Y Combinator

es6.js

const Y = (f) => {
  const something = x => f(v => x(x)(v));
  // const something = x => f(x(x));  // actually it's this, but will cause stask overflow
  
  return something(something);
};

const factorial = Y(function f(fac) {
  return function almost(n) {
    return (n == 0 ? 1 : n * fac(n - 1));
  }
});

factorial(5);  // 120

// Explanations
// There are two 'proper' factorial functions above.
// One is being referenced by variable factorial, it's the return value of Y(fn), according to Y's definition,
// factorial references to something(something);
// The other is referenced in line 11, fac, it's used in subsequent computations, in line 2, the function passed in as fac is x(x), it
// actually has equivalent of something(something), because x is something.
// Our computation starts with something(something)(5), is equivalent to
// something(something) -> f(v => something(something)(v)) -> f(something(something))
// But why something(something) is the key here? It looks weird.

verbose.js

const Y = (f) => {
  const something = function something(x) {
    return f(function another_fac(v) {
      return x(x)(v) ;
    });
  };

  return something(something);
};

const factorial = Y(function f(fac) {
  return function almost(n) {
    return (n == 0 ? 1 : n * fac(n - 1));
  }
});

factorial(5);  // 120