[EM] The mathematical troll: monotone resistant set methods attempts

[Closed Limelike Curves]

I don't have a proof of this, but I strongly suspect there is no such
(deterministic) method. Potentially there is a randomized method? Something
about this reminds me of the approach/justification to maximal lotteries,
but I couldn't tell you for sure what it is.

On Tue, Jul 1, 2025 at 3:17 PM Kristofer Munsterhjelm via Election-Methods <
election-methods at lists.electorama.com> wrote:

> So I keep returning to my attempts to make a monotone resistant set
> method. I find some approach that *almost* works, but then there's some
> small snag that makes it not work after all -- so close but always out
> of reach.
>
> But I thought I'd document two of my attempts in case anybody finds
> something I've overlooked. This post is about the first one; I'll do the
> other later.
>
> Define the disqualification relation of order k, A~(k)~>B, as holding if
> A has more than 1/|S| of the first preferences on every subelection that
> contains A and B and have k candidates or fewer in total. E.g. the
> disqualification relation on order 2 is just pairwise comparison (I'm
> assuming full ranking, I can deal with equal-rank and truncation once
> I've found a method that works for full ranking).
>
> Then, if we have an election of n candidates where the disqualification
> relation is cyclical on every order below n, then the "Borda order"
> combined with walking along the cycle is monotone and elects from the
> resistant set.
>
> Say we have a three-candidate election with a Condorcet cycle, and the
> cycle order is A>B>C>A. Then if we walk along the cycle, A's order is A
> B C, B's order is B C A, and C's order is C A B - just listing the
> candidates in order of the cycle, starting with our chosen candidate.
> Then the Borda order is:
>         score(A) = 2 * fpA + 1 * fpB + 0 * fpC
> i.e. we add the first preferences of each candidate as we walk along the
> cycle, weighting them by Borda coefficients. So, similarly for B, B's
> order is B C A,
>         score(B) = 2 * fpB + 1 * fpC + 0 * fpA
> and
>         score(C) = 2 * fpC + 1 * fpA + 0 * fpB.
>
> For three candidates, this method reproduces fpA-fpC's behavior when we
> have a Condorcet cycle, because fpA-fpC's score for an A>B>C>A cycle is
>         score(A) = 1 * fpA + 0 * fpB - 1 * fpC
> and that's just the Borda coefficients with each value subtracted by one.
>
> So, the generalization then is, if we have a four-candidate election and
>         A~(3)~>B
>         B~(3)~>C
>         C~(3)~>D
>         D~(3)~>A
> then
>         score(A) = 3 * fpA + 2 * fpB + 1 * fpC + 0 * fpD
>         score(B) = 3 * fpB + 2 * fpC + 1 * fpD + 0 * fpA
>         etc.
> elects from the resistant set and is monotone.
>
> But what do we do when things aren't that neat? E.g. we have a
> four-candidate election with a Condorcet cycle that gets resolved when
> we go to disqualifications of order three? I could never figure that
> out, and the problem is further complicated by that raising A may break
> B~(3)~>C so that we don't see a cycle anymore.
>
> -km
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