[next] [prev] [up] Date: Mon, 30 Oct 95 13:47:59 +0100
[next] [prev] [up] From: Michiel Boland <boland@sci.kun.nl >
~~~ ~~~ [up] Subject: Exchanging just four edges in antislice impossible?

Hello all,

can anyone provide an easy proof of the fact that it is
impossible to exchange just four edges using just antislice
moves, whilst leaving everything else fixed? (We're talking
about the 3x3 cube of course.)

Another way of putting it: why are the 2xH and 4-dot patterns
not in the antislice group?

I have thought about this a little, but not hard enough to find
an answer. I looked it up in Singmaster's Notes but could not
find a satisfying explanation either.

Here is some more background.

The antislice group is contained in the group of all positions
that are symmetric under `cube half-turns' (the subgroup of M
containing I,(FB)(LR),(FB)(UD) and (UD)(LR)). This group has
(8*4*12*8*4*3*2^2)/2 = 73728 elements.

It can be shown that in the antislice group, the orientation of
the corners is determined by the edge positions [I am willing to
explain this, but it is much easier visualized than written
down], which means that the antislice group contains at most
73728/3=24576 elements.

But apparently the antislice group contains just 6144 elements,
which is a factor 4 below the abovementioned number. This factor
4 is explained by the fact above, which I am trying to prove.
Michiel Boland <boland@sci.kun.nl>
University of Nijmegen
The Netherlands

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