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The Polynomial from Beyond Space
Oct. 9th, 2007 02:27 pmAs far as I've bothered to run the exhaustive search, the number of ways of making a tetrahedron out of N colours of perspex rods such that you never have two rods of the same colour coming out of a single vertex, but not requiring you to use all N colours, is N (N-1) (N-2) (N^3-9N^2+29N-32).
This doesn't seem a very natural polynomial to me.
There seems to be a unique-up-to-A5 three-colouring of the edges of a dodecahedron and a unique-up-to-S4 three-colouring of the edges of a cube; there's an obvious unique-up-to-S3 three-colouring of the edges of a tetrahedron. Obviously you can't three-colour the edges of an octahedron or an icosahedron, since four or five edges meet at a vertex; it looks as if there are maybe two group-theoretically-distinct four-colourings of the octahedron, and roughly 1560 five-colourings of the icosahedron.
As an obviously related question, where can I get some (fifty would be a good number) two-inch solid black rubber balls? I have already found a supplier of six-foot lengths of half-inch perspex rod in three distinct fluorescent colours ...
What's a material that can both be cast and machined? To get the holes in the balls in the right places I'd need the use of a drill-press and a couple of oddly-shaped jigs, and the easy way to make the jigs seems to be to use a ball as a mould to cast a hemispherical hole in something castable, then orient the casting properly (this is the tough part, I don't know what the name for the kind of contraption of adjustable vices I'm thinking of would be), drill a hole in the correct place with the drill-press, and insert a spare bit of half-inch dowelling to fit into the first hole drilled so that the second and third (and fourth and fifth for the icosahedron) end up in the right places. Or is drilling holes in rubber balls likely to be a source of frustration, horrible stench, and an unreasonable amount of time spent cleaning melted rubber off drill-bits?
This doesn't seem a very natural polynomial to me.
There seems to be a unique-up-to-A5 three-colouring of the edges of a dodecahedron and a unique-up-to-S4 three-colouring of the edges of a cube; there's an obvious unique-up-to-S3 three-colouring of the edges of a tetrahedron. Obviously you can't three-colour the edges of an octahedron or an icosahedron, since four or five edges meet at a vertex; it looks as if there are maybe two group-theoretically-distinct four-colourings of the octahedron, and roughly 1560 five-colourings of the icosahedron.
As an obviously related question, where can I get some (fifty would be a good number) two-inch solid black rubber balls? I have already found a supplier of six-foot lengths of half-inch perspex rod in three distinct fluorescent colours ...
What's a material that can both be cast and machined? To get the holes in the balls in the right places I'd need the use of a drill-press and a couple of oddly-shaped jigs, and the easy way to make the jigs seems to be to use a ball as a mould to cast a hemispherical hole in something castable, then orient the casting properly (this is the tough part, I don't know what the name for the kind of contraption of adjustable vices I'm thinking of would be), drill a hole in the correct place with the drill-press, and insert a spare bit of half-inch dowelling to fit into the first hole drilled so that the second and third (and fourth and fifth for the icosahedron) end up in the right places. Or is drilling holes in rubber balls likely to be a source of frustration, horrible stench, and an unreasonable amount of time spent cleaning melted rubber off drill-bits?
