# 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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