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? ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
