[next] [prev] [up] Date: Tue, 27 Jun 95 23:22:50 -0400
[next] [prev] [up] From: Jerry Bryan <BRYAN@wvnvm.wvnet.edu >
~~~ ~~~ [up] Subject: Constructing K-symmetric Cubes

This is a followup on several recent messages concerning the
question of K-symmetric cubes, where K is one of the
98 subgroups of M.

We recall that a permutation is a special kind of function,
namely a one-to-one and onto function on a set. A very
common technique used with functions is to restrict the
domain to a (usually proper) subset of the original domain.
In the paradigm of a function as a general rule, the general rule
is applied to the subset of the domain to obtain the restriction
of the function. In the paradigm of a function as a set of ordered
pairs, the restriction is simply a (usually proper) subset of the
set of ordered pairs. A restriction of a permutation is usually
not a permutation (certainly not a permutation on the original
domain), but it is still a function.

We will treat a permutation on the cube as a set of restrictions
(functions) of the several cubicles. We take as our first example
the function UFL->UFL. Unlike cycle notation, it is not
assumed that the other cubicles are fixed; rather, the other
function values are undefined (e.g., URF->?).

Let X be any function (not necessarily a permutation) whose domain
is some subset of the cubicles. We define Symm(X) in the standard
fashion -- Symm(X) is the set of all m in M such that
m'Xm=X. If X is the function UFL->UFL, we have Symm(X)=AT4, not
Symm(X)=M as you might expect. AT4 is a subgroup in Dan's taxonomy
containing six elements, and which has an axis of symmetry along
the UFL-DBR axis.

This definition of Symm(X) perhaps requires a minor bit of
justification. In a function composition such as FG (left-to-
right notation) or G(F(x)) (right-to-left "calculus" notation),
it is sometimes taken as a convention that the range of F must
match the domain of G. But we can also take the restriction of
G to the intersection of the range of F with the domain of G,
and we do so. Having done so, Symm(X) is well defined.

We wish to build an M-symmetric permutation containing the
function UFL->UFL, but Symm(UFL->UFL)=AT4 is not a very
auspicious start. Rather, we define the conditions under which
a function is K-symmetric is follows. A function X is K-symmetric
if the union of the K-conjugates k'Xk is a function.

Given this definition, the only M-symmetric function on UFL is
in fact UFL->UFL, so we really have made a good start. Furthermore,
any M-symmetric permutation that contains UFL->UFL must also
contain all the M-conjugates of UFL->UFL, and the union of the
M-conjugates is simply the identity permutation on the corners.

The fact that Symm(UFL->UFL)=AT4 can be of some benefit in our
investigations. In particular, the fact that |AT4|=6 means that
there are 8 M-conjugates, so taking all the M-conjugates of
UFL->UFL means that all 8 corners are specified.

Our next example will be UFL->LUF (a twist of the corner). In
this case, we have Symm(X)=ET4. ET4 is the subgroup of M in
Dan's taxonomy which contains 3 elements including the identity
plus the 1/3 and 2/3 rotations around the UFL-DBR axis.

Of more import, UFL->LUF is not M-symmetric. However, it is both
C-symmetric (C is the set of 24 rotations) and H-symmetric
(H is the set of 12 even rotations and 12 odd reflections). As
an aside, we note that it is not A-symmetric, where A is the set
of 24 even rotations and reflections. But since it is both
C-symmetric and H-symmetric, there is not a unique largest
subgroup K for which we can say it is K-symmetric.

It is easy to see that UFL->LUF is not M-symmetric. The set of
M-conjugates contains both UFL->LUF and UFL->FLU, so the union
of the M-conjugates is not a function. We can see the same thing
from the fact that |ET4|=3. Since |ET4|=3, there are 16 M-conjugates,
but there are only 8 corners to represent the 16 M-conjugates.

Since UFL->LUF is C-symmetric, let's see if we can build a
C-symmetric permutation. There are 8 C-conjugates (a good
start!), and the 8 C-conjugates twist each of the 8 corners
by 1/3 in the same direction. Hence, this is a C-symmetric
but not M-symmetric permutation. Of course, it is an "illegal"
position in the sense that it is not in the same orbit
as Start.

In the same manner, we can build a K-symmetric permutation for
any K. We start with a K-symmetric function on a single
cubicle. (A function which is K-symmetric is L-symmetric for
any L which is a subgroup of K). We include all
K-conjugates. If the cube is completely specified, we stop.
Otherwise, we choose another K-symmetric function for any
previously unspecified cubicle, add in the new K-conjugates,
and so forth, repeating until the entire permutation is

Needless to say, this construction process suffers from not
preserving orbit. Additional steps must be taken to assure that
the constructed position is in the desired orbit (usually, the
Start orbit). And some orbits do not have representatives from
some subgroups, for example it is well known that there are
no C-symmetric but not M-symmetric permutations in the Start

To use Dan's adjectives, this process very quickly can become
long, mechanical, and boring. But I now see how to build a
K-symmetric permutation for any K.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax
837 Chestnut Ridge Road                              BRYAN@WVNVM
Morgantown, WV 26505                                 BRYAN@WVNVM.WVNET.EDU

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