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Topic: Compiling numerical iterations
Replies: 7   Last Post: Mar 5, 2013 4:15 AM

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Ray Koopman

Posts: 3,383
Registered: 12/7/04
Re: Compiling numerical iterations
Posted: Feb 27, 2013 3:04 AM
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On Feb 25, 10:09 pm, firlefranz <> wrote:
> Thanks a lot! To be honest, some of the commands Ray is using, I've never seen before. I stopped using mathematica before version 5 came out.
> Coming back to Peters statement of exporting the code from Mathematica to C. How this can be done starting from my or Ray's code? There is an automated C-code-gernerator implemented in Mathematica 9, am I right?
> As a step forward to the first little peace of code, here is another try, which is not really optimized. The be concrete, I try to simulate the autocorrelation function of a random walk, which is doing a step at none equally distant time steps. This has to been done for a long random walk for many particles. Instead of doing an average over many random walks and calculate one autocorrelation function I want to simulate many correlation functions and make an average over them. Since the time steps are non equal, I wrote a sub-function, which creates a new time axis and taking the necessary value for the random walk from the first table.
> Here is what I come up with. It's running in a reasonable time for one particle, but for a real statistic ensemble, I have to do it over 1.000.000 particles for a long time. Optimizing this or (probably better) exporting it to C would hopefully help a lot. So how to export it?
> << code omitted >>

Before spending more time on your real problem, you might want to
work at understanding why the code I posted for your example problem
necessarily gives the same corr as your code does, and then why the
following, which is faster, is also correct, even tho its corr can
not be compared to the other two because it generates the random
walks differently.

num = 10^3;
AbsoluteTiming@Length[ tab = FoldList[Plus,0,
RandomInteger[BinomialDistribution[1000,1/2],num-1]-500] ]
AbsoluteTiming@Length[ corr = #.#[[1]]& @
Most@Partition[tab,num*9/10,1]/(num*10*9/4) ]

{0.010728, 1000}
{0.010363, 100}

There are other ways to speed things up even more,
but that will suffice for now.

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