Copeland's method: Difference between revisions
Add more criterion failures.
(Categorize criterion compliances and failures, add independence of covered alternatives failure) |
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==Criterion failures==
=== Schwartz ===
In this election:
{{ballots|
1: A > B3 > B1 > B2 > C
1: B3 > B1 > B2 > C > A
2: C > A > B2 > B1 > B3
}}
The Schwartz winner is C, but Copeland elects A.
=== Independence of clones ===
Copeland's method is vulnerable to crowding. This alternative is [[w:Independence_of_clones_criterion#Copeland|due to Wikipedia]].
First consider the election
{{ballots|
1: A>B>C
1: B>C>A
2: C>A>B}}
C is the Condorcet winner and thus also the Copeland winner. Now clone B into B1, B2, and B3:
{{ballots|
1: A > B3 > B1 > B2 > C
1: B3 > B1 > B2 > C > A
2: C > A > B2 > B1 > B3
}}
The Copeland winner changes to A.
Since the Copeland winner is unique in both cases, every method that elects from the Copeland set must also fail clone independence.
=== Independence of covered alternatives ===
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