First uncategorized solution for Minesweeper by Juge_Ti - python coding challenges

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First solution in Uncategorized category for Minesweeper by Juge_Ti

def checkio(mat):
    if mat == [[-1]*10]*10:
        return [False, 0, 0]
    newmat = [x + [10] for x in mat] + [[10]*11]
    for i, row in enumerate(mat):
        for j, x in enumerate(row):
            neighbours = [newmat[i + k//3 - 1][j + k%3 - 1] for k in range(9)]
            neighbours[4] = 10
            if x == -1 and 0 in neighbours:
                return [False, i, j]
            if 0 < x < 9 and -1 in neighbours:
                k = neighbours.index(-1)
                if neighbours.count(9) == x:
                    return [False, i + k//3 - 1, j + k%3 - 1]
                if neighbours.count(9) + neighbours.count(-1) == x:
                    return [True, i + k//3 - 1, j + k%3 - 1]

if __name__ == '__main__':
    #These are just examples
    checkio([
        [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
        [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
        [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
        [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
        [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
        [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
        [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
        [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
        [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
        [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
    ])  # [False, 0, 0]
    checkio([
        [0, 2, -1, -1, -1, -1, -1, -1, -1, -1],
        [0, 2, -1, -1, -1, -1, -1, -1, -1, -1],
        [0, 1, 1, 1, -1, -1, -1, -1, -1, -1],
        [0, 0, 0, 1, -1, -1, -1, -1, -1, -1],
        [0, 1, 1, 2, -1, -1, -1, -1, -1, -1],
        [0, 1, -1, -1, -1, -1, -1, -1, -1, -1],
        [0, 1, -1, -1, -1, -1, -1, -1, -1, -1],
        [2, 1, -1, -1, -1, -1, -1, -1, -1, -1],
        [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
        [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
    ])  # [True, 0, 2]

Sept. 13, 2013

Comments:

nickie on Jan. 25, 2014, 12:30 a.m.
I believe we have the same algorithm with a few rather technical differences, affecting performance (I guess your solution is faster) and readability (I guess mine is more clear). The only algorithmically non-trivial difference is your "if" statement in line 9, which handles the case of an unknown cell with a 0 in its neighbors. Checking this separately is redundant, as it will be covered by your "if" statement in line 13, when the 0 cell is reached. Other than that, the similarity of our solutions lies in the basic principles, which I explain here (in pseudocode) for those who care to read it: <pre class='brush: python'>for each cell in the field: if the cell contains a number X with 0 <= X <= 8: find its neighbors if there is no unknown neighbor: continue if X = the number of neighboring mines: pick any of the unknown neighbors and declare that it's empty if X = the number of neighboring mines plus unknown neighbors: pick any of the unknown neighbors and declare that it's a mine </pre> Now, the technical differences between our solutions are: <ol> <li>You're returning the upper left corner immediately by checking if the whole field contains -1, while I'm returning it in the end, if it's a plausible answer and I realize I have nothing better to do.</li> <li>You're adding an extra row (to the bottom of the field) and an extra column (to the right), using the value 10 as a sentinel which correctly handles the field's borders when looking for neighbors. What puzzled me for a minute is that you don't have to add anything to the top or to the left of the field: if an index becomes -1, Python automatically takes the last value which is the other end's sentinel (OMG!). On the other hand, I'm being more verbose and check if my indices are inside the field's bounds.</li> <li>You are using range(9) and modulo 3 to access the neighbors. This gives arguably shorter code (but forces you to treat 4 as a special case) but makes it less readable. On the other hand, I'm enumerating the neighbors explicitly.</li> </ol> Well done...

veky on March 6, 2014, 4:52 p.m.
Juge_Ti is the perfect glue: people who have nothing in common often think they have similar solution as him. :-D

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