[EM] Friendly Voting: Some Criteria Compliance Proof Sketches

[Forest Simmons]

OK, here's what I would like to call Friendly Approval, a decisive approval
score method based on friendly ideas:-)

For each ballot B, let f(B) be B's favored first place choice.

For each candidate X  let A(X) be the number of ballots B that rank X, for
which X has a short beatpath to f(B).
This A(X) is X's base level approval. If argmax A(X) has only one member,
that member is to be elected.

Otherwise, break the tie with E(X), defined as the number of ballots B that
rank X for which X has a beatpath to f(B) of at most one step.

If there are still tied candidates, break the tie with C(X) defined as the
number of ballots B for which X is ranked on B and covers (or weakly
covers) f(B).

If there are still tied candidates, break the tie with F(X) defined as the
number of ballots for which X=f(B).

In other words elect argmax S(X), where S(X) is the sum given by

A(X)/epsilon^3+E(X)/epsilon^2
+C(X)/epsilon+F(X)

For a weaker approval method, add to S(X) the term G(X)/epsilon^4, where
G(X) is the number of ballots B on which X is ranked for which X has a
finite beatpath to f(B).

-Forest


On Sun, Oct 16, 2022, 2:26 AM Kristofer Munsterhjelm <km_elmet at t-online.de>
wrote:

> On 10/16/22 03:08, Forest Simmons wrote:
> > Suppose de granted each candidate one free bullet ballot .... that would
> > keep non-Smith candidates from being in the tied-for-win set ... but it
> > creates a non-scaling problem.
> >
> > However, if we grant each candidate a gratuitous bullet ballot with
> > positive weight epsilon, and shrink epsilon until further shrinkage
> > stops changing the winner, then the method becomes scale invariant.
>
> Yeah, I thought about that. Basically you can add an epsilon to every
> candidate's first preference count and then let epsilon go to zero. This
> is equivalent to counting sum over friends A: fpA - sum over defeaters
> C: fpC as a two-vector whose first element is just the sum and the
> second is the number of friends minus the number of defeaters, and then
> using leximax.
>
> The problem is that this fails clone independence. Suppose A and B are
> tied even given the tiebreaker above, and let C be some friend of A
> who's not a friend of B. Clone C, then A wins.
>
> -km
>
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