# 6x6 Minesweeper grid with all threes

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

• Note that 16 is the minimum possible. Commented May 27, 2021 at 6:43
• Any proof for that? (that 16 is the minimum possible) Commented May 27, 2021 at 9:11
• I don't have a proof, but I have an efficient program that finds these grids. From many attempts it always finds 16. Commented May 27, 2021 at 10:19
• I can confirm that 16 is the minimum number of mines to make every non-mine cell have value 3. Commented May 27, 2021 at 12:07
• @Flater Such cells are equivalent to having value zero (no neighbouring mines), and I just checked and it makes no difference. Apart from the completely empty board, 16 mines is also the minimum where every non-mine cell has exactly 0 or 3 neighbouring mines. Commented May 28, 2021 at 12:54

I think this arrangement of mines will work (red squares are mines)

• That is correct! Can you find another, completely different, solution? Commented May 27, 2021 at 10:19

Apart from the solution that hexomino found, there is another solution:

According to my computer program, there are no other solutions up to symmetry (so 4 solutions if we count the rotated/reflected pattern as distinct).

• yep that's the other solution. Interestingly my program mostly finds this one and very rarely the other one. It must a strong local minima. Commented May 27, 2021 at 13:06
• @DmitryKamenetsky My program does an exhaustive search. The 7x7 board, 25 mines, all empty cells value 4 is an interesting case with a unique solution. Commented May 27, 2021 at 13:15
• Cool, I will see if I can find it. Commented May 27, 2021 at 13:18
• I found the 25 mine case for 7x7. Very nice and symmetric. Feel free top post it as another puzzle. Commented May 28, 2021 at 1:44
• @JaapScherphuis Oh of course! Commented May 29, 2021 at 15:44

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.

• I wonder what forces the centre to be unique? Commented May 27, 2021 at 23:16