Re: Preliminary Droop-fit proportionality results
Toby Pereira via Election-Methods <[email protected]> Tue, 2 Jun 2026 21:11:29 +0000 (UTC)
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I see yes, thanks. In that case, Schulze STV seems to do pretty terribly, =
not just a bit worse than other STV methods. And looking at your whole list=
, considerably worse than things like (Bloc) Borda! It's definitely not bro=
ken in your simulation?
Toby
On Tuesday, 2 June 2026 at 20:33:08 BST, Kristofer Munsterhjelm <km-elm=
[email protected]> wrote: =20
=20
On 2026-06-02 16:54, Toby Pereira wrote:
> Interestingly Kristofer put the two methods together suggesting the=20
> simulation ran them as if they were they same method. I don't really=20
> know the difference between all the STV methods, but QPQ also got a=20
> score 0f 0.998 for 9 seats, suggesting it's up there as well. Do we know=
=20
> why Schulze outperformed the others for 5 seats but not 9?
I put Meek and Warren together because their errors and VSE values were=20
identical for every run, even though I implemented them as distinct=20
methods. They seem to be *very* close in practice.
As for Schulze on the five-seater, STV-ME is a different method than=20
Schulze STV. STV-ME is the following generalization of BTR-IRV:
=C2=A0=C2=A0=C2=A0 Do STV k-seat STV, but when a candidate needs to be elim=
inated, do an=20
"unlucky loser" election containing the k+1 candidates with the fewest=20
first preference votes, using the base method in question. (The=20
remaining candidates are eliminated from the unlucky loser election=20
before it is run.) Then eliminate the loser of that election, i.e. the=20
candidate ranked last by the base method.
So STV-ME(Schulze) is not Schulze STV, it's this BTR-IRV generalization=20
with Schulze as the method used to call the loser. In the single-winner=20
case, with a base method that passes the majority criterion, STV-ME=20
reduces to BTR-IRV.
For 5 seats, we have
=C2=A0 =C2=A0 =C2=A0 =C2=A0 Schulze STV=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 0.21
=C2=A0 =C2=A0 =C2=A0 =C2=A0 STV=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 0.94
=C2=A0 =C2=A0 =C2=A0 =C2=A0 QPQ (d'Hondt)=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0=
=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 0.94
=C2=A0 =C2=A0 =C2=A0 =C2=A0 Meek/Warren STV=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=
=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 0.94
=C2=A0 =C2=A0 =C2=A0 =C2=A0 STV-ME(Schulze)=C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=
=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 0.96
Schulze STV proper is still not all that great.
I guess STV-ME does better because its loser selection is less=20
susceptible to center squeeze-like problems; but that doesn't explain=20
why its less clearly an improvement with two seats than with five; if=20
the elimination process is the problem, then you'd expect that it would=20
beat STV more decisively the fewer seats you have.
In a naive combinatorial sense, 5-of-10 is the toughest because the=20
number of possible outcomes is maximized. But I don't know if that holds=20
for the proportionality problem as such.
