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Re: April 17th Draft of the Voting GR



On Tue, Apr 22, 2003 at 10:24:10AM +0200, Jochen Voss wrote:
> I think we are loosing the track again.  What is the problem you are
> trying to solve here?  I think that your draft in
> <87smsf3ewg.fsf@glaurung.green-gryphon.com> was really ok.

Hmm...  that says

<       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.

But after defeats are dropped you need to find the revised set of
transitive defeats.  But, the specification for transitive defeats is
in step 4 in that draft.

How about this (I've split apart what were steps 4 and 5 into smaller
steps, and changed the numbering to fit):

______________________________________________________________________

Replace A.6 with:

   A.6 Vote Counting

     1. Each voter's ballot ranks the options being voted on.  Not all
        options need be ranked.  Ranked options are considered
        preferred to all unranked options.  Voters may rank options
        equally.  Unranked options are considered to be ranked equally
        with one another.  Details of how ballots may be filled out
        will be included in the Call For Votes.
     2. If the ballot has a quorum requirement R any options other
        than the default option which do not receive at least R votes
        ranking that option above the default option are dropped from
        consideration.
     3. Any (non-default) option which does not defeat the default option
        by its required majority ratio is dropped from consideration.
        a. Given two options A and B, V(A,B) is the number of voters
           who prefer option A over option B.
        b. An option A defeats the default option D by a majority
           ratio N, if V(A,D) is strictly greater than N * V(D,A).
        c. If a supermajority of S:1 is required for A, its majority ratio
           is S; otherwise, its majority ratio is 1.
     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).
     5. From the list of [undropped] pairwise defeats, we generate a
        set of transitive defeats.
        a. 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.
     6. We construct the Schwartz set from the set of transitive defeats.
        a. An option A is in the Schwartz set if for all options B,
           either A transitively defeats B, or B does not transitively
           defeat A.
     7. 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.
        a. A defeat (A,X) is weaker than a defeat (B,Y) if V(A,X)
           is less than V(B,Y).  Also, (A,X) is weaker than (B,Y) if
           V(A,X) is equal to V(B,Y) and V(X,A) is greater than V(Y,B).
        b. A weakest defeat is a defeat that has no other defeat weaker
           than it.  There may be more than one such defeat.
     8. If there are no defeats within the Schwartz set, then the winner
        is chosen from the options in the Schwartz set.  If there is
        only one such option, it is the winner. If there are multiple
        options, the elector with the casting vote chooses which of those
        options wins.  

______________________________________________________________________


-- 
Raul



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