[EMAIL PROTECTED]
> If I un-comment any line in this program below the line where I
> commented " all OK up to this point " This program locks up my
> computer.
>
> Windows task manager will show "Not Responding" for Python in the
> Applications tab and in the Performance tabe the CPU usage will be
> locked at %100.
>
> I've experienced the same problem on 2 different computers, one running
> 2000 pro. the other running XP home eddition. both computers run
> Python 2.4.2
>
> I'm just wondering if any one else has noticed any problems with
> working with large numbers in Python ind if there is anything that can
> work around this issue.
>
> Thankd for reading
> David
>
> def factor(n):
> d = 2
> factors = [ ]
> while n > 1:
> if n % d == 0:
> factors.append(d)
> n = n/d
> else:
> d = d + 1
> print factors
You primary problem is that this is a horridly inefficient way to
factor, taking time propotional to n's largest prime divisor (which
may be n).
> factor (12)
> factor (123)
> factor (1234)
> factor (12345)
> factor (123456)
> factor (1234567)
> factor (12345678)
> factor (123456789)
> factor (1234567898)
> factor (12345678987)
> factor (123456789876)
> factor (1234567898765) # all OK up to this point
> #factor (12345678987654) # locks up computer if I run this line
It doesn't lock up for me, using Python 2.3.5 or 2.4.2 on Windows (XP
Pro SP2, but the specific flavor of Windows shouldn't matter). I ran
it from a DOS box, and while it was plugging away on 12345678987654,
hitting Ctrl+C stopped it.
If you let it continue running, and you & your computer were immortal
(something worth shooting for :-)), it would eventually print the
factorization. Since
12345678987654 = 2 * 3 * 2057613164609
the loop would have to go around over 2 trillion times to find the
final 2057613164609 prime factor. A simple enormous improvement is to
get out of the loop when d*d > n. Then n must be prime or 1. That
would slash the worst-care runtime from being proportional to n to
being proportional to sqrt(n).
> ...
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