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

  • $\begingroup$ Note that 16 is the minimum possible. $\endgroup$ Commented May 27, 2021 at 6:43
  • $\begingroup$ Any proof for that? (that 16 is the minimum possible) $\endgroup$
    – justhalf
    Commented May 27, 2021 at 9:11
  • 3
    $\begingroup$ I don't have a proof, but I have an efficient program that finds these grids. From many attempts it always finds 16. $\endgroup$ Commented May 27, 2021 at 10:19
  • 1
    $\begingroup$ I can confirm that 16 is the minimum number of mines to make every non-mine cell have value 3. $\endgroup$ Commented May 27, 2021 at 12:07
  • 1
    $\begingroup$ @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. $\endgroup$ Commented May 28, 2021 at 12:54

4 Answers 4


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

enter image description here

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

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

enter image description here

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).

  • $\begingroup$ 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. $\endgroup$ Commented May 27, 2021 at 13:06
  • 5
    $\begingroup$ @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. $\endgroup$ Commented May 27, 2021 at 13:15
  • $\begingroup$ Cool, I will see if I can find it. $\endgroup$ Commented May 27, 2021 at 13:18
  • $\begingroup$ I found the 25 mine case for 7x7. Very nice and symmetric. Feel free top post it as another puzzle. $\endgroup$ Commented May 28, 2021 at 1:44
  • $\begingroup$ @JaapScherphuis Oh of course! $\endgroup$ 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

first answer with corners clear

I also found/verified the other known solution...

duplicate of @hexomino 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.

  • 1
    $\begingroup$ I wonder what forces the centre to be unique? $\endgroup$ Commented May 27, 2021 at 23:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.