[EM] Yee diagrams and Condorcet

[fsimmons at pcc.edu]

Here's how to prove it with a non-euclidean metric.

1. Assume that we are dealing with a metric (like all of the L_p metrics) whose "balls" are symmetric 
with respect to the origin.

2.  Suppose that AB and DE are line segments with the same midpoint C, and define the set

pi={x| d(x,A)=s(x,B)}

(analogous to the perpendicular bisector plane).

It follows that this pi still divides space into two sides (the A side and the B side), and that D and E will 
be on opposite sides of pi (unless they are both on the boundary set pi itself).

3.  From this it follows that a distribution with center symmetry through C will have the same weight on 
both sides of pi.

4.  If we get pi' by using A' and B' instead of A and B, where A'B' is parallel to AB and with the same 
length
d(A,B)=d(A',B'), then if  A' is closer to C than B', it follows that the A' side of pi' has all of the weight of 
the A side of pi, plus all of the weight between pi and pi'.  The additional weight was lost from the B side 
of pi.  therefore, the A' side has more weight than the B' side of pi'.

5. Therefore in a pairwise contest the candidate A' nearer to C gets more votes than the more distant 
candidate B'.



----- Original Message -----
From: Jameson Quinn 
> That proof assumes a euclidean distance metric. With a non-
> Euclidean one,
> the "planes" could have kinks in them. I believe I have heard 
> that the
> result still holds with, for instance, a city-block metric, but 
> I cannot
> intuitively demonstrate it to myself by imagining volumes and 
> planes as in
> this proof.
> 
> JQ