[EM] Meek Webster method?

[Kristofer Munsterhjelm]

The RangeVoting page says that Webster minimizes, for all parties or 
candidates k,

	sum over k: P_k * (S_k / P_k - S/P)^2

where S_k is the number of seats given to the kth option, P_k is the 
number of voters given to that option, S is the total number of seats, 
and P is the total number of people (voters).

Would it be possible to make a Meek type method out of this? I'm going 
to deal with the party list version here, but doing it with candidates 
(where a candidate can get no more than a single seat) would be rather 
easy too.

As one may recall, a Meek method has a weighting for each candidate. The 
candidate receives a number of votes proportional to that weighting, 
then passes the rest on to the next candidate in the ranking, for each 
ballot.

So, perhaps something like: Each party has a weighting and a count (as 
in Meek). Start with the weighting being 1, which means that each 
party's count (P_k) is equal to the number of first preference votes it got.

Then determine the party seat numbers by running ordinary Webster or 
Sainte-Lague on the counts.

Now, some parties will be safely within their win regions; that is, 
lowering their P_k wouldn't change their S_k, but it would lower their 
(S_k / P_k - S/P)^2. Say that these parties have an excess, and lower 
the weight of the party with the greatest excess, ever so slightly[1]. 
To keep the total number of votes constant, the other parties' weights 
may also have to be adjusted in turn to compensate.
The result is that voters who voted for the excess party transfer some 
of their ballot power to the next candidate in line, again like Meek.

Recalculate the counts and repeat the above until every party is more or 
less stable, i.e. the weights oscillate around a fixed point.

Could this work? I have no idea what criteria it would pass. It would 
probably not be monotone.

The idea is similar to ordinary Meek: when a group makes a party win x 
seats, the number of voters who didn't have to actually vote for the 
party can transfer their votes to the second party they preferred. But 
that might give the second party a seat at the expense of some other 
party that now has a great excess, and so on.

It's a bit trickier than with ordinary Meek, however. In ordinary Meek, 
the quota doesn't change that much. Here, there's no explicit quota, 
only a sliding fitness function. Thus, I don't know if Meek will work in 
this context, or if it'll be proportional. I *do* know that determining 
the party weights that correspond to the global minimum of the sum above 
fails to produce proportional results.

(To make a candidate method instead of a party list method, just give 
each "party" a maximum of a single seat. Sainte-Laguë then no longer 
really works, so you'd have to use Webster to determine S_k.)

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[1] You could probably determine how much it has to be lowered. If you 
go one party at a time, then assume the worst case that every voter for 
the party in question votes for the party closest to the "margin", after 
it. That will give some idea of how much reweighting you can do in one go.