[EM] Fw: IBCM, Tideman, Schulze

[MIKE OSSIPOFF]

You'd written:

> > > But the SSD definition uses Schwartz sets. And the
> > > definition of Schwartz sets uses beat paths.
> >
> > The Schwartz set can be defined in terms of innermost
> > unbeaten sets. The Schwartz set is the set of candidates
> > who are in innermost unbeaten sets. But my SSD definition
> > yesterday avoided mentioning the Schwartz set, and only
> > spoke of innermost unbeaten sets.
>
>You wrote (23 July 2000):
> > That isn't simple. This is simple:
> >
> > 1. An unbeaten set is a set of candidates none of whom are
> >    beaten by anyone outside that set.
> > 2. An innermost unbeaten set is an unbeaten set that doesn't
> >    contain a smaller unbeaten set.
> > 3. Drop the weakest defeat that is among an innermost unbeaten
> >    set. Repeat till there's an unbeaten candidate.
>
>It is dangerous to believe that you can make an election method
>simpler simply by using the term "innermost unbeaten set" instead
>of "Schwartz set" and simply by avoiding explaining properly how
>this set is actually calculated.

I do claim that it's simpler and more natural for people to
speak of an innermost unbeaten set. I defined SSD without
naming the Schwartz set, because it seems simpler to just use
innermost unbeaten sets directly, rather than using them to
define the Schwartz set and then using the Schwartz set. I'm
sure that's simpler.

As for explaining how it's calculated, I'm not asking the
man-in-the-street, or the initiative committee, or the
legislator to calculate the SSD winner. I'm merely asking them
to agree that it makes sense to do as SSD's definition says.

I don't think the calculation details need be explained in
the method definition. It's enough that the definition is clear,
natural, and unambiguous.

>
>***
>
>In so far as SSD and Schulze differ only when there are pairwise
>ties, it is necessary to consider examples with pairwise ties to
>decide which method is better. But when there are pairwise ties
>then SSD violates independence from clones.

I don't deny that Schulze is better than SSD for small committee
elections. It's been said that Schulze is more decisive, in
addition to doing better by the clone criterion. However none
of that matters in public elections, where there won't be pair-ties
or equal defeats, and where both methods will pick the same winner.
If they both pick the same winner under public election conditions,
then why not propose whichever one has more natural & obvious
motivation & justification?



>
>You wrote (23 July 2000):
> > It's _obvious_ that the members of an innermost unbeaten set
> > are uniquely deserving of winning.
>
>This statement demonstrates that Steve erred when he suggested
>that you don't consider the Schwartz set to be important (26 Feb
>2000). When you say that "it's _obvious_ that the members of an
>innermost unbeaten set are uniquely deserving of winning" then
>this includes that you consider the Schwartz criterion to be
>_very_ important.

Ok, maybe I worded that too strongly. It is obvious that they're
uniquely deserving of winning, when one hears the definition
of SSD. When one is looking at the election from the perspective
of the diagram used in defining SSD, sure it's obvious that
the members of an unbeaten set are uniquely deserving of winning.

But if someone shows that Tideman can choose outside the Schwartz
set, then that wouldn't convince me that Tideman isn't as good
Schulze, or that Tideman isn't as good as SSD in public elections.
Maybe that would count for Schulze, and against Tideman. That
I admit. But it wouldn't necessarily be decisive. There are
other considerations, and each method is favored by different
considerations. If Tideman can choose outside the Schwartz set,
that would tend to make Tideman & Schulze look more nearly equal
to me.

Tideman gives an impression of special soundness and stability
when it solves every cycle. Are Schulze & SSD missing important
relationships among the defeats when they don't do that?

Mike Ossipoff



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