Haskell 具有自定义类型的教会数字

Question

我正在尝试解决 Church 数字解析器我有一个自定义类型，它区分变量、lambda 和应用程序

type Var = String

data Term =
    Variable Var
  | Lambda   Var  Term
  | Apply    Term Term
  deriving Show

我有一个函数叫做 church 我可以为不同的教堂变体手动定义一个案例。那么让我们说：

church 0 Lambda "f" (Lambda "x" (Variable "x"))
church 1 应该输出 Lambda "f" (Lambda "x" (Apply (Variable "f") (Variable "x")))
church 2 应该输出 Lambda "f" (Lambda "x" (Apply (Variable "f") (Apply (Variable "f") (Variable "x"))))

等等。

我试过递归调用教会函数：

church :: Int -> Term
church 0 = Lambda "f" (Lambda "x" (Variable  "x"))
church i = Apply (church(i -1)) (Apply (Variable "f") (Variable "x"))

然而，Lambda "f" (Lambda "x"

的部分也在不断重复

我尝试过的另一种方法是

church :: Int -> Term
church 0 = Lambda "f" (Lambda "x" (Variable  "x"))
church i = Apply (church (i -1)) (Apply (Variable "f") (Variable "x"))

然而，这也会产生带有复制的 lambda 的结果。我在这里错过了什么吗？怎么只重复申请部分(Apply (Variable "f") (Variable "x"))

Answer 1

所有教堂数字都以 \f -> (\i -> …) 开头，这意味着我们的函数应该如下所示：

church :: Int -> Term
church n = Lambda f (Lambda x (<strong>go n</strong>))

我们还需要实施的地方 go :: Int -> Term。因此，这意味着对于 go 我们必须找到一个看起来像这样的函数：

go 0 = Variable "x"
go 1 = Apply (Variable ("f")) (Variable "x")
go 2 = Apply (Variable "x") (Apply (Variable ("f")) (Variable "x"))

所以对于 go <i>n</i>，我们因此必须 return a Variable "x" 包裹在 n 项 Apply (Variable "x")

因此实现应该如下所示：

go :: Int -> Term
go 0 = Variable "x"
go n = … (go (n-1))

我离开实施 … 作为练习。

Haskell Church Numerals with custom types