From hoey@aic.nrl.navy.mil Tue Dec 28 14:17:02 1993 Received: from Sun0.AIC.NRL.Navy.Mil by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA24468; Tue, 28 Dec 93 14:17:02 EST Received: from sun30.aic.nrl.navy.mil by Sun0.AIC.NRL.Navy.Mil (4.1/SMI-4.0) id AA19858; Tue, 28 Dec 93 14:16:50 EST Return-Path: Received: by sun30.aic.nrl.navy.mil; Tue, 28 Dec 93 14:16:49 EST Date: Tue, 28 Dec 93 14:16:49 EST From: hoey@aic.nrl.navy.mil Message-Id: <9312281916.AA25640@sun30.aic.nrl.navy.mil> To: Cube-Lovers@life.ai.mit.edu Cc: Jerry Bryan Subject: Re: Some Additional Distances in the Edge Group In his message of Fri, 17 Dec 1993 00:54:00 EST, Jerry Bryan makes some observations on the distances between the following positions in the edge group: I = Solved, P = Pons Asinorum (or Mirror), E = All edges flipped, and PE = P E = Pons Asinorum with all edges flipped. [I _will_ continue to use permutation multiplication as we have done so in this group since its inception. I realize that this agrees with some textbooks and is backwards from others, but it would be far more confusing to write these functionally all the time.] Jerry's brute-force search has shown that d(I,PE)=15, and he notes that conjugation by E shows us that d(P,E)=15 as well. He concludes: > I have the sensation in describing this that the Edge group is > square, with Start and Mirror-Image-of-Edges-Flipped 180 degrees > apart, and Mirror-Image-of-Start and Edges-Flipped at the other > two corners of the square. Well, it's not quite a square, since d(I,P)=12 and d(I,E)=9, according to Jerry's message of Wed, 8 Dec 1993 10:02:15 EST. Conjugation will similarly show that d(E,PE)=12 and d(P,PE)=9. So we are dealing with a rectangle. The sides of the rectangle are 9 and 12, and the diagonal is 15: a most fortuitous set of numbers, in that we can actually embed such a rectangle in the Euclidean plane! We can map the positions of the edge group to 4-tuples of distances. For any position X, let f(X)=(d(I,X), d(E,X), d(P,X), d(PE,X)). If f(X)=(a,b,c,d), then conjugation shows us that f(X E)=(b,a,d,c), f(X P)=(c,d,a,b), and F(X PE)=(d,c,b,a). So the set of quadruples has the symmetries of the rectangle. We know f(I)=(0,9,12,15). What is more, the earlier results on symmetry show us that I is at a local maximum distance from E, P, and PE. So, letting I1 be the unique (up to M-conjugacy) position adjacent to I, we have F(I1)=(1,8,11,14). (This destroys Euclidean embeddability.) An analogous result holds for the unique neighbor of each corner of the rectangle. We also have Jerry's results of Wed, 8 Dec 1993 22:41:28 EST and 23:16:50 EST that H (the 6-H pattern) and HE=H E are at distances 8 and 13 from start, respectively. Since H is an M-conjugate of P H, this gives us f(H)=(8,13,8,13). [Note: there are two distinct M-conjugates of H, call them H and Hbar. This distinction is important when we compose permutations: H H = I, but H Hbar = P. So we have to be careful when conflating M-conjugates.] We can by symmetry find f(H1)=(7,12,7,12) for H's unique neighbor H1. What quadruples are possible? If f(X)=(a,b,c,d), and X is not one of the eight corners and neighbors, we have max(2,9-b,12-c,15-d) <= a <= min(14,9+b,12+c) with constraints on b, c, and d from symmetry. A quick hack tells me there are 7836 such quadruples. I wonder how many of them are realized? If it's fairly few, I would like to see a diagram of quadruples, with lines between those quadruples that represent adjacent positions (adjacent quadruples differ by at most one in each coordinate). Maybe with the number of positions for each quadruple, too. I have an idea that such a diagram might tell us something about the problem, or at least look pretty. Dan Hoey Hoey@AIC.NRL.Navy.Mil