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

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
594810f028c0303b75339ad5	Y combinator	1	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 the factorials. The factorial(N) function is already given to you.

--hints--

Y sollte eine Funktion zurückgeben.

assert.equal(typeof Y((f) => (n) => n), 'function');

factorial(1) sollte 1 zurückgeben.

assert.equal(factorial(1), 1);

factorial(2) sollte 2 zurückgeben.

assert.equal(factorial(2), 2);

factorial(3) sollte 6 zurückgeben.

assert.equal(factorial(3), 6);

factorial(4) sollte 24 zurückgeben.

assert.equal(factorial(4), 24);

factorial(10) sollte 3628800 zurückgeben.

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)));