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 Kommentar

Jan
Jan am 28 Mai 2021
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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 Akzeptierte Antwort

Jan
Jan am 28 Mai 2021
Bearbeitet: Jan am 28 Mai 2021

0 Stimmen

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 Kommentare

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 am 1 Jun. 2021
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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Jan
am 1 Jun. 2021

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