ee1e8abd87
* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
80 lines
2.1 KiB
Markdown
80 lines
2.1 KiB
Markdown
---
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id: 594810f028c0303b75339ad5
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title: 和组合
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challengeType: 5
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videoUrl: ''
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dashedName: y-combinator
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---
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# --description--
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<p>在严格的<a href='https://en.wikipedia.org/wiki/Functional programming' title='wp：函数式编程'>函数编程</a>和<a href='https://en.wikipedia.org/wiki/lambda calculus' title='wp：lambda演算'>lambda演算中</a> ，函数（lambda表达式）没有状态，只允许引用封闭函数的参数。这排除了递归函数的通常定义，其中函数与变量的状态相关联，并且该变量的状态在函数体中使用。 </p><p> <a href='http://mvanier.livejournal.com/2897.html'>Y组合</a>器本身是一个无状态函数，当应用于另一个无状态函数时，它返回函数的递归版本。 Y组合器是这类函数中最简单的一种，称为<a href='https://en.wikipedia.org/wiki/Fixed-point combinator' title='wp：定点组合器'>定点组合器</a> 。 </p>任务： <pre> <code>Define the stateless Y combinator function and use it to compute <a href="https://en.wikipedia.org/wiki/Factorial" title="wp: factorial">factorial</a>.</code> </pre><p> <code>factorial(N)</code>功能已经给你了。另见<a href='http://vimeo.com/45140590'>Jim Weirich：功能编程中的冒险</a> 。 </p>
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# --hints--
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Y必须返回一个函数
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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）必须返回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）必须返回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）必须返回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）必须返回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）必须返回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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