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

[Filip Ejlak]

I take that back -  even if it always chooses an uncovered candidate, the
method is not totally independent of covered alternatives.
In this 4-candidate cycle example:

8: A>B>C>D
9: B>D>A>C
12: C>D>A>B

Landau set is {A,B,D} as B covers C (and gets their 1st preferences added
to the score), but C owns all their 1st preferences at the expense of D.
Eliminating C in the first round doesn't happen because D would gain too
much at the expense of B. A gets eliminated instead, and eventually B wins.

If Landau//[This Method] was used instead, D would be the winner.


niedz., 23 kwi 2023 o 14:43 Filip Ejlak <tersander at gmail.com> napisał(a):

> Thank you! But a little clarification - in the score definition I meant "*pairwise
> beaten by all candidates that pairwise beat X*" as in "*Y has to be
> beaten by each and every candidate that beats X*", so it's just a
> covering relation. Still, it does seem that the uncovered candidates should
> be the last to be eliminated.
>
>
> niedz., 23 kwi 2023 o 01:03 Forest Simmons <forest.simmons21 at gmail.com>
> napisał(a):
>
>> 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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>>> info
>>>
>>
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