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

On 2024-02-19 18:13, Kristofer Munsterhjelm wrote: > 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. I now think that my claim that this is monotone if the social ordering is provided by a method that passes strong monotonicity - is wrong. Closed: if that's right, then what I said in my mail to you about how this could be used with Range to get monotonicity is also wrong. Clone independence might still be preserved, though; I haven't checked. More to come once I've investigated further. -km