Program for Fibonacci numbers

Program for Fibonacci numbers

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.

Method 1 ( Use recursion ):-

int fib(int n)

{

   if (n <= 1)

      return n;

   return fib(n-1) + fib(n-2);

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

Method 2 ( Use Dynamic Programming )

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

Method 3 ( Space Optimized Method 2 )

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;

}

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

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.

The matrix representation gives the following closed expression for the Fibonacci numbers:

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 of geeksforgeeks)

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

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.

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

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