# speeding up nested for loops

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sasha on 8 Aug 2014
Edited: per isakson on 8 Aug 2014
I have a couple of nested for loops that take a very long time to calculate. Basically they look like:
for x= 1:Nx
for y = 1:Ny
f(x,y) = A(x+1,y+1) + B(x-1,y+1) + C(x+1,y-1) + D(x-1,y-1);
end
end
Ideally I would like to be able to compute this in parallel because f(x+1,y) doesn't not depend on f(x,y) and the same for the y values. But it appears that I can't do nested parfor loops. Is there any way I can do this calculation in parallel?
(A, B, C, and D are all defined previously in the code)
per isakson on 8 Aug 2014
Edited: per isakson on 8 Aug 2014
This will cause a error
x = 1;
B(x-1,y+1)
What are the sizes of A, B, C and D?

Nir Rattner on 8 Aug 2014
Edited: Nir Rattner on 8 Aug 2014
I'm assuming A, B, C, and D are functions considering the 0 valued arguments.
Typically, vectorizing your code should be a first step before using “parfor”. It would be best to simply vectorize the functions themselves, but if you can’t then you can use “meshgrid” and “arrayfun”:
[x, y] = meshgrid(1 : Nx, 1 : Ny);
Aout = arrayfun(@A, x + 1, y + 1);
Bout = arrayfun(@B, x - 1, y + 1);
Cout = arrayfun(@C, x + 1, y - 1);
Dout = arrayfun(@D, x - 1, y - 1);
f = Aout + Bout + Cout + Dout;
If you do prefer to use “parfor”, then you can coalesce the indices of the loops:
parfor xy = 1 : (Nx * Ny)
x = xy / (Nx * Ny);
y = xy % (Nx * Ny);
f(x,y) = A(x+1,y+1) + B(x-1,y+1) + C(x+1,y-1) + D(x-1,y-1);
end
sasha on 8 Aug 2014
Each of my functions is an Nx by Ny array already. I want to calculate terms in a new array using specific values from the already defined ones.

A Jenkins on 8 Aug 2014
What about vectorizing it? Just replace x with 1:Nx, and y with 1:Ny to get rid of the for loops completely:
f = A((1:Ny)+1,(1:Ny)+1) + B((1:Ny)-1,(1:Ny)+1) + C((1:Ny)+1,(1:Ny)-1) + D((1:Ny)-1,(1:Ny)-1)
(Also how is your code handling zero indices? For example when x and y are 1, you would be getting B(0,2)...)
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sasha on 8 Aug 2014
Edited: sasha on 8 Aug 2014
Ok, I was running a few test runs to see. It seems to work if I type:
f(2:Nx-1, 2:Ny-1) = A((2:Nx-1)+1, (2:Ny-1)+1) +...
f(1, 1:Ny) = 0;
f(Nx, 1:Ny) = 0;
f(1:Nx, 1) = 0;
f(1:Nx, Ny) = 0;
The only thing I want to make sure of before using this for my code is that the x value in all of the functions will be the same and also the same as the x value in f. And the y's will also be the same. Basically, will using this notation give the correct argument for f, and will the arguments for A, B, C, and D be all the same (up to the +-1)?
If this solves it, thanks!

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