[EM] A Little Help From My Friends

[Forest Simmons]

Let's call the candidates that do not beat X the "friends" of X ...
[...kind of minimal friends, really].

If every candidate is a friend of X, then we say X covers the set of
candidates.

If every candidate is a friend of a friend of X, then X and her other
friends together cover the set of candidates.

Etcetera ...

For each candidate X let f(X) be the minimum number of proper friends (i.e.
friends beyond self) that X requires in order to cover the rest of the
candidates.

If X together with seven of her other friends cover all of the candidates,
but X cannot cover all of the candidates with the help of only six friends,
then f(X)=7.

If f(X)=0, then X is unbeaten so it seems natural (in a single  winner
election) that if f(X)=0 and f(Y)>0, then Y should not be elected.

This idea is called the Condorcet Criterion.

In general, it seems reasonable that if there is some candidate X such that
f(X)=n, then Y should not be elected if f(Y)>n.

In other words,  the winner of a single winner election should be a
candidate that minimizes f(X).

Let's call this imperative the "Friendly Criterion," or the FC.

Obviously, the FC is a generalization of the CC that we just mentioned a
minute ago.

Suppose we accept the Friendly Criterion, but it turns out that argmin f(X)
has several members. How do we narrow it down to one of them?

Why not elect the candidate from among those that minimize f(X), the one
that is treated with most deference by her friends (the friends that helped
her minimally cover the rest of the candidates)?

In other words, elect from argmin f(X) the X whose friends not only refrain
from beating her, but don't even think about it!

Suppose that Y has a friend that lacks only ten votes of beating her, but
none of X's friends come within even forty votes of beating her. Then Y's
friends are not as friendly to her as X's friends are to X. Therefore, Y
should not win.

This degree of friendliness idea applied repeatedly is enough to
narrow down the
set argmin f(X) to the one candidate X with the least danger from being
beaten by one of her so called "friends".

That's an election method we can call, "A Little Help From My Friends".

I hope you liked it.

-Forest

P.S. Keep reading if you prefer Set Theory to plain English.

Let T=argmin f(X), and let n be the common value of f(t) for each candidate
t in the tied set T that we have to narrow down in order to get to a unique
winner.

For each tied candidate t in T, let F(t) be the set of n friends of t that
together with t minimally cover the entire set of candidates.

For each t in T, let m(t) be the min pairwise support for t in the pairwise
contests between t and the members of F(t).

Elect argmax m(t).
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