python 解数独 多种解法

时间:2024-01-20 19:16:37

方法一:回溯法

回溯法是解决数独问题的常用方法。其基本思想是在数独的空格中填入数字,如果填写了一个错误的数字,就回溯到前一个空格重新填写,直到找到正确的解。

具体实现如下:

def solve_sudoku(board):
    # 找到未填的空格
    row, col = find_empty(board)
    
    # 如果没有未填的空格,则说明已经解决了数独问题,返回 True
    if row is None:
        return True
    
    # 在当前空格中填入数字 1 到 9
    for i in range(1, 10):
        if valid(board, row, col, i):
            # 如果填写的数字是有效的,则继续填写下一个空格
            board[row][col] = i
            if solve_sudoku(board):
                return True
            # 如果填写的数字导致后面无法找到正确的解,则回溯到上一个空格
            board[row][col] = 0
    
    # 如果在当前空格中填写了所有可能的数字都无法找到正确的解,则返回 False
    return False

其中,find_empty() 函数用于找到未填的空格,valid() 函数用于判断在当前空格填写的数字是否有效。

方法二:Dancing Links 算法

Dancing Links 算法是一种高效的求解精确覆盖问题的算法,其可以用于解数独问题。

具体实现如下:

class DancingLinks:
    def __init__(self, matrix):
        self.row = len(matrix)
        self.col = len(matrix[0])
        self.matrix = matrix
        self.header = [Node() for _ in range(self.col)]
        self.solution = []
        
        for i in range(self.col):
            self.header[i].right = self.header[(i + 1) % self.col]
            self.header[i].left = self.header[(i - 1 + self.col) % self.col]
        
        for i in range(self.row):
            prev = None
            for j in range(self.col):
                if matrix[i][j] == 1:
                    node = Node(i, j)
                    if prev is None:
                        prev = node
                    self.header[j].size += 1
                    node.down = self.header[j]
                    node.up = self.header[j].up
                    self.header[j].up.down = node
                    self.header[j].up = node
                    prev.right = node
                    node.left = prev
                    prev = node
    
    def solve(self):
        if self.header[0].right == self.header[0]:
            return True
        
        column = self.get_min_column()
        self.cover(column)
        
        node = column.down
        while node != column:
            self.solution.append(node)
            j = node.right
            while j != node:
                self.cover(j.down)
                j = j.right
            
            if self.solve():
                return True
            
            node = self.solution.pop()
            column = node.column
            j = node.left
            while j != node:
                self.uncover(j.down)
                j = j.left
        
        self.uncover(column)
        return False
    
    def cover(self, column):
        column.right.left = column.left
        column.left.right = column.right
        node = column.down
        while node != column:
            j = node.right
            while j != node:
                j.down.up = j.up
                j.up.down = j.down
                self.header[j.column].size -= 1
                j = j.right
            node = node.down
                
    def uncover(self, column):
        node = column.up
        while node != column:
            j = node.left
            while j != node:
                j.down.up = j
                j.up.down = j
                self.header[j.column].size += 1
                j = j.left
            node = node.up
        column.right.left = column
        column.left.right = column
    
    def get_min_column(self):
        min_column = self.header[0]
        for col in self.header:
            if col.size < min_column.size:
                min_column = col
        return min_column

class Node:
    def __init__(self, row=None, column=None):
        self.up = None
        self.down = None
        self.left = None
        self.right = None
        self.row = row
        self.column = column

其中,DancingLinks 类用于实现 Dancing Links 算法,solve() 方法用于求解数独问题,cover()uncover() 方法用于在 Dancing Links 矩阵中删除和恢复某一列及其对应的行,get_min_column() 方法用于找到 Dancing Links 矩阵中未覆盖的节点数最少的列。在 Dancing Links 矩阵中,每一行表示数独中某一个空格可以填写的数字,每一列表示数独中某一个空格。