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12:13 AM
@acl I'd like that. If you don't mind, wait until I'm around so that the order of answers is generally chronological. I'll be very curious to see if anyone has a faster solution. It certainly wouldn't surprise me.
 
@Mr.Wizard I'll FGITW you and link to the chat transcript =)
 
You'd have the right; you were working on it too. Just don't post my answer. ;-)
 
@Mr.Wizard I did mean your answer ;)
 
blah! damn toad steals my... oh wait. :-P
 
:D
 
12:18 AM
This could end up with the devs having to get involved... ;->
 
Well, you do know they won't do anything, right? ;)
 
O rly? (* rushes to destroy rm -rf acount *)
 
you can't :) Try it...
 
Okay, delete then. :-P
(I know your rep is too high)
 
Even delete won't work... it puts it in a queue to the devs
 
12:20 AM
I asked if it was possible to delete my own account when I was first made a moderator. I was told: "Don't."
@rm-rf Actually that's good to know. Seemed kind of dangerous.
 
I think you can't merge/delete any account if it is over 2k rep, IIRC.
Also, you can't merge two accounts if a linked account (or that account) is a moderator anywhere on the network
 
Interesting nuggets you bring.
 
@Mr.Wizard All hell would break loose if a cranky SO mod went about merging random accounts into Jon Skeet =)
Merge all the top 20 users into one giant monster!
 
I thought that's what Jon Skeet already was...
 
Alright, I gotta run now... see you later!
 
12:27 AM
Bye!
 
 
1 hour later…
1:29 AM
@any mod: can we have "licence" as a synonym for license? Not all of us accept all of Webster's spelling reforms (here the meaning of the word is changed as well as the spelling).
 
1:44 AM
@OleksandrR. done
 
 
6 hours later…
7:16 AM
@halirutan the points have been pre-computed and loaded in from a text file
 
 
2 hours later…
9:17 AM
@halirutan are you here??
or Hi All, may I ask you a favor?
 
 
1 hour later…
10:44 AM
@daniele I'm not really here, no.
Maybe in 4 hours.
 
11:12 AM
@halirutan ok, thank you. I just want to ask you few questions:
1) My input is made by a matrix of 0 and 1 (that in the code I ask you to help me to implement where respectively 1 and 2). If I want to change the code, can I write the following one - after Count[{nb,#]&/@ can I write {1,2}, instead of {0,1}??
countRedBlack[a_, i_Integer?Positive, j_Integer?Positive] :=
With[{nb = Flatten@a[[i - 1 ;; i + 1, j - 1 ;; j + 1]]},
Count[nb, #] & /@ {0, 1}]
and then write:
evaluateCell[a_, i_, j_] :=
Module[{numWhite, numBlack}, {numWhite, numBlack} =
countRedBlack[a, i, j];
Which[numWhite == numBlack, 2, numWhite > numBlack, 0,
numBlack > numWhite, 1]]
What do you think?
2) If I would like to get the last step number matrix of my randomeLife, how can I do?
I mean, I need to count the number of white, black and red cell into the last step of my simulation. How can I do??
 
 
2 hours later…
12:57 PM
@J.M. you around?
 
@belisarius Now I am.
 
@J.M. Hi
 
How might I help you?
 
I wrote a small program for a kid that is able to translate many trig expressions
to a form involving only one trig function
 
@belisarius That question in main from a few days ago?
 
1:05 PM
no, no, a "real" kid. A cousin's son
I used a part of the answer I gave the other day
an example
trigExpress[Tan[x] Sin[x] - Cot[2 x], tan]
results in
(-1 + tan[x]^2 - (2 tan[x]^3)/Sqrt[1 + tan[x]^2])/(2 tan[x])
using only Tan[x]
nothing brilliant
but I saw this kind of q posted often here
I wonder if you do remember some q were I can paste my code instead of asking an on purpose question
@J.M. and ping just in case
 
@belisarius Hmm, let me check...
 
@J.M. Nice one. Thanks!
 
