How to avoid nested for loop to make the code faster?

Hi everyone!
Do you have some suggestions to improve the following code?
As it is now, it is very slow…
B_2 = zeros(d,d);
for m = 1:M
for n = m+1 : 2*N
B_2 = B_2 + 4*N^(-1)* A(n) * B(n-m);
end
end
where A(n) and B(n) are 2x2 real, symmetric and full rank, matrices for all possible value of n.
Thanks in advance!

6 comentarios

madhan ravi
madhan ravi el 14 de Mzo. de 2019
Editada: madhan ravi el 14 de Mzo. de 2019
Without knowing the values of variables... M , N?
Stef
Stef el 14 de Mzo. de 2019
Thanks for the comment!
I've just simplified the code, since the problem seems to be in the nested loop not in the operations inside the loop.
Bests!
Adam
Adam el 14 de Mzo. de 2019
Editada: Adam el 14 de Mzo. de 2019
I doubt this will gain much speed, but since N appears to be constant you should pull
4*N^(-1)
out of the loops and calculate it once before the loops because it shouldn't change.
Also if A and B are 2x2 matrices how does B(n-m) work? There are only 4 valid indices for a 2x2 matrix so your loop cannot be very long if it only produces values of n and n-m between 1 and 4?
Stephen23
Stephen23 el 14 de Mzo. de 2019
@Stef: please tell us the values of M, N, and d.
Stef
Stef el 14 de Mzo. de 2019
@Adam The matrices A(n) and B(n) are 2x2 matrices for every possible value of n (e.g. A(1) is a 2x2 matrix). In particular, for each n, I evaluate A(n) and B(n) as outer multiplication of two pre-stored, 2x1 column vectors, i.e. A(n) = u(n)*v(n).'
Stef
Stef el 14 de Mzo. de 2019
d = 2,
N = 10^5 (could be also 10^6),
M is of the order of sqrt(M), but it can change...

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 Respuesta aceptada

Doug Mercer
Doug Mercer el 14 de Mzo. de 2019
If you can generate A and B in a vectorized manner (i.e., for inputs x=[x_1, x_2, ..., x_k] the function A(x) returns a 2x2xk 3D array where a(:, :, i) is the 2x2 array corresponding to x_i) then the following would vectorize the inner loop.
function B_2 = stef()
d = 2;
N = 10^5;
M = sqrt(N);
B_2 = zeros(d,d);
for m = 1:M
n_iter = m+1:2*N;
B_2 = B_2 + 4*N^(-1)*sum(A(n_iter).*B(n_iter - m), 3);
end
function b = B(x)
b = rand(2, 2, length(x));
function a = A(x)
a = rand(2, 2, length(x));

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