[EM] Preliminary Droop-fit proportionality results

[Kristofer Munsterhjelm]

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