Wolfram Mathematica

I mean, functions like that one simply get their symbol defined when the kernel sends them to the FE, just like you would get it defined if you did bla[FrontEnd`whatever], but nothing

is actually "loaded" in the kernel

rm -rf

If you do Names["FrontEnd*"]` in a fresh mma session, the function that John used is missing... I think that's what halirutan is referring to: How do we find hidden functions that are loaded/defined only when called

Rojo

12:19 AM

Ok, what I'm saying is that

I suspect

that they aren't defined when called. It's only the FE that knows them, never the kernel

No DeclarePackage

running kernel code

rm -rf

Ah, I see what you mean... So essentially Names only tells you what the kernel knows?

Rojo

No stub symbol registered to run some code

That's my guess

halirutan

@Rojo hmm..

Rojo

It would be interesting to ask if there's any FrontEnd`UndocumentedFrontEndFunctionsList

halirutan

Ok, do we have a way to proof your assumption?

Rojo

12:23 AM

Humm

halirutan

One way would be to listen to the communication which happens on $FrontEnd // FullForm

Rojo

Yeah

or dig into the output of

DeclarePackage["bs`", {"bs`bs"}];
Trace@Information@"bs`*"

halirutan

@Rojo What is this?

Rojo

@halirutan Nothing useful

@halirutan What's easy to check is that

DeclarePackage["asf`", {"gfs"}];
Names["asf`gfs"]

DeclarePackage actually defines the symbols

halirutan

yep

Rojo

12:38 AM

so if it is not how I guessed then there's some mechanism I'm not aware of

P. Fonseca

12:59 AM

V9

Any news from the wtc?

halirutan

@Rojo Still here?

Rojo

Leaving for the cinema

Leave a meessabe, beep

halirutan

Yep.. Have fun

J. M.

1:53 AM

@chris I'm not very unfamiliar with it, yes. :) Well, it is quick-and-dirty; it has no error estimate, for one, and you know how sensitive the business of inverting Laplace transforms is...

@chris If you're going to use it, be prepared to keep increasing the precision of your numbers (and increasing the number of evaluation points), just to make sure the digits agree.

(Damn, I should really write more stuff on that blog; I'm starting to accumulate lots of undocumented routines again...)

@halirutan I'm not surprised that most people I talk to are many time zones away... :)

halirutan

@J.M. So I assume I can say "Good morning" ;-)

J. M.

2:37 AM

@halirutan Yep, good morning. Bit cloudy on my end, tho.

:)

halirutan

@J.M. Still dark here and time for bed..

J. M.

In which case, guten Nacht. :)

halirutan

Yep

Rojo

7:22 AM

@halirutan Back

halirutan

9:14 AM

@Rojo Good morning.

Daniele Ricci

9:36 AM

@halirutan good morning

and good mornig to everyone

I know that i'm boring, but can i ask to everyone here in the chat, if do you know how to create run a rule 30 CA with 3 colors?

Daniele Ricci

11:51 AM

I'm alone here??

@halirutan i have tested my model and I have a problem

the problem is that the number of cell white is always bigger than black cell

halirutan

@DanieleRicci Ok, and?

Daniele Ricci

ant the reason is because the starting matrix (the input) has a number of white cell bigger than black cell

and for this reason the number of white cell will be always bigger than black cell

and it's not good form my result

halirutan

Ok, and you want to achieve what?

Daniele Ricci

my purpose is to achieve a mixed result

i mean i want to have times in which i have black cells winners and times in which i want to have white cells winners

halirutan

Ok, so I guess you want your start-matrix random, but not really random because you want to fix the number of black and white cells?

Daniele Ricci

11:55 AM

i have my start matrix

that is not randome

it comes from another simulation

from rule 30

do you have Skype?

halirutan

Hmm, but if your input-matrix is always given, you cannot do much about the outcome of your CA.

Daniele Ricci

i know

halirutan

So what exactly you want to know now?

Daniele Ricci

my input matrix change following other input (that change) put in a rule 30 simulation

i wanted to tell you the result of my simulation

so now I have to change and see again my input matrix

halirutan

@DanieleRicci ahh, so that white cells are always more than black cells was an information and not a question?

Daniele Ricci

12:09 PM

ye

yep

halirutan

Ok, then I got it now..

Daniele Ricci

maybe you can help me:

to change my input

in the sense:

I want to run more time the rule 30 and at each step get and save in a matrix the last step

for example 1000 times

halirutan

What do you take as input for the rule 30 ca?

