# Talk:Schulze method

I added FBC to the list of violated criteria. No one has yet taken an issue with my proof at [1]. -KVenzke

## Contents

## move from CSSD[edit source]

I moved CSSD to Schulze because this has also been done at WikiPedia and because I want to do this before some newbie does this in the same manner as at WikiPedia and destroys all the connections to the "talk" page, the "history", and the "watchlists". MarkusSchulze 04:27, 22 May 2005 (PDT)

## electionmethods.org[edit source]

This site now displays a page claiming that studying election methods is pointless because one of the collaborators apparently believes some wacky stuff, so the other author has decided that he's representative of all left-wing folks. --ClaireConnelly 00:18, 29 Nov 2005 (PST)

## Example 4[edit source]

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[edit source]

From there, the Schulze method operates as follows to select a winner (or create a ranked list):

Calculate the Schwartz set based only onundroppeddefeats. 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. 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. Kristomun (talk) 21:56, 25 February 2020 (UTC)