Given an integer N. The task is to write a program to find the parity of the given number.
Note: Parity of a number is used to define if the total number of set-bits(1-bit in binary representation) in a number is even or odd. If the total number of set-bits in the binary representation of a number is even then the number is said to have even parity, otherwise, it will have odd parity.
Examples:
Input : N = 13
Output : Odd Parity
Explanation:
Binary representation of 13 is (1101)
Input : N = 9 (1001)
Output : Even Parity
Method 1
The parity of a number represented by 32-bits can be efficiently calculated by performing the following operations.
Let the given number be x, then perform the below operations:
- y = x^(x>>1)
- y = y^(y>>2)
- y = y^(y>>4)
- y = y^(y>>8)
- y = y^(y>>16)
Now, the rightmost bit in y will represent the parity of x. If the rightmost bit is 1, then x will have odd parity and if it is 0 then x will have even parity.
So, in order to extract the last bit of y, perform bit-wise AND operation of y with 1.
Why does this work?
Consider that we want to find the parity of n = 150 = 1001 0110 (in binary).
1. Let’s divide this number into two parts and xor them and assign it to n: n = n ^ (n >> 4) = 1001 ^ 0110 = 1111.
Dissimilar bits result in a 1 bit in the result while similar bits result in a 0. We have basically considered all 8 bits to arrive at this intermediate result. So, effectively we have nullified even parities within the number.
Now repeat step 1 again until you end up with a single bit.
n = n ^ (n >> 2) = 11 ^ 11 = 00
n = n ^ (n >> 1) = 0 ^ 0 = 0
Final result = n & 1 = 0
Another example:
n = 1000 0101
n = n ^ (n >> 4) = 1000 ^ 0101 = 1101
n = n ^ (n >> 2) = 11 ^ 01 = 10
n = n ^ (n >> 1) = 1 ^ 0 = 1
Final result = n & 1 = 1
if(y&1==1)
odd Parity
else
even ParityBelow is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool findParity(int x)
{
int y = x ^ (x >> 1);
y = y ^ (y >> 2);
y = y ^ (y >> 4);
y = y ^ (y >> 8);
y = y ^ (y >> 16);
if (y & 1)
return 1;
return 0;
}
int main()
{
(findParity(9)==0)?cout<<"Even Parity\n":
cout<<"Odd Parity\n";
(findParity(13)==0)?cout<<"Even Parity\n":
cout<<"Odd Parity\n";
return 0;
}
|
Java
import java.io.*;
class GFG
{
static boolean findParity(int x)
{
int y = x ^ (x >> 1);
y = y ^ (y >> 2);
y = y ^ (y >> 4);
y = y ^ (y >> 8);
y = y ^ (y >> 16);
if ((y & 1) > 0)
return true;
return false;
}
public static void main (String[] args)
{
if((findParity(9) == false))
System.out.println("Even Parity");
else
System.out.println("Odd Parity");
if(findParity(13) == false)
System.out.println("Even Parity");
else
System.out.println("Odd Parity");
}
}
|
Python3
def findParity(x):
y = x ^ (x >> 1);
y = y ^ (y >> 2);
y = y ^ (y >> 4);
y = y ^ (y >> 8);
y = y ^ (y >> 16);
if (y & 1):
return 1;
return 0;
if(findParity(9) == 0):
print("Even Parity");
else:
print("Odd Parity\n");
if(findParity(13) == 0):
print("Even Parity");
else:
print("Odd Parity");
|
C#
using System;
class GFG
{
static bool findParity(int x)
{
int y = x ^ (x >> 1);
y = y ^ (y >> 2);
y = y ^ (y >> 4);
y = y ^ (y >> 8);
y = y ^ (y >> 16);
if ((y & 1) > 0)
return true;
return false;
}
public static void Main ()
{
if((findParity(9) == false))
Console.WriteLine("Even Parity");
else
Console.WriteLine("Odd Parity");
if(findParity(13) == false)
Console.WriteLine("Even Parity");
else
Console.WriteLine("Odd Parity");
}
}
|
PHP
<?php
function findParity($x)
{
$y = $x ^ ($x >> 1);
$y = $y ^ ($y >> 2);
$y = $y ^ ($y >> 4);
$y = $y ^ ($y >> 8);
$y = $y ^ ($y >> 16);
if ($y & 1)
return 1;
return 0;
}
(findParity(9) == 0) ?
print("Even Parity\n"):
print("Odd Parity\n");
(findParity(13) == 0) ?
print("Even Parity\n"):
print("Odd Parity\n");
?>
|
Javascript
<script>
function findParity(x) {
let y = x ^ (x >> 1);
y = y ^ (y >> 2);
y = y ^ (y >> 4);
y = y ^ (y >> 8);
y = y ^ (y >> 16);
if (y & 1)
return 1;
return 0;
}
(findParity(9) == 0) ? document.write("Even Parity<br>") :
document.write("Odd Parity<br>");
(findParity(13) == 0) ? document.write("Even Parity<br>") :
document.write("Odd Parity<br>");
</script>
|
OutputEven Parity
Odd Parity
Time Complexity: O(1)
Auxiliary Space: O(1)
Method 2
We know that a number with odd parity has an odd number of set bits (1 in binary) in its binary representation, and an even parity number has an even number of 1’s in its binary representation.
Hence we can simply count the number of 1’s by the following code:
C++
#include <bits/stdc++.h>
using namespace std;
int main()
{
int number = 7;
bool oddParity = false;
while(number)
{
oddParity = !oddParity;
number &= (number-1);
}
cout<<oddParity<<endl;
return 0;
}
|
Java
import java.util.*;
public class Main {
public static void main(String[] args) {
int number = 7;
boolean oddParity = false;
while (number != 0) {
oddParity = !oddParity;
number &= (number - 1);
}
if(oddParity) System.out.println(1);
else System.out.println(0);
}
}
|
Python3
number = 7
oddParity = False
while number:
oddParity = not oddParity
number &= (number-1)
if oddParity:
print(1)
else:
print(0)
|
Javascript
let number = 7;
let oddParity = false;
while (number) {
oddParity = !oddParity;
number &= (number-1);
}
if (oddParity) {
console.log(1);
} else {
console.log(0);
}
|
C#
using System;
class MainClass {
static void Main()
{
int number = 7;
bool oddParity = false;
while (number != 0) {
oddParity = !oddParity;
number &= (number - 1);
}
if (oddParity) {
Console.WriteLine(1);
}
else {
Console.WriteLine(0);
}
}
}
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Note: The number of times the above while loop runs is equal to the number of set bits in the number.
References:
1. Bit Twiddling Hacks from a Stanford University professor