Difficulty (approximate): ★★★☆☆

Contains: Double mine, triple mine

Now you might be wondering: If this puzzle contains not a single mine, then why does it contain mines? Well, that's because there isn't a single mine in this puzzle. (yes, I know, corny joke. You can downvote now. [not actually, please don't downvote]) Also, since it seems that I've been performing the Day 6 gimmick well (numbered tiles that need to be filled into the grid), there are also going to be missing numbers. Here is the puzzle:

2 8 15 6
10 18 8
4 10 14 6
14 12
12 8 7 7 5
4 4

Now, there are three tiles that are "clearings" tiles that are filled in neither with a mine nor a number, which you will also need to deduce where they are. Here are the numbers that are missing:



















There are 23 double mines and 8 triple mines

  • 1
    $\begingroup$ I've been trying to solve this and keep hitting a contradiction around the 8 and 4 bottom-centre. $\endgroup$
    – fljx
    Commented Nov 22, 2023 at 16:58

1 Answer 1


The completed grid:

enter image description here

Solve path:

Start with basic forced entries:

  • The 15 at the top must be surrounded by triple-mines.
  • The 6 top-right must have triple-mines in all its free neighbours.
  • The 2 top-left must have a double-mine in its single free neighbour. enter image description here

  • The 5 bottom-right must have one double-mine and one triple-mine adjacent. That is the eighth and final triple-mine so there are none anywhere else.
  • The missing 19 needs at least seven neighbouring squares that could contain mines, and there is only one possible location.
  • That location has three adjacent triple mines, so we require ten more mines in the other five spaces. So the rest of the spaces around the 19 must be filled with double-mines.
  • That completes the 10 at R2C2, so the remaining neighbours must be clear. (And contain digits 2 and 9).
  • That means the 4 at R4C1 must have double-mines in its two free neighbours.
  • And (with all the triple-mines elsewhere) the 14 at R4C6, the 14 at R5C4 and the 12 at R5C5 must have all have double-mines in all their remaining free neighbours. enter image description here

  • At this point there is an obvious contradiction in the bottom-centre. The 4 at R8C5 needs two double-mines to complete, but that would give us ten mines around the 8 above. So either the 4 should be a 2, or the 8 should be a 10.
    Let's ignore this for now, and see where we can get to.

  • That completes all the numbers in the top-right, so the remaining neighbours must be clear, making R1C6 a 12, and R3C8 a 7.
  • The 7 at R7C6 just needs a triple-mine to complete. That triple-mine must also be adjacent to the 5 in R7C8, so must be in one of the blue squares.
  • That means R8C6 must be clear, and has only one adjacent unknown. We have already placed the missing "2", so there is no available number and this must be one of the "clearing" squares.
  • So the "clearing" squares must be zeros that would auto-clear in a normal minesweeper game. So R8C7 cannot be a mine, and (by a repeat of the logic above) with R8C8 must the other two clearing squares.
  • So the final triple mine is in R6C7. And R6C8 is a double-mine to complete the 5.
  • And that means R5C8 must be the final 9. enter image description here

  • We have only three more double-mines to place, which must all be adjacent to the 12 in R7C3.
  • So R7C1 must be clear.
  • So the 4 in R8C1 must have double-mines in its other two neighbouring squares.
  • That completes 6 in R6C1 so its remaining neighbours must be clear making R6C2 a 10 and R7C1 a 4. enter image description here

  • Finally, we have two double-mines and an 8 to place, and the only way the 8 works is in R8C3, which gives us the final grid.
  • And we can resolve the contradiction by saying that the 8 in R7C5 should be a 10.

  • $\endgroup$

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