From cube-lovers-errors@mc.lcs.mit.edu Fri Mar 12 00:15:29 1999 Return-Path: Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38]) by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id AAA16151 for ; Fri, 12 Mar 1999 00:15:29 -0500 (EST) Precedence: bulk Errors-To: cube-lovers-errors@mc.lcs.mit.edu Date: Wed, 10 Mar 1999 22:19:21 -0500 (Eastern Standard Time) From: Jerry Bryan Subject: Re : Re: Edges only, Ignoring Flips, Face Turn Metric In-Reply-To: <199903050409.XAA24597@cauchy.math.brown.edu> To: michael reid Cc: Cube Lovers Message-Id: On Thu, 04 Mar 1999 23:09:00 -0500 michael reid wrote: > jerry writes > > > I have completed a God's Algorithm run in the face turn metric for the > > group consisting of edges only ignoring flips. The size of the group is > > therefore 12! The results are as follows: > [ ... ] > > very interesting. i hope that you'll also do the quarter turn metric. > I have completed a run out to 11q (took a long time). Regrettably, the diameter proved to be greater than 11q. I now have a run in progress out to 12q. It's going *very* slowly. The problem I described where my method is inefficient calculating the tail of the distribution is even worse for the quarter turn metric than for the face turn metric for this particular problem. Also, to calculate to 11q or 12q I have to store all the positions out to 6q, which I can do. I don't think I can store out to 7q on the computer I have. Even if I could, a run to 13q or 14q would be too slow, I think. I know from parity considerations that the diameter is greater than 12q, so in some ways my run to 12q is a fool's errand. That is, there are less than 12!/2 odd positions through 11q, so there must be at least a few at 13q. My only hope is that all the even positions will show up by the time I get to 12q. If so, I would know that the rest of the odd ones must be at 13q. Otherwise, I am doomed for now. I have an idea of how to approach the inefficiency in the tail. Since I am calculating ends-with for each position I calculate, I know for sure for each position I calculate which quarter-turns go further from Start and which go closer. The idea is that once I get to the tail of the distribution, I once again begin storing calculated positions in memory (those which are at the maximal distance which I am able to calculate). From there, I continue further from Start in a more traditional fashion, leaping one level at a time rather than many levels at a time. This works because I have knowledge of which quarter turns go closer to Start, and hence I don't have to worry about comparing against those positions closer to Start which I am not able to store. If I had time to put this plan into action, the run time for the tail of the distribution should be only a few minutes or a few seconds. ---------------------------------------- Jerry Bryan jbryan@pstcc.cc.tn.us