From @mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU Sun Feb 13 16:59:40 1994 Return-Path: <@mitvma.mit.edu,@WVNVM.WVNET.EDU:BRYAN@WVNVM.WVNET.EDU> Received: from mitvma.mit.edu by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA17813; Sun, 13 Feb 94 16:59:40 EST Message-Id: <9402132159.AA17813@life.ai.mit.edu> Received: from MITVMA.MIT.EDU by mitvma.mit.edu (IBM VM SMTP V2R2) with BSMTP id 1093; Sun, 13 Feb 94 16:59:35 EST Received: from WVNVM.WVNET.EDU (NJE origin MAILER@WVNVM) by MITVMA.MIT.EDU (LMail V1.1d/1.7f) with BSMTP id 4178; Sun, 13 Feb 1994 16:59:35 -0500 Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.1d/1.7f) with BSMTP id 1509; Sun, 13 Feb 1994 16:59:25 -0500 X-Acknowledge-To: Date: Sun, 13 Feb 1994 16:59:22 EST From: "Jerry Bryan" To: "Cube Lovers List" Subject: Re: Some Proposed Terminology In-Reply-To: Message of 01/21/94 at 18:32:15 from hoey@AIC.NRL.Navy.Mil On 01/21/94 at 18:32:15 hoey@AIC.NRL.Navy.Mil said: >I welcome Jerry Bryan's efforts to >improve the terminology of the groups associated with Rubik's cube. >But there is some additional clarification I think is necessary. >> Let G\C be the corners with edges without centers group. I intend >> for the notation to indicate G reduced by C, where C is the rotation >> group for the cube.... >> Let GC\C be the corners without edges without centers group.... >> Let GE\C be the edges without corners without centers group.... >First, these are not, strictly speaking, groups. Well, you can make >them groups, by defining what the group operation is. But I don't >know any way of doing that without losing the symmetrical nature of >the problem. >Second, I would suggest that G/C, GC/C, and GE/C are more standard >names for these objects. The elements are nominally 24-element sets, >each of which is an equivalence class when two positions are >considered equivalent when they differ by their position with respect >to the corners. The classes are called the cosets of C in G, GC, and >GE, respectively. Dan Hoey's criticism's are quite valid. I will attempt to repair the damage as follows: 1) accept the Gx/C notation in lieu of Gx\C, 2) define an operation within Gx/C such that Gx/C is a group, and 3) use Gx/C as a model for cubes without centers in such a way that the symmetrical nature of the problem is retained. Let C be the set of twenty-four whole cube rotations of the cube, and let G be the standard 3x3x3 cube group. We observe that if X is a cube in G, then c'Xc is also a cube in G for every c in C. We could call this operation C-conjugancy. However, there is seldom (if ever) any reason to speak of C-conjugancy. That is, C is just a subset of M, the set of forty-eight whole cube rotations and reflections. Indeed, C is half of M, and the other half of M is the reflection of C. Hence, M-conjugancy of the form m'Xm is more powerful than C-conjugancy, and there is normally no reason to speak of C-conjugancy. I only bring it up to emphasize that if X is in G, then c'Xc is in G. On the other hand, if I understand correctly the model most people use for G, elements of the form Xc or cX are not in G except for the trivial case where c=I. The problem is that C is considered to move the centers, but G is generated by Q, the set of quarter-turns of the faces, and Q does not move the centers. For example, there is a c in C such that F=c'Rc, but there is not a c in C such that F=Rc or F=cR. And indeed, neither Rc nor cR are in G at all unless c=I. As we said, G is generated as G=, where Q is the set of quarter-turns Q={F,B,U,D,L,R,F',B',U',D',L',R'}. Elements of Q move the corners and edges, but Q is the identity on the centers. C, on the other hand, is generally considered to move the centers. Hence, the group generated as is a supergroup of G, and there are elements of the supergroup which are not in G. (This supergroup, by the way, is not The Supergroup. The Supergroup is generated by Q alone, but with orientations of the (otherwise fixed) centers considered.) Therefore, our first order of business is to make C into a sub-group of G. We observe that since the elements of Q are the identity on the centers, the primary function of the centers is to provide a frame of reference. But we can provide a frame of reference without the centers actually being there. For example, consider the group GC consisting of