Daniele Ricci

a vector of 1 and 0

the dimension is a row made of 9 numbers (0 and 1)

9x1

so for example i have

r = CellularAutomaton[30, q4, v];
q4= {0,1,1,1,0,0,1,1,0};
v=1000.

halirutan

12:26 PM

And you want to use the output of this in your own CA?

Daniele Ricci

yep

halirutan

ok.

This here

Tuples[{0, 1}, 9]

gives you all possibilities of a vector of length 9 with 0/1 as elements.

So what you could do is, you run rule 30 exactly 1000 times on each of those vectors. This gives you 512 input matrices for your own CA

inputMatrices = Map[
   Function[input, CellularAutomaton[30, input, 1000]],
   Tuples[{0, 1}, 9]];

Daniele Ricci

one second let me think :)

ok let's say

ok let's do in the following way

I have to produce 2 different matrixes and then put them together in my CA code

then

the difference btw the 2 matrixes is the starting vector

the starting vector is a mix of fixed numbers:
the central part is the result of one my database, 5 values made by 1 and 0;
the sides (2 values right and 2 value left) are random (0,1 or 0,1 and 2).

now i want to save in a each matrix the result of this simulation using the rule 30, with the central vector that is not changing and the sides that are random.

then i want to put together the matrixes and use them in my CA

how can I do?

halirutan

12:52 PM

The question is, if you can calculate all possible input vectors, why do you want to make it random?

Daniele Ricci

because i need to have in the central part of the input vector my numbers, because those numbers are the starting point of my studying

halirutan

Say you want to build all possibilities of vectors with 2 elements left and right that are a combination of {0,1,2} and a central vector consisting of all posibilities of 5 numbers {0,1}:

Daniele Ricci

yep and use them as input for the simulation in rule 30

halirutan

Let's do it step by step so you really understand it.

Daniele Ricci

then save the result of each simulation in a matrix to use then in my CA model

halirutan

12:56 PM

All possibilities of 2 numbers which are either 0,1 or 2 is

Tuples[{0, 1, 2}, 2]

Daniele Ricci

ok

halirutan

{{0, 0}, {0, 1}, {0, 2}, {1, 0}, {1, 1}, {1, 2}, {2, 0}, {2, 1}, {2,
2}}

This is what you want for your left and right side.

Daniele Ricci

yep

halirutan

you central vector is Tuples[{0, 1}, 5]

{{0,0,0,0,0},{0,0,0,0,1},{0,0,0,1,0},{0,0,0,1,1},{0,0,1,0,0},{0,0,1,0,1},{0,0,1,1,0},{0,0,1,1,1},{0,1,0,0,0},{0,1,0,0,1},{0,1,0,1,0},{0,1,0,1,1},{0,1,1,0,0},{0,1,1,0,1},{0,1,1,1,0},{0,1,1,1,1},{1,0,0,0,0},{1,0,0,0,1},{1,0,0,1,0},{1,0,0,1,1},{1,0,1,0,0},{1,0,1,0,1},{1,0,1,1,0},{1,0,1,1,1},{1,1,0,0,0},{1,1,0,0,1},{1,1,0,1,0},{1,1,0,1,1},{1,1,1,0,0},{1,1,1,0,1},{1,1,1,1,0},{1,1,1,1,1}}

Daniele Ricci

ok

halirutan

12:58 PM

lets move to another chatroom so we don't annoy the folks here:

Daniele Ricci

ok

halirutan

zeroth halirutan

halirutan

2:49 PM

Run 10k, RUUUUN, I'm coming!

J. M.

3:45 PM

@halirutan I'm wagering you'll be part of the 10k club by Tuesday or so...

chris

@J.M. When I run the code from your blog I get (i.stack.imgur.com/yCi6l.png) instead of tpfto.files.wordpress.com/2012/02/firemap.png. What am I missing?

(other than the ColorTable!)

@J.M. I am interested in modifying the code of your blog to produce 2D Gaussian random fields on the sphere following mathematica.stackexchange.com/questions/4829/…

J. M.

@chris Hmm, you used the routine in the Compile[] docs?

chris

@J.M. yes: this is my (your) code !Mathematica graphics

J. M.

@chris Quite odd. Unfortunately, my notebooks are in a different computer. :( I won't be able to debug/test things out until tomorrow...

chris

@J.M. no worries. Just curious.

Am I right it assuming it could be modified to produce equal area Gaussian Random Noise?

J. M.

3:58 PM

Don't worry, I'll ping you once I am able to carry out tests...

