hihi, all - rich's first nit about the set of reachable positions: indeed each move is a transposition in S16, but it is easy to see that the only way for the blank space to get back to its original position requires an even number of them, so all of those permutations are even, and my previous note was about a proof (i think) that all even permutations are possible, if the outline is thick enough (as the square is) then the set of all positions is A15 in S16 fixing 16 with the added transpositions making a transitive group on all 16 positions, with overall size therefore (15!/2) x 16, though it is not A16 (as rich said) the key to the proof that you can get all of A15 is to show that in any 2x2 subsquare, with any one of the four unit squares fixed, you can get the 3-cycle for the other three unit squares (almost): pick a 2x2 subsquare, and one of the unit squares in it that is not a corner of the entire square - then it is easy to move the blank spot to a position adjacent to the fixed unit square in that 2x2 subsquare without disturbing any of the unit squares in it - move the fixed one out into the blank spot, cycle the other three unit squares, move the fixed one back, and reverse the moves to get the blank spot back to position 16 this is a lot of 3-cycles, i think easily enough of them to generate all of A15 (i do remember having a real argument for that, but i can't remember what it was) more later, cal