[EM] Two questions

[Kristofer Munsterhjelm]

Kevin Venzke wrote:
> 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.

>> 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.


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.