[EM] The critical importance of Precinct Summability.

[Closed Limelike Curves]

I suspect the uncovered set might be slightly better because it's a close
approximation of the bipartisan set that isn't too hard to explain. Maximal
lotteries also have some very nice strategy-resistance properties.

On this topic and the lack of focus on proportional representation
mentioned elsewhere, I think it would be super useful to have some kind of
strongly-summable PR algorithm. ElectoWiki claims Ebert's method is summable
<https://electowiki.org/wiki/Summability_criterion#Multi-winner_generalizations_and_results>,
but the link is broken and Ebert has some big issues (e.g. negative
response and Pareto inefficiency).

On Thu, Aug 8, 2024 at 7:04 AM Kristofer Munsterhjelm <
km-elmet at munsterhjelm.no> wrote:

> On 2024-08-08 13:59, robert bristow-johnson wrote:
> >
> >
> >
> > RP, Schulze, MiniMax needs N(N-1) tallies.
> >
> > Condorcet-Plurality or Condorcet-TTR needs N^2 tallies.
> >
> > Since cycles are rare and not predictable, I'm just not as worried as
> > you scholars are about the differences in Condorcet methods regarding
> > resistance to strategy.
>
> That's a fair point.
>
> > 3 years ago I was plugging BTR-IRV because it was a simple modification
> > to IRV, already in use.  But since I have been convinced that the law
> > should be a two-method Condorcet system.  The law should say what it
> > means and means what it says.  In the most pedestrian language possible.
> >
> > Last year we introduced H.424 which was Condorcet-Plurality.  Next year
> > (new legislative session) I hope to persuad e Bob and Carol to introduce
> > another Condorcet RCV bill, this time maybe Condorcet-TTR which might be
> > better, almost identical to Condorcet-IRV, but without all the
> > round-by-round baggage.
>
> What do you think of minmax? No round-by-round, just "whoever does best
> one-on-one against his toughest rival". Too foreign for people used to
> top count methods like Plurality and TTR?
>
> Copeland,TTR would give you Smith. But it might be too complicated
> (count number of pairwise wins per candidate, more is better, then if
> there's a tie, choose the two tied candidates with most first
> preferences and elect the one who beats the other pairwise).
>
> -km
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>
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