Re: [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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