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

Kristofer Munsterhjelm km_elmet at t-online.de
Wed Oct 5 13:57:51 PDT 2016

```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.
```