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2021-01-31 21:48
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求用python代码解奇偶数独和堡垒数独

标准数独概念如下:

现有求解代码:

import datetime

sudoku2 = [
    [0, 4, 0,    5, 1, 7,    0, 6, 0],
    [2, 0, 0,    0, 4, 0,    0, 0, 9],
    [0, 0, 3,    0, 0, 0,    5, 0, 0],

    [5, 0, 0,    4, 0, 1,    0, 0, 7],
    [9, 2, 0,    0, 0, 0,    0, 8, 5],
    [7, 0, 0,    9, 0, 5,    0, 0, 1],

    [0, 0, 8,    0, 0, 0,    7, 0, 0],
    [6, 0, 0,    0, 7, 0,    0, 0, 8],
    [0, 5, 0,    8, 6, 9,    0, 3, 0]
]

def wr_sudoku(board):  # 判断数独是否符合条件
    # 判断一行是否有效
    for i in range(9):
        for j in board[i]:
            if (j != '.') and (board[i].count(j) > 1):
                return False
        # 判断一列是否有效
        column = [k[i] for k in board]
        for n in column:
            if (n != '.') and (board[i].count[n] > 1):
                return False
    # 判断九宫格是否有效
    for i in range(3):
        for j in range(3):
            grid = [tem[j * 3:(j + 1) * 3] for tem in board[i * 3:(i + 1) * 3]]
            merge_str = grid[0] + grid[1] + grid[2]  # 合并为一个list[]
            for m in merge_str:
                if (m != '.') and (merge_str.count(m) > 1):
                    return False
    return True


class fill_sudoku(object):
    def __init__(self, board):
        self.b = board
        self.t = 0

    def check(self, x, y, value):  # 检查每行每列及每九宫是否有相同项
        for row_item in self.b[x]:
            if row_item == value:
                return False
        for row_all in self.b:
            if row_all[y] == value:
                return False
        row, col = int(x / 3) * 3, int(y / 3) * 3
        row3col3 = self.b[row][col:col + 3] + self.b[row + 1][col:col + 3] + self.b[row + 2][col:col + 3]
        for row3col3_item in row3col3:
            if row3col3_item == value:
                return False
        return True

    def get_next(self, x, y):  # 得到下一个未填项
        for next_soulu in range(y + 1, 9):
            if self.b[x][next_soulu] == 0:
                return x, next_soulu
        for row_n in range(x + 1, 9):
            for col_n in range(0, 9):
                if self.b[row_n][col_n] == 0:
                    return row_n, col_n
        return -1, -1  # 若无下一个未填项,返回-1

    def try_it(self, x, y):  # 主循环
        if self.b[x][y] == 0:
            for i in range(1, 10):  # 从1到9尝试
                self.t += 1
                if self.check(x, y, i):  # 符合 行列宫均无条件 的
                    self.b[x][y] = i  # 将符合条件的填入0格
                    next_x, next_y = self.get_next(x, y)  # 得到下一个0格
                    if next_x == -1:  # 如果无下一个0格
                        return True  # 返回True
                    else:  # 如果有下一个0格,递归判断下一个0格直到填满数独
                        end = self.try_it(next_x, next_y)
                        if not end:  # 在递归过程中存在不符合条件的,即 使try_it函数返回None的项
                            self.b[x][y] = 0  # 回朔到上一层继续
                        else:
                            return True
    def start(self):
        begin = datetime.datetime.now()
        if self.b[0][0] == 0:
            self.try_it(0, 0)
        else:
            x, y = self.get_next(0, 0)
            self.try_it(x, y)
        for i in self.b:
            print(i)
        end = datetime.datetime.now()
        print('cost time: ', end - begin)
        print('times: ', self.t)
        return
s = fill_sudoku(sudoku2)
s.start()

求改编代码解奇偶数独和堡垒数独

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2条回答 默认 最新

  • 天际的海浪 2021-01-31 23:14
    已采纳
    # 堡垒数独
    
    import datetime
    
    sudoku2 = [
        [0, 0, 0,    0, 0, 0,    0, 0, 0],
        [0, 0, 0,    0, 0, 0,    0, 0, 0],
        [0, 5, 7,    1, 2, 3,    6, 4, 0],
    
        [0, 7, 0,    0, 0, 0,    0, 2, 0],
        [0, 0, 0,    9, 6, 8,    0, 0, 0],
        [0, 0, 0,    0, 0, 5,    0, 0, 0],
    
        [3, 9, 1,    4, 5, 6,    8, 7, 2],
        [0, 0, 0,    0, 9, 0,    0, 0, 0],
        [0, 0, 4,    7, 3, 0,    0, 0, 0]
    ]
    
    
    class fill_sudoku(object):
        def __init__(self, board):
            self.b = board
            self.t = 0
    
