Vectorize a simple code

Hello, Is there more vectorized way to write this code ? ( and avoid the loops)

for j=[1:size(C,1)];
    for k=1:size(C,2)
        C(j,k)=sum(sum(A([1:j],[1:k]).*B([j:-1:1],[k:-1:1])));
    end
end

Thank you in advance

EDIT

I modified the code

j=1:size(C,1);
k=1:size(C,2);
B2=rot90(B,2);
C(j,k)=sum(sum(A(1:j,1:k).*B2(end+1-j:end,end+1-k:end)));

It's normal that I have the same value on all the matrix C ?

P.S: I understood that conv2 is the more performant function but I just want to understand where is the problem (why (j, k) doesn't vary) and what's the solution (without loop)

Answer 1

[Jan]

Note: See https://www.mathworks.com/matlabcentral/answers/35676-why-not-use-square-brackets : Even omitting the unneeded square brackets will accelerate the copde already.

A = rand(56);  % If I understand your inputs correctly:
B = rand(56);
tic;
for k = 1:100
  C = zeros(size(A));
  for j=[1:size(C,1)]
    for k=1:size(C,2)
      C(j,k)=sum(sum(A([1:j],[1:k]).*B([j:-1:1],[k:-1:1])));
    end
  end
end
toc

tic;
for k = 1:100
  C = zeros(size(A));
  for j=1:size(C,1)
    for k=1:size(C,2)
      C(j,k)=sum(sum(A(1:j,1:k).*B(j:-1:1,k:-1:1)));
    end
  end
end
tic

tic;
for k = 1:100
  C = zeros(size(A));
  for k = 1:size(C,2)
    BB = B(end:-1:1, k:-1:1);
    for j = 1:size(C,1)
      C(j, k) = sum(sum(A(1:j, 1:k) .* BB(end-j+1:end, :)));
    end
  end
end
toc

Elapsed time is 5.530599 seconds.
Elapsed time is 4.280142 seconds.
Elapsed time is 3.919369 seconds.

I assume conv2 is a much better approach, although I cannot include the reverted parts of B yet.

But it is worth to know this detail in general: avoid unneeded square brackets. You see a corresponding MLint warning in the editor also: the small orange mark under the [.

[term nv]

Interesting informations ! Thank you

[term nv]

@Jan-Simon could you please see my EDIT ?

Answer 2

[Honglei Chen]

I don't know what dimensions you have in C, but this looks like

D = conv2(A,B);

Maybe you can use

D = D(1:size(C,1),1:size(C,2));

to extract the portions you want.

HTH

[term nv]

Thank you for the answer.

If I perform a convolution for a matrix that size is (56,56) and a second one that has the same size, the size of convolution will be 56+56-1=111. But if I need only a portion (56,56), It wont be faster to calculate by the mathematic formula of convolution the result?

[Stephen23]

"It wont be faster to calculate by the mathematic formula of convolution the result?"

Most likely conv2 would be faster, simpler, less buggy, more efficient,...