bsxfun implementation in solving a min. optimization task

问题

I really need help with this one.

I have to matrices L1 and L2, both are (500x3) of size.

First of all, I compute the difference of every element of each column of L1 from L2 as follows:

lib1 = bsxfun(@minus, L1(:,1)',L2(:,1));
lib1=lib1(:);
lib2 = bsxfun(@minus, L1(:,2)',L2(:,2));
lib2=lib2(:);
lib3 = bsxfun(@minus, L1(:,3)',L2(:,3));
lib3=lib3(:);
LBR = [lib1 lib2 lib3];

The result is this matrix LBR. Then I have a min-problem to solve:

[d,p] = min((LBR(:,1) - var1).^2 + (LBR(:,2) - var2).^2 + (LBR(:,3) - var3).^2);

Which returns the point p where this min-problem is fulfied. Finally I can go back to my matrices L1 and L2 to find the index-positions of the values which satisfy this min-problem. I done this as follows:

[minindex_alongL2, minindex_alongL1] = ind2sub(size(L1),p);

This is OK. But what I need now is:

I have to multiply , take the tensor-product, also called Kronecker product of a vector called alpha to LBR, alpha is given as follows:

alpha = 0:0.1:2;

And, this Kronecker product I have computed as follows:

val = bsxfun(@times,LBR,permute(alpha,[3 1 2]));
LBR = reshape(permute(val,[1 3 2]),size(val,1)*size(val,3),[]);

what I need now is: I need to solve the same minproblem:

[d,p] = min((LBR(:,1) - var1).^2 + (LBR(:,2) - var2).^2 + (LBR(:,3) - var3).^2);

but, this time, in addition of finding the index-positions and values from L1 and L2 which satisfies this min-problem, I need to find the index position of the single value from the alpha vector which has been multiplied and which fulfills the min-problem. I don't have idea how can I do this so any help will be very appreciated!

Thanks in advance!

Ps: I can post the L1 and L2 matrices if needed.

回答1:

I believe you need this correction in your code -

[minindex_alongL2, minindex_alongL1] = ind2sub([size(L2,1) size(L1,1)],p)

For the solution, you need to add the size of p into the index finding in the last step as the vector whose min is calculated has the "added influence" of alpha -

[minindex_alongL2, minindex_alongL1,minindex_alongalpha] = ind2sub([size(L2,1) size(L1,1) numel(alpha)],p)

minindex_alongalpha might be of your interest.

来源：https://stackoverflow.com/questions/24181012/bsxfun-implementation-in-solving-a-min-optimization-task

标签

matlab

min

matrix-multiplication

bsxfun