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3

Via integer linear programming, I found a solution that uses and all such kings can exit in Initial configuration: Covered squares: Animation:


6

My answer is The solution is drawn here: Note that


2

Posted here by user @bof on Chess Stack Exchange: Under the specific rules of this question, Re7++ and Rc5++ don't count (or do you mean they don't count as two checkmates each?), so that would be 103.


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