[EM] UD Landau (correction)

Forest Simmons forest.simmons21 at gmail.com
Sat Jul 30 05:17:06 PDT 2022


It must be the 40 degree Celsius heat wave frying my brain: D(X) is
supposed to be the candidates defeating X, not defeated by X. And so D'(X)
should be the set of candidates that do not defeat X.

I apologize... right mental picture ... wrong verbal description!

I have made this same error before ... in the context of Decloned Copeland.

So let me run this past you ...

Let FX) be the First place preference total of all candidates that do not
defeat X. If no candidate defeated X, then F(X) would just be the total
number of ballots ... which suggests electing the candidate X that
maximizes F(X).

Anything wrong with that?

If not, then it is (arguably) the best UD Landau efficient method we could
offer for public proposal (un my not overly humble opinion).

Two more days of high temperatures!

-Forest



El vie., 29 de jul. de 2022 8:30 p. m., Forest Simmons <
forest.simmons21 at gmail.com> escribió:

> Kristofer,
>
> [For the record I am cc-ing this to the EM list.]
>
> While pondering the geometry of our agenda deas for Landau efficient
> methods, the following idea for a  Universal Domain Landau Efficient method
> came into focus:
>
> For each alternative X let D(X) be the set of alternatives that defeat X
> pairwise, and let D'(X) be its complement. Note that X is in D'(X).
>
> Our idea is that the closer D(X) is to being engulfed by D'X), the closer
> X is to being undefeated.
>
> So how do we measure how much D'(X) would have to be enlarged in order to
> swallow up D(X)?
>
> By the distance from the member of D(X) that is furthest from the
> unswollen D'(X).
>
> The distance of an alternative Y to the set D'(X) is defined as
> dist(Y, D'(X))=Min over Z in D'(X) of dist(Y,Z), i.e. the distance from Y
> to its nearest neighbor Z in D'(X).
>
> So to engulf D(X) the frontier of D'(X) has to be extended outward a
> distance of Extend(X)=Max over Y in D(X) of dist(Y,D'(X)).
>
> The less extension needed, the better, so we want to elect
> argmin(Extend(X))
>
> The only thing left is to pick an appropriate distance metric.
>
> If we want to remain in the Universal Domain category, it seems to me that
> the best gauge of dist(Y, Z) is the disappointment incurred by moving from
> alternative Y to Z, in other words, the number of ballots that prefer Y
> over Z.
>
> Note that if A covers B, then it is (in general) easier for D(A) to engulf
> D'(A) than for D(B) to engulf D'(B), so this method naturally elects
> uncovered candidates.
>
> There is a natural way to break ties that makes this Landau property snug
> ... by expanding/extending the engulfing frontier gradually, while keeping
> track of how much expansion is needed to swallow up the respective
> candidates one by one ... but for now we won't worry about the details.
>
> Does this make sense?
>
> Any obvious defects?
>
> The basic idea that came from pondering our agenda covering Landau  method
> (in connection with its Banks efficient mimic) is that it minimizes the max
> distance to D'(X) or "theall(X) union {X}" instead of minimizing the max
> distance to X itself as per standard MinMax. Similarly, the Banks version
> minimizes the max distance to a dense subset of D'(X) [a chain that covers
> D'(X].
>
> ["dense" because every member of D'(X) is just one small step away from
> the chain]
>
> So the topology of digraphs guides our intuition as always!
>
> -Forest
>
>
>
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