Oh, good example. I mean, the FPC scores are different - 16:17:15 and 16:15:17 respectively - but it unfortunately still gives a nonmonotonic result (chaging A to C). Standard FPC behaves in a more logical way here, changing B to C as B gets lowered. I don't know, perhaps there could exist some middle ground between sequential elimination and lack thereof? (not Smith//FPC, it is still susceptible to clones.) Or it can be an entirely wrong tree to be barking up... Although for me this idea still looks temptingly nice. It is my impression that all elimination methods are not monotonic, though > I certainly can't prove that. > > In this case, if I've understood it correctly, try > > 7 A>B>C > 2 B>A>C > 7 B>C>A > 8 C>A>B > > Pairwise we have a cycle: A>B>C>A, so FPC scores are A:9; B:8; C:7. Thus, > the method eliminates C and A beats B and A wins. > > Now suppose the two B>A>C voters increase their appreciation of A and vote > A>B>C instead. > > 9 A>B>C > 7 B>C>A > 8 C>A>B > > We've got the same pairwise cycle, but now A gets eliminated first and B > wins. > > Did I calculate correctly? Is this the type of monotonicity you were > asking about? > > Jim Faran > ------------------------------ > *From:* Election-Methods <election-methods-bounces@lists.electorama.com> > on behalf of Filip Ejlak <tersander@gmail.com> > *Sent:* Monday, April 17, 2023 12:19 PM > *To:* election-methods@lists.electorama.com < > election-methods@lists.electorama.com> > *Subject:* [EM] a simple, great ISDA-compliant method I've never seen > mentioned > > Hello, everyone, new to the list here, > > what I really like about First preference Copeland is that it implies a > nice method for calculating a sort of "Condorcet score": just count 1st > preferences of all candidates (including candidate X) that do not beat X > pairwise. Although it is highly resistant to burial, FPC is not cloneproof > and fails ISDA. > > However, there is an easy way to fix these two problems: just do > sequential loser elimination using FPC score! > (And when there is a tie at the bottom, break it by applying the same > method to the candidates involved in the tie and finding the loser.) > > This method could be named FPCE (First Preference Copeland Elimination) or > ISCR (Instant Simmons-Copeland Runoff). It is cloneproof (because even if > there are clones which hurt one another, eventually there is only one of > them left) and ISDA-compliant (because all candidates outside the Smith set > get eliminated before any Smith set member is eliminated). I think its > burial resistance might be weakened, though, as it fails DMTBR. > > Could anyone help with proving/disproving monotonicity, or even LIIA? I > applied the method to the 5-candidate Smith set from the Wikipedia article > about Schulze and there turned out to be an LIIA-compliant order: ADCEB. > I'm curious if it was by accident or not. > > Filip Ejlak >

On 4/17/23 22:42, Filip Ejlak wrote: > Oh, good example. I mean, the FPC scores are different - 16:17:15 and > 16:15:17 respectively - but it unfortunately still gives a nonmonotonic > result (chaging A to C). > > Standard FPC behaves in a more logical way here, changing B to C as B > gets lowered. I don't know, perhaps there could exist some middle ground > between sequential elimination and lack thereof? (not Smith//FPC, it is > still susceptible to clones.) Or it can be an entirely wrong tree to be > barking up... Although for me this idea still looks temptingly nice. First of all, welcome! It's always nice to see new people on the list :-) The usual middle point between elimination and none is below-average elimination like what Carey does to be monotone in the three-candidate case. However, you would probably lose ISDA this way, and it doesn't necessarily work for larger number of candidates. Other ideas I've been thinking about include an idea of somehow making A's score the maximum through some elimination sequence that always favors A. E.g. let X be a sequence of the candidates except for A and two others, so that when you eliminate one after another (in the order of the sequence) then the (k+1)-th candidate in the sequence is never ranked first by first preference Copeland once the k first candidates are eliminated. Then A's score is the score in the final first preference Copeland election after everybody but A and the two others are eliminated, when the sequence is chosen to maximize A's score in this way, where the idea is that the max operator forces monotonicity. But I never managed to make that concrete. Perhaps something along those lines could work, though, but it would be much more hairy than plain FPC. -km