[EM] Resistant Set Questions

[Kristofer Munsterhjelm]

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.

(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.

> 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:

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).
> 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:

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.

> 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.

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.

-km