在递归迷宫中回溯； Python

Question

我正在编写一个迷宫求解算法，一旦你撞到“墙”，我就会遇到回溯问题

到目前为止，我的代码检查了一些基本的递归情况，例如是否已到达终点或是否已检查所有点。 我也做了它，以便有一个列表来跟踪解决方案“路径”并相应地添加和删除点。 因此，每次添加一个点时，它都会检查其是否有效并检查该点的上方、左侧、右侧和下方等。如果都无效，则执行此语句

    else:
    path.pop()
    return solve_maze(maze,path,end) 

这将删除点并回溯并检查函数开头的以下条件。

square = list(path[-1])

#Base Case

if end in path:
    return True

elif square == None:
    return False

elif maze[square[0]][square[1]] == "X":
    path.pop()
    return solve_maze(maze,path,end)

elif tuple(square) in path:
    path.pop()
    return solve_maze(maze,path,end)

但是，当我执行最后一行时，它只会删除我路径中的所有点，并且出现索引错误。

关于如何在遇到死胡同后回溯的任何建议？

Answer 1

这是一个评论量很大的解决方案：

def backtrack(i,j):
   # if matrix[i][j] is visited, then do nothing
   # if it is a wall, then do not insert it in the path from the first place
   if visit[i][j] != 0 or matrix[i][j] == 'X':
      return
   # mark the index as visited as to not visit it again
   visit[i][j] = 1
   # add the index to the path
   path.append((i,j))
   # if it is the GOAL, then store this path somewhere, 
   # and pop the last index to traverse other paths that reaches the GOAL
   if matrix[i][j] == 'G':
      # any paths that reach the GOAL are appended here
      ans.append(list(path))
      path.pop()
      return
   # loop on the neighbors. if they are valid indices, do the same work again
   for k in range(3):
      nwi = i + row[k]
      nwj = j + col[k]
      if valid(nwi, nwj):
         backtrack(nwi, nwj)
   # after finishing this index, just pop because we do not need it any more
   path.pop()
   return

def valid(i,j):
   return (i >= 0 and i <=2) and (j >= 0 and j <= 2)

# just a dummy matrix to start with
matrix = [['.', '.', '.'], ['.', 'X', '.'], ['.', 'G', 'X']]
# an array that marks indices as visited or not
visit = [[0,0,0],[0,0,0],[0,0,0]]
# the path that is used in the back tracking
path = []
# this will be a list of lists containing ALL the paths that can reach the GOAL
ans = []
# these are the neighbors indices
row = [-1, 1, 0, 0]
col = [0, 0, 1, -1]

backtrack(0,0)

if len(ans) == 0: 
   print("no path found!")
else:
   print(ans)

我的解决方案将在名为ans的列表列表中为您提供可以到达目标的所有路径。 如果ans的长度为零，则找不到路径。

如果您不理解代码中写的注释或无论如何感到困惑，请不要犹豫，评论下来，我会回复。

注意：（编辑）

我假设了迷宫的起始状态和函数内容本身，因为您没有提供任何测试用例或完整的函数本身。