Talk:Schulze method: Difference between revisions
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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)
: Schulze doesn't drop candidates, just defeats. E.g. suppose the initial Schwartz set is {ABCD}. Then that means there must be a cycle among these candidates. Schulze drops the weakest defeat involving these four candidates. This resolves a part of the cycle (e.g. suppose the four-cycle has an embedded ABCA cycle, and it breaks C>A; then A>B and B>C is transitive). As a consequence, someone gets booted out of the Schwartz set (e.g. if there's an ABCA cycle and D beats C as well, then C gets booted out of the Schwartz set since no undropped defeat has C beating someone else). Rinse and repeat until everybody but one candidate is booted out, or the candidates remaining all tie each other. If it's the latter, then you have a tie for victory, otherwise the remaining candidate wins. [[User:Kristomun|Kristomun]] ([[User talk:Kristomun|talk]]) 21:56, 25 February 2020 (UTC)
== Clearer Path Heuristic Procedure Section ==
I think the Path Heuristic Procedure section could be a lot clearer if it was written in plain English, rather than math/logic notation. Unless I'm misunderstanding something, I think the following should have an equivalent effect:
You can find the strength of a path from candidate X to candidate Y by creating a sequence of candidates with the following 4 properties:
# The sequence starts with candidate X.
# The sequence ends with candidate Y.
# Each candidate in the sequence is preferred by more voters than the next candidate.
# There are no repeated candidates.
The strength of any given path is equivalent to the smallest margin of victory between two sequential candidates on that path, unless no valid sequence can be constructed, in which case the path from candidate X to candidate Y is said to have a strength of 0.
Candidate X is only better than candidate Y if the strongest path from X to Y is greater than the strongest path from Y to X.
Candidate X can only be a potential winner if their strongest path to every other candidate is at least as strong as that candidate's path to them.
--[[User:Epsilon Rose|Epsilon Rose]] ([[User talk:Epsilon Rose|talk]]) 20:30, 26 March 2021 (UTC)
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