Python 9x9 数独游戏：规则生成，随机空位，挑战你的逻辑！

以下是一个简单的 Python 数独游戏实现：

import random

def generate_random_sequence():
    # 生成一个随机数列
    sequence = []
    empty_spaces = random.randint(15, 26)
    for i in range(9):
        row = []
        for j in range(9):
            if empty_spaces > 0:
                row.append(' ')
                empty_spaces -= 1
            else:
                # 生成符合规则的数字
                valid_numbers = [x for x in range(1, 10) if x not in row and x not in [sequence[k][j] for k in range(i)]]
                if valid_numbers:
                    row.append(random.choice(valid_numbers))
                else:
                    # 如果没有符合规则的数字，则重新生成数列
                    return generate_random_sequence()
        sequence.append(row)
    return sequence

def print_sequence(sequence):
    # 打印数列
    for row in sequence:
        print(' '.join([str(x) for x in row]))

def play_game():
    print('按S开始游戏，按C结束游戏，按R重新生成数列')
    while True:
        choice = input().lower()
        if choice == 's':
            sequence = generate_random_sequence()
            print_sequence(sequence)
            while True:
                # 填空位为可输入空格
                row_index = int(input('请输入行号: ')) - 1
                col_index = int(input('请输入列号: ')) - 1
                if sequence[row_index][col_index] == ' ':
                    number = int(input('请输入数字: '))
                    if number not in sequence[row_index] and number not in [sequence[i][col_index] for i in range(9)]:
                        sequence[row_index][col_index] = number
                        print_sequence(sequence)
                    else:
                        print('请输入符合规则的数字！')
                else:
                    print('该位置已经有数字了！')
        elif choice == 'r':
            sequence = generate_random_sequence()
            print_sequence(sequence)
        elif choice == 'c':
            break
        else:
            print('无效的选择！')

play_game()

注意，这只是一个简单的实现，可能还有一些细节需要进一步完善，比如输入的数字是否在合法范围内、用户输入的行号和列号是否在合法范围内等。