Solution to Exercise 4.18

[LucasGdosR]

// Fibonacci
// The first function receives n as an argument
// It applies the fib function recursively, passing n as an argument, as well as the initial arguments (k = 1, fib1 = 1, fib2 = 1)
(n => (fib => fib(fib, n, 2, 1, 1))
      // The fib function is then defined as ft, with parameters n, k, fib1, and fib2
      // Establish the base cases: n === 1 or n === 2
      ((ft, n, k, fib1, fib2) => n === 1
                  ? 1
                  : n === 2
                  ? 1
                  :
                  // Iterate until k equals n. Notice k starts at 2, and gets incremented every iteration
                  k === n
                  // When k reaches n, return the accumulated fib2
                  ? fib2
                  // Otherwise, accumulate the sum as the new fib2
                  : ft(ft, n, k + 1, fib2, fib1 + fib2)));

// Is even
function f(x) {
    return ((is_even, is_odd) => is_even(is_even, is_odd, x))
           ((is_ev, is_od, n) => n === 0 ? true : is_od(is_ev, is_od, n - 1),
           (is_ev, is_od, n) => n === 0 ? false : is_ev(is_ev, is_od, n - 1));
}