On Tue, May 17, 2011 at 2:04 PM, Wolfram Hinderer <wolfram.hinde...@googlemail.com> wrote: > On 17 Mai, 20:56, geremy condra <debat...@gmail.com> wrote: >> On Tue, May 17, 2011 at 10:19 AM, Jussi Piitulainen >> >> <jpiit...@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) > > ;-)
Ok, that's amusing. It does hide the interaction with the golden ratio though, which is what I find so fascinating about the earlier one. Geremy Condra -- http://mail.python.org/mailman/listinfo/python-list
