Precision issue in python

mukesh tiwari mukeshtiwari.iiitm at gmail.com
Sat Feb 20 16:13:07 CET 2010


On Feb 20, 5:44 pm, Mark Dickinson <dicki... at gmail.com> wrote:
> On Feb 20, 11:17 am, mukesh tiwari <mukeshtiwari.ii... at gmail.com>
> wrote:
>
> > Hello everyone. I think it is  related to the precision with double
> > arithmetic so i posted here.I am trying with this problem (https://www.spoj.pl/problems/CALCULAT) and the problem say that "Note : for
> > all test cases whose N>=100, its K<=15." I know precision of doubles
> > in c is 16 digits. Could some one please help me with this precision
> > issue.I used stirling (http://en.wikipedia.org/wiki/
> > Stirling's_approximation) to calculate the first k digits of N.
> > Thank you.
>
> If I understand you correctly, you're trying to compute the first k
> digits in the decimal expansion of N!, with bounds of k <= 15 and 100
> <= N < 10**8.  Is that right?
>
> Python's floats simply don't give you enough precision for this:
> you'd need to find the fractional part of log10(N!) with >= 15 digits
> of precision.  But the integral part needs ~ log10(log10(N!)) digits,
> so you end up needing to compute log10(N!) with at least 15 +
> log10(log10(N!)) ~ 15 + log10(N) + log10(log10(N)) digits (plus a few
> extra digits for safety);  so that's at least 25 digits needed for N
> close to 10**8.
>
> The decimal module would get you the results you need (are you allowed
> imports?).  Or you could roll your own log implementation based on
> integer arithmetic.
>
> --
> Mark

Yes i am trying to first k digits of N!.I will try your method.
Thank you



More information about the Python-list mailing list