Re: April 17th Draft of the Voting GR

[Raul Miller <moth@magenta.com>]

>  > I've also made explicit that only undropped defeats are used in
>  > determining the schwartz set.

On Fri, Apr 18, 2003 at 04:05:06PM -0500, Manoj Srivastava wrote:
> 	Whoa there. I think A transitively defeats option C if a
>  defeats B and B defeats C, whether or not the defeat is dropped.

Why do you think that?

> 	I think we should separate the definition of terminology of
>  defeats from the details of constructions of the Schwartz set. 

Hmm.. but we need to change the Schwartz Set to correspond to the changes
introduced by dropped defeats.  Since the only thing about constructing
the Schwartz Set which directly depends on defeats is transitive defeats
you'll still need something logically equivalent to my proposal.

>      4. From the list of undropped options, we generate a list of
>         pairwise defeats:
>         a. An option A defeats an option B, if V(A,B) is strictly greater
>            than V(B,A). 
>         b. (A,B) is a defeat of option B if option A defeats option B.
>         c. An option A transitively defeats an option C if A
>            defeats C or if there is some other option B where A
>            defeats B AND B transitively defeats C.
>      5. We construct the Schwartz set from the list of pairwise
>         defeats:
> 
>          -  An option A is in the Schwartz set if for all options B,
>             considering only the undropped defeats, either A
>             transitively defeats B, or B does not transitively defeat
>             A.
> 
>         If there are defeats between options in the Schwartz set, we
>         drop the weakest such defeats from the list of pairwise
>         defeats, and return to step 5.

If, as you said, you think that "A transitively defeats option C if
a defeats B and B defeats C, whether or not the defeat is dropped."
what does the above mean?

Thanks,

-- 
Raul