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:

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.

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.

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).

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.

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

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

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

That does answer my question, thanks.

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.

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.

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.

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

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.

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

On 2026-05-10 22:25, Gustav Thorzen via Election-Methods wrote:

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.

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.

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.

[snip]

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.

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