Dear Steve Barney, you wrote (3 Jan 2003): > Please show me the arithmetic of determining the Kemeny outcome in your > example, and explain why the higher score of 311 for ADBC is, apparently, > better than 313 for DABC. Since we are measuring "distance" from unanimity, > I thought the lower score would be the better. > > Markus Schulze wrote (2 Jan 2003): > > Steve Barney wrote (2 Jan 2003): > > > Now, is Kemeny's Rule, as defined by my matrix for 3 candidate > > > tallies, the same as the Condorcet method which is described on > > > the EM website <http://electionmethods.org/>? > > > > Nope! > > > > Example: > > > > A:B=52:48 > > A:C=53:47 > > A:D=49:51 > > B:C=56:44 > > B:D=45:55 > > C:D=54:46 > > > > My beat path method chooses the ranking ADBC. However, the > > Kemeny score of ADBC is only 311 while the Kemeny score of > > DABC is 313. Or as Forest Simmons would say: "The beat path > > method is not locally Kemeny optimal."
The ranking ADBC contains the pairwise defeats A > D, A > B, A > C, D > B, D > C, and B > C. When you add the numbers of voters who agree to the corresponding pairwise defeats, then you get 49 + 52 + 53 + 55 + 46 + 56 = 311. Therefore, the Kemeny score of the ranking ADBC is 311. The ranking DABC contains the pairwise defeats D > A, D > B, D > C, A > B, A > C, and B > C. When you add the numbers of voters who agree to the corresponding pairwise defeats, then you get 51 + 55 + 46 + 52 + 53 + 56 = 313. Therefore, the Kemeny score of the ranking DABC is 313. As the ranking DABC has a higher (= better) Kemeny score than the ranking ADBC, the ranking ADBC cannot be the Kemeny outcome. [I guess that there are different ways to define and to calculate the Kemeny score of a given ranking. However, all definitions must lead to the same Kemeny outcome. Especially it doesn't make any difference whether (1) the absolute number of voters who agree to a given pairwise defeat or (2) the margin of this pairwise defeat is being used.] Markus Schulze ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
