Here is a definition. If X is any cube, then the function B
is defined as B(X) = min (C[i] R[j] X R[j] C[k]) for i in {1 to 24},
j in {1 to 2}, and k in {1 to 24}, and where C is the set of rotations
of the cube and R is the set of reflections of the cube. Then, two
cubes X and Y are said to be B-equivalent if B(X)=B(Y). Note that
the B function calculates the "canonical form" or the "representative
element" of an equivalence class under M-conjugation and rotation of
the cube in space.

Dan Hoey already gave a thorough analysis of this situation. I would
like to provide an alternative analysis which I hope to be equivalent
to Dan's. My analysis will be fairly informal as compared to Dan's.

Here are a couple of preliminaries. First, multiplication of
permutations is generally not commutative. For example, if X is any
cube, then it is generally not the case that m'Xm=mXm', where m
is in M, the set of all cube rotations composed with reflections.
However, we can calculate all M-conjugates of X as B[n]=m[n]' X m[n]
for n in {1 to 48}, or we can calculate all M-conjugates of X as
B[n]=m[n] X m[n]'. Either way, B will be the same set of cubes. It
will be in a different order, but it will be the same set. The reason
is that the set of all m[n] is the same as the set of all m[n]',
namely just M, but m[n] is in a different order than m[n]'. This means
that as long as we are calculating all M-conjugates as opposed to
a specific M-conjugate, we can sort of "violate" the normal rules
about multiplication commutivity.

Second, if X is any cube, consider the set of all rotations of
X, namely B[i] = X c[i] for i in {1 to 24}, and where c[i] is in C,
the set of all cube rotations. Having generated the set of all
rotations of X, we can rotate as many times as we wish, for example
B[j] = X c[i] c[j] for i in {1 to 24} and j in {1 to 24}, or even
B[k] = X c[i] c[j] c[k] for i in {1 to 24}, j in {1 to 24}, and
k in {1 to 24}. No matter how many times we multiply, B will be the
same set, it will just be in a different order. Conversely, if we
have any number of adjacent rotations in the multiplication, we can
eliminate all but one rotation, and B will be the same set, and again
will just be in a different order.

With the preliminaries out of the way, we note that the set of all
M-conjugates of X is generated as B[i]=m[i]' X m[i] for i in
{1 to 48}.  But we can also generate the same set in a different order
as B[i]=m[i] X m[i]' for i in {1 to 48}.  We can decompose M and
calculate all M-conjugates as B[i,j]=c[i] r[j] X r[j]' c[i]' for i in
{1 to 24} and j in {1 to 2}.

But r[1]'=r[1] (r[1] is the identity) and r[2]'=r[2] (the reflection
is its own complement).  So we have B[i,j]=c[i] r[j] X r[j] c[i]'
for i in {1 to 24} and j in {1 to 2}.  The set of all c[i] is the
same as the set of all c[i]' (just in a different order), so we
define i' as the index for which c[i']=c[i]'.  Hence the calculation
of an M-conjugate can be written as B[i,j]=c[i] r[j] X r[j] c[i']
for i in {1 to 24} and j in {1 to 2}.

Finally, we wish to multiply the M-conjugate by the set of all rotations,
so we have B[i,j,k]=c[i] r[j] X r[j] c[i'] c[k] for i in {1 to 24},
j in {1 to 2}, and k in {1 to 24}. But as we noted in our second
preliminary note, we can collapse multiple rotations into one, and we
have B[i,j,k]=c[i] r[j] X r[j] c[k], and B will be the set of all
M-conjugates of X multiplied by all rotations. I guess I am overusing
the letter "B" a bit, because the "B" function is simply the minimum
of the "B" matrix. But in any case, we have shown that the "B" loop
in my program is simply calculating the set of all M-conjugates multiplied
by all rotations. This is the exact result already proven by Dan Hoey,
but I found the above derivation a little easier to follow.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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?