Let's say I have a $D$-dimensional sphere with a radius $R$. I want to plot $N$ number of points evenly distributed (equidistant apart from each other) on the surface of the sphere. It doesn't matter where those points are exactly, just that they are ROUGHLY equidistant from each other. How would I do this?
How to plot $N$ points on the surface of a $D$-dimensional sphere roughly equidistant apart?
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euclidean-geometry
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1Not sure whether it meets any of your hidden requirements (like speed), but [this](http://mathworld.wolfram.com/SpherePointPicking.html) maybe be useful. – 2012-10-03
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0For the last method on that page, how would I generate Gaussian random variables so that the radius of the sphere is always the same for each generated point? – 2012-10-03
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2Generate vector $\{Z_1,Z_2,\ldots, Z_{d+1}\}$ then divide each of its components by the Euclidean length of that vector and multiply by $R$. This rescales the vector to have length $R$. – 2012-10-03
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0I guess N coordinates must be generated with any Gaussian random and after that resulting vector have to be normalized and multiplied on the R. – 2012-10-03
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4To have N points drawn uniformly on a region is quite different to have N points approximately equidistant on that region – 2012-10-03
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0RElated: http://math.stackexchange.com/questions/31619/well-separated-points-on-sphere and http://math.stackexchange.com/questions/165819/how-to-tile-a-sphere-with-points-at-an-even-density – 2012-10-03
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0Sasha, when I generate those vectors, they will determine the randomness of the point locations. So, what is a good way to generate them so that for, let's say 2 points, they are on opposite poles? But please provide a method that works with any number of dimensions and any number of points. I'd prefer it to not be a table-lookup. – 2012-10-03
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0Related: * [Which tessellation of the sphere yields a constant density of vertices?](http://math.stackexchange.com/q/9846/856) * [well separated points on sphere](http://math.stackexchange.com/q/31619/856) * [Optimal distribution of points over the surface of a sphere](http://math.stackexchange.com/q/66365/856) * [How to tile a sphere with points at an even density?](http://math.stackexchange.com/q/165819/856) * [Maximum total distance between points on a sphere](http://math.stackexchange.com/q/191273/856) – 2013-02-13
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0I would try to minimize a potential function of the form $\sum_{i
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0I knew someone who needed to do this, and modeled the points as negative charges, so that they repulsed as far apart from each other as possible; essentially equidistant. Unfortunately, I just got the broad overview, and wouldn't know how to implement something like that. – 2015-02-26
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0What exactly does one mean with 'roughly' equidistant? – 2016-12-29
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0approximately / more or less – 2016-12-30
1 Answers
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If $N \le D + 1$ all points must be located on $N-1$ dimension equilateral triangle.
UPDATE
So, I thought today at this problem and invented method, contains following steps:
- Generating $N$ random points on sphere $R$ (that is generating points with coordinates with Gaussian distribution).
- Building Convex Hull (or triangulation on hypersphere even better) with generated on first step points. This step can be solved with MIConvexHull library if you familar with C#, similar library on your favorite language or your own code.
- Using Genetic algorithm, Simulated annealing or another method of global optimization. This method then have to be applied to variance value of all edge lengths from convex hull from step 2.
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0What is an example of a 4D sphere with one equilateral triangle? How can it be generalised to NDimensional Sphere and triangle? – 2012-10-03
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0You mean if N = D, right? The simplex (http://en.wikipedia.org/wiki/Simplex) will solve the problem if N = D, but I need a solution for any N and any D. – 2012-10-03
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0No, I mean $N \le D + 1$, that is all simplexes with such dimensions. For example, for 3D sphere segment there are segment (2 vertices), triangle (3 vertices) and tetrahedron (4 vertices). But you are right: this method not suitable for $N > D + 1$ case. – 2012-10-03