I would assume (though without any real proof) that a program could be written that could solve minesweeper in linear time (as the board gets bigger linearly, if the mines/area ratio stays the same). Rather hoping there is a clever way to do this, but I will not be so surprised if there isn't. The question is whether or not I am correct in saying that this is a non-unique solution.How would someone go about doing this? Assume that the first "click" will never be a bomb, and that the number of mines and the area are both known. The rule of the 2 square is again broken (too little mines) Then, the rule of the 2 square is broken (too many mines)Ĭase 2) They are both in the right column of the square. The two mines cannot be adjacent to each other for the following reason:Ĭase 1) They are both in the left column of the square. Likewise, at least one of the bottom must be filled, in order to fill the required amount of mines. If you fill two, then the rule of the 3 square is broken. In the top section of the square, exactly 1 mine must be filled, so that rule of the square with a three is not broken. Basic logic shows that that is the only way to solve it. The bottom right 4 squares require a pattern of two squares with mines, and two without, forming the diagonals of the square. Here is a screenshot of the final position: From the research I have done, it appears thatī) A method to determine this solution without any guesses I appear to have encountered a position in which it is impossible to logically determine whether or not a square is a mine.
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