How to do a fast multidimensional matrix vector multiplication?

Question

A is a 3D N*N*L matrix, x is a N*1 vector, on which I need to do the following operation:

for i=1:L
    res(i)=x'*squeeze(A(:,:,i))*x
end

I hope to use most efficient vectorized method instead of a for loop. Please anyone give me some suggestions?

I suggest being more specific: you don't need matrix-vector multiplication, but computing a quadratic form. But it's possible that it only makes a difference for me:) — Andras Deak -- Слава Україні, Nov 01 '15 at 20:37
I can't try it right now, but you might be after something like `sum(sum(bsxfun(@times,permute(bsxfun(@times,permute(A,[3 1 2]),reshape(x,[1 1 N])),[1 3 2]),reshape(x,[1 1 N])),2),2)` — Andras Deak -- Слава Україні, Nov 01 '15 at 20:44
@J.Andrew I have rolled your question back to its previous state. This is because you updated your question making both answers, including the one you accepted as "helpful" out of date. If you have a new question, please open a new question rather than expanding on this one. — Adriaan, Nov 04 '15 at 20:56

score 5 · Accepted Answer · edited May 23 '17 at 12:29

5

With bsxfun -

sum(reshape(bsxfun(@times,x,bsxfun(@times,A,x.')),[],L),1)

With matrix-multiplication-fun -

reshape(x*x.',1,[])*reshape(A,[],L)

edited May 23 '17 at 12:29

Community

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answered Nov 01 '15 at 20:46

Divakar

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Regarding the second: "are you a wizard?";) – Andras Deak -- Слава Україні Nov 01 '15 at 21:00
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You seem like a `fun` guy to be around. – Adriaan Nov 01 '15 at 21:03
@Adriaan I blame the *fun-tools* hanging around :) – Divakar Nov 01 '15 at 21:05
I like the 2nd one,it seems very efficient. – J. Andrew Nov 02 '15 at 15:13
I don't know what matlab is doing inside of reshape, I am wondering why it is more efficient for the 2nd compared to the first with bsxfun? thanks! – J. Andrew Nov 02 '15 at 15:15

score 2 · Answer 2 · answered Nov 01 '15 at 20:47

2

N=10;L=5;
A = rand(N,N,L);x=rand(N,1);
C = sum(sum(bsxfun(@times,permute(bsxfun(@times,permute(A,[3 1 2]),reshape(x,[1 1 N])),[1 3 2]),reshape(x,[1 1 N])),2),2);
C = squeeze(C(:,1,:));

Thanks to @AndrasDeak, though you did miss the last squeeze call.

answered Nov 01 '15 at 20:47

Adriaan

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I'm quite surprised it works:) Sadly (and as expected), @Divakar's answer is much more elegant. – Andras Deak -- Слава Україні Nov 01 '15 at 20:48
That is exactly what we expected of course. I tried using `permute(x,[4 3 1 2])`, which started of correctly, but ended in a 4D matrix I wasn't sure on how to smack back into shape – Adriaan Nov 01 '15 at 20:49

How to do a fast multidimensional matrix vector multiplication?

2 Answers2