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

[Forest Simmons]

We've all had ruined days from monotonicity fails, but these are steps that
ultimately pay off ... amazing how that works out!

Don't get discouraged.

Remember, if the method is Condorcet efficient, the Yee diagrams that show
so much monotonicity dysphoria in IRV ... these diagrams cannot detect any
defect whatsoever in your method!

On Mon, Apr 24, 2023, 1:50 PM Filip Ejlak <tersander at gmail.com> wrote:

> (disclaimer: unfortunately there exists a 5-candidate cycle example that
> breaks monotonicity and my day is ruined, I show the example later)
>
> > 8: A>B>C>D
> > 9: B>D>A>C
> > 12: C>D>A>B
>
> So, in the first round we got B covering C, so B score includes C's 1st
> preferences:
> A score: 8
> B score: 9+12=21
> C score: 12
> D score: 0
>
> If A got eliminated, we would have:
> A score: 0 (*-8*)
> B score: 29 (*+8*) - receives votes from A, now a Condorcet winner
> C score: 12 (+0)
> D score: 0 (+0)
>
> With B eliminated, we would have:
> A score: 8 (+0)
> B score: 0 (*-21*)
> C score: 12 (no change)
> D score: 9 (*+9*) - receives B's votes
>
> With C eliminated, we would have:
> A score: 8 (+0)
> B score: 9 (*-12*) - loses C's votes, as they go to D which isn't covered
> by B
> C score: 0 (-12)
> D score: 12 (*+12*) - receives C's votes
>
> With D eliminated, we would have:
> A score: 29 (*+21*) - now a Condorcet winner, covers everyone
> B score: 21 (+0)
> C score: 12 (+0)
> D score: 0 (*+0*)
>
> The greatest increases and decreases are bolded. We calculate the
> penalties for each elimination:
> - A: 8+8=16
> - B: 9+21=30
> - C: 12+12=24
> - D: 21+0=21
>
> Eliminating A would create the smallest distortion of the scores, so A
> gets eliminated.
> Now we are left with B, C and D; B is the Condorcet winner.
> What happened was that B had a good starting position thanks to covering
> C, so the method preferred a choice that maintained this position.
>
> Now, the monotonicity counterexample:
>
> 17 ABCDE
> 16 BCDEA
> 15 CDEAB
> 14 DEABC
> 13 EABCD
>
> Here E gets eliminated first, and eventually D wins. However, if we change
> ABCDE to ABDCE and EABCD to EABDC, then C stops pairwise-beating D, which
> means that A will cover C if E gets eliminated. That discourages the method
> from eliminating E first (as A's score growth would be too big) and D gets
> eliminated instead.
>
> So even though I think I do like it a bit more than the Condorcet-IRV
> hybrids... unfortunately it doesn't fully do what I intended it to do.
> I guess the elimination approach actually *is* unsuitable for the
> monotonicity goal here.
>
>
>
> niedz., 23 kwi 2023 o 22:38 Kristofer Munsterhjelm <km_elmet at t-online.de>
> napisał(a):
>
>> On 4/23/23 21:45, Filip Ejlak wrote:
>> > 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.
>>
>> That's not surprising -- under mild assumptions, you can't have
>> monotonicity and independence of covered candidates.
>>
>> I haven't quite gathered how the elimination process works. Could you
>> give an example election, e.g. the 4-candidate cycle example above?
>>
>> -km
>>
>> ----
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