From there the solution should be straightforward (e.g. two intersecting 3-cycles). Finally, it seems clear that this entire problem --and all the subsequent discussion-- maps directly onto a virtually identical problem on the 4by cube (i.e. simply be removing the center planes). >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. Again, perhaps I'm missing the point, but if you don't care about how the flipping comes out, the A->B->C->A 3-cycle is certainly doable: For example: [WARNING: EVEN MORE BORING STUFF AHEAD!! ;] (I have no idea how to show this notationally, so I'll try pictorially.) | 1] 2] V 3] X|A|X|C|X X|Y|X|C|X ->Z|Z|Z|Z|Z X|X|X|X|X X|Y|X|X|X X|Y|X|X|X X|X|X|X|X X|Y|X|X|X X|Y|X|X|X X|X|X|X|X X|Y|X|X|X X|Y|X|X|X X|X|X|B|X X|A|X|B|X X|A|X|B|X 4] 5] 6] Z|A|Z|Z|Z X|B|X|X|X X|B|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|B|X Z|Z|Z|A|Z ->?|?|?|?|? ^ | [Rotate Face one-half turn] >--- 7] \ 8] 9] X|B|X|X|X \ X|B|X|X|X Z|C|Z|Z|Z X|X|X|X|X | X|X|X|X|X X|X|X|X|X X|X|X|X|X | X|X|X|X|X X|X|X|X|X X|X|X|X|X V X|X|X|X|X X|X|X|X|X ?|?|?|?|? Z|Z|Z|C|Z<- X|X|X|B|X [Rotate next- [Rotate Face to-bottom one-half turn] plane 1/4 Turn] <---- 10] 11] \ 12] Z|C|Z|Z|Z Z|C|Z|Z|Z \ Z|C|Z|Z|Z X|X|X|X|X X|X|X|X|X | X|X|X|X|X X|X|X|X|X X|X|X|X|X | X|X|X|X|X X|X|X|X|X X|X|X|X|X ^ X|X|X|X|X ->?|?|?|?|? ?|?|?|?|? X|X|X|A|X<- [Rotate next- to-bottom plane 1/4 turn] | 13] V 14] 15] Z|Z|Z|Z|Z X|Y|X|B|X<- X|C|X|B|X X|Y|X|X|X X|Y|X|X|X X|X|X|X|X X|Y|X|X|X X|Y|X|X|X X|X|X|X|X X|Y|X|X|X X|Y|X|X|X X|X|X|X|X X|C|X|A|X X|C|X|A|X X|X|X|A|X ^ | Cub.E.D