@chris Well, Perlin noise is a very different beastie. I'm not sure how much of the setup will carry over to Gaussian noise.

chris

I am after the equal area feature...

In healpix (what my collegues use) its quite sophisticated. You seem to get this feature in a less painful manner.

J. M.

I'll need to mull it over, I think. Maybe even peek at the code behind healpix and see what can be carried over to Mathematica...

chris

@J.M. I have another unrelated question. Your function topoFake, is it possible to integrate it into the ColorData[] framework?

Rojo

@halirutan Mornin

J. M.

@chris That's one good question. I don't know if one could safely modify curated data functions like that...

chris

4:05 PM

I have a ColorTable (which I got from xmedcon and looks like this !Mathematica graphics) and was used by ESA com to make this image lastro.epfl.ch/files/content/sites/lastro/files/shared/sciact/… which I would like to integrate as well as your

@J.M. I would be useful though woudn't it: one would only need to remember its name.

J. M.

@chris What format is the color table in? I might be able to figure how to turn it into a Blend[] function...

chris

@J.M. (b.t.w. I use MMA 8.0.4)

its is a table of char. I use this code `Clear[fcol, fcolc];
pal = Import["~/Yorick/Gist/xmedcon/xmedcon.fits", "RawData"]/255.;
Do[fcol[i - 1] =
Function[{x, y, z},
RGBColor @@ ( #[Mod[z*199, 200] + 1] & /@ (Interpolation /@
pal[[i]])) // Release];
fcolc[i - 1] =
Function[z,
RGBColor @@ ( #[Mod[z*199, 200] + 1] & /@ (Interpolation /@
pal[[i]])) // Release];, {i, 1, 96}]`

`ContourPlot[Sin[x] Sin[y] , {x, -4, 4}, {y, -3, 3},
ColorFunction -> fcolc[64]]`

J. M.

Is there any place where I could download xmedcon.fits without having to install Yorick?

chris

dl.dropbox.com/u/659996/xmedcon.fits

(its not even part of yorick!)

J. M.

Thanks, I'll take a look.

@chris May I ask what you'll use topoFake[] for? :)

chris

4:13 PM

No idea yet! I just looked nice ;-)

In yorick there is a palette called earth

may be make "fake earth" mollweide views.

This xmedcon includes all these palettes. (i.stack.imgur.com/f3dYn.png)

There are lots of gray scales for bones I believe ;-)

J. M.

@chris I was basically browsing around this site and decided to combine all the nice-looking features of the "realistic Earth" gradients. Hence, topoFake[].

@chris Ah, the bone() map of MATLAB, I did a while back...

chris

The yorick earth ColorTable looks like this (sourceforge.net/projects/yorick/screenshots/42193/182/137)

István Zachar

4:45 PM

Hello

I had a strange experience with Definition:

f[x_] := Sin[x];
g[x_] := Tan[x];
Definition /@ {f, g}

The Map returns a syntactically invalid output.

Does anyone have any idea what is going on?

The result is so hidious I did not want to post it as a question without first consulting with someone.

chris

@IstvánZachar it seems to be a print problem.

Definition prints Something

J. M.

@chris I have the file now. Unfortunately, I need to step out, so I'll only be able to look at the gradient later. I now have two IOUs. :)

As soon as I figure out the gradient, and what's happening with the spherical Perlin noise, I'll ping you.

So, see y'all later.

chris

@J.M. thanks.

Daniele Ricci

7:00 PM

@halirutan when you come back, please, write me

:)

Rojo

@IstvánZachar It seems Definition formatting doesn't deal well when it's not at the highest level. I would just use Print/@Definition/@{f, g};

Or ToString/@Definition/@{f, g}

Daniele Ricci

7:16 PM

or to everyone

maybe you can hel pe

help me

is anyone here?

Rojo

Alo

@DanieleRicci what's going on

Daniele Ricci

I need to know why my code has this following problem

cl.ly/image/2n2F1G2a1G1r

i mean why the code that i have developed with @halirutan give me this previous problem

the code is easy

Rojo

@DanieleRicci I have no idea of your previous talks with halirutan, so you will have to explain

