From walts@federal.unisys.com Tue Apr 30 07:42:15 1996 Received: from www.han.federal.unisys.com by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA12966; Tue, 30 Apr 96 07:42:15 EDT Received: from homer.MCLN.Federal.Unisys.COM by www.han.federal.unisys.com (8.6.12/mls/8.0) id HAA14387; Tue, 30 Apr 1996 07:42:12 -0400 Return-Path: Received: from h3-105.MCLN.Federal.Unisys.COM by homer.MCLN.Federal.Unisys.COM (8.6.12/mls/4.1) id HAA22489; Tue, 30 Apr 1996 07:45:42 -0400 Message-Id: <318626DD.291E@federal.unisys.com> Date: Tue, 30 Apr 1996 07:42:38 -0700 From: Walt Smith X-Mailer: Mozilla 2.0 (Win16; I) Mime-Version: 1.0 To: cube-lovers@ai.mit.edu Subject: Rubiks Revenge X-Url: http://home.netscape.com/ Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Rubiks Revenge presents several difficult cases when the last few pieces are reached. There has been discussion on this over the years but it does not seem to have reached closure. Mark Longridges list of best known solutions, (http://admin.dis.on.ca:80/~cubeman/cmoves.txt) lists some of these but here are some improved solutions of my own and additional operators that are needed to solve it. If the center four are done on all sides and the edge pieces are paired up, it can be can be treated like a 3x3x3 Rubik Cube. Several cases come up at the end that can not occur on a 3x3x3. If the corners are done and the last few edges are left for last, four cases occur. Here are my methods that are shorter than any others I have seen. Case 1. The last three edges can be solved with 3x3x3 techniques. Case 2. Two edge pairs are swapped. (swaps RF and RB) d2 R2 d2 rR2 d2 r2 (7) This is based on combining the following two sequences: (d2 R2)2 d2 and (d2 r2)2 Each of these is useful in itself. This is shorter than Marks Longridges p4. (Marks p4 is mislabeled Opp. It should be Adj and p5 should be labeled Opp) Case 3. A single edge pair is flipped. (flips BL) L2 d1 R2 d1 R2 d3 L2 u3 B2 u2 B2 u3 B2 R2 B1 r3 B3 R2 B1 r1 B1 (21) This is shown in four groups because it proceeds in stages. The second move fixes the parity, the first along with 3rd through 12th fix the faces and the last group of moves fix the edges. This is shorter than Marks p3. Case 4. Both Case 2 and Case 3 exist. The flipped edge pair might be one of the swapped edge pairs or it might not be. Obviously this can be solved by using the techniques of Case 2 and 3 applied separately. I have always thought that it should be possible to find an operator that is shorter than the sum of these two and possibly shorter that Case 3. I have not done as much study on this case as the others. If you want to solve the corners last (avoids Case 3 and 4), you may still need to solve Case 2 but you may also need to swap two corner pieces. This can be done by applying Case 2 then fixing the edges then corners. This will take about 40 turns. The following does it quicker. (swaps LDB and LDF with the bottom cubie faces remaining bottom faces.) R3 D1 L1 D3 R1 D2 L3 D1 L1 D2 L1 U3 r2 F2 r2 fF2 r2 f2 U1 L2 (21) It is listed in three groups to make it easier to memorize as it proceed in three stages. This is shorter than the sequence in Mark Jeays on-line solution at http://qlink.queensu.ca/~4mj/rr.html ( or link from Mark Longridges home http://admin.dis.on.ca:80/~cubeman/ ) I do not have any books on Rubiks Revenge. If someone out there does, please see how these solutions compare. If anyone has shorter solutions, either that they have developed or found in books please submit them so Mark L. can improve his list (and so I can improve my technique). Also Marks p1 does not work. Can anyone fix it? Walter Smith near Washington D.C. walts@federal.unisys.com