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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