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<DIV><FONT face=Arial size=2><SPAN class=051053200-17012005>From
Wikipedia:</SPAN></FONT></DIV>
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<H3 id=siteSub><FONT face=Arial size=2>In </FONT><A title="Voting system"
href="http://en.wikipedia.org/wiki/Voting_system"><FONT face=Arial size=2>voting
systems</FONT></A><FONT face=Arial size=2>, 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 </FONT><A title="Condorcet winner"
href="http://en.wikipedia.org/wiki/Condorcet_winner"><FONT face=Arial
size=2>Condorcet winner</FONT></A><FONT face=Arial size=2>. However, when the
electorate is conflicted (as in </FONT><A title="Voting paradox"
href="http://en.wikipedia.org/wiki/Voting_paradox"><FONT face=Arial
size=2>Condorcet's paradox</FONT></A><FONT face=Arial size=2>), the set has at
least one cycle of candidates for whom A beats B, B beats C, and C beats A. See
also </FONT><A title="Schwartz set"
href="http://en.wikipedia.org/wiki/Schwartz_set"><FONT face=Arial
size=2>Schwartz set</FONT></A><FONT face=Arial size=2>.</FONT></H3>
<DIV><SPAN class=051053200-17012005><FONT face=Arial size=2>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)?</FONT></SPAN></DIV>
<DIV><SPAN class=051053200-17012005><FONT face=Arial
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=051053200-17012005><FONT face=Arial size=2>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. </FONT></SPAN></DIV>
<DIV><SPAN class=051053200-17012005><FONT face=Arial
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=051053200-17012005><FONT face=Arial size=2>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.</FONT></SPAN></DIV>
<DIV><SPAN class=051053200-17012005><FONT face=Arial
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=051053200-17012005><FONT face=Arial size=2>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).</FONT></SPAN></DIV>
<DIV><SPAN class=051053200-17012005><FONT face=Arial
size=2></FONT></SPAN> </DIV>
<DIV><SPAN class=051053200-17012005><FONT face=Arial size=2>Or is there
something I'm missing?</FONT></SPAN></DIV>
<DIV><SPAN class=051053200-17012005></SPAN><SPAN
class=051053200-17012005></SPAN><SPAN class=051053200-17012005></SPAN><SPAN
class=051053200-17012005></SPAN><SPAN class=051053200-17012005><FONT face=Arial
size=2></FONT></SPAN></SPAN></FONT> </DIV></DIV></DIV>
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<DT class=tqpQuote><!--StartFragment -->
<DT class=tqpQuote>The great tragedy of Science - the slaying of a beautiful
hypothesis by an ugly fact.</DT>
<DD><FONT size=-1><FONT face=Arial>Thomas H. Huxley</FONT><A class=tqpAuthor
href="http://www.quotationspage.com/quotes/Thomas_H._Huxley"
target=_blank></A></FONT></DD></DIV>
<DT class=tqpQuote><FONT size=2>----------------------</FONT>
<DT class=tqpQuote><FONT size=2>Paul Kislanko</FONT></DT>
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