0
$\begingroup$

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.

$\endgroup$
3
  • 2
    $\begingroup$ −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, 2023 at 14:17
  • $\begingroup$ @msh210 Ah my bad, I'll fix that quickly $\endgroup$
    – CrSb0001
    Nov 2, 2023 at 14:18
  • 1
    $\begingroup$ rot13(Gur gjb $4v$f ba gur gbc evtug zhfg or pbzcyrgryl pbirerq va $v$-zvarf.) $\endgroup$
    – Someone
    Nov 2, 2023 at 14:59

1 Answer 1

1
$\begingroup$

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 - -
$\endgroup$
4
  • 1
    $\begingroup$ didn't hide table behind spoiler cause I wasn't sure how to. $\endgroup$ Nov 2, 2023 at 15:00
  • $\begingroup$ ah dang it, it's not uniquely solvable again :( but yeah good job $\endgroup$
    – CrSb0001
    Nov 2, 2023 at 15:16
  • 1
    $\begingroup$ @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 $\endgroup$ Nov 2, 2023 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. $\endgroup$ Nov 2, 2023 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.