# [EM] Best Copeland-like Method Ever ? (small correction)

Forest Simmons fsimmons at pcc.edu
Thu Dec 10 16:51:42 PST 2020

``` I wish I could say typo but in reality it's more of a "thinko" ... see
below ...

On Thursday, December 10, 2020, Forest Simmons <fsimmons at pcc.edu> wrote:

> Let BOT(X) be the set of alternatives Beaten Or Tied by X pairwise.
>
> Definition of diameter: the diameter of a set is the distance between the
> two furthest apart points of the set.
>
> Here's the simplest version of Copeland that is clone free:
>
> Elect the candidate X that maximizes the diameter of BOT(X)
>
> Why do I call this a version of Copeland? Compare it with this version of
> Copeland:
>
> Elect the candidate X that maximizes the cardinality of BOT(X).
>
> The only difference is that one maximizes cardinality while the other
> maximizes the diameter.
>
> So why is one clone dependent and the other clone free?
>
> The key idea is this: if you replace an alternative in BOT(X) with a clone
> set, the cardinality of BOT(X)  definitely increases, while its diameter
> increase is relatively small except when it started at zero,  i.e. the case
> when X was the Condorcet winner.  In that case the winner of the method
> restricted to the Clone set becomes the new overall winner.
>

In the above paragraph replace all instances of "winner" with "loser"

> The proof that this new (diameter based) version is monotone follows
> mutatis-mutandis from the proof of the original (cardinality based) version.
>
> And the same goes for the proof of Landau compliance ... that both
> versions always elect uncovered alternatives.
>
> So what remains to do?
>
> The immediate priority is to construct convenient and appropriate metrics
> for different spaces of alternatives.
>
> After that, how to generalize it to multi-winner proportional represention
> and to non-zero entropy methods.
>
> It seems to me that there are great opportunities for exploration and
> discovery ahead of us for years to come!
>
>  in my humble opinion it is better to estimate distance between
> alternatives separately from any kind of preference schedules.
>
> One should keep in mind that distorted estmates of distance can affect
> outcomes perversely.
>
> So, for example if you ask voters to specify which pair of candidates is
> most disparate, it is important to either accept only one answer per
> customer, or count multiple answers fractionally.
>
> If you have to estimate distances from ranks or ratings, here is a simple
> approach that will work well in large diverse electorates assuming sincere
> ratings or rankings:
>
> Let d(x, y) be the percentage of the ballots on which x and y appear at
> opposite extremes counted fractionally. For rankings that means one is
> ranked Top (or equal top) and the other Bottom ( above no alternative).
>
> So if ballot B is one of N ballots, and Top(B) and Bottom(B) have
> respective cardinalities of k and m, then each pair (x, y) in the Cartesian
> product of Top(B) and Bottom(B) gets a contribution from ballot B to its
> distance estimate d(x, y) in the amount of one hundred percent over the
> product of N, k, and m.
>
> We have our work cut out for us!
>
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