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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?