Daniele Ricci

yep

my code is a 2d CA

with a my rule

the rule is the followinf

following

countRedBlack[a_, i_, j_] :=
With[{nb = Flatten@a[[i - 1 ;; i + 1, j - 1 ;; j + 1]]},
Count[nb, #] & /@ {1, 2}];
evaluateCell[a_, i_, j_] :=
Module[{numRed, numBlack}, {numRed, numBlack} =
countRedBlack[a, i, j];
Which[numRed == numBlack, 0, numRed > numBlack, 1, True, 2]];
evaluateAll[A_] :=
With[{paddedA = ArrayPad[A, 1]},
Table[evaluateCell[paddedA, i, j], {i, 2, Length[A] + 1}, {j, 2,
Length[A[[1]]] + 1}]];
result = Function[input, FixedPoint[evaluateAll, input]] /@
inMatrices2;
makeFrames[A_, n_] :=

randomLife = makeFrames[inMatrices2, 10];
animate[randomLife]
FixedPoint[evaluateAll, inMatrices2] // Flatten // Tally

if in my grid i have the number of red cell bigger than black cell, it will return red cell

and viceversa

Rojo

I just ran that, without thinking

and got errors like

ArrayPad::arr: First argument inMatrices2 to ArrayPad should be an array.

inMatrices2 hasn't been defined

Daniele Ricci

7:21 PM

inMatrices2 =
Function[mat,
Table[If[RandomReal[] > .8, 2, j], {i, mat}, {j, i}]] /@
inMatrices;

Rojo

Now inMatrices is undefined

Daniele Ricci

the whole code is the following

:

allVecs =
Flatten /@
Tuples[{Tuples[{0, 1}, 2], {{1, 1, 1, 1, 1}},
Tuples[{0, 1}, 2]}]; inMatrices =
CellularAutomaton[30, #, 10] & /@ allVecs;
inMatrices2 =
Function[mat,
Table[If[RandomReal[] > 0.01, 2, j], {i, mat}, {j, i}]] /@
inMatrices;

countRedBlack[a_, i_, j_] :=
With[{nb = Flatten@a[[i - 1 ;; i + 1, j - 1 ;; j + 1]]},
Count[nb, #] & /@ {0, 1}];
evaluateCell[a_, i_, j_] :=
Module[{numRed, numBlack}, {numRed, numBlack} =
countRedBlack[a, i, j];
Which[numRed == numBlack, 2, numRed > numBlack, 0,

the problem is that

Rojo

Ok, now it worked

Daniele Ricci

if you see the grid

for example

if the grid is all white

then

it should be always white, because i don't have red or black cell

Rojo

Can you explain in short how it should work?

Daniele Ricci

7:24 PM

while the frame is starting to became all red

Rojo

Hey chris

Daniele Ricci

without any reason

don't consider the first part of the code

because is the input

consider the code just from

countRedBlack[a_, i_, j_] :=
With[{nb = Flatten@a[[i - 1 ;; i + 1, j - 1 ;; j + 1]]},
Count[nb, #] & /@ {0, 1}];
evaluateCell[a_, i_, j_] :=
Module[{numRed, numBlack}, {numRed, numBlack} =
countRedBlack[a, i, j];
Which[numRed == numBlack, 2, numRed > numBlack, 0,
numBlack > numRed, 1]];
evaluateAll[A_] :=
With[{paddedA = ArrayPad[A, 1]},
Table[evaluateCell[paddedA, i, j], {i, 2, Length[A] + 1}, {j, 2,
Length[A[[1]]] + 1}]];

result = Function[input, FixedPoint[evaluateAll, input]] /@
inMatrices2;

if you use as inMatrices2 a randomInteger matrix

made by 0,1 and 2

you can obtain the same result

Rojo

What should countRedBlack do?

Sometimes, explaining and documenting in comments what each function does makes you discover the bugs

Daniele Ricci

count the numbers of neighbors

Rojo

It takes 3 arguments

a, i and j. A is the matrix and i,j the position?

Daniele Ricci

7:27 PM

yep

Rojo

0 -> red, 1 -> black, 2 -> white, right?

Daniele Ricci

yes

Rojo

Ok, I see what you mean

I'm taking a look

Daniele Ricci

thanks

Rojo

Oh

I see

In evaluateAll, you are padding with 0's, you should pad with whites=2, right?

Daniele Ricci

7:38 PM

well

yes

Rojo

Which simply means, change the ArrayPad line to ArrayPad[A, 1, 2]

Daniele Ricci

because 2 is the neutral state

ok let me try

should go right know

5 minutes

Rojo

@DanieleRicci Ok, good luck

Btw, it's your intention to count not only the neighbours but also the very same square, right?

Daniele Ricci

thanks

what is the very same square??

Rojo

I mean

countRedBlack[matrix, 3, 3], for example

counts the number of blacks and reds in the neighbouring places

in 3,3

AND

Daniele Ricci

7:51 PM

and?

