[EM] Defensive strategy for Condorcet methods KM

[Kevin Venzke]

Hi Kristofer,

Quick one:

--- En date de : Mer 15.6.11, Kristofer Munsterhjelm <km_elmet at lavabit.com> a écrit :
> I haven't tested FPC (since my reimplementation of JGA's
> strategy ideas was done before I moved to a more modular
> design for Quadelect), but as far as I recall, the really
> standout systems as far as strategy was concerned, were
> Condorcet-IRV hybrids (C//IRV, Smith,IRV, Smith//IRV, but
> interestingly not Landau//IRV or Landau,IRV). Next to those,
> at some distance, was a sorta-system that's like Ranked
> Pairs, only in that winners are ranked above losers in the
> sorted order and otherwise the order is fixed, so the system
> isn't neutral.
> ("Winners are ranked above losers" means that if A beat B,
> A>B is reached before B>A.)
> That method is a worst-case (I think) estimate for Ranked
> Pairs-type methods. Making the sorting order depend on, say,
> the winner's FPP score, could be interesting, but I don't
> think that would be monotone (but hey, let's try to find
> burial resistance before we find monotone burial resistance
> :-)
> 
> I also read a paper about trying to find the method that
> would minimize burial opportunities. It reached the
> conclusion that in the three candidate case, one should
> elect the candidate that beat the FPP winner. The author
> further said that he didn't know how to generalize it beyond
> three candidates.
> I think it was this one: http://www.nhh.no/Admin/Public/DWSDownload.aspx?File=%2FFiles%2FFiler%2Finstitutter%2Ffor%2Fdp%2F2008%2F1108.pdf
> "Condorcet Methods - When, Why and How? ".

Ah, very interesting. I will try implementing "BPW" myself and see if
I can get anything.

> And from your other mail:
> 
> > Also am I mistaken or is FPC with three candidates
> identical to C//IRV?
> > The candidate with the lowest penalty is going to be
> the one who is
> > defeated by the "FPL." If you were to eliminate the
> FPL, the candidate
> > with the best FPC score would beat the remaining
> candidate.
> 
> Hm, let's see. The three candidates are A, B, and C, and
> they're in a cycle. Say that C is the candidate with the
> least first place votes. In other words, f(A) + f(B) >
> f(A) + f(C) and f(A) + f(B) > f(B) + f(C), since f(A) +
> f(B) + f(C) sums to the number of voters, which is a
> constant for each election.
> 
> A's penalty is f(B) + f(C).
> B's penalty is f(A) + f(C).
> C's penalty if f(A) + f(B).

er... Is this right? I thought your penalty in the 3c case would have
to be just a single candidate's first preferences.

I think I am probably right here. If you draw a triangle with ABC,
you have two cycle possibilities. In both cycles IRV elects the winner
between A and B. And in both cycles, that same candidate is beaten
by the "FPL" C.

Kevin Venzke