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There is an edited section at the bottom to clarify some questions which have come up in the comments.

I have some code which generates (according to two coordinates) two $m\times n$ arrays and then performs element wise multiplication of the two arrays to then get the average.

part[da_, db_, array_] := Mean@Flatten@(array[[span[db], span[da]]]* array[[span[-db], span[-da]]]);
span[x_] := Switch[Sign[x], -1, 1 ;; x - 1, 0, 1 ;; -1, 1, x + 1 ;; -1];    

SeedRandom[1]
array = Developer`ToPackedArray[RandomChoice[{-1, 1}, {1000, 1000}]];
p = {2,3};
q = {-6,8};

diff=p-q;
d1=diff[[1]];
d2=diff[[2]];

test = part[d1,d2,array]// AbsoluteTiming

This gives me

{0.0647411, -(919/493520)}

I need to run this code for a lot of different points - in the order of 500 million. Since all I am doing in the function "part" is generating

a1=array[span[db], span[da]] ]

and

a2=array[span[-db], span[-da]] ]

and element wise multiplying them

a1*a2

I really feel that I could use the power of my graphics card (GTX 1070) and 4 CPU cores (i7 7700k) to really get a massive speed boost.

As a current bench mark (on my slower 2 core laptop), I get the following for 1000 different points

pointsList = Partition[RandomInteger[1000, {2000, 2}], 2];

compute = ParallelTable[
    diff = pointsList [[i, 2]] - pointsList [[i, 1]]; 
    N@part[diff[[1]], diff[[2]], array],
 {i, Length[pointsList]}]; // AbsoluteTiming

This gives me

{18.8524, Null}

which would roughly make it ~ 110 days for all 500 million points. I honestly believe this could be done in a matter of hours using a CUDA implementation. Since I have never used CUDA in mathematica and will be teaching myself - I could in principal take some time and figure this out myself. However, due to the fact that I need the results of these computations for a paper I am writing, It would really be great to use a working code now, and figure it out later.

EDIT

To give better clarity to my question here is the full code with explanation of what is happening.

I have 14 sets of arrays. Each is 2048 x 2048, with entries which can be real values. For the sake of explanation, this is what I call

SeedRandom[1]
array = Developer`ToPackedArray[RandomChoice[{-1, 1}, {2048,2048}]];

Next I feed a list of points. This list also does not change. I generate my list according to the following:

solns[r_] := DeleteCases[{ToRules[Reduce[(x)^2 + (y)^2 == r, {x, y}, Integers]]}[[All, All, 2]], n_?(#[[2]] < 0 &)] /. {n_ /; n < 0, 0} ->Sequence[];
makedata[{x_, y_}] := ({x^2 + y^2, x, y})

makeRdata[m_] := Module[{allR, allSolns},
  allR = DeleteCases[Table[If[SquaresR[2, i] != 0, i, 0], {i, m}], 0];
  allSolns = Flatten[Table[DeleteCases[solns[allR[[i]]], n_?(#[[1]] < 0 &&#[[2]] == 0 &)], {i, Length[allR]}], 1];
  Table[makedata[allSolns[[i]]], {i, Length[allSolns]}]
];

max = 2048^2+2048^2;
pntsList = makeRdata[max];

Once I have now created all the points I will be using (if you plan on test running this code, I suggest using max = 100^2+100^2), I feed them into part, to create a nested list called t1.

t1 = ParallelTable[{N@Sqrt@dat[[j, 1]], N@part[dat[[j, 2]], dat[[j, 3]],array]}, {j, Length[dat]}] // AbsoluteTiming

Once the nested list t1 has been generated, I will write it to a file, and then re run the same code, but this time for a new array - still exactly the same size. For this new array, pntsList does not change, and so it will only have to be generated once.

In summary,I create all the points to be fed into part before hand. Once I start feeding them into part the array does not change. It is all the same array with the same dimensions. Only after all the points have been proccessed and the result written to a file. Do I rerun the code for a new array - still the same size- and I feed into it the exact same points. I repeat this for 14 different arrays.

