Alexander Belopolsky <belopol...@users.sourceforge.net> added the comment:
Here is one more datapoint.
$ ./python.exe -m timeit -s "import factorial4 as fm;
fm.partial_product = fm.partial_product; f = fm.factorial " "f(10000)"
10 loops, best of 3: 66.1 msec per loop
[32794 refs]
$ ./python.exe -m timeit -s "import factorial4 as fm;
fm.partial_product = fm.partial_product3; f = fm.factorial "
"f(10000)"
10 loops, best of 3: 63 msec per loop
[32794 refs]
$ ./python.exe -m timeit -s "import factorial4 as fm;
fm.partial_product = fm.partial_product2; f = fm.factorial "
"f(10000)"
10 loops, best of 3: 43.3 msec per loop
partial_product3 multiplies adjacent numbers instead of first by last.
I am not sure it reproduces the order of multiplication in the
recursive version exactly, but it does show that the order of
multiplication matters a lot.
I wonder if one could write an elegant recursive version that would
multiply first by last in partial_product.
----------
Added file: http://bugs.python.org/file17313/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 naive_factorial(n):
"""Naive implementation of factorial: product([1, ..., n])
>>> naive_factorial(4)
24
"""
return product(range(1, n+1), 1)
def factorial(n):
"""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)
return r << (n - count_bits(n))
def loop(n):
p = r = 1
for i in range(n.bit_length() - 2, -1, -1):
m = n >> i
if m > 2:
p *= partial_product(((m >> 1) + 1) >> 1, (m - 1) >> 1)
r *= p
return p, r
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 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:
n >>= 1
for k in range(n):
a[k] *= a[n-k]
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 count_bits(n):
count = 0
while n:
n &= n - 1
count += 1
return count
if __name__ == '__main__':
#print(partial_product(10, 20))
import doctest
doctest.testmod()
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