[EM] Preliminary Droop-fit proportionality results
Toby Pereira
tdp201b at yahoo.co.uk
Tue Jun 2 14:11:29 PDT 2026
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 broken in your simulation?
Toby
On Tuesday, 2 June 2026 at 20:33:08 BST, Kristofer Munsterhjelm <km-elmet at munsterhjelm.no> wrote:
On 2026-06-02 16:54, Toby Pereira wrote:
> Interestingly Kristofer put the two methods together suggesting the
> simulation ran them as if they were they same method. I don't really
> know the difference between all the STV methods, but QPQ also got a
> score 0f 0.998 for 9 seats, suggesting it's up there as well. Do we know
> why Schulze outperformed the others for 5 seats but not 9?
I put Meek and Warren together because their errors and VSE values were
identical for every run, even though I implemented them as distinct
methods. They seem to be *very* close in practice.
As for Schulze on the five-seater, STV-ME is a different method than
Schulze STV. STV-ME is the following generalization of BTR-IRV:
Do STV k-seat STV, but when a candidate needs to be eliminated, do an
"unlucky loser" election containing the k+1 candidates with the fewest
first preference votes, using the base method in question. (The
remaining candidates are eliminated from the unlucky loser election
before it is run.) Then eliminate the loser of that election, i.e. the
candidate ranked last by the base method.
So STV-ME(Schulze) is not Schulze STV, it's this BTR-IRV generalization
with Schulze as the method used to call the loser. In the single-winner
case, with a base method that passes the majority criterion, STV-ME
reduces to BTR-IRV.
For 5 seats, we have
Schulze STV 0.21
STV 0.94
QPQ (d'Hondt) 0.94
Meek/Warren STV 0.94
STV-ME(Schulze) 0.96
Schulze STV proper is still not all that great.
I guess STV-ME does better because its loser selection is less
susceptible to center squeeze-like problems; but that doesn't explain
why its less clearly an improvement with two seats than with five; if
the elimination process is the problem, then you'd expect that it would
beat STV more decisively the fewer seats you have.
In a naive combinatorial sense, 5-of-10 is the toughest because the
number of possible outcomes is maximized. But I don't know if that holds
for the proportionality problem as such.
-km
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20260602/96720b1b/attachment.htm>
More information about the Election-Methods
mailing list