Fast Fibonacci recursion

Question

I\'m trying to recall an algorithm on Fibonacci recursion. The following:

public int fibonacci(int n)  {
  if(n == 0)
    return 0;
  else if(n == 1)
    ret         


        

Answer 1

This kind of problems are linear recurrence types and they are solved fastest via fast matrix exponentiation. Here's the blogpost that describes this kind of approach concisely.

Answer 2

An example in JavaScript that uses recursion and a lazily initialized cache for added efficiency:

var cache = {};

function fibonacciOf (n) {
  if(n === 0) return 0;
  if(n === 1) return 1;
  var previous = cache[n-1] || fibonacciOf(n-1);
  cache[n-1] = previous;
  return previous + fibonacciOf(n-2);
};

Answer 3

I found interesting article about fibonacci problem

here the code snippet

# Returns F(n)
def fibonacci(n):
    if n < 0:
        raise ValueError("Negative arguments not implemented")
    return _fib(n)[0]


# Returns a tuple (F(n), F(n+1))
def _fib(n):
    if n == 0:
        return (0, 1)
    else:
        a, b = _fib(n // 2)
        c = a * (2 * b - a)
        d = b * b + a * a
        if n % 2 == 0:
            return (c, d)
        else:
            return (d, c + d)

# added iterative version base on C# example
def iterFib(n):
    a = 0
    b = 1
    i=31
    while i>=0:
        d = a * (b * 2 - a)
        e = a * a + b * b
        a = d
        b = e
        if ((n >> i) & 1) != 0:
            c = a + b;
            a = b
            b = c
        i=i-1
    return a