On Tue, Nov 12, 2002 at 06:03:54PM -0500, Raul Miller wrote:
> Please let me know of any flaws in the following partial draft:
These are mostly fairly minor.
> A.6 Vote Counting
> 1. Each ballot orders the options being voted on in the order
> specified by the voter. Any options unranked by the voter are
> treated as being equal to any other unranked options and below
> all options ranked by the voter.
> 2. Options which do not defeat the default option are eliminated.
> Definition: Option A defeats option B if more voters prefer option
> A over option B than prefer option B over option A.
Definition: A voter prefers option A over option B if he gives option
A a higher ranking than than option B, or gives option A a ranking and
does not give option B a ranking.
This might be being a bit pedantic, *shrug*.
> 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.
Does this mean:
3. If an option has a quorum requirement, Q, that option must be preferred over
the default option by at least Q people.
or
3. If an option, O, has a quorum requirement, Q, then option O must be
preferred over the default option by at least Q more people than prefer
the default option over option O.
> 5. If one remaining option defeats any other remaining options,
> that option wins.
s/any/all/
> 6. If more than one option remains after the above steps, we use
> Cloneproof Schultz Sequential Dropping to eliminate any cyclic
> ambiguities and then pick the winner. These represent a procedure
> and must be carried out in the specified order:
%s/Schultz/Schwartz/g
> ii. Unless this would eliminate all options in the Schultz set,
> the options which have the weakest defeat are eliminated.
> iii. If eliminating the weakest defeat would eliminate all options
> in the Schultz set, a tie exists and the person with the
> casting vote picks from among these options.
Hey, wait a minute. You eliminate *defeats*, not *options* here. That is, if
you've eliminated options E and F, and are left with A, B, C and D you might
have:
A > B > C > A (> == beats)
A, B > D > C
Set = A,B,C,D
eliminating, say, B > C gives you:
A > B > D > C > A
A > D
Set = A,B,C,D
with no *options* eliminated if you then eliminate, say, B > D:
A > B
A > D > C > A
Set = A,C,D
but then A>B is out of the running for elimination since B isn't in
the Schwartz set. So if you then eliminate _everything_, you have:
A > B
A,B,C,D > E,F
Set = A,C,D (since nothing beats those options, and everything else is
beaten)
So I don't think this rule is right: eliminating all defeats _can't_
reduce the Schwartz set. Even if you did drop A>B, then you'd end up
with a Schwartz set of A,B,C,D, not an empty set.
> ii. Unless this would eliminate all options in the Schultz set,
> the options which have the weakest defeat are eliminated.
> iii. If eliminating the weakest defeat would eliminate all options
> in the Schultz set, a tie exists and the person with the
> casting vote picks from among these options.
ii. If there are defeats remaining between options in the Schwartz set,
the weakest defeats are eliminated.
iii. If there are no defeats amongst options in the Schwartz set,
the outcome of the vote is a tie amongst the options in the Schwartz
set, and the elector with a casting vote picks the winner from
amongst the tied options.
> iv. Otherwise, a new schultz set is found, with those weakest
> defeats eliminated,
%s/schultz/Schwartz/
> v. If this new schultz set contains only one option, that option
> wins.
This can happen with the initial Schwartz set too, so should really be
before (ii).
> vi. Otherwise, these steps (i-vi) are repeated with this new
> schultz 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.''
Attachment:
pgpYd01qxmfnN.pgp
Description: PGP signature