Rojo

Just checking it was your intention

since you said it should count the number of neighbors

Daniele Ricci

yep

Daniele Ricci

8:06 PM

ok now it works

@halirutan

@Rojo help me in the while

Rojo

?

DRM

@Rojo you help me to find the problem

now the last problem i have is the following

Rojo

@DRM Thanks? please? :P

DRM

I have to put together 2 matrices

Rojo

How?

DRM

8:10 PM

which have the same number of rows, but different numbers of columns

i can put them or side by side

or, i will appreciate to put them one column of Matrix A alternate to one column of Matrix B

how can i do??

Rojo

Side by side, try Join

with a last argument of 2

To alternate, first thing that comes through mind is to transpose and alternate the rows, and transpose again

Transpose@Riffle[Transpose@mat1, Transpose@mat2]

but there might be a better way

Perhaps also something like first joining them and then shuffling

Anyway, that should work

DRM

not it doesn't work because i have the following error:

Part::take: Cannot take positions 1 through 3 in 2. >>

Rojo

What code?

DRM

i want to put this Total Matrix as input of the code that you saw before

so i produce these 2 matrix in this way:
allVecsA =
Flatten /@
Tuples[{Tuples[{0, 1}, 2], {q0},
Tuples[{0, 1}, 2]}]; inMatricesLTI =
CellularAutomaton[30, #, 10] & /@ allVecsA;
inMatrices2A =
Function[mat,
Table[If[RandomReal[] > .8, 2, j], {i, mat}, {j, i}]] /@
inMatricesA;

allVecsB =
Flatten /@
Tuples[{Tuples[{0, 1}, 2], {q6},
Tuples[{0, 1}, 2]}]; inMatricesB =
CellularAutomaton[30, #, 10] & /@ allVecsB;
inMatrices2B =
Function[mat,
Table[If[RandomReal[] > .8, 2, j], {i, mat}, {j, i}]] /@
inMatricesB;

Rojo

Just give me what code you are using to join the matrices

DRM

8:18 PM

q0 and q6 are 2 row vectors of 0 and 1 with dimension respectively 5 and 7

to join the matrix I use

inMatricesTotal=Join[inMatrices2A,inMatrices2B]

Rojo

You mean Dimensions@q0 =={1, 5}?

DRM

yep

Rojo

Yeah, add a last argument to Join. Join[inMAtrices2A, inMAtrices2B, 2]

Otherwise you are joining them one below the other

DRM

always the same error then when i run:

result = Function[input, FixedPoint[evaluateAll, input]] /@
inMatricesTotal;

Part::take: Cannot take positions 1 through 3 in 2. >>

Part::take: Cannot take positions 2 through 4 in 2. >>

Part::take: Cannot take positions 3 through 5 in 2. >>

General::stop: Further output of Part::take will be suppressed during this calculation. >>

Rojo

Check the joining

if it's done correctly, then I answered your question

if it isn't, then investigate your problem before asking

DRM

8:24 PM

Join[{inMAtrices2A}, {inMAtrices2B}, 2]

??

any suggestion?

it's 14 hours that i'm in front of this pc

just a little bit tired

Rojo

Run this

(A1 = Array[a1, {3, 3}]) // MatrixForm
(A2 = Array[a2, {3, 4}]) // MatrixForm
(Join[A1, A2, 2]) // MatrixForm

@DanieleRicci

DRM

done

Rojo

:)

DRM

but please see it

and run it

q0={0, 1, 1, 1, 0};
q6={1, 0, 1, 1, 1, 1, 1, 0}

allVecsA =
Flatten /@
Tuples[{Tuples[{0, 1}, 2], {q0},
Tuples[{0, 1}, 2]}]; inMatricesLTI =
CellularAutomaton[30, #, 10] & /@ allVecsA;
inMatrices2A =
Function[mat,
Table[If[RandomReal[] > .8, 2, j], {i, mat}, {j, i}]] /@
inMatricesA;

allVecsB =
Flatten /@
Tuples[{Tuples[{0, 1}, 2], {q6},
Tuples[{0, 1}, 2]}]; inMatricesB =
CellularAutomaton[30, #, 10] & /@ allVecsB;
inMatrices2B =
Function[mat,
Table[If[RandomReal[] > .8, 2, j], {i, mat}, {j, i}]] /@

look at the resulting matrix inMatricesTotal

Rojo

8:43 PM

inMatricesA undefined

but

just check the dimensions of what you are trying to join

DRM

In[868]:= Dimensions[inMatricesA]

Out[868]= {16, 11, 9}

Rojo

What about inMatrices2A

and 2B

DRM