From: Jim Mahoney To: ERCO@compuserve.com Cc: CUBE-LOVERS@ai.mit.edu Subject: Re: 4x4x4 solution -- [Digest v23 #159] >Your problem is one of parity. You have two edges cubies swapped >(this swap is visible) and two face center (centre) cubies of the >same color swapped (this swap is invisible). You have to have an Edwin> I've had this problem as well.... If I understand you Edwin> correctly, this problem simply doesn't occur anymore as Edwin> soon as you number (or mark in any other way) the center Edwin> pieces which a) makes solving the cube a bit more difficult Edwin> b) makes sure that you'll always get back to the original Edwin> configuration of center pieces. This isn't quite true, at least not on the 4x4x4. While it is true that parity is the question at hand, and also that on the 4x4x4 cube a quarter of a central slice performs an odd permutation on the edges which is otherwise "invisible", it is *not* true that marking the centers will help. The reason is that a quarter turn on a center slice of the 4x4x4 performs a cyclic rearrangement of 4 edges - an odd permutation - while at the same time rearranges *two* sets of 4 central pieces - an even permutation of the centers. Thus parity does not prohibit swapping two edges while leaving the centers untouched. Moreover, in fact there are move sequences which will exchange two edges without disturbing the position of any other piece, corner or center - though I don't have any on hand which are short. If there's interest, though, I can produce a move sequence to exchange two 4x4x4 edges while leaving all corners and centers in their original positions. A cross-section looks like this. A quarter turn cycles the four E's, the four C1's, and the four C2's. This is an odd permutation of the E's but an even permutation of the C's. (All the C's are corners, and can be put into each other's positions with a combination of face and center turns.) E C1 C2 E C2 C1 C1 C2 E C2 C1 E A full account of parity and possible 4x4x4 moves gives 4x4x4 type , how many , parity after: 1/4 face turn , 1/4 center turn - --------------------------------------------------------------------- corners | 8 | odd | even (untouched) edges | 24=2x(12 edges) | even (8 move) | odd centers | 24=4x(6 faces) | odd | even (8 move) Thus to solve a 4x4x4 cube you must have made both (1) an even total number of moves on the faces (to restore the corners and centers to even parity), as well as (2) an even total number of moves on the center slices, to restore the edges to even parity. The parity constraints on the 5x5x5 are a bit different. In that case there are two types of edges (the one in the middle of an edge vs the ones next to the corners) and three types of centers. Each has its own parity change under each different slice. A bit of playing around shows that any central slice move which cycles 4 edges must also cycle several kinds of centers. At least one of those center cycles is odd. Therefore on the 5x5x5 you cannot exchange a pair of edges without also exchanging two centers somewhere. So marking where the centers go will help on the 5x5x5. Regards, Jim Mahoney mahoney@marlboro.edu Physics & Astronomy Marlboro College, Marlboro, VT 05344 ------------------------------ End of Cube-Lovers Digest *************************