Talk:Schulze method: Difference between revisions
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(Question about tie in fourth example) |
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== Example 4 ==
Example 4 ends with «Possible Schulze rankings are B > C > D > A, B > D > A > C, B > D > C > A, D > A > B > C, D > B > A > C, and D > B > C > A.». Perhaps someone could elaborate where to go from here when you really want a unique winner and not a tie?
== Possible mistake for Schwartz set heuristic ==
From there, the Schulze method operates as follows to select a winner (or create a ranked list):
Calculate the Schwartz set based only on '''undropped''' defeats.
If there are no defeats among the members of that set then they (plural in the case of a tie) win and the count ends.
Otherwise, drop the weakest defeat among the candidates of that set. Go to 1.
The part I am seeing a possible mistake in is the "undropped" part. As I understand it, the Schwartz set heuristic operates by eliminating everyone outside the Schwartz set, dropping the weakest defeat if there is any, and repeating until no more defeats can be dropped, at which point all of the remaining candidates are tied to win. An example where the two definitions might diverge is: if there is a Schwartz set of 7 candidates, and the weakest defeat among them is removed, this may result in an even smaller Schwartz set of 3 candidates being possible; but if the Schwartz set can only be calculated from undropped defeats, then this would not be allowed. [[User:BetterVotingAdvocacy|BetterVotingAdvocacy]] ([[User talk:BetterVotingAdvocacy|talk]]) 19:39, 25 February 2020 (UTC)
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