On Tue, Nov 19, 2002 at 11:58:35AM -0500, Raul Miller wrote:
> It's probably worth comparing the strategies possible with this draft [...]
I'm going to ignore the fact you meant wrt quorums not supermajorities.
Consider 100 voters, a constitutional amendment, A, and a set of
conscientious objectors. The objectors vote:
30 D A
This should, surely, be enough to stop "A" winning, no? Isn't that the point
of a supermajority requirement?
If the remaining 70 people vote:
70 A D
they'll succeed. If, however, they can introduce a cycle amongst A, D and
something else, and have D:A 90:70 be the weakest defeat, like so:
Introducing the new option, B, no supermajority requirement. They convince
a bunch of the conscientious objectors to support it.
20 B D A
10 D A
70 A B D
A defeats B (100:0), B defeats D (90:10), D defeats A (90:70). "D beats A"
is then the weakest defeat, and A wins, in spite of the 30 person bloc
trying to ensure its defeat.
I *believe* that 30 of 100 people voting "D" as their first option can
successfully ensure that a supermajority requiring option they're opposed
to will be defeated, but why should they be required to insincerely say
they wouldn't be satisfied with B in order to exercise their prerogative
as a superminority?
> 2. We drop the weakest defeats from the Schwartz set until there
> are no more defeats in the Schwartz set:
> a. An option A is in the Schwartz set if for all options B,
> either A transitively defeats B, or B does not transitively
> defeat A.
> h. A defeat is in the Schwartz set if both of its options are
> in the Schwartz set.
> i. A defeat (R,S) is dropped by making N(S,R) the same as N(R,S).
> Once a defeat is dropped it must stay dropped.
If you have a Schwartz set {A,B,C} and eliminate a defeat, can you then end
up putting some {D} into the Schwartz set? If D is not in the Schwartz set
then there's some option X which transitively defeats D, but D does not
transitively defeat D. For X to transitively defeat D, there must be a Y
(possibly Y = X), that defeats D directly. Can Y be transitively defeated
by D? Can the defeat chain from X to Y be broken?
X > .... > Z > .. > Y > D > Z
Possibly X = Z, and/or Z = Y. Possibly Y=Z=D (ie, D's not in a loop
like that). If Z is in the Schwartz set, D would be, so Z isn't in the
Schwartz set. But in that case for whatever Z' it is that defeats Z,
that defeat won't be dropped, nor will any of the Z .. Y defeats, so
D will stay out of the Schwartz set.
Dunno if that makes sense to anyone else, don't care. I was wondering
if the "Propositions are between members of the Schwartz set" rule from
previously was necessary, and couldn't convince myself it wasn't at
the time.
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.''
