[EM] Friendly Voting: Some Criteria Compliance Proof Sketches

[Kristofer Munsterhjelm]

On 13.10.2022 02:38, Forest Simmons wrote:
> 
> 
> On Wed, Oct 12, 2022, 3:35 AM Kristofer Munsterhjelm 
> <km_elmet at t-online.de <mailto:km_elmet at t-online.de>> wrote:
> 
>     Does this come with the same caveat as in Friendly Cover that if
>     someone
>     has no first preference, then the compliance may be failed? E.g.
>     suppose
>     a bunch of nobodies are ranked first (enough so that they're not in
>     Smith), then every viable candidate's first preference is zero.
> 
> 
> Yeah, but if even one Smith member is in the support of lottery L, then 
> no non-Smith winner has a finite Sum S(X).
> 
> That's why all my suggestions for Lare designed to take care of that 
> problem.

My hunch is that since the strength of first preferences (against 
burial) comes from that X>W voters can't change who they vote first by 
burying, then a burial resistant extension should be based on first 
preferences of some subset of the candidates. (If we take the subset to 
be just every candidates, we get the usual Friendly Cover/Voting patterns.)

If so, then always using the subset of every possible candidate will 
lead to the ISDA/pathological Smith/Landau failure shown above. So when 
comparing X to W, we have two possibilities: either a fixed-cardinality 
subset (probably containing both X and W) or a variable-cardinality one.

IRV is "essentially" (if you handwave enough) doing the latter, with its 
path dependence producing nonmonotonicity and all of the usual flaws. 
Since variable-cardinality subsets seem to lead directly to summability 
violations, it's kind of a no-go. But perhaps there is a way to do the 
former and still retain burial resistance?

There are two problems.

First, suppose we always use subsets of three. Then Condorcet cycle 
analogs may occur - e.g. A's fpA-fpC score beats B's in the subset 
(A,B,C), B's beats C's in the subset (B, C, D), C's beats D's in the 
subset (C, D, A), and D's beats A's in the subset (D, A, B). The usual 
fix would be to use something like Ranked Pairs over these orderings, 
but I can't see how to prove that burial resistance will be preserved by 
this.

Second, if we clone A into A1...An, then we can engineer the cloning to 
set the A clones' scores to arbitrary values for fpA-fpC restricted to 
the subset (A1,...,An). So using the result for subsets containing only 
clones to determine whether A or B should win would fail clone independence.

So, not entirely easy. But perhaps it will give you some ideas for 
lotteries along the "restricted subset" path :-)

Maybe the easiest way is to just come up with something that passes 
DMTCBR and is based on the fixed subset pattern, and see if its burial 
resistance is robust. A possible clue here is that fpA-fpC restricted to 
every (assume cardinality three) subset containing A will have A 
outscore B and C even after burial. But then again, that's also true of 
pairwise victories...

Or it might be possible to do something along the lines of: if A ties B, 
then A's tiebreaker is based on the relation of A's max-scoring friend 
wrt B's -- because in the case that A covers B, then B can't possibly 
win. As long as the "relation of A's max-scoring friend wrt B's" also 
takes into consideration the tiebreakers of these max-scoring friends.

> Also apologies for not checking with you before dubbing my version 
> "Friendly Voting."
> I started out just trying to do generalized median voting, and was 
> surprised when the final simplified version turned out to be "Friendly" ; -)

That's no trouble at all :-)

I definitely won't complain if Friendly turns out to be a really good 
method!

-km