        KH = (
            (0,2),(0,4),(0,6)
        )
        PD = ( (0,-1),(-1,0),(0,1),(1,0) )
    
        def check(self, x, y, value):  # 检查每行每列及每九宫是否有相同项
            for row_item in self.b[x]:
                if row_item == value:
                    return False
            for row_all in self.b:
                if row_all[y] == value:
                    return False
            row, col = int(x / 3) * 3, int(y / 3) * 3
            row3col3 = self.b[row][col:col + 3] + self.b[row + 1][col:col + 3] + self.b[row + 2][col:col + 3]
            for row3col3_item in row3col3:
                if row3col3_item == value:
                    return False
    
            if (x,y) in self.KH:
                for dx,dy in self.PD:
                    tx = x + dx
                    ty = y + dy
                    if 0<=tx<=8 and 0<=ty<=8 and (tx,ty) not in self.KH and value <= self.b[tx][ty]:
                        return False
    
            return True
    
        def get_next(self, x, y):  # 得到下一个未填项
            for next_soulu in range(y + 1, 9):
                if self.b[x][next_soulu] == 0:
                    return x, next_soulu
            for row_n in range(x + 1, 9):
                for col_n in range(0, 9):
                    if self.b[row_n][col_n] == 0:
                        return row_n, col_n
            return -1, -1  # 若无下一个未填项,返回-1
    
        def try_it(self, x, y):  # 主循环
            if self.b[x][y] == 0:
                for i in range(1, 10):  # 从1到9尝试
                    self.t += 1
                    if self.check(x, y, i):  # 符合 行列宫均无条件 的
                        self.b[x][y] = i  # 将符合条件的填入0格
                        next_x, next_y = self.get_next(x, y)  # 得到下一个0格
                        if next_x == -1:  # 如果无下一个0格
                            return True  # 返回True
                        else:  # 如果有下一个0格,递归判断下一个0格直到填满数独
                            end = self.try_it(next_x, next_y)
                            if not end:  # 在递归过程中存在不符合条件的,即 使try_it函数返回None的项
                                self.b[x][y] = 0  # 回朔到上一层继续
                            else:
                                return True
        def start(self):
            begin = datetime.datetime.now()
            if self.b[0][0] == 0:
                self.try_it(0, 0)
            else:
                x, y = self.get_next(0, 0)
                self.try_it(x, y)
            for i in self.b:
                print(i)
            end = datetime.datetime.now()
            print('cost time: ', end - begin)
            print('times: ', self.t)
            return
    s = fill_sudoku(sudoku2)
    s.start()
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  • 天际的海浪 2021-01-31 22:40
    # 奇偶数独
    
    import datetime
    
    sudoku2 = [
        [0, 0, 5,    1, 0, 6,    2, 0, 0],
        [0, 4, 2,    0, 0, 0,    1, 3, 0],
        [6, 8, 0,    0, 0, 0,    0, 5, 9],
    
        [1, 0, 0,    0, 0, 0,    0, 0, 4],
        [0, 0, 0,    0, 0, 0,    0, 0, 0],
        [9, 0, 0,    0, 0, 0,    0, 0, 7],
    
        [5, 7, 0,    0, 0, 0,    0, 6, 1],
        [0, 6, 9,    0, 0, 0,    3, 8, 0],
        [0, 0, 3,    6, 0, 5,    7, 0, 0]
    ]
    
    
    class fill_sudoku(object):
        def __init__(self, board):
            self.b = board
            self.t = 0
    
        KH = (
            (2,2),(2,3),(3,2),(3,3),
            (2,5),(2,6),(3,5),(3,6),
            (5,2),(5,3),(6,2),(6,3),
            (5,5),(5,6),(6,5),(6,6)
        )
    
        def check(self, x, y, value):  # 检查每行每列及每九宫是否有相同项
            for row_item in self.b[x]:
                if row_item == value:
                    return False
            for row_all in self.b:
                if row_all[y] == value:
                    return False
            row, col = int(x / 3) * 3, int(y / 3) * 3
            row3col3 = self.b[row][col:col + 3] + self.b[row + 1][col:col + 3] + self.b[row + 2][col:col + 3]
            for row3col3_item in row3col3:
                if row3col3_item == value:
                    return False
    
            if (x,y) in self.KH:
                p = self.KH.index((x,y))//4*4
                for tx,ty in self.KH[p:p+4]:
                    item = self.b[tx][ty]
                    if item != 0 and item%2 != value%2:
                        return False
    
            return True
    
        def get_next(self, x, y):  # 得到下一个未填项
            for next_soulu in range(y + 1, 9):
                if self.b[x][next_soulu] == 0:
                    return x, next_soulu
            for row_n in range(x + 1, 9):
                for col_n in range(0, 9):
                    if self.b[row_n][col_n] == 0:
                        return row_n, col_n
            return -1, -1  # 若无下一个未填项,返回-1
    
        def try_it(self, x, y):  # 主循环
            if self.b[x][y] == 0:
                for i in range(1, 10):  # 从1到9尝试
                    self.t += 1
                    if self.check(x, y, i):  # 符合 行列宫均无条件 的
                        self.b[x][y] = i  # 将符合条件的填入0格
                        next_x, next_y = self.get_next(x, y)  # 得到下一个0格
                        if next_x == -1:  # 如果无下一个0格
                            return True  # 返回True
                        else:  # 如果有下一个0格,递归判断下一个0格直到填满数独
                            end = self.try_it(next_x, next_y)
                            if not end:  # 在递归过程中存在不符合条件的,即 使try_it函数返回None的项
                                self.b[x][y] = 0  # 回朔到上一层继续
                            else:
                                return True
        def start(self):
            begin = datetime.datetime.now()
            if self.b[0][0] == 0:
                self.try_it(0, 0)
            else:
                x, y = self.get_next(0, 0)
                self.try_it(x, y)
            for i in self.b:
                print(i)
            end = datetime.datetime.now()
            print('cost time: ', end - begin)
            print('times: ', self.t)
            return
    s = fill_sudoku(sudoku2)
    s.start()
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