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 <[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
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