nth root

Mark Dickinson dickinsm at gmail.com
Sat Jan 31 13:24:56 CET 2009


On Jan 31, 5:43 am, "Tim Roberts" <t.robe... at cqu.edu.au> wrote:
> Dan,
>
> Thanks - you're probably right - just my intuition said to me that rather than calculating that the 13th root of 4021503534212915433093809093996098953996019232
> is 3221.2904208350265....
> there must be a quicker way of finding out its between 3221 and 3222....
>
> ....but perhaps not.

I don't think you'll find anything much quicker than n**(1./13)
(though I hope
that if you're doing this millions of time then you're precomputing
the 1./13
rather than redoing the division every single time.

What happens behind the scenes here is that your integer is
immediately
converted to a float, then the system math library is used for the
power operation.  The integer -> float conversion is probably quite
significant, timewise.

I'd also be a bit worried about accuracy.   Is it important to you
that the
integer part of the result is *exactly* right, or is it okay if
(n**13)**(1./13) sometimes comes out as slightly less than n, or if
(n**13-1)**(1./13) sometimes comes out as n?

Mark



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