From cube-lovers-errors@mc.lcs.mit.edu Fri Aug 1 20:34:49 1997 Return-Path: Received: from sun30.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.1/mc) with SMTP id UAA01174; Fri, 1 Aug 1997 20:34:49 -0400 (EDT) Precedence: bulk Errors-To: cube-lovers-errors@mc.lcs.mit.edu Mail-from: From Hoey@AIC.NRL.Navy.Mil Fri Aug 1 20:32:48 1997 Date: Fri, 1 Aug 1997 20:31:56 -0400 Message-Id: <199708020031.UAA28142@sun30.aic.nrl.navy.mil> From: Dan Hoey To: Cube-Lovers@ai.mit.edu, ponder@austin.ibm.com In-Reply-To: <199708010138.VAA24279@sun30.aic.nrl.navy.mil> (message from Dan Hoey on Thu, 31 Jul 1997 21:38:18 -0400) Subject: Re: puzzle to be simulated Continuing on my quest for Ponderesque cube-planes made of a square torus, possibly cut on the bias, I notice that any square torus will have k=a^2+b^2 faces for some a>b; if b=0 there is no bias. Carl Ponder suggests (if I understand it correctly) cutting each face into nine facets, and permuting the puzzle by turning a face together with the 16 neighboring facets. As with Rubik's cube, the edge facets move in pairs (called edge cubies). The corner facets move in quadruplets (called corner cubies). There are k corner cubies that can apparently achieve any of four orientations (twist) each, and 2k edge cubies that can apparently achieve any of two orientations (flip) each. The face center cubies can achieve any of four orientations (twist) each but never move. So the "constructible" group size--the size of the group before we consider parity and orientation constraints--is k! (2k)! 16^k, or k! (2k)! 64^k for the supergroup. But if a+b is even, we can shade the faces in a checkerboard, and the shades never change when we turn the faces. So in this case, the edges never flip, and the corners have only two orientations. The checkerboard-constructible group size is then k! (2k)! 2^k, or k! (2k)! 8^k for the supergroup. Everyone who knows Rubik's cube will suspect (and everyone who understands orientation theory will know!) that the corner orientations must sum to zero (mod 4) and the edge orientations must sum to zero (mod 2). If a+b is even, there is only a corner orientation constraint (mod 2). [See my article of 23 September 1982 for a sketch of orientation theory. Essentially, if the orientation group of a kind of piece is Abelian then there is an orientation constraint of the order of that orientation group.] The permutation parity constraint is also familiar to anyone who knows the cube. The edge permutation parity must equal the corner permutation parity, and in the supergroup the parity of the face center twist must also be equal (mod 180 degrees). So we should find groups of size k! (2k)! 2^f(k), where f(k)=4k - 4 a+b odd, or f(k)= k - 2 a+b even for the permutation group, and f(k)=6k - 5 a+b odd, or f(k)=3k - 3 a+b even for the supergroup. I've used GAP to find the group sizes for (a,b) = (2,0), (2,1), (3,0), (3,1), (3,2), (4,0), (4,1), and the group sizes agree except for (2,0) and (2,1). The group is smaller than expected by a factor of 5040 for the (2,0) permutation group, 20160 for the (2,0) supergroup, 6 for the (2,1) permutation group, and 12 for the (2,1) supergroup. I'm not too surprised about the (2,0) groups (for instance, all four corner cubies move cyclically on every turn!) but I don't see why (2,1) is pathological. Maybe it's one of those special group things that happen for just one permutation group. By the way, I suggest that a simulation of these should not try to map them to a curved torus, but to a toroidal tesselation of the plane. Then when you turn one piece, you see a lattice of other pieces turning in synchrony. Dan Hoey Hoey@AIC.NRL.Navy.Mil