[EM] Decloned Copeland Borda Hybrid

[Forest Simmons]

On each ballot B identify the anti-favorite a'(B) as the candidate a' among
those bottom listed on ballot B that is bottom listed the most on the rest
of the ballots.

[A candidate is bottom listed on a ballot if it is ranked over no candidate
on that ballot]

Elect the candidate X that, on the most ballots B, either is ranked ahead
of or pairwise defeats a'(B).

In other words, elect argmax N(X), where N(X) is the cardinality of the set
of ballots
{B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.

The key to monotonicity is that if some candidate X is uniquely raised on
some ballot B, then for any other candidate Y, the difference N(X)-N(Y)
does not decrease.

The only non-trivial case is where X starts in the a'(B) position and swaps
positions with some Z that was ranked above it.

This definitely adds a point to N(X) because X is newly ranked ahead of
a'(B). This will also add a point to Y if Z is pairwise defeated by Y.

In any case, N(X)-N(Y) does not decrease.

 Does this method satisfy the Condorcet Criterion?

Well, if X pairwise defeats all other candidates, then N(X) is equal to the
total number of ballots minus the number of ballots on which X is in the a'
position, which must be fewer than half of the ballots, otherwise X would
be the Condorcet loser.

Is it possible for N(Y) to be greater than this N(X) while losing pairwise
to the CW  X?

Perhaps ... if so, then this method is not a stand alone Condorcet method.

It would be a valuable task for someone to test its Condorcet efficiency
experimentally.

Thanks!

-Forest
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