Speed up nested for loop

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Francesco Lisi
Francesco Lisi 2021-5-28
评论： Jan 2021-6-1
Hi,
I have to speed up the following code.
N is an integer of the order 10^4, l is of the order of (N)^(1/4), c and v are column complex vectors with N elements.
B = 0;
for m = 1:l
for n = m+1 : N
B = B + 2*real(c(n)*conj(v(n))* conj(c(n-m))*v(n-m));
end
end
Thank you for your help.
Francesco
  1 个评论
Jan
Jan 2021-5-28
Please provide some inputs values, maybe produced by rand(). It is hard to improve the code without running it. And inveting default inputs might be misleading, if we oversee an important detail.

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Jan
Jan 2021-5-28
编辑：Jan 2021-5-28
Maybe:
% [UNTESTED CODE] Please provide inputs for testing
B = 0;
for m = 1:l
B = B + sum(2 * real(c(m+1 : N) .* conj(v(m+1 : N)) .* ...
conj(c(1 : N-m)) .* v(1 : N-m)));
end
Does this work? Then:
B = 0;
c_cv = c .* conj(v);
cc_v = conj(c_cv);
for m = 1:l
B = B + 2 * sum(real(c_cv(m+1 : N) .* cc_v(1 : N-m)));
end
  2 个评论
Francesco Lisi
Francesco Lisi 2021-6-1
Thank you Jan for your fast reply!
The code you provided is faster indeed. Here is the updated script
y=c.*conj(v);
yc=conj(y);
B_est = sum(abs(y).^2);
for m = 1:l
B_est = B_est + 2*sum(real(y(m+1:N) .* yc(1:N-m)));
end
Is there a way to speed up this one?
As input you can use complex arrays generated as randn(N,1)+1i*randn(N,1). I use this function in a Monte Carlo simulation with 10^3 iteration and this function is the slowest one.
Thank you again for your help.
Francesco
Jan
Jan 2021-6-1
A further improvement:
c_cv = c .* conj(v);
cc_v = conj(c_cv);
B = 0; % sum(abs(y).^2) did not appear in the original question
for m = 1:l
B = B + real(c_cv(m+1 : N).' * cc_v(1 : N-m));
end
B = B * 2;
Letting the sum() be done by the dot product saves some time, because the optimized BLAS library is used.

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