n isomorphic points on a sphere

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Strilanc
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n isomorphic points on a sphere

Postby Strilanc » Thu Apr 17, 2008 6:37 am UTC

I read a post about minimum potential of n electrons on a sphere, and started thinking about this.

You have n points you want to place on a sphere. However, you may only place them such that all the points are isomorphic to each other. In other words, for any points p and q, there exists an automorphism on the set of points which labels q as p. So for example you could place the points at the vertices of a 2-dimensional polygon (for any number of points), or use the corners of a cube (for 8 points).

Also, you want to maximize the average distance between the points. So just placing them at the vertices of a polygon won't always be optimal. What strategy do you use? How high can you get that average? What values of n work really well?

Solutions can be posted here, since the puzzle has no well-defined goal.
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Re: n isomorphic points on a sphere

Postby Nimz » Thu Apr 17, 2008 11:26 am UTC

The maximum average distance between points would probably be the vertices of the platonic solids when n=4, 6, 8, 12, and 20. When n=2, opposite poles would do the trick, and when n=3, an equilateral triangle on the equator should work. Of course, n=1 is trivial.

Other than those values of n, I would probably start with something like the way atoms in a molecule line up - e.g. a square on the equator and a point on a pole for n=5 - and make small perturbations from there. Or, probably equivalently, remove one vertex at a time from a platonic solid and use that as the starting point for perturbation... at least for n<20.

[edit] seems my intuition was flawed in some spots. I just read the other thread and there are some obvious counterexamples, like the n=8 one, where one square is rotated 45 degrees. This still preserves the automorphism requirement, since a 90 degree rotation of the whole thing in the same direction would be an automorphism. Also, square pyramidal isn't as good as trigonal bipyramidal for n=5, as noted on that thread. Still, I think the approach of perturbations holds some promise - something along the lines of the calculus of variations. Done right, variational calculus can give all the extrema for a given variable. You just have to look at which are maxima, which are minima, and which are local and which are global.
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Re: n isomorphic points on a sphere

Postby skeptical scientist » Fri Apr 18, 2008 4:40 am UTC

Assume that the origin is in the convex hull of your n points, and they aren't all in a plane. In that case, your convex hull will make an isogonal polyhedron, and the dual will be an isohedral polyhedron. There is a finite list of these, so you can check which has the greatest average distance.

If the origin is not in the interior convex hull, then if x is in your set, throw in -x. Now the origin is in your convex hull, and it has the same symmetry, so your new shape is on the list, and you can get a finite list of all the possibilities for the original configuration by going back to the old shape.

If all of your points are in a plane, then they are either at the vertices of a regular polygon, or an irregular isogonal polygon.

For small n, here's what I think you get:

n shape
3 vertices of a triangle about the equator
4 vertices of a tetrahedron
5 pentagon about the equator
6 octahedron
7 heptagon about the equator
8 square antiprism
9 nonagon about the equator
10 pentagonal antiprism
11 undecagon about the equator
12 icosahedron
etc.

It's easy to continue this list, because if n is even, it's the dual of the "best" (closest to sphere-shaped, aka the one that comes in dice sets) fair n-sided die, and if n is odd, it's a n-gon about the equator, since the symmetry requirement forces you to get the vertices of an n-gon.
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Re: n isomorphic points on a sphere

Postby Nimz » Fri Apr 18, 2008 5:59 am UTC

Ah. I misinterpreted one of the conditions. Not only should there be a nontrivial automorphism of the points (my original interpretation), but the group of automorphisms should act transitively on the points. That is, if a and b are ANY two points, there is an automorphism that maps a to b.
skeptical scientist wrote:n shape
3 vertices of a triangle about the equator
4 vertices of a tetrahedron
5 pentagon about the equator
6 octahedron
7 heptagon about the equator
8 square antiprism
9 nonagon about the equator
10 pentagonal antiprism
11 undecagon about the equator
12 icosahedron
etc.

It's easy to continue this list, because if n is even, it's the dual of the "best" (closest to sphere-shaped, aka the one that comes in dice sets) fair n-sided die, and if n is odd, it's a n-gon about the equator, since the symmetry requirement forces you to get the vertices of an n-gon.
If n is 8, wouldn't the method you give for extending the list give you a cube and not a square antiprism? By my reckoning, the dual of a square antiprism is some kind of square bipyramid with the bases out of sync. Would that really be a better fair 8-sided die than an octahedron?
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Re: n isomorphic points on a sphere

Postby skeptical scientist » Fri Apr 18, 2008 7:26 am UTC

Nimz wrote:By my reckoning, the dual of a square antiprism is some kind of square bipyramid with the bases out of sync. Would that really be a better fair 8-sided die than an octahedron?

Hmm, I'm not sure. It's certainly not the one they make. Why is it that an 8 sided die is always an octahedron, but a 10 sided die always has the two pentagonal pyramids 36 degrees apart instead of perfectly aligned? Maybe they just like to use the platonic solid when there's one available, although either would work to give a fair 8-sided die.
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Re: n isomorphic points on a sphere

Postby Seraph » Sat Apr 19, 2008 2:51 am UTC

Hmm, I'm not sure. It's certainly not the one they make. Why is it that an 8 sided die is always an octahedron, but a 10 sided die always has the two pentagonal pyramids 36 degrees apart instead of perfectly aligned? Maybe they just like to use the platonic solid when there's one available, although either would work to give a fair 8-sided die.


The reason 10 sided dice are made like that is because the natural way to read one is to take the number off the side that is facing up. If you didn't have the 36 degree rotation on the 10 sided die then you wouldn't have an "up" face.

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Re: n isomorphic points on a sphere

Postby skeptical scientist » Sun Apr 20, 2008 3:05 am UTC

Seraph wrote:The reason 10 sided dice are made like that is because the natural way to read one is to take the number off the side that is facing up. If you didn't have the 36 degree rotation on the 10 sided die then you wouldn't have an "up" face.

Ah, true. And if you used the dual of a square anti-prism* instead of an octahedron, you'd have the same problem.

*Not sure what it's called.
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Re: n isomorphic points on a sphere

Postby Macbi » Sun Apr 20, 2008 7:47 am UTC

The dual of an antiprism is a trapezohedron. I've never understood the way the faces fit together, they look like they should stick out.
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Re: n isomorphic points on a sphere

Postby skeptical scientist » Sun Apr 20, 2008 4:10 pm UTC

Macbi wrote:The dual of an antiprism is a trapezohedron. I've never understood the way the faces fit together, they look like they should stick out.

The way I think of it is imagine the faces as infinite planes. They have to cut each other somewhere, so you get a solid with 4-sided faces arranged like that.
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Re: n isomorphic points on a sphere

Postby Macbi » Sun Apr 20, 2008 6:18 pm UTC

Ah, that helps! Thank you.
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