FpA-fpC: Difference between revisions
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* If the Smith set is size two, then the winner is according to majority rule. |
* If the Smith set is size two, then the winner is according to majority rule. |
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* If the Smith set is size three, then for each candidate, assume without loss of generality that the candidate is A in an A>B>C>A cycle. A's score is A's first preferences minus C's first preferences. The candidate with the highest score wins. |
* If the Smith set is size three, then for each candidate, assume without loss of generality that the candidate is A in an A>B>C>A cycle. A's score is A's first preferences minus C's first preferences. The candidate with the highest score wins. |
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A more concise variant that breaks size-two Smith sets differently is:<ref>{{cite web|url=http://lists.electorama.com/pipermail/election-methods-electorama.com/2022-January/003409.html|title=A more elegant three-candidate fpA-fpC phrasing, inspired by Heaviside formulation|website=Election-methods mailing list archives|date=2022-01-20|last=Munsterhjelm|first=K.}}</ref> |
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* Let A be the candidate whose score is to be evaluated. Then A's score is the sum, over all candidates B who A pairwise beats, B's first preferences plus two times A's first preferences. |
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* The candidate with the highest score wins. |
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This method shares the strategy resistance of Smith-IRV hybrids, such as [[chicken dilemma]] compliance and [[dominant mutual third burial resistance]]; yet, unlike them, is monotone. It is open (not obvious) how to extend the method to more than three candidates in a way that retains both monotonicity and strategy resistance.<ref group="fn" name="Footnote">Any generalization will preserve its [[chicken dilemma]] compliance, as that criterion is only defined on three-candidate elections. However, this is not true of [[dominant mutual third burial resistance]].</ref> |
This method shares the strategy resistance of Smith-IRV hybrids, such as [[chicken dilemma]] compliance and [[dominant mutual third burial resistance]]; yet, unlike them, is monotone. It is open (not obvious) how to extend the method to more than three candidates in a way that retains both monotonicity and strategy resistance.<ref group="fn" name="Footnote">Any generalization will preserve its [[chicken dilemma]] compliance, as that criterion is only defined on three-candidate elections. However, this is not true of [[dominant mutual third burial resistance]].</ref> |
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Revision as of 18:20, 14 February 2022
fpA-fpC (for first preference A minus first preference C) is a three-candidate Condorcet method based on first preference Copeland.[1] Its election cases are:
- If there's a Condorcet winner, then that candidate wins.
- If the Smith set is size two, then the winner is according to majority rule.
- If the Smith set is size three, then for each candidate, assume without loss of generality that the candidate is A in an A>B>C>A cycle. A's score is A's first preferences minus C's first preferences. The candidate with the highest score wins.
This method shares the strategy resistance of Smith-IRV hybrids, such as chicken dilemma compliance and dominant mutual third burial resistance; yet, unlike them, is monotone. It is open (not obvious) how to extend the method to more than three candidates in a way that retains both monotonicity and strategy resistance.[fn 1]
It produces similar results to Condorcet,IFPP.
This page is a stub - please add to it.
Notes
- ↑ Any generalization will preserve its chicken dilemma compliance, as that criterion is only defined on three-candidate elections. However, this is not true of dominant mutual third burial resistance.
References
- ↑ Munsterhjelm, K. (2016-02-07). "Strategy-resistant monotone methods". Election-methods mailing list archives.