# Tag Info

6

This is what I'd do: You say most of the pixels are dark, and thus uninteresting, but some of them are bright. So I'd start by summing all images up to find the "bad" pixels: files = FileNames[ "*.png"]; totalBrightness = 0.0; Monitor[Do[ totalBrightness = ImageData[Import[f]] + totalBrightness, {f, files}], f]; meanBrightness = ...

4

The data FileNames["*.png"] (* {"image_01.png", "image_02.png", "image_03.png", \ "image_04.png", "image_05.png", "image_06.png", "image_07.png", \ "image_08.png", "image_09.png", "image_10.png"} *) All at once If there are no memory constraints, you can load all in a single array (read below for other cases). data = ImageData[Import[#], "Byte"] & ...

4

What I am most interested in is described in Oleksandr's answer. Here's something else I would also try: There are several built-in functions that take advantage of parallelization without any special settings (and without using the Parallel tools framework). I would like to know how well these scale to a high number of cores. Examples: Matrix ...

8

4 TB memory? N-body of course; dir = 2; T = 10; k = 1; l = 10; n = l^3; v = 6; q[i_] := (-1)^i m[i_] := RandomReal[5] Rem[o_] := l (o - IntegerPart[o]) eqns = Table[ {D[Subscript[r, i][t], {t, 2}] m[i] == Sum[ Normalize[Subscript[r, i][t] - Subscript[r, j][t]] k q[ i] q[j]/(Subscript[r, i][t] - Subscript[r, ...

3

Pieter, here is one suggestion: - try a FEM calculation on a huge and complicatd structure, eg. stress around clusters of cracks :)

4

To take advantage of that kind of memory, you really want to do some parallel processing. Mathematica's parallel processing focuses on data-parallelism, or more simply, embarrassingly parallel problems. So you might try out various Monte Carlo simulations. I don't recommend trying to reproduce full MPI functionality with LinkCreate et al. In the case ...

4

The more I've thought about this question, the more my answer (above) has changed. Now my answer is this: Assuming there is a high-bandwidth connection to the Wolfram server, choose a problem that relies on Mathematica's superior handling of curated data. Create an enormous problem that relies on curated financial data, geographic data, biological data, ...

1

You could just try to find the next largest prime number http://www.iflscience.com/editors-blog/largest-ever-prime-number-found-gimps https://www.youtube.com/watch?v=tlpYjrbujG0

20

I've always wondered about the scalability of MathLink (now officially "Wolfram Symbolic Transfer Protocol"). This is the protocol used by Mathematica to communicate between the front end and the kernel, and the basis of the Parallel` package. It has quite low bandwidth and high latency relative to, for example, MPI libraries. I also wonder how many MathLink ...

13

I am sure you can easily install also Linux on it and then you could contact Vladyslav Shtabovenko, the current maintainer of FeynCalc (https://github.com/vsht) and ask him about hard problems in High Energy Physics he would like to benchmark on such a King-Kong machine. Either him or somebody else could also provide you with more complicated examples of ...

11

I would choose a problem that exploits the unique power of Mathematica, in particular the natural functions involving graph theory, symbolic math, graphics, and the high compute power you have available. So I would choose some image recognition and clustering problem such as: Take some large number of images ($\sim\!\!10^8$) and perform deep learning on ...

2

I post this as a modest gain in efficiency. Consider the partition $N$ into distinct integers in ascending order. Let $T_n$ be the largest triangular numberless than $N$. So if there can be a partition of $m>n$ integers then $N=\sum_{j=1}^m a_j$. Now, $N=T_n+\delta<T_{n+1}$ where $0\leq\delta<n+1$. As $a_1\ge1$ and $a_{i+1}-a_i\ge1$ then \$N\ge ...

3

This, based on stackoverflow iterative partitions lets you generate those partitions incrementally and will let you discard those which contain duplicates as you go. First some test data to make sure this works Length[IntegerPartitions[20]] which returns 627 and Length[Select[IntegerPartitions[20], DuplicateFreeQ]] which returns 64. Now a test case ...

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