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 <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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