simont | (no subject)

(no subject)

Oh yes, and I've just added some more Stuff to my website. It's now getting to the point where I might have to fiddle with the front page so that it has a usable index; all this chatty text introducing each of my subpages is fine if you're reading it because you want to know about me, but even I'm increasingly finding it a pain when I actually want to find a specific page.

Still. For anyone with any interest in polyhedra, or even simply with a desire to look at some pretty pictures or see what weird mathematical things I've been up to recently when I should have been lying on my sofa watching DVDs, you can now go and look at http://www.chiark.greenend.org.uk/~sgtatham/polyhedra/.

Now, with any luck, I ought to be able to actually relax for a bit. These momentary obsessions tend to take me over totally for a week or so, but then leave me in peace after that…

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I don't believe that your 60-vertex solid really *is* that different from buckminsterfullerene. I can clearly see hexagons and pentagons in there, you just have an extra vertex in the centre of each face connected to all the corners. I wonder what happens if you start from an an example pentagon, pretend it's a flat face and remove the internal lines crossing the face, then move outwards to the hexagons and pentagons surrounding that, etc. Do you get anything that makes any sense?

I think one of the reasons you always get triangles with the vertex construction is that you've given yourself no maximum limit on how many edges can meet at any given vertex - thus of course as the number of points goes up, the face size will decompose down to a triangle.

When you're talking atoms, you can't approximate them as points, you'd have to approximate them as points *which must have a specific number of connections* - do you get anything interesting if try to build this kind of restriction into your models?

I think there's a standard result that means a solid of this type needs twelve 5-edge vertices if the rest are all going to be 6-edge, otherwise you get a flat triangular lattice rather than a closed solid. So yes, there are twelve "pentagons" visible, and I think this would continue to be the case on any solid of this type as you increase the number of points, no matter how many 6-edge vertices go in between.

In the 60-point case: if you remove each 5-edge vertex and replace the five faces surrounding it with a pentagon, I don't think you get anything particularly interesting. If you look at the visible 5-edge vertices:

you see that actually they aren't even consistently separated. The two "pentagons" I've marked in the upper left end up contiguous; the central one is separated from both of them by a single layer of triangles (not enough to form hexagons); and the central and lower ones touch at a vertex rather than an edge. This is unquestionably an irregular polyhedron, not a trivial knockoff of buckminsterfullerene.

Hum, so it is. In which case it seems distinctly unintuitive that you should end up with an irregular arrangement from your repelling points - I'd have expected bilateral symmetry if nothing else.

In atomic terms, it's pretty much the case that you want to maximise the angle between bonds as well as maximising the distance between atoms. I wonder if you'd get to buckminsterfullerene if you built that aspect in. Probably not without restricting the number of bonds per atom as well, in which case you've probably built in enough assumptions to make it trivial.

There might be bilateral symmetry; I haven't actually looked closely enough to be completely sure. (One thing I haven't yet arranged is a viewing utility that would allow me to rotate one of these solids at whim so I could look at it from all angles. The maths is easy enough, but the UI work sounds too much like effort.)

I've been vaguely wondering about taking one of these point sets (no need to use the output polyhedron) and attempting to find all its symmetries algorithmically. There aren't that many possibilities: simply attempt to map each point on to each other point, and then find other points the same distance from those two candidate points and rotate to try to bring another one into alignment, then see whether you've mapped every point on to (or close to) another. It would in principle take O(N^3) time, but in practice more like O(N^2) times a small constant since there wouldn't be many possibilities for the third candidate point; and you'd end up outputting an exhaustive list of permutations of the points, which would of course form a closed group.

If I could figure out a way to take that set of permutations and compute some sort of canonical characterisation of the group they describe (so that you could tell two point sets had the same symmetry group if and only if the output of this program was identical for them), I would try this, and see just how much symmetry there is in these solids. I noticed in particular that the 13-point set does appear to have two axes of reflective symmetry, so I broadly agree with your intuition that some symmetry is to be expected.

That just reminds me of The Crystal Maze

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