[EM] Party lists and candidate multiwinner elections

[Kristofer Munsterhjelm]

On 10/20/2014 10:01 PM, Kathy Dopp wrote:
> Hmm. With only 2 seats, I would think group A would get both since
> they are so overwhelmingly larger than groups B and C.

It would do so if you were to use party list (e.g. Webster). But the 
property says that if you run a multiwinner election (where one 
candidate can only be elected once, e.g. STV) then if k parties were 
elected in the party list election, these parties will be elected as 
individual candidates in the multiwinner election. Toby had another way 
of phrasing it: it's like a party list election, except each party only 
fields a single candidate.

If each party only fields a single party, you can't do any better than A 
and B. Clearly, given these Plurality ballots, {A, B} is better than {A, 
C} or {B, C}.

Anyway, if you were to do it using Webster (proper party list, i.e. 
every party fields enough candidates), then you would get:

12000 voters: Party A
300 voters: Party B
10 voters: Party C

Divisor is 6000.
A gets 2 seats, B and C get none. (Note here that k=1 and the same 
result holds - A would win a Plurality election as a candidate.)

> There ideally should be 3 seats in this case.  IMO, the number of
> seats should range flexibly within certain values, allowing the
> number of seats to slightly change in order to make the
> proportionality fairer.  Are there any systems like that?

I don't know of any that are, but I think that it would be relatively 
easy to incorporate into Condorcet-like systems such as Schulze STV. If 
you let the seats range between say, 5 and 10 seats, then you simply do 
every contest of candidate subsets of cardinality between 5 and 10 
instead of limiting yourself to only 5-candidate ones (say).

Clustering methods could also be adjusted that way. Just replace the 
constraint that the number of clusters (each corresponding to a 
different elected candidate) must be exactly 5 (or whatnot) with two 
inequality constraints saying the number of clusters can't be less than 
5 or greater than 10.

Ordinary STV would have some difficulties here. For party list, like 
Webster, one could calculate the disproportionality of each party (by 
the chi-squared test or similar) and then choose the seat number that 
gives least disproportionality in total, since Webster optimizes the 
chi-squared metric. Warren gives other measures that different divisor 
methods optimize, so the same should in principle be possible for, say, 
D'Hondt/Jefferson. But for such methods, the sequential algorithm 
(Sainte-Laguë or D'Hondt) would probably be better than the 
divide-and-round algorithm since the former lets you assign one seat at 
a time, thus having a lower time complexity given that you have to go 
through all seat numbers to find the one with the least 
disproportionality anyway.