Number of calls for nth Fibonacci number

Question

Consider the following code snippet:

int fib(int N)
{
   if(N<2) return 1;
   return (fib(N-1) + fib(N-2));
}

Given that fib

Answer 1

phi is a constant

position = ceil(log((n - 0.5)*sqrt(5))/log(phi));

n is the fibonacci number... position will give you the which fibonacci number is n

for example given 13 , position will be 7 - 0 1 1 2 3 5 8 13

using this position just calculate the fibonacci number at position-1 or any position you want relative to given fibonacci number.

Previous Fibo Num = floor((pow(phi,position-1)/sqrt(5))+0.5);

floor((pow(phi, position)/sqrt(5))+0.5) - is the standard formula for calculating Nth fibonacci num (Note - This is not an approximation)

I have just reverse this formula to calculate the position and use the position - 1 to calculate the previous fibonacci number.

Ref - http://itpian.com/Coding/20951-Given-the-Nth-fib-no-and-find-the--N-1-th-fib-number-without-calculating-from-the-beginning---------.aspx

Answer 2

Interesting question, I can't give you a formula, but I wrote a Ruby program to do it, it works on numbers I figured out on paper, and it should work for any.

#!/usr/bin/ruby
#find out how many times fib() would need to be called

def howmany(n)
    a = [ ]
    a.push n-1
    a.push n-2
    while a.select{|n| n > 2}.length > 0
        a.map! do |n|
            n > 2 ? [n-1,n-2] : n
        end
        a.flatten!
    end
    a.length
end

.

>> howmany(10)
=> 55

It's slow.. I'm figuring out 35 right now, I'll edit when it finishes.

Edit:

>> howmany(35)
=> 9227465

Answer 3

Let f(n) be the number of calls made to calculate fib(n).

• If n < 2 then f(n) = 1.
• Otherwise, f(n) = 1 + f(n - 1) + f(n - 2).

So, f is at least O(fib(n)). In fact, f(n) is 2 * fib(n) - 1. We show this by induction:

• Base cases (n < 2, that is, n = 0 or n = 1):
  • f(n) = 1 = 2 * 1 - 1 = 2 * fib(n) - 1.
• Induction step (n >= 2):
  • f(n + 1) = f(n) + f(n - 1) + 1
  • f(n + 1) = 2 * fib(n) - 1 + 2 * fib(n - 1) - 1 + 1
  • f(n + 1) = 2 * fib(n + 1) - 1

There exist efficient ways to calculate any Fibonacci term. Thus the same holds for f(n).

Answer 4

Is there any relation for this problem ?

There is a close-form equation for the nth fibonacci number: http://en.wikipedia.org/wiki/Fibonacci_number#Closed_form_expression

In the pseudocode you posted, the number of calls satisfies the recurrence relation

x(n) = x(n-1) + x(n-2) +1   # for n>=2
x(1) = 1
x(0) = 1

This is almost same as the Fibonacci recurrence relation. Proof by induction can show that the number of calls to fib made by fib(n) is equal to 2*fib(n)-1, for n>=0.

Of course, the calculation can be sped up by using the closed form expression, or by adding code to memorize previously computed values.

Answer 5

As mentioned above, you need to solve the following recurring equation: K(n)=K(n-1)+K(n-2)+1

Let's write it for n-1: K(n-1)=K(n-2)+K(n-3)+1

Now, subtract the second one from the first one: K(n)-K(n-1) = K(n-1) - K(n-3),

or

K(n) - 2*K(n-1) + K(n-3) = 0.

The respective characteristic equation will be: x^3 - 2*x^2 + 1 = 0.

It has the following roots: 1, (1+sqrt(5))/2, (1-sqrt(5))/2

Thus for any real A,B,C the following function K(n) = A*(1)^n + B*((1+sqrt(5))/2)^n + C*((1-sqrt(5))/2)^n

will be a solution for your equation.

To find A,B,C you need to define several initial values K(0), K(1), K(2) and solve the system of equations.

Answer 6

This is a classic problem for solving with Recurrence Relations.

Specifically, the fibonacci problem has the following parameters:

f(0) = 1
f(1) = 1
f(n) = f(n-1) + f(n-2)

Once you master solving recurrences, you'll have no problem reaching the solution (which, incidently, is exactly the same as fib(n)).