Python写一个数独游戏,代码实现
时间: 2024-02-16 09:06:21 浏览: 104
Python数独游戏源代码
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好的,以下是一个简单的Python数独游戏实现:
```python
board = [
[0, 0, 0, 0, 0, 0, 6, 8, 0],
[0, 0, 0, 0, 7, 3, 0, 0, 9],
[3, 0, 9, 0, 0, 0, 0, 4, 5],
[4, 9, 0, 0, 0, 0, 0, 0, 0],
[8, 0, 3, 0, 5, 0, 9, 0, 2],
[0, 0, 0, 0, 0, 0, 0, 3, 6],
[9, 6, 0, 0, 0, 0, 3, 0, 8],
[7, 0, 0, 6, 8, 0, 0, 0, 0],
[0, 2, 8, 0, 0, 0, 0, 0, 0]
]
def print_board(board):
for i in range(len(board)):
if i % 3 == 0 and i != 0:
print("- - - - - - - - - - - - ")
for j in range(len(board[0])):
if j % 3 == 0 and j != 0:
print(" | ", end="")
if j == 8:
print(board[i][j])
else:
print(str(board[i][j]) + " ", end="")
def find_empty(board):
for i in range(len(board)):
for j in range(len(board[0])):
if board[i][j] == 0:
return (i, j)
return None
def valid(board, num, pos):
# Check row
for i in range(len(board[0])):
if board[pos[0]][i] == num and pos[1] != i:
return False
# Check column
for i in range(len(board)):
if board[i][pos[1]] == num and pos[0] != i:
return False
# Check box
box_x = pos[1] // 3
box_y = pos[0] // 3
for i in range(box_y*3, box_y*3 + 3):
for j in range(box_x * 3, box_x*3 + 3):
if board[i][j] == num and (i,j) != pos:
return False
return True
def solve(board):
find = find_empty(board)
if not find:
return True
else:
row, col = find
for i in range(1,10):
if valid(board, i, (row, col)):
board[row][col] = i
if solve(board):
return True
board[row][col] = 0
return False
print_board(board)
print("-------------------")
solve(board)
print_board(board)
```
这个程序首先定义了一个 9x9 的数独矩阵。然后,它定义了一个函数 `print_board()` 来打印数独矩阵, `find_empty()` 函数来查找空格子, `valid()` 函数来检查数独是否有效,以及 `solve()` 函数来解决数独游戏。
最后,它打印出最终的数独矩阵。
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