I may get back to this in greater detail later, but some notes for now (yes, I'm writing late again):

On 09/03/2013 11:07 PM, Vidar Wahlberg wrote:
On Mon, Sep 02, 2013 at 10:18:36AM +0200, Kristofer Munsterhjelm wrote:
Here's a short post (since I don't have as much time as I would
like) with an idea of how to make Sainte-Lague even more like STV. I
started thinking about it as part of my thinking that "perhaps
pairwise multiwinner methods will always be too complex"; and so I
tried to include some Condorcet compliance here as well.

I hope I'm not misunderstanding you, but since you're mentioning
"pairwise multiwinner methods" and parties I'll write a bit about an
idea I've been playing with recently.

Pairwise multiwinner methods could be any method that fills a council using a pairwise setup, i.e. a generalization of a single-winner pairwise method. But there's a more specific pattern that turns up in many generalizations of pairwise single-winner methods, and I was referring to that pattern here.

In this particular generalization, you make a "virtual" single-winner election where each winner candidate is a particular assembly. For STV methods like CPO-STV or Schulze STV, each such winner is then a list of candidates, so there are n choose k of them. (These methods tend to be computationally expensive). For my "CPO version" of the "eliminate parties that don't have a chance" system, each winner is a list of parties that are part of the outcome (i.e. not eliminated), so there are 2^(num parties) virtual candidates. My very complex nameless pairwise Sainte-Lague/Approval system also has n choose k winners and thus is utterly infeasible for a national count unless it can be mathematically reduced to something more like my CPO-SL.

In any case, these systems then define a pairwise function, call it f({A}, {B}), where A and B are virtual candidates, and the pairwise score of {A} against {B} is f({A}, {B}), while the pairwise score of {B} against {A} is f({B}, {A}). Note that this is the score prior to any thresholding (margins, wv, etc).

Then you just determine which virtual candidate wins and parse that back into an assembly. That assembly wins.

(A final note: Schulze STV bases its f({A}, {B}) on an inner function which one may call g({A}, {B}), that is only defined when the two sets differ by one candidate. f({A}, {B}) extrapolates this into every one-on-one by a strongest-path logic similar to that of the Schulze method itself.)

I'm a fan of Condorcet methods, and notably Ranked Pairs. I wanted to
try modifying RP in a way so it could be used for party-list elections,
giving a result where the party most people agree on being the best
party wins the most seats, rather than the party that have the most
first preference votes. Party having the most first preference votes may
of course also be the party that most people agree on being the best
party.

[snip]

Now the idea was that if some voters expressed a second preference, that
should cause the second preference to win more seats, but not at the
expense of the first priority, only at the expense of the other parties.
So I made every vote for FRP have H as second preference, while leaving
all other votes have no second preference. This gave me _almost_ the
result I expected:
    A: 46 seats
   SV: 12 seats
   RV:  2 seats
   SP:  9 seats
  KRF:  9 seats
    V:  8 seats
    H: 40 seats
  FRP: 41 seats
KYST:  1 seat
   PP:  1 seat

As expected, H won seats from the other parties, but to my surprise,
FRP also won more seats, even though no votes ranked FRP higher than in
the previous run, and it was the exact same amount of votes.
I haven't dug deep down into the code yet to figure out why it benefited
FRP to add H as second preference.

I think there are two problems here.

First, because of the sequential nature, this method must pass house monotonicity (as does ordinary Sainte-Laguë). That means that for any k, the outcome for k-1 seats is a subset of the outcome for k seats. But let's do a quick and dirty LCR example again:

48: L>C>R
42: R>C>L
10: C>R>L

where L is left-wing, C is center, and R is right-wing.

If you're electing just one seat, then C should win; anything else would be unfair to a majority. But if you're picking two, then if you give the first seat to C, giving the second to L will bias the assembly to L and giving the second to R will bias the assembly to R. So the right outcome for two seats would be {LR}. But {C} is not a subset of {LR}, so house monotonicity is not desirable. You might argue that {C1, C2} would fix the problem, but that would just push the problem itself into the three-seat case.

Second, a voter may gain undue power with additional preferences. Say a voter's preference is H > FRP. Then when a H seat is chosen, that will deweight his preference for H over AP (say), but it won't deweight his preference for FRP over AP. Thus some of his pairwise preferences get counted at full strength even though he got his first choice.

If you want to go down this sequential deweighting route, I think you should instead deweight the ballots themselves. So say H gets a seat. Then everybody who voted for H first should have his ballot deweighted, including later preferences (e.g. FRP > AP). That method isn't summable, but it's better[1]. You'd end up with something somewhat similar to Forest Simmons and Olli Salmi's "D'Hondt without lists", but with Sainte-Laguë instead of D'Hondt. See http://lists.electorama.com/pipermail/election-methods-electorama.com/2002-August/008561.html .

For those especially interested, the code (still using the D language)
is located here:
https://github.com/canidae/voting/blob/master/slrp.d
(Rough code, minimalistic RP, likely buggy, etc.)


Any thoughts? And is it something like this you're talking about,
Kristofer, or did I misunderstand you?

[1] Making a strongly summable setwise proportional method would make a lot of EM members pay attention, I think. I suspect it's impossible, but I only have a heuristic argument for it. By "strongly summable" I mean something that is summable with O(n^k) numbers of O(log(v)) digits each, n being the number of candidates and v the number of voters, for some constant k.
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