freeCodeCamp/curriculum/challenges/italian/10-coding-interview-prep/rosetta-code/y-combinator.md

2.2 KiB

id title challengeType forumTopicId dashedName
594810f028c0303b75339ad5 Y combinator 5 302345 y-combinator

--description--

In strict functional programming 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. 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.

--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)));