From:

~~~ Subject:

> 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.)