[EM] Using Schulze Election Method to elect a flexible amount of winners

[Juho Laatu]

I guess those mentioned examples are about the worst ones. If the number of elected representatives is high, the symmetry/proportionality errors may easily get lost in the overall noise. The described problem is still there. There could be a need e.g. to change the n-1'th candidate when electing the n'th candidate, just as in the original examples. But since we are now talking about the last allocated seats, maybe nobody cares. The errors probably get smaller if counted as % of error, but may stay about at the same level if counted as number of representatives that should have been changed.

I don't have any exact numbers or proper estimates though. Maybe your calculations showed that the frequency of errors gets lower, or that the amount of distortion per error (as %) gets lower.

Juho


P.S. This is also related to the highest averages vs. largest remainder question.



> On 05 Oct 2016, at 23:57, Kristofer Munsterhjelm <km_elmet at t-online.de> wrote:
> 
> On 10/05/2016 06:00 PM, Juho Laatu wrote:
>> One interesting question is, if you want proportional representation,
>> should you use the proportional ranking approach. The simplest
>> example in maybe the following one.
>> 
>> 40: A>B>C 
>> 10: B>A>C
>> 10: B>C>A
>> 40: C>B>A
>> 
>> B is the Condorcet winner, so it would be a good choice if we will
>> elect one winner. But if we elect two winners, then electing A and C
>> could be a good choice. B would not be elected although B was elected
>> if the number of elected candidates was one lower.
>> 
>> One thus needs to decide if one wants to have an ordering of the
>> candidates (e.g. if one picks them one by one without possibility to
>> change the already elected candidates), or if one wants the
>> proportionality to be accurate (in the way described above). One can
>> not have both.
> 
> The example I usually trot out is LCR (exact numbers vary):
> 
> 20: L>C>R
> 18: R>C>L
> 3: C>R>L
> 
> For a single-winner election, C should win. For two winners, the set
> should be L and R, otherwise the outcome will be biased in favor of
> whoever the not-centrist ends up being.
> 
> LCR is also a center-squeeze scenario. Plurality elects L and IRV elects R.
> 
> You're right that you can't have both. Imperfect symmetry is the price
> you pay for "house monotonicity" (that n-1 of the n winners in an n-seat
> election are the winners of the n-1 seat election). I seem to recall
> reading somewhere (while investigating approximation methods to the
> k-median and facility location problems) about how close you can get to
> the optimum while being house monotone. However, I don't recall what the
> bounds were, other than that I was surprised they weren't worse.