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