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

[Kristofer Munsterhjelm]

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