[EM] A Comparison of the Two Known Monotone, Clone Free Methods for Electing Uncovered Alternatives

[Kristofer Munsterhjelm]

fsimmons at pcc.edu wrote:
>> Where's a good place to find out more about the Landau set? Is
>> it really
>> possible to have a monotone, clone free method that is
>> independent of non-Landau
>> alternatives?
> 
> It turns out that there are several versions of covering, depending on how ties
> are treated.  All of them including the Landau set are the same when there are
> no pairwise ties (except with self).
> 
> In July of this year I gave an example that shows that no decent deterministic
> monotone method can be independent from covered alternatives.  The example
> applies to the Landau version of uncovered.  So neither Ranked Pairs nor
> Beatpath nor Range restricted to Landau can monotone.

Then consider Smith,IRV. To find a full ranking of Smith,IRV, we take 
the Smith set (ordering all members above all others), then we "break 
ties" in this ordering by the IRV ordering.

Could we do something similar with Ranked Pairs (or Beatpath or Range)? 
The method wouldn't pass independence from covered alternatives, but 
you've already established that is impossible to satisfactorily 
reconcile with monotonicity anyway.

(checks using his simulator)

It doesn't seem to work for Beatpath, though. In your first case, we 
have the following orderings:

Landau: A = B = C > D
Schulze: D > B > C > A
Landau,Schulze: B > C > A > D.

In the second, where B is raised,

Landau: D = B = C > A
Schulze: D > B > C > A
Landau,Schulze: D > B > C > A

Thus, a method where that example would work would have to not rank D 
first, and rank either B or C before A. E.g. Landau,Borda would work:

Landau: A = B = C > D
Borda: C > D = A > B
Landau,Borda: C > A > B > D

then

Landau: D = B = C > A
Borda: B > C > D > A
Landau,Borda: B > C > D > A

but this is of course not cloneproof. Apparently, Copeland also elects 
from the uncovered set, but it is not cloneproof either (and quite 
indecisive).