Copeland's method: Difference between revisions
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== Criteria == |
== Criteria == |
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Copeland's method passes the [[Smith criterion]] because any candidate in the Smith set by definition beats everybody outside of the Smith set, but no candidate outside of it does so. Thus, every candidate in the Smith set must have a greater Copeland score than any candidate outside of it. Furthermore, the Copeland ranking of candidates (the ordering of candidates based on Copeland score) is a [[Smith set ranking]], since every candidate in the nth Smith set defeats every candidate in the (n+1)-th Smith set, and thus a candidate from the former set always has a greater Copeland score than a candidate from the latter. |
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| ⚫ | (Example showing Smith members having only 2 points more than non-Smith members: Suppose there are two candidates, one of whom is the Condorcet winner, and thus the only candidate in the Smith set. The CW has one victory and no defeats for a Copeland score of 1, while the other candidate has no victories and one defeat for a score of -1.) |
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| ⚫ | Copeland's method also passes [[ISDA]]; the |
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| ⚫ | Copeland's method also passes [[ISDA]]; the first paragraph proves that all candidates in the Smith set must have higher Copeland scores than all candidates not in the Smith set, and since by definition candidates in the Smith set have a pairwise victory (and thus no pairwise defeat) against every candidate not in the Smith set, adding or removing any number of candidates not in the Smith set will only result in every candidate in the Smith set having that number of pairwise victories added or subtracted from their total (with no change to their number of pairwise defeats); since the original Copeland winner must have had a higher Copeland score than all other Smith set candidates in order to win, they will still have a higher Copeland score and thus still win. |
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The Copeland ranking of candidates (the ordering of candidates based on Copeland score) is a [[Smith set ranking]]. This is because in general, a candidate in the n-th Smith set (if n is 1, this is the Smith set. If n is 2, this is the secondary Smith set, which is the set of candidates that would be in the Smith set if all the candidates in the Smith set were eliminated from the election. If n is 3, this is the tertiary Smith set, the set of candidates that would be in the Smith set ignoring all candidates in the Smith set and secondary Smith set, etc.) will have pairwise victories against at least all candidates in k-th Smith sets (for any value of k which is greater than n), and have pairwise defeats against at most all candidates in j-th Smith sets (for any non-negative value of j which is smaller than n) and all but two candidates in the n-th Smith set, which reduces to no defeats against any candidates in the n-th Smith set when the n-th Smith set is smaller than 3 candidates (since they can't be beaten by themselves, and must not be beaten by everyone else in their Smith set in order to be in it, and there are guaranteed to be no pairwise defeats for candidates in a given Smith set against other candidates in that same Smith set when that Smith set is either one or two candidates large), while a candidate in any k-th Smith set will have pairwise victories against at most all candidates other than themselves in k-th Smith sets, and will have pairwise defeats against at least all candidates in n-th or j-th Smith sets. Mathematically, this can be represented as minimal/maximal Copeland scores of (K-J-N+1) and (K-J-N-1) respectively, with K being the number of candidates in all Smith sets "after" the n-th Smith set, J for all Smith sets "before" the n-th Smith set, and N for the candidates in the n-th Smith set. Therefore, the Copeland score of a candidate in the n-th Smith set is guaranteed to be at least 2 points higher than a candidate in any of the k-th (after or below n) Smith sets. Similar reasoning shows that a candidate in any of the j-th (before or above n) Smith sets is at least 2 points higher than any candidate in the n-th Smith set. |
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| ⚫ | Further, Copeland always elects from the [[uncovered set]], and the Copeland ranking is an uncovered set ranking. This is because when one candidate covers another, the former candidate pairwise beats all candidates pairwise beaten by the latter candidate, and also either pairwise beats the latter candidate or beats someone who beats the latter candidate. Because of this, the covering candidate will have a minimum Copeland score of ((number of candidates beaten by latter candidate) + 1) - (number of candidates beating former candidate)), and the covered candidate will have a maximal Copeland score of ((number of candidates beaten by latter candidate) - ((number of candidates beating former candidate) + 1), resulting in the covering candidate having at least 2 more points than the covered candidate. This type of logic can be used to simplify the above Smith set-related proofs too. |
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| ⚫ | (Example showing Smith members having only 2 points more than non-Smith members: Suppose there are two candidates, one of whom is the Condorcet winner, and thus the only candidate in the Smith set. The CW has one victory and no defeats for a Copeland score of 1, while the other candidate has no victories and one defeat for a score of -1.) |
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== Generalizations== |
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| ⚫ | Further, Copeland always elects from the [[ |
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Zoghi et al have developed a multi-armed bandit variant of Copeland's method. It can be used to determine a winner in a multi-armed bandit setting, even if a Condorcet winner does not necessarily exist.<ref>{{Cite journal|last=Zoghi|first=Masrour|last2=Karnin|first2=Zohar|last3=Whiteson|first3=Shimon|last4=de Rijke|first4=Maarten|date=2015-05-31|title=Copeland Dueling Bandits|url=http://arxiv.org/abs/1506.00312|journal=arXiv:1506.00312 [cs]}}</ref> |
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== See also == |
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Revision as of 13:00, 15 April 2020
Copeland's method is a Smith-efficient[1] Condorcet method in which the winner is determined by finding the candidate with the most (pairwise victories minus pairwise defeats), known as their Copeland score. It was invented by Ramon Llull in his 1299 treatise Ars Electionis, but his form only counted pairwise victories and not defeats (which could lead to a different result in the case of a pairwise tie).[2]
(Some alternative versions of Copeland don't count pairwise defeats, and others give each candidate in a pairwise tie half a point each.)
