Can you place 16 mines on a 6x6 Minesweeper grid such that each number produced is a 3? Bonus: can you find multiple solutions that are not rotations or reflections of each other? Good luck!
Related question: Paint Eleven Squares
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This is equivalent to asking how to place kings on a 6x6 board so that each empty square is attacked by 3 of those kings. This is a special case of the problem posed here. That web site gives hexomino's 16-king solution for the 6x6, among
3, 5, 8, 12, 16, 29 (sic), 27
-king solutions for the 2x2, ..., 8x8 boards.
After seeing hexomino's solution and your comment, I started playing with (manually generated, but automatically checked) ideas in Excel, with the goal of
"I wonder if there's a solution with a corner clear?"
and quickly found
I also found/verified the other known solution...
But other avenues explored have
so far led to dead-ends / contradictions - cells that need more adjacent mines next to clusters of other cells that can't have any more mines next to them etc. or getting edited into something approaching a duplicate of the other known answers...
which would be fairly obvious to @jaap-scherphuis, who I now see posted while I was typing up mine...
I also noticed that
The two solutions are identical in the central 4x4 box, implying this part of the solution is unique.