From mreid@ptc.com Mon Feb 20 16:38:19 1995 Return-Path: Received: from ptc.com (poster.ptc.com) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA05190; Mon, 20 Feb 95 16:38:19 EST Received: from ducie by ptc.com (5.0/SMI-SVR4-NN) id AA04548; Mon, 20 Feb 95 16:36:20 EST Received: by ducie (1.38.193.4/sendmail.28-May-87) id AA17023; Mon, 20 Feb 1995 16:51:10 -0500 Date: Mon, 20 Feb 1995 16:51:10 -0500 From: mreid@ptc.com (michael reid) Message-Id: <9502202151.AA17023@ducie> To: cube-lovers@ai.mit.edu, mouse@collatz.mcrcim.mcgill.edu Subject: Re: Qturn Lengths of M-Symmetric Positions Content-Length: 1407 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 related. 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. mike