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

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Jan
Jan 2021년 5월 28일
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 2021년 5월 28일
편집: Jan 2021년 5월 28일

0 개 추천

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

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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 2021년 6월 1일
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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