Arie Groeneveld wrote: > It looks to me that the subgroup is the alternating group of the > permutation group, > meaning they are even permutations. By saying this, one can ask 'are the > odd permutation > arrangements of the puzzle unsolvable?'
The generating moves are even permutations, e.g. C. 3 2 1 0 4 5 +---+---+-+-+ |2 1|3 0|4|5| +---+---+-+-+ so any odd permutation of the solution is unsolvable (like the trick version of the 15 puzzle). I think it is probably the full alternating group, although I am not completely sure. Since, as Raul suggests, you can generate the complete set of solutions and so decide the shortest solution. Using group theory alone, you need to apply techniques used in word problem questions. I guess this would be useful if the "cube" were bigger. Best wishes, John ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
