[EM] Taking advantage of cycle proof conditions

[fsimmons at pcc.edu]

----- 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