-km
=20
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<html><head></head><body><div class=3D"ydpd090390eyahoo-style-wrap" style=
=3D"font-family:Helvetica Neue, Helvetica, Arial, sans-serif;font-size:13px=
;"><div></div>
<div dir=3D"ltr" data-setdir=3D"false">I see yes, thanks. In that c=
ase, Schulze STV seems to do pretty terribly, not just a bit worse than oth=
er STV methods. And looking at your whole list, considerably worse than thi=
ngs like (Bloc) Borda! It's definitely not broken in your simulation?</div>=
<div dir=3D"ltr" data-setdir=3D"false"><br></div><div dir=3D"ltr" data-setd=
ir=3D"false">Toby</div><div><br></div>
=20
</div><div id=3D"ydp60f28aa2yahoo_quoted_1247257313" class=3D"ydp60=
f28aa2yahoo_quoted">
<div style=3D"font-family:'Helvetica Neue', Helvetica, Arial, s=
ans-serif;font-size:13px;color:#26282a;">
=20
<div>
On Tuesday, 2 June 2026 at 20:33:08 BST, Kristofer =
Munsterhjelm <[email protected]> wrote:
</div>
<div><br></div>
<div><br></div>
=20
=20
<div><div dir=3D"ltr">On 2026-06-02 16:54, Toby Pereira wro=
te:<br></div><div dir=3D"ltr">> Interestingly Kristofer put the two meth=
ods together suggesting the <br></div><div dir=3D"ltr">> simulation ran =
them as if they were they same method. I don't really <br></div><div dir=3D=
"ltr">> know the difference between all the STV methods, but QPQ also go=
t a <br></div><div dir=3D"ltr">> score 0f 0.998 for 9 seats, suggesting =
it's up there as well. Do we know <br></div><div dir=3D"ltr">> why Schul=
ze outperformed the others for 5 seats but not 9?<br></div><div dir=3D"ltr"=
><br></div><div dir=3D"ltr">I put Meek and Warren together because their er=
rors and VSE values were <br></div><div dir=3D"ltr">identical for every run=
, even though I implemented them as distinct <br></div><div dir=3D"ltr">met=
hods. They seem to be *very* close in practice.<br></div><div dir=3D"ltr"><=
br></div><div dir=3D"ltr">As for Schulze on the five-seater, STV-ME is a di=
fferent method than <br></div><div dir=3D"ltr">Schulze STV. STV-ME is the f=
ollowing generalization of BTR-IRV:<br></div><div dir=3D"ltr"><br></div><di=
v dir=3D"ltr"> Do STV k-seat STV, but when a candidate ne=
eds to be eliminated, do an <br></div><div dir=3D"ltr">"unlucky loser" elec=
tion containing the k+1 candidates with the fewest <br></div><div dir=3D"lt=
r">first preference votes, using the base method in question. (The <br></di=
v><div dir=3D"ltr">remaining candidates are eliminated from the unlucky los=
er election <br></div><div dir=3D"ltr">before it is run.) Then eliminate th=
e loser of that election, i.e. the <br></div><div dir=3D"ltr">candidate ran=
ked last by the base method.<br></div><div dir=3D"ltr"><br></div><div dir=
=3D"ltr">So STV-ME(Schulze) is not Schulze STV, it's this BTR-IRV generaliz=
ation <br></div><div dir=3D"ltr">with Schulze as the method used to call th=
e loser. In the single-winner <br></div><div dir=3D"ltr">case, with a base =
method that passes the majority criterion, STV-ME <br></div><div dir=3D"ltr=
">reduces to BTR-IRV.<br></div><div dir=3D"ltr"><br></div><div dir=3D"ltr">=
For 5 seats, we have<br></div><div dir=3D"ltr"><br></div><div dir=3D"ltr">&=
nbsp; Schulze STV &=
nbsp; 0.21<br></div><div dir=3D"ltr">&nb=
sp; STV &nbs=
p; 0.94<br></div><d=
iv dir=3D"ltr"> QPQ (d'Hondt) &nbs=
p; 0.94<br></div><div dir=
=3D"ltr"> Meek/Warren STV &=
nbsp; 0.94<br></div><div dir=3D"ltr">&nb=
sp; STV-ME(Schulze) =
0.96<br></div><div dir=3D"ltr"><br></div><div =
dir=3D"ltr">Schulze STV proper is still not all that great.<br></div><div d=
ir=3D"ltr"><br></div><div dir=3D"ltr">I guess STV-ME does better because it=
s loser selection is less <br></div><div dir=3D"ltr">susceptible to center =
squeeze-like problems; but that doesn't explain <br></div><div dir=3D"ltr">=
why its less clearly an improvement with two seats than with five; if <br><=
/div><div dir=3D"ltr">the elimination process is the problem, then you'd ex=
pect that it would <br></div><div dir=3D"ltr">beat STV more decisively the =
fewer seats you have.<br></div><div dir=3D"ltr"><br></div><div dir=3D"ltr">=
In a naive combinatorial sense, 5-of-10 is the toughest because the <br></d=
iv><div dir=3D"ltr">number of possible outcomes is maximized. But I don't k=
now if that holds <br></div><div dir=3D"ltr">for the proportionality proble=
m as such.<br></div><div dir=3D"ltr"><br></div><div dir=3D"ltr">-km<br></di=
v></div>
</div>
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