javascript中的y-combinator

Question

我已经在js中建立了一个y-combinator

const y = f => { const g = self => x => f(self(self))(x); return g(g);}

我这样简化了这段代码

 const y = f => { const g = self => f(self(self)); return g(g);}

这得到了无限递归。 这两个版本有什么区别？

Answer 1

如果您不了解两者之间的区别，那么您实际上构建了它们，我会感到惊讶。 然而，证明两者之间差异的最好方法是遵循他们的评估。

const y = f => {
  const g = self => x => f(self(self))(x)
  return g(g)
}

y (z) ...
// (self => x => z(self(self))(x)) (self => x => z(self(self))(x)) ...
// returns:
// x => z((self => x1 => z(self(self))(x1))(self => x2 => z(self(self))(x2)))(x)

好的，所以y(z) （其中z是某个函数，没关系）返回一个函数x => ... 在我们应用该功能之前，评估到此为止。

现在，将其与您的第二个定义进行比较

const y = f => {
  const g = self => f(self(self))
  return g(g)
}

y (z) ...
// (self => z(self(self))) (self => z(self(self)))
// z((self => z(self(self)))(self => z(self(self)))) ...
// z(z((self => z(self(self)))(self => z(self(self))))) ...
// z(z(z((self => z(self(self)))(self => z(self(self)))))) ...
// z(z(z(z((self => z(self(self)))(self => z(self(self))))))) ...
// ... and on and on

因此y (z)永远不会终止-至少在使用应用顺序评估的JavaScript中-在应用调用函数之前先评估函数参数

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备用Y组合器

在这里，我们可以从头开始构建一个Y组合器

// standard definition
const Y = f => f (Y (f))

// prevent immediate infinite recursion in applicative order language (JS)
const Y = f => f (x => Y (f) (x))

// remove reference to self using U combinator
const U = f => f (f)
const Y = U (h => f => f (x => h (h) (f) (x)))

让我们测试一下

 const U = f => f (f) const Y = U (h => f => f (x => h (h) (f) (x))) // range :: Int -> Int -> [Int] const range = Y (f => acc => x => y => x > y ? acc : f ([...acc,x]) (x + 1) (y)) ([]) // fibonacci :: Int -> Int const fibonacci = Y (f => a => b => x => x === 0 ? a : f (b) (a + b) (x - 1)) (0) (1) console.log(range(0)(10).map(fibonacci)) // [ 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 ] 

或我最近的最爱

 // simplified Y const Y = f => x => f (Y (f)) (x) // range :: Int -> Int -> [Int] const range = Y (f => acc => x => y => x > y ? acc : f ([...acc,x]) (x + 1) (y)) ([]) // fibonacci :: Int -> Int const fibonacci = Y (f => a => b => x => x === 0 ? a : f (b) (a + b) (x - 1)) (0) (1) console.log(range(0)(10).map(fibonacci)) // [ 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 ]