[EM] Tanking advantage of cycle proof conditions

[Chris Benham]

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?  If the method fails the Condorcet criterion,
does it meet Favourite Betrayal?

Does it ( like the Condorcet method Raynaud that also works by eliminating pairwise losers) fail mono-raise?

http://wiki.electorama.com/wiki/Raynaud

I can't make an example of it failing the Plurality criterion.   Does it meet that criterion?

Chris Benham