Talk:N-queens minimum and knights and bishops: Difference between revisions
Talk:N-queens minimum and knights and bishops (view source)
Revision as of 00:16, 26 April 2022
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: Although I realized this some time ago and several sources simply state the minimum number of bishops to be N for an N x N board I haven't been able to find a mathematical proof of this. In other words how do you know that there isn't a solution which only uses N-1 bishops?
:: If you number the diagonals (in one or either directions) it can be seen there must be at least one bishop on each diagonal. So the minimum number is the (number of rows) == (number of diagonals parallel to one direction).
: Incidentally, although I've written a Go solution, the only one I've posted so far is Wren *. The former is, of course, much quicker but still far too slow for my liking.
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