在Haskell中写一个传递闭包关系函数

Question

我应该为 return 关系的传递闭包函数编写一个函数。这是我到目前为止写的：

relTrans :: [(Integer, Integer)] -> [(Integer, Integer)]
relTrans rel  |  rel == remove_dups(rel ++ relComp (remove_dups rel) (remove_dups rel))     =     rel
              |  otherwise                                                                  =     relTrans remove_dups(rel ++ relComp (remove_dups rel) (remove_dups rel))

这里提到的作为remove_dups和relComp的函数如下-

relComp :: [(Integer, Integer)] -> [(Integer, Integer)] -> [(Integer, Integer)]
relComp r1 r2 = [(a,c) | (a,k1)<-r1, (k2,c)<-r2, k1 == k2]

remove_dups :: [a] -> [a]
remove_dups [] = []
remove_dups (x:xs) = x : remove_dups (removeElem x XS)

removeElem :: a -> [a] -> [a]
removeElem n [] = []
removeElem n (m:zs) = if (n == m) then removeElem n zs else m:(removeElem n zs)

但是，我不断收到以下错误：-

    * Couldn't match expected type `[(Integer, Integer)]
                                    -> [(Integer, Integer)]'
                  with actual type `[(Integer, Integer)]'
    * The function `relTrans' is applied to two arguments,
      but its type `[(Integer, Integer)] -> [(Integer, Integer)]'
      has only one
      In the expression:
        relTrans
          remove_dups (rel ++ relComp (remove_dups rel) (remove_dups rel))
      In an equation for `relTrans':
          relTrans rel
            | rel
                == remove_dups (rel ++ relComp (remove_dups rel) (remove_dups rel))
            = rel
            | otherwise
            = relTrans
                remove_dups (rel ++ relComp (remove_dups rel) (remove_dups rel))
    |       
183 |               |  otherwise                                                                  =     relTrans remove_dups(rel ++ relComp (remove_dups rel) (remove_dups rel))
    |                                                                                                   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

还有这个错误-

    * Couldn't match expected type `[(Integer, Integer)]'
                  with actual type `[a0] -> [a0]'
    * Probable cause: `remove_dups' is applied to too few arguments
      In the first argument of `relTrans', namely `remove_dups'
      In the expression:
        relTrans
          remove_dups (rel ++ relComp (remove_dups rel) (remove_dups rel))
      In an equation for `relTrans':
          relTrans rel
            | rel
                == remove_dups (rel ++ relComp (remove_dups rel) (remove_dups rel))
            = rel
            | otherwise
            = relTrans
                remove_dups (rel ++ relComp (remove_dups rel) (remove_dups rel))
    |       
183 |               |  otherwise                                                                  =     relTrans remove_dups(rel ++ relComp (remove_dups rel) (remove_dups rel))
    |                                                                                                            ^^^^^^^^^^^

有人可以帮我解决为什么会出现这些错误吗？

Answer 1

您似乎想在这里使用 C 风格的函数应用程序：

  relTrans remove_dups(rel ++ relComp (remove_dups rel) (remove_dups rel))
  --             here ^

In Haskell 将 relTrans 函数应用于两个参数：remove_dups 和 (rel ++ ...)，但是 relTrans 只有一个参数，所以你得到一个错误。

你可能打算写这个：

  relTrans (remove_dups (rel ++ relComp (remove_dups rel) (remove_dups rel)))

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