From: Yoann Guillot <john-rubytalk@...>
Date: 2006-05-10T09:58:58+09:00
Subject: Re: Exponential calcs with very large exponents

On Wed, May 10, 2006 at 05:15:06AM +0900, doug meharry wrote:
> for bit j = k - 2 downto 0
> begin
>   y = y * y mod n   /* square */
>   if e(j) == 1 then
    if e[j] == 1
>     y = y * x mod n  /* multiply */
> end
> return y
> 
> 
> Thanks for looking,
> 
> Doug

Integers in ruby has a [] method that returns the bit at the specified
offset (0 for lsb)

-- 
Yoann