Challenge 1 from https://bartoszmilewski.com/2014/11/24/types-and-functions/ => A memoize function takes a function f and returns a function that takes a value x. When x is passed the first time it returns a function that would take the next x. If the x passed was passed before it would compute f(x) and return the result. The solution here is do…

memorize.js

const head = x => _ => x;
const tail = _ => y => y;
const arr = x => xs => f => f(x)(xs);
const none = () => null;

const memorize = f => {
  const bind = xs => _x => y => {
    if (xs === none) {
      return bind(arr({arg: y, ret: f(y)})(_x))(none);
    }
    if (xs(head).arg === y) {
      return xs(head).ret;
    }
    return bind(xs(tail))(arr(xs(head))(_x))(y);
  };
  return bind(none)(none);
};

const fib = n => {
  if (n <= 0) {
    return 1;
  }
  return n * fib(n - 1);
};

((/* main */) => {
  let memid = memorize(fib);
  
  memid = memid(3);
  memid = memid(4);
  memid = memid(1);
  memid = memid(2);

  console.log(memid(4))
})()