[EM] Resistant Set Questions

[Kristofer Munsterhjelm]

On 2026-05-10 22:25, Gustav Thorzen via Election-Methods wrote:
> On Sun, 10 May 2026 20:36:08 +0200
> Kristofer Munsterhjelm <km-elmet at munsterhjelm.no> wrote:

>> This would make the resistant set less decisive than either completion
>> of the equal-rank. For instance, you could have a situation like:
>>
>> 31: A>B>C
>> 31: B>A>C
>> 28: C>A>B
>>
>> Both A and B disqualify C, so the resistant set does not contain C. But if
>>
>> 62: A=B>C
>> 28: C>A>B
>>
>> Then if I understood your definition correctly, neither A nor B gets any
>> first-preference votes, so they can't disqualify C on {A,B,C}, and hence
>> they don't disqualify C at all, which leads C to be part of the
>> resistant set. More generally, no matter how we break the tie:
> 
> Well, my intension was to have both A and B disqualify C in that case,
> but you make a good case that if we generalize the candidates rank this way,
> then we should change the disqualification definition to something like,
> A~>C if A is top ranked and C is not on enough rankorder in each sub-election.

My usual objection to counting everybody's first preferences at once is 
that it could break antisymmetry, e.g.

100: A=B=C

if you just count this as 100 votes for A, 100 for B, and 100 for C, 
then they all pass the quota of 100/3 in the full election (and 100/2 in 
every pairwise matchup), so you have A~>B and B~>A at the same time.

You say "C is not on enough rankorder in each sub-election" which 
*would* solve the problem above (since if both A and B have above 100/3, 
then B has "enough" votes to not be on the losing end). But this is a 
restriction of the ordinary resistant set definition, e.g. with a 
Condorcet cycle example:

32: A>B>C
32: B>C>A
26: C>A>B

Here A~>B because A exceeds the quota of 30 in the full election and 
also beats B pairwise. But if we were to say B also needs to be below 
quota, then that disqualification would fail.

The "union of resistant sets possible by completing equal-rank in *some* 
way" solution works both in the

62: A=B>C
28: C>A>B

and the

100: A=B=C

examples; in the former, there exists a way to break A~>C and another to 
break B~>C, so C is never part of the resistant set (while A and B oth 
are); in the latter, it's possible to break equal-rank so that any one 
candidate is part of the resistant set; hence the union set is {A, B, 
C}, which makes sense.

However, it's very complicated. Perhaps I could try play with polytopes 
a bit to see if anything more user-friendly pops out.

> Then the top ranked vs not top ranked seems like the simplest to use.
> Or maybe multiplying the ratio with the rank,
> A being 2:nd place in the subelection requireing 2 times the ratio,
> and 3:rd place requireing 3 times the ratio,
> likely it still needs to be combined with some for of A>C requrement as well.
> I will have to check if this throws a spanner in the works for the method I
> proposed in my other post.

I think involving lower ranks would weaken resistance unless additional 
constraints are added. The idea of A~>B being unmanipulable is that for 
a voter whose honest preference is B>A, the following holds for any 
sub-election:
	- If the sub-election's set S contains only one of B and A (or none of 
them), then ranking B higher has no effect; this voter's ballot has no 
effect on either A~>B or B~>A in this sub-election.
	- If S contains both A and B, but also some C that the voter ranks 
higher than B, then burying A has no effect and compromising for B has 
no effect on A~>B, though it may have an effect on B~>A.
	- If S contains both A and B and nobody the voter prefers to B, then 
the voter is already contributing maximally to B and minimally to A. 
Lowering A or raising B won't change anything.

However, if you start using a different positional system than 
Plurality, i.e. count second preferences in a subelection, then it's 
possible for a voter who votes something like B>A>X to weaken A by 
burying A, B>X>A with the aim of denying A the partial credit of being 
ranked second instead of last.

Now, globally speaking, something like IRV *does* care about second 
preferences for A in the full election; it's not Plurality, after all. 
But I think that it's easier to retain the desirable properties of 
resistant set methods by limiting oneself to counting first-preference 
votes in any given sub-election, and then come up with ways to combine 
these or consider patterns across multiple sub-elections if one wishes 
to add more complex logic.

>>> 2) Regardless if we stick to the original or the generalization,
>>> if a candidate a satisfy the following in all their sub-elections,
>>> v(a>{rest of the k candidates}) > n/k,
>>> then is a the only member of the Resistant set?
>>> And if yes, is it known whether that is a sufficient and/or a neccecary condition?
>>
>> That's a super CW; it's sufficient but not necessary. Suppose A~>B and
>> B~>C but A~/~>C. Then A is the only member of the resistant set. But in
>> {A,C}, A might not exceed the n/k quota, i.e. A might not beat C
>> pairwise. Here's an example:
> 
> That is even better,
> confirming it is a sufficient condidtion for the immunity to both burial and compromise scenario.
> Not supprised it was not a neccecary condition for a single set member,
> but the super CW sufficiency is what I hoped to figure out using this answer,
> so that is very nice to know I will only have to search for neccecity for super CW.

It's not strictly speaking sufficient, though. Consider a method that 
elects the super CW when one exists, and otherwise just does minmax. Say 
that A is the super CW. Then, although B>A voters may not be able to 
break A~>B, it is possible that they can break A~>C. If they accomplish 
that, then A won't be the super CW anymore; and if minmax's winner is 
then B, then their strategy will have paid off.

