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.