# Message #591

From: David Vanderschel <DvdS@Austin.RR.com>

Subject: Re: [MC4D] Something interesting and strange about permutations

Date: Sat, 27 Sep 2008 09:19:37 +0000

On Friday, September 26, "David Smith" <djs314djs314@yahoo.com> wrote:

>— In 4D_Cubing@yahoogroups.com, David Vanderschel <DvdS@…> wrote:

>> David Smith, you might want to consider extending your

>> permutation counting exercise to include also puzzles

>> with mirroring allowed. …

>Yes, I will definitely work on this new problem. By the

>way, from my point of view the super-supercube, while being

>a generalization of the regular cube, is the most elegant

>and simplest form of the Rubik’s Cube. In d dimensions,

>it’s a hypercube subdivided into n^d hypercubies, each of

>which is uniquely identifiable in any position or orientation.

>Any 1 x n^(d-1) group of hypercubies can rotate freely

>around the point at the center of the group of hypercubies

>in a manner which brings each hypercubie’s position to

>a position previously occupied by a hypercubie.

Actually, this is how I have always viewed the order-3

puzzles (any dimension). And, yes, the view applies

as well to puzzles of order higher than 3. What you

are describing as "1 x n^(d-1) group of hypercubies"

is what I call a "slice". I distinguish between

external slices and internal slices.

My lack of interest stemmed from what I perceived as a

lack of practical physical realizablity, since the

group permutes what amount to stickers that can never

be visible. But I must agree that it can be simulated

in an understandable way.

>If we include reflections, they can occur about any

>(n-1)-dimensional space which contains the point at

>the center of the group of hypercubies and is

>orthogonal to the faces of the hypercubies it

>intersects.

You can also reflect about hyperplanes that are on

diagonals. The reflection planes must be such that

the puzzle transforms onto itself, but that does not

mean they must be axis-aligned. E.g., for the

3-puzzle, a plane of reflection can intersect the

locations of 4 corners in such a way that 2 pairs are

diagonal from each other across opposing faces and 2

other pairs are adjacent along opposing edges. (The

last 2 pairs are what I call "triagonal" from one

another, diagonal within the 4-point rectangle.)

>> >MC2D moves only do interesting things with odd

>> >numbers of reflections,

>> ? I do not understand the above statement. Why is a

>> 4-cycle (Rotate corner positions.) not interesting?

>> [(1,2)(3,4)] * [(2,4)] = [(4,3,2,1)]

>You are definitely correct here David, but your example

>is unfortunately not (it actually contains 3

>reflections).

Perhaps we are not together on the permutation

notation. I am using cycle representation. If you

want to think about it geometrically, assign indices

to the corner positions in clockwise order starting at

the upper left hand corner. Then [(1,2)(3,4)] is a

reflection about the y-axis, [(2,4)] is a reflection

about the NW->SE diagonal, and [(4,3,2,1)] is a

4-cycle rotating all four corners counter clockwise.

>An even number of reflections can only produce an even

>permutation of the cubies,

Not true. Reflections can be odd or even and there is

one of each in the two I composed above. (Fairly

obvious, with 2 2-cycles in one and only 1 2-cycle in

the other.) So the product, a 4-cycle, is odd.

Furthermore, the product is not mirroring. (It is

true that the composition of two reflecting

permutations is always a non-reflecting

reorientation (or identity if the same reflection is

used twice).)

>and an odd number of reflections, an odd

>permutation. (For example, reflecting two adjacent

>sides of MC2D would create a 3-cycle of the corners,

>which is not possible with an odd number of

>reflections. Of course, you wrote a very lengthy

>reply which was very accurate, so this small error is

>understandable! :)

I suppose yours is somewhat larger since you implied

that I had erred when I had not; but I will forgive

you anyway! And please do not hesitate to attempt to

correct me if, in future, it again looks as if I have

erred. Next time it may be so, and I would not wish

to go uncorrected. I do err frequently because I

often have complex thoughts. I learn by being

corrected, and I occasionally assert something I am

not quite sure about with the hope that someone will

point me in the right direction if I turn out to be

confused.

After I posted my previous message it occurred to me

that I missed yet another possibility and an even

larger group. We might call it a SUPER–super-

supercube. I had written,

>If you try to flip a slice of the 3-puzzle in 3D, you

>do wind up (uselessly) with it being outside in; but

>it is not reflected.

I now regret that "(uselessly)" parenthesis. Let us

take the attitude that an order-m n-puzzle is an m^n

stack of nD cubies. (My n is your d and my m is your

n. I prefer to stick with n for the dimension of the

puzzle.) Also let us imagine that _every_ one of the

2n facelets on each cubie has a sticker. Use a normal

scheme for assigning colored stickers to the visible

facelets. I am willing to make all the invisible

stickers be black; but it would not be a necessity.

In the order-3 cases, it is certainly OK as each cubie

is uniquely identified by the combination of colored

stickers it bears. Now I want to say that, in

addition to our familiar twists, you can remove a

slice, flip it over with respect to the axis in which

it is ‘flat’, and replace it. (I can imagine no real

mechanism, but you can easily simulate the 3D cases

with a pile of cubical blocks. Just a little tedious

to reorient the whole cube or to remove a slice and

replace it flipped - but certainly possible.) Now

this is not mirror reflection; but it is a new kind of

permutation. The colored stickers can be turned to

face inwards and have valid possible internal

positions. Flipping of slices need not be limited to

external slices.

Imagine that all 2n*m^n facelet positions are indexed

relative to fixed spatial coordinates and we identify

the stickers on them in the initial position by the

position they occupy. The permitted alterations to

the pile can be seen as permutations of the 2n*m^n

stickers. There must be some sense in which the

familiar group of the regular order-m n-puzzle is

still in there as a subgroup; but I am not sure yet

how to characterize that, since there are extra

stickers. (May require a quotient.) We now have the

ability to turn over any slice relative to the

dimension in which it has thickness 1. When the

puzzle is scrambled, many initially-visible colored

stickers may no longer be visible; but they can still

be accessed by appropriate manipulation.

Anyone for slice-swapping? ;-)

Regards,

David V.