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

F_{n}= F_{n-1}+ F_{n-2}

with seed values

F_{0}= 0 and F_{1}= 1.

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

For n = 9 Output:34

Following are different methods to get the nth Fibonacci number.

**Method 1 ( Use recursion ) **

A simple method that is a direct recursive implementation mathematical recurrence relation given above.

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).

**Method 2 ( Use Dynamic Programming )**

We can avoid the repeated work done is the method 1 by storing the Fibonacci numbers calculated so far.

Output:

34

*Time Complexity:* O(n)

*Extra Space: *O(n)

**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.

*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)

*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} = F_{n+1}F_{n-1} – F_{n}^{2}

Moreover, since A^{n}A^{m} = A^{n+m} for any square matrix A, the following identities can be derived (they are obtained form two different coefficients of the matrix product)

F_{m}F_{n} + F_{m-1}F_{n-1} = F_{m+n-1}

By putting n = n+1,

F_{m}F_{n+1} + F_{m-1}F_{n} = F_{m+n}

Putting m = n

F_{2n-1} = F_{n}^{2} + F_{n-1}^{2}

F_{2n} = (F_{n-1} + F_{n+1})F_{n} = (2F_{n-1} + F_{n})F_{n} (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.

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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