I wish to propose some terminology and definitions to make certain
concepts a bit more precise. For example, when we are talking about
"corners only", it is not always clear whether we are talking about
"corners with centers without edges" or "corners without centers
without edges". In this note, I have tried to be consistent with
previous usage on the list, but I welcome any historical corrections
that might be deemed necessary.

Let G be the standard cube group for the 3x3x3 cube, and let |G| be
the size G.  Hence, we have |G| = (8!(3^8)/3 * 12!(2^12)/2) / 2,
which is the famous 4.3 * 10^19.

Let GC be the corners with centers without edges group for the 3x3x3
cube, and let |GC| be the size of GC. Hence, we have
|GC| = 8!(3^8)/3. (I welcome a suggestion other than "GC" as the
name for this group. I did not find one in the archives.) This group
could be modeled by removing the edge labels from a standard
3x3x3 cube.

Let GE be the edges with centers without corners group for the 3x3x3
cube, and let |GE| be the size of GE. Hence, we have
|GE| = 12!(2^12)/2. (As before, I welcome a suggestion other than
"GE" for the name for this group.) This group could be modeled by
removing the corner labels from a standard 3x3x3 cube.

Note that |G| = |GC| * |GE| / 2.

Let G\C be the corners with edges without centers group. I intend
for the notation to indicate G reduced by C, where C is the rotation
group for the cube. It should be the case that |G\C| = |G| / 24,
but I want to return to this question a little later. This group
could be modeled by removing the center labels from a standard
3x3x3 cube.

Let GC\C be the corners without edges without centers group. This is
the 2x2x2 cube. We should have |GC\C| = |GC| / 24, but again I want
to return to this question a little later. In addition to being the
2x2x2 cube, this group could be modeled by removing the center and
edge labels from a standard 3x3x3 cube.

Let GE\C be the edges without corners without centers group. We
should have |GE\C| = |GE| / 24, but again I want to return to this
question a little later. This group could be modeled by removing
the center and corner labels from a standard 3x3x3 cube.

Let G\M be the set of M-conjugate classes for G. In this case,
|G\M| is approximately 48 times smaller than |G|. I believe that
when Dan Hoey asked in 1984 the question "how big is G, really?",
that he was really asking how big is G\M, and that he was asking
for the approximation to be resolved to an exact number.

Let GC\M be the set of M-conjugate classes for GC. In this case,
|GC\M| is approximately 48 times smaller than |GC|.

Let GE\M be the set of M-conjugate classes for GE. In this case,
|GE\M| is approximately 48 times smaller than |GE|.

Recall that B is the function which calculates the canonical form
for a cube under the composed operations of M-conjugation plus
rotation. My programs calculate equivalence classes under B.

Let G\B be the set of B-classes for G. Let GC\B be the set of B-classes
for GC. Let GE\B be the set of B-classes for GE. So far, my programs
have built complete search trees for GC\B and GE\B.

Let Gx denote any of G, GC, and GE. Then, we have Gx\B=(Gx\C)\M=(Gx\M)\C.
In English, we can decompose B into a multiplication by C and M (in
either order).

Also, note that Gx\C=(Gx\C)\C=((Gx\C)\C)\C=.... Similarly,
(G\M)\C=((G\M)\C)\C=.... In English, having reduced once by C, we can
reduce again by C as many times as we wish, but we simply get the same
set back again each time.

This notation can help us address the question of whether B actually
accomplishes a "times 48" or a "times 1152" reduction in the size of
the cube. If we are dealing with Gx, then Gx\B is a "times
1152" reduction. However, information is lost. For example, consider
GC and GC\B. GC is "corners plus centers", and B-reduction of GC
removes the centers and calculates M-conjugates of the corners.
But you really don't have the same problem any more because the centers
are gone. If on the other hand we are dealing with Gx\C, then (Gx\C)\B
is a "times 48" reduction. All we have really done is calculate
M-conjugates. The reduction by the C that is composed into B is duplicate
effort which accomplishes nothing.

I have come to realize that my program for the 2x2x2 actually models
GC (corners with centers without edges) rather than GC\C (corners
without centers without edges). My program does not explicitly encode
the centers. However, it encodes all eight corner cubies, and when
it makes qturns, any of the eight cubies can move. Hence, rotational
information is encoded, even if the centers themselves are not explicitly
encoded. If I wanted to model GC\C, I would have had to either model
only seven of the cubies, or else modeled all eight but moved only seven
of them. Since what I really wanted was (GC\C)\M, and since what I had
was GC, I had to invent this funny B thing, where GC\B=(GC\C)\M. If I
had been clever enough to model GC\C in the first place, I never would
have had to invent B. Similar comments apply to my model for the edges.

To convince yourself that eight corner cubies model GC and seven corner
cubies model GC\C, just think about calculating |GC| and |GC\C|.
For |GC|, there are eight ways to pick the first cubie, seven ways to
pick the second cubie, and so forth yielding the familiar
|GC|=8!*(3^8)/3. For |GC\C|, we let one of the cubies be fixed,
then there are seven ways to pick the second cubie, and so forth yielding
|GC\C|=7!*(3^7)/3, and |GC| = |GC\C| * 24. Hence, the "corners of the
3x3x3" problem is 24 times larger than the "2x2x2" problem.

I will discuss the "times 24" reduction that is accomplished by
reducing by C in a followup note.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?