From dik@cwi.nl Tue Jan 7 16:13:50 1992
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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.)