[EM] Resistant Set Questions

[Gustav Thorzen]

On Sun, 10 May 2026 20:36:08 +0200
Kristofer Munsterhjelm <km-elmet at munsterhjelm.no> wrote:

> Before I reply, I'll just mention my notation, which may not be obvious. 
> In the context of the resistant set, I use A ~> B to indicate that A 
> disqualifies B, and A ~/~> B to indicate that A *doesn't* disqualify B. 
> A super CW is a candidate A who disqualifies everybody else, i.e. A ~> B 
> for every other B.  

I had started to suspect that is what was ment by the notation,
but this clears thing up.
Thanks for making it clear.

> (Some of what I say might be redundant - I haven't checked your most 
> recent post yet.)
> 
> On 2026-05-07 02:26, Gustav Thorzen via Election-Methods wrote:  
> > So I got qurious about the Resistant set and tried to read up on it,
> > but the electowiki page and its references left some thing unclear,
> > and other things I was wondering about left out.  
>   
> > I am assuming complete rankorders are provided by and/or infered from the ballots,
> > using some not-insane inference rule (like every unmentioned candidates ranked equally last).
> > I also assume that outside of question (0),
> > we have a total of m candidates, n voters, and k candidates in each sub-election (k<=m).  
> 
> You can use any base rule you want with only small increases in 
> manipulability. See for instance 
> https://github.com/kristomu/voting-methods/tree/master/top_vote_path, 
> with TVP and IFPP being slightly worse than IRV and Smith,IRV (which 
> seems to be the current champion - at least among methods that are 
> commonly known here).
> 
> You could even do something like, have the base method be "elect the 
> candidate with the fewest first preferences" ("Worst Plurality"), and 
> then do Resistant,Worst Plurality, and you'd still get around IRV/TVP 
> level manipulation resistance.
> 
> The core strategy resistance part is that if A~>B, then B is out of the 
> resistant set, and there's nothing that voters who prefer B to A can do 
> to break that disqualification, because they are already contributing 
> maximally to weakening A~>B by their initial (honest) preference of B 
> over A. In any subelection containing both A and B, even their honest 
> vote will be counted for B instead of A, and they can't make it count 
> more for B or less for A by strategizing.
>   
> > 0) In each sub-election, does the definition of the Resistant set have
> > k as the total number of candidates (no?)
> > or as number of candidates in that specific sub-election (yes?)?  
> 
> Suppose you have an election like:
> 
> 1: A>B>C>D
> 1: C>D>A>B
> 
> The subelection with set {A,B,C} is just
> 
> 1: A>B>C
> 1: C>A>B
> 
> Each subelection is defined by its set of continuing candidates (here 
> {A,B,C}) and the number of candidates is the cardinality of the set 
> (here, three). So if I understand you correctly, k is the number of 
> candidates in that specific sub-election, not the total number of 
> candidates in the election that it is a subelection of.  

That does answer my question, thanks.

> > 1) The set appears to be defined only for strict rankorder ballots,
> > does the following generalization make sense?
> > If we assign each candidate a rank equal to
> > the number of candidate ranked the same or higher (self included),
> > and and apply the definition,
> > do we still get its desireable properties.
> > 
> > And as a clarifying example,
> > if a=b>c>d=e=f>g then
> > a and b are considered 2:nd ranked,
> > c 3:d ranked,
> > d and e and f 6:th ranked,
> > g 7:th ranked,
> > and if a>b=c then
> > a is 1:st ranked,
> > b and c 3:rd ranked,  
> 
> 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.  

> 62-n: A>B>C
> n:    B>A>C
> 28:   C>A>B
> 
> C is ejected from the resistant set.
> 
> So a better, but much harder to use definition might be: the resistant 
> set with equal-rank should be the union of all resistant sets where the 
> equal-rank tie is broken in some way or other.  

> That said, you should still preserve the non-cyclical property with your 
> definition, but it could incentivize random tiebreak strategy (e.g. 
> voting A>B>C instead of A=B>C because you want to disqualify C).  
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.

> > 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.

> 31: A>B>C
> 31: B>C>A
> 28: C>A>B
>   
> A>B>C>A cycle, the full election's quota is 30. So A~>B because A has   
> more than a third of the first preferences in the full election and 
> beats B pairwise. Similarly B~>C. But since C beats A pairwise, we don't 
> have A~>C.
> 
> The resistant set is A because both B and C are disqualified by someone 
> else.
> 
> Hence the disqualification relation is intransitive. A significant part 
> of my effort designing TVP (the monotone resistant set method) involved 
> finding a transitive relation that implied disqualification.  

Which also explicitly clarifies a candidate can be disqualified by other non-members.

> > 3) This one is probably wrong if (2) is wrong, but in any case,
> > If I create the smalest non-empty set of all candidates a satisfying  
> > v(a>{rest of the k candidates}) > n/k for each subelection  
> > where a is the only set member,
> > then is this new set the same as the Resistant set when
> > strict rankorder ballots are used?
> > 
> > 4) If we create a new set as the intersection (set theory)
> > of the Resistant set and the MB-Smith set (that is candidates member of both),
> > does NewSet//M satisfy Majority criterion and fail Monotonicity,
> > since Resistant//M fails Monotonicity if M satisfies Majority.
> > Here MB-Smith means the smalest non-empty set where each member
> > Majority-Beats each non-member, unlike the regular Smith set,
> > where each member simply Plurality-Beats non-member.
> > MB-Smith is therefore a superset of the regular Smith, or PB-Smith, set.  
> 
> Let's call the new set the MB-Smith-Resistant set or MSR set. Then run 
> Ejlak's example:
> 
> We first have
> 
> 7: A>B>C
> 7: B>C>A
> 6: C>A>B
> 
> which is an A>B>C>A cycle where the resistant set is {A} since A~>B,   
> B~>C; and the Smith set is {A, B, C}, so the MSR set is {A}.  
> 
> Now raise A to the top on one of the BCA ballots to get
> 
> 8: A>B>C
> 6: B>C>A
> 6: C>A>B
> 
> The resistant set is {A, C} and the Smith set is {A, B, C}, so the MSR 
> set is {A, C}. But C beats A pairwise 12 to 8, so the majority criterion 
> requires that C wins if B is eliminated.  

> 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 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.

Thanks for the answers
Gustav