# [EM] Geometric Condorcet cycle example, improved

fsimmons at pcc.edu fsimmons at pcc.edu
Wed Dec 30 14:31:30 PST 2009

```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: A>C>B
y:B>C>A
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: A>B>C
y: B>C>A
z: C>A>B

which leads to a Condorcet cycle whenever no candidate has a majority.

```