XNOR of two numbers

XNOR(Exclusive-NOR) gate is a digital logic gate that takes two inputs and gives one output. Its function is the logical complement of the Exclusive-OR (XOR) gate. The output is TRUE if both inputs are the same and FALSE if the inputs are different. The truth table of the XNOR gate is given below.

A	B	Output
1	1	1
1	0	0
0	1	0
0	0	1

Problem Statement

Given two numbers x and y. Find the XNOR of the two numbers.

Sample Example 1

Input: x = 12, y = 5

Output: 6

Explanation

(12)10 = (1100)2
(5)10 = (101)2
XNOR = (110)2 = (6)10

Sampe Example 2

Input: x = 16, y = 16

Output: 31

Explanation

(16)10 = (10000)2
(16)10 = (10000)2
XNOR = (11111)2 = (31)10

Approach 1: Brute Force Approach

The brute force approach will be to check each bit of both numbers and compare if they are the same or not. If they are the same, put 1 else 0.

Pseudocode

procedure xnor (x, y)
   if x > y then
      swap(x,y)
   end if
   if x == 0 and y == 0 then
      ans = 1
   end if
   while x
      x_rem = x & 1
      y_rem = y & 1
      if x_rem == y_rem then
         ans = ans | (1 << count)
      end if
      count = count + 1
      x = x >> 1
      y = y >> 1
end procedure

Example: C++ Implementation

In the following program, check if bits of both x and y are same and then set bits in the answer.

#include <bits/stdc++.h>
using namespace std;

// Function to swap values of two variables
void swap(int *a, int *b){
   int temp = *a;
   *a = *b;
   *b = temp;
}

// Function to find the XNOR of two numbers
int xnor(int x, int y){

   // Placing the lower number in variable x
   if (x > y){
      swap(x, y);
   }
   
   // Base Condition
   if (x == 0 && y == 0){
      return 1;
   }
   
   // Cnt for counting the bit position Ans stores ans sets the bits of XNOR operation
   int cnt = 0, ans = 0;
   
   // executing loop for all the set bits in the lower number
   while (x){
   
      // Gets the last bit of x and y
      int x_rem = x & 1, y_rem = y & 1;
      
      // If last bits of x and y are same
      if (x_rem == y_rem){
         ans |= (1 << cnt);
      }
      
      // Increase counter for bit position and right shift both x and y
      cnt++;
      x = x >> 1;
      y = y >> 1;
   }
   return ans;
}
int main(){
   int x = 10, y = 11;
   cout << "XNOR of " << x << " and " << y << " = " << xnor(x, y);
   return 0;
}

Output

XNOR of 10 and 11 = 14

Time Complexity: O(logn) as traversal is done on all the bits of number which are logn.

Space Complexity: O(1)

Approach 2

XNOR is the inverse of XOR operation and it’s truth table is also the inverse of XOR table. Thus, toggling the bits of the higher number i.e. setting 1 to 0 and 0 to 1 and then XORing with the lower number will yield XNOR number.

Sampe Example 1

Input: x = 12, y = 5

Output: 6

Explanation

(12)10 = (1100)2
(5)10 = (101)2
Toggled bits of 12 = 0011
0011 ^ 101 = 0110

Sampe Example 2

Input: x = 12, y = 31

Output: 12

Explanation

(12)10 = (1100)2
(31)10 = (11111)2
Toggled bits of 31 = 00000
00000 ^ 1100 = 01100

Pseudocode

procedure toggle (n)
   temp = 1
   while temp <= n
      n = n ^ temp
      temp = temp << 1
end procedure

procedure xnor (x, y)
   max_num = max(x,y)
   min_num = min(x,y)
   toggle (max_num)
   ans = max_num ^ min_num
end procedure

Example: C++ Implementation

In the following program, all the bits of higher number are toggled and then XORed with the lower number.

#include <bits/stdc++.h>
using namespace std;

// Function to toggle all bits of a number
void toggle(int &num){
   int temp = 1;
   
   // Execute loop until set bit of temp cross MST of num
   while (temp <= num){
   
      // Toggle bit of num corresponding to set bit in temp
      num ^= temp;
      
      // Move set bit of temp to left
      temp <<= 1;
   }
}

// Function to find the XNOR of two numbers
int xnor(int x, int y){

   // Finding max and min number
   int max_num = max(x, y), min_num = min(x, y);
   
   // Togglinf the max number
   toggle(max_num);
   
   // XORing toggled max num and min num
   return max_num ^ min_num;
}
int main(){
   int x = 5, y = 15;
   cout << "XNOR of " << x << " and " << y << " = " << xnor(x, y);
   return 0;
}

Output

XNOR of 5 and 15 = 5

Time Complexity: O(logn) due to traversal in toggle() function

Space Complexity: O(1)

Approach 3: Bit Mask

Logically XNOR is inverse of XOR, but upon performing complement of XOR, leading zeroes are also inverted. It is avoided using a bit mask that removes all the unnecessary leading bits.

Sample Example 1

Input: x = 12, y = 5

Output: 6

Explanation

(12)10 = (1100)2
(5)10 = (101)2
1100 ^ 101 = 1001
Inverse of 1001 = 0110

Sample Example 2

Input: x = 12, y = 31

Output: 12

Explanation

(12)10 = (1100)2
(31)10 = (11111)2
1100 ^ 11111 = 10011
Inverse of 10011 = 01100

Psuedocode

Procedure xnor (x, y)
   bit_count = log2 (maximum of a and y)
   mask = (1 << bit_count) - 1
   ans = inverse(x xor y) and mask
end procedure

Example: C++ Implementation

In the following program, bit mask is used to obtain only the required bits from inverse of x xor y.

#include <bits/stdc++.h>
using namespace std;

// Function to find the XNOR of two numbers
int xnor(int x, int y){

   // Maximum number of bits used in both the numbers
   int bit_count = log2(max(x, y));
   int mask = (1 << bit_count) - 1;
   
   // Inverse of XOR operation
   int inv_xor = ~(x ^ y);
   
   // GEtting the required bits and removing the inversion of leading zeroes.
   return inv_xor & mask;
}
int main(){
   int x = 10, y = 10;
   cout << "XNOR of " << x << " and " << y << " = " << xnor(x, y);
   return 0;
}

Output

XNOR of 10 and 10 = 7

Conclusion

In conclusion, XNOR of two numbers can be found using various approaches and time complexities ranging from O(logn) which brute force to O(1) which is most oprimized. Applying bit operations is cheaper thus the brute force complexity is logarithmic.