[EM] Sequential elimination method that turns out to be a generalization of fpA-fpC

[Forest Simmons]

Very Good ... and nice explanation!

On Sat, Apr 22, 2023, 11:11 AM Filip Ejlak <tersander at gmail.com> wrote:

> Ok, after trying out a few things I found something that looks promising
> in terms of passing all of ISDA, DMTCBR, Cloneproofness and, I hope,
> Monotonicity. The scoring rule is similar to Friendly Cover (but without
> subtracting fpC); the elimination rule is a bit atypical.
>
> The definition:
>
> "*Let X's score be the sum of 1st preferences of candidates that are
> pairwise beaten by all candidates that pairwise beat X.*
>

In friendly parlance ... X's score is the sum of the first preference
counts of all the friends of all of the enemies of X.

Now suppose X' covers X.  Then enemies(X') is a subset of enemies(X), and
so friends of enemies of X' is a subset of friends of enemies of X ...
which gives X' a lower score than X.

So argminScore(X) always contains an uncovered candidate ... which you
probably already mentioned ... I haven't read all of the posts in this
thread yet (sorry about that).

-fws

* In each round eliminate such a candidate that the sum of the eliminated
> candidate's score and the greatest score growth caused by the elimination
> is minimized*."
>
> The growth-minimizing part works for at least two reasons: 1) It
> reinforces the behaviour we would expect in a situation with no cycles,
> when an elimination shouldn't cause a positive score growth (if an
> eliminated candidate was worse than X, then its 1st preferences were
> already included in X's score). 2) In cycle situations it should benefit
> candidates that already have a high score (because letting a low-score
> candidate achieve some score level is less preferable to letting a
> high-score candidate achieve such level).
>
> Also notice that the eliminated candidate's score is equal to the greatest
> score decrease caused by the elimination, so there is a nice symmetry
> between these two things we sum. It also means we can express the
> elimination rule in a more elegant way ("*eliminate such a candidate that
> the greatest score distance change between any two candidates is minimized*"),
> but perhaps the former definition makes it more clear why it works.
>
> It seems to me that the incentive to eliminate a non-Smith candidate
> because of a low score should always outweigh the incentive to eliminate a
> Smith candidate because of a low score growth, so all non-Smith candidates
> should be eliminated first.
>
> Proof that the 3-candidate case is equivalent to fpA-fpC:
>
> If there is no cycle, it is trivial as both methods are Condorcet. If
> there is an A>B>C>A cycle, then we look at the possible eliminations. Let
> "a" be the number of A's 1st preferences, "b" of B, "c" of C.
>
> - if C is eliminated, A's score increases by b+c and A wins with B
>
> - if A is eliminated, B's score increases by a+c and B wins with C
>
> - if B is eliminated, C's score increases by a+b and C wins with A
>
> We add eliminated scores to score growths in order to find the optimal
> elimination. If 2c+b is the lowest, A wins. If 2a+c is the lowest, B wins.
> If 2b+a is the lowest, C wins.
>
> Notice that min(2c+b, 2a+c, 2b+a) = min(2c+b-(a+b+c), 2a+c-(a+b+c),
> 2b+a-(a+b+c)) = min(c-a, a-b, b-c) = max(a-c, b-a, c-b). So we get the
> fpA-fpC formula, QED.
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