The Fibonacci numbers are the numbers in the following integer sequence.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……..

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation

Fn = Fn-1 + Fn-2
with seed values

F0 = 0 and F1 = 1.

Write a function int fib(int n) that returns Fn. For example, if n = 0, then fib() should return 0. If n = 1, then it should return 1. For n > 1, it should return Fn-1 + Fn-2

For n = 9
Output:34
Following are different methods to get the nth Fibonacci number.

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Method 1 ( Use recursion )
A simple method that is a direct recursive implementation mathematical recurrence relation given above.

c program1 
//Fibonacci Series using Recursion
#include<stdio.h>
int fib(int n)
{
   if (n <= 1)
      return n;
   return fib(n-1) + fib(n-2);
}
 
int main ()
{
  int n = 9;
  printf("%d", fib(n));
  getchar();
  return 0;
}

Output

34

Time Complexity: T(n) = T(n-1) + T(n-2) which is exponential.

We can observe that this implementation does a lot of repeated work (see the following recursion tree). So this is a bad implementation for nth Fibonacci number.

                         fib(5)   
                     /             \     
               fib(4)                fib(3)   
             /      \                /     \
         fib(3)      fib(2)         fib(2)    fib(1)
        /     \        /    \       /    \  
  fib(2)   fib(1)  fib(1) fib(0) fib(1) fib(0)
  /    \
fib(1) fib(0)

Extra Space: O(n) if we consider the function call stack size, otherwise O(1).

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Method 2 ( Use Dynamic Programming )
We can avoid the repeated work done is the method 1 by storing the Fibonacci numbers calculated so far.

c program2 
//Fibonacci Series using Dynamic Programming
#include<stdio.h>
 
int fib(int n)
{
  /* Declare an array to store Fibonacci numbers. */
  int f[n+1];
  int i;
 
  /* 0th and 1st number of the series are 0 and 1*/
  f[0] = 0;
  f[1] = 1;
 
  for (i = 2; i <= n; i++)
  {
      /* Add the previous 2 numbers in the series
         and store it */
      f[i] = f[i-1] + f[i-2];
  }
 
  return f[n];
}
 
int main ()
{
  int n = 9;
  printf("%d", fib(n));
  getchar();
  return 0;
}

Output:

34

Time Complexity: O(n)
Extra Space: O(n)

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Method 3 ( Space Optimized Method 2 )
We can optimize the space used in method 2 by storing the previous two numbers only because that is all we need to get the next Fibonacci number in series.

c program3 
// Fibonacci Series using Space Optimized Method
#include<stdio.h>
int fib(int n)
{
  int a = 0, b = 1, c, i;
  if( n == 0)
    return a;
  for (i = 2; i <= n; i++)
  {
     c = a + b;
     a = b;
     b = c;
  }
  return b;
}
 
int main ()
{
  int n = 9;
  printf("%d", fib(n));
  getchar();
  return 0;
}

Time Complexity: O(n)
Extra Space: O(1)

[ad type=”banner”] Method 4 ( Using power of the matrix {{1,1},{1,0}} )
This another O(n) which relies on the fact that if we n times multiply the matrix M = {{1,1},{1,0}} to itself (in other words calculate power(M, n )), then we get the (n+1)th Fibonacci number as the element at row and column (0, 0) in the resultant matrix.

 

c program4 
#include <stdio.h>
 
/* Helper function that multiplies 2 matrices F and M of size 2*2, and
  puts the multiplication result back to F[][] */
void multiply(int F[2][2], int M[2][2]);
 
/* Helper function that calculates F[][] raise to the power n and puts the
  result in F[][]
  Note that this function is designed only for fib() and won't work as general
  power function */
void power(int F[2][2], int n);
 
int fib(int n)
{
  int F[2][2] = {{1,1},{1,0}};
  if (n == 0)
      return 0;
  power(F, n-1);
 
  return F[0][0];
}
 
void multiply(int F[2][2], int M[2][2])
{
  int x =  F[0][0]*M[0][0] + F[0][1]*M[1][0];
  int y =  F[0][0]*M[0][1] + F[0][1]*M[1][1];
  int z =  F[1][0]*M[0][0] + F[1][1]*M[1][0];
  int w =  F[1][0]*M[0][1] + F[1][1]*M[1][1];
 
