ast wrote: > > "Chris Angelico" <ros...@gmail.com> a écrit dans le message de > news:mailman.15058.1413968065.18130.python-l...@python.org... >> On Wed, Oct 22, 2014 at 7:27 PM, ast <nom...@invalid.com> wrote: >>> If i am writing (-1)**1000 on a python program, will the >>> interpreter do (-1)*(-1)*...*(-1) or something clever ? >>> >>> In fact i have (-1)**N with N an integer potentially big. >> >> Exponentiation is far more efficient than the naive implementation of >> iterated multiplication. > > In the very particular case of (-1)**N, I belive that Python check > the odd or even parity of N and provides the result accordingly. > > I tried: >>>> (-1)**1000000000000000000000000000000000 > 1 >>>> (-1)**1000000000000000000000000000000001 > -1 > > and it is instantaneous
Not instantaneous once you defeat the peephole optimizer by introducing a variable: $ python3 -m timeit '(-1)**10000000000000000000000000000000001' 10000000 loops, best of 3: 0.0356 usec per loop $ python3 -m timeit -s'a = 10000000000000000000000000000000001' '(-1)**a' 100000 loops, best of 3: 3.23 usec per loop When you increase the exponent you might discern a pattern: $ python3 -m timeit -s 'a = 10**10' '(-1)**a' 1000000 loops, best of 3: 1.42 usec per loop $ python3 -m timeit -s 'a = 10**100' '(-1)**a' 100000 loops, best of 3: 11.6 usec per loop $ python3 -m timeit -s 'a = 10**1000' '(-1)**a' 10000 loops, best of 3: 101 usec per loop $ python3 -m timeit -s 'a = 10**10000' '(-1)**a' 1000 loops, best of 3: 992 usec per loop 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 -- https://mail.python.org/mailman/listinfo/python-list
