# Message #1651

From: David Smith <djs314djs314@yahoo.com>

Subject: Re: [MC4D] Generalized corner orientation theorem

Date: Tue, 03 May 2011 19:44:33 -0700

Hi Brandon,

No apologies necessary! I just had the time available, and made use of it.

It appears that you have found some extra puzzles I had not considered!

However, I believe that when the corner orientation cannot be handled by

the theorem, it is usually simple to figure out; i.e. all orientations or no

orientations. But I could be wrong.

My theorem should apply to any vertex or edge-twisting puzzle (with an

important exception you made me aware of, see below). I can’t see why

it doesn’t apply to the puzzle in the first link you posted, the 1.2.2. I’ll note

that if it is possible to rotate a single corner, then of course it is easy to

calculate the number of orientations; it will be the maximum with no

restrictions.

You said,

> violates your first theorem because the permutation+orientation

> formula for it is ((20!/2)^3)

I’m not sure what you mean by "first theorem". Also you mention that the

"permutation+ orientation formula" for it is (20!/2)^3. I’m not certain if you’re just

talking about the corners, and why you added the word ‘permutation’. Where

did you get this formula? Upon analyzing it, it appears that you meant,

‘(20!/2)^30’, because there are 30 edges. I can tell you this: From looking at

the image alone, and assuming that the only type of twist one can perform is

the one displayed, then it appears that the corners can have 20!/2 permutations,

and the corners can have 3^20/3 orientations.

> Also, the Helicopter cube has the /3 orientation restriction but does

> not have the /2 permutation restriction.

"

This seems perfectly fine and consistent with my formula.

Your image here was very insightful:

http://users.skynet.be/gelatinbrain/Applets/Magic%20Polyhedra/icosa_v4.gif

I simply hadn’t considered the (most likely obvious) fact that corners can have

more stickers than the number of dimensions! I have to admit, I made some

claims about my theorem holding for almost all puzzles, but I was just considering

what I’d seen on the Cubing group. I was unaware of all of these variants!

In the case of a corner having at least n stickers if it’s corner sphere is

n-dimensional, I need to make the following changes:

Definition 1: A corner will be an n-dimensional hypersphere, n >= 3, floating at

a fixed position in n-dimensional space with at least n stickers on its surface.

Condition 3: The stickers on a corner must all lie on the vertices of a regular

polytope, and this polytope cannot lie in an (n-1)-dimensional hyperplane which

bisects the corner into two equal halves.

Generalized Corner Orientation Theorem: If a permutation

puzzle’s corners

satisfy Condition 1 by using k n-dimensional hyperspheres as corners, each

having s stickers, 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:

((order of rotation group of the regular (n-1)-dimensional polytope with

s vertices)^k)/3

if n = 3 or 4

and

(order of rotation group of the regular (n-1)-dimensional polytope with

s vertices)^k

if n >= 5.

The complex phrase, "order of rotation group…" could be simplified, as

there are only so many regular polytopes of each dimension. For 3-dimensional

puzzles, it would equal s, which corresponds to n!/2 if n = 3 and s = 3.

> ((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…

No, I have it correctly. I think you are just making a simple oversight, as I had

been doing with the Klein’s Quartic counts! :) I believe you’re making the mistake

of thinking in terms of permutations, hence your reasoning for the term, "k!/2".

My formula concerns only orientations. It needs to be multiplied by the permutation

count (which is simple, but as I said before depends upon the other piece families

due to parity interaction) to finish the calculation for that piece. The n!/2 term comes

from the number of ways one can rotate an n-dimensional simplex. We raise it

to the power of k because there are k corners. Really, the only two outstanding

features of this theorem are that for dimensions greater than 4, there is no

division by three (no corner restriction), and that it can be generalized so far as

I have done.

> 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.

My theorem should apply to vertex-twisting and edge-twisting puzzles,

but it is important to keep in mind that it only works for corners, i.e. pieces

with at least as many stickers as dimensions. It shouldn’t matter if we are

twisting about the vertex, as the other surrounding vertices are moved by the

rotation. For duals, we would have to recalculate the corner orientation; we

can’t interchange corners for centers and centers for corners. It’s a

good idea however.

I hope this has been helpful. Thanks again for welcoming me! It’s great

to be back.

All the best,

David

— On Tue, 5/3/11, Brandon Enright <bmenrigh@ucsd.edu> wrote:

From: Brandon Enright <bmenrigh@ucsd.edu>

Subject: Re: [MC4D] Generalized corner orientation theorem

To: 4D_Cubing@yahoogroups.com

Cc: djs314djs314@yahoo.com, bmenrigh@ucsd.edu

Date: Tuesday, May 3, 2011, 8:33 PM

—–BEGIN PGP SIGNED MESSAGE—–

Hash: SHA1

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

—–BEGIN PGP SIGNATURE—–

Version: GnuPG v2.0.17 (GNU/Linux)

iEYEARECAAYFAk3AnvUACgkQqaGPzAsl94LyZQCgjWLiA3asxRBq1y/3RO63/sJS

1kEAoLBuyhQWEOudsOn7mSuVqNJxRLT2

=7D6o

—–END PGP SIGNATURE—–