On Sun, Nov 17, 2002 at 10:42:40PM +1000, Anthony Towns wrote: > "in the order of the voter's preference." perhaps.
Ok. > "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." I'm going to go with the latter, as it eliminates the risk of causing something to be approved because you voted against it. > > 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" Ok. > Alternatively, and IMO, simpler: ... Simpler doesn't count here, because you fail to handle pairwise ties. I am tempted to use your definition of V(X,Y), but I'd prefer the thing being defined be something more intrinsically meaningful than "V". > Probably a good idea to number the definitions, actually. That should not be necessary. Each definition should uniquely define a term, so it should be possible to refer to a definition as "the definition of ____". > Uh, you're not eliminating votes, you're eliminating propositions. Good point. > 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). I agree. I'll try to write it this way. Your observation that we're "eliminate propositions" (as opposed to options or votes) should help quite a bit. Thanks, -- Raul -- To UNSUBSCRIBE, email to [EMAIL PROTECTED] with a subject of "unsubscribe". Trouble? Contact [EMAIL PROTECTED]
