[EM] Possible resistant set cloneproofing template

[Kristofer Munsterhjelm]

Here's a cloneproofing idea I had. I'm basically just trying to 
generalize the IRV approach as much as possible to see if I can get a 
hang of how to do cloneproofing in a more general sense.

Suppose we have two elections, call them e_A and e_AC, before and after 
cloning A. Also suppose that we have two associated social orders (p_A, 
p_AC), that are equal except that the clones are ranked next to each other.

E.g. p_A might be B>A>C, and then after cloning A into A1, A2, A3, p_AC 
is B>A1_A2>A3>C.

Now consider the following method: repeatedly eliminate the lowest 
ranking remaining candidate, according to the social ordering, who does 
not disqualify someone else according to the remaining sub-elections.

That is, once a candidate X is eliminated, the disqualification 
relations are updated as if X never was a part of the race.

Stop eliminating when only a single candidate is left; that candidate is 
the winner.

This method is not very good since it obviously fails neutrality. But it 
passes the "proper order property" of my 2023-12-12 post on the 
resistant set, so it elects from the resistant set.

I *think* it passes clone independence, but I'm not sure. Here's my 
reasoning:

If we clone a candidate A, then the only thing that can happen is that 
some disqualification A ~> B is broken for all the A clones, due to A_x 
~(A_1, A_2, ..., B)~> B being false, i.e. in some sub-election featuring 
more than one A clone, none of the A clones individually have greater 
than 1/k support.

Suppose that B ~> C before cloning. Then B has greater than 1/k support 
in every sub-election involving both him and C. Cloning A doesn't change 
B's first preferences anywhere, so this is still true.

Suppose that B ~> A before cloning. Again, cloning A can't affect B's 
first preferences in any sub-election, and it can only lower each 
individual A clone's first preference count. Thus we still have B ~> A_x 
for all clones A_x.

So now let's consider the method/template.

I'll call a "restricted" disqualification one that operates on the 
currently remaining candidates, and for convenience, I'll use A ~> B for 
both unrestricted and restricted qualification - it should be clear 
which is which.

Suppose the winner according to e_A is B, and A is cloned. Then in any 
given round, we know that B is not eliminated, because A being cloned 
can't affect B ~> C or B ~> A and so can't affect it when restricted to 
continued candidates either.

Since B won before cloning, we know that in each round, B is protected 
from elimination either by being higher in the ranking or by 
disqualifying someone. Introducing clones of A doesn't change this: it 
can't change the restricted disqualifications, and it can't put the A 
clones higher in the ranking than B if they weren't already.

That takes care of teaming and crowding. As for vote-splitting:

Suppose the winner according to e_A is A, and A is cloned. Then in each 
round, either an A clone, or someone else is eliminated. The elimination 
order of the non-A candidates doesn't change since cloning A can't 
change B ~> C or B ~> A for any sub-election. Cloning A can't make A 
last longer than before either, since it can never establish A ~> X 
where there were no such disqualification before.

So eventually every A clone but one is eliminated. At that point, all 
restricted disqualifications by A that protect A from further 
eliminations are active again. (Basically, this is the same reasoning as 
for IRV: every A clone but one is eliminated, and then A can't be 
eliminated because he wasn't before.)

Does that sound right?




Here's Joshua Boehme's example, altered slightly to not have equal-rank:

8: A>B
6: B>A

4: A2>A1>B
4: A2>A1>B
3: B>A1>A2
3: B>A2>A1

Just for fun, let the orders be B>A and B>A1>A2 respectively, failing 
majority!

Before cloning, A ~> B so A is protected from elimination. B is 
eliminated and A wins.

After cloning: there are no disqualifications, so A2 is eliminated due 
to being ranked last in the order. This activates A1 ~> B. Second round: 
A1 is protected from elimination, and B is eliminated. Then A1 wins.

(Another interesting question: suppose the external ordering is provided 
by a method that is strongly monotone, where raising A can only move A 
up the ordering, with no other candidates changing places. Is this 
method then monotone? The simplest strongly monotone method is Range.)

-km