[EM] extending fpA-fpC

[Kristofer Munsterhjelm]

A very late reply, but:

On 26/12/2020 22.25, Kevin Venzke wrote:
> Hi Kristofer,
> 
> Le jeudi 24 décembre 2020 à 20:56:20 UTC−6, Kristofer Munsterhjelm <km_elmet at t-online.de> a écrit : 
>> About fpA-fpC, one way to get an idea of what a four-candidate fpA-fpC
>> method may be like could be to investigate elections where clone
>> independence demands that some candidate wins (e.g. you can get to the
>> election by cloning A, and B wins in the original three-candidate
>> fpA-fpC election; then A must win in the new election).
> 
> I took a look at this problem. You can use a different route to get
> to the same 3-candidate scores, while trying to make sense of what the
> fpA-fpC metric represents. I think it's a measure of doubt as to whether
> the A>B win is genuine. An A-top ballot is definitely a genuine
> expression of A>B. A C-top ballot may not be. A B-top ballot is
> irrelevant. (A C>A=B or C>B>A vote would also be irrelevant. There could
> be a possible refinement to be found in this point, though so far I
> haven't had much luck.)
> 
> My best attempt to extend the method has better monotonicity than
> C//IRV, has a few Plurality failures, and appears to satisfy Schwartz
> without imposing it explicitly... I'm not sure if that's correct though.
> It could be that with a large number of candidates it becomes possible
> to make a scenario that fails Schwartz.

I think this method also fails DMTBR, if I've implemented it right.

Before:

1: A>C>D>B
1: A>D>B>C
1: B>C>D>A
1: C>A>B>D
1: D>B>A>C

A wins. Now let a C>A voter bury A:

1: A>C>D>B
1: A>D>B>C
1: B>C>D>A
1: C>B>D>A  (A is buried here)
1: D>B>A>C

and A and C tie for first.

Replacing min and max with leximin and leximax where possible helps, but
doesn't seem to fix that particular problem instance.