[EM] ISDA + mono-add-top + mono-raise

Kristofer Munsterhjelm km_elmet at t-online.de
Sun Sep 10 11:37:39 PDT 2017

I've been working more on extending my linear Condorcet method search to 
four and three candidates methods, and what I've found strongly suggests 
that the following criteria are incompatible, at least for linear methods:

Resolvability (Tideman's version), ISDA, mono-add-top, and mono-raise.

The only "method" passing the three latter ones that I've found so far 
is plain Smith: every candidate in the Smith set ties for first, and 
every candidate not in it ties for last. But that method clearly fails 
resolvability. And I suspect that since I can't get the three criteria 
plus resolvability to give a non-empty feasible region, and since the 
methods are anonymous and neutral, all the resolvability-failing methods 
will end up being equivalent to Smith[1].

There are a few caveats, however:

- What I've checked is ISDA, not Smith. ISDA is stronger than Smith, so 
it might be that resolvability + Smith + MAT + mono-raise are 
compatible, just that ISDA isn't.

- Furthermore, just because there are no linear Condorcet methods with 
all four of these, there do not necessarily have to be no Condorcet 
methods at all. I would have to check if the impossibility results for 
the linear methods (in the form of an empty feasible region for a 
particular LP) can be turned into a general impossibility proof.

So I'd have to check this more before I say anything really definitive. 
But as that'll take time, and this list is pretty quiet at the moment, I 
thought I'd just let the people here know what I've found out so far :-)

I might try to find a strategy resistant four-candidate method that 
passes monotonicity first. There may also be possible to extend linear 
Condorcet so that say, minmax also is describable as one, since LPs can 
also handle min and max operators (in certain directions).


[1] This because the linear nature of linear Condorcet within each cycle 
scenario means that the ranking will stay the same for all candidates 
within the cycle no matter what the ballots are, if the method fails 
resolvability within that cycle. But since one can rotate the cycle by 
relabeling the candidates, that means that every candidate's score must 
be the same, which thus gives a perfect tie among them.

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