast the results of future chess matches based on the results of past matches. This post contains a description of my approach. The corresponding MATLAB code can be found here. For a non-technical account of my experiences in the competition, see my earlier post.

The base model

The basic model underlying my approach was inspired strongly by the TrueSkill model, as well as by the winner and runner-up of an earlier chess rating contest on Kaggle. The final model was programmed in Matlab, but some of the early experimentation was done using the Infer.NET package, which is definitely worth having a look at. Warning: The discussion in this section is somewhat technical.

The basic statistical model assumed for the result of a match between a white player A and a black player B is the familiar ordered probit model:

$d = s_{A} - s_{B} + \gamma + \epsilon$, $\epsilon \sim N(0,\sigma^{2})$

if $d$ > 1 : A wins

if -1 < $d$ < 1 : A and B draw

if $d$ < -1 : B wins

Here $d$ can be seen as the _performance difference_ between A and B in this match, $s_{A}$ and $s_{B}$ as the skills of player A and B, $\gamma$ as the advantage of playing white, and $\epsilon$ as a random error term.

Given a set of match results, we will infer the skills of all players by means of factorized approximate Bayesian inference. $\gamma$ and $\sigma^{2}$ are estimated using approximate maximum likelihood.

We specify an independent normal prior for each player’s skill $s_{i}$, having mean $\mu_{i}$ and a variance of 1 (determined by cross validation). The means $\mu_{i}$ can be initialized to 0 and will be set to a weighted average of the posterior means of the skills of each players’ opponents at each iteration. The effect of this is to shrink the skills of the players to those of their opponents, as was first done by Yannis Sismanis.

Given a set of match results $r$, the skills have the following posterior density

$p(s|r) \propto \prod_{i=1}^{players} \phi(s_{i}-\mu_{i}) \prod_{j=1}^{matches} \psi_{j}(s_{W_{j}},s_{B_{j}})$

where $\phi()$ is the standard normal distribution function, $\psi_{j}()$ is the likelihood term due to the match result $r_{j}$, and $W_{j}$ and $B_{j}$ identify the white and black player in match $j$. This posterior distribution is of a very high dimension and is not of any standard form, which makes it intractable for exact inference. Approximate Bayesian inference can solve this problem by approximating the above density by a product of univariate normal densities.

$\tilde{p}(s|r) \propto \prod_{i=1}^{players} \phi(s_{i}-\mu_{i}) \prod_{j=1}^{matches} \phi([s_{W_{j}}-m_{j,1}]/\sqrt{v_{j,1}}) \phi([s_{B_{j}}-m_{j,2}]/\sqrt{v_{j,2}})$

There exist various ways of obtaining the mean and variance terms ($m_{j,1}$, $v_{j,1}$, $m_{j,2}$, $v_{j,2}$) in this pseudo-posterior. The two methods I tried were expectation propagation, as is used in the TrueSkill model and Laplace approximation, as used here in a similar context. For the current model and data set the results for both methods were practically the same. The advantage of the Laplace approximation is that it is easier to apply when we change the ordered probit specification to something else like a (ordered or multinomial) logit model. However, since the ordered probit specification provided the best fit, my final submission was made using this specification in combination with expectation propagation. Both methods can only be applied directly when we know which of the two players was playing white. This wasn’t the case for part of the data. My solution to this problem was to calculate the likelihood terms for both the case that the first player is white as well as the case that the second player is white, after which I weight the likelihood terms of both cases by their respective posterior probabilities. This is the natural thing to do when using the Laplace approximation, but it also works well with expectation propagation.

In estimating the skills of the players we would like to assign more importance to matches that have occurred recently than to matches that were played long ago. The main innovation in my approach is to do this by replacing the pseudo posterior above with a weighted version:

$\tilde{p}_{w}(s|r) \propto \prod_{i=1}^{players} \phi(s_{i}-\mu_{i}) \prod_{j=1}^{matches} \phi([s_{W_{j}}-m_{j,1}]/\sqrt{v_{j,1}})^{w_{j,1}} \phi([s_{B_{j}}-m_{j,2}]/\sqrt{v_{j,2}})^{w_{j,2}}$

for weights $w_{j,1}$ and $w_{j,2}$ between zero and one. Since the normal distribution is a member of the exponential family this does not change the functional form of the posterior. Because of this, the weights can be incorporated quite naturally into the expectation propagation algorithm. An advantage of using this weighting scheme in combination with factorized approximate inference is that each match may now have a different weight for each of the two players. This is not possible using more conventional weighting methods like the one used to win the first Kaggle chess competition.

