[next] [prev] [up] Date: Mon, 20 Feb 95 16:51:10 -0500
[next] [prev] [up] From: michael reid <mreid@ptc.com >
~~~ [prev] [up] Subject: Re: Qturn Lengths of M-Symmetric Positions

der mouse writes

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].

we've had this discussion before. pons asinorum does not "commute with
everything". however, it does commute with "slices" and with cube
symmetries. (which is all you're really using here.)

e. (RL')(FB')(RL')     a surprise to me
Me too.  [ ... ]

the only reason i'm not surprised here is that i've read the archives.
dan hoey gave this maneuver on jan 7 1981. also, it was recently
mentioned by chris worrell (on dec 16 1994).

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?

er, i think he means that "many" (i.e. 4) of the 5 maneuvers are closely

however, i disagree with jerry's conjecture that the maneuvers for pons
asinorum composed with superflip will be closely related.

i guess we'll know for sure fairly soon.


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