Algorithmic Asset Voting: Difference between revisions
→The multi-winner Smith Set and Smith-efficient cycle resolution
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Some optional assumptions are:
* When a negotiator is indifferent between certain outcomes (i.e. because their voters equally ranked those outcomes), they use their assets to help pick the socially best of those outcomes. As an example, if the voters cast rated ballots <blockquote>49 A5 B4 3 A5 B5
*
*
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'''26 A
'''26.5 C
'''26.5 D'''
21 S
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If there are 51 voters who prefer Party A and 49 for Party B, then in a 5-winner election, the 49 B voters can always ensure two of their candidates are in the final winner set.
Chicken dilemma:
{| class="wikitable"
!Number
!Ballots
|-
|34
|A>B>C
|-
|25
|B>A>C
|-
|40
|C
|}
C has the most 1st choices, so A>C preferring voters could shift their votes to A, at which point no voter can improve the winner. Note that if voters had shifted their votes to B to beat C first, then some special steps would need to be applied to elect the Condorcet winner A. <ref>https://www.reddit.com/r/EndFPTP/comments/ewgjss/comment/fg2fd63</ref>
{| class="wikitable"
!Number
!Ballots
|-
|1
|A
|-
|34
|A>B>C
|-
|25
|B>A>C
|-
|40
|C
|}
<br />
== Explanation of how Asset Voting is, under certain assumptions, a Condorcet method (and how this enables it to be done as an algorithm) ==
[[File:Asset Condorcet Winner example modified.png|thumb|400x400px|Algorithmic Asset to elect a Condorcet winner. ]]
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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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It may be possible to make AAV always elect proportionally from Droop semi-solid coalitions. If so, then this may greatly speed up computation of the result, as only winner sets that satisfy all semi-solid coalitions need be considered; discovering semi-solid coalitions can be done by checking ballots and observing which candidates are ranked higher than others and by which voters. <ref>https://www.removeddit.com/r/EndFPTP/comments/euiup2/a_new_pr_concept_of_semisolid_coalitions/</ref> Semi-solid coalitions may overlap.
It's important to keep in mind that when discussing coalitions in the context of voter preferences as recorded on ballots, the loosest possible definition would be groups of voters who mutually prefer at least one candidate above at least one other candidate, though there may be disagreement on other candidates. Thus, even if the only thing two groups of voters have in common is that they mutually rank some candidate last, they are still a coalition in the sense that they mutually prefer all other candidates to that last-ranked candidate.
== References ==
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