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Date: Wed, 22 Oct 2014 05:43:02 -0500
From: Tim Chase <python.list@tim.thechases.com>
To: Peter Otten <__peter__@web.de>
Subject: Re: (-1)**1000
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Xref: csiph.com comp.lang.python:79430

On 2014-10-22 12:29, Peter Otten wrote:
> That looks like log(a) while a parity check takes constant time:
> 
> $ python3 -m timeit -s 'a = 10**10' 'a & 1'
> 10000000 loops, best of 3: 0.124 usec per loop
> $ python3 -m timeit -s 'a = 10**100' 'a & 1'
> 10000000 loops, best of 3: 0.124 usec per loop
> $ python3 -m timeit -s 'a = 10**1000' 'a & 1'
> 10000000 loops, best of 3: 0.122 usec per loop

Just for the record, this is a one-bit even/odd check (which is
useful & fast in this sign-of-large-exponent case), not a parity check
(which typically counts the number of "1" bits, adds the parity bit,
and asserts the result is even)

-tkc