Algorithmic Asset Voting: Difference between revisions
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Note that in a pairwise match-up between two winner sets, a third winner set can actually emerge as the winner. This is because some voters may prefer some candidates from one set and some from the other. For example, in a matchup between a set with only the majority's preferred candidates and another set with mostly candidates with almost no support, but one candidate whom a quota prefer, the final winner will actually be a new winner set where most of the candidates are the majority's preferred candidates and the final candidate is the quota's preferred candidate.
== The multi-winner Smith Set and Smith-efficient cycle resolution ==
<blockquote>
8 D>A>B>C
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Initializing an algorithmic Asset negotiation, D and D2 have the most votes (1st choices). But within each Hare Quota, ABC and A2 B2 C2 are in a Condorcet cycle. Therefore, once a few negotiating moves (pairwise comparisons) have been done, D and D2 won’t be in any of the 2-winner winner sets the negotiators cycle through. For example, if the negotiators are currently supporting (B, B2), and D and D2 attempt to gather enough support to win, 12 ballots prefer B to 8 for D or D2, and the same for B2. So the Smith Set here is all of the outcomes in the Condorcet cycles, which is a proper subset of all possible 2-winner sets.[https://forum.electionscience.org/t/free-and-private-cardinal-polling/536/11]</blockquote>
Some reasoning for why this method is most likely Droop-proportional and even Hagenbach-Bischoff-proportional: mathematically, if any candidate has an HB quota, no matter how the votes are arranged among other candidates, the quota-preferred candidates can guarantee they are
It's possible to use another optimal or sequential PR method on the Smith Set winner sets to decide the final winner set. With optimal PR methods, this can be done by just picking the best winner set in the Smith Set, while with sequential PR methods, one way is to at times eliminate all candidates from contention whose election would guaranteeably result in a non-Smith winner set being the final result. As an example, suppose [[Sequentially Spent Score]] is used to determine the winner among the Smith Set winner sets in the following 2-winner example using scored ballots, with the max score being 10 and the rankings inferred from the scores: <blockquote>8 D:10>A:3>B:2>C:1
7 B:10>C:9>A:8>D:5
5 C:10>B:9>A:8>D:5
8 D2:10>A2:3>B2:2>C2:1
7 B2:10>C2:9>A2:8>D2:5
5 C2:10>B2:9>A2:8>D2:5</blockquote>Scores for each candidate are D and D2 140, A/A2 120, B/B2 131, and C/C2 121
D and D2 would be the winners in regular SSS, [[Sequential Monroe|SMV]], and [[Reweighted Range Voting|RRV]] because they each have the highest scores in their respective Hare Quotas. But the Smith Set here is all winner sets with at least 1 candidate from (A, B, C) and at least 1 candidate from (A2, B2, C2) since 12 voters prefer each of them over 8 voters for D and D2 each, which means D and D2 must not win in order for the winner set to satisfy the Smith criterion (this can be discovered using a combinatorics calculator such as https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html). Therefore, one way of applying SSS here would be to run SSS but first eliminate D and D2 to ensure they can't win, making one of B or B2 tie for the first seat. WLOG supposing B won, then A and C must be eliminated, because if one of them were to win the second seat, then it'd be impossible for 1 of (A2, B2, C2) to win, violating the Smith criterion. Then B2 wins the second seat, making the final winner set (B, B2). In the single-winner case, using cardinal PR methods to select from the Smith Set is equivalent to using [[Smith//Approval]] or [[Smith//Score]]. If a method such as [[STV]] is used to decide the winning set among the Smith winner sets, then it may even be necessary at times to prevent the elimination of a candidate who, if eliminated, would guarantee that a non-Smith winner set would be the final result.
== Resisting Favorite Betrayal and burying ==
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