[EM] Best Copeland-like Method Ever?

Forest Simmons fsimmons at pcc.edu
Thu Dec 10 16:25:28 PST 2020

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

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.

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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