4 candidates, 4 voters: manipulable  475/768  (0.62), 9% tied
4 candidates, 5 voters: manipulable 1165/4608 (0.25), <0.1% tied

4 candidates, 4 voters: manipulable 1129/2304 (0.49), 22% tied
4 candidates, 5 voters: manipulable  215/1152 (0.19), <0.1% tied

4 candidates, 4 voters: manipulable 1189/2304 (0.52), 20% tied
4 candidates, 5 voters: manipulable 1585/6912 (0.23), <0.1% tied

4 candidates, 4 voters: manipulable  379/576  (0.66), 7% tied
4 candidates, 5 voters: manipulable 3655/6912 (0.53), 3% tied

Define the fpA-fpC scores for three candidates like this: - If W is the CW, then W gets score equal to |V|, the number of voters; and the other two get score 0 - If the Smith set is size 2, both get |V|/2 and the final candidate gets 0 - Otherwise the scores are as in fpA-fpC. If the cycle is A=B>C>A, then B is counted as if A beat him pairwise. Now define minx(fpA-fpC) as the following: - fpA-fpC restricted to a three-set is the fpA-fpC election with every candidate but those three eliminated. - A's score is the vector of three-candidate fpA-fpC scores of every three-set that includes A. - The candidate with the greatest leximin score wins. Clearly, this method is monotone because raising A can't decrease A's score on any three-set containing A, due to three-candidate fpA-fpC being monotone. And it's summable because we only need the Condorcet matrix, as well as three Plurality count values for every three-set. Can anyone find a DMTBR counterexample for minx(fpA-fpC)? I haven't been able to for either four or five candidates. Since DMTBR plus Condorcet and either monotonicity or summability is a high bar in itself, I want to be *very* sure before proclaiming that I've accomplished something spectacular. Even proving plain DMT is hard! I think the logic is something like: a candidate outside of the DMT set will have very low scores when matched up against someone *in* the DMT set. No amount of extremely good scores elsewhere can compensate for that when the metric is leximin. On the other hand, someone *in* the DMT set will not get these very low scores against other candidates. A test election: 3: A>B>C>D 1: B>C>D>A 2: C>D>A>B 1: C>D>B>A The sorted score arrays, according to my implementation, are: A: (0, 0, 0) B: (-2, -2, 7) C: (0, 2, 7) D: (0, 0, 2) so the social order is C>D>A>B. My manipulability check gives the following exact results for Smith,minx(fpA-fpC): 4 candidates, 4 voters: manipulable 475/768 (0.62), 9% tied 4 candidates, 5 voters: manipulable 1165/4608 (0.25), <0.1% tied To compare, here's Smith,IRV: 4 candidates, 4 voters: manipulable 1129/2304 (0.49), 22% tied 4 candidates, 5 voters: manipulable 215/1152 (0.19), <0.1% tied Smith,IFPP: 4 candidates, 4 voters: manipulable 1189/2304 (0.52), 20% tied 4 candidates, 5 voters: manipulable 1585/6912 (0.23), <0.1% tied and Smith,Ext-Minmax: 4 candidates, 4 voters: manipulable 379/576 (0.66), 7% tied 4 candidates, 5 voters: manipulable 3655/6912 (0.53), 3% tied With 4 voters, tie-breaking matters a lot (the nitty-gritty as it were). But with five, there's a threshold of sorts with strategy resistant methods having 20%-30% susceptibility and not so strategy resistant ones having 50% or more, and minx(fpA-fpC) is on the right side of this divide.

Hi Kristofer, This method is a little tricky to double-check so I have a little less confidence than usual in offering scenarios... Le samedi 16 avril 2022, 06:05:45 UTC−5, Kristofer Munsterhjelm <km_elmet@t-online.de> a écrit : > Define the fpA-fpC scores for three candidates like this: >     - If W is the CW, then W gets score equal to |V|, the number of voters; > and the other two get score 0 >     - If the Smith set is size 2, both get |V|/2 and the final candidate gets 0 >     - Otherwise the scores are as in fpA-fpC. If the cycle is A=B>C>A, then > B is counted as if A beat him pairwise. That last sentence is an interesting choice. Since Schwartz is {B} I might treat B the same as a CW. Also what about when there are two ties? A=B, B=C, C>A. It's a size 3 Smith set. > Now define minx(fpA-fpC) as the following: >     - fpA-fpC restricted to a three-set is the fpA-fpC election with every > candidate but those three eliminated. >     - A's score is the vector of three-candidate fpA-fpC scores of every > three-set that includes A. >     - The candidate with the greatest leximin score wins. >  > Clearly, this method is monotone because raising A can't decrease A's > score on any three-set containing A, due to three-candidate fpA-fpC > being monotone. And it's summable because we only need the Condorcet > matrix, as well as three Plurality count values for every three-set. Hmm, but try this one: 0.345: C>D>B>A 0.320: A>C>D>B 0.243: B>A>C>D 0.090: D>B>C>A  -->  D>C>B>A It seems to me this moves the win from C to A. Defeats are A>C>B>A, A>D, C>D, D>B. They're unchanged by the vote. Changing B>C to C>B seems to help A by reducing B's effective first pref count for the three-set {A,B,C}. > Can anyone find a DMTBR counterexample for minx(fpA-fpC)? I haven't been > able to for either four or five candidates. Since DMTBR plus Condorcet > and either monotonicity or summability is a high bar in itself, I want > to be *very* sure before proclaiming that I've accomplished something > spectacular. How about this: 0.389: D>A>C>B 0.334: A>D>B>C  -->  A>B>C>D 0.162: B>A>D>C 0.114: C>B>D>A Initially D is the CW and has over a third of first prefs. I think this vote change elects A. Lowering D creates a B>D win. The other wins are A>B, A>C, C>B, D>A, D>C. D's worst score becomes -.107 (from {B,C,D}) while A's worst is -.055 (from {A,B,D}). Let me know if something seems off. Kevin

On 17.04.2022 07:46, Kevin Venzke wrote: > Hi Kristofer