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Sudoku Solver

I wanted to make my own sudoku solver to challenge myself. Im not a sudoku player so my approach is a brute force scan of possible combinations sort-of. I just know the basic rules:
  • Numbers 1-9 are allowed.
  • Numbers in the same row cannot be repeated.
  • Numbers in the same column cannot be repeated.
  • Numbers in the 3x3 square cannot be repeated.
  • The first thing i did was to build a some classes that calculates the possible values a cell can have if it's empty, based on the constraints. I came up with 3 classes:
  • Board that stores the entire board.
  • BoardSlice that stores a slice of a board. An object of this type is returned when a Board is sliced (method __getitem__).
  • Cell that stores the value of a single cell and calculates all possible values a cell can take.
  • The class Cell receives a board, the coordinates on the board, and the value that holds. Also has the method options that uses python set data structure to calculate the posibilites. If you look at the following snippet you can see that the method options generates the sets: options that contains all possible options (1-9), row that contains all the numbers that are in the same row, column that contains all the numbers that are in the same column and square that contains all the numbers that are in the same 3x3 square. The return value is options without all the used values.
    class Cell:
        def __init__(self, b, i, j, value):
            self.b = b
            self.value = value
            self.i = i
            self.j = j
        def options(self):
            if self.value != 0:
                return {self.value}
            options = set(range(1, 10))
            row = set(map(lambda x: x.value, self.b[self.i]))
            column = set(map(lambda x: x.value, self.b[:][self.j]))
            def to_square(k): return slice((k // 3) * 3, (k // 3) * 3 + 3)
            square = set(
                map(lambda x: x.value,
            return options - row - column - square - {0}
    To make easier the implementation of the square I used the class BoardSlice that contains a slice of a board and implements the magic method __getitem__.
    class BoardSlice:
        def __init__(self, board_slice):
            self.board_slice = board_slice
        def __getitem__(self, items):
            if type(items) == slice:
                return (el for row in self.board_slice for el in row[items])
            if type(items) == int:
                return (row[items] for row in self.board_slice)
            raise KeyError
    The base class: Board contains the board and a copy method that copies all the values and creates a new Board object. This is necessary to avoid messing with object references and have a clean object when needed.
    class Board:
        def __init__(self, board):
            self.board = [[Cell(self, i, j, value)
                           for (j, value) in enumerate(row)] for (i, row) in enumerate(board)]
        def copy(self):
            return Board(((cell.value for cell in row) for row in self.board))
        def __getitem__(self, items):
            if type(items) == int:
                return self.board[items]
            if type(items) == slice:
                return BoardSlice(self.board[items])
            raise KeyError
        def __repr__(self):
            return repr(self.board)
    With these tools the next step is to solve the problem! My idea was to generate a mixed iterative-recursive algorithm. The first pass will be iterative, and if needed, the second pass will be recursive.
    Iterative pass
    Iterates over the whole board and calculates the options that each cell can have. If a cell has only one option set that option on the cell and set a flag to repeat the iterative pass, if has 0 options return None meaning that the board has no solutions, and if has more than one option store the options for the recursive pass. If the loop ends and we found that no cell has more than one option then we solved the board! The idea of this first step is to solve an _easy_ board quickly.
    Recursive pass
    If the iterative pass ends and we found that a cell has more than one option then we try all that options and call solve again! If solve returns a board that means we've found the solution! If solve returns None (back at the iterative passs) we have to try with another options.
    The class is pretty straightforward.
    class SudokuSolver:
        def solve(board):
            b = board.copy()
            # First pass: Iterative
            board_map = {}
            exhaust = False
            while not exhaust:
                exhaust = True
                for i in range(9):
                    for j in range(9):
                        cell = b[i][j]
                        if cell.value == 0:
                            options = cell.options()
                            if len(options) == 1:
                                cell.value = options.pop()
                                exhaust = False
                            elif len(options) == 0:
                                return None
                            elif len(board_map) == 0:
                                board_map[(i, j)] = options
            # Second pass: Recursive
            for ((i, j), options) in board_map.items():
                for op in options:
                    b[i][j].value = op
                    solved = SudokuSolver.solve(b)
                    if solved:
                        return solved
                return None
            return b
    Actually my implementation is not a brute force algorithm, is a search algorithm, that searches the path to solving a board. Because it doesn't try all values on all cells nonsensically, it rather tries _some_ options for a given cell and advances to the next option as _soon_ as it detects that it's not the correct path.


    Take a look at the source code.