From: William Morgan <wmorgan-ruby-talk@...>
Date: 2005-04-29T06:59:30+09:00
Subject: Re: TIMTOWTDI letter permutation

Excerpts from Trans's mail of 28 Apr 2005 (EDT):
> Don't leave us hang'n! What's this groovy i.th solution?

The traditional solution breaks i into a sum of multiples of the
factorials of 1 to n-1 (the "pattern"), then successively pulls out
those multiples as indexes into the string.

Kinda hard to explain, but here's how you do it:

def permutation(s, i)
  s = s.dup
  pat = (1 .. s.length).map do |j|
    r = i % j
    i /= j
    r
  end 
  pat.reverse.map { |j| s.slice!(j).chr }.join
end

(Note that the first element of the pattern is always 0.)

-- 
William <wmorgan-ruby-talk@masanjin.net>