[EM] Rehabilitating Copeland

[Forest Simmons]

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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