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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:

Jim Weirich: Adventures in Functional Programming.

--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)));

2.2 KiB Raw Blame History