[next] [prev] [up] Date: Mon, 20 Feb 95 15:36:54 -0500
[next] [prev] [up] From: der Mouse <mouse@collatz.mcrcim.mcgill.edu >
[next] [prev] [up] Subject: Re: Qturn Lengths of M-Symmetric Positions
2. The length of Pons Asinorum is of course 12 qturns.  [...]
      d. (FB')(RRLL)         a surprise to me

Surprising, but explicable. Write PA as (RRLL)(UUDD)(FFBB). Since PA
commutes with everything, [PA](FB') = (FB')[PA]. (I am writing X Y to
mean sequence X followed by sequence Y.) Note also that [PA](F'B) =
(RRLL)(UUDD)(FB'). But then [PA] = [PA](FB')(F'B) = (FB')[PA](F'B) =
(FB')(RRLL)(UUDD)(FB') gives us a length-12 process for PA whose first
half is what you found.

e. (RL')(FB')(RL')     a surprise to me

Me too. By elimination, its second half must be M-equivalent to the
first half, since we can look at these five half-processes as equally
being M-representatives of the reversals of the PA second halves, and
the other four first halves' second halves account for the other four.
(Got that? :-) In fact, each first half is M-equivalent to the
reversal of the corresponding second half, which is pleasing.

After juggling letters for a while, I've been unable to "justify" this
process the way I did the one above, which leads me to suspect it may
be a fundamentally new process for PA. Amazing, the things you can
find when you're not looking for them. :-)

3. The length of Pons Asinorum composed with Superflip is 20 qturns.
[...] I expect we will find that many (or all) of [the midway
positions for this] are really closely related, differing only by
commuting in fairly trivial ways, just as do the five half-way
positions for Pons Asinorum.

Does this mean you see the fifth process for PA as a trivial
commutation of the known PA process? How?

der Mouse


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