# [EM] Smith//3Score (was MAM vs Schulze)

Forest Simmons fsimmons at pcc.edu
Sun Oct 9 15:19:03 PDT 2016

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> Now, how do we adapt this to general rankings? We assume that equal top
> rankings and equal bottom or multiple truncations are allowed.
>
> For each ballot on which a candidate is ranked above bottom but below top
> that candidate receives one point.  For each ballot on which the candidate
> is ranked top or equal top that candidate receives two points.
>
> The Smith candidate with the greatest number of points wins.
>
> [End of definition]
>
> Note that the method does satisfy CD unlike Smith//ImplicitApproval.
> Jameson's idea of three slot scores makes it work.
>
> How does it do on burial?
>

Note that for three candidates and complete rankings the method is just
Smith//Borda.  But Smith//Borda is extremely clone dependent.  The three
category (top, middle, bottom) point system is key to making the method
clone proof.

For public proposal it is important to have a non-cryptic definition of
Smith.

The smallest set each of whose members pairwise beats every candidate
outside the set is succinct, but probably (by itself) too cryptic for the

I think the idea of a beat-path is very intuitive.  I remember that ten or
fifteen years ago Michael said that he had success explaining Schulze to
various people on the street from the beat-path perspective.

If A beats B beats C beats D, then that chain of defeats is called a
beat-path from A to D.

The human brain is adept at generalizing this kind of definition to a path
with more or fewer steps..

For the purpose of defining the Smith set we don't even need to talk about
the strength of a beat-path.

Candidate X is a member of Smith iff for each candidate Y (other than X
itself), there is a beat-path from X to Y.

In typical public elections we expect Smith to be a cycle of three.  In
that case Smith//3Score is the same as Max Covering 3Score.  In general the
latter is preferable because no matter how large Smith may be it continues
to elect an uncovered candidate.

For public consumption it is mainly a question of ease of understanding.

Which is easier to understand a sequence of successive pairwise defeats or
a covering relation.

Another way to elect the same candidate when the top cycle has only three
members is to list the candidates by their point totals (highest total at
the top of the list) and starting with the bottom candidate move up the
list replacing pairwise losers with winners.

In other words the winner will be the highest listed candidate that beats
both the champion from below and every (if any) candidate above it in the
list.

I would be perfectly happy with any of these proposals, though there is an
obvious soft spot in my heart for the covering version.

Another reason I like the covering version is that various strengthenings
of the covering relation can be put to good use to refine this method.
More on that in another post.

Forest
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