, and the 2x2x2 cube could be called GC=because there are no Face-centers. The "standard Singmaster model" (my terminology) would be written as G[C,E,F] = . (Well, I think Singmaster would write it as G[C,E,F] =, since I think he prefers to accept H turns as single moves.) However, I tend to work with G[C,E] =instead. I consider G[C,E] to be equivalent to G[C,E,F] for most purposes because G fixes the Face-centers, as does M-conjugation. I have described this equivalence before as the Face-centers simply providing a frame of reference that can be provided in other ways. However, when you step outside the friendly confines of G=, it does start to matter whether the Face-centers are there or not. As an example important to this discussion, if you consider CG=, then it makes a considerable difference whether you are talking about CGC,E or CGC,E,F. For example, G[C,E] =can be simulated on a real cube by removing the color tabs from the Face-centers, by restricting yourself to Q moves only (no whole cube rotations or slices), and by declaring the cube solved only when the Up color is up and the Front color is Front. Notice that with the Face centers absent, you can make the cube look solved even when it isn't. It will be rotated instead, but it won't be solved. This model may seem a little simple-minded. Why are no rotations allowed, and why don't you count it as solved when it looks solved? But computers are simple-minded. My programs only consider things equal when they are literally equal, and equivalence is something I have to program in. As an example I have used before, consider G[C]=, modeled in the real world by a 2x2x2 pocket cube or by removing both the edge and Face-center color tabs from a 3x3x3 cube. Take a solved cube in GC and perform RL'. The cube will still look solved, but it will be rotated. The memory cells in my program will not be the same for I as for RL', but I want to treat them as equivalent, as would nearly everybody with a real world 2x2x2 cube in their hands. This is where I have claimed before that a model that treats RL' the same as I is G[C]/C[C]. The idea is that G[C]/C[C] is a group with the identity being C[C] itself (i.e., rotating the cube is "doing nothing".) The proof is fairly simple. From each element (coset) of G[C]/C[C], pick the unique permutation that fixes a particular corner, say UFR, and form a new set G[C]* containing the one element chosen from each coset. The elements of G[C]/C[C] are sets (namely cosets), but the elements of G[C]* are permutations which are also in G[C]. In particular, G[C]* =. Hence, G[C]* is a group. Note that the generators of G[C]* are the twists of those faces which are diagonally opposed to the corner fixed by the selection function from G[C]/C[C] to G[C]*. Hence, the generators fix the same corner as the selection function, showing that is really the same set as G[C]*, namely the set of all cubes in G[C] for which the UFR corner is fixed. Finally, there is an obvious isomorphism between G[C]/C[C] and . Namely, to multiply two cosets, map each to via the selection function, perform the multiplication there using standard cube multiplication, and map the product back to a coset. Hence, G[C]/C[C] is a group. A similar argument applies to G[E]/C[E] except that we have to fix an edge cubie instead of a corner cubie. A similar argument applies to G[C,E]/C[C,E] except we have to fix an edge cubie and restrict C to even permutations. Dan calls the set of even rotations E, so let's call it G[C,E]/E[C,E]. (Still wish we had letters whose use did not conflict so blatantly.) But when we started, we were talking about CG/C, not about G/C. However, notice that when our model does not include Face-centers, we have =,=, and=. (I mean that the groups are equal, not that the Cayley graphs are the same.) Hence, speaking generically of the first two cases, we have C is in G, CG=G, and both CG/C and G/C are groups. In the last case, we have to say E is in G, EG=G, and EG/E is a group. But we can go one step further. Since there are no Face-centers, we can admit Slice moves or C as generators (it doesn't matter which), and we no longer have to restrict ourselves to even rotations. We can say G+C,E=and we will have C is in G+, CG+=G+, and CG+/C is a group which is the same size as EG/E. (G+ is twice as big as G, of course.) I guess this must mean that CC, CE, and CC,E are all normal subgroups of their respective CG's, but that CC,F, CE,F, and CC,E,F are not. That should not be surprising. Having the Face-centers there only as a frame of reference and never moving is not the same as having them there and really moving (as when you rotate the entire cube). After joining Cube-Lovers, I discovered that others had solved God's algorithm for the 2x2x2 long before me. I was expecting my solution to be 24*48 times smaller than theirs because I was using cosets of C and M-conjugates. But my solution was only 48 times smaller than theirs. By taking both cosets and M-conjugates I really had reduced by 24*48 times. However, everybody else who worked on the problem had modeled it as something like , fixing a corner. (Any other corner would do as well. There are eight conjugate groups, any of which would do as well as any other.) is 24 times smaller than in the first place, and as I said earlier, is equivalent to for most purposes anyway because of the fixed Face-centers. Hence, everybody else had a 24 times head start on me. (At the time, Dan suggested that I was increasing the size of the problem by 24 and then reducing it by 24*48 for a net reduction of 48. But that would only be true if the model were . Since the model was , there really was a reduction of 24*48. But does not really model the 2x2x2 cube, and is 24 times larger than it needs to be in the first place if you are trying to model the 2x2x2.) Modeling cubes without centers such as the 2x2x2 is trickier than it looks because of the requirement that rotations be treated as equivalent. I did it by using cosets of rotations; everybody else did it by fixing a corner. But before I realized all this, I went on a Quixotic chase for a model which would simultaneously be a true model for a 2x2x2 cube and would retain the cubic symmetry of the problem (whatever that means). There are articles in the archives concerning this subject, with the conclusion that no such model is possible because any true model would be isomorphic to , which does not have "cubic symmetry". I guess the "cubic symmetry of the problem" means that you should use M-conjugate classes. Recall that when I solved I used what Dan calls W-conjugate classes because W is the symmetry group for, and W-conjugate classes reduced the size of the problem by four times. This leads me to a question. The way I modeled the 2x2x2, I used M-conjugate classes of cosets and reduced the size of the problem by 48 times. If I were going to model, I would be very inclined not to use M-conjugate classes, but rather to use a subgroup of M which was the symmetry group of . The subgroup would have less than 48 elements, and I would get less than a 48 times reduction in the size of the problem. But a fixed corner model such as is isomorphic to a coset model such as /CC, and M-conjugates are appropriate to the coset model. Therefore, my analysis of the situation is obviously very flawed. Can anybody see what is wrong? = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = 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