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Henry
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How do I efficiently calculate a scalar distance with 3d geometry?

Asked by Henry
on 29 Jul 2013
I have a piece of code that is called many times (~5e5) during a time stepping solution. Having run profiler, the following line is slowing everything down:
r2 = sqrt((x-x2).*(x-x2)+(y-y2).*(y-y2)+(z-z2).*(z-z2));
Where x, y, z are (500, 1) and x2, y2, z2 are scalars.
Any suggestions?
Thanks!

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2 Answers

Answer by Matt J
on 29 Jul 2013

This might speed things up,
Haven't used it myself, though.

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It should also help (a little) if you don't compute x-x2 etc... twice. So in other words, you would first do
dx=x-x2;
dy=y-y2;
dz=z-z2;
and then
r2 = sqrt(dx.*dx +dy.*dy +dz.*dz);
or whatever...
DNorm2 requires matrices as input and creating [dx, dy, dz] at first wastes too much time. So if the data could be organized as [500 x 3] matrix instead of three vectors, DNorm2 would be an option.

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Answer by Richard Brown on 29 Jul 2013
Edited by Richard Brown on 29 Jul 2013

You should use hypot. It also has better numerical stability.
edit sorry, you're in 3D. That obviously won't work. You can use hypot as
r = hypot((x-x2) + 1i*(y-y2), z - z2);
This may be slower than what you currently have though. If you use (x - x2).^2 instead of (x - x2) .* (x - x2) you'll probably get a small performance boost. Can you avoid calculating the square root and work with squared distance instead?

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