From: Joe Cheng <code@...>
Date: 2004-10-05T01:54:54+09:00
Subject: Re: [SOLUTION] Secret Santas (#2)

Carlos wrote:
> I see! :) Nothing trivial here.

Your solution becomes viable again if you build up a circular list of 
santa->recipient (each person in the list gives to the next).  That's 
what I did for my solution.  So the possible outputs (depending on what 
sorting tiebreaker is used) of the A-B-C example are:

ABCA
ACBA
BACB
BCAB
CABC
CBAC

I don't know if I would call it "trivial", but there it is.

I have to give Niklas credit for setting maximum randomness as one of 
his goals.  I just went for speed, simplicity, and correctness, which 
was much less challenging.