python中任意位整数的奇偶校验

Question

Lot of solutions suggest using xor with right shift, as described here https://www.geeksforgeeks.org/finding-the-parity-of-a-number-efficiently/

def findParity(x):
    x = x ^ (x >> 16); 
    x = x ^ (x >> 8); 
    x = x ^ (x >> 4);
    x = x ^ (x >> 2); 
    x = x ^ (x >> 1); 
    return x & 1; 

But they assume 32 or 64 bit or some 2^n bits integer. In python integer could have any number of bits. For instance i = 7, has only 3 bits.

i = 7
print(len(bin(i)) - 2)

Any suggestions on how to calculate parity using xor and right shift for arbitrary number of bits ?

Answer 1

You can use a loop to dynamically change the length of the parity check:

def parity(num):
    length = math.ceil(math.log2(math.ceil(math.log2(num)+1)))
    for i in range(length-1, -1, -1):
        print(2**i)
        num^=(num >> (2**i))
    return num&1

You will need to use log twice because you first have to find the length of the number then you need log that many operations.