The use of a weighted likelihood in a Bayesian framework is an ad hoc solution, but can be viewed as a way of performing approximate inference in a model where the skills vary over time according to some stochastic process. An alternative solution would be to assume that this stochastic process is a (possibly mean-reverting) random walk, in which case we could use a forward-backward algorithm similar to the Kalman filter. However, for this particular problem the weighting approach performed slightly better.

After trying multiple options, the weight function chosen was $w_{j,1}=\exp(-0.012 \times l_{j,1}-0.02 \times t_{j}-0.1 \times q_{j})$, with $l_{j,1}$ the number of matches played by this player between the current month and the end of the sample, $t_{j}$ the number of months in the same period, and $q_{j}$ an indicator variable equal to one if match $j$ is from the tertiary data set, which was of lower quality. The coefficients in this function were determined by cross-validation. There were other weighting schemes that showed some promise, such as overweighting those matches with players close in skill level to player B, when estimating the skill of player A for predicting his/her result against B. Alternatively, we could overweight those matches containing players that regularly played against B, as we can be more certain about their strength in relation to B than for players that have never played against this player. Due to time constraints I was unable to explore these possibilities further. The combination of approximate inference with likelihood weighting may be an interesting topic for future research.

Post-processing

The predictions of the base model scored very well on the leaderboard of the competition, but they were not yet good enough to put me in first place. It was at this time that I realized that the match schedule itself contained useful information for predicting the results, something that had already been noticed by some of the other competitors. In chess, most tournaments are played according to the Swiss system, in which in each round players are paired with other players that have achieved a comparable performance in earlier rounds. This means that if in a given tournament player A has encountered better opponents than player B, this most likely means that player A has won a larger percentage of his/her matches in that tournament.

In order to incorporate the information present in the match schedule, I generated out-of-sample predictions for the last 1.5 years of data using a rolling 3-month prediction window. (i.e. predicting months 127-129 using months 1-126, predicting months 130-132 using months 1-129 etc.) I then performed two post-processing steps using these predictions and the realized match outcomes, the first using standard logistic regression and the second using a locally weighted variant of logistic regression. These post-processing steps used a large number of different variables as can be seen in the code below, but the most important variables were:

• the predictions of the base model
• the posterior means of the skills of A and B
• the number of matches played by these players
• the posterior means of the skills of the opponents encountered by A and B
• the variation in the quality of the opponents
• the average predicted win percentage over all matches in the same month for these players
• the predictions of a random forest using these variables

By comparing the quality of the opponents of A and B in a given tournament we may predict the result of the match between A and B. However, the data only indicated the month in which each match was played and not the tournament, and some players appear to have played in multiple tournaments in the same month. What finally pulled me ahead of the competition may have been the addition of a variable that weighs the skills of the opponents of B by their rooted pagerank to A on the match graph. The idea behind this was that if a player C, who played against B, is close to A on the match graph, the match between B and C most likely occurred in the same tournament as the match between A and B.

In order to make the locally weighted logistic regression computationally feasible, the post-processing procedure first allocates the matches in the test set into a small number of cluster points for which the logistic regression is performed. For the final prediction we can then use local interpolation between the parameter estimates at the cluster centers using Gaussian processes. Using this approach the post-processing was sufficiently fast to allow me to quickly try many different settings and variables.

Conclusions

I would like to thank both the organizers and the competitors for a great competition. Much of the contest came down to how to use the information in the match schedule. Although interesting in its own right, this was less than ideal for the original goal of finding a good rating system. Despite of this I hope that the competition, and my contribution to it, was useful and that it will help to advance the science of rating systems.

Winning the Deloitte/Fide chess comp

The following is a cross-post of my contribution to the Kaggle blog, made after winning the Deloitte/FIDE Chess Rating Challenge: a contest for developing a new predictive system for ranking chess players, sponsored by Deloitte Analytics Australia and the world chess federation FIDE.

My name is Tim Salimans and I am a PhD candidate in Econometrics at Erasmus University Rotterdam. For my job I constantly work with data, models, and algorithms, and the Kaggle competitions seemed like a fun way of using these skills in a competitive and social environment. The Deloitte/FIDE Chess Rating Challenge was the first Kaggle contest I entered and I was very fortunate to end up taking first place. During the same period I also used Kaggle-in-class to host a prediction contest for an undergraduate course in Econometrics for which I was the teaching assistant. Both proved to be a lot of fun.

