Hello Raul,
On Sat, Nov 16, 2002 at 05:44:40PM -0500, Raul Miller wrote:
> On Sat, Nov 16, 2002 at 09:31:08PM +0100, Jochen Voss wrote:
> > Is the default option supposed to be in the list of options for which
> > we do "Cloneproof Schwartz Sequential Dropping" below? Maybe we
> > should replace "Options" by "Non-default options" in this paragraph?
>
> For an option to be in the Schwartz set, it must either tie or
> beat at least one other option in the Schwartz set.
>
> The default option can never be in the Schwartz set, because any
> option which does not defeat the default option is eliminated
> before the first Schwartz set is created..
Ok, this is a proof.
But the change I suggested above would make a (theoretical) difference
nevertheless: if the default option could survive step 2 it could
become the winner in step 5. If the default option cannot survive
step 2, it never could become the winner.
Do we intend that the default option actually can win the vote?
Am I correct that this only could happen via step 5?
Jochen
--
Omm
(0)-(0)
http://www.mathematik.uni-kl.de/~wwwstoch/voss/privat.html
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