[next] [prev] [up] Date: Fri, 17 Dec 93 11:21:51 -0500 (EST)
[next] [prev] [up] From: Jerry Bryan <BRYAN%WVNVM.BITNET@mitvma.mit.edu >
[next] ~~~ [up] Subject: Size of the Cube Group

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        6624

The 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?

[next] [prev] [up] [top] [help]