Re: [EM] Possibly making Sainte-Lague even more STV-like
On Wed, Sep 04, 2013 at 12:14:36AM +0200, Kristofer Munsterhjelm wrote:
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.
Very well explained. Some quick thoughts this early morning:
I fully agree with you that it would make much more sense for {LR} to
win in a 2-seat election.
If you were to elect some of the seats using a quota, then resolve any
remaining seats using a pairwise method, I do see some issues once you
get more parties into the equation, let's assume a 4-seat election:
28: ABDFEC
1: BCAFED
27: CBDFEA
11: DBFACE
7: EBDCAF
26: FBDAEC
B is despite receiving very few first preference vote, still a candidate
that every voter rank high, but depending on how you'll weight down the
votes after distributing the directly (quota) won seats, B may end up
winning no seats at all. That's not quite fortunate either.
Not saying that the monotonocity criterion must be fulfilled. In your
example I would agree that in a 2-seat election then L and R should win
each their seat (and C should win in a 1-seat election), but in my
example above I would expect B to win at least one seat, regardless of
the amount of available seats (with the possible exception of 3 seats,
where 3 parties stand out, but still not enough to reach the quota).
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
.
Interesting read, and I absolutely see your point about subsequent
preferences remaining at full strength.
Some months ago I believe I actually tried this approach (that is,
reweighting the entire vote and not just the vote strength for parties
who already won some seats), but I scrapped the idea and the code after
receiving some peculiar results. I can't rule out that the
implementation was flawed, I might give this another shot.
--
Regards,
Vidar Wahlberg
Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Possibly making Sainte-Lague even more STV-like
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. 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. So I implemented a basic Ranked Pair algorithm, but with a tiny modification: When you compare two parties against each other, you divide the amount of votes that prefer party A over party B with the Sainte-Laguë divisor (2 * seats + 1), and do the same for votes that prefer party B over party A. In other words, if 3 votes prefer party A over party B, 2 votes prefer party B over party A and neither parties have won any seats, then you'll just compare 3 vs. 2. If party A won one seat already while party B won no seats so far then it'll be 3 / 3 vs. 2. When I feed the algorithm with the result from the Norwegian 2005 election (where all votes have only one preference and consider all other parties equally bad), then as expected, the result will be as if only using normal Sainte-Laguë. This was the result: A: 56 seats SV: 15 seats RV: 2 seats SP: 11 seats KRF: 12 seats V: 10 seats H: 24 seats FRP: 37 seats KYST: 1 seat PP: 1 seat 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. 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? -- Regards, Vidar Wahlberg Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] Possibly making Sainte-Lague even more STV-like
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: LCR
42: RCL
10: CRL
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
