[英]Python for loop range function causing an infinite loop
我正在构建一个数独函数来学习如何在python中编码。 我似乎用for循环创建了一个无限循环,但我不明白如何。 代码试图查看数独板的每个空方块,并检查数独规则是否允许值counter
。 如果允许计数器,则更新板并且功能移动到下一个空方块。 如果不允许计数器,则计数器增加1并再次测试。
我遇到的问题是当计数器大于9时。当发生这种情况时,我想查看原始板上的前一个方块(命名为难题)并删除此方块中的值。 该函数应该设置计数器等于前一个方格+1中的值,并调用自身再次运行。
实质上,该函数正在测试每个空方块的可能值,直到找到一个值,然后移动到下一个方块。 如果没有可能的值,该功能将回溯,删除最后一个方块并再次尝试运行。
当计数器大于9时,我的问题似乎发生在else条件下。这部分函数导致无限循环,反复打印出'no'。
我假设我的函数卡在while循环中,但我不知道为什么。
puzzleBoard =[[1,2,3,4,5,6,7,8,9],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0]]
def solvePuzzle():
#start by looking at each square in the 9x9 sudoku grid and check if that square is empty (=0)
for i in range(9):
for j in range(9):
counter = 1
topX = 3*(i//3)
topY = 3*(j//3)
# while the board at index [i][j] is empty check if the value of 'counter' fits in the square and adheres to the sudoku rules
# if counter is not an allowed value increment counter by 1 and try again
while puzzleBoard[i][j] ==0:
if counter < 10:
row = all([counter != puzzleBoard[i][x] for x in range(9)])
column = all([counter != puzzleBoard[y][j] for y in range(9)])
box = all([counter != puzzleBoard[x][y] for x in range(topX, topX+3) for y in range(topY, topY+3)])
if row and column and box == True:
puzzleBoard[i][j]=counter
uploadBoard()
else:
counter = counter + 1
# if counter is larger than ten set the previous square ([i][j-1]) equal to zero, set the counter equal to one more than the previous squares value, and call the solvePuzzle function again.
else:
for k in range(i,0,-1):
for l in range(j-1,0,-1):
if puzzle[k][l]==0:
counter = puzzleBoard[k][l] + 1
puzzleBoard[k][l]=0
solvePuzzle()
return
else:
print("no")
我能够找到答案。 代码有一些问题,但主要问题是在较低的else
语句counter = puzzleBoard[k][l] + 1
然后再次调用该函数,它将变量counter
重置为1。
我能够通过创建一个全局变量countholder
并修改else语句来说明countholder = puzzleBoard[k][l] + 1
来解决这个问题。
完整的工作代码如下所示:
puzzleBoard =[[0,2,0,0,0,0,0,0,0],[0,0,0,6,0,0,0,0,3],
[0,7,4,0,8,0,0,0,0],[0,0,0,0,0,3,0,0,2],
[0,8,0,0,4,0,0,1,0],[6,0,0,5,0,0,0,0,0],
[0,0,0,0,1,0,7,8,0],[5,0,0,0,0,9,0,0,0],
[0,0,0,0,0,0,0,4,0]]
puzzle =[[0,2,0,0,0,0,0,0,0],[0,0,0,6,0,0,0,0,3],
[0,7,4,0,8,0,0,0,0],[0,0,0,0,0,3,0,0,2],
[0,8,0,0,4,0,0,1,0],[6,0,0,5,0,0,0,0,0],
[0,0,0,0,1,0,7,8,0],[5,0,0,0,0,9,0,0,0],
[0,0,0,0,0,0,0,4,0]]
countholder = 1
def solvePuzzle():
#start by looking at each square in the 9x9 sudoku grid and check if that square is empty (=0)
for i in range(9):
for j in range(9):
global countholder
counter = countholder
topX = 3*(i//3)
topY = 3*(j//3)
# while the board at index [i][j] is empty check if the value of 'counter' fits in the square and adheres to the sudoku rules
# if counter is not an allowed value increment counter by 1 and try again
while puzzleBoard[i][j] ==0:
if counter < 10:
row = all([counter != puzzleBoard[i][x] for x in range(9)])
column = all([counter != puzzleBoard[y][j] for y in range(9)])
box = all([counter != puzzleBoard[x][y] for x in range(topX, topX+3) for y in range(topY, topY+3)])
if (row and column and box) == True:
puzzleBoard[i][j]=counter
print(puzzleBoard)
countholder = 1
else:
counter = counter + 1
# if counter is larger than ten set the previous square ([i][j-1]) equal to zero, set the counter equal to one more than the previous squares value, and call the solvePuzzle function again.
else:
run_One = True
for k in range(i,-1,-1):
if run_One == True:
run_One = False
for l in range(j,0,-1):
if l == 0:
print(run_One)
elif puzzle[k][l-1]==0:
countholder = puzzleBoard[k][l-1] + 1
puzzleBoard[k][l-1]=0
solvePuzzle()
return
else:
continue
else:
for l in range(8,-1,-1):
if puzzle[k][l]==0:
countholder = puzzleBoard[k][l] + 1
puzzleBoard[k][l]=0
solvePuzzle()
return
else:
continue
solvePuzzle()
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