# Nth XOR Fibonacci number

Given three integers a, b and N where a and b are the first two terms of the XOR Fibonacci series and the task is to find the Nth term.
The Nth term of the XOR Fibonacci series is defined as F(N) = F(N – 1) ^ F(N – 2) where ^ is the bitwise XOR.

Examples:

Input: a = 1, b = 2, N = 5
Output: 3
F(0) = 1
F(1) = 2
F(2) = 1 ^ 2 = 3
F(3) = 2 ^ 3 = 1
F(4) = 1 ^ 3 = 2
F(5) = 1 ^ 2 = 3

Input: a = 5, b = 11, N = 1000001
Output: 14

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Since, a ^ a = 0 and it is given that

F(0) = a and F(1) = b
Now, F(2) = F(0) ^ F(1) = a ^ b
And, F(3) = F(1) ^ F(2) = b ^ (a ^ b) = a
F(4) = a ^ b ^ a = b
F(5) = a ^ b
F(6) = a
F(7) = b
F(8) = a ^ b

It can be observed that the answer repeats itself after every 3 numbers. So the answer is F(N % 3) where F(0) = a, F(1) = b and F(2) = a ^ b.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include using namespace std;    // Function to return the nth XOR Fibonacci number int nthXorFib(int n, int a, int b) {     if (n == 0)         return a;     if (n == 1)         return b;     if (n == 2)         return (a ^ b);        return nthXorFib(n % 3, a, b); }    // Driver code int main() {     int a = 1, b = 2, n = 10;        cout << nthXorFib(n, a, b);        return 0; }

## Java

 // Java implementation of the above approach  class GFG {                // Function to return the      // nth XOR Fibonacci number      static int nthXorFib(int n, int a, int b)      {          if (n == 0)              return a;          if (n == 1)              return b;          if (n == 2)              return (a ^ b);                 return nthXorFib(n % 3, a, b);      }             // Driver code      public static void main (String[] args)      {          int a = 1, b = 2, n = 10;                 System.out.println(nthXorFib(n, a, b));      }  }    // This code is contributed by AnkitRai01

## Python3

 # Python3 implementation of the approach     # Function to return # the nth XOR Fibonacci number  def nthXorFib(n, a, b):     if n == 0 :          return a      if n == 1 :          return b      if n == 2 :          return a ^ b         return nthXorFib(n % 3, a, b)     # Driver code  a = 1 b = 2 n = 10 print(nthXorFib(n, a, b))     # This code is contributed by divyamohan123

## C#

 // C# implementation of the above approach  using System;        class GFG {                // Function to return the      // nth XOR Fibonacci number      static int nthXorFib(int n, int a, int b)      {          if (n == 0)              return a;          if (n == 1)              return b;          if (n == 2)              return (a ^ b);                 return nthXorFib(n % 3, a, b);      }             // Driver code      public static void Main (String[] args)      {          int a = 1, b = 2, n = 10;                 Console.WriteLine(nthXorFib(n, a, b));      }  }    // This code is contributed by Princi Singh

Output:

2

Time Complexity: O(1)

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.