一个棘手的haskell问题

Question

我的Haskell课程有这个任务,但我觉得很难.如果你能帮助一点.你有一个迷宫

maze = ["x xxx",
        "x   x",
        "x x x",
        "x x  ",
        "xxxxx"]

而且你只能穿过空间.你从(0,1)开始,函数必须返回一个字符串,其中包含逃避迷宫的方向:

f - forward
r- turn right
l - turn left 

如果你有一个选择,你总是喜欢向前,向前向左.对于当前的例子,答案是ffllffrffrfflf

提前致谢

data Direction = N | W | S | E deriving (Show,Eq)

maze = ["x xxx",
        "x   x",
        "x x x",
        "x x  ",
        "xxxxx"]

d = 's'
pos = (0,1)

fpath d pos | fst pos == (length maze - 1) = ""
            | snd  (pos) ==0 || (snd ( pos ) == ((length (maze!!0))-1)) = ""
            | rightPossible d pos = "r" ++ ( fpath (rightRotate d) pos )
            | forwardPossible d  pos = "f" ++ ( fpath d (nstep d pos) )
            | True = "l" ++ fpath (leftRotate d) pos
            where nstep :: Direction -> (Int, Int) -> (Int, Int) {-next step-}
                  nstep N (x,y) = (x-1,y)
                  nstep W  (x,y) = (x,y-1) 
                  nstep S (x,y) = (x+1,y) 
                  nstep E  (x,y) = (x,y+1)

                  rightPossible :: Direction -> (Int, Int) -> Bool 
                  rightPossible N (x,y)= (maze !! x)!! (y+1) == ' '
                  rightPossible W (x,y)= (maze !! (x-1))!! y   == ' '              
                  rightPossible S (x,y)= (maze !! x)!! (y-1) == ' '                   
                  rightPossible E (x,y)= (maze !! (x+1))!! y   == ' '

                  rightRotate :: Direction -> Direction 
                  rightRotate N = E
                  rightRotate W = N
                  rightRotate S = W
                  rightRotate E = S

                  forwardPossible :: Direction -> (Int, Int) -> Bool 
                  forwardPossible N (x,y)= ((maze !! (x-1))!! y) == ' ' 
                  forwardPossible W (x,y)= ((maze !! x)!! (y-1)) == ' '
                  forwardPossible S (x,y)= ((maze !! (x+1))!! y) == ' '
                  forwardPossible E (x,y)= ((maze !! x)!! (y+1)) == ' '

                  leftRotate :: Direction -> Direction
                  leftRotate N = W
                  leftRotate W = S
                  leftRotate S = E
                  leftRotate E = N

Answer 1

我看到的第一件事是,你有一个优先问题.表达式(maze !! x)!! y-1被解析为,((maze !! x)!! y)-1而您希望将其解析为(maze !! x)!! (y-1).添加大括号来解决此问题.

添加此内容后,您的代码会编译,尽管您的算法似乎已被破坏.也许其他人可以帮助你.

一些编码建议:

• 在适当的位置添加类型签名以简化调试.(如果类型失败,编译器更可能在正确的位置显示错误)

• 使用模式匹配而不是额外的case语句.代替

  nstep d (x,y)   {-next step-}
                  | d == 'n' = (x-1,y)
                  | d == 'w' = (x,y-1)
                  | d == 's' = (x+1,y)
                  | d == 'e' = (x,y+1)
  

  写

  nstep 'n' (x,y) = (x-1,y)
  nstep 'w' (x,y) = (x,y-1)
  nstep 's' (x,y) = (x+1,y)
  nstep 'e' (x,y) = (x,y+1)
  

• 编写自己的data类型而不是依赖于字符.例如,您可以为方向创建自己的数据类型:

  data Direction = N | W | S | E deriving (Show,Eq)