在这种情况下如何再次使用结果

Question

我的任务是：

Implement the function puzz. It should simulate the game similar to 15 Puzzle. In our case, we have 25 squares, where 24 squares are occupied by tiles with big case letters from 'A' to 'X'. One tile is free and is denoted by ' '. In each move, you can move a tile (denoted by its letter) into the free space. The function gets the original configuration and a sequence of valid moves. It should output the resulting configuration.

我是新手Haskell，我只能做第一步。

最后一步后的结果必须是 ["ABCDE", "FGHIJ", "KLMNO", "PQRST ", "UVWX_"]。如何保存第一步的结果并用它来做第二步，然后下一步，然后再迭代一些？

s1 = ["AC DE",
      "FBHIJ",
      "KGLNO",
      "PQMRS",
      "UVWXT"]

p1 = "CBGLMRST"

puzz :: Result -> [Char] -> Result
puzz s1 p1 = 
   let i = 0
       p = p1 !! i
       change r = [ if c == ' ' then p else if c == p then ' ' else c | c <- r ]
       out = [ change r | r <- s1  ]
    in out

Answer 1

puzz s1 p1 中的 s1 和 p1 不是您的常量 s1 和 p1。它们只是模板。例如 puzz [" B"] "B" returns ["BB "] 和 p1 变成 [" B"] 并且 s1 在那个函数调用中变成 "B"。
不要尝试迭代（我用 i 删除了部分代码）。使用递归。 Haskell里面没有变量，只有常量和输入参数。由于参照完整性，输入参数（由模板表示）在每次调用中都应该不同（接近尾声）。
别忘了完成算法。

(p:p1) - Template of input parameter which says that there is a nonempty list and the first character will be p and tail will be p1.
puzz s1 [] = s1 - When second parametr is empty list, then return first parameter (end).
puzz out p1 - Calling puzz for the changed s1 as out and p1 from (p:p1) (recursion).

s1 = ["AC DE","FBHIJ","KGLNO","PQMRS","UVWXT"]

p1 = "CBGLMRST"

puzz :: [String] -> [Char]-> [String]
puzz s1 (p:p1) =  let                  
                  change r = [ if c == ' ' then p else if c == p then ' ' else c | c <- r ]
                  out = [ change r | r <- s1  ]
                in puzz out p1
puzz s1 [] = s1

输出：

   puzz s1 p1
=> ["ABCDE","FGHIJ","KLMNO","PQRST","UVWX "]

在这种情况下如何再次使用结果

How to use result again in this case

iteration

haskell