Vectorize a double loop

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AC on 10 May 2021
Commented: Walter Roberson on 13 May 2021
Hi everyone
I'm tryng to vectorize the folowing piece of code:
n=30;
d=3;
a=1;
b=2;
cell=3;
for ki = 2:n-1
for kj = 2:n-1
M((cell-1)*n^2+(ki-1)*n+kj,(cell-1)*n^2+(ki-1)*n+kj-1) = 1/d^2;
M((cell-1)*n^2+(ki-1)*n+kj,(cell-1)*n^2+(ki-1)*n+kj) = -4/d^2-a;
M((cell-1)*n^2+(ki-1)*n+kj,(cell-1)*n^2+(ki-1)*n+kj+1) = 1/d^2;
M((cell-1)*n^2+(ki-1)*n+kj,(cell-1)*n^2+(ki-1)*n+kj-n) = 1/d^2;
M((cell-1)*n^2+(ki-1)*n+kj,(cell-1)*n^2+(ki-1)*n+kj+n) = 1/d^2;
C((cell-1)*n^2+(ki-1)*n+kj,1) = b;
end
end
How can I do it? I will appreciate any help! Thanks!
Walter Roberson on 13 May 2021
If you feel that your post is unclear, then since you are the one who wrote it, you should clarify it.

Bob Thompson on 10 May 2021
Does this work? I haven't been able to test it.
M((cell-1)*n^2+(1:n-2)*n+(2:n-1),(cell-1)*n^2+(1:n-2)*n+(2:n-1)-1) = 1/d^2;
M((cell-1)*n^2+(1:n-2)*n+(2:n-1),(cell-1)*n^2+(1:n-2)*n+(2:n-1)) = -4/d^2-a;
M((cell-1)*n^2+(1:n-2)*n+(2:n-1),(cell-1)*n^2+(1:n-2)*n+(2:n-1)+1) = 1/d^2;
M((cell-1)*n^2+(1:n-2)*n+(2:n-1),(cell-1)*n^2+(1:n-2)*n+(2:n-1)-n) = 1/d^2;
M((cell-1)*n^2+(1:n-2)*n+(2:n-1)j,(cell-1)*n^2+(1:n-2)*n+(2:n-1)+n) = 1/d^2;
C((cell-1)*n^2+(1:n-2)*n+(2:n-1),1) = b;
AC on 10 May 2021
Thank you for your answer! I tried this, but didn't work. When using the double for loop, I get (n-2)*(n-2) combinations of ki's and kj's. But (cell-1)*n^2+(1:n-2)*n+(2:n-1) is a vector of length (n-2). I need a vector of length (n-2)*(n-2)

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