cube centers and corners. You can model this group by removing the edge labels from a physical cube. Establish the cube at Start and perform RL'. The corners will be rotated forward, and will be positioned properly with respect to each other, but the cube is clearly not solved. You can tell that the cube is not at Start because the corners are not aligned properly with the centers. Now, do the same thing except remove both the edge and center labels. If you perform RL' at Start, the cube "looks" solved but rotated forward. However, we can adopt the convention that the cube is solved only if the Up color is Up, the Front color is Front, etc. With this convention in place, RL' is clearly seen not to be solved; it is two moves from Start. The convention provides the fixed frame of reference. Furthermore, RL' (which is in GC) is equal to an element of C, and indeed all elements of C are in GC, as are all elements of the form Xc or cX for c in C and X in GC. Hence, we have =. Similar comments apply to GE, the group of edges and centers, except that processes composed from elements of Q to accomplish rotations in C are not quite so short in GE as they are in GC. G, the full 3x3x3 cube group consisting of corners, edges, and centers is a bit more difficult. The problem is that if X is in G, then objects of the form Xc or cX are in G only if c is even. Twelve elements of C are even and twelve are odd. Indeed, C[even] is a sub-group of C, but C[odd] is not. We will deal with this situation (as circumstances require) in two different ways. One is simply to restrict ourselves to C[even] when dealing with G. The other is to define a new group we will call GS. In our model for G in which the centers are implied by a frame of reference convention rather than by actual physical centers, we can easily add slice moves to the standard face moves. If the centers were physically present, then the slice moves would move the centers, but without the physical centers there is no problem. If S is the set of slice moves, then GS is generated as . GS is essentially G with parity restrictions removed. Hence we observe that |G|=|GC|*|GE|/2, |GS|=|GC|*|GE|, and |GS|=|G|*2. Also, if X is in G or in GS, then elements of the form cX or Xc are in GS for all c in C. In those occasions where we are willing to think of GS rather than G, we can use C rather than C[even]. At this point, we can say that GS/C, G/C[even], GC/C, and GE/C are cosets of C in GS, C[even] in G, C in GC, and C in GE, respectively. To be a little more conformant with standard coset notation, we will write cube elements as lower case letters for the remainder of this note, and hence for a particular cube x a coset of C is denoted as Cx={y: y=cx} or xC={y: y=xc}. Now, we propose a group operator for the cosets: Cx Cy = C(xy) and xC yC = (xy)C. Showing that we have a group is easy. I originally included a proof in this note, but there is a proof in Chapter 8 of Frey and Singmaster's _Handbook of Cubik Math_. Hence, I will defer to their proof instead. According to Frey and Singmaster, G/C is called the factor group of C in G, or the quotient group of G by C. Of most significance to us right now is the fact that the identity of the factor group is Ci or iC, where i is the identity of G. But Ci or iC is just C. Hence, the identity of the factor group is C. This justifies our identification of G/C with a centerless cube. In English, it means that we can rotate a centerless cube in space without changing anything. I think this would comply with most people's intuitive sense of what it means for a cube to be centerless. Finally, as to whether this model retains the "symmetrical nature of the problem", I will have to leave that as an open question, depending on precisely what we mean by "symmetrical". It seems to me that this model does a better job of being "symmetrical" than a model which includes only seven corner cubies or only eleven edge cubies, but maybe not. What does "symmetrical" mean when it comes to centerless cubes? = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) (304) 293-5192 Associate Director, WVNET (304) 293-5540 fax 837 Chestnut Ridge Road BRYAN@WVNVM Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU If you don't have time to do it right today, what makes you think you are going to have time to do it over again tomorrow?