<div dir="ltr">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.</div><br><div class="gmail_quote gmail_quote_container"><div dir="ltr" class="gmail_attr">On Tue, Jul 1, 2025 at 3:17 PM Kristofer Munsterhjelm via Election-Methods <<a href="mailto:election-methods@lists.electorama.com">election-methods@lists.electorama.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">So I keep returning to my attempts to make a monotone resistant set <br>
method. I find some approach that *almost* works, but then there's some <br>
small snag that makes it not work after all -- so close but always out <br>
of reach.<br>
<br>
But I thought I'd document two of my attempts in case anybody finds <br>
something I've overlooked. This post is about the first one; I'll do the <br>
other later.<br>
<br>
Define the disqualification relation of order k, A~(k)~>B, as holding if <br>
A has more than 1/|S| of the first preferences on every subelection that <br>
contains A and B and have k candidates or fewer in total. E.g. the <br>
disqualification relation on order 2 is just pairwise comparison (I'm <br>
assuming full ranking, I can deal with equal-rank and truncation once <br>
I've found a method that works for full ranking).<br>
<br>
Then, if we have an election of n candidates where the disqualification <br>
relation is cyclical on every order below n, then the "Borda order" <br>
combined with walking along the cycle is monotone and elects from the <br>
resistant set.<br>
<br>
Say we have a three-candidate election with a Condorcet cycle, and the <br>
cycle order is A>B>C>A. Then if we walk along the cycle, A's order is A <br>
B C, B's order is B C A, and C's order is C A B - just listing the <br>
candidates in order of the cycle, starting with our chosen candidate. <br>
Then the Borda order is:<br>
score(A) = 2 * fpA + 1 * fpB + 0 * fpC<br>
i.e. we add the first preferences of each candidate as we walk along the <br>
cycle, weighting them by Borda coefficients. So, similarly for B, B's <br>
order is B C A,<br>
score(B) = 2 * fpB + 1 * fpC + 0 * fpA<br>
and<br>
score(C) = 2 * fpC + 1 * fpA + 0 * fpB.<br>
<br>
For three candidates, this method reproduces fpA-fpC's behavior when we <br>
have a Condorcet cycle, because fpA-fpC's score for an A>B>C>A cycle is<br>
score(A) = 1 * fpA + 0 * fpB - 1 * fpC<br>
and that's just the Borda coefficients with each value subtracted by one.<br>
<br>
So, the generalization then is, if we have a four-candidate election and<br>
A~(3)~>B<br>
B~(3)~>C<br>
C~(3)~>D<br>
D~(3)~>A<br>
then<br>
score(A) = 3 * fpA + 2 * fpB + 1 * fpC + 0 * fpD<br>
score(B) = 3 * fpB + 2 * fpC + 1 * fpD + 0 * fpA<br>
etc.<br>
elects from the resistant set and is monotone.<br>
<br>
But what do we do when things aren't that neat? E.g. we have a <br>
four-candidate election with a Condorcet cycle that gets resolved when <br>
we go to disqualifications of order three? I could never figure that <br>
out, and the problem is further complicated by that raising A may break <br>
B~(3)~>C so that we don't see a cycle anymore.<br>
<br>
-km<br>
----<br>
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</blockquote></div>