[next] [prev] [up] Date: Mon, 24 May 82 17:19:00 -0400 (EDT)
[next] [prev] [up] From: Allan C. Wechsler <ACW@MIT-AI >
~~~ ~~~ [up] Subject: |4^3| = 1.7*10^55
The corner group of 4^3 is exactly like that of the 3^3.  It has
8!*3^7 elements.

Now to calculate the order of the whole group, we have to find
the order of the corner stabilizer group. That's the group of
moves that leave the corners fixed.

SO let's start by thinking about the edge group of the corner
stabilizer. I personally have a tool that exchanges two edge cubies
without moving corners. Since the edge group is transitive, I can
exchange any two edges without moving corners.

You cannot flip any edge of the 4^3 without moving it. So there's no
edge-flippage in the 3^3 sense on the 4^3. That means that the edge
group in the corner stabilizer has order 24!.

Now all there is left is compute the order of the edge AND corner
stabilizer. The inside twists (slices) are center-even. The outside
twists (faces) are center-odd, but they are also corner-odd, and so if
we want to bring the corners home we have to make an even number of
outside twists. That means that all the center-permutations in the
edge-and-corner stabilizer are even. But how many of these can be

I have a tool that exchanges two pairs of centers. I think (but I
haven't yet proved) that the center group is 4-transitive, so that
ANY two pairs of centers can be exchanged. That means that all the
even permutations of centers can be done without perturbing edges and
corners. Hence the size of the edge-and-corner stabilizer is 24!/2.

As for supergroupiness, Dave Plummer just pointed out to me that
once you know a center's position, you know its orientation, since
center cubies always keep one corner pointing to the middle of the
face. So 4^3 has no supergroup.

So the order of the 4^3 group is 8!*3^7*24!*24!/2.  In numbers, if you



or about 1.7*10^55.

Now that number is a little deceptive, because it includes whole-cube
rotations. The 4^3 has no nice fixed reference frame like the 3^3 has.
If you don't want to count whole-cube rotations you have to divide by
24, to get


or about 7.1*10^53.

Finally, we have to face the fact that the center cubies are
in six nonintradistinguishable sets of four. (All the edge cubies are
distinguishable by color or orientation). So we have to divide our
last result by 4!^6/2. That leaves


or about 7.4*10^45 distinguishable color patterns. Remember that these
do not form a group.

I leave it as an exercise for Hoey and Saxe to find a lower bound for
the diameter of the group.


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