Richard D. Moores, 09.11.2010 12:07:
That sqrt(n) works for speeding up the finding of primes, but here we
want to use int(n/2) (and why didn't I think of that?), which results
in about a 2x speedup. See<http://tutoree7.pastebin.com/dyRC8vuX>.
NO! Use int(n/2)+1 . I'll correct that in
<http://tutoree7.pastebin.com/dyRC8vuX> and report back.
See<http://tutoree7.pastebin.com/eZPaVpwG>
But now I'll have to redo again using Stefan's ingenious suggestion.
Here's what I wrote using Stefan's suggestion:
def proper_divisors_sum(n):
pd_list = []
for x in range(1, int(n**.5)+1):
if n % x == 0:
factor = int(x)
factor2 = int(n/factor)
pd_list.extend([factor, factor2])
pd_list = list(set(pd_list))
pd_list.sort()
pd_list = pd_list[:-1]
return sum(pd_list)
Have a look at divmod():
http://docs.python.org/library/functions.html#divmod
Note that you don't need to convert 'x' to an int. It already has that type
(see the docs on range()). Also, int(n/factor) is the same as n//factor here.
Stefan
_______________________________________________
Tutor maillist - Tutor@python.org
To unsubscribe or change subscription options:
http://mail.python.org/mailman/listinfo/tutor
