Alexander Belopolsky <belopol...@users.sourceforge.net> added the comment:
On Thu, May 13, 2010 at 5:31 PM, Mark Dickinson <rep...@bugs.python.org> wrote:
> And why are we trying to speed up the pure Python factorial code here?
I would expect that for large factorials the performance will be
determined by the number of long multiplications and the size of
multiplicands.
>Â I don't imagine that those speed differences are going to translate well to
>C.
The differences between recursive and non-recursive versions are not
likely to translate well, but the difference (if any) between the
order of multiplication most likely will.
In any case, I am attaching fixed version of factorial4.
$ ./python.exe -m timeit -s "from factorial4 import f0 as f" "f(10000)"
10 loops, best of 3: 65.5 msec per loop
$ ./python.exe -m timeit -s "from factorial4 import f1 as f" "f(10000)"
10 loops, best of 3: 66.9 msec per loop
$ ./python.exe -m timeit -s "from factorial4 import f2 as f" "f(10000)"
10 loops, best of 3: 56.5 msec per loop
$ ./python.exe -m timeit -s "from factorial4 import f3 as f" "f(10000)"
10 loops, best of 3: 63 msec per loop
----------
Added file: http://bugs.python.org/file17320/factorial4.py
_______________________________________
Python tracker <rep...@bugs.python.org>
<http://bugs.python.org/issue8692>
_______________________________________
import functools
import operator
product = functools.partial(functools.reduce, operator.mul)
def partial_product(j, i):
if i == j:
return j << 1 | 1
if i == j + 1:
return (j << 1 | 1) * (i << 1 | 1)
l = i + j >> 1
return partial_product(j, l) * partial_product(l + 1, i)
def naive_factorial(n):
"""Naive implementation of factorial: product([1, ..., n])
>>> naive_factorial(4)
24
"""
return product(range(1, n+1), 1)
def factorial(n, pp=partial_product):
"""Implementation of Binary-Split Factorial algorithm
See http://www.luschny.de/math/factorial/binarysplitfact.html
>>> for n in range(20):
... assert(factorial(n) == naive_factorial(n))
>>> import math
>>> assert(factorial(100) == math.factorial(100))
"""
_, r = loop(n, pp)
return r << (n - count_bits(n))
def loop(n, pp):
p = r = 1
for i in range(n.bit_length() - 2, -1, -1):
m = n >> i
if m > 2:
p *= pp(((m >> 1) + 1) >> 1, (m - 1) >> 1)
r *= p
return p, r
def partial_product1(j, i):
return product((l << 1 | 1 for l in range(j, i + 1)), 1)
def partial_product2(j, i):
a = [l << 1 | 1 for l in range(j, i + 1)]
n = i - j + 1
p = 1
while n > 1:
for k in range(n>>1):
a[k] *= a[n-k-1]
n = (n>>1) + (n&1)
return a[0]
def partial_product3(j, i):
a = [l << 1 | 1 for l in range(j, i + 1)]
while 1:
n = len(a)
if n == 1:
return a[0]
a = [a[k<<1] * a[k<<1|1] for k in range(n>>1)] + a[(n >> 1)<<1:]
def partial_product4(j, i):
if i == j:
return j << 1 | 1
if i == j + 1:
return (j << 1 | 1) * (i << 1 | 1)
return (j << 1 | 1) * (i << 1 | 1) * partial_product4(j + 1, i - 1)
def count_bits(n):
count = 0
while n:
n &= n - 1
count += 1
return count
def f0(n):
return factorial(n)
def f1(n):
return factorial(n, partial_product1)
def f2(n):
return factorial(n, partial_product2)
def f3(n):
return factorial(n, partial_product3)
if __name__ == '__main__':
for n in list(range(12, 20)) + [100, 101, 200]:
assert(f0(n) == f1(n) == f2(n) == f3(n))
import doctest
doctest.testmod()
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