[EM] My Matrix for Kemeny's Rule, n=3

[barnes99]

Markus:

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.

SB

>From: Markus Schulze <markus.schulze at alumni.tu-berlin.de>
>To: election-methods-list <election-methods-list at eskimo.com>
>Subject: RE: [EM] My Matrix for Kemeny's Rule, n=3
>
>Dear SB,
>
>you wrote (2 Jan 2002):
>> 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."
>
>Markus Schulze

Steve Barney

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