From mouse@collatz.mcrcim.mcgill.edu Mon Feb 20 15:37:03 1995 Return-Path: Received: from Collatz.McRCIM.McGill.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA01812; Mon, 20 Feb 95 15:37:03 EST Received: (root@localhost) by 23249 on Collatz.McRCIM.McGill.EDU (8.6.8.1 Mouse 1.0) id PAA23249 for cube-lovers@ai.mit.edu; Mon, 20 Feb 1995 15:36:54 -0500 Date: Mon, 20 Feb 1995 15:36:54 -0500 From: der Mouse Message-Id: <199502202036.PAA23249@Collatz.McRCIM.McGill.EDU> To: cube-lovers@ai.mit.edu 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 mouse@collatz.mcrcim.mcgill.edu