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How to perform a sum over matrices in several dimensions?

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Henrik
Henrik el 13 de En. de 2014
Comentada: Henrik el 13 de En. de 2014
Hello I have a rather large sum that I want to evaluate in MATLAB. Symbolically it looks like this:
huge_sum(qx,qy)=sum_n(sum_m(sum_kx(sum_ky(sum_i(sum_j( M1(n,m,kx,ky)*M2(n,kx,ky,i,j)*M3(m,kx+qx,ky+qy,i,j)*M4(qx,qy,i,j)
) ) ) ) ) ),
where each of the M's is a matrix. The indices run from 1 to
qx,qy,kx,ky: 25
i,j,m,n: 16
My current implementation is rather slow; it's simply a bunch of nested for loops, defining a matrix,M5(m,n,kx,ky,i,j), which I then sum for each value of qx.
The problem is that this code will take around 100 days to run, which is of course too much! Can anyone suggest a smarter way to do this?
  5 comentarios
Walter Roberson
Walter Roberson el 13 de En. de 2014
Do some factoring. Your M1 and M2 indices are independent of qx and qy, so you can calculate the M1 * M2 part independently.
Henrik
Henrik el 13 de En. de 2014
Calculating the matrices M1-M4 is rather quick and is done beforehand. I know it's the calculation of M5 which is slow; at the moment I calculate its values one entry at a time. I would like to vectorize it somehow, but I can't see how to do that when the matrices have different sizes..
@Walter thanks, but I actually realized I made a mistake in what I wrote here - M1 depends on qx and qy as well.

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Respuesta aceptada

Matt J
Matt J el 13 de En. de 2014
Editada: Matt J el 13 de En. de 2014
Warning. Not tested.
op=@(A,B,dim) squeeze(sum(bsxfun(@times,A,B),dim))
M2=reshape(M2,16,1,25,25,256); %M2(n,1,kx,ky,z)
T=op(M1,M2,1); %T(m,kx,ky,z)
M3=M3(:,:,:,:); %M3(m,kx+qx,ky+qy,z)
M4=M4(:,:,:); %M4(qx,qy,z)
sx=size(M3,2);
sy=size(M3,3);
Tr=reshape(T,16,25,25,1,1,256);
M3r=reshape(M3,16,1,1,sx,sy,256);
U=op(Tr,M3r,1); %U(kx,ky,kx+qx,ky+qy,z)
U=permute(U,[1,3,2,4,5]); %U(kx,kx+qx,ky,ky+qy,z)
U=reshape(U,25*sx,25*sy, 256);
K=1:25;
for qx=K
idx=sub2ind([25,sx],K,K+qx);
for qy=K
idy=sub2ind([25,sy],K,K+qy);
u=reshape(U(idx,idy,:)[],256);
v=reshape(M3(qx,qy,:),[],256);
M5(qx,qy)=sum(u,1)*v.';
end
end
  1 comentario
Henrik
Henrik el 13 de En. de 2014
I can't say I understand exactly how this code works, but it seems to be doing the right thing. I will try and think of a way to test it. Thanks for the effort!

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