Chess rating systems

The first thing to do when tackling a new problem is to look up what other people have done before you. Since Kaggle had already organized an earlier chess rating competition, the blog posts of the winners were a logical place to start. After reading those posts and some of the academic literature, I found that chess rating systems usually work by assuming that each player’s characteristics can be described by a single rating number. The predicted result for a match between two players is then taken to be some function of the difference between their ratings. Yannis Sismanis, the winner of the first competition, used a logistic curve for this purpose and estimated the rating numbers by minimizing a regularized version of the model fit. Jeremy Howard, the runner-up, instead used the TrueSkill model, which uses a Gaussian cumulative density function and estimates the ratings using approximate Bayesian inference.

I decided to start out with the TrueSkill model and to extend it by shrinking each player’s rating to that of their recent opponents, similar to what Yannis Sismanis had done in the first competition. In addition, I introduced weights into the algorithm which allowed me to put most emphasis on the matches that were played most recently. After some initial experimentation using the excellent Infer.NET package, I programmed everything in Matlab.

Using the match schedule

The predictions of my base model scored very well on the leaderboard of the competition, but they were not yet good enough to put me in first place. It was at this time that I realized that the match schedule itself contained useful information for predicting the results, something that had already been noticed by some of the other competitors. In chess, most tournaments are organized using to the Swiss system, in which in each round players are paired with other players that have achieved a comparable performance in earlier rounds. If in a Swiss system tournament player A has encountered better opponents than player B, this most likely means that player A has won a larger percentage of his/her matches in that tournament.

In order to incorporate the information present in the match schedule, I generated out-of-sample predictions for the last 1.5 years of the data using a rolling 3-month prediction window. I then performed two post-processing steps using these predictions and the realized match outcomes. The first step used standard logistic regression and the second step used a locally weighted variant of logistic regression. The most important variables used in the post-processing procedure were:

•  the predictions of the base model
•  the ratings of the players
•  the number of matches played by each player
•  the ratings of the opponents encountered by each player
•  the variation in the quality of the opponents encountered
•  the average predicted win percentage over all matches in the same month for each player
•  the predictions of a random forest using these variables

This post-processing dramatically improved my score and put me well ahead of the competition for some time. Later, other competitors made similar improvements and the final weeks of the competition were very exciting. After a long weekend away towards the end of the competition I came back to find that I had been surpassed on the leaderboard by team PlanetThanet. By tweaking my approach I was able to crawl back up during the next few days, after which I had to leave for a conference in the USA. Upon arrival I learned that I was again surpassed, now by Shang Tsung. Only by making my last submissions from my hotel room in St. Louis was I finally able to secure first place.

Conclusions

Much of the contest came down to how to use the information in the match schedule. Although interesting in its own right, this was less than ideal for the original goal of finding a good rating system. To my relief, the follow-up data set that Jeff Sonas made available showed that my model also makes good predictions without using this information. Finally, I would like to thank both the organizers and the competitors for a great competition!


text:
# Question on Impulse-Momentum theory

A 5kg stone kicked by a student undergoes a velocity change from 6m/s to 10m/s in 4 seconds;

---- What is the momentum change of the stone within the time interval?

---- What impulse was impacted on the body during this time interval?

For the first question : $$p_2-p_1= M V_1 - MV_2 = M(V_1-V_2)= 5\textrm{ kg}(10-6)\textrm{m s}^{-1} = 20 \textrm{ kg m s}^{-1}$$

Is that correct for the first question?

As for the second : I'm stuck. Help.

$$\textrm{impulse}=I = F\cdot t$$ $$F = (mv-mu)/t$$ $$F = m(v-u)/t$$ $$F\cdot t = m(v-u)$$ $$\Rightarrow I = 5\textrm{ kg}(10-6)\textrm{ m s}^{-1} = 20 \textrm{ kg m s}^{-1}$$

is that correct? Because there has been no need to use time (4 seconds)......

or : $$F = ma = 5 \textrm{ kg} \frac{(10-6)\textrm{ m s}^{-1}}{4\textrm{s}}= 5\textrm{ N}$$

$$\textrm{impulse}= I = F\cdot T= 5\textrm{ N}\cdot 4 \textrm{ s} = 20\textrm{ N s},$$

still the same 20.

Thanks

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Yes, your thinking is correct. The impulse imparted is equal to both the applied force times the time and to the change in momentum, $$I=F\Delta t=\Delta p.$$ (This is exactly Newton's second law.) The time taken by the kick is superfluous in your question.