If instead of Simmons's choice 4:2:2:2 for the city populations, we use
4:4-k:k:4-k then
* with k=2 we get Simmons populations and 60:40 majorities.
* with k=1 we get 7:4 majorities (or more)
* when k drops to 0 we get 66.66% majorities for every pair,
which is maximum possible, and city C becomes
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 preference
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 rememb
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=
This point-set also works:
A=(1,0) B=(0,4) C=(3,5) D=(9,2)
--
Warren D. Smith
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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