Proponents argue that this method is more understandable to the general populace, which is generally familiar with the sporting equivalent. In many team sports, the teams with the greatest number of victories in regular season matchups make it to the playoffs.
This method often leads to ties in cases when there are multiple members of the Smith set; specifically, there must be at least five or more candidates in the Smith set in order for Copeland to not produce a tie for winner unless there are pairwise ties. Critics argue that it also puts too much emphasis on the quantity of pairwise victories rather than the magnitude of those victories (or conversely, of the defeats).
Example:
25 A>B>C 40 B>C>A 35 C>A>B
A beats B beats C beats A, so there is a Condorcet cycle between all candidates. Each candidate has one pairwise victory and one defeat, so their Copeland scores are all 0, thus there is a tie. This example demonstrates why Copeland is almost never used for actual elections; it can guarantee someone in the Smith set will win, but says much less about who.
Criteria
Copeland's method passes the Smith criterion because any candidate in the Smith set by definition beats everybody outside of the Smith set, but no candidate outside of it does so. Thus, every candidate in the Smith set must have a greater Copeland score than any candidate outside of it. Furthermore, the Copeland ranking of candidates (the ordering of candidates based on Copeland score) is a Smith set ranking, since every candidate in the nth Smith set defeats every candidate in the (n+1)-th Smith set, and thus a candidate from the former set always has a greater Copeland score than a candidate from the latter.
(Example showing Smith members having only 2 points more than non-Smith members: Suppose there are two candidates, one of whom is the Condorcet winner, and thus the only candidate in the Smith set. The CW has one victory and no defeats for a Copeland score of 1, while the other candidate has no victories and one defeat for a score of -1.)
Copeland's method also passes ISDA; the first paragraph proves that all candidates in the Smith set must have higher Copeland scores than all candidates not in the Smith set, and since by definition candidates in the Smith set have a pairwise victory (and thus no pairwise defeat) against every candidate not in the Smith set, adding or removing any number of candidates not in the Smith set will only result in every candidate in the Smith set having that number of pairwise victories added or subtracted from their total (with no change to their number of pairwise defeats); since the original Copeland winner must have had a higher Copeland score than all other Smith set candidates in order to win, they will still have a higher Copeland score and thus still win.
Further, Copeland always elects from the uncovered set, and the Copeland ranking is an uncovered set ranking. This is because when one candidate covers another, the former candidate pairwise beats all candidates pairwise beaten by the latter candidate, and also either pairwise beats the latter candidate or beats someone who beats the latter candidate. Because of this, the covering candidate will have a minimum Copeland score of ((number of candidates beaten by latter candidate) + 1) - (number of candidates beating former candidate)), and the covered candidate will have a maximal Copeland score of ((number of candidates beaten by latter candidate) - ((number of candidates beating former candidate) + 1), resulting in the covering candidate having at least 2 more points than the covered candidate. This type of logic can be used to simplify the above Smith set-related proofs too.
Generalizations
Zoghi et al have developed a multi-armed bandit variant of Copeland's method. It can be used to determine a winner in a multi-armed bandit setting, even if a Condorcet winner does not necessarily exist.[3]
See also
- E Stensholt, "Nonmonotonicity in AV"; Electoral Reform Society Voting matters - Issue 15, June 2002 (online).
- A.H. Copeland, A 'reasonable' social welfare function, Seminar on Mathematics in Social Sciences, University of Michigan, 1951.
- V.R. Merlin, and D.G. Saari, "Copeland Method. II. Manipulation, Monotonicity, and Paradoxes"; Journal of Economic Theory; Vol. 72, No. 1; January, 1997; 148-172.
- D.G. Saari. and V.R. Merlin, 'The Copeland Method. I. Relationships and the Dictionary'; Economic Theory; Vol. 8, No. l; June, 1996; 51-76.
References
- ↑ http://dss.in.tum.de/files/brandt-research/choicesets.pdf "The Copeland set C is given by [...] i.e., the set of alternatives with maximal Copeland score." "Theorem 1. The Copeland set [...] [is] contained in the Smith set."
- ↑ Colomer, Josep (2013). "Ramon Llull: From Ars Electionis to Social Choice Theory". Social Choice and Welfare. 40 (2): 317-328. doi:10.1007/s00355-011-0598-2.
- ↑ Zoghi, Masrour; Karnin, Zohar; Whiteson, Shimon; de Rijke, Maarten (2015-05-31). "Copeland Dueling Bandits". arXiv:1506.00312 [cs].