[EM] methods based on cycle proof conditions

[Jameson Quinn]

How about "Condorcet-compliant Range-2 with Runoffs": use 3 levels, 0
through 2 (these numbers matter). Find the largest number of approvals for
any candidate. Any candidate with a range score higher than that number
makes it into the runoff. This cannot eliminate the CW (or indeed, any
member of the Smith set. To extend this property to more than two levels,
you have to implicityly collapse to 2 levels first.) Then use a
condorcet-compliant method for the runoff.

JQ

2010/6/1 <fsimmons at pcc.edu>

> A couple of other possibilities for methods based on cycle proof
> conditions:
>
> I.  BDR or "Bucklin Done Right:"
>
> Use 4 levels, say, zero through three.  First eliminate all candidates
> defeated
> pairwise with a defeat ratio of 3 to 1.  Then collapse the top two levels,
> and
> eliminate all candidates that suffer a defeat ratio of 2 to 1.  If any
> candidates are left, among these elect the one with the greatest number of
> positive ratings.
>
> II.  SSCPE or "Six Slot Cycle Proof Elimination"
>
> Use six levels, zero through five.  First eliminate all candidates with a
> pairwise defeat ratio of five to one.  Then allowing only those ballot
> comparisons with strength 2 or greater (i.e. the preferred candidate is
> rated at
> least two levels above the other), eliminate all candidates with a defeat
> ratio
> of two to one.  Then allowing only comparisons with strength three or
> greater,
> eliminate all candidates beaten with a defeat ratio of one to one, i.e. all
> defeated candidates.  If there are two or more undefeated candidates, elect
> the
> one with the greatest number of positive ratings.
>
> III.  SPE or Strong Preference Elimination:
>
> Use 2n levels.  First eliminate all of the candidates that are defeated
> when the
> only ballot preferences counted are of strength n or greater, i.e. the
> rating of
> the preferred one is at least n levels greater than the rating of the
> other.  If
> there are two or more unbeaten candidates, collapse the bottom three levels
> to
> zero, decrement the other levels by two, and decrement 2n to 2(n-1) and
> repeat
> the process recursively.
>
> The idea of SPE is that the most important eliminations are done by strong
> preferences, and weaker preferences are invoked only to break ties.  More
> than
> one tie breaker step is needed only to ensure the technical compliance with
> Pareto.  Random ballot could be used as a tie breaker just as well where
> ever
> voters are not allergic to it.
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