Message #1649
From: Brandon Enright <bmenrigh@ucsd.edu>
Subject: Re: [MC4D] Generalized corner orientation theorem
Date: Wed, 04 May 2011 00:33:50 +0000
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Hi David,
This is an interesting post as I have thought about corners for 3D
puzzles quite a bit.
First, my apologies for spending 15 minutes replying to your 3.5 hour
email. I don’t know how you and Melinda and some others manage to
find the time to write such long, thoughtful, and comprehensive emails.
The rest of my email is inline.
On Tue, 3 May 2011 16:41:52 -0700 (PDT)
David Smith <djs314djs314@yahoo.com> wrote:
> […setting up of proof and list of conditions snipped…]
I have a few questions regarding your conditions. I can’t tell in your
conditions if you have tried to or succeeded in excluding
vertex-twisting or edge-twisting puzzles. For example, I know
Gelatinbrain’s 1.2.2
(http://users.skynet.be/gelatinbrain/Applets/Magic%20Polyhedra/dodeca_v3.gif)
violates your first theorem because the permutation+orientation
formula for it is ((20!/2)^3)
Also, the Helicopter cube has the /3 orientation restriction but does
not have the /2 permutation restriction.
> Generalized Corner Orientation Theorem: If a permutation puzzle’s
> corners satisfy Condition 1 by using k n-dimensional hyperspheres as
> corners, and the isomorphic versions of the puzzle’s corners and
> moves satisfy Conditions 2 through 5, then the number of orientations
> the corners of the puzzle can achieve is:
>
> ((n!/2)^k)/3 if n = 3 or 4
>
> and
>
> (n!/2)^k if n >= 5.
Hmm, did you swap n and k in these?
That is, k is the number of corners and they are restricted to even
permutations and if they are 3 or 4 dimensional than the last corner’s
orientation is uniquely defined by the others.
To me that would be:
((k!/2)^n)/3 if n = 3 or 4
and
(k!/2)^n if n >= 5.
If this isn’t the case they maybe I don’t follow you after all…
>
> The proof relies heavily on others’ work and thus I cannot really
> claim it as my own. But neither the authors of The Rubik Tesseract
> nor An n-dimensional Rubik Cube applied their results to anything
> other than the 3^n cube, so I made the generalization.
>
> It took me over 3.5 hours to write this email (I had no formulation
> of the theorem in my mind before I started; I had to come up with the
> entire sphere/point/simplex idea as I wrote, and there were many
> parts rewritten or rephrased). Given that, I hope that this post has
> been interesting to some of you! I would appreciate any comments or
> questions you have for me. Thanks to Roice for the suggestion!
>
> All the best,
> David
On the topic of vertex-twisting puzzles, what about duals? For
example, Gelatinbrain’s 2.2.6
(http://users.skynet.be/gelatinbrain/Applets/Magic%20Polyhedra/icosa_v4.gif)
is the dual of the Penultimate (1.1.7 /
http://users.skynet.be/gelatinbrain/Applets/Magic%20Polyhedra/dodeca_f6.gif)
It is nice to be able to think of centers with super-stickers as
corners and corners and centers. Using the 1.1.7 <-> 2.2.6 dual
relationship is troublesome though because the number of orientations
of a 1.1.7 center / 2.2.6 corner is 5 and isn’t tied to the
dimensionality of the puzzle. Perhaps this is how you’ve avoided
vertex-twisting puzzles by using definition 1 where the number of
stickers matches the number of dimensions? That doesn’t handle the
Helicopter cube though.
Best,
Brandon
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