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

fsimmons at pcc.edu fsimmons at pcc.edu
Sat Dec 4 11:32:50 PST 2010

Jobst has also pointed out that, like Copeland, the "Condorcet Lottery" is not touched by my example, 
since it gives equal probability to all three candidates in the top cycle.

I want to go in a slightly different direction with this post.  Covering is a nice relation because it is transitive 
and cycle free.  The trouble is that in general there can be two alternatives neither of which covers the 
other,  When that is not the case (i.e. in the case where for each X and Y, either X covers Y or else Y 
covers X) we can easily agree to elect the alternative that covers each of the others.

This leads to the idea of "almost covering."  We can say that X almost covers Y iff every alternative that is 
beaten or almost beaten by Y is also beaten or almost beaten by X, too.  

The if "almost" is defined appropriately, for each X and Y, either X will almost cover Y or Y will almost cover 

If "almost" is defined too loosely, then every alternative will almost cover every other alternative, so the 
relation is not useful.  On the other hand, if it is defined too strictly, then there may be a pair of alternatives 
for which neither almost covers the other.

Accordingly, for each whole number k let's say that "X  k-covers  Y "  iff  for each alternative that Y beats or 
comes within k votes of beating, X also beats or comes within k votes of beating.  

Then let k be the smallest whole number such that for every pair of alternatives at least one will k-cover the 

This k is somewhere between zero and the total number of voters, because if N is the total number of voters, 
then every alternative N-covers every other alternative.

With k defined as above, we can say that X almost covers Y iff  X  k-covers Y.

Note that the almost covering relation is transitive, and that each two alternatives are comparable.  

Of course, in general we can expect that there will be alternatives X and Y such that each almost covers the 
other, so electing a candidate that almost covers all of the others is not decisive.  Instead we have an 
equivalence class of potential winners each of which almost covers every alternative.

So like Copeland, this basic method needs a tie breaker.  One way is to pick at random from the top 
equivalence class.

Another way would be to eliminate all alternatives outside of the top equivalence class, calculate a new k for 
the remaining alternatives, and repeat the process (recursively).  It's not clear that this procedure would 
preserve monotonicity.

Any other ideas?.  

----- Original Message -----
From: 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).

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