From: Regis d'Aubarede <lists@...>
Date: 2013-04-15T04:57:33+09:00
Subject: Re: Looking for more elegant solutions.

Hello,

Maybe this one is a little more elegant :


def f( accu,already, n, times )
  if n > times
    accu << already.dup
  else
    ((1..times).to_a - already-[n]).each do |i|
      f( accu, already+[i], n+1, times )
    end
    accu
  end
end
p f([],[],1,4)

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