XNOR of two numbers

The XNOR (XNOR) gate is a digital logic gate that accepts two inputs and gives one output. Its function is the logical complement of the exclusive OR (XOR) gate. The output is TRUE if the two inputs are the same; FALSE if the inputs are different. The truth table of the XNOR gate is given below.

one	B	Output
1	1	1
1	0	0
0	1	0
0	0	1

Problem Statement

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

Example Example 1

Input: x = 12, y = 5

Output: 6

illustrate

(12)<sub>10</sub> = (1100)<sub>2</sub>
(5)<sub>10</sub> = (101)<sub>2</sub>
XNOR = (110)2 = (6)<sub>10</sub>

The Chinese translation of

Sampe Example 2

is:

Sampe Example 2

Input: x = 16, y = 16

Output: 31

illustrate

(16)<sub>10</sub> = (10000)<sub>2</sub>
(16)<sub>10</sub> = (10000)<sub>2</sub>
XNOR = (11111)<sub>2</sub> = (31)<sub>10</sub>

Method One: Violence Method

The brute force method is to check every bit of the two numbers and compare whether they are the same. If they are the same, add 1, otherwise add 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 the bits of x and y are the same, then set the 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), because the traversal is performed on all logn bits.

Space complexity: O(1)

Method Two

XNOR is the inverse operation of the XOR operation, and its truth table is also the inverse operation of the XOR table. So switching the bits of the higher number, i.e. setting 1 to 0 and 0 to 1, then XORing with the lower number will produce an XNOR number.

Example 1

Input: x = 12, y = 5

Output: 6

illustrate

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

The Chinese translation of

Sampe Example 2

is:

Sampe Example 2

Input: x = 12, y = 31

Output: 12

illustrate

(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 program below, all bits of the 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 the toggle() function

Space complexity: O(1)

Method 3: Bit Mask

Logically speaking, XNOR is the inverse operation of XOR, but when performing a complement operation on XOR, leading zeros will also be inverted. To avoid this, a bit mask can be used to remove all unnecessary leading bits.

Example Example 1

Input: x = 12, y = 5

Output: 6

illustrate

(12)<sub>10</sub> = (1100)<sub>2</sub>
(5)<sub>10</sub> = (101)<sub>2</sub>
1100 ^ 101 = 1001
Inverse of 1001 = 0110

Example 2

Input: x = 12, y = 31

Output: 12

illustrate

(12)<sub>10</sub> = (1100)<sub>2</sub>
(31)<sub>10</sub> = (11111)<sub>2</sub>
1100 ^ 11111 = 10011
Inverse of 10011 = 01100

pseudocode

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, a bit mask is used to get only the required bits from the 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

in conclusion

In summary, the XNOR of two numbers can be found using a variety of methods and time complexity, ranging from O(logn) brute force methods to O(1) optimal methods. Applying bitwise operations is cheaper, so the complexity of the brute force method is logarithmic.

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