各位数独游戏用python写谁会可否看看
哪位数独游戏用python写出内容是实现游戏与破解谁能给我写一个正确的上网查了一个但是有问题改了好长时间没改对
# coding=utf-8
import datetime
class solution(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 = x / 3 * 3, 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('\ncost time:', end - begin)
print('times:', self.t)
return
s=solution([[8, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 3, 6, 0, 0, 0, 0, 0],
[0, 7, 0, 0, 9, 0, 2, 0, 0],
[0, 5, 0, 0, 0, 7, 0, 0, 0],
[0, 0, 0, 8, 4, 5, 7, 0, 0],
[0, 0, 0, 1, 0, 0, 0, 3, 0],
[0, 0, 1, 0, 0, 0, 0, 6, 8],
[0, 0, 8, 5, 0, 0, 0, 1, 0],
[0, 9, 0, 0, 0, 0, 4, 0, 0]])
s.start()