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

[Filip Ejlak]

(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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