[EM] Rehabilitating Copeland

[Forest Simmons]

Here's the Enhanced Copeland method we've been leading up to:

Elect the candidate that needs the "littlest help from its friends" to
cover its enemies.

Explanation:

Last time we learned that if C is a candidate tied for Copeland winner,
then C's friends will cover its enemies ... in other words, every enemy of
C is a friend of a friend of C ... in other words C will be a Landau
candidate ... in other words,  our Enhanced Copeland Winner is the Landau
candidate that requires the least help from its friends to befriend its
enemies.

Operationally we proceed as follows:

1. Each Landau candidate C (including all tied Copeland Winners) designates
a set of candidates H(C) as its "helpers".

2. A Deliberative Assembly checks the validity of these helper claims, and
eliminates any candidate C whose enemies are not covered by H(C).

3. Next the Deliberative Aassembly narrows down the candidate pool to
T=argmin #H(C) ... the set of candidates tied for the fewest (designated)
helpers needed to defeat all of its enemies.

4.From among the members of the  set T tied for fewest needed (designated)
helpers, elect the candidate C whose Max Pairwise Opposition from an enemy
of C to a designated helper of C, is minimal.

This is the candidate C that requires the fewest helper friends (and the
least help from those so designated) in order to befriend all of its
enemies (especially the one hardest for a designated helper to befriend).

In practice, no candidate in a public election will need more than one
helper, so the method simplifies to electing the candidate C whose
designated helper H most easily befriends the enemy E of C having the most
Pairwise Opposition against H.

The less Pairwise Opposition that E has against H the less help (defined as
H's Pairwise Support against E) it takes to befriend E.

So C is the candidate that needs the least help in befriending its enemies.


Next time some examples ...

On Tue, Mar 14, 2023, 10:11 PM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> I've always had a soft spot in my heart for Copeland ... perhaps because
> it's such a friendly method. Let me explain ...
>
> Remember Kristofer's wry definition of friendship ... whoever doesn't beat
> you is your friend?
>
> By that definition the Copeland winner is simply the candidate with the
> most friends.
>
> And everybody is a friend of the Condorcet Winner (the unbeaten candidate
> if there is one) ... so Copeland is Condorcet efficient.
>
> Simple and compelling ... as far as it goes ... but it doesn't always go
> the whole distance ... what if there are two or three candidates tied for
> most friends?
>
> Various Copeland tie breaking proposals have been suggested over the years
> .... but none of them seem to naturally continue the basic friendship theme
> that inspired Copeland in the first place.
>
> Here's a friendship idea that might help:
>
> A set of candidates can have friends: you are a friend of a set if you
> don't beat every member of the set.
>
> Another way to say this ... is that you are a friend of a set iff the set
> covers you.
>
> This friendship/ covering relation leads to a natural hierarchy of
> candidates ...
>
> A Condorcet candidate has everybody as a friend ... and therefore covers
> everybody  ... without any help.
>
> A candidate  X that covers everybody with the help of only one friend F is
> the next level. Together X and F form a set X+F that cannot be beaten ...
> everybody is friendly to X+F just like everybody is friendly to a Condorcet
> Candidate, when there is one.
>
> So different candidates need different amounts of help from their friends
> to cover their "enemies", the candidates that are not their friends.
>
> The fewer friends you need to cover all of your enemies, the stronger you
> are.
>
> A Condorcet Winner has lots of friends and zero enemies.
>
> At the other extreme ... a Condorcet Loser has no friends to help cover
> his enemies .... which is unfortunate because all of the other candidates
> are his enemies.
>
> What about a Copeland candidate C tied for most friends?  Let F be the set
> of friends of C.  Then C and F together cover the enemies of C.
>
> Any time a candidate C with some of her friends F together cover all of
> her enemies, we say that C is a Landau candidate. So every max score
> Copeland candidate is a Landau candidate .... which makes Copeland a
> "Landau efficient" method.
>
> So the bigger the Copeland score of a candidate the fewer her enemies and
> the fewer friends she needs to help cover those enemies.
>
> This brings us to another imperfection of Copeland that we will take care
> of with the help of our friends:
>
> Suppose candidate F is a friend of candidate C and of nobody else. Then
> unscrupulous C supporters might be tempted to enter a whole "team" T of
> clones of F into the race to increase C's Copeland score and nobody else's.
>
> How can we effectively counter this kind of "Teaming" temptation?
>
> A natural way is to see how far we can pare down the friendly help while
> still covering the enemies of C .... just how many friends does C actually
> need to cover her enemies?
>
> This doesn't completely resolve the tie problem ... but it reduces the
> covering sets down to manageable size.
>
> In fact, the most amazing and useful thing I have learned in these
> friendly explorations, is that in public elections ... for all practical
> purposes ... it is 99.999... percent sure that at least one candidate C
> will have a friend F which is all the help she needs to cover her enemies.
>
> This fact hugely simplifies tie breaking and, at the same time, completely
> erases the clone teaming problem!
>
> To be continued ....
>
> -Forest
>
>
>
>
>
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