--- "Simmons, Forest" <[EMAIL PROTECTED]> wrote: > David Cary asked ... > > Does geometric consistency mean here that the candidates and voters > can be represented by points in some space so that a voter prefers > candidate A over Candidate B iff the voter is "closer" to A than to > B? > > Basically, yes, any metric on the candidates would do, though the > triangle inequality is not needed. >
So using the standard metric in the Cartesian plane, with three candidates positioned: A at (0,0) B at (4,4) C at (8,0) (A and C are furthest apart) A voter at the following positions would have the preferences: V1 at (1,1) prefers A>B>C V2 at (3,-3) prefers A>C>B V3 at (3,3) prefers B>A>C V4 at (5,3) prefers B>C>A V5 at (5,-3) prefers C>A>B V6 at (7,1) prefers C>B>A V7 at (2,-2) prefers A>(B=C) V8 at (4,3) prefers B>(A=C) V9 at (6,-2) prefers C>(A=B) With such a full range of preferences possible, it would seem that just about any ordinally ranked ballot is possible, and with an appropriate distribution of voters, just about any election result with 3 candidates is geometrically consistent. That includes a result with A beating B beating C beating A in pair-wise elections that does not produce a Condorcet winner. For example 2 voters at V1, 3 voters at V4, and 4 voters at V5. This example seems to contradict what I understood to be Forest's earlier claim that a geometrically consistent election with 3 candidates always produced a Condorcet winner. Am I missing something? __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com ---- election-methods mailing list - see http://electorama.com/em for list info