From cube-lovers-errors@oolong.camellia.org Wed Jun 25 17:18:25 1997 Return-Path: cube-lovers-errors@oolong.camellia.org Received: from oolong.camellia.org (localhost [127.0.0.1]) by oolong.camellia.org (8.6.12/8.6.12) with SMTP id RAA09126; Wed, 25 Jun 1997 17:18:25 -0400 Precedence: bulk Errors-To: cube-lovers-errors@oolong.camellia.org Date: Wed, 25 Jun 1997 22:14:32 BST From: David Singmaster Computing & Maths South Bank Univ To: cube-lovers@ai.mit.edu CC: notation@vax.sbu.ac.uk, for@vax.sbu.ac.uk, 4^3@vax.sbu.ac.uk, and@vax.sbu.ac.uk, 5^3@vax.sbu.ac.uk Message-ID: <009B653D.9EE57540.331@vax.sbu.ac.uk> Subject: notation for 4^3 and 5^3 David Barr has described some moves for the 5^3 using his own notation. In the early 1980s, I extended my 3^3 notation to the 4^3 and this can be used on the 5^3 as well. I described this in my Cubic Circular but perhaps it would be useful to give it here. I will describe the situation for the 4^3. On the 5^3, we don't ever have to make a slice move of just a central layer. Further, a combination of 4^3 and 3^3 processes will solve the 5^3, so we don't really need to label the central layers. Consider the four layers from L to R. I denote the inner layers by l and r. So the four layers are: LlrR. Similarly we have four layers: FfbB and UudD. To describe a piece requires more effort than before, but each piec lies in three layers and we can describe the piece by these layers. E.g. FUR is a corner piece; FUr is an edge piece, lying on the FU edge - but there are two of these and they are distinguished as being FUr and FUl (I usually give the layers in clockwise order, but it is not essential and there are times when it is more informative to use the other order.); Fur is a face-centre piece, the one in the upper right corner of the inner four cells of the F face. If you have been paying attention, you will ask about fur. This is one of the body-centre cells, invisible to you unless you make a transparent cube! Using the standard notation of [F, R] = FRF'R', we find a number of easy 3-cycles. [[F,R],L] = (FLU,ULB,RFU) [[F,R],l] = (FlU,UlB,DfR) (I've copied this from my Circular, but I wonder if it's right as I thought there'd be some symmetry with the preceding??) [[F,r],l] = (Flu,Ulb,Drb) In theory, these and a careful consideration of parity are sufficient to solve the 4^3 and the 5^3. However, the parity problem is a bit awkward. In my original approach, I got the corners in place and then all edges except leaving the four edges along the FU and BU edges. Examine the parity of these carefully. If they are in an odd permutation, apply r^2 D^2 l' D^2 r^2 which 4-cycles these four edges and moves some centres. Once the parity is corrected, there is little difficulty restoring the rest of the cube. For the 5^3, once you have paired up the edges, one can solve the central edges by treating the 5^3 as a 3^3 with fat slices. To correct a single pair of edges, one can use the following. rrDDl'DDrr rrD'RR [[R,U],l'] RRDrr = rrDDl'DR'UR'U'l'URU'lRDrr = (UBl,UBr) (Ful,Ubl,Bdl,Ufr) (Fdl,Ufl,Bul,Ubr). This messes up some centres, but they are not too hard to restore. Indeed applying rrUUr (uurrll)^2 r'UUrr wil correct the F and B centres disturbed by the above, leaving a 180 degree rotation of the four U centres. After I had developed the notation and solution, a Peter Lees pointed out an unexpected feature. The exchange of upper and lower case letters is a duality. The dual of URF is urf, while the dual of URf is urF. This gives us a Pricniple of Duality: The dual of a sequence of moves is the same process on the dual pieces. E.g., we had [[F,R],l] = (FUl,UBl,DRf), so [[f,r],L] = (fuL,ubL,drF). This duality allows one to transfer a number of 3^3 processes to 4^3 processes and to solve the invisible interior part of the cube! By always moving an outer layer with its inner layer, one is obviously simulating the 2^3. However, if one always moves, say R and l together, one is also simulating the 2^3 in eight copies! Ooops, one wants to move R and l' together. If one moves R and l together, I think you get eight versions of the 2^3, but each is a reflection of its neighbours! If you are tired of thinking about God's Algorithm on the 3^3, try the 4^3. I'm not even sure how to count moves. E.g., to do r, one normally does Rr and then R', so does r count as one move or two? Likewise, does Rr count as one move or two? Enough for now. DAVID SINGMASTER, Professor of Mathematics and Metagrobologist School of Computing, Information Systems and Mathematics Southbank University, London, SE1 0AA, UK. Tel: 0171-815 7411; fax: 0171-815 7499; email: zingmast or David.Singmaster @sbu.ac.uk