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 meMe 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.