From:

Subject:

In rsponse to my request for info about the Hungarian Rings,

Dik.Winter@cwi.nl writes:

} You don't nood commutators for it, cycles are sufficient (because there

Dik.Winter@cwi.nl writes:

} You don't nood commutators for it, cycles are sufficient (because there

} are so many similar colored beads)....

My apologies --- I meant to say "cycles" when I said I had found lots of

them... And I hate to seem dense, but but I'm still stuck...

} ... If I remember right one useful move

} is: turn right ring clockwise two beads, turn left clockwise two beads,

} turn right anti-clockwise two beads, turn left anti-clockwise two beads.

} Using them properly will solve the rings.

The 'properly' is the part I'm finding hard. There seem to be LOTS of

cycles, but even with that big choice I can't see, quite, how to solve

the thing.

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.

[*] since five is the distance between the intersections, if the

rotate is a muiltiple of 5 the intersections interact, things get a

little different: it makes a *two* cycle! In the diagram above

[with A five away from C], the move just _swaps_ A & C [and the A'

and C' at the lower intersection, too, of course].

Given that my rings are totally non-symmetrically messed up, I can't

figure out a plan for making forward progress. I can do lots of

diffent cycles, but I can't manage to get the rings set up so that the

cycle at both intersections is useful: if I try to fix something at the

top intersection I invariably mess up something at the bottom one.

Thanks again for you patience with my rantings. I feel like I'm

overlooking something simple [since this wasn't supposed to be all that

hard a puzzle], but I don't see what it is ...

/Bernie\