>Therefore candidate C wins all pairwise defeats via beat paths.
Yes, I checked, and Schulze, SSD, & SD choose as you described in
that example. I'd missed a few possibilities when I'd concluded that
, under many-voter conditions, SD & SSD choose the same, and that
SD chooses the same as Schulze.
>Suppose that candidate D is substituted with a set of clones
>with D1 > D2 > D3 > D1. Suppose that all the pairwise defeats
>between two clones are larger than 65:35. Then the SD winner of
>my yesterday's example is changed from candidate D to candidate C.
>Therefore SD violates independence from clones.
Sure, but isn't every clone example an example with some identical
entries in the pairwise defeat table?
>
>******
>
>You wrote (27 Jun 2000):
> > But of course till it's demonstrated that SD doesn't have a
> > many-voters problem, I'll only propose Tideman instead.
>
>Does that mean that you don't promote SSD any more? What is
>the reason for that change in your opinion?
I'm concerned because I haven't heard any definite statements about
which BC complying methods are always monotonic, and which ones
are monotonic as long as there aren't identical entries in the pairwise
defeats table. I assume that Tideman(m) is probably always monotonic,
or else Tideman wouldn't like it. But it doesn't meet BC. But
if Tideman(m) is always monotonic, that suggests hope that
Tideman(wv) is too--though it doesn't prove it.
Does anyone know which of the following methods are always monotonic?
And which are monotonic as long as there aren't any identical entries
in the pairwise defeat table?:
Tideman(wv), SSD, SD, Schulze's method.
Mike Ossipoff
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