Re: Nov 16 draft of voting mechanics

[Raul Miller <moth@debian.org>]

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