From @mail.uunet.ca:mark.longridge@canrem.com Wed Sep 13 02:15:29 1995 Return-Path: <@mail.uunet.ca:mark.longridge@canrem.com> Received: from seraph.uunet.ca (uunet.ca) by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA01159; Wed, 13 Sep 95 02:15:29 EDT Received: from portnoy.canrem.com ([198.133.42.17]) by mail.uunet.ca with SMTP id <210967-2>; Wed, 13 Sep 1995 02:17:54 -0400 Received: from canrem.com by portnoy.canrem.com (4.1/SMI-4.1) id AA14615; Wed, 13 Sep 95 02:11:18 EDT Received: by canrem.com (PCB-UUCP 1.1f) id 1F4769; Wed, 13 Sep 95 01:47:39 -0500 To: cube-lovers@life.ai.mit.edu Reply-To: CRSO.Cube@canrem.com Sender: CRSO.Cube@canrem.com Subject: Alexander's Star From: mark.longridge@canrem.com (Mark Longridge) Message-Id: <60.1229.5834.0C1F4769@canrem.com> Date: Wed, 13 Sep 1995 02:21:00 -0400 Organization: CRS Online (Toronto, Ontario) > From: pbeck@pica.army.mil (Peter Beck) > Subject: ranking > > noticed you did not include > square-1 or alexanders star > > thanks for the compilation Very little literature exists on Alexander's Star. Ideal Toy published a solution booklet, and Adam Alexander (the puzzle's inventor) wrote a book, and David Singmaster wrote his analysis in one of his Cubic Circulars. Dr. Singmaster also mentions Alexander's Star in Rubik's Cubik Compendium in a chapter about variations on Rubik's cube. The Ideal booklet mentions that Alexander's Star has 24 start positions, and Dr. Singmaster states there are 12 start positions, each one with it's own mirror reflection, again giving 24 start positions in total. Alexander's Star is akin to an "edge-only" Dodecahedron (Megaminx) without centres. It has 30 edges, that is 15 pairs of distinct 2 coloured edges. To calculate the number of permutations of N objects with N1 objects which are like, N2 objects which are like ... Nrth objects which are like we use the formula: N! -------------------- N1! * N2! * ... Nr! In the case of Alexander's Star it is a bit more complicated because we must contend with edge flips and no centres, however I believe the correct calculation is.... 30! / 2^15 * 2^29 / 60 = 72,431,714,252,715,638,411,621,302,272,000,000 approx = 7.2 * 10^34 I think the term for this is approximately 72 decillion. (decillion = 10^33) Here is a bit of an explanation.... There are 30 total objects, with 15 different pairs of 2 like edges. 29 edges may be flipped in any fashion but the 30th edge is forced. The Great Dodecahedron has 60 orientations in space. Here is another way to calculate the same number... Pick one edge and lock it in position and orientation. We still have 29 edges which can move in position and orientation. There are still 28 edges which can be flipped freely, the 29th edge being forced. We still have 15 different pairs of 2 like edges. 29! / 2^15 * 2^28 = 7.2 * 10^34 (approx) To be completely clear the calculation is really 29!/2 / 2^15/2 * 2^28 ....because only half of the 29! arrangements are possible and only half of the 2^15 arrangements are possible but both the /2's cancel. On the Great Cosmic Ranking List this puzzle has less combinations than Rubik's Revenge, but more than the VIP sphere. I know of no Square 1 calculations whatsoever, but would be interested in seeing anyone's calculations. -> Mark <-