From reid@math.berkeley.edu Mon May 11 20:31:30 1992 Return-Path: Received: from math.berkeley.edu by life.ai.mit.edu (4.1/AI-4.10) id AA17723; Mon, 11 May 92 20:31:30 EDT Received: from jacobi.berkeley.edu.berkeley.edu by math.berkeley.edu (4.1/1.33(math)) id AA22739; Mon, 11 May 92 17:31:10 PDT Date: Mon, 11 May 92 17:31:10 PDT From: reid@math.berkeley.edu (michael reid) Message-Id: <9205120031.AA22739@math.berkeley.edu> To: Dik.Winter@cwi.nl, cube-lovers@life.ai.mit.edu Subject: Re: More on the Cube (2x2x2 in this case). > Singmaster states that the diameter of the group for the 2x2x2 cube is not > known. in his "cubic circular," issue(s) 5/6 (pages 26, 27) he gives some info about this, although it is (at least) third hand information, and therefore not necessarily reliable. > I just did calculate it. ... > If we allow half-turns: > 1 with 0 moves > 9 with 1 moves > 54 with 2 moves > 321 with 3 moves > 1847 with 4 moves > 9992 with 5 moves > 50136 with 6 moves > 227536 with 7 moves > 870072 with 8 moves > 1887748 with 9 moves > 623800 with 10 moves > 2644 with 11 moves he gives the same figures, so they are probably correct. > If we do not allow half-turns: > 1 with 0 moves > 6 with 1 moves > 27 with 2 moves > 120 with 3 moves > 534 with 4 moves > 2256 with 5 moves > 8969 with 6 moves > 33058 with 7 moves > 114149 with 8 moves > 360508 with 9 moves > 930588 with 10 moves > 1350852 with 11 moves > 782536 with 12 moves > 90280 with 13 moves > 276 with 14 moves he does not give these, but he does mention that the diameter is 14. > BTW, calculation did not take very long, only a few (<3) minutes on an FPS singmaster says that the calculation took "over 51 hours of computer time"! ouch! this was 10 years ago, though. (what's 51 hours / 3 minutes ?) the "unix news item" from which singmaster apparently got his info was included in a cube-lovers message. it's in the archives, cube-mail-3, sept 15, 1981 in a message from "ISAACS at SRI-KL". dik's results show that the corners of the 3x3x3 can be "solved" (i.e. positioned correctly with respect to one another) in 11 face (respectively 14 quarter) turns. it would be nice to know if they can be solved with respect to the centers within 11 face (respectively 14 quarter) turns. this seems likely. mike