python中的运行时错误：“超出最大递归深度”

Question

我有一个程序可以搜索代表迷宫的2D列表：

####################################
#S#  ##  ######## # #      #     # #
# #   #             # #        #   #
#   # ##### ## ###### # #######  # #
### # ##    ##      # # #     #### #
#   #    #  #######   #   ###    #E#
####################################

我理解递归错误是什么，但我不知道为什么这个代码会导致它，因为它应该只是导致找到＆＃34; E＆＃34;。有谁知道这可能会产生错误？

def solve(x,y):
    mazeList = loadMaze("sample.maze")    

    if mazeList[y][x] == "E":
        return "YOU'VE SOLVED THE MAZE!"
    elif mazeList[y][x+1] == " ":  #right
        mazeList[y][x+1] = ">"
        solve(x+1,y)
    elif mazeList[y+1][x] == " ":  #down
        mazeList[y+1][x] = "v"
        solve(x,y+1)    
    elif mazeList[y][x-1] == " ":  #left
        mazeList[y][x-1] = "<"
        solve(x-1,y)    
    elif mazeList[y-1][x] == " ":  #up
        mazeList[y-1][x] = "^"
        solve(x,y-1)    

Answer 1

每次调用函数时，都会重新加载mazeList 。

所以在solve()开始时，你就会回到起始状态，然后你最终会在圈子里跑。

使用关键字参数将mazeList传递给递归调用，将其默认为None并仅在迷宫仍为None时加载：

def solve(x, y, mazeList=None):
    if mazeList is None:
        mazeList = loadMaze("sample.maze")

并将mazeList传递给递归调用。

下一个问题是你永远不会返回递归调用;当您从solve()内拨打solve()时，仍需要返回结果：

def solve(x, y, mazeList=None):
    if mazeList is None:
        mazeList = loadMaze("sample.maze")

    if mazeList[y][x] == "E":
        return "YOU'VE SOLVED THE MAZE!"
    elif mazeList[y][x+1] == " ":  #right
        mazeList[y][x+1] = ">"
        return solve(x+1,y,mazeList)
    elif mazeList[y+1][x] == " ":  #down
        mazeList[y+1][x] = "v"
        return solve(x,y+1,mazeList)    
    elif mazeList[y][x-1] == " ":  #left
        mazeList[y][x-1] = "<"
        return solve(x-1,y,mazeList)    
    elif mazeList[y-1][x] == " ":  #up
        mazeList[y-1][x] = "^"
        return solve(x,y-1,mazeList)    

你还是会用这种技巧把自己画在角落里;要递归地解决迷宫问题，你需要尝试所有路径，而不仅仅是一个路径，并给每个递归调用一个迷宫的副本，只选择一个选定的路径。

您也始终测试 next 单元格，但从不考虑下一个单元格可能是目标;你永远不会将移到 E，因为该单元格不等于' '，所以它不是移动候选者。

以下版本可以解决你的迷宫：

directions = (
    (1, 0, '>'),
    (0, 1, 'v'),
    (-1, 0, '<'),
    (0, -1, '^'),
)

def solve(x, y, mazeList=None):
    if mazeList is None:
        mazeList = loadMaze("sample.maze")

    for dx, dy, char in directions:
        nx, ny = x + dx, y + dy

        if mazeList[ny][nx] == "E":
            return "YOU'VE SOLVED THE MAZE!"

        if mazeList[ny][nx] == " ":
            new_maze = [m[:] for m in mazeList]
            new_maze[ny][nx] = char
            result = solve(nx, ny, new_maze)
            if result is not None:
                return result

分别对每个方向进行测试变得乏味，因此我将其替换为一系列方向的循环;每个元组都是x的变化，y是在该方向上移动时使用的字符。

演示，打印出已解决的迷宫：

>>> def loadMaze(ignored):
...     maze = '''\
... ####################################
... #S#  ##  ######## # #      #     # #
... # #   #             # #        #   #
... #   # ##### ## ###### # #######  # #
... ### # ##    ##      # # #     #### #
... #   #    #  #######   #   ###    #E#
... ####################################
... '''
...     return [list(m) for m in maze.splitlines()]
... 
>>> directions = (
...     (1, 0, '>'),
...     (0, 1, 'v'),
...     (-1, 0, '<'),
...     (0, -1, '^'),
... )
>>> 
>>> def solve(x, y, mazeList=None):
...     if mazeList is None:
...         mazeList = loadMaze("sample.maze")   
...     for dx, dy, char in directions:
...         nx, ny = x + dx, y + dy
...         if mazeList[ny][nx] == "E":
...             print '\n'.join([''.join(m) for m in mazeList])
...             return "YOU'VE SOLVED THE MAZE!"
...         if mazeList[ny][nx] == " ":
...             new_maze = [m[:] for m in mazeList]
...             new_maze[ny][nx] = char
...             result = solve(nx, ny, new_maze)
...             if result is not None:
...                 return result
... 
>>> solve(1, 1)
####################################
#S#  ##  ######## # #^>>>>>#  ^>># #
#v#^>>#    ^>>>     #^#   v>>>>#v>>#
#v>>#v#####^##v######^# #######  #v#
### #v##^>>>##v>>>>>#^# #     ####v#
#   #v>>>#  #######v>>#   ###    #E#
####################################
"YOU'VE SOLVED THE MAZE!"