Note that the poundstone example has the candidate sites offset from the centers of support. It turns out that this is inevitable in a three candidate example, i.e. if the respective supports are centered and concentrated around the three respective candidates and preferences respect distances, then there can be no Condoret Cycle.
So my four candidate geometrical example of a Condorcet cycle is minimal if you require the supporters to be concentrated around the candidate positions. > Date: Tue, 15 Dec 2009 11:10:53 -0500 > From: Warren Smith > To: election-methods > Subject: [EM] Simmons Condorcet Cycle geometric example > Message-ID: > > Content-Type: text/plain; charset=ISO-8859-1 > > > Here's a natural scenario that yields an exact Condorcet Tie: > > A together with 39 supporters at the point (0,2) > B together with 19 supporters at (0,0) > C together with 19 supporters at (1,0) > D together with 19 supporters at (4,2) > > D is a Condorcet loser. > A beats B beats C beats A, 60 to 40 in every case. > > Think of four cities represented by A, B, C, and D respectively, > with > the pairwise distances in miles calculated from the above planar > coordinates: > > A B C D > D 4 20^(1/2) 14^(1/2) > C 5^(1/2) 1 > B 2 > > Assuming voters prefer candidates closer to them, the ballots are: > > 40: A>B>C>D > 20: B>C>A>D > 20: C>B>A>D > 20: D>C>A>B > > ---- See http://rangevoting.org/CondorcetCycles.html > and there is a pretty picture there, basically taken from > Poundstone's book, > showing how a Condorcet cycle can arise in a geometrical manner > like this. > His cycle maybe is not as nice mathematically as Simmons', but > it works. ---- Election-Methods mailing list - see http://electorama.com/em for list info
