Algorithmic Asset Voting: Difference between revisions
→The multi-winner Smith Set and Smith-efficient cycle resolution
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The use of the [[KP transform]] on rated ballots and then converting those Approval ballots to ranked ballots allows voters to submit, for example, a rated ballot A5 B4 C3, and have it treated either as 0.2 votes A>B, 0.2 votes B>C, and 0.4 votes A>C (score for preferred candidate minus score for less preferred candidate divided by max score yields the number of votes in each matchup) or 1 vote A>B, 1 vote B>C, 1 vote A>C.
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. Example: <blockquote>5-winner election:
51: Party A candidates
20: Party B candidates
20: Party C
9: split between random candidates, call them D, E, F, G
Winner set (5 A’s) vs. (B, D, E, F, G)
I’d argue that the winning winner set of the pairwise matchup here is actually (4 A’s, B). The B voters have enough votes to force B to be one of the top 5 candidates in a iterated cumulative voting election or an Asset negotiation, and no voters can force an improvement from their perspective from there.
Edit: To be clear no improvement based on the candidates in either set is possible. If we take (4 A’s, B) and compare it to all remaining candidates, it’s clear the C voters can force 1 C to be among the top 5 as well, and that would be the equilibrium outcome, so the final Condorcet PR winner set is 3-1-1 A-B-C.<ref>https://forum.electionscience.org/t/monroe-pr-doesnt-work-properly/528/9?u=assetvotingadvocacy</ref></blockquote>
== The multi-winner Smith Set and Smith-efficient cycle resolution ==
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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 [[Sequentially Spent Score|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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