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In 1984, Dan Hoey posed a question as follows:

This discussion of symmetry recalls a question I have meant to propose

to Cube-Lovers for some time: How many positions are there in Rubik's

Cube? We know from Ideal that the number is somewhat over three

billion. Most cube lovers will tell you a number of about 43

quintillion. But I really don't see why we should count twelve

distinct positions at one quarter-twist from solved--all twelve are

essentially the same position. So the question, suitably rephrased, is

of the number of positions that are distinct up to conjugacy in M, the

48-element symmetry group of the cube. I think this is an interesting

question, but I don't see any particularly easy way of answering it.

My best guess is that it involves a case-by-case analysis of the 98

subgroups of M, or at least the 33 conjugacy classes of those

subgroups. In ``Symmetry and Local Maxima'', Jim Saxe and I examined

five of the classes, which we called M, C, AM, H, and T.Even finding the numbers for the pocket cube is a little tricky. If we

limit ourselves to symmetry in S, I believe the pocket cube has 2

positions with a six-element symmetry group, 160 positions with a

three-element symmetry group, 3882 positions with a two-element

symmetry group, and 3670116 positions with a one-element symmetry

group, for 613062 positions distinct up to S-conjugacy. But the

numbers for M-conjugacy are still elusive; I am not even sure how to

deal with factoring out whole-cube moves in the analysis. I hope to

find time to write a program for it.I expanded my pocket cube program to deal with the corner group of

Rubik's cube. This group is 24 times as large as the group of the

pocket cube, having 3^7 * 8! = 88179840 elements. The number of

elements P(N) and local maxima L(N) at each (quarter-twist) distance N

from solved are given below.N P(N) L(N) 0 1 0 1 12 0 2 114 0 3 924 0 4 6539 0 5 39528 0 6 199926 114 7 806136 600 8 2761740 17916 9 8656152 10200 10 22334112 35040 11 32420448 818112 12 18780864 9654240 13 2166720 2127264 14 6624 6624The alert reader will notice that rows 10 through 14 contain values

exactly 24 times as large as those for the pocket cube. This is not

surprising, given that the groups are identical except for the position

of the entire assembly in space, and each generator of the corner cube

is identical to the inverse of the corresponding generator for the

opposite face except for the whole-cube position. Thus when solving a

corner-cube position at 10 qtw or more from solved, it can be solved as

a pocket cube, making the choice between opposite faces in such a way

that the whole-cube position comes out right with no extra moves.

I wish to propose an answer to Dan's question. I will propose an

approximation then (hopefully) the exact answer.

The approximation is simply 4.3*(10^19) / 1152, or about

3.7*(10^16). 1152=24*24*2, and is based on my version of Dan's

M symmetry group. I remain convinced that my version of M is

isomorphic to Dan's, but the subject deserves some more thought

and discussion.

But we can do better. We already know (under my version of M) how

many equivalence classes there are for the corner group (namely,

77,802). But each of the equivalence classes for the corners can

be rotated 24 ways with respect to the centers, so we have

77,802*24. We also already know (under my version of M) how many

equivalence classes there are for the edge group (namely

851,625,008). But each of the equivalence classes for the edges

can be rotated 24 ways with respect to the centers, so we have

851,625,008*24. Hence, we have

(77,802*24) * (851,625,008*24) = 38,164,682,230,511,620

This figure is gratifyingly close to 3.7*(10^16), and I believe it

is the correct answer to Dan's question. It is slightly larger

than the approximation because some of the equivalence classes

have fewer than 1152 elements, and consequently there are a few

more equivalence classes than the approximation suggests.

However, the alert reader should have noticed a problem. Why did I

not divide by 2 to take into account the fact that odd edge

permutations can only occur with odd corner permutations and vice

versa? Actually, I did, but the division by 2 cancelled. The reason

it canceled is slightly tricky. Also, remember that we are talking

about equivalence classes, not specific cube configurations. Any

equivalence class has both even and odd members, depending on how

the members are rotated. Hence, any corner equivalence class can be

matched up with any edge equivalence class, assuming the rotations

are compatible. But you still have to worry about "dividing by 2",

as follows.

Let G be the number of states of the whole cube without M, namely

the 4.3*(10^19) figure, and similarly let C be the number of states of

the corners without M and let E be the number of states of the edges

without M. Then, we have the trivial relation G = C * E / 2.

Here, the division by 2 does properly reflect the odd/even parity

of the corners vs. the edges.

Let Gm = G / (24*24*2), Cm = C / (24*24*2), and Em = E / (24*24*2). Hence, G = Gm * (24*24*2), C = Cm * (24*24*2), and E = Em * (24*24*2). What I have available (approximately) is Cm and Em, and what I want is Gm. Hence,

Gm = G / (24*24*2) Gm = (C * E / 2) / (24*24*2) Gm = ((Cm * (24*24*2)) * (Em * (24*24*2)) / 2) / (24*24*2) Gm = (Cm*24) * (Em*24)

Therefore, I replace Cm by the real figure for the number of corner

equivalence classes, replace Em by the real figure for the number

of equivalence classes, and Gm becomes the real figure for the total

states of the cube. The "division by 2" is in the formula, but it is

invisible because of all the cancellations.

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