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Contains: Complex mines ($i$ mines, where $i:=\sqrt{-1}$), Anti mines (-1 mines)

Note: This is in conjunction with my Minesweeper puzzles

Sorry for not posting a Minesweeper puzzle yesterday. I will try my best to get Day 10 and 11 out tomorrow

So the gimmick today is that there are complex mines! What is a complex mine, you may ask? Well, a complex mine is $i$ mines, where$$i:=\sqrt{-1}$$The reason we are including anti mines here (where an anti mine is equal to (-1) mines, you can learn more about them on Day 16 of the Minesweeper Advent Calendar here) The general definition of an anti-mine is a mine that counts (-1) mines towards a tile, which in some cases can result as a tile having a total of 0 mines that the tile "sees". (a total of $n$ mines added to $n$ anti-mines, where the total of anti mines "seeable" by a tile is the same of the total of regular mines (single, double, triple mines) "seeable" by a mine) Here is the puzzle:
4i 4i
-6+i -3+5i
-2 6i 5i
-2 -6 -4+3i -1+3i 5i
-1+3i -2+3i
-4+i -8 -3+i -2+2i
-3+2i -3+2i -5
-2+i -5

If I have counted correctly, there are 18 complex mines and 17 anti mines.

Note that there are no numbers that have been removed to make this gimmick more easily understandable.

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    $\begingroup$ −1. This puzzle doesn't explain what an anti mine is, nor what it means to take the square root of a mine, and as far as I can tell they're not standard terms. Solving it seems to be impossible without that info. $\endgroup$
    – msh210
    Nov 2 at 14:17
  • $\begingroup$ @msh210 Ah my bad, I'll fix that quickly $\endgroup$
    – CrSb0001
    Nov 2 at 14:18
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    $\begingroup$ rot13(Gur gjb $4v$f ba gur gbc evtug zhfg or pbzcyrgryl pbirerq va $v$-zvarf.) $\endgroup$
    – Someone
    Nov 2 at 14:59

1 Answer 1

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Plot twist!

It is not uniquely solvable, the top left corner could be either an additional "i" on just the -3+5i and -6+i respectively, or the two in the middle could have one "i" and one "-1". The ol' minesweeper non determinism-eroo
i 4i 4i i
- -6+i -3+5i i i i i
-2 - - i i 6i i 5i
-2 -6 - -4+3i -1+3i 5i i i
- - - - i -1+3i i -2+3i
-4+i - -8 - -3+i -2+2i - -
i - - - -3+2i -3+2i - -5
-2+i - -5 - i i - -
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    $\begingroup$ didn't hide table behind spoiler cause I wasn't sure how to. $\endgroup$
    – Going hamateur
    Nov 2 at 15:00
  • $\begingroup$ ah dang it, it's not uniquely solvable again :( but yeah good job $\endgroup$
    – CrSb0001
    Nov 2 at 15:16
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    $\begingroup$ @CrSb0001 I recommend you go through and solve your puzzles yourself before posting. It's the easiest way to check both difficulty and whether the solution is unique $\endgroup$
    – Beastly Gerbil
    Nov 2 at 16:17
  • $\begingroup$ for example an easy fix would be putting either a -1+2i or a -2+i in the 3rd square and adjusting the -6+i and -3+5i accordingly. $\endgroup$
    – Going hamateur
    Nov 2 at 17:42

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