Calculate XOR from 1 to n.

2.9

Given a number n, the task is to find the XOR from 1 to n.

Examples :

Input : n = 6
Output : 7
// 1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6  = 7

Input : n = 7
Output : 0
// 1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6 ^ 7 = 0

Method 1 (Naive Approach):
1- Initialize result as 0.
1- Traverse all numbers from 1 to n.
2- Do XOR of numbers one by one with result.
3- At the end, return result.

Method 2 (Efficient method) :
1- Find the remainder of n by moduling it with 4.
2- If rem = 0, then xor will be same as n.
3- If rem = 1, then xor will be 1.
4- If rem = 2, then xor will be n+1.
5- If rem = 3 ,then xor will be 0.

C/C++

// C++ program to find XOR of numbers
// from 1 to n.
#include <bits/stdc++.h>
using namespace std;

// Function to calculate xor
long computeXOR(const int n)
{
    // Modulus operator are expensive on most of the 
    // computers. n & 3 will be equivalent to n % 4.   

    switch(n & 3) // n % 4 
    {
    case 0: return n;     // if n is multiple of 4
    case 1: return 1;     // If n % 4 gives remainder 1  
    case 2: return n + 1; // If n % 4 gives remainder 2    
    case 3: return 0;     // If n % 4 gives remainder 3  
    }
}

// Driver code
int main()
{
    // your code goes here
    int n = 5;
    cout <<    computeXOR(n);
    return 0;
}

Java

// Java program to find XOR of numbers
// from 1 to n.

class GFG 
{
    // Method to calculate xor
    static int computeXOR(int n)
    {
        // If n is a multiple of 4
        if (n % 4 == 0)
            return n;
     
        // If n%4 gives remainder 1
        if (n % 4 == 1)
            return 1;
     
        // If n%4 gives remainder 2
        if (n % 4 == 2)
            return n + 1;
     
        // If n%4 gives remainder 3
        return 0;
    }
    
    // Driver method
    public static void main (String[] args)
    {
         int n = 5;
         System.out.println(computeXOR(n));
    }
}


Output:

1

How does this work?
When we do XOR of numbers, we get 0 as XOR value just before a multiple of 4. This keeps repeating before every multiple of 4.

Number Binary-Repr  XOR-from-1-to-n
1         1           [0001]
2        10           [0011]
3        11           [0000]  <----- We get a 0
4       100           [0100]  <----- Equals to n
5       101           [0001]
6       110           [0111]
7       111           [0000]  <----- We get 0
8      1000           [1000]  <----- Equals to n
9      1001           [0001]
10     1010           [1011]
11     1011           [0000] <------ We get 0
12     1100           [1100] <------ Equals to n

This article is contributed by Sahil Chhabra(KILLER). 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.

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