Date: Tue, 07 Jan 92 15:02:18 -0500 (EST)
From: Bernie Cosell <cosell@bbn.com >
Subject: Re: Hungarian Rings solution?

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,
} 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\