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Date: Thu, 19 May 2011 17:03:45 -0700
Subject: Re: Faster Recursive Fibonacci Numbers
From: geremy condra <debatem1@gmail.com>
To: "Steven D'Aprano" <steve+comp.lang.python@pearwood.info>
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On Thu, May 19, 2011 at 3:47 PM, Steven D'Aprano
<steve+comp.lang.python@pearwood.info> wrote:
> On Tue, 17 May 2011 10:02:21 -0700, geremy condra wrote:
>
>> or O(1):
>>
>> =F6 =3D (1 + sqrt(5)) / 2
>> def fib(n):
>> =A0 =A0 numerator =3D (=F6**n) - (1 - =F6)**n
>> =A0 =A0 denominator =3D sqrt(5)
>> =A0 =A0 return round(numerator/denominator)
>
> I'd just like to point out that, strictly speaking, it's only O(1) if you
> assume that exponentiation is O(1). Computer scientists often like to
> make this simplifying assumption, and it might even be true for floats,
> but for long ints and any numeric data types with unlimited precision, it
> won't be.

Great point, and something I should have paid attention to. Thanks.

Geremy Condra