From: Christian Neukirchen <chneukirchen@...>
Date: 2005-12-15T05:27:58+09:00
Subject: Re: ruby beats them all

Edward Faulkner <ef@alum.mit.edu> writes:

> On Thu, Dec 15, 2005 at 04:27:08AM +0900, Christian Neukirchen wrote:
>> def fib(n)
>>   (1..n-2).inject([1, 1]) { |(a, b), n| [b, a+b] }.last
>> end
>
> If we're talking about elegance, I prefer this one.  It reads just
> like the definition of the sequence:
>
> def fib(n)
>     n > 1 ? fib(n-1) + fib(n-2) : n
> end

Yes.  But Ruby is not tail-recursive (neither is your method), and
this solution wont work for bigger values.

> You can even make it run efficiently by memoizing it.  :-)

If it was about efficiency, I'd calculate them with the golden mean...

> -Ed
-- 
Christian Neukirchen  <chneukirchen@gmail.com>  http://chneukirchen.org