送交者: lakeblue 于 December 05, 2009 21:16:13:
回答: 还是可以用初等办法证明的 由 只是没想到而已 于 December 05, 2009 15:15:36:
My reasoning:
For an NxN board,each square is either occupied by a marker or not, so there are altogether 2^(NxN)possible ways,including both even and odd sums.
For each row and each column,the number of ways with odd sums equals that with even sums,so the number of ways for each row and each column should be reduced by half to keep only odd sums. Thus,it seems that 2^(NxN)/(2^Nx2^N) is the answer.
However, once all the N rows and N-1 columns are restricted to contain only odd numbers, the last column is ensured to have an odd number of markers.So the answer is
2^(NxN)/(2^Nx2^(N-!))=2^[(N-1)x(N-1)] for any integer N.