Here's a thought about how to get resistant set compliance and unrestricted domain. If it works, then that's a new combination: no other method that I know of has all three. And it would show that monotonicity is not incompatible with Resistant, despite IRV making it seem like they are. Let A inclusively disqualify B on a set S of candidates if, for every subelection corresponding to a subset of S, A disqualifies B. Let there be a disqualification path from A to X_n, A ==> X_1 ==> X_2 ==> ... ==> X_n if A inclusively disqualifies X_1 on {A, X_1, ..., X_n} X_1 inclusively disqualifies X_2 on {X_1, ..., X_n} and so on up to X_(n-1) inclusively disqualifies X_n on {X_(n-1), X_n}, i.e. X_(n-1) beats X_n pairwise. Call the set of candidates {A, X_1, ..., X_n} the path set for this disqualification path. The method is then simply: Elect the candidate with the longest disqualification path to someone else. Break ties by your monotone method of choice. Monotonicity compliance: Suppose A is the winner. Raising A can't break A~(S)~>B for any set S, so it can't break the first step of the disqualification path. Nor can it affect any of the subsequent steps because they are defined over sets that don't include A, and thus A's position makes no difference. So A's path can't get shorter. Nor can anybody else's path get longer, because raising A can't establish disqualifications, only break them. Resistant set: suppose B has a disqualification path to X_n but is not in the resistant set. By definition of the resistant set, there exists some A who disqualifies B in the full election. Thus A also inclusively disqualifies B on {A, B, ..., X_n}, where {B, ..., X_n} is B's path set for the disqualification path. Hence A has a longer disqualification path than B, so B doesn't win. It's definitely not cloneproof, though. Teaming: Suppose A and B tie for first, and B's disqualification path doesn't contain A. Clone A so that one of them inclusively disqualifies the others on A's longest path set (e.g. every voter ranks the clones in the same order). Then A's path set gets longer but B's doesn't, making A win. Crowding: Ditto, let A tie B and have B's path contain C, then we clone C similarly. Vote-splitting: Suppose A and B tie for first, A's longest path set doesn't contain B, and B's longest disqualification path is to A. Clone A so that none of the clones disqualify each other on A's longest path set, but one of them beats the others pairwise. Then B's path gets longer but A's doesn't. (For clone independence, we need another score than "longest", but one that preserves the property that if A ~> B, then B's score is less than A's.) What'd ya think? -km

On 2024-02-21 19:05, Kristofer Munsterhjelm wrote: > Here's a thought about how to get resistant set compliance and > unrestricted domain. If it works, then that's a new combination: no > other method that I know of has all three. And it would show that > monotonicity is not incompatible with Resistant, despite IRV making it > seem like they are. Replying to myself because I think this is busted. The problem is that the induction step doesn't hold - it's possible for B's disqualification path to contain A, and then A can't simply prepend his own disqualification of B and automatically get a longer path than B. Hence the resistant set compliance proof fails.[1] I suspect that the monotonicity proof also fails; unless I'm missing something, it could be possible that raising A could establish A ~(x)~> E for some subelection x, and so allow B's path to grow from, say, B ==> ... ==> A to B ==> ... ==> A ==> E without affecting the length of A's path. I have some ideas for other monotone methods, but I really need to test them. If they turn out not to work, I have an idea of how I can use integer programming to find a template for monotone four candidate elections that pass resistant and monotonicity, and hopefully learn from or try to generalize from that. There's just a lot of work between here and there. -km [1] A simple patch is to require that for each step of the path, the current head node is not disqualified by any candidate lower down the path, on the current path set and all subsets of it. But that doesn't fix the monotonicity problem.