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Topic: Efficiently computing large numbers of vector norms
Replies: 16   Last Post: Nov 15, 2012 8:32 AM

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Matt J

Posts: 4,996
Registered: 11/28/09
Re: Efficiently computing large numbers of vector norms
Posted: Nov 8, 2012 8:05 PM
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"Harry Commin" wrote in message <k7elhj$as1$1@newscl01ah.mathworks.com>...
> I am frequently faced with problems where I need to find the norm (squared) of row vectors having the following structure:
> A(:,i)'*X
> where A is (N x Q) complex, and solutions must be found for all i = 1,2,...,Q. Doing this efficiently turns out to be extraordinarily important in my code. In this simple case, I find the following to be pretty swift (compared to a for loop):
> sum(abs(A'*X).^2,2)
> However, I become stumped with a slightly more complicated case:
> kron(A1(:,i),A2(:,j))'*X2

So first, you could do this simultaneously for all i and j without looping by reorganizing as follows

result = sum( abs( kron(A1',A2')*X2 ).^2 , 2) ;

As Bruno says, though, efficient algorithms are available for matrix multiplications with a Kronecker product and my KronProd class implements them,


In your case, this would lead to

result = sum( abs( KronProd(A2',A1')*X2 ).^2 , 2) ;

Note also that if X also happens to be a Kronecker product,

X2= kron(XX2,XX1);

then the whole thing gets even simpler/faster

rownorms = @(M) sum(abs(M).^2);

result=kron(rownorms(A2'*XX2) , rownorms(A1'*XX1));

> where now a fast solution is needed for all i = 1,2,...,Q1 and j = 1,2,...,Q2. Here, I resort to:
> Z = zeros(Q1,Q2);
> for j = 1:Q2
> Z(:,j) = sum(abs(kron(A1,A2(:,j))'*X2).^2,2);
> end
> Is there a neat way to speed this up? Is there a way to further speed up my sum(abs(A'*X).^2,2)? Efficiency really is critical here!

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