我正在创建一个基于文本的Sudoku求解器,每次运行代码时,我都遇到RecursionError错误。我以为我的代码出了点问题,所以我增加了递归深度,并且可以正常工作,我只是不确定如何重写函数,以便摆脱递归深度错误。
def backtrack (self):
'''Goes back in the grid and checks for valid numbers'''
del self.history[len(self.history) - 1] # goes back to the last spot, not current
self.pos = self.history[len(self.history) - 1] # reassign current position
for numbers in range(9):
if self.valid(numbers + 1) and (numbers + 1) != self.board[self.pos[0]][self.pos[1]]: # valid number but not the same as before
self.board[self.pos[0]][self.pos[1]] = numbers + 1
return True
self.board[self.pos[0]][self.pos[1]] = 0 #reset this position to 0
return False
def solve(self): #recursive, once you get to the end of the board it's solved
'''solves the Sudoku board, backtrack alg'''
empty = self.find_empty()
if not empty:
return None
if empty: #if there's an empty spot on the grid:
for nums in range(9): #try all numbers on a specific spot
if self.valid(nums+1): #theres no numbers on the column, row, or grid
self.board[self.pos[0]][self.pos[1]] = nums+1
break
elif nums == 8: #reached end of for loop, no number fits in the grid
while self.backtrack() == False: #keep going until we can plug in a number
if self.backtrack() == True:
break
self.solve() #recursive process
board = Sudoku([
[7, 8, 0, 4, 0, 0, 1, 2, 0],
[6, 0, 0, 0, 7, 5, 0, 0, 9],
[0, 0, 0, 6, 0, 1, 0, 7, 8],
[0, 0, 7, 0, 4, 0, 2, 6, 0],
[0, 0, 1, 0, 5, 0, 9, 3, 0],
[9, 0, 4, 0, 6, 0, 0, 0, 5],
[0, 7, 0, 3, 0, 0, 0, 1, 2],
[1, 2, 0, 0, 0, 7, 4, 0, 0],
[0, 4, 9, 2, 0, 6, 0, 0, 7]
])
board.solve()
为澄清起见,self.history是一个元组列表,它记住我们迭代经过的所有0,而self.pos是我们要检查的当前网格。我增加了递归限制,它解决了一半以上的开发板,而以前只解决了一半的开发板,但是我对如何重写递归部分有0的想法。我知道这有点多,但是可以得到帮助!
Error Log:
File "C:/Users/User/Desktop/Sudoku/sudoko_alg.py", line 26, in on_column
for i in range (9):
RecursionError: maximum recursion depth exceeded in comparison
Process finished with exit code 1
我建议将您的算法重新设计为迭代。
# Verty rough sketch!
states = [Sudoku(initial_numbers)] #a list with the starting configuration
for state in iter(states):
if state.is_solved():
print("success!")
break
states += state.get_next_states()
您的代码的问题是,每次在self.solve()
中对板进行更改时,都会发出对self.solve()的新调用。 self.solve()
从不向父级self.solve()
调用返回值,因此直到代码的最后,所有函数调用都不会退出。
我相信您打算做的是,这样每次添加一个值时,都会对self.solve()
进行新的调用。并且,每当发现一个值无效时,就会将某个指示符(即False
)返回到先前的self.solve()
调用。在此实现中,最多将有81个对self.solve()
的递归调用。在您当前的体系结构中,可能有多达9 ^ 81个递归调用,因此RecursionError
随您在每次后续调用中快速耗尽堆栈中的可用空间。
[为了解决,我建议您修改代码,以便如果存在有效的板卡组合,则self.solve()
返回True
,否则返回False
,并在每次添加值时对self.solve()
进行递归调用。