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86 lines
1.8 KiB
Markdown
86 lines
1.8 KiB
Markdown
---
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id: 594810f028c0303b75339ad5
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title: Y 组合器
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challengeType: 1
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forumTopicId: 302345
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dashedName: y-combinator
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---
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# --description--
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In strict <a href="https://www.freecodecamp.org/news/the-principles-of-functional-programming/" target="_blank" rel="noopener noreferrer nofollow">functional programming</a> and the lambda calculus, functions (lambda expressions) don't have state and are only allowed to refer to arguments of enclosing functions. This rules out the usual definition of a recursive function wherein a function is associated with the state of a variable and this variable's state is used in the body of the function.
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Y 组合子本身是一个无状态函数，当应用于另一个无状态函数时，返回该函数的递归版本。 Y 组合子是这类功能中最简单的一类，称为不动点组合子。
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# --instructions--
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定义无状态的 Y 组合子函数并使用它来计算阶乘。 `factorial(N)` 函数已经定义好了。
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# --hints--
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Y 应该返回一个函数。
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```js
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assert.equal(typeof Y((f) => (n) => n), 'function');
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```
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factorial(1) 应该返回 1。
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```js
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assert.equal(factorial(1), 1);
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```
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factorial(2) 应该返回 2。
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```js
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assert.equal(factorial(2), 2);
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```
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factorial(3) 应该返回 6。
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```js
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assert.equal(factorial(3), 6);
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```
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factorial(4) 应该返回 24。
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```js
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assert.equal(factorial(4), 24);
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```
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factorial(10) 应该返回 3628800。
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```js
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assert.equal(factorial(10), 3628800);
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```
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# --seed--
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## --after-user-code--
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```js
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var factorial = Y(f => n => (n > 1 ? n * f(n - 1) : 1));
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```
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## --seed-contents--
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```js
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function Y(f) {
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return function() {
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};
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}
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var factorial = Y(function(f) {
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return function (n) {
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return n > 1 ? n * f(n - 1) : 1;
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};
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});
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```
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# --solutions--
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```js
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var Y = f => (x => x(x))(y => f(x => y(y)(x)));
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```
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