Faster Recursive Fibonacci Numbers

[rusi]

On May 18, 2:04 am, Wolfram Hinderer <wolfram.hinde... at googlemail.com>
wrote:
> On 17 Mai, 20:56, geremy condra <debat... at gmail.com> wrote:
>
>
>
> > On Tue, May 17, 2011 at 10:19 AM, Jussi Piitulainen
>
> > <jpiit... at ling.helsinki.fi> wrote:
> > > geremy condra writes:
>
> > >> or O(1):
>
> > >> ö = (1 + sqrt(5)) / 2
> > >> def fib(n):
> > >>     numerator = (ö**n) - (1 - ö)**n
> > >>     denominator = sqrt(5)
> > >>     return round(numerator/denominator)
>
> > >> Testing indicates that it's faster somewhere around 7 or so.
>
> > > And increasingly inaccurate from 71 on.
>
> > Yup. That's floating point for you. For larger values you could just
> > add a linear search at the bottom using the 5f**2 +/- 4 rule, which
> > would still be quite fast out to about 10 times that. The decimal
> > module gets you a tiny bit further, and after that it's time to just
> > use Dijkstra's, like rusi suggested. In any event, I still think this
> > formulation is the most fun ;).
>
> I think you can write it even more funny
>
> def fib(n):
>     return round(((.5 + .5 * 5 ** .5) ** n -  (.5 - .5 * 5 ** .5) **
> n) * 5 ** -.5)
>
> ;-)

VOW!
Tour de Force - Thanks
[I am going to trouble some class of students with that :-) ]