On Sat, Nov 16, 2002 at 05:23:13AM +0100, Matthias Urlichs wrote:
> > v. If this new Schwartz set contains only one option, that
> > option wins.
> >
> As per the definition, there is no such thing as a one-option Schwartz
> set, so actually you'll have to restart with step 5. You might therefore
> replace the last two sections with a simple
] i. All options not in the Schwartz set are eliminated.
]
] Definition: An option C is in the Schwartz set if there is no
] other option D such that D transitively defeats C AND C does
] not transitively defeat D.
Uh, if C is the Condorcet winner, or becomes the Condorcet winner after
ignoring some defeats of it, the Schwartz set has only one option.
Consider the vote:
20 A B C D
30 B C A D
40 C A B D
the weakest defeat is C > A (50:40), which gets eliminated leaving A >
B and B > C (and A,B,C > D), and the second Schwartz set is {A}.
Cheers,
aj
--
Anthony Towns <[EMAIL PROTECTED]> <http://azure.humbug.org.au/~aj/>
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``If you don't do it now, you'll be one year older when you do.''
