Re: [EM] Geometric Condorcet cycle example, improved
Actually it's the symmetry property of metrics (d(p,q)=d(q,p)), not the triangle inequality, that guarantees the absence of a Condorcet cycle in the triangle case: Suppose that A, B, and C are the only candidates, and that voters are concentrated very near their favorites. Assume that preferences are determined by distances, i.e. a voter prefers a nearer candidate over a more distant one. Without loss in generality, assume that d(A,B) is the greatest of the three distances. Then we have the preference profile: x: ACB y:BCA z: C Suppose, by way of contradiction, that we have a Condorcet cycle. The without loss in generality suppose that the cycle is A beats B beats C beats A. The first step in this beat path ( A beats B) implies (x+z)y, while the second step (B beats C) implies y(x+z). We got into this contradiction by assuming the existence of a cycle. So that assumption is untenable. We have made tacit use of the symmetry property of distance by assuming that the longest side of the triangle was agreed upon by both A and B, i.e. A thought B was the most distant candidate, and B considered A to be the most distant candidate. To show that this property is essential, suppose that we use taxicab distance is a grid of one way streets, and that A, B, and C are located three of the corners [say (0,0), (1,0), and (0,1)] of a block about which the one way streets have a counterclockwise orientation. Then (by taxi) it is quicker to go from A to B than from A to C, from B to C than from B to A, and from C to A than from C to B. The preference profile has to be of the form... x: ABC y: BCA z: CAB which leads to a Condorcet cycle whenever no candidate has a majority. Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Geometric Condorcet cycle example, improved
Hi folks, I just recalled that four years ago I constructed a sophisticated example which is somewhat similar: http://lists.electorama.com/htdig.cgi/election-methods-electorama.com/2005-May/015982.html Happy New Year! Jobst Warren Smith schrieb: This point-set also works: A=(1,0) B=(0,4) C=(3,5) D=(9,2) Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Geometric Condorcet cycle example, improved
Warren,Thanks for upgrading my example to be valid for the L^1 through L^infinity cases. Of course, it's still true that four points are needed, since the proof that there is no example like this with three relies only on the triangle inequality, which is satisfied for any metric.Jobst,I remember that example, and I'm sure that it was a subconscious influence on my trapezoid example. Certainly, your point in the last paragraph Aren't such examples a good reason to always consider more than just onepiece of the information (like approval or ranks or defeats or directsupport) and instead combine them to get the whole picture?is the most important lesson to learn from such examples.Guten Rutsch,Forest - Original Message -From: Jobst Heitzig Date: Tuesday, December 29, 2009 4:51 amSubject: Re: [EM] Geometric Condorcet cycle example, improvedTo: Warren Smith Cc: election-methods , Forest W Simmons Hi folks, I just recalled that four years ago I constructed a sophisticated example which is somewhat similar: http://lists.electorama.com/htdig.cgi/election-methods- electorama.com/2005-May/015982.html Happy New Year! Jobst Warren Smith schrieb: This point-set also works: A=(1,0) B=(0,4) C=(3,5) D=(9,2) Election-Methods mailing list - see http://electorama.com/em for list info
