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# Re: Nov 19 draft of voting amendment

```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 <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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