[EM] Geometric Condorcet cycle example, improved

[Warren Smith]

F.Simmons: 4 towns A,B,C,D with
respective town populations in proportion 2:1:1:1.
Preference profiles

40%  A>B>C>D
20%  B>C>A>D
20%  C>B>A>D
20%  D>C>A>B

Which makes D a Condorcet loser and creates the cycle
A beats B beats C beats A.

Warren D. Smith:
Each of the following lines gives a set of coordinates for the 4 towns
in the plane,
such that the above preference profile happens, using EITHER
Euclidean(L2), Taxicab(L1) or Linfinity (or any Lp) distance metric
(all work).
A=(4,0) B=(0,2) C=(2,5) D=(8,7)
A=(4,0) B=(8,2) C=(6,5) D=(0,7)
A=(4,7) B=(0,5) C=(2,2) D=(8,0)
A=(4,7) B=(8,5) C=(6,2) D=(0,0)

All 4 point-sets are really the same point set (up to coordinate-reflections).
This point set is unique if x and y coords are to be integers in [0,8] and [0,7]
respectively (I did an exhaustive search).

Unless I screwed up.  Caveat emptor.

-- 
Warren D. Smith
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