From dik@cwi.nl Tue Jan 7 16:13:50 1992 Return-Path: Received: from charon.cwi.nl by life.ai.mit.edu (4.1/AI-4.10) id AA07462; Tue, 7 Jan 92 16:13:50 EST Received: by charon.cwi.nl with SMTP; Tue, 7 Jan 1992 22:13:46 +0100 Received: by boring.cwi.nl ; Tue, 7 Jan 1992 22:13:43 +0100 Date: Tue, 7 Jan 1992 22:13:43 +0100 From: Dik.Winter@cwi.nl Message-Id: <9201072113.AA06247@boring.cwi.nl> To: cosell@bbn.com, cube-lovers@life.ai.mit.edu Subject: Re: Hungarian Rings solution? > Dik.Winter@cwi.nl writes: > } You don't nood commutators for it, cycles are sufficient (because there > } are so many similar colored beads).... I have already been chastised that what I described are commutators. Of course they are. Not only is my thinking bad late at night, but apparently my spelling is atrocious :-). > As far as I can tell, basically ANY set of ring-turns that has a total > movement of zero seems to define a pretty small cycle. For example, > the sequence LnA RnA LnC RnC, for n not a multiple of 5[*], does a > three-bead cycle: if you look at the upper intersection: > A C > Intersection ---> C ======> B > B A > Where 'A' and 'B' are each n beads away from the intersection [and by > changing theorder of L/R you reverse the cycle, and by interchanging A > and C you move the cycle to the other side of the intersection. > BUT: the problem is that this isn't really a 3-cycle, but rahter _two_ > 3-cycles: you also make a central-symmetric move of the beads at the > bottom intersection. True. But if you prefix the move by a (series of moves) that makes the upper three of an identical color (and postfix by its inverse), you will not see the difference between a true cycle. At least that is how I always did the final part. (Correctly coloring the two lobes is in fact easy; you better start with that.)