This combinatorics question may now have some interest: HOW MANY orderings of the (N-1)*N/2 entries in the lower triangle in an NxN matrix, are there, such that each one may be zeroed via an adjacent-coordinate 2x2 rotation, successively, without destroying the previously-created zeros? The going-up-columns raster scan (starting at leftmost column) and the diagonal raster (going southeast, on longer and long diagonals) ordering are two answers, but there evidently are many more possible orderings. (Here I am just considering multiplying by the 2x2 rotation from the left. I we also allow multiplications by 2x2s on the right, that'd be more.) The question is equivalent to the following. How many ways can you fill in that triangle with the integers 1,2,3,...,(N-1)*N/2 in such a way that: if some cell contains integer K, then each cell on the same row containing an integer<K, must be accompanied by a "mate" -- the cell immediately above it -- also with contents <K. ? A crude rough "probabilistic" heuristic estimate is as follows. Consider choosing one of the [(N-1)*N/2]! orderings at random. The chance it satisfies the "mate conditions" seems to be about C^(-N*N) for some constant C>1, perhaps C is about 2. So therefore I estimate the number of allowable orderings is about [(N-1)*N/2]! * C^(-N*N) for some constant C>1. Note this grows to infinity with N rather rapidly regardless of the value of C. But one might be able to get an exact not approximate formula. I would suggest determining the #orderings as a function of N for small N by computer enumeration. By hand: N #ord 1 1 2 1 3 1 4 ? -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)