[EM] Simmons Condorcet Cycle geometric example

[Warren Smith]

> 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.

-- 
Warren D. Smith
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and
math.temple.edu/~wds/homepage/works.html