haskell:否定正规形式的函数得到“非详尽模式”异常

时间:2016-02-14 23:22:34

标签: haskell non-exhaustive-patterns

-- | data type definition of WFF: well formed formula
data Wff = Var String 
        | Not Wff
        | And Wff Wff 
        | Or Wff Wff
        | Imply Wff Wff

-- | Negation norm form  nnf function
--   precondition: φ is implication free
--   postcondition: NNF (φ) computes a NNF for φ
nnf :: Wff -> Wff
nnf (Var p) = Var p
nnf (Not (Not p)) = (nnf p)
nnf (And p q) = And (nnf p) (nnf q)
nnf (Or p q) = Or (nnf p) (nnf q)
nnf (Not (And p q)) = Or (nnf(Not p)) (nnf(Not q))
nnf (Not (Or p q)) = And (nnf(Not p)) (nnf(Not q))

测试用例:¬( p Q

(*** Exception:: Non-exhaustive patterns in function nnf

但是,如果我将nnf (Not p) = Not (nnf p)添加到该函数中,它将显示

Pattern match(es) are overlapped
In an equation for ‘nnf’:
    nnf (Not (Not p)) = ...
    nnf (Not (And p q)) = ...
    nnf (Not (Or p q)) = ...

我想知道我做错了什么?

1 个答案:

答案 0 :(得分:2)

您只是将线路插入错误的位置。 nnf (Not p) = ...是一个否定的全能。如果您稍后添加其他处理更具体否定的条款,例如Not (And p q),则它们可能不再触发。

包罗万象的条款需要最后,即

nnf (Var p) = Var p
nnf (Not (Not p)) = (nnf p)
nnf (And p q) = And (nnf p) (nnf q)
nnf (Or p q) = Or (nnf p) (nnf q)
nnf (Not (And p q)) = Or (nnf $ Not p) (nnf $ Not q)
nnf (Not (Or p q)) = And (nnf $ Not p) (nnf $ Not q)
nnf (Not p) = Not $ nnf p