  F[0][0] = x;
  F[0][1] = y;
  F[1][0] = z;
  F[1][1] = w;
}
 
void power(int F[2][2], int n)
{
  int i;
  int M[2][2] = {{1,1},{1,0}};
 
  // n - 1 times multiply the matrix to {{1,0},{0,1}}
  for (i = 2; i <= n; i++)
      multiply(F, M);
}
 
/* Driver program to test above function */
int main()
{
  int n = 9;
  printf("%d", fib(n));
  getchar();
  return 0;
}

Time Complexity: O(n)
Extra Space: O(1)
Method 5 ( Optimized Method 4 )
The method 4 can be optimized to work in O(Logn) time complexity. We can do recursive multiplication to get power(M, n) in the prevous method (Similar to the optimization done in this post)

C program5 
#include <stdio.h>
 
void multiply(int F[2][2], int M[2][2]);
 
void power(int F[2][2], int n);
 
/* function that returns nth Fibonacci number */
int fib(int n)
{
  int F[2][2] = {{1,1},{1,0}};
  if (n == 0)
    return 0;
  power(F, n-1);
  return F[0][0];
}
 
/* Optimized version of power() in method 4 */
void power(int F[2][2], int n)
{
  if( n == 0 || n == 1)
      return;
  int M[2][2] = {{1,1},{1,0}};
 
  power(F, n/2);
  multiply(F, F);
 
  if (n%2 != 0)
     multiply(F, M);
}
 
void multiply(int F[2][2], int M[2][2])
{
  int x =  F[0][0]*M[0][0] + F[0][1]*M[1][0];
  int y =  F[0][0]*M[0][1] + F[0][1]*M[1][1];
  int z =  F[1][0]*M[0][0] + F[1][1]*M[1][0];
  int w =  F[1][0]*M[0][1] + F[1][1]*M[1][1];
 
  F[0][0] = x;
  F[0][1] = y;
  F[1][0] = z;
  F[1][1] = w;
}
 
/* Driver program to test above function */
int main()
{
  int n = 9;
  printf("%d", fib(9));
  getchar();
  return 0;
}

Time Complexity: O(Logn)
Extra Space: O(Logn) if we consider the function call stack size, otherwise O(1).

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Method 6 (O(Log n) Time)
Below is one more interesting recurrence formula that can be used to find n’th Fibonacci Number in O(Log n) time.

If n is even then k = n/2:
F(n) = [2*F(k-1) + F(k)]*F(k)

If n is odd then k = (n + 1)/2
F(n) = F(k)*F(k) + F(k-1)*F(k-1)

How does this formula work?
The formula can be derived from above matrix equation.
fibonaccimatrix

Taking determinant on both sides, we get
(-1)n = Fn+1Fn-1 – Fn2
Moreover, since AnAm = An+m for any square matrix A, the following identities can be derived (they are obtained form two different coefficients of the matrix product)

FmFn + Fm-1Fn-1 = Fm+n-1

By putting n = n+1,

FmFn+1 + Fm-1Fn = Fm+n

Putting m = n

F2n-1 = Fn2 + Fn-12

F2n = (Fn-1 + Fn+1)Fn = (2Fn-1 + Fn)Fn (Source: Wiki)

To get the formula to be proved, we simply need to do following
If n is even, we can put k = n/2
If n is odd, we can put k = (n+1)/2

Below is C++ implementation of above idea.

C Program 6 
// C++ Program to find n'th fibonacci Number in
// with O(Log n) arithmatic operations
#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 1000;
 
// Create an array for memoization
int f[MAX] = {0};
 
// Returns n'th fuibonacci number using table f[]
int fib(int n)
{
    // Base cases
    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return (f[n] = 1);
 
    // If fib(n) is already computed
    if (f[n])
        return f[n];
 
    int k = (n & 1)? (n+1)/2 : n/2;
 
    // Applyting above formula [Note value n&1 is 1
    // if n is odd, else 0.
    f[n] = (n & 1)? (fib(k)*fib(k) + fib(k-1)*fib(k-1))
           : (2*fib(k-1) + fib(k))*fib(k);
 
    return f[n];
}
 
/* Driver program to test above function */
int main()
{
    int n = 9;
    printf("%d ", fib(n));
    return 0;
}

Output :

34

Time complexity of this solution is O(Log n) as we divide the problem to half in every recursive call.

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