From hoey@aic.nrl.navy.mil Tue Jan 4 21:36:19 1994 Return-Path: Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA23585; Tue, 4 Jan 94 21:36:19 EST Received: by Sun0.AIC.NRL.Navy.Mil (4.1/SMI-4.0) id AA07661; Tue, 4 Jan 94 21:36:18 EST Date: Tue, 4 Jan 94 21:36:18 EST From: hoey@aic.nrl.navy.mil (Dan Hoey) Message-Id: <9401050236.AA07661@Sun0.AIC.NRL.Navy.Mil> To: "Jerry Bryan" , Cube-Lovers@ai.mit.edu Subject: Re: Which is the Real Start? Jerry Bryan has some more questions. > Take a standard 3x3x3 Rubik's cube, and remove the corner and > center labels to make an Edges Cube.... Scramble..., how will > the cubemeister distinguish Start from Pons Asinorum? > ... if you identify the identity with Start, you are in the > disquieting situation of having a group with two distinct identities > (grin!). The problem is that we would not be dealing with a _group_ then, but a collection of cosets of M. Just as in the edge `group', we deal with either 1) a less-symmetric group in which one of the edges never moves, or 2) a larger group in which we distinguish positions that differ by rigid motions of the cube, or 3) a non-group in which we consider cosets--equivalence classes of group #2, where group elements that differ by rigid motions are equivalent. You have got a lot of mileage out of working with group #2 to save duplication among symmetries, then reducing to non-group #3. But what you lose is the group structure of the object you are studying. Instead, you have to work in the large group and then deduce information about the cosets. All in all, though, I'm very glad of it, for the lost symmetries of group #1 were sorely missed. For most of the other questions, mouse@collatz.mcrcim.mcgill.edu provides satisfactory answers. However, strictly speaking we should not call an equivalence class to be a group element (unless it is a coset of a normal subgroup, and neither C nor M is normal in the large group). I'll admit I've also abused the term when considering distances in the ``edge group'', as if all 24 rotations of a position were the same element of some group. But when we start dealing with the distinction between fixed and movable cubes I think we need to start being more careful. [ mouse also mentions that quarter-turn ``usually doesn't include slice turns on the 3-Cube, but on the 4-Cube and higher, they must of necessity be included.'' I'll take that as an argument for eccentric slabism: a QT rotates any 1xNxN slab except a central slab of an odd-edged cube. As opposed to cutism, where a QT consists of a rotation of part of the cube with respect to the other. ] Other questions: > ...since Start and Pons Asinorum differ only by a simple > reflection, why had not my version of M-conjugation declared them > to be equivalent? Your versino treats positions X,Y for which m'Xmc=Y (m in M, c in C) as equivalent. If you instead determine when m'Xmn=Y (m,n in M) you would find them equivalent. This is equivalent to changing the loop in your version of M-conjugacy. > For j = 1 to 24 for k = 1 to 24 for m = 1 to 2 > for i = 1 to 24 Bj,k,m(i) = Qm(Pj(A(Qm(Pk(i))))) so that the two occurrences of Qm need not be the same. > (I speak of "my version of M-conjugation", but the question is no > different if you look at Dan Hoey's original M-conjugation). No, I didn't use M-conjugation except for a cube with a fixed orientation in space [or equivalently, with face centers]. So in the original concept of M-conjugation that Jim Saxe and I put together, Start and Pons Asinorum don't just differ by a reflection. > I found Dan Hoey's postings about the four special states of the > Edge Group to be delightful.... However, [without the results on > distances] if we identified the group as being rectangular, would we > be led to saying which of the four special states were diagonally > opposed without the computer search? Without the search, I might be > tempted to say that Start and Pons Asinorum were diagonally opposed. Well, really the `group' is in the shape of a sphenoid, a word I learned yesterday for a tetrahedron whose three pairs of opposite edges are equal. [Or equivalently, a tetrahedron whose edges are face diagonals of a rectangular prism.] But it might be more accurate to consider it as a large ball of string with a bunch of symmetries. Calling it a rectangle or sphenoid may lead us to ignore the structure that is not representable in Euclidean space. Dan Hoey Hoey@AIC.NRL.Navy.Mil