Parity of integer with arbitrary bits in python

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 ?

Did you not try a loop testing bit 0 and shifting one bit right each iteration until the value is 0? Very easy to extend to doing eight bits at a time looking the lowest 8 bits up in a lookup table giving parity for that value. — DisappointedByUnaccountableMod, Sep 15 '20 at 15:04
Yeah that solution works, but it is not nlog(n) time solution. It takes O(n) steps. Look-up table is relatively complex solution. — developer developer, Sep 15 '20 at 15:08
What? Any other requirements not written into your question? Please share the best code you’ve written as a [mre]. — DisappointedByUnaccountableMod, Sep 15 '20 at 15:11

mousetail · Accepted Answer · 2020-09-15T15:19:49.670

0

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.

edited Sep 15 '20 at 15:19

answered Sep 15 '20 at 15:09

mousetail

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perfect, i was exploring in the same direction . I had overlooked the fact that log2 should be on the length of the bits, not on the value of integer. – developer developer Sep 15 '20 at 15:19

Parity of integer with arbitrary bits in python

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