Re: April 17th Draft of the Voting GR

[Raul Miller <moth@magenta.com>]

On Fri, Apr 18, 2003 at 12:42:44AM -0400, Anthony DeRobertis wrote:
> >      5. If there are defeats between options in the Schwartz set, we
> >         drop the weakest such defeats, and return to step 4.
> >            a. A defeat (A,X) is weaker than a defeat (B,Y) if V(A,X)
> 
> The normal comma things. Also, what are A, X, B, and Y? Just arbitrary
> options? And is a defeat just an arbitrary pairing of two options (I
> don't think so, but it's not clear. Can we define a defeat somewhere?)

The following was meant to answer this question:

<quote>
     4. We construct the Schwartz set based on undropped options and
        defeats:
           a. An option A defeats an option B, if V(A,B) is strictly
              greater than V(B,A).
</quote>

However, given that "defeats" is a verb and we're using "defeat" as a
noun, maybe it would be clearer to say:

     4. We construct the Schwartz set based on undropped options and
        defeats:
           a. The vote tallies comparing option A and option B are a
              defeat of option B if V(A,B) is strictly greater than
              V(B,A).

Is that clearer?

Jochen: with this change is it clearer what is meant by dropping
a defeat?

Thanks,

-- 
Raul