Data Structures and Algorithms (Fibonacci series)- Tirthankar Pal - MBA from IIT Kharagpur, GATE, GMAT, IIT Written Test, Interview were a part of MBA Entrance, B.S. in Computer Science from NIELIT

 

Write a recursive function to find the Fibonacci series

 

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 

F₀ = 0 and F₁ = 1.

 

Given a number n, print n-th Fibonacci Number. 

Examples: 

Input  : n = 2

Output : 1

 

Input  : n = 9

Output : 34

Recommended Problem

Nth Fibonacci Number

 

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

The 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 is given above.

// 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: Exponential, as every function calls two other functions.

If the original recursion tree were to be implemented then this would have been the tree but now for n times the recursion function is called

Original tree for recursion

                          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)

Optimized tree for recursion for code above

    fib(5) 

    fib(4)

    fib(3)

    fib(2)

    fib(1)

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 in method 1 by storing the Fibonacci numbers calculated so far. 

//Fibonacci Series using Dynamic Programming #include<stdio.h>  int fib(int n) {   /* Declare an array to store Fibonacci numbers. */   int f[n+2];   // 1 extra to handle case, n = 0   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) for given n

Auxiliary 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. 

// 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; }

Output

34

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

 

Method 4: Using power of the matrix {{1, 1}, {1, 0}}
This is another O(n) that 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: 

 

#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; }

Output

 34

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

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

#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; }

Output

34

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 the above matrix equation. 

Taking determinant on both sides, we get

 

(-1)ⁿ = F_n+1F_n-1 - F_n²

 

Moreover, since AⁿA^m = A^n+m for any square matrix A,

the following identities can be derived (they are obtained

from two different coefficients of the matrix product)

 

F_mF_n + F_m-1F_n-1 = F_{m+n-1         ---------------------------(1)}

 

By putting n = n+1 in equation(1),

F_mF_n+1 + F_m-1F_n = F_{m+n             --------------------------(2)}

 

Putting m = n in equation(1).

F_2n-1 = F_n² + F_n-1²

Putting m = n in equation(2)

 

F_2n = (F_n-1 + F_n+1)F_n = (2F_n-1 + F_n)F_n (Source: Wiki)   --------

( By putting Fn+1 = Fn + Fn-1 )

To get the formula to be proved, we simply need to do the following

If n is even, we can put k = n/2

If n is odd, we can put k = (n+1)/2

Below is the implementation of the above idea.  

 

// C++ Program to find n'th fibonacci Number in // with O(Log n) arithmetic operations #include <bits/stdc++.h> using namespace std;   const int MAX = 1000;   // Create an array for memoization int f[MAX] = {0};   // Returns n'th fibonacci 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;       // Applying 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: O(Log n), as we divide the problem in half in every recursive call.

 

Method 7: (Another approach(Using Binet’s formula))
In this method, we directly implement the formula for the nth term in the Fibonacci series. 
F_n = {[(√5 + 1)/2] ^ n} / √5 

Note: Above Formula gives correct result only upto for n<71. Because as we move forward from n>=71 , rounding error becomes significantly large . Although , using floor function instead of round function will give correct result for n=71 . But after from n=72 , it also fails.

Example: For N=72 , Correct result is 498454011879264 but above formula gives 498454011879265.

// C Program to find n'th fibonacci Number #include<stdio.h> #include<math.h> int fib(int n) {   double phi = (1 + sqrt(5)) / 2;   return round(pow(phi, n) / sqrt(5)); } int main () {   int n = 9;   printf("%d", fib(n));   return 0; }

Output

34

Time Complexity: O(logn), this is because calculating phi^n takes logn time
Auxiliary Space: O(1)

 

Method 8: DP using memoization(Top down approach)

 

We can avoid the repeated work done in method 1 by storing the Fibonacci numbers calculated so far. We just need to store all the values in an array.

#include <bits/stdc++.h> using namespace std; int dp[10]; int fib(int n) {     if (n <= 1)         return n;       // temporary variables to store     //  values of fib(n-1) & fib(n-2)     int first, second;       if (dp[n - 1] != -1)         first = dp[n - 1];     else         first = fib(n - 1);       if (dp[n - 2] != -1)         second = dp[n - 2];     else         second = fib(n - 2);       // memoization     return dp[n] = first + second; }   // Driver Code int main() {     int n = 9;       memset(dp, -1, sizeof(dp));       cout << fib(n);     getchar();     return 0;       // This code is contributed by Bhavneet Singh }

Output

34

Tirthankar Pal

MBA from IIT Kharagpur with GATE, GMAT, IIT Kharagpur Written Test, and Interview

2 year PGDM (E-Business) from Welingkar, Mumbai

4 years of Bachelor of Science (Hons) in Computer Science from the National Institute of Electronics and Information Technology

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10 years of Experience in Software Development out of that 6 years 8 months in Wipro

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