Hi,
I have number of functions that I use to find roots to Neumann problem. One of the functions contains nested for loops which runs forever.
Please see below:
The function consists of three nested for loops as:
function [Lsn] = coefFourier_ll_log(k,ll)
N=length(ll);
ll_n=coefFourier1(k*ll);
lambda=zeros(1,N-1); lambda(N/2+1:N-1)=-0.5./(1:N/2-1);
lambda(1:N/2-1)=fliplr(lambda(N/2+1:N-1));
M=N/2-1; Lsn=zeros(N-1,N-1);
for s=-M:M
for n=-M:M
rmin=max(-M,max(-M-n,s-M));
rmax=min(s+M,min(M-n,M));
for r=rmin:rmax
Lsn(s+M+1,n+M+1)=Lsn(s+M+1,n+M+1)+ll_n(s-r+M+1)*ll_n(n+r+M+1)*lambda(r+M+1);
end
end
end
Lsn=0.5*Lsn;
This function receives k(double), ll(1D array) and returns Lsn(Matrix). For the given time profile, the system size is 512, which means M is 255.
How I can vectorize it to increase its speed? I need to run this group of functions for couple of hundreds times and eventually with even bigger system size.
Thank you for your time.

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Jan
Jan 2018년 3월 21일

0 개 추천

Start with a slightly cleaned version:
function Lsn = coefFourier_ll_log(k,ll)
N = length(ll);
ll_n = coefFourier1(k*ll);
lambda = zeros(1, N-1);
lambda(N/2+1:N-1) = -0.5 ./ (1:N/2-1);
lambda(N/2-1:-1:1) = lambda(N/2+1:N-1); % Instead of FLIPLR
M = N/2-1;
M1 = M + 1;
Lsn = zeros(N-1, N-1);
for s = -M:M
for n = -M:M
rmin = max(-M, max(-M-n, s-M));
rmax = min(s+M, min(M-n, M));
Lsn(s+M1, n+M1) = Lsn(s+M1, n+M1) + ...
sum(ll_n(s+M1-rmin:s+M1-rmax) .* ...
ll_n(n+M1+rmin:n+M1+rmax) .* ...
lambda(M1+rmin:M1+rmax));
end
end
Lsn = 0.5 * Lsn;
Here the inner loop from rmin:rmax way vectorized. Please post the timing before and after the changes measure by tic/toc. The profiler disables the JIT acceleration, such that it is less useful. Providing some test input of a relevant size (e.g. created by rand) would allow us to run the tests by our own.

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Thank you for your effort Jan. I have deleted the function unrelated to the for-loop issue, created a random vector of complex numbers for testing purposes.
Unfortunately, I am getting matrix dimensions must agree error.
Here is the code that I run:
N=256;
ll_n=rand(1,N) + 1i*rand(1,N);
tic
lambda=zeros(1,N-1); lambda(N/2+1:N-1)=-0.5./(1:N/2-1);
lambda(1:N/2-1)=fliplr(lambda(N/2+1:N-1));
M=N/2-1; Lsn=zeros(N-1,N-1);
for s=-M:M
for n=-M:M
rmin=max(-M,max(-M-n,s-M));
rmax=min(s+M,min(M-n,M));
for r=rmin:rmax
Lsn(s+M+1,n+M+1)=Lsn(s+M+1,n+M+1)+ll_n(s-r+M+1)*ll_n(n+r+M+1)*lambda(r+M+1);
end
end
end
toc
tic
lambda = zeros(1, N-1);
lambda(N/2+1:N-1) = -0.5 ./ (1:N/2-1);
lambda(N/2-1:-1:1) = lambda(N/2+1:N-1); % Instead of FLIPLR
M = N/2-1;
M1 = M + 1;
Lsn = zeros(N-1, N-1);
for s = -M:M
for n = -M:M
rmin = max(-M, max(-M-n, s-M));
rmax = min(s+M, min(M-n, M));
Lsn(s+M1, n+M1) = Lsn(s+M1, n+M1) + ...
sum(ll_n(s+M1-rmin:s+M1-rmax) .* ...
ll_n(n+M1+rmin:n+M1+rmax) .* ...
lambda(M1+rmin:M1+rmax));
end
end
toc
I figured what is causing the error. When rmin=-1 and rmax=0 this term
ll_n(s+M1-rmin:s+M1-rmax)
returns empty vector. I am trying to figure how i can solve it without using an if statement.
@Turker Topal: I made a typo. This works and replies the same value:
tic
lambda = zeros(1, N-1);
lambda(N/2+1:N-1) = -0.5 ./ (1:N/2-1);
lambda(N/2-1:-1:1) = lambda(N/2+1:N-1); % Instead of FLIPLR
M = N/2-1;
M1 = M + 1;
Lsn2 = zeros(N-1, N-1);
for n = -M:M
c = min(M-n, M);
d = max(-M, -M-n);
for s = -M:M
rmin = max(s-M, d);
rmax = min(s+M, c);
Lsn2(s+M1, n+M1) = Lsn2(s+M1, n+M1) + ...
sum(ll_n(s-rmin+M1:-1:s-rmax+M1) .* ...
ll_n(n+rmin+M1:n+rmax+M1) .* ...
lambda(rmin+M1:rmax+M1));
end
end
toc
For your test input, in runs in 0.92 sec compared to 3.39 sec of the original version.
Thank you Jan. Do you think it is possible to use Ngrid for the first two loops? Actually, the first two matrices have sort of repeating form. For example;
M = 1
rmin =
0 -1 -1
0 -1 -1
0 0 0
rmax =
0 0 0
1 1 0
1 1 0
M=2
rmin =
0 -1 -2 -2 -2
0 -1 -2 -2 -2
0 -1 -2 -2 -2
0 -1 -1 -1 -1
0 0 0 0 0
rmax =
0 0 0 0 0
1 1 1 1 0
2 2 2 1 0
2 2 2 1 0
2 2 2 1 0
Jan
Jan 2018년 3월 22일
While rmin and rmax have a specific structure, which could be exploited, the actual indices are e.g. s-rmin+M1 and change with each iteration. Therefore I do not see a way to use the pattern in the outer loops.
It would be useful if you explain, what the operation does. Maybe it is much faster to calculate it by conv or conv2.

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