[EM] My Matrix for Kemeny's Rule, n=3
Markus Schulze
markus.schulze at alumni.tu-berlin.de
Sat Jan 4 03:45:41 PST 2003
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
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