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
hadn't been eliminated already.
> 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
> all votes,
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 <[EMAIL PROTECTED]> <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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