codes instead of 'for' loops to speed up

Hi to all,

I have 3 'for' loop which make the excutation to be time-consuming. Is there any way to speed up? The code is:

 ...
 for i= 0:P
     for j=0:P
         ss=0;
         for k=P:N-1
	         ss=ss+x(k-j+1,1)*x(k-i+1,1);
         end
         R(i+1,j+1)=ss;
     end
 end
 ...

Thanks,

What are P, N and X? Specifically what is the size of X?

Hi Matt! I'm happy to see you back again.

Hello Jan!

Answer 1

as Matt has mentioned already, the question does not contain enough information to be answered exhaustively. The simple approach to vectorize the innermost loop is most likely not efficient:

for i = 0:P
  for j = 0:P
    R(i+1, j+1) = sum(x(P-j+1:N-j, 1) .* x(P-i+1:N-i, 1));
  end
end

Now the creation of the intermediate vectors x(P-j+1:N-j, 1) takes too much time - but this is a guess only, which I cannot prove due to the missing details in the question.

But you could try to store one of the vectors in a temporary variable:

for i = 0:P
  t1 = x(P-i+1:N-i, 1);
  for j = 0:P
    R(i+1, j+1) = sum(x(P-j+1:N-j, 1) .* t1);
  end
end

Then you should exploit the symmetric structure of the result:

for i = 0:P
  t1 = x(P-i+1:N-i, 1);
  R(i+1, i+1) = sum(t1 .* t1);  % Diagonal element

   for j = i+1:P
      t2          = sum(x(P-j+1:N-j, 1) .* t1);
      R(i+1, j+1) = t2;
      R(j+1, i+1) = t2;
    end
  end

Again: I cannot test this code. Please post requests for optimizing code always with a set of usual test data. Most likely the SUM(vector1 .* vector2) approach is slower than a lean and clean loop. Therefore I assume that exploiting the symmetry is a more useful strategy.

Omitting the 2nd index in "x(k, 1)" saves time, if x is a vector. But perhaps x is a function and the 2nd input is important.

Thanks for your pefect answer.