Message #1659

From: David Smith <djs314djs314@yahoo.com>
Subject: Re: [MC4D] Generalized corner orientation theorem
Date: Wed, 04 May 2011 06:11:39 -0700

Hi Andrey,

You are absolutely right, my theorem does not apply for as many puzzles as I
claimed.  Thank you for pointing this out!  But it is correct as written, and is very
useful in determining orientations.  For practical purposes, It can be modified to
handle different cases such as the ones you mentioned, but I haven’t included
all of the variations.  It is related to a much stronger property of corner orientations
in general, as I mentioned in the beginning of my post, and I’m sure that I can modify
it as needed to apply it to any puzzle.  It is also useful to check if the conditions
apply, and if not then one knows how to fix it.

Of course, when you can rotate a single corner, the theorem does not apply. :)
But those cases are easy to handle, so the practicality of the theorem is
not affected.  I find your examples of the factor being something other than 3 to
be very interesting; I haven’t worked with them too much and will study them
later to see why the theorem does not apply as-is, and how to modify it to work.

Also, this statement was obviously in error:

((order of rotation group of the regular (n-1)-dimensional polytope with
s vertices)^k)/3

if n = 3 or 4

with the division by 3.  I’ll fix this later when I get home.

An easy way to see if my theorem applies as-is is first to make sure that the
number of stickers is equal to the number of dimensions.  That is the first
condition that more unique puzzles are likely to fail.  But the most restrictive
condition is this:

every corner would be required
to eventually occupy each position in every possible permutation of its stickers
given by the rotations of the corner.

This is basically the key condition that Rubik-like puzzles are likely to fail to
qualify for the theorem, and studying why it fails can often provide the insight
as to how to modify the theorem to make it work.  An easy way to check
if this condition holds is to rotate two adjacent faces, and see if you can get
a corner to achieve all of its orientations on it home square.  If it can, then
the above requirement would be satisfied, by using conjugates.

I thank you very much for pointing out the inaccuracy in my statement that
my theorem works for almost every puzzle!  With modifications, I believe it can
handle them all.  Later I will study the examples you presented, and also give
a brief overview of my general orientation-counting technique.

Thanks again, and best wishes!

All the best,
David

— On Wed, 5/4/11, Andrey <andreyastrelin@yahoo.com> wrote:

From: Andrey <andreyastrelin@yahoo.com>
Subject: Re: [MC4D] Generalized corner orientation theorem
To: 4D_Cubing@yahoogroups.com
Date: Wednesday, May 4, 2011, 12:32 AM

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.

Regards,

Andrey