# How to choose $x$ evenly distributed points from within an n-ball

I would like to know how to choose $x$ evenly distributed points from within an n-ball. I think a formal way of defining this is that we want to choose $x$ points from within the n-ball such that we maximize the closest distance between any two points. As a result, it seems, all points should be evenly spaced and many should be located at the surface. What is an algorithm to generate such a set?

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Here is an algorithm for studying the 2D sphere to get a feel for potential theory techniques, and here is a stackoverflow thread discussing the general problem. –  Eugene Shvarts Jan 26 '13 at 1:42
This question is almost an exact duplicate, except it's for points on the sphere instead of the ball. The answers there may still be helpful. –  Rahul Jan 26 '13 at 1:46
Ok, so far what I'm getting from those questions is that only approximate solutions exist for this type of problem. –  Matt Munson Jan 26 '13 at 2:03

Suppose the closest distance between two points is $d$. This implies that $x$ spheres of radius $d/2$ centered at your points can fit inside a sphere of radius $1+d/2$ without overlapping. Equivalently, $x$ spheres of radius $d/(2+d)$ fit in a unit sphere, and maximizing $d$ is equivalent to maximizing $d/(2+d)$.
So your problem amounts to finding the densest packing of $x$ spheres in a sphere. There are some (approximate) precomputed solutions for $x\le51$ in three dimensions, but there is probably no closed-form solution in general. I guess this is not an answer to your question of what is an algorithm to generate such sets, but searching for "sphere packing algorithms" may find you some useful references.
Cool. But what is the values of $x$ for a given value of $d$? And where did you get $1+d/2$? –  Matt Munson Jan 26 '13 at 3:18
@Matt: Finding the largest $x$ for a given $d$ is just as hard as finding the largest $d$ for a given $x$; as I said there is no known explicit formula. You get $1+d/2$ as follows: If a point lies within distance $1$ of the origin, then a sphere of radius $d/2$ around it may not be contained in the unit sphere around the origin, but it will be contained in a sphere of radius $1+d/2$ around the origin. –  Rahul Jan 26 '13 at 4:45
@Matt: Right, I somehow assumed you meant the unit $n$-ball in your question. It's just a matter of scaling in the end. –  Rahul Jan 26 '13 at 9:19