Message #1653

From: Andrey <>
Subject: Re: [MC4D] Generalized corner orientation theorem
Date: Wed, 04 May 2011 04:32:35 -0000

Hi David,
It looks like your result works for hypercubes, dodecahedron and 120-cell and for the most 2D Magic Tile with fixed centers (for which we use n=3, not 2). But there are puzzles with 2D and 3D stickers where all corner orientations are possible, or where the factor is different from 3.
1) Everybody knows that you can twist single corner in pyraminx :)
2) For face-turn octahedron number of possible orientations is only 1/2 of "order^k". But it doesnt’s satisfy conditions: either you should paint only 4 its faces - and corners will be 2-colored, or not all permutations of sticker colors will be reachable (each 4-colored corner may be only on two orientations).
3) in "hemidodecahedron" from Magic Tile you also can twist one corner. Number of corner orientations there will be 6^10/2 (puzzle is non-orientable so corner piece has 6 possible orientations)
4) in {6,3,3} puzzle with 8 colors I saw a corner sticker twisted by 3-cycle - so all 12^28 corner orientations are possible.
5) It looks like for the face-turn icosahedron number of corner orientations is 5^12/5 (I never tried it, but you can see it just by the picture - select 4 faces without common corners and count sum of corner rotations during twists - it always will be zero).

I afraid that there are too many possible permutation groups even for corner stickers to have some common theorem about them. May be, something like "you always can change orientation of one corner to any possible state so that only two corners will be twisted (this and one more) - and other k-2 corners will save their orientation" will work? I’m not sure even in this.