1
$\begingroup$

I need a code for thermodynamics research that generates a matrix of 32768 by 32768 elements.

The code is the followng:

s0[MicroNumber_, k_] := If[RealDigits[MicroNumber, 2, 15][[1]][[k]] == 0, -1, 1]
s0[MicroNumber_, 0] = 1;
s0[MicroNumber_, 16] = 1;
s = Table[s0[i - 1, j], {i, 1, 32768}, {j, 0, 16}];

AbsoluteTiming[With[{MicroElements = s}, Table[If[Abs[Sum[(MicroElements[[i]]- MicroElements[[j]])[[d]], {d, 1, 
    16}]] == 2, 1, 0], {i, 1, 32768}, {j, 1, 32768}]]]

This code takes a very long time to run and I am trying to make it run faster. Any ideas on how to make it more efficent? I tried using ParallelTable instead of Table but it didn't help.

$\endgroup$
1
  • $\begingroup$ Are you sure that you want to check Abs[Sum[(MicroElements[[i]]- MicroElements[[j]])[[d]], {d, 1, 16}]] == 2 and not Sum[Abs[MicroElements[[i, d]] - MicroElements[[j, d]]], {d, 1, 16}] == 2? $\endgroup$ Commented May 21, 2022 at 17:40

2 Answers 2

4
$\begingroup$

I suspect you only want to optimize the last line, i.e. I haven't bothered speeding up s. In what follows I'll use the first 1000 x 1000 submatrix to show some comparative timings.

Your current implementation:

With[{MicroElements = s}, 
   Table[If[Abs[Sum[(MicroElements[[i]] - MicroElements[[j]])[[d]], {d, 1,16}]] == 2, 1, 0],
      {i, 1, 1000}, {j, 1, 1000}]]; // AbsoluteTiming
(*{5.44532,Null}*)

The main reason your implementation is slow is the use of Sum, you should instead use Total since you're trying to sum over the entire list.

With[{MicroElements = s}, 
   Table[If[Abs[Total[(MicroElements[[i]] - MicroElements[[j]])]] == 2, 1,0],
   {i, 1, 1000}, {j, 1, 1000}]]; // AbsoluteTiming
(*{0.681037, Null}*)

We can go a bit further by compiling the function:

cf = Compile[{{x, _Integer, 1}, {y, _Integer, 1}}, 
  Boole[Abs[Total[x - y]] == 2], CompilationTarget -> "C", 
  RuntimeOptions -> "Speed", RuntimeAttributes -> {Listable}, 
  Parallelization -> True]

This seems to help very slightly, but can be used with DistanceMatrix and DistanceFunction to give fairly clean code:

AbsoluteTiming[DistanceMatrix[s[[;;1000]], s[[;;1000]], DistanceFunction -> cf];]
(*{0.469302,Null}*)

Since we specified Listable in our compiled function above, we could also consider using the following:

AbsoluteTiming[Map[cf[s[[;; 1000]], #] &, s[[;; 1000]]];]
(*{0.160225,Null}*)

Since we came this far, we might as-well compile all the way:

cf2 = Compile[{{mat, _Integer, 2}}, 
  Table[Boole[Abs[Total[i - j]] == 2], {i, mat}, {j, mat}], 
  CompilationTarget -> "C", RuntimeOptions -> "Speed"]

cf2[s[[;;1000]]];//AbsoluteTiming
(*{0.105187,Null}*)

As for the actual calculation, the methods above appear to exhaust the memory on my machine (32768 x 32768 is a large array!), but some simple scaling suggests it'll take ~2 mins to complete

With[{timings = {2^#, First[AbsoluteTiming[cf2[s[[;; 2^#]]];]]} & /@ Range[9, 14]},
 Echo[ListLogLogPlot[timings]];
 2^(Fit[Log[2, timings], {1, x}, x] /. x -> Range[9, 15])
 ]

(*{0.0275781, 0.109413, 0.434082, 1.72217, 6.83248, 27.107, 107.544}*)

enter image description here

Finally, note that your matrix is actually symmetric - so you might even consider using Subsets to only compute the upper-triangular part of the matrix, as demonstrated here.

$\endgroup$
4
$\begingroup$

This seems to produce the correct result and is much faster for small values of n.

n = 32768;

x = Total[2 RealDigits[Range[0, n - 1], 2, 15][[All, 1]] - 1, {2}] + 1;
nf = Nearest[x -> "Index"];
B = SparseArray[
   Join @@ MapThread[
      Thread[{#1, Complement[#2, #3]}] &,
      {Range[Length[x]], nf[x, {\[Infinity], 2}], 
       nf[x, {\[Infinity], 1}]}
      ] -> 1,
   {n, n}
   ];

Note that x is effectively the same as Total[s[[All,1;;16]],2]. So you only have to compute only differences of scalars instead of sums of differences of vectors.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.