[EM] An odd case for Ranked-Pairs

[Eric Gorr]

>Dear Eric,
>
>you wrote (14 Jan 2003):
>>  This example was recently brought to my attention.  Consider:
>>
>>     A>C>B>F>D>E
>>     B>C>E>F>D>A
>>     D>B>A>F>E>C
>>     E>A>B>C>F>D
>>     E>D>A>B>C>F
>>     F>C>D>A>B>E
>>
>>  The pairwise matrix is:
>>
>>     0 4 4 2 3 4
>>     2 0 4 3 4 5
>>     2 2 0 4 3 4
>>     4 3 2 0 3 2
>>     3 2 3 3 0 3
>>     2 1 2 4 3 0
>>
>>  Now, with Ranked-Pairs, the only kept-defeat will be B:F.
>>
>>  However, once all of the defeats have been considered, there will be
>>  no kept-defeats for <someone>:E, which allows E to participate in a
>>  tie (according to my computations).
>
>No oddities. Due to my computations, the defeat B:E=4:2 will be locked
>in any case. Therefore, candidate E cannot be a (decisive or random)
>Ranked Pairs winner.

Would you care to point to the algorithm you used that would pick B:E 
as a kept-defeat.

-- 
== Eric Gorr ========= http://www.ericgorr.net ========= ICQ:9293199 ===
"Therefore the considerations of the intelligent always include both
benefit and harm." - Sun Tzu
== Insults, like violence, are the last refuge of the incompetent... ===

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