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Hybrid Theory

Focusing on just A.6 again, in this draft:

(*) Weakest defeats can now be eliminated: before a defeat of the default
option is eliminated, all options which fail to meet their supermajority
requirements are deleted.
(*) When artificial supermajority defeats are eliminated the corresponding
option is also deleted.
(*) Terms expressed using mathematical shorthand have been named.
(*) Other minor fixes


   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.
        Options left unranked by the voter are considered to be ranked
        equally with one another and below any ranked options.  The other
        details of how ballots may be filled out will be included in
        the Call For Votes.

     2. We construct the Schwartz set based on undropped options and

          a. An option A is in the Schwartz set if A has not been dropped
             and if for all options B, either A transitively defeats B,
             or B does not transitively defeat A.

          b. 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.

          c. An option A defeats an option B, if the strength
             N(A,B)*V(A,B) is larger than N(B,A)*V(B,A), and if the
             (A,B) defeat has not been dropped.

          d. Given two options A and B, the votes V(A,B) is the number
             of voters who prefer option A over option B.

          e. If a majority of n:1 is required for A, and if B is the
             default option, majority requirement N(B,A) is n.  In all
             other cases, N(B,A) is 1.

     3. If any options involved in the weakest defeats between options
        in the Schwartz set are options in superdefeats, we drop the
        corresponding superdefeated options then return to step 2.

          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 larger 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.

          c. An option B is superdefeated by an option A if A defeats B,
             and if V(A,B) is not larger than V(B,A).

          c. Given a defeat (A,B) the options A and B are involved in
             this defeat.

          d. An option A is dropped by dropping all defeats which involve
             A and also stipulating that option A is not a member of
             the Schwartz set.

     4. If there are defeats between members of the Schwartz set, we
        drop the weakest defeats then return to step 2.

     5. If there are no defeats within the Schwartz set, then the winner
        is chosen from the undropped options in the Schwartz set where
        at least Q voters ranked that option above default option, where
        Q is the quorum requirement for the ballot. If there is only one
        such option, it is the winner. If there are multiple options, the
        elector with a casting vote chooses which of those options wins.

     "RATIONALE": Options which voters rank above the default option are
     options they find acceptable.  Options ranked below the default
     option are unacceptable options.  Supermajority options require
     some approximation of unanimity before they can be accepted.


This changes a few of the outcomes for tests posted in 

Here's the test cases which have changed:

A requires 2:1 majority; D is the default option

A defeats B 4:1
B defeats D 4:1
D superdefeats A 4:3

eliminate A

B defeats D 4:1

B wins
in previous draft, this was a tie between B and D
the requirement was that D did not win

Z is the default option
40 A
25 B
35 ZBA

B defeats A 60:40
A defeats Z 40:35
Z defeats B 35:25

eliminate 35:25

B defeats A 60:40
A defeats Z 40:35

B wins
in previous draft, Z won
the requirement was that A not win

Quorum of n, no supermajorities, D is the default option:
25 DAB
30 BDA
35 ABD

B defeats D 65:25
A defeats B 60:30
D defeats A 55:35

eliminate 55:35

B defeats D 65:25
A defeats B 60:30

A wins unless quorum is not met in which case D wins.
In the previous draft B won unless quorum was not met in which case D wins.

Additionally, here's the result of the test proposed by Clinton Mead in

A requires 3:1 majority; D is the default option

A defeats B 6:1
D superdefeats A 6:5
B defeats D 5:2

eliminate A

B defeats D 5:2

B wins
in the previous draft, D won
the requirement is that B wins


If there's any ambiguities or problems with this, please let me know.
[And please bear with me if I seem to be asking inane questions in the
course of understanding a possible problem.]



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