2008/1/1, Steve Eppley <[EMAIL PROTECTED]>: > > 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