Talk:Schwartz set

message: The article says: "If there are any weak Condorcet winners in the election, then the Schwartz set consists of exactly all of them." This statement is incorrect. Example:
* A pairwise beats B.
* B pairwise beats C.
* C pairwise beats A.
* A pairwise ties with D.
* B pairwise ties with D.
* C pairwise ties with D.
Only candidate D is a weak Condorcet winner. However, the Schwartz set is { A, B, C, D }. Markus Schulze 20:07, 17 January 2006 (UTC)
replies: text: You're right. I'll withdraw the statement. -- Dissident (Talk) 19:24, 18 January 2006 (UTC)
thread_title:  Weak Condorcet winners 

message: " 3 voters preferring candidate A to B to C
     1 voter preferring candidate B to C to A
     1 voter preferring candidate C to A to B
     1 voter preferring candidate C to B to A
     
     then we have A pairwise beating B, B pairwise beating C, and A tying with C in their pairwise comparison, making A the only member of the Schwartz set, 
     while the Smith set on the other hand consists of all the candidates"
-
This statement is doubly inaccurate. Firstly, B does not beat C. B is preferrred twice to C, and C is likewise preferred twice to B. They tie. To see this more clearly (and this type of illustration should be used in whatever example is eventually decided upon):
Preference Tally (The value in each cell denotes how many voters prefer the candidate in the cell's column to the candidate in the cells row):
          A   B   C
       A| - | 4 | 3 |
       B| 2 | - | 2 |
       C| 3 | 2 | - |
       
Preference Result (In each cell 1 and 0 represent the truth value of the statement that the candidate in the associated column beats the candidate in the associated row. 1 indicates that it is true, 0 indicates that the contrary is true. T indicates that they tie.):
          A   B   C
       A| - | 1 | T |
       B| 0 | - | T |
       C| T | T | - |
Clearly A and C Tie, B and C tie, and A beats B. So A could not be the "only member of the Schwartz set", because A ties with C, and A and C are both unbeaten by any other candidate. Even so, the example still appears to  be good, provided that the summarizing statement is changed to reflect reality and that the preference tally and preference result tables are added, so as to prevent the reader from having to tally the results himself.
--Comiscuous
:You are mistaken. The example clearly shows that 4 people prefer B to C and only 2 people prefer C to B. -- Dissident (Talk) 01:56, 25 July 2006 (UTC)
thread_title:  Example given is incorrect 

message: ThurnerRupert removed the fact that the Schulze method satisfies the Schwartz criterion . His argument: "please state all complying methods, like Llull-Approval Voting, etc. or leave it out." However, the Schulze method is the only single-winner election method that satisfies the Schwartz criterion and that has its own Wikipedia article. Markus Schulze 11:26, 28 July 2011 (UTC)
replies: text: Late, but: pretty much every Smith-efficient system can be made Schwartz-efficient, by replacing the words "less than" with "less than or equal to" at some point in the definition. The two sets only differ in the case of tied votes. –Maximum Limelihood Estimator 16:54, 15 April 2024 (UTC)
thread_title:  Schulze method