[EM] Definite Approval/Disapproval

[Forest Simmons]

El sáb., 7 de may. de 2022 2:41 a. m., Kristofer Munsterhjelm <
km_elmet at t-online.de> escribió:

> On 07.05.2022 07:20, Forest Simmons wrote:
> > That's a good template for all of us to imitate in our outreach to the
> > proletariat!
>
> "Arise ye workers from your slumbers", eh? :-)
>
> > Let me see how well I can imitate it for my latest attempt in the
> > fpA-gpC vein:
> >
> > Candidates are arranged into
> > one-on-one match-ups like runoffs where the candidate who would win a
> > runoff with only the two wins. If there is a candidate who wins every
> > runoff he's part of, he is elected.
> >
> > Otherwise candidates get reward value for every matchup they win or tie,
> > but lose value for every matchup they lose.
> >
> > The candidate with the greatest net reward value is elected.
> >
> > Specifically, a candidate's prize/reward value is proportional to its
> > estimated formidability as an opponent in these matchups. You get your
> > opponent's prize value when you defeat or tie him in a matchup.
> > Otherwise, you pay out that value.
> >
> > The formidability of a candidate is gauged as a function of its
> > respective first and last place showings on the ballots.
>
> Yeah, it would have to be something like that. For fpA - sum fpC:
>
> A candidate starts out being as strong as his first preference votes,
> but he loses strength by being beaten by other candidates, according to
> *their* first preference votes. The strongest candidate wins.
>
> On an aside, your definition makes me think of an interesting recursive
> primitive for Condorcet-compliant methods. Suppose X's score is f(X) and
> let f(X) be some monotone nondecreasing function of f of all the
> candidates X beat (or the strongest such candidate). Then there exist
> very broad subsets of monotone nondecreasing functions so that the CW
> always wins by f score (e.g. sum or max, possibly any p-norm for p>=1 as
> long as f(X) >= 0 for all X). Similar broad proofs can probably be had
> for the case where f(X) is some monotone nonincreasing function of f of
> the candidates who beat X pairwise. However, cycles would require
> finding equilibrium points, which can be pretty hard to explain.
>
> -km
>

The formidability measure by rights would depend on how well the candidate
was situated vis-à-vis the projected outcome of the election by the method
that depends on the candidate formidabilities ... highly self-referential!

A common DSV dilemma. Recursion would help if the result of removing one
candidate or one ballot could be incorporated profitably into the
formidability calculation.

>
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