From: Christoph <chr_news@...>
Date: 2002-04-06T02:54:34+09:00
Subject: Re: Fibonacci Number Generators


<sinara@blade.nagaokaut.ac.jp> wrote in
....
> Here I show some examples inadequate for newbie's question :-)
>
> 1. The parallel assignment makes an Array object, so it costs
> a little. It may be better to avoid that and use the temporary
> variable which belongs to the outside of the iterator block:
>
>   def fib2(n)
>     return n if n < 2
>     f1, f2 = 0, 1
>     tmp = nil
>     (n-1).times {
>       tmp = f1; f1 = f2; f2 += tmp # swap
>     }
>     f2
>   end
>
> This resembles the original fib(n) as a result.
>
> 2. If you want to calculate fib(n) repeatedly, cashing the
> intermediate results is advantageous:
>
>   def (Fib = [0, 1]).[](n)
>     super || Fib[n] = Fib[n-1] + Fib[n-2]
>   end
>

You can also combine 1 & 2 (this avoids the unnecessary recursive super
call's)

----
Fib = [0,1]

def Fib.[](n)
   super || (
    for i in length..n do
      self[i] = super(i-1)+super(i-2)
    end
    super)
end
----

/Christoph