From mreid@ptc.com Thu Apr 20 15:46:58 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 AA21486; Thu, 20 Apr 95 15:46:58 EDT Received: from ducie by ptc.com (5.0/SMI-SVR4-NN) id AA11510; Thu, 20 Apr 95 15:45:01 EDT Received: by ducie (1.38.193.4/sendmail.28-May-87) id AA19008; Thu, 20 Apr 1995 16:03:31 -0400 Date: Thu, 20 Apr 1995 16:03:31 -0400 From: mreid@ptc.com (michael reid) Message-Id: <9504202003.AA19008@ducie> To: cube-lovers@ai.mit.edu Subject: correction and an interesting example Content-Length: 1221 [ i sent a similar message several days ago, but it appears to have gotten lost. my apologies if anyone already got this. ] i wrote > in the same way, local maxima (within the antislice group) in the > 90 degree antislice metric are local maxima in the full cube group > (quarter turn metric). this isn't necessarily true. one must check that the minimal maneuvers (within the antislice group) for such positions are also minimal in the full group. the position i mentioned > (FB') (RL') (U'D) (R2L2) = [ ... ] is quickly checked to require 10 quarter turns, so indeed it is locally maximal. here's an example i found of a locally maximal position whose inverse is not locally maximal: A = B U2 F2 R U' R' B' R' U F2 U2 (15q) this position has six symmetries, generated by the cube rotation C_UFR and central reflection. using these symmetries we can give minimal maneuvers which end with a half turn of any face, and thus with any of the twelve quarter turns. the same is not true of its inverse, and we can easily check that there is no minimal maneuver for A which begins with the quarter turn B'. equivalently, the position A^-1 B' requires 16 quarter turns. mike