Full resistant set compliance avoids this particular strategy because no 
matter what the B>A voters do, they can't break A~>B, so the only thing 
they can accomplish by breaking A~>C is to introduce C into the 
resistant set, which doesn't help them get B elected.

>> Let's call the new set the MB-Smith-Resistant set or MSR set.

[snip]

>> So the impossibility example still holds. If a method passes
>> (MB-)Smith-resistant, then if it also passes monotonicity, it must fail
>> majority. But it reveals nothing about any other combination. I think
>> both "pass majority, fail monotonicity" and "fail both" are possible;
>> for instance, I'd imagine MSR//Plurality passes majority but fails
>> monotonicity, and that MSR//Antiplurality fails both.
> 
> Which also leaves it unclear if M satisfies Monotonicity but fails Majority
> can still leave MSR//M satisfying Monotonicity while failing Majority.
> I suspected the MB-Smith intersection would cause an autofail
> for Monotonicity from MB-Smith implying Majority,
> but I suppose I will have to go into details to determine that then.

I think I was wrong here. Suppose that A is a majority winner. Then A is 
also a super CW. Proof: In any sub-election with set S containing A, A 
is listed first on more than half the ballots. Half is greater than 
1/|S|. Hence, on that sub-election, A disqualifies everybody else. Since 
this holds for every sub-election, A disqualifies everybody else.

I think the MB-Smith set is also just {A} when A is a majority winner, 
because then A beats everybody else pairwise by a majority.

So Resistant//M must pass majority, because the resistant set consists 
of just the majority candidate when one exists. And thus so must MSR//M. 
So I was wrong when I expected Resistant//Antiplurality to fail majority.

Monotone resistant methods are very hard to create. I wouldn't expect 
Resistant//M to pass monotonicity for any simple M, i.e. ones not 
deliberately designed to make this possible.


I think where I got turned around is that I confused M passing majority 
with Resistant//M passing majority. If M passes majority, that means 
that when the resistant set is S, and the sub-election S has a majority 
winner (i.e. when everybody but the candidates in S are eliminated, 
there's a majority candidate among those who remain), then that majority 
winner must be elected. This property is not the same as the majority 
criterion, and is more like what IRV proponents call "majority support".

You could possibly get something Smith-y by doing the following:
	- Consider all sub-elections of cardinality two (i.e. pairwise 
matchups). Determine the resistant set based only on these, and 
eliminate everybody outside it. If only one candidate is left, elect 
that candidate.
	- Otherwise consider all sub-elections of cardinality no greater than 
three between the remaining candidates (i.e. pairwise matchups and 
three-candidate sub-elections). Determine the resistant set based only 
on these. Eliminate everybody outside it, if only one candidate's left, 
elect that candidate.
	- Keep increasing the maximum cardinality like this.
	- If you get to the end (cardinality = number of remaining candidates), 
use some completion rule.

That should reduce some of the IIA-like problems but I don't think it's 
monotone.

>> I think the problem is similar to Copeland: the set membership
>> definition is too entangled, for lack of a better term, with what
>> candidates outside the set are doing, to be easily handled in isolation.
>>
>> That is, consider something like the Smith set. Removing candidates
>> outside the Smith set from that set doesn't change who's in it, so you
>> can get ISDA by doing something like Smith//X. But with something like
>> Copeland (the set of winning candidates by the Copeland method),
>> removing candidates outside the set *can* change the composition of the
>> set. So you can't get "independence from Copeland losers" by just doing
>> Copeland//X, and attempting to do so may lead to basic criterion failures.
>>
>> And I suspect the Resistant set is like this, too. It makes sense, since
>> to have A~>B, A must be above quota in every sub-election, some of which
>> may contain some other candidate C; then it's possible that A
>> disqualifies B in every sub-election except for those containing C, and
>> that B~>C, and then removing C makes A disqualify B.
> 
> Yeah, I think so to, mostly becuase the participation of a disqualified candidate
> can disqualify other candidates, which seems to be in line with how many (most?)
> burial scenarios work.
> 
> Would be really nice though to figure out a/the neccecary condition for
> when electing from the Resistant set when its composed of only one candidate
> leads the election being immune to both burial and compromise.
> Especially so if that candidate being the CW is only a
> sufficient rather the neccecary condition.

If you want the most strictly necessary condition, it's probably going 
to depend on the size of the strategic factions. The resistant set quota 
of 1/|S| is intended to ensure that the disqualification relation is 
always acyclical. But if e.g. A is the CW and there are too few C>A 
voters to get a cycle going even if they bury A below B, then burial 
can't get C elected if the method passes Condorcet.

 From the other direction, IRV has better resistance to coalitional 
manipulation under impartial culture with four candidates than IFPP 
does, even though both methods elect from the resistant set. Thus there 
must exist elections where one can improve on the resistant set, i.e. 
where there's more than one candidate in the resistant set, but electing 
a particular one according to some logic is strategy-proof. I don't know 
what that logic is, but I can observe that in the finite voter case with 
three candidates impartial culture, and infinite voter case with four, 
IRV does better than the worst-case resistant set method does.

Re "CW only being sufficient", I do know that in the full rank-order 
case, if the base method passes the majority criterion, "elect the CW if 
he exists, otherwise do the base method" is never coalitionally 
manipulable on more elections than the base method itself. Francois 
Durand proved this. So there's no downside (in the full rank-order case) 
to being Condorcet compliant as far as coalitional manipulability goes.

-km