# Wait, so how many mines are there? A tetromino minesweeper

Here's another tetromino minesweeper. I have bolded where the rules differ between this one and my first tetromino minesweeper

## Rules:

• 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.
• Two of each tetromino is used. Rotation and reflection are allowed.
• Tetrominoes may not touch orthogonally (on a side). They may touch diagonally.

Also, there are some question marks in the grid. Each represents a different number. (Credit to Avi's second Trichain for the idea!)

Here is the puzzle. All the available tetrominos are included for reference.

CSV:

,,,3,,,,,,,
2,,,,,,,,,?,
,3,,,,?,,,,,
,,,6,,,,?,,,1
,,,,3,,4,,,,
,,5,,,,,,6,,
,,,,2,,2,,,,
2,,,2,,,,?,,,
,,,,,?,,,,1,
,?,,,,,,,,,2
,,,,,,,?,,,


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

First:

The 6 in the upper left can have some cells shaded to prevent a run of 5 or more. The tetromino of the 2 in the bottom right needs to satisfy the 1, so the top of the 1 must go unused.
So with that, we can place the first L. We can also rule out some cells near that 3: it only needs one more cell, and that cell can't be isolated.

Now, an interesting step:

The last cell for that 3 also couldn't be the one to the lower right - it wouldn't be directly cut off, but you wouldn't be able to place a tetromino there without breaking the 2 to its lower right. With some more extendability logic, that leads here:

Now, what happens if we don't use that cell in the center, between the 4 and the 2? Both the shaded cell to its right, and the topmost group of shaded cells, will become forced L tetrominoes. This is a problem, because we can only have two L tetrominoes -- so that cell must be used.

Some more logic sprouts off of the same area:

Look at the 2 just below-left of the center. There are only three cells that could be filled there. If we ignore the topmost of those three, we break the other 2 nearby -- so that one must be filled, and that finally satisfies our 3!

And now if we use the top-left cell of the 6, we have too many Ls again. So we must block that cell off, and now some more deductions lead us to placing both the Ss and the Ts!

And hey, wait a second...

looking at that 1 in the upper right, it must have a tetromino too. Which means we now have all 10 tetromino locations.

Region ④ must specifically have the other L tetromino. So none of the others can be Ls.

And finally, since the ? clue in the top right must be a 4, the we must place the L to prevent the other ? nearby from being a 4. And the puzzle is solved!

The final grid looks like this:

EDIT: Here's the promised write-up:

To start us off, OP gives us a couple of things for free:

* Any 6 can have at most 2 pieces next to it. Since the 3 is empty, we know the other break must be on the opposite side of the 6.
* The passages between numbers in the middle are narrow, so some of the squares can be marked empty on the basis of no tetromino being able to fit in there.
* In the bottom right, there must be at least 1 square with a tetromino on the bottom row. The only way to fit an entire tetromino there is the L piece as shown.

The square to southwest of the other 6 (starred) must be empty. If it weren't, the it would take two more L-pieces to satisfy the nearby 4. Again avoiding a third L, we get more progress around the 6:

At the top left, at least one of the two squares adjacent to both the 2 and 3 (starred in the picture below) must be empty. This means there must be a filled square next to the 2 on the top row.

This also gives us a couple more squares, which define the break around the nearby six.

With the break around the six known, we see that only one of the top squares around the for can be filled.

This puts two filled squares next to a two, and with the constraint around the 5 we can progress downwards.

At this point, we get a lot of "free progress", with very little deduction needed:

At the top right, we notice that the 1 must use up the remaining an L-piece (we don't know how, yet), so the remaining incomplete ones must be some other shapes.

With everything else in place, we get the position of the final L-piece only by solving the question marks.

Here's one more progress picture from along the way: