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

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Henry 2013-7-29

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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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Answer 1

Matt J 2013-7-29

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This might speed things up,

http://www.mathworks.com/matlabcentral/fileexchange/29035-dnorm2

Haven't used it myself, though.

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Matt J 2013-7-30

编辑：Matt J 2013-7-30

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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...

Jan 2013-7-30

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 2

Richard Brown 2013-7-29

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编辑：Richard Brown 2013-7-29

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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?