[EM] Correction. Big CS fault?

[Forest Simmons]

On Wed, 18 Dec 2002, Forest Simmons wrote:

>
> By the way, it turns out that in the three candidate case, if the
> preference ballots are generated in this way, regardless of the metric
> used in step 2, a CW is assured; there can be no cycle.
>
> So somewhere in this 5 step process the cyclical
> "contradiction" is eliminated automatically.
>
> All Condorcet methods give the same result!
>
> Whether the same can be said for four or more candidates remains to be
> seen.
>

The same cannot be said for four or more candidates as the following
counterexample shows:

Put candidates A,B,C, and D at the vertices of a tetrahedron
whose respective edges AB, AC, AD, BC, BD, and DC have lengths of
5,7,8,9,4, and 6, respectively.

If A, B, C, and D have 5, 4, 3, and 1 avid supporters, respectively,
located at the exact same positions as their favorite candidates, then
there are four factions as follows:

5 ABCD
4 BDAC
3 CDAB
1 DBCA

Then A beats B beats C beats D beats A.

This shows that "beat cycles" cannot be eliminated by assigning
preferences to voters based on the location of their favorites in
candidate space.

So I withdraw my suggestion of using candidate space in that particular
manner.

My new suggestion based on moving candidate vectors into voter space via
the "most natural" isomorphism, looks much more promising.

See my posting under the title "Candidate-Space Method" dated December 26,
2002.

Forest

P.S. The minimum total distance criterion would give the win to A in the
above example. IRV also picks A. Who would win under the rules of Ranked
Pairs?

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