[EM] Resistant, monotonicity, and UD

[Kristofer Munsterhjelm]

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