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Nth XOR Fibonacci number

  • Difficulty Level : Medium
  • Last Updated : 27 Apr, 2021

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:
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 
 

 

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 <bits/stdc++.h>
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

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the nth XOR Fibonacci number
function 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
    let a = 1, b = 2, n = 10;
 
    document.write(nthXorFib(n, a, b));
 
</script>
Output: 
2

 

Time Complexity: O(1)
 


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