Here is a simple proof of concept, attempting to find a smallest set of scenarios which can differentiate a set of election methods. Here I've used full, strict rankings with three candidates and four blocs, and I've used a rather short list of 17 methods. The result is that 7 scenarios will differentiate them all. Note that certain methods become identical with such parameters: MinMax includes MMPO (of course), MAMPO, and numerous others. DSC includes DAC. IRV includes King of the Hill (KOTH). C//A includes MinMax(AWP) and MDDA. Approval elimination runoff (AER) includes DMC. C//FPP includes BTRIRV. Antiplurality is equivalent to implicit Approval. fpA-fpC includes C//IFPP.* (* But actually, I think these two may be the exact same method. That would be interesting not just because the formulations of the logic look very different, but also because the motivations behind them weren't the same. fpA-fpC was discovered based on a search for low-manipulability methods (if I remember correctly what Kristofer did) while IFPP seems intended to fix the monotonicity issue of IRV and nothing more.) The scenarios are evaluated in order, so that as we go on fewer and fewer distinctions can be made. Pairs of methods split at any step aren't considered further. --- Scenario 1 0.4770: B>C>A 0.2471: A>B>C 0.1960: A>C>B 0.0796: C>A>B Elect A: Stensholt BPW, IRV, C//IRV, IFPP, fpA-fpC, declonedCopeland Elect B: Approval elimination runoff (AER), Borda, FPP, C//FPP, DSC, MinMax Elect C: Antiplurality, Bucklin, C//A, ChainRunoff, TACC It's hard to completely generalize these groups, but a number of "anti-burial" methods pick A. Many approval-oriented methods pick C. The B group is rather diverse, but includes several FP-emphasizing methods. Scenario 2 0.3222: A>C>B 0.3048: B>A>C 0.2100: C>B>A 0.1628: C>A>B Elect A: AER, C//A, Antiplurality, Borda, Bucklin, BPW, IRV, C//IRV, dc-Copeland, MinMax Elect B: ChainRunoff Elect C: FPP, C//FPP, DSC, IFPP, fpA-fpC, TACC The C group unites some intuitively FP-emphasizing methods, and almost every other method is in the A group. Chain Runoff's result of B is unique. Any unique result would differentiate a method in a single step. TACC is also grouped with completely different methods from before, and so is completely differentiated. Of note, this scenario lets us differentiate now between BPW, IRV, and C//IRV on on one side, and IFPP and fpA-fpC on the other. Scenario 3 0.4119: A>C>B 0.2785: B>C>A 0.2431: C>A>B 0.0664: A>B>C Elect A: Borda, ChainRunoff, DSC, FPP, IFPP, IRV Elect C: AER, Antiplurality, BPW, Bucklin, C//A, C//FPP, C//IRV, fpA-fpC, dc-Copeland, MinMax, TACC Here C is the Condorcet winner. This establishes splits especially between non-Condorcet methods and their Condorcet versions. Additionally Borda gets split off from MinMax and AER; DSC gets split from C//FPP; and IRV splits from BPW and dc-Copeland. Scenario 4 0.4012: B>C>A 0.2855: A>B>C 0.1776: A>C>B 0.1355: C>A>B Elect A: FPP, C//FPP, ChainRunoff, IRV, C//IRV, dc-Copeland, DSC, IFPP, fpA-fpC, MinMax   Elect B: AER, Borda Elect C: Antiplurality, BPW, Bucklin, C//A, TACC This scenario splits AER (elect B) from MinMax (elect A). And it splits BPW (elect C) from C//IRV and dc-Copeland (elect A). Scenario 5 0.3202: A>B>C 0.2788: C>B>A 0.2145: B>A>C 0.1863: C>A>B Elect A: most of the methods Elect B: Antiplurality, Bucklin     Elect C: FPP, IFPP This splits off C//A from Antiplurality and Bucklin. And it splits up FPP and DSC. Scenario 6 0.3904: A>B>C 0.2525: C>B>A 0.2054: A>B>C 0.1516: B>A>C Elect A: everything except Antiplurality Elect B: Antiplurality A is majority favorite. All this does is split Antiplurality and Bucklin. Scenario 7 0.3911: C>A>B 0.2984: B>C>A 0.2814: A>B>C 0.0289: A>C>B Elect A: TACC Elect B: BPW, dc-Copeland   Elect C: all other methods There is no CW, but almost all methods agree on C winning. This scenario just splits up C//IRV and dc-Copeland. --- And that's the end. If we add more methods and allow truncation or equal-ranking, the number of scenarios needed doesn't grow very fast. What would have been quite useful is if the small number of scenarios could be translated into a "method DNA" setting, where the number of possible scenarios is relatively large despite offering less flexibility in expressiveness. Unfortunately that doesn't seem very feasible because in the DNA setting we can define the "movement" from one scenario to another, and thereby evaluate criteria, which we can't do here. A slightly different approach could be to focus on finding a few scenarios that each independently differentiate as many methods as possible, while also correlating as little as possible with the other scenarios' differentiations. Then these several scenarios could be used as a litmus test to gauge what a proposed rank method is similar to, without having to implement it. Kevin

On 14.02.2022 02:55, Kevin Venzke wrote: > A slightly different approach could be to focus on finding a few scenarios that > each independently differentiate as many methods as possible, while also > correlating as little as possible with the other scenarios' differentiations. > Then these several scenarios could be used as a litmus test to gauge what a > proposed rank method is similar to, without having to implement it. For my minimal strategy simulations, something with very few candidates would be nice... but I suspect that the three-candidate minimal strategy method would be very close to fpA-fpC anyway. I also tried to do something like this in quadelect, by using the mean Kendall-tau distances between voting methods' results to give the distance between voting methods, and then constructing a dendrogram from it with hierarchical clustering. I got some strange results, so perhaps the "winner agreement distance" would be better, i.e. the distance between methods is the chance that they disagree for a randomly chosen ballot. And then use a more plausible model than impartial culture - say IAC or a spatial model - for the ballot randomness. In any case, if you do that, you could try to find an election set that distinguishes between the left branch and right branch at every level of the tree. This could possibly work as a more principled way to group methods together. On the other hand, there's no guarantee that there exist a single election that will divide the sets in the same way that the clustering method does (unless that's made a constraint on the clustering process itself). With only a few methods (say 9 or 16), it should be feasible to brute force such an assignment: separate methods into two (or three) equally sized clusters and check if there's a single election that tells them apart. You could do that by random checking or by integer programming. The former is (much) easier to program; the latter either definitely returns a no (no such election exists) or a yes (here's the separating election you want). -km