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[英]Y-Combinator factorial in javascript works for numbers not for the Church numerals.
[英]y-combinator in javascript
我已經在js中建立了一個y-combinator
const y = f => { const g = self => x => f(self(self))(x); return g(g);}
我這樣簡化了這段代碼
const y = f => { const g = self => f(self(self)); return g(g);}
這得到了無限遞歸。 這兩個版本有什么區別?
如果您不了解兩者之間的區別,那么您實際上構建了它們,我會感到驚訝。 然而,證明兩者之間差異的最好方法是遵循他們的評估。
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
是某個函數,沒關系)返回一個函數x => ...
在我們應用該功能之前,評估到此為止。
現在,將其與您的第二個定義進行比較
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)
永遠不會終止-至少在使用應用順序評估的JavaScript中-在應用調用函數之前先評估函數參數
備用Y組合器
在這里,我們可以從頭開始構建一個Y組合器
// 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)))
讓我們測試一下
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 ]
或我最近的最愛
// 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 ]
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