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Can you detect where the mines are in this 8x8 Minesweeper grid containing single, double, triple, and anti-mines?

1 3 4
-1 21
5 3 1 9
3 -1
3 -5 -4 -3
13 4 5
20 12
5 5

Notes:

  1. Anti-mines are worth (-1) mines: there are 17 of them

  2. Single mines are worth (1) mine: there are 3 of them

  3. Double mines are worth (2) mines: there are 5 of them

  4. Triple mines are worth (3) mines: there are 16 of them

  5. Partial answers are by any and all means definitely allowed, however, you must figure out the location of all 42 mines to get the green checkmark.

  6. I will post a hint in 72 hours if no one has managed to get a total of 21 mines filled out (half the grid).

  7. I have checked, and there should only be one unique solution.

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16
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    $\begingroup$ Is a solution required, or can I just post a hidden final grid? $\endgroup$
    – Someone
    Oct 24, 2023 at 21:17
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    $\begingroup$ But $17+2+7+17=43\neq 44$ and $22$ numbered spaces in $8\times 8=64$ grid should give at most $42$ mines. $\endgroup$ Oct 24, 2023 at 21:23
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    $\begingroup$ Also the 1 on the top middle cannot be solved with the 21 next to it; 6 -2 is still 4 not 1 $\endgroup$
    – Stevo
    Oct 24, 2023 at 21:27
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    $\begingroup$ Not trying to nitpick. But now the mines add up to 41 still not 42. I actually like this idea and had made good progress (i.stack.imgur.com/iVHqW.png) but I can say with certainty there are at least 2 solutions, and the numbers dont add up, so downvoting for now. Should always try to solve a puzzle yourself before posting to check the solution and solution path is unique, I will ofc upvote instead if the problems get fixed, but this is flawed rn $\endgroup$ Oct 24, 2023 at 23:19
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    $\begingroup$ Still infeasible. $\endgroup$
    – RobPratt
    Oct 24, 2023 at 23:34

1 Answer 1

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Although it is not necessarily unique, one solution of the puzzle is:

enter image description here

As far as I can tell, the other solutions

Switch the (-1,2) in rows 3-4, column 1 or switch the (2,3) in rows 6-7, column 1

I approached and solved this problem by

Realizing 21 must consist of 7 3s. The additional neighbors of 9 must be -1s (since the sum is already 12). By similar logic, I worked out which pairs of mines were needed for the remaining numbers until it all filled in!

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    $\begingroup$ @CrSb0001 I'm not sure how to hide comments so I added a brief explanation on how I solved it in the body of my post! $\endgroup$
    – Bryce
    Oct 25, 2023 at 1:46
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    $\begingroup$ This solution does not satisfy the required counts in notes 1 through 4. If you ignore those, there are 276 solutions. $\endgroup$
    – RobPratt
    Oct 25, 2023 at 1:58
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    $\begingroup$ Ahh I didn’t notice that while solving @RobPratt. Do you have the combination of 1/2/3/-1 mines for all the solutions? I’d be interested to see how much they varied if at all $\endgroup$
    – Bryce
    Oct 25, 2023 at 4:29
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    $\begingroup$ Here are the ranges of counts for the different values in [-1,3]: $$-1: [15,17], 0: [0,4], 1: [1,6], 2: [2,10], 3: [14,18]$$ $\endgroup$
    – RobPratt
    Oct 25, 2023 at 13:49

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