[EM] Logic/Jargon question

[Paul Kislanko]

>From Wikipedia:

In  <http://en.wikipedia.org/wiki/Voting_system> voting systems, the Smith
set is the smallest set of candidates in a particular election who, when
paired off in pairwise elections, can beat all other candidates outside the
set. Ideally, this set consists of only one candidate, the
<http://en.wikipedia.org/wiki/Condorcet_winner> Condorcet winner. However,
when the electorate is conflicted (as in
<http://en.wikipedia.org/wiki/Voting_paradox> Condorcet's paradox), the set
has at least one cycle of candidates for whom A beats B, B beats C, and C
beats A. See also  <http://en.wikipedia.org/wiki/Schwartz_set> Schwartz set.

If there are N candidates, how can the size of the Smith set be smaller than
N-1 if it is not exactly 1 (i.e. there is a Condorcet winner)?
 
If there's no CW, then disregarding ties there can be only one candidate who
pairwise-loses to all of the others, so candidates for the Smith set are all
who pairwise defeat that one. 
 
Suppose there are two members of the Smith set's complement. Then one would
have pairwise-beaten the other, and therefore would not have pairwise-lost
to anybody outside of the Smith set, which would make it a part of the Smith
set.
 
I can see how there can be "a Smith partition", but THE Smith set just seems
to be by definition a partitioning that excludes either N-1 candidates (when
there is a Condorcet winner) or 1 candidate (when there isn't a CW).
 
Or is there something I'm missing?
 
  

The great tragedy of Science - the slaying of a beautiful hypothesis by an
ugly fact.
Thomas H. Huxley <http://www.quotationspage.com/quotes/Thomas_H._Huxley> 
---------------------- 

Paul Kislanko
 
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20050116/d70c417c/attachment-0003.htm>