[Election-Methods] Smith + mono-add-top?

[Juho]

Yes, this approach nicely follows the original idea of the method.  
Just make small random differences if there are none.

Juho


On Jan 1, 2008, at 19:43 , Diego Santos wrote:

> 2008/1/1, Steve Eppley <SEppley at alumni.caltech.edu>:
> I think the method Diego Santos is considering can elect outside the
> Smith set (a.k.a. top cycle), depending on the tie-breaker.  Here's an
> example with 21 voters and 4 candidates:
>
>     4    4    4    3    3    3
>    ---  ---  ---  ---  ---  ---
>     A    B    C    D    D    D
>     B    C    A    A    B    C
>     C    A    B    B    C    A
>     D    D    D    C    A    B
>
> {A,B,C} is a set of clones in a "vicious" cycle. (By vicious, I  
> mean all
> margins in the cycle are large.  I think Mike Ossipoff may have been
> first to use the term, many years ago.)  What makes this scenario very
> rare (assuming many voters) is that the margins in the vicious  
> cycle are
> equal:
>
>    A over B by (4+4+3+3) - (4+3) = 7
>    B over C by (4+4+3+3) - (4+3) = 7
>    C over A by (4+4+3+3) - (4+3) = 7
>
> The Smith set is {A,B,C}.  Can D win?  If I understand Diego's
> definition, D is not eliminated since the margin in D's pairwise  
> defeats
> is smallest (12 - 9 = 3).  I think A and B and C are also not  
> eliminated
> since there's a tie in their cycle's margins.  Thus the set of
> non-eliminated candidates is {A,B,C,D}.  Among {A,B,C,D} there is no
> Condorcet winner.  So, a tiebreaker must select from {A,B,C,D}.  If  
> the
> tiebreaker can select outside the Smith set, D can be elected.   
> Typical
> tiebreakers (Random, Random Voter's Ballot, Chairperson's Choice) can
> select outside the Smith set.
>
>
> A possible tiebraker can be: "if no Condorcet Winner exists among  
> non-eliminated candidates, reuse this method with one of equal  
> margins 'pseudo-augmented'" selected at random.
>
> In Steve's example, we can select, for instance, B win over C as  
> "pseudo-augmented" (marked with an asterisk):
>
> A(7): C(7,7*)
> B(7): A(7,7)
> C(7*): B(7*,7)                    eliminated
> D(3): A(3,7), B(3,7), C(3,7*)
>
> Then a member of the "vicious circle" is disqualified.
>
> New set: {A, B, D}, and A wins. But uses of this tiebraker would be  
> too rare.
> ________________________________
> Diego Renato dos Santos
> ----
> Election-Methods mailing list - see http://electorama.com/em for  
> list info

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