我一直在编写这个程序，在继续之前测试每个部分的工作情况。然而，当我完成并把所有东西放在一起时，我无法得到解决方案。我试图通过为每个方块创建一个可能的数字列表来解决数独，并根据现有的方块移除它们。我假设当一个正方形中只有一个可能的数字时，这就是解决方案。它会循环，直到它完成。
我一直在查看我的代码半小时但仍然没有运气。我插入raw_input("")以查看是否有任何问题。我发现它在开始时取得了一些进展，但随后又停止了。
所以我打印了一个坐标的可能数字以及过程中的某个地方，每个可能性都被删除了。
以下是代码的样子：
# Create the Sodoku grid
grid = [[3,2,0,1,6,0,8,0,9],
[0,7,8,9,0,3,1,2,6],
[6,0,0,8,0,0,4,5,3],
[7,1,0,4,0,0,0,6,2],
[5,4,0,0,0,0,0,0,7],
[0,0,0,2,0,5,3,1,0],
[0,5,9,7,4,0,2,0,8],
[2,0,7,5,0,9,0,0,0],
[8,6,4,0,0,0,0,9,5],]
# Create possibilities
possible = {}
for y in range(9):
for x in range(9):
possible[(y,x)] = [1,2,3,4,5,6,7,8,9]
# A function that returns the row it is in.
def check_row(y,x):
return grid[y]
# A function that returns the column it is in.
def check_column(y,x):
column = []
for hops in range(9):
column.append(grid[hops][x])
return column
# A function that returns the square it is in.
# -------------
# 1| 0 | 1 | 2 |
# -------------
# 2| 3 | 4 | 5 |
# -------------
# 3| 6 | 7 | 8 |
# -------------
# 1 2 3
def check_square(they,thex):
square0 = []
square1 = []
square2 = []
square3 = []
square4 = []
square5 = []
square6 = []
square7 = []
square8 = []
for y in range(3):
for x in range(3):
square0.append([y,x])
for y in range(3):
for x in range(3,6):
square1.append([y,x])
for y in range(3):
for x in range(6,9):
square2.append([y,x])
for y in range(3,6):
for x in range(3):
square3.append([y,x])
for y in range(3,6):
for x in range(3,6):
square4.append([y,x])
for y in range(3,6):
for x in range(6,9):
square5.append([y,x])
for y in range(6,9):
for x in range(3):
square6.append([y,x])
for y in range(6,9):
for x in range(3,6):
square7.append([y,x])
for y in range(6,9):
for x in range(6,9):
square8.append([y,x])
tests = [square0,
square1,
square2,
square3,
square4,
square5,
square6,
square7,
square8]
square_list = []
def list_of_grid(result):
for cood in result:
[they,thex] = cood
square_list.append(grid[they][thex])
# Check which square it of and print the list of grid
for test in tests:
if [they,thex] in test:
list_of_grid(test)
return square_list
# Function that eliminates row possibilities
def elim_row(y, x):
get_rid_of = []
for element in check_row(y, x):
if element != 0:
get_rid_of.append(element)
for stuff in get_rid_of:
try:
if stuff in possible[(y,x)]:
possible[(y,x)] = []
else:
possible[(y,x)].remove(stuff)
except ValueError:
pass
# Funciton that eliminates column possibilites
def elim_column(y, x):
get_rid_of = []
for element in check_column(y, x):
if element != 0:
get_rid_of.append(element)
for stuff in get_rid_of:
try:
if stuff in possible[(y,x)]:
possible[(y,x)] = []
else:
possible[(y,x)].remove(stuff)
except ValueError:
pass
# Function that eliminates square possibilites
def elim_square(y, x):
get_rid_of = []
for element in check_square(y, x):
if element != 0:
get_rid_of.append(element)
for stuff in get_rid_of:
try:
if stuff in possible[(y,x)]:
possible[(y,x)] = []
else:
possible[(y,x)].remove(stuff)
except ValueError:
pass
# Check if done:
def done():
empty = 0
for y in range(9):
for x in range(9):
if grid[y][x] == 0:
empty += 1
if empty == 0:
return True
else:
return False
# print grid
if __name__ == "__main__":
# Go through each row, column and square and delete possibilites
while done != True:
raw_input("")
for cood in possible.keys():
(y, x) = cood
elim_row(y,x)
elim_column(y,x)
elim_square(y,x)
# Check if len of possible == 1
if len(possible[cood]) == 1:
grid[y][x] = possible[cood][0]
print possible[(0,2)]
for rows in grid:
print rows
答案 0 :(得分：1)
您永远不会致电 done()。您只测试函数对象是否永远不等于True：
while done != True:
函数对象永远不会等于True。不要在这里测试相等性，只需调用函数：
while not done():
接下来，在elim_row()中，只要您循环覆盖值，就可以清除可能的值：
for stuff in get_rid_of:
try:
if stuff in possible[(y,x)]:
possible[(y,x)] = []
else:
possible[(y,x)].remove(stuff)
except ValueError:
pass
将possible[(y,x)]设置为行中任何值的空值而不是0.您在其他2个elim_函数中执行相同操作。
您可能想要使用：
for stuff in get_rid_of:
if stuff in possible[(y,x)]:
possible[(y,x)].remove(stuff)
这将很快清除你的可能性。