From cube-lovers-errors@mc.lcs.mit.edu Sat Nov 22 22:56:58 1997 Return-Path: Received: from sun30.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.1/mc) with SMTP id WAA07129; Sat, 22 Nov 1997 22:56:57 -0500 (EST) Precedence: bulk Errors-To: cube-lovers-errors@mc.lcs.mit.edu Mail-from: From cube-lovers-request@life.ai.mit.edu Sat Nov 22 17:53:28 1997 From: roger.broadie@iclweb.com (Roger Broadie) To: Cc: "Tenie Remmel" Subject: Re: Rubiks Revenge moves Date: Sat, 22 Nov 1997 22:51:33 -0000 Message-Id: <19971122224927.AAA7296@home> Tenie Remmel wrote (19 November 1997 ) > Is there an easy way to cycle three adjacent top edges on the > Rubiks Revenge? I can't find one shorter than 62 moves, but if > there was a short one I could simplify my solution greatly. > > . b c . . a b . > a . . . => c . . . > . . . . . . . . > . . . . . . . . Rather than just throw a few more solutions into the pot, I'd like to start with some comments on the sort of process everyone, including me, seems to use to deliver 3-cycles of edge pieces in the 4x4x4. It is of the general form [P, TQT'] where the square brackets are used to show a commutator, that is, [A,B] means ABA'B'. In this process P and Q are turns of layers that are parallel to one another, and T is a turn of a layer transverse to P and Q. For instance, P and Q could be L and r and T could be U (capitals for the outer layers, lower case for the neighbouring inner layers, with sense parallel to the corresponding outer layer). That gives [L, UrU'] == L UrU'.L' Ur'U' which is a (not especially appealing) 3-cycle of edges. In fact any process of this form is a 3-cycle provided it takes one piece from the layer Q into the layer P. That will happen if T is a quarter turn in either sense - I haven't found anything useful with T as a half turn. But P and Q can be any power. The reason that processes of this form are 3-cycles is simple. If two permutations intersect at only one element, then their commutator is a 3-cycle. Thus if A = (...a1, x, a2...) and B = (...b1, x, b2...) then [A,B] -> (a1, x, b1) If you do just UrU' you will find there is a line of displaced pieces along the intersection Ub, but no other displaced piece in any of the layers parallel to r. Any of these pieces can be picked out to form part of a 3-cycle by selecting the layer that is parallel to r and contains the piece and using a turn of that layer as the component P of the commutator, with UrU' forming the component TQT'. In general, if all of P, Q and T are outer layers we will have a 3-cycle of corner pieces, if two are outer layers and one an inner layer we will have a 3-cycle of edge pieces, if one is an outer layer and the other two inner layers we will have a 3-cycle of centre pieces, and if all three are inner layers we will have done nothing visible to the cube, but in fact there will have been an invisible 3-cycle of the pieces of the imaginary internal 2x2x2. We can derive the last of these cases from the first quite neatly applying a fascinating concept called evisceration, which I recently met trawling through the archives. It was first quoted from David Singmaster's Cubic Circular by Stan Isaacs on 26 May 1983 and our present acting moderator also discussed it on 1 June 1983. If you turn a cube inside out by changing each outer layer in a process into an inner and vice-versa (i.e. capitals to lower case), then, in the effect of the process, you will interchange corner pieces with the pieces of the internal 2x2x2, and edge pieces with centre pieces. Making P, Q and T all to be outer layers gives just a 3-cycle of corner pieces; therefore applying evisceration takes that cycle into one on the pieces of the internal 2x2x2. Singmaster's Notes on Rubik's Magic Cube, the fifth edition, interprets processes of the type [P, TQT'] as [P,[T,Q]]. This expands to P TQT'Q'.P' QTQ'T', but the sequence Q'P'Q in the centre reduces to just P', giving the same expansion as before. Of course, the two components of the commutators TQT' and [T,Q] have different total effects, but what they have in common is that they put the same single piece into P. We can look at them both as being sort of like a mono-operation. Let's call it a "monopop": each process pops a piece into P; you then turn P, then reverse the pop operation, which extracts a different piece, and finally restore P. It's relatively straightforward to use this form to design specific processes. Say we want to move an edge piece from ULf to FLd and keep the third member of the 3-cycle in the top layer. Then we can take P to be L to achieve the required part of the cycle. We now know that Q must be in r or l. Let's take l. The transverse move in T has to take a piece from l into the point of intersection of the two components of the commutator, FLd. So it must be in F. Playing with F and l shows that the following does the job. [L, FlF'] == L FlF'. L' Fl'F' -> (ULf, FLd, UBl) If we'd taken Q to be in r we'd have needed a bit more care to keep the third piece of the 3-cycle in the top layer, but [L, F'r^2F] does, putting it at UBr. If we want to move a piece along a diagonal - from ULf to FLu, say - we need to use the other component of the commutator, TQT'. Thus we can build 3-cycles which include ULf to FLu around the component U'FU. For instance [f', U'FU] -> (ULf, LFu, RUf) With a clear head and a good following wind it's possible to work out these processes on the fly. They also transform nicely into another process of the same type by cycling the elements, which has the effect of conjugating the original process. Thus the last process can be dealt with as follows U[f', U'FU]U' = [Uf'U', F] -> (UFr, LFu, BUr) This cycling procedure comes from Singmaster. Let's now think about top-layer edge processes. I'll denote the pieces like this. X a1 a2 X d2 o o b1 d1 o o b2 X c2 c1 X The purpose of the numbering in pairs is to emphasize that the processes come in pairs. Each process has a twin created by changing each inner layer turn into its next-door inner neighbour. Thus the simplest U process of the general type we're using is of this form: [l, F'LF] -> (a1, c2, d1) Its twin is [r', F'LF] -> (a2, c1, d2) In the twin process, each edge piece is changed into its next door neighbour. We want to capture this regularity. I will therefore represent this pair by [M', F'LF] -> (a*, c', d) In this representation, M is either r or l', the asterisked piece defines the layer that contains M and primed letters denote a piece with the opposite suffix number to the asterisked piece. Obviously, these suffixes are closely related to flip in a 3x3x3 and the assignment of the numbers is arbitrary. Some assignments are more helpful than others in a particular context, and the method used in the diagram above is the obvious one of giving the same number ("flipperty"?) to the pieces in the positions that a single piece moves into during a complete U turn. Here, then is a complete set of top layer 3-cycles of edge pieces, to within a reflection. It comes from a fairly systematic search I did for processes of the type [P, TQT'] that can be conjugated by at most one turn into a top-layer process. They are oriented in a way I find easy to do. I will leave them as commutators, because it is very easy to perform the full set of turns from them. The T/T' sequence remains constant for both halves and the only adjustment needed is to invert P and Q the second time around. Inverses are also easy to perform, since all one has to do is read off the second component first. (a*, c', d) [M', F'LF] (a*, b, c) F2 [D R2 D', M2] F2 (c*, a, b') R2 [M' D' M, U2] R2 (d1, b2, b1) (Bb)' [U l U', R2] (Bb) (d2, b1, b2) (Ff) [U r' U', R2] (Ff)' (d2, c2, d1) (Bb) [L2, D l D'] (Bb)' (Tenie's question) (d1, c1, d2) f' [L2, D r' D'] f The last two pairs could have been left in the M form if say N was introduced to represent either f or b'. But keeping them separate lets us save a wrist movement for the first three by combining the inner and neighbouring outer layers for a turn relative to the central cut. That won't work for the final process, since the F layer is already included in the 3-cycle and can't be amalgamated with the f layer. All these process can be directly transferred to the 3x3x3, using the one single central slice as M (or l' or r). The primes then correspond to flipping the edge piece relative to the top surface. Roger Broadie 22 November 1997