(Very conveniently, there are not that many questions tagged ... :D )
 
1:17 PM
@J.M. A lame tag, certainly
 
Oh sure, but at least they're gathered up. :)
 
yep. that's the purpose of tags. To pile up things
I ever wonder why the SE network doesn't use a tagging hierarchy, like a library. Anyway, no librarian around I guess
 
@belisarius It's categories all the way down... :D
 
@J.M. if the tags were categorized, it would make a huge improvement on searching the site
a thing Google can't do
right now
 
I have a good feeling improving the built-in search is not too high on their priorities...
How many years have people been grumbling about the blasted thing? :)
 
1:28 PM
@J.M. The know they can't compete with big G
but they could integrate Google search on the site
 
@belisarius "could" is the operative word... oh well.
 
instead of the poor function they use now
see ya. thanks for the link
 
Okay, see you later.
 
1:48 PM
@DanieleRicci 1) Yes, this is possible. I would suggest to use 0 for black, 1 for white and 2 for the neutral state. This is because in images and colors 0 is always black. Look at GrayLevel[0] for instance. Therefore I would use
Which[numWhite == numBlack, 2, numWhite > numBlack, 1, numBlack > numWhite, 0]
2) Very easy
Currently you use NestList inside the makeFrames function to apply evaluatedAll over and over again.
The problem here is, that have to specify the number of steps you want to make.
@DanieleRicci There is a far more better solution, because what you basically want is to apply evaluateAll over and over again until your matrix does not change anymore
The function for this is called FixedPoint
Therefore, get the last step in your calculation, you only have to call:
FixedPoint[evaluateAll, A]
And if you want to create all frames until the end, you should do something like
makeFrames[A_, n_] :=
  Map[ArrayPlot[#, Mesh -> False, ColorFunction -> "RedBlueTones"] &,
   FixedPointList[evaluateAll, A]];
 
2:38 PM
Could anyone please tell me what the PrecisionGoal is in NDSolve if one simply omits it? I thought it might be infinity or WorkingPrecision/2 but neither of these seems consistent with the answers I get when I compare against explicity setting PrecisionGoal->Infinity or (wp/2). However if I use PrecisionGoal->Automatic this does line up with WorkingPrecision/2 (as suggested in docs). But what omit when it's simply omitted from the options?
 
@fpghost It takes whatever is the setting that appears in the defaults: Options[NDSolve].
 
thanks!
 
hhh
3:19 PM
I should get 2x1 -matrix, I try to do simple matrix multplication (2x2).(2x1) ---> (2x1)

http://pastie.org/5052699
I get a tensor, not 2x1 matrix, anyone ideas why?
 
 
1 hour later…
4:43 PM
@hhh your AA is not a matrix. See the structure with AA // MatrixForm
 
5:07 PM
When I use NDSolve to compute 2 solutions to a particular ODE and then do some numerical integration on an integrand dependent on these solutions Mathematica does it, however if I wrap a Module around these three commands Module[NDSolve[...]; NDSolve[...]; Nintegrate[something] then Mathematica dies of memory outage. Why can I do it command by command, but not as a Module?
 
 
3 hours later…
acl
8:17 PM
@Mr.Wizard I posted a question you might like to answer!
 
8:32 PM
@acl He lost:
The fastest solution is: tiltedhamiltonian[nwells_ /; EvenQ@nwells, theta_] := Module[{size = nwells, centre = nwells/2, bdiag, boffdiag}, boffdiag = SparseArray[{}, size - 2, -0.5]; bdiag = RandomReal[99, size - 1]; SparseArray[{{i_, i_} :> bdiag[[i]]}, size - {1, 1}] + Sum[DiagonalMatrix[boffdiag, i], {i, {-1, 1}}] ]... hurr hurr :D — rm -rf 1 min ago
 
acl
@rm-rf lol
 
hhh
8:55 PM
@rm-rf Thank you! I had messed up with a' and a. Now solved it, got a recursive eq by accident due to the mistake.
 
hhh
9:31 PM
Solve[{F2*c2 + F1*c1 - a*(2*g*h)^(1/2)*N/(A*h) == 0}, N] // Simplify

<-- how can I get the simplest form of it?
For some reason, ti does not simplify the h -terms.
 

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