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
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.
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 :
- 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.
- 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.
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.
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.
# Python3 program to illustrate Compute the
# parity of a number using XOR
# Generating the look-up table while
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))
return (P6(0, table), P6(1, table),
P6(1, table), P6(0, table))
# LOOK_UP is the macro expansion to
# generate the table
table =  * 256
# 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)
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 email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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