Fast product of large matrices

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P.C.
P.C. 2020-12-12
评论： Matt J 2020-12-14
Hello,
I need to calculate a matrix product like the following a lot of times:
A=rand(2^18,500)
AT=transpose(A)
x=rand(2^18,1)
B=AT*spdiags(x)*A
This should give a 500x500 matrix B (above matrices are just a random example). I need to do this quite often for changing x but constant A and it would take too long if I do it as above, is there any way to do it faster? I think the expression comes from the sum
B(k,l)=sum(A(:,k).*A(:,l).*x)
written in matrix form.
Thanks for your help!
  2 个评论
KALYAN ACHARJYA
KALYAN ACHARJYA 2020-12-13
Apart from your question, is the dimensionality OK for matrix multiplication?
Matt J
Matt J 2020-12-13
编辑：Matt J 2020-12-13
So A is not sparse? What do you plan to do with B downstream in your code?

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回答（1 个）

Matt J
Matt J 2020-12-13
编辑：Matt J 2020-12-13
If see about a 25% speed-up when done as B=AT*(x.*A),
N=2^18;
A=rand(N,500);
AT=A.';
x=rand(N,1);
tic;
B=AT*spdiags(x,0,N,N)*A;
toc%Elapsed time is 1.594161 seconds.
tic;
B=AT*(x.*A);
toc%Elapsed time is 1.195049 seconds.
I don't think you can hope for better than 50% speed-up. In the very best case, when x=ones(N,1), the compute time you are stuck with is,
tic;
B=AT*A;
toc;%Elapsed time is 0.899292 seconds.
  1 个评论
Matt J
Matt J 2020-12-14
If you have the Parallel Computing Toolbox and a decent grpahics card, you could get some speed-up by doing this using gpuArrays.

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