# Tag Info

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Short answer: Longer answer, no spoiler tags: To be systematic about it, first we can update the numbers in squares that are adjacent to both blue squares and flags. By "update the numbers" I mean decrease the numbers by however many flags are adjacent to the square. So, for example, the square above the $\color{red}{A}$ decreases from $2$ to $1$ because ...

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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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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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It is better to start in the corner in Minesweeper. The reason for this is simple: corners are the most likely place for logically unsolvable positions. The more corners you remove, the greater the chance you'll be able to solve the puzzle, regardless of mine density. Since your starting move can't be a mine, starting there reduces the chance of being stuck ...

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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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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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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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Small notes Remember that the number of remaining mines is the clue number minus the number of surrounding mines. This is important in patterns like 1-2-1; 3-2-3, for instance, also works if there are two mines next to each 3, since they both need 1 mine to be solved. A one by itself on a corner always indicates that there is a mine A two by itself next to ...

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There is no contradiction here. The base rules of Minesweeper just state that there are mines, and that the number of mines adjacent to any square is indicated. A guarantee of uniqueness or logical deductibility is not required. Any variation on those rules is solely an implementation choice. While some implementations do guarantee a logically unique ...

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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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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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In general, when looking at these situations, you want to click on the squares that are the least likely to contain mines. This will invariably give you more information about your current situation. Here's an example of a game that can't be beaten without strict testing, which I will use to demonstrate what I mean. There are two mines remaining. I ...

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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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Given your current situation, I believe that I would therefore guess that Diagram:

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Here is another way of completing this puzzle:

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If you're going by Microsoft Minesweeper rules (or any where the number of mines are known), and assuming we aren't limited by the artificial limitation Microsoft provides on board creation, the answer is which happens to be and is a trivial solution. Once you mark the opening move, which will be an 8, the board is solved, either automatically in the case ...

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It feels like there's more slack in here for other solutions 1. 2. 3.

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A MineSweeper Crossword: A Making of I created this a puzzle a while ago, but @humn suggested that a "making of" wrap-up post would not be amiss, as such, here it is (I tried to keep it fairly spoiler free for those that wish to solve the puzzle at a later date).This was written to the best of my recollection (some things may have happened slightly ...

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There are four regions of unknown tiles. Call the three at the bottom A (size 5), B (size 2), C (size 2) from left to right, and the one at top right D. Region C Regions A and B So now region D At this point we have the following unresolvable unknowns: To be more explicit about what is guaranteed:

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Here is what I have: . . .

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