# Re: Nov 16 draft of voting mechanics

```On Sat, Nov 16, 2002 at 12:48:07PM -0500, Raul Miller wrote:
>   A.6 Vote Counting
>     1. Each ballot orders the options being voted on in the order
>        specified by the voter. [...]

"in the order of the voter's preference." perhaps.

>     2. Options which do not defeat the default option are eliminated.

>     3. If an option has a quorum requirement, that option must defeat
>        the default option by the number of votes specified in the quorum
>        requirement, or the option is eliminated.

"If an option has a quorum requirement, Q, that option must have been
preferred to the default option by at least Q voters."

"If an option has a quorum requirement, Q, that option must have been
preferred to the default option by at least Q more voters than preferred
the default option to it."

ie, if Q = 10, does "O defeats D, 10:9" satisfy the quorum requirement,
or do you need 19:9? Another option would be to just add up the number
of votes that expressed a preference for O wrt D, which would be a more
usual quorum requirement (ie, the number of people present to vote),
and would allow "D defeats O, 6:4" pass the quorum requirement, if it

>     4. If an option has a supermajority requirement, [...]
>
>     5. If one remaining option defeats all other remaining options,
>        that option wins.

>     6. If more than one option remains after the above steps, we use
>        Cloneproof Schwartz Sequential Dropping to eliminate any cyclic
>        ambiguities and then pick the winner.  This procedure and must
>        be carried out in the following order:
>
>        i. All options not in the Schwartz set are eliminated.

>           Definition: An option C is in the Schwartz set if there is no
>           other option D such that D transitively defeats C AND C does
>           not transitively defeat D.

>           Definition: An option F transitively defeats an option G if F
>           defeats G or if there is some other option H where H defeats
>           G AND F transitively defeats H.

>        ii. Unless this would eliminate all options in the Schwartz set,
>            the weakest propositions are eliminated.

Uh, eliminating propositions can never eliminate all options from the
Schwartz set. The condition is "unless there are no propositions in the
Schwartz set"

>            Definition: A proposition is a pair of options J and K
>            from the Schwartz set which are considered along with
>            the number of voters who prefer J to K and the number
>            of voters who prefer K to J.

>            Definition: The dominant strength of a proposition is the
>            count of votes in a proposition which is not smaller than
>            the other vote count in that proposition.

>            Definition: a weak proposition is a proposition which
>            has a dominant strength greater than zero and no larger
>            than that of any other proposition.

>            Definition: a weakest proposition is a weak proposition where
>            the vote count in the proposition which is not larger than
>            the other vote count is also no smaller than that of any
>            other weak proposition.

Alternatively, and IMO, simpler:

Definition: A proposition is a pair of options, J and K from
the Schwartz set, such that J defeats K.

Definition: V(X,Y) is the number of voters who prefer option X to
option Y.

Definition: A proposition J,K is weaker than a proposition L,M
if V(J,K) is less than V(L,M); or V(J,K) and V(L,M) are equal
and V(K,J) is greater than V(M,L).

Defintiion: A weakest proposition is a proposition that has no
other proposition weaker than it. There may be more than one
such proposition.

Probably a good idea to number the definitions, actually.

>            Definition: A proposition is eliminated by treating both
>            of its vote counts as zero from this point forward.

>        iii. If eliminating the weakest propositions would eliminate all
>             votes represented in the Schwartz set,

See above.

>                                                    a tie exists and the
>             person with the casting vote picks from among the options
>             represented in this Schwartz set.
>
>        iv. If eliminating the weakest propositions would not eliminate

Uh, you're not eliminating votes, you're eliminating propositions. I think
you can just say:

>                       a new Schwartz set is found based on the newly
>            revised set of propositions.

here, no ifs or buts.

>        v. If this new set of propositions allows one option to
>           defeat all other options in the Schwartz set, that
>           option wins.

It seems simpler to not have either (5) or (6.v) but just to say "If
the Scwartz set has a single option, it is the winner." as step (6.i).

>        vi. Otherwise, these steps (i-vi) are repeated with this new
>            Schwartz set.

Cheers,
aj

--
Anthony Towns <aj@humbug.org.au> <http://azure.humbug.org.au/~aj/>
I don't speak for anyone save myself. GPG signed mail preferred.

``If you don't do it now, you'll be one year older when you do.''
```

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