----- Original Message ----- From: Chris Benham Date: Tuesday, June 1, 2010 12:14 pm Subject: Tanking advantage of cycle proof conditions To: EM Cc: Forest Simmons
> > Forest Simmons wrote (29 May 2010): > > > Here's a four slot method that takes advantage of the > impossibility of beat > >cycles under certain conditions:. > > > > > >Use range style ballots with four levels: 0, 1, 2, and 3. > > > >(1) First eliminate all candidates that are pairwise defeated > by a ratio greater > >than 3/1. > > > > > >(2) Then eliminate all of the candidates that are pairwise > defeated by a ratio > >greater than 2/1 based on only those comparisons that involve > an extreme rating, > >i.e. 3 beats a 0, 1, or 2, while 1, 2, or 3 beats a 0, but > don't count a 2 as > >beating a 1, since neither 1 not 2 is an extreme rating on our > four slot ballot. > > > > > >(3) Finally, eliminate all of the candidates that are pairwise > defeated by any > >ratio greater than 1/1 on the basis of comparisons that involve > a rating > >difference of at least two, i.e. 3 vs. 0 or 1, and 2 or 3 vs. > 0, while > >considering 3 vs. 2, 2 vs. 1, and 1 vs. 0 to be too weak for > this final > >elimination decision that is based on a mere 1/1 defeat ratio cutoff. > > > >The candidate that remains is the winner. > > > > > > > >If there is a pairwise tie in step three, use the middle two > levels to resolve > >it, which is the same as electing the tied candidate with the > greatest number of > >ratings strictly above one. > > > > > >None of the three elimination steps can eliminate all of the > candidates because > >the elimination conditions are cycle proof. Furthermore, (with > the tie breaker > >in place) the third step will eliminate all of the remaining > candidates except one. > > > >Notice that the ballot comparisons get progressively stronger > as we go from step > >one to step three, while the defeat ratio requirements get > weaker, (from 3/1 to > >2/1 to 1/1) but stay strong enough at each step to prevent cycles. > > > >Isn't that cool? > > > Forest, > > It is certainly elegant and interesting. > > Wouldn't it be possible for step (2) to eliminate a Condorcet > winner? Yes, it could eliminate a "low utility CW." ... > > Does it ( like the Condorcet method Raynaud that also works by > eliminating pairwise losers) fail mono-raise? If it does, it would be for a different reason. If a Raynaud winner gets more support, it can move ahead in the sequence before its strongest challengers have been eliminated. This method doesn't have that problem. However, increasing support for a winner W of this method could help it to eliminate a candidate X so early that X is no longer around to eliminate Y which ends up eliminating W in the last stage. Suppose that before the increased support for W, it was candidate Z that eliminated X, and that later W eliminated Z. Then we have a cycle: W beats Z beats X beats Y beats W, which cannot happen if the eliminations are truly cycle proof. I'm still not sure about the answers to your other questions (FBC and Plurality). Forest ---- Election-Methods mailing list - see http://electorama.com/em for list info