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  • $\begingroup$ Is pointsList in your application also a random array that you will generate in the GPU? $\endgroup$ – Eisbär Jun 2 '17 at 9:24
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    $\begingroup$ No, pointsList is a 2 level predefined nested list of specific coordinates which I have already computed. I will be looping through the main list, and every sub-list will contain a set of these points which will be fed into the function. However, if it works out that the GPU can generate all these points before hand and thus compute the entire code in the GPU, that would be even better (my guess). It may be easier if I present my full code (not long at all), as it may elucidate better what I am doing and thus facilitate a CUDA solution of it. $\endgroup$ – Luca Pontiggia Jun 2 '17 at 9:39
  • $\begingroup$ Is array the same for each computation, or does it change? Is array always of dimension 1000 x 1000? $\endgroup$ – Carl Woll Jun 3 '17 at 1:12
  • $\begingroup$ @CarlWoll The size of the array changes each time. Since the array is determined by the points fed into the function. One only knows the size of it has been split into a1 and a2. Does it make a difference if the size changes all the time? $\endgroup$ – Luca Pontiggia Jun 4 '17 at 19:36
  • $\begingroup$ In your example that takes 18.8524 seconds, you have pointsList and array. Clearly in this example array is not changing. In your ultimate application, is pointsList a list of 500 million points and you compute part with the same array, or do you do 1000 points for 500,000 different array objects? Also, in your ultimate application, is array of dimension 1000 x 1000? $\endgroup$ – Carl Woll Jun 5 '17 at 2:13
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I would use ListCorrelate and rely on Mathematica's internal parallel implementation. Here is your code:

part[da_, db_, array_] := Mean @ Flatten @ (array[[span[db],span[da]]]*array[[span[-db],span[-da]]]);
span[x_]:=Switch[Sign[x],-1,1;;x-1,0,1;;-1,1,x+1;;-1];

Let's use a smaller array example:

array = Developer`ToPackedArray @ RandomChoice[{-1, 1}, {100, 100}];

Here is a table of all possible indices:

r1 = Table[part[da, db, array], {da, -99, 99}, {db, -99, 99}]; //AbsoluteTiming

{0.768523, Null}

And, here is an equivalent ListCorrelate implementation:

count[dim_] := Outer[
    Times,
    Join[Range[dim], Reverse[Range[dim-1]]],
    Join[Range[dim], Reverse[Range[dim-1]]]
]

r2 = ListCorrelate[array, array, {-1, 1}, 0] / count[100] //Transpose; //AbsoluteTiming

{0.025539, Null}

They are equal:

r1 === r2

True

Now, let's consider a larger array:

array = Developer`ToPackedArray @ RandomChoice[{-1, 1}, {2048, 2048}];

I won't bother timing the OP code, but here is the timing for ListCorrelate:

ListCorrelate[array, array, {-1, 1}, 0] / count[2048] //Transpose; //AbsoluteTiming

{11.6008, Null}

This is slow because rational numbers can't be packed. If an answer using reals is acceptable, you could use:

ListCorrelate[array, array, {-1, 1}, 0] / N[count[2048]] //Transpose; //AbsoluteTiming

{1.15444, Null}

So, you should be able to do all 14 array objects in less than 170 seconds using rationals, or 15 seconds using reals.

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  • $\begingroup$ I need to see if I can translate this into the output I require for my analysis, but for now - before I'm happy to accept this as an answer - the part function you wrote has Total instead of Mean. That Mean really has to be there as I am getting the average. The way you put it wouldn't be a problem, other then the fact that I don't know what to divide the total by to get the mean. In my part function you know what to divide by - the length of the flattened array of a1*a2. In ListCorrelate I wouldn't know how to do this off hand. $\endgroup$ – Luca Pontiggia Jun 5 '17 at 19:14
  • $\begingroup$ From reading up on ListCorrelate, I have a feeling I need to define my new functions fand g where f remains Times and g will become an average instead of the default Plus $\endgroup$ – Luca Pontiggia Jun 5 '17 at 19:40
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    $\begingroup$ Sorry, I forgot that I used Total instead of Mean during development of my answer. I'll rewrite it, but in the mean time, you just need to divide by Outer[Times,#,#]&[Join[Range[2048],Reverse@Range[2047]]]. This will be slow because rational numbers can't be packed. If reals are allowed, then use N before dividing. $\endgroup$ – Carl Woll Jun 5 '17 at 20:02
  • $\begingroup$ +1 To that! Awesome. Okay, I need to massage slightly the output of ListCorrelate (in what I am doing, there are certain pairs of {da,db} which are not included). Also, I need to make sure I pair the correct rsq=da^2+db^2 with the relevant {da,db} as I will be gathering all correlations using the rsq as the identifier. However I have an idea how to do this. If all works out. I will be very happy, and slightly ashamed at the same time that Mathematica - with a single function - was able to beat my code(which i have been optimizing for some time) by several orders of magnitude. $\endgroup$ – Luca Pontiggia Jun 5 '17 at 20:31
  • $\begingroup$ This is incredible, I've managed to edit the output in my desired format and it matches with my previous code! I can definitely say my question is answered even though it wasn't in the manner I initially intended. I must however ask one last thing. Would this only work on a square array? I also have a few 1800 x 6840 arrays that I would love to use this on. $\endgroup$ – Luca Pontiggia Jun 6 '17 at 1:01

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