This is a minesweeper puzzle with a tetromino twist. The goal is to place mines in the grid, following a few constraints.


  • A number indicates how many adjacent (including diagonally adjacent) cells have mines in them.
  • Mines cannot go in numbered cells.
  • Mines must be grouped into tetromino shapes.
  • Only one of each tetromino is used. Rotation but not reflection is allowed.
  • Tetrominos may not touch orthogonally (on a side). They may touch diagonally.

A simple puzzle is below, with all the available tetrominos included for reference.

enter image description here



Checkmark goes to the first solution to show a logical path.

I called this the "amphitheater" because I imagine the numbers around the edges as an audience, the numbers in the center as performers, and the number on the bottom as a stagehand. So they're in an amphitheater. Yes, I have great naming ideas. Thank you very much.




The techniques are mostly from minesweeper. Here are the steps:

Nothing around the 0.

There must be one mine between the 3 and the 1.

There must be two mines between the two 2's.

There must be one mine between the 1 and the 2.

The mine must be on the right most place.

There must be two mines between the upper 2 and 5.
By calculation, there must also be two mines between the lower 2 and 5.

If the upper right corner of the 4 is a mine, then it quickly leads to a contradiction. Thus the upper left coner of the 4 is a mine.

Fill in the mines near the 2.

The left of the upper 3 cannot be a mine, so the lower left corner and the lower of the upper 3 are mines.

It's now easy to finish the rest.

  • $\begingroup$ Doesn't your 3 in the 7th row have 4 mines around it? $\endgroup$ – Jens Oct 8 '20 at 1:28
  • $\begingroup$ Only 1 of each tetromino. You've used two S tetrominos, so this is not valid. Also, Jens has a valid complaint. $\endgroup$ – bobble Oct 8 '20 at 1:31
  • $\begingroup$ @Jens Sorry, I was a bit hurry toward the end... Thank you for pointing it out. I corrected my solution. $\endgroup$ – WhatsUp Oct 8 '20 at 1:34
  • $\begingroup$ You've got the check, but please expand your steps with some explanations. $\endgroup$ – bobble Oct 8 '20 at 1:34
  • 1
    $\begingroup$ Got me by 22 seconds!!! Nice job! $\endgroup$ – Jeremy Dover Oct 8 '20 at 1:53

I believe the answer is:



Starting in the upper left corner, block off (pink) all of the squares around the 0. The square diagonally above and right of the 3 in the upper right corner must be shaded, for if not it blocks the squares above the 3 from being shaded, leaving only two candidates. Once this square is shaded, we can block off all other squares around the 1, forcing the I tetromino to lie above the 3, like so:

Now looking at the two in the upper middle:

If the square to its upper right is shaded, then the two to its right limits the block size to 2. So we must have the lower two squares shaded, which must extend a square further out, making this tetromino one of the Ls. If the L were to come out to the right, it would block the squares above and left of the top 5 from being shaded, which would force there to be another I tetromino to be to the right of the 5s. So it must tail to the left. Thus we have:


Looking at the 2 above the 5s:

We clearly cannot shade the squares to its upper right, since that would force the 2 to have at least three shaded neighbors. More over, we cannot shade the square to its right, since this would again force another I tetromino down the right side. So we must get the other L tetromino around the 5s. This fills four of the five squares adjacent to the top 5, and the fifth cannot be the square to its left, since we would need to shade the remaining square to get the tetromino out. This then forces the remaining squares around the lower 5 to be shaded, and also fills out our bottom right 2. Our progress thus far:


Now look at the 3 under the 1:

We cannot shade the square to its left, for if we did, we would have to use a Z tetromino which would need two shaded squares around the 1 to the left. Now, based on the fact that we've already placed all of the tetrominos with three squares in a row, we can conclude this is a Z tetromino escaping right around the 6. This fills the 3 adjacent to the T tetromino, and forces the remaining squares around the 6 to be shaded, which must be the other Z tetromino. Finally, there is only one place to be the square tetromino, which completes the puzzle.


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