[EM] Two questions

[Kevin Venzke]

Hi Kristofer,

--- En date de : Jeu 24.6.10, Kristofer Munsterhjelm <km-elmet at broadpark.no> a écrit :
> > Hi Kristofer,
> > 
> > --- En date de : Mar 22.6.10, Kristofer Munsterhjelm
> <km-elmet at broadpark.no>
> a écrit :
> >> Here are two questions regarding
> >> criterion compliance:
> >> 
> >> First, does ordinary Copeland (one point for a
> win, nothing
> >> for a tie or loss) pass Smith?
> > 
> > I believe so. Suppose that there are x candidates in
> the Smith set and
> > y candidates outside of it. Then everyone in the Smith
> set will have
> > a score of at least y+1, and everyone outside the
> Smith set can have a
> > score of at most y-1.
> 
> Does that mean that second order Copeland passes Landau? If
> not, I suppose the counterexample would involve a "winner"
> that is victorious against a few candidates that beat
> many... but I'm not certain.

I don't know. I've never really thought about either of these.

> >> Second, could anyone give me an example of
> Copeland failing
> >> mono-add-top?
> > 
> > I can't produce the ballots off the top of my head,
> but suppose A is
> > winning with a score of say 5 while B has a score of
> 4. Then you add
> > new ballots of the form A>B>... that cause B to
> win two additional contests with no other effect. I would
> assume for simplicity that the
> > Smith set is unchanged (containing A and B in both
> elections).
> 
> Yes, that works, and it let me find the bug (actually
> mis-specification) in my criterion compliance program. The
> problem was that it didn't consider situations where there
> were ties in either outcome, and Copeland is very well known
> for producing lots of ties.
> 
> I've fixed that, and now it easily finds Copeland failures
> - for instance, this will work:
> 
> 1: A > C > B
> 2: B > A > C
> 1: C > B > A
> 
> which gives B = A > C, then we add a ballot ranking A
> top:
> 
> 1: A > B > C
> 
> 1: A > C > B
> 2: B > A > C
> 1: C > B > A
> 
> and we get B > A > C, QED.
> 
> Incidentally, this means that it is theoretically possible
> (at least given that disproof) that a more decisive version
> of Copeland can pass mono-add-top. That method would give
> the same results as Copeland, except it would resolve ties.
> *But* in that case, we know that it must elect B in the
> first example (i.e. have ordering B > A > C);
> otherwise, the disproof above can still be applied; and if
> we could find an example where, say, A = B > C and we can
> make both a mono-add-top failure for A and B, then that
> would prove that no "more decisive" version of Copeland
> could pass, because however it resolves the tie, we can
> counter with the appropriate disproof.

That seems to make sense to me...

> I'll test Smith,minmax(margins) next, which seems to be a
> method where it's harder to find a mono-add-top failure. Can
> you think of a way of constructing one? IIRC,
> Minmax(margins) fails Plurality and so we can't use
> Woodall's proof.

I take it there is a Woodall proof that shows the incompatibility of
Mono-add-top, Plurality, and Smith? I don't remember off the top of my
head.

The question is what criterion could we possibly use to pick a candidate,
which implies Smith, and which doesn't change its mind when you alter
pairwise contests (other than those of the winner) which could very well
alter the membership of the Smith set?

It's difficult to imagine. Altering the other contests can turn a
different candidate into the Condorcet winner, or at least evict the
current winner from the Smith set. So whoever we elect must always be
immune to eviction from the Smith set by altering other contests. Is
it possible?

If the Smith set has more than one member then in the best case our 
winner has a single pairwise loss. But sometimes it must be inevitable
that new ballots can turn the winner of this loss into the CW. So we
needed to pick that candidate to start with. But I already assumed the
"best case" for the original winner; I have trouble seeing what we could
ever do to find a "more stable" winner who doesn't have the same weakness.

Kevin