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Comment by rossant

3 days ago

Among the 2^n configurations, how many are solvable?

1 comment

rossant

Reply

raymondtana  6 hours ago

In the n-by-n case when n is even, all of them are possible :)

In the n-by-n case when n is odd , that's not the case... it breaks into a few equivalence classes that you can separate out by looking at the "mod 2 sums" across each of the rows and columns.

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