From kon@bach.stanford.edu Tue May 14 21:31:35 1991 Return-Path: Received: from bach.Stanford.EDU by life.ai.mit.edu (4.1/AI-4.10) id AA03214; Tue, 14 May 91 21:31:35 EDT Received: by bach.Stanford.EDU (4.1/inc-1.0) id AA00208; Tue, 14 May 91 18:31:33 PDT Date: Tue, 14 May 91 18:31:33 PDT From: kon@bach.stanford.edu (Ronnie Kon) Message-Id: <9105150131.AA00208@bach.Stanford.EDU> To: dn1l+@andrew.cmu.edu, ncramer@bbn.com Subject: Re: ARGGHHH!! [was: 5by cubes] Cc: Cube-Lovers@life.ai.mit.edu > >(Sorry.) The right answer should be: > >The state of the cube is not: > >X|O|X|X|X X|A|X|C|X >X|X|X|X|X X|X|X|X|X >X|X|X|X|X But rather: X|X|X|X|X >X|X|X|X|X X|X|X|X|X >X|X|X|O|X X|X|X|B|X > >Where cubie "C" just "looks" like it's in the right place. > >You need an operator that rotates A->B->C->A. (Left as an exercise; hints >available on request.) > >This will very likely leave an inconvenient number of edges flipped. For >the answer to _this_ problem, see my last post. ;) I think you must be wrong here (but would love to be proved wrong--I'm no mathematician so group theory is very much beyond me). Proof #1: We hold the cube with the red face on top, and the yellow face in front (colors obviously don't matter, but I find it easier to discuss using them). We will assign a parity to the edge cubies, being defined by holding the cube such that the red face of the cubie is on top and the yellow in front. If the cubie is on the left as we look at it in this position it is parity 0, on the right it is parity 1. There are only two operations available which affect the cubie we are interested in: rotating the front face 90deg; and rotating the slice the cubie is in 90deg. It is easy to see that neither of these moves alters the parity (assume the cubie's frame of reference, and think of rotating the rest of the cube around it--it is clear that it will not end up on the other side). Therefore the move C->A in the above is impossible. Proof #2: Take apart the order 4 cube (my falls apart depressingly easilly) and try to reassemble it with the two edges exchanged. It will not fit, as they are mirror images of each other. Note that you get an apparant parity reversal by flipping the cubies, but this does not actually move anything. In other words, no amount of flipping and moving will allow you to end up moving A->B->C->A. That's why I solve edges first. Ronnie