Compute the parity of a number using XOR and table look-up

Parity of a number refers to whether it contains an odd or even number of 1-bits. The number has “odd parity”, if it contains odd number of 1-bits and is “even parity” if it contains even number of 1-bits.

1 --> parity of the set is odd
0 --> parity of the set is even

Examples:

Input : 254
Output : Odd Parity
Explanation : Binary of 254 is 11111110. 
There are 7 ones. Thus, parity is odd.

Input : 1742346774
Output : Even

Method 1 : (Naive approach)
We have already discussed this method here.

Method 2 : (Efficient)
Pr-requisites : Table look up, X-OR magic

If we break a number S into two parts S1 and S2 such S = S1S2. If we know parity of S1 and S2, we can compute parity of S using below facts :

  1. If S1 and S2 have the same parity, i.e. they both have an even number of bits or an odd number of bits, their union S will have an even number of bits.
  2. Therefore parity of S is XOR of parities of S1 and S2

The idea is to create a look up table to store parities of all 8 bit numbers. Then compute parity of whole number by dividing it into 8 bit numbers and using above facts.

Steps:

1. Create a look-up table for 8-bit numbers ( 0 to 255 )
   Parity of 0 is 0.
   Parity of 1 is 1.
   .
   .
   .
   Parity of 255 is 0.
2. Break the number into 8-bit chunks
   while performing XOR operations.
3. Check for the result in the table for
    the 8-bit number.

Since a 32 bit or 64 bit number contains constant number of bytes, the above steps take O(1) time.

Example :

1. Take 32-bit number : 1742346774

2. Calculate Binary of the number : 
   01100111110110100001101000010110

3. Split the 32-bit binary representation into 
  16-bit chunks :
0110011111011010 | 0001101000010110 

4. Compute X-OR :
  0110011111011010
^ 0001101000010110
___________________
= 0111110111001100

5. Split the 16-bit binary representation 
   into 8-bit chunks : 01111101 | 11001100

6. Again, Compute X-OR :
  01111101
^ 11001100
___________________
= 10110001
10110001 is 177 in decimal. Check
 for its parity in look-up table :
Even number of 1 = Even parity.

Thus, Parity of 1742346774 is even.

Below is the implementation that works for both 32 bit and 64 bit numbers.

C++

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// CPP program to illustrate Compute the parity of a
// number using XOR
#include <bits/stdc++.h>
  
// Generating the look-up table while pre-processing
#define P2(n) n, n ^ 1, n ^ 1, n
#define P4(n) P2(n), P2(n ^ 1), P2(n ^ 1), P2(n)
#define P6(n) P4(n), P4(n ^ 1), P4(n ^ 1), P4(n)
#define LOOK_UP P6(0), P6(1), P6(1), P6(0)
  
// LOOK_UP is the macro expansion to generate the table
unsigned int table[256] = { LOOK_UP };
  
// Function to find the parity
int Parity(int num)
{
    // Number is considered to be of 32 bits
    int max = 16;
  
    // Dividing the number into 8-bit
    // chunks while performing X-OR
    while (max >= 8) {
        num = num ^ (num >> max);
        max = max / 2;
    }
  
    // Masking the number with 0xff (11111111)
    // to produce valid 8-bit result
    return table[num & 0xff];
}
  
// Driver code
int main()
{
    unsigned int num = 1742346774;
  
    // Result is 1 for odd parity, 0 for even parity
    bool result = Parity(num);
  
    // Printing the desired result
    result ? std::cout << "Odd Parity" :
             std::cout << "Even Parity";
  
    return 0;
}

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Python3

# Python3 program to illustrate Compute the
# parity of a number using XOR

# Generating the look-up table while
# pre-processing
def P2(n, table):
table.extend([n, n ^ 1, n ^ 1, n])
def P4(n, table):
return (P2(n, table), P2(n ^ 1, table),
P2(n ^ 1, table), P2(n, table))
def P6(n, table):
return (P4(n, table), P4(n ^ 1, table),
P4(n ^ 1, table), P4(n, table))
def LOOK_UP(table):
return (P6(0, table), P6(1, table),
P6(1, table), P6(0, table))

# LOOK_UP is the macro expansion to
# generate the table
table = [0] * 256
LOOK_UP(table)

# Function to find the parity
def Parity(num) :

# Number is considered to be
# of 32 bits
max = 16

# Dividing the number o 8-bit
# chunks while performing X-OR
while (max >= 8):
num = num ^ (num >> max)
max = max // 2

# Masking the number with 0xff (11111111)
# to produce valid 8-bit result
return table[num & 0xff]

# Driver code
if __name__ ==”__main__”:
num = 1742346774

# Result is 1 for odd parity,
# 0 for even parity
result = Parity(num)
print(“Odd Parity”) if result else print(“Even Parity”)

# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

PHP

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<?php
// PHP program to illustrate
// Compute the parity of a
// number using XOR
  
/* Generating the look-up 
table while pre-processing
#define P2(n) n, n ^ 1, n ^ 1, n
#define P4(n) P2(n), P2(n ^ 1), 
              P2(n ^ 1), P2(n)
#define P6(n) P4(n), P4(n ^ 1), 
              P4(n ^ 1), P4(n)
#define LOOK_UP P6(0), P6(1), 
                P6(1), P6(0)
  
LOOK_UP is the macro expansion
to generate the table
$table = array(LOOK_UP );
*/
  
// Function to find
// the parity
function Parity($num)
{
    global $table;
      
    // Number is considered 
    // to be of 32 bits
    $max = 16;
  
    // Dividing the number 
    // into 8-bit chunks 
    // while performing X-OR
    while ($max >= 8) 
    {
        $num = $num ^ ($num >> $max);
        $max = (int)$max / 2;
    }
  
    // Masking the number with 
    // 0xff (11111111) to produce
    // valid 8-bit result
    return $table[$num & 0xff];
}
  
// Driver code
$num = 1742346774;
  
// Result is 1 for odd 
// parity, 0 for even parity
$result = Parity($num);
  
// Printing the desired result
if($result == true)
        echo "Odd Parity" ;
    else
        echo"Even Parity";
  
// This code is contributed by ajit
?>

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Output:

Even Parity

Time Complexity : O(1). Note that a 32 bit or 64 bit number has fixed number of bytes (4 in case of 32 bits and 8 in case of 64 bits).

This article is contributed by Rohit Thapliyal. 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 write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



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Improved By : jit_t, SHUBHAMSINGH10