From boland@sci.kun.nl Mon Oct 30 07:48:01 1995 Return-Path: Received: from wn1.sci.kun.nl by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA06686; Mon, 30 Oct 95 07:48:01 EST Received: from canteclaer.sci.kun.nl by wn1.sci.kun.nl via canteclaer.sci.kun.nl [131.174.132.34] with SMTP id NAA19553 (8.6.10/2.14) for ; Mon, 30 Oct 1995 13:48:00 +0100 Message-Id: <199510301248.NAA19553@wn1.sci.kun.nl> To: cube-lovers@ai.mit.edu Subject: Exchanging just four edges in antislice impossible? Date: Mon, 30 Oct 95 13:47:59 +0100 From: Michiel Boland 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 University of Nijmegen The Netherlands