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I just found a weird-looking sudoku I left years ago in my basement:

Weird sudoku puzzle

I think there was a specific name for this type of puzzle, but I forgot to write it down and now I can't remember it!

The only other thing I found is a message on the back of the puzzle saying that every correct circle has been placed.

So, what's the name of this puzzle?

Hint:

Just some nice coloring of mine ☺ Weird sudoku puzzle (colorized, 2023)


If needed, here's a text version of the puzzle:

F x31
THIS IS A   _  _  _    _  _  _    _  _  _

    =F 20  (1) _  4   (5) 6  _    8 (_) _
    =F 30   _  9  _    _  _  _    7 (_) 4
    =F 14  (_) 8  5    _  4  _    9  _  _

    =F 19   _  _  3    _  1  2   (_) 4  7
    =F 4    _  5 (_)   _  _ (4)   _  1  _
    =F 7    _  4  _    9  _  _   (_) 2  _

    =F 9    4  _  _    _  9  5    _  _  3
    =F 9    6  _  _   (2) _  _    _  7  _
    =F 18   _  _  _    _  _ (3)   1  9  6
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    $\begingroup$ The Sudoku is solvable. Can't for the life of me figure out the next bit. $\endgroup$ Dec 10, 2023 at 23:55

2 Answers 2

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This is a:

MINESWOKU puzzle!

Because:

The sudoku is easily solved using only basic deductions:
enter image description here

Now we have a number of circled digits that can be treated as minesweeper clues, allowing us to flag a number of cells.
(Marking mines with grey background, and known safe cells with green)

The 5 in R1C4 must be surrounded by mines.
Then the only way to get a sum of 20 in the first row is by flagging 8 and 2.

In row five, the only way to get a sum of 4 is as 1+3, so the rest of that row is safe.
And we can mark 8 and 9 safe in row six.

In row two, we have a total of 9 in marked mines, so need 21 more from (9,8,7,4,3) which can only be 9,8,4.
enter image description here

The 5 in R2C8 needs two more mines. They cannot be the 9 and the 6, as that would exceed the row three sum of 14.
So we must have the 1 and one of (6,9) as mines.
That means the other three unknowns around R4C7 must all be mines.
And that gives us a 7 in row three, so we know the 6 must be the last mine in this row.

The 4 in R5C6 needs two more mines. We cannot have both the 5 and 6 in row six (row sum of 7), so the 1 in R4C5 must be a mine.
And we only need 1 or 2 more in row six, so everything else is safe.
enter image description here

Row four needs 12 more from (9,8,7,6,3) which can only be 9,3.

The 1 in R6C3 must be a mine to complete the 2 above. Then the other mine in row six must be the 6. enter image description here

The 8 in row seven cannot be a mine, because that would make all other cells except 1 in that row safe, which would make the uncircled 2 above a correct clue.
So (still in row seven) to avoid making the 2 in R6C8 a correct clue, either both the 2 and 3 are mines or neither are.
If both are mines, we need 4 to be a mine to complete the row.
If neither are mines, we need the 5 to be a mine to complete the 3 in R6C7, and again need the 4 to complete the row.
So we know the 4 is a mine, with either (2,3) or 5. And the rest of that row is safe. enter image description here

The 9 in row eight cannot be a mine, because that would make the rest of that row safe, leaving too few possible mines for the 3 in R9C6.
For the same reason, the 6 in that row cannot be a mine, as the only way to complete the row would then be the 3, again making the rest of the row safe.
And, the 7 in that row cannot be a mine, as the 2 is circled.

The 9 in row nine cannot be a mine, because that would make the rest of the row safe, requiring three mines in the row above the 3 in R9C6, which would exceed the total for row eight.
enter image description here

From here, I struggled to find a logical path forward.
If the 8 in row 8 is a mine, then (after a chain of mine placements) the uncircled 3 in R8C2 ends up being a correct clue.
So the 8 must be safe, and the 5 must be a mine. Now the 1 and 4 in row eight cannot both be mines, so the 7 and 1 in row nine must be for the 3 in R9C6.
In row nine, that leaves us needing 10 more from (8,6,5,4,2) which must be (8,2) or (6,4). So the 5 is safe.

But from there, I can find three apparently valid solutions, so what have I missed?:
The intended solution (applying A1Z26 to the sum of mines in each column to get the solution):
enter image description here

Two other solutions to the minesweeper puzzle:
enter image description here
enter image description here

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  • $\begingroup$ Awesome solve! You didn't miss anything; these are the 3 possible solutions but because of c6 (as you correctly pointed it out), only one of them could be compatible with the "name of the puzzle" clue on top. $\endgroup$
    – Jujustum
    Dec 12, 2023 at 15:40
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This is a

MINESWOKU

My completed sudoku looks like

solved sudoku

Then we

treat the circled cells as minesweeper clues, with numbers at the side indicating the sum of flagged mines in each row. Note we also need to use the total number of mines (31) given to obtain a unique solution

To get the following grid:

At this point you might as well refer to fljx's answer

From here we

sum the mines along columns and convert to letters with A=1, B=2, ... to give:
MIN ESW OKU

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    $\begingroup$ I think you have to take into account the negative constraint: if this were correct, for example the 2 in R6C8 would also have been circled. OP says that every correct circle has been placed ... $\endgroup$ Dec 12, 2023 at 12:00
  • $\begingroup$ Nice solution. I tried this approach (along with several other dead-ends), but stopped when I got a contradiction between the top left 1 and the second row total. $\endgroup$
    – fljx
    Dec 12, 2023 at 12:07
  • $\begingroup$ @sarsaparilla The way the column totals produce something intelligible makes me think this is probably the right solution, with a couple of errors/typos in the puzzle. $\endgroup$
    – fljx
    Dec 12, 2023 at 12:07
  • $\begingroup$ Oh, the second flag is indeed wrong; it’s supposed to be 30 and not 28! I’ll still +1 the answer because the reasoning is here - although you do have to take into account the negative constraint as pointed out by sarasparilla. I also really like that you found an unintended yet intelligible solution! $\endgroup$
    – Jujustum
    Dec 12, 2023 at 13:38

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