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A fast way to perform sparse matrix-free product with a vector

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Dear all,
I was wondering if it is possible to increase the performance of a matrix-free product of a sparse matrix defined by 3 vectors (rows, columns and values) with another vector {B}. That is:
A(l,m)=v and performing {C}=[A]*{B}
So far I have the following strategy:
function C = mfree_times(l,m,v,B)
aux = B(m);
prod = v.*aux;
C = accumarray(l,prod);
end
My intentions are to increase this performance as much as possible. The bottleneck is majorly given by accumarray function. I am already using l, m and v as gpuArrays. Please share your thoughts. Thanks!

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Paulo Ribeiro
Paulo Ribeiro on 5 Sep 2018
Thanks Bruno. I tried with:
C = accumarray(l,prod,[],[],[],true);
and with the "false" flag also. Since {C} is a full vector the performance using "true" was worse. And it makes no difference with the "false" flag or no flag at all.
The accumarray performs quite fast on the GPU but over so many iterations it becomes a bottleneck.
James Tursa
James Tursa on 6 Sep 2018
@Paulo: So, based on your replies, your variables appear to have nothing to do with CSR format sparse matrices ... you simply have 3-tuples of row,column,value to use. My answer below is based on that assumption.

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Accepted Answer

James Tursa
James Tursa on 6 Sep 2018
Edited: James Tursa on 6 Sep 2018
Assuming the vectors are 3-tuples of row, column, and value, and that all of the variables involved are full double vectors, here is a naive mex routine to do the multiplication and produce a full vector result. If you are working with R2018a and compiling with the -R2018a option then replace the mxGetPr calls with mxGetDoubles calls.
/* C = mfree_timesx(r,c,v,B) does the equivalent of:
*
* C = sparse(r,c,v,numel(B),numel(B)) * B
*
* where
*
* r,c,v are 3-tuples of values from an MxM matrix:
* r = row index (1-based)
* c = column index (1-based)
* v = values
* B = column vector of size Mx1
* C = column vector of size Mx1
*
* It is assumed that the indexing contained in the r and c vectors are
* all within an MxM square matrix (i.e., are integers between 1 and M).
* (The code does *not* check for this. A violation will bomb this routine)
*
* Duplicate indexing within the r and c vectors is allowed, and is handled
* by simply adding the results together (same as what would happen if you
* did C = sparse(r,c,v,numel(B),numel(B)) * B).
*
* All inputs must be full real double vectors.
*
* The output is a full real double column vector.
*
* Programmer: James Tursa
*
*/
/* Includes ----------------------------------------------------------- */
#include "mex.h"
/* Gateway ------------------------------------------------------------ */
void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[])
{
mwSize i, n, m;
double *Ar, *Ac, *Av, *B, *C;
if( nrhs != 4 ) {
mexErrMsgTxt("Need exactly four double input vectors");
}
if( nlhs > 1 ) {
mexErrMsgTxt("Too many outputs");
}
if( !(mxIsDouble(prhs[0]) && mxIsDouble(prhs[1]) && mxIsDouble(prhs[2]) && mxIsDouble(prhs[3])) ||
mxIsComplex(prhs[0]) || mxIsComplex(prhs[1]) || mxIsComplex(prhs[2]) || mxIsComplex(prhs[3]) ||
mxIsSparse(prhs[0]) || mxIsSparse(prhs[1]) || mxIsSparse(prhs[2]) || mxIsSparse(prhs[3]) ) {
mexErrMsgTxt("All inputs must be full real double vectors");
}
n = mxGetNumberOfElements(prhs[0]);
if( mxGetNumberOfElements(prhs[1]) != n || mxGetNumberOfElements(prhs[2]) != n ) {
mexErrMsgTxt("The first three inputs must be the same size");
}
m = mxGetNumberOfElements(prhs[3]);
if( mxGetM(prhs[3]) != m ) {
mexErrMsgTxt("The last input must be a column vector");
}
Ar = mxGetPr(prhs[0]);
Ac = mxGetPr(prhs[1]);
Av = mxGetPr(prhs[2]);
B = mxGetPr(prhs[3]);
plhs[0] = mxCreateDoubleMatrix(m,1,mxREAL);
C = mxGetPr(plhs[0]);
for( i=0; i<n; i++ ) {
C[(mwSize)*Ar++ - 1] += *Av++ * B[(mwSize)*Ac++ - 1];
}
}
Some sample test code:
N = 5000;
r = ceil(N*rand(N*N,1));
c = ceil(N*rand(N*N,1));
v = rand(N*N,1);
B = rand(N,1);
disp(' ')
disp('mex routine')
tic
x = mfree_timesx(r,c,v,B);
toc
disp('sparse')
tic
s = sparse(r,c,v,numel(B),numel(B))*B;
toc
disp('accumarray')
tic
aux = B(c);
prod = v.*aux;
C = accumarray(r,prod);
toc
disp(' ')
disp('mex vs sparse')
max(abs(x-s))
disp('mex vs accumarray')
max(abs(x-C))
disp('sparse vs accumarray')
max(abs(s-C))
disp(' ')
and results:
mex routine
Elapsed time is 0.150372 seconds.
sparse
Elapsed time is 5.806035 seconds.
accumarray
Elapsed time is 0.666030 seconds.
mex vs sparse
ans =
1.2960e-011
mex vs accumarray
ans =
0
sparse vs accumarray
ans =
1.2960e-011

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James Tursa
James Tursa on 6 Sep 2018
Yes, that small loop does all of the real work. Everything else is just overhead. If you want I can add in code to check the individual r and c indexes to make sure they are legit.
Also, if these variables are really mxGPUArray variables, then you will need to replace all of the mxETC( ) functions with their mxGPU( ) counterparts. E.g., use (double *)mxGPUGetData(etc) instead of mxGetPr(etc).
Paulo Ribeiro
Paulo Ribeiro on 7 Sep 2018
Dear James,
A GPU version for the 3 tuples and B will be more than welcome. Do you think it is manageable? Many thanks for your attention and work on this thread. Glad to find an expert :)
Best regards,
Paulo
Paulo Ribeiro
Paulo Ribeiro on 8 Sep 2018
Here are some values after @James contributions:
Problem: finite element, 2D elasticity, 116.000 nodes. Setup: i7 7700HQ, NVIDIA GTX 1050, Linux (gcc compiler); Solver: matrix-free conjugate gradient, as described on this thread
Solver time for CPU with MEX compilation (mfree_timesx) = 42s;
Solver time for CPU with mfree_times (no MEX) = 317s;
Solver time for GPU with mfree_times (no MEX) = 82s.
Really impressive results with MEX on CPU

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More Answers (1)

Bruno Luong
Bruno Luong on 5 Sep 2018
Can you give the detail about how you iterate ?
There seems no room to improve the basic block mfree_times
If you output is sparse and you add them, and internally it requires new memory allocation and merge sorted array.
Better building all collection triplets then call ACCUMARRAY once.

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Bruno Luong
Bruno Luong on 6 Sep 2018
It seems index complement of "l" has no true impact on CG algorithm. If it is a small vector with large value, you might want to squeeze as consecutive vector (using UNIQUE command) and work on the subspace of the new value.
That prevent CG work unnecessary large vector/matrix.
Then after converging, you can recover original index of the solution if that is matter to use the result.
Paulo Ribeiro
Paulo Ribeiro on 6 Sep 2018
Bruno,
After several attempts, I am really out of options. Vectors l,m and v are very large, up to 1e9. I imagine two possible scenarios:
1. Fast convergence and reduced number of iterations; 2. Translate this code to Fortran or C/C++.
Thanks!
Bruno Luong
Bruno Luong on 6 Sep 2018
It seems that your problem is memory and not inneficient calculation.
If the matrix is sparse (a lot of zero) you should consider my suggestion if squeezing subspace.
By do try to reduce memory by using storage the indexes with INT32, matrix values with SINGLE, or using TALL ARRAY for more efficient HD cache.

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