On Sun, Nov 17, 2002 at 09:38:08AM -0500, Raul Miller wrote:
> On Sun, Nov 17, 2002 at 12:50:59PM +0100, Jochen Voss wrote:
> > A B C D X
> > A - 27 24 29 29
> > B 22 - 25 29 28
> > C 25 24 - 30 32
> > D 21 21 19 - 30
> > X 21 22 18 18 -
> > the Schwartz set is { A, B, C }
> > proposition (A,C) is weakest -> eliminated
> > proposition (B,C) is weakest -> eliminated
>
> Minor nit: it's (C,A) which is weakest, as (A,C) is not a proposition.
Why not? To quote from your Nov 17 draft:
Definition: A proposition is a pair of options with a
non-zero preference (S,T) for at least one of the options.
The proposition is a defeat for one of the options unless
the preference (S,T) equals the preference (T,S).
Which part of this definition does not apply?
> > A B C
> > A - 27 0
> > B 22 - 0
> > C 0 0 -
> > the Schwartz set is { A, C }
>
> Hmm.. I forgot to eliminate options with no votes for them.
Be careful here: in this example my program agrees with Anthony
Towns implementation, both get a tie between A and C. If we
eliminate options with no votes for them (C in this case) we
change the result: now the option A becomes the winner.
It may be possible, that we want that change. But we should
know that this is a deviantion from our former implementation.
Maybe the electionmethods website can clarify this?
I hope this helps,
Jochen
--
Omm
(0)-(0)
http://www.mathematik.uni-kl.de/~wwwstoch/voss/privat.html
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