[EM] Re: Sprucing up vs. Condorcet Lottery vs. immunity: The "twisted prism" example

[Ted Stern]

On 6 Jan 2005 at 01:32 PST, Jobst Heitzig wrote:
> Dear Forest!
>
> Your sprucing up technique is a very nice idea since it can simplify the 
> tallying of those methods which fulfil beat-clone-proofness and 
> uncoveredness. However, some of which you wrote has confused me 
> completely: Did I understand you right in that you claim that the 
> technique should reduce everything to the 3-candidate case? I wonder how 
> that could possibly be when, in general, there is neither a proper beat 
> clone set nor a covered element! Second, I don't understand how the 
> method I posted yesterday under the name "Condorcet Lottery" could be 
> the same as anything based on Random Ballot???
>
> In order to study these questions I provide the following example which 
> is also interesting with respect to immunity:
>
>
> The "twisted prism" example:
> ----------------------------
> Six candidates A1,A2,A3,B1,B2,B3, 15 voters. Sincere preferences:
> 4 A1>B2>A2>B1>B3>A3
> 4 A2>B3>A3>B2>B1>A1
> 4 A3>B1>A1>B3>B2>A2
> 1 B1>B3>B2>A1>A2>A3
> 1 B2>B1>B3>A2>A3>A1
> 1 B3>B2>B1>A3>A1>A2
> Defeats:
> A1>A2>A3>A1, strength 10 (the "upper clockwise" 3-cycle)
> B1<B2<B3<B1, strength 10 (the "lower counter-clockwise" 3-cycle)
> Bi>Ai, strength 9 (the 3 "straight upward" beats)
> Ai>Bj (i!=j), strength 8 (the 6 "diagonal downward" beats)

Jobst, you made a small mistake.  I find the pairwise matrix to be the
following:

    A1  A2  A3  B1  B2  B3
A1   -  10   5   4   8   8
A2   5   -  10   8   4   8
A3  10   5   -   8   8   4
B1  11   7   7   -   5  10
B2   7  11   7  10   -   5
B3   7   7  11   5  10   -

So your defeat rankings should actually be

        Bi>Ai       , strength 11 (the 3 "straight upward" beats)

        A1>A2>A3>A1 , strength 10 (the "upper clockwise" 3-cycle)

        B1<B2<B3<B1 , strength 10 (the "lower counter-clockwise" 3-cycle)

        Ai>Bj (i!=j), strength 8  (the 6 "diagonal downward" beats)

I think other aspects of this example still hold, though.

Is there an automated way to find the Dutta set?  I've found this reference:

   Dutta's Minimal Covering Set and Shapley's Saddles
   John Duggan and Michel Le Breton
   Journal of Economic Theory, 1996, vol. 70, issue 1, pages 257-265

but I don't have access to the article to see if it contains an algorithm.

Ted
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