2.2 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
594810f028c0303b75339ad5 | Y combinator | 5 | 302345 | y-combinator |
--description--
In strict [functional programming](https://en.wikipedia.org/wiki/Functional programming "wp: functional programming") and the [lambda calculus](https://en.wikipedia.org/wiki/lambda calculus "wp: 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. The Y combinator is itself a stateless function that, when applied to another stateless function, returns a recursive version of the function. The Y combinator is the simplest of the class of such functions, called [fixed-point combinators](https://en.wikipedia.org/wiki/Fixed-point combinator "wp: fixed-point combinator").
--instructions--
Define the stateless Y combinator function and use it to compute factorial. The factorial(N)
function is already given to you. See also:
--hints--
Y should return a function.
assert.equal(typeof Y((f) => (n) => n), 'function');
factorial(1) should return 1.
assert.equal(factorial(1), 1);
factorial(2) should return 2.
assert.equal(factorial(2), 2);
factorial(3) should return 6.
assert.equal(factorial(3), 6);
factorial(4) should return 24.
assert.equal(factorial(4), 24);
factorial(10) should return 3628800.
assert.equal(factorial(10), 3628800);
--seed--
--after-user-code--
var factorial = Y(f => n => (n > 1 ? n * f(n - 1) : 1));
--seed-contents--
function Y(f) {
return function() {
};
}
var factorial = Y(function(f) {
return function (n) {
return n > 1 ? n * f(n - 1) : 1;
};
});
--solutions--
var Y = f => (x => x(x))(y => f(x => y(y)(x)));