# Tag Info

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Assuming standard Minesweeper rules, here’s one solution (with $X$ = a mine): EDIT: In response to Euphoric in the comments, I solved this purely by logical deduction with a bit of educated guessing to make things easier on me. But if you really want to know how I did it, here’s a rigorous solution:

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Don't trust the (formerly) accepted answer. :-) (It depends, in part, on a mistake you've made earlier: there's a third flag next to a couple of 2s along the left side. I added a green square to mark the bogus flag in the picture below.) Despite that, you have several break-ins left to use: the flag that's next to the mistaken flag has a 1 next to it. the 1-...

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Many Minesweeper games, including the one that ships with Windows, do not allow the first square you click to be a mine. The lower-left-hand corner is not necessarily safe, but if it's the first square you click, then it will be safe.

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Although the puzzle is most likely to be solved without a computer, and we already have a winner, here are all 16 solutions, just for the record: There are some symmetries in there, of course. Whether rotations and flips should count as "different boards" is a matter of interpretation. Found with the following (quick and dirty) Java program that jus ...

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I think that yes, it is possible. You can create a custom grid with a ratio of more than 8 bombs per empty square, then, by the pigeonhole principle, there exists a bomb with no empty square next to it, since each empty square can be adjacent to at most 8 bombs. I don't have my own picture but I found this on the internet: Source

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There are 50 different possible ways that the unknown mines next to the revealed region could be configured: Here, the green cells are clear (no mines), while the X's around the perimeter indicate the different ways the mines could potentially be placed. If we consider each of these to be of equal probability (probably not quite true, because the total ...

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The crossed bomb corresponds to a square where a flag had been placed while no bomb was on it. Therefore it is not actually a bomb and this solves the problem.

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The 2 down-right from your bad flag should have tipped you off; it is touching two bombs you had already flagged. That would show the one next to it (which you triggered) to be a bomb; the two under that one would clear the one left of it (above the 4), leaving just the four bombs around the 4. You could also have figured out the 4's surroundings just from ...

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I think this arrangement of mines will work (red squares are mines)

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Yes: Consider the four boxed cells here. The 3 above C tells you that at least one of B and C has a mine. But the 3 above B can only accept one more mine! Therefore there is exactly one mine in {B,C}... and so A is safe, and D is a mine. (You can also do a similar trick with the 2 and 1 just to the right of the 3s!)

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My answer is: This was found:

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(I did this without looking at the CW. I claim no credit for imposing arbitrary restrictions on myself, but it means any mistakes are my own :-).) The final grid is as follows: which obeys the following constraint: If we interpret we get "First", "Second", etc., are of course When groups of cells are referred to collectively, Note: in the course of ...

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I (finally) finished it ! Blue squares are 'colored', and green squares are 'not colored'. This took a few days (I think I started less than a day after the problem was published), and I worked around 6-8 hours (roughly) on this (I'm slow). Every single pixel here was hand-placed in paint (please don't tell me how many pixels I colored - I don't think I ...

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Apart from the solution that hexomino found, there is another solution: According to my computer program, there are no other solutions up to symmetry (so 4 solutions if we count the rotated/reflected pattern as distinct).

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As GentlePurpleRain says in their excellent answer, there are $50$ different possible placements for the mines in the squares around the solved region. However they make the assumption that each of these possibilities is equally likely. This is not correct. The possibilities that GentlePurpleRain lists contain either $4$, $5$, $6$, or $7$ mines in the ...

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This answer, although it does address the question itself, is more of an interesting observation coming from analysing the situation by expressing the system in terms of simple linear equations (eg/ using capitals for the reds, you might say A+B=1 to say that exactly one of A and B can be a mine, or F+G+H+I=2 because there must be exactly two mines amongst ...

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The most you can do according to these new rules is 8 tiles, with the following grid: 1 2 3 3 3 2 1 2 # # # # # 2 3 # ? ? ? # 3 3 # ? 7 ? # 3 3 # ? ? ? # 3 2 # # # # # 2 1 2 3 3 3 2 1 # represents an already-discovered mine. To show that this is maximal, look at the "information-conveying" tiles required by the instructions. We'll call these "useful tiles"...

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Code: Some sample outputs: .

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First: Now, an interesting step: Some more logic sprouts off of the same area: And hey, wait a second...

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This is equivalent to asking how to place kings on a 6x6 board so that each empty square is attacked by 3 of those kings. This is a special case of the problem posed here. That web site gives hexomino's 16-king solution for the 6x6, among -king solutions for the 2x2, ..., 8x8 boards.

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I can get this much: which is to say, Here's how it goes. Then After this [EDITED to add:] Here's a sketch of how to prove that Note that I am writing this after reading other people's solutions :-) but I haven't read them in much depth and am not deliberately copying anything. So, first of all, What about the top region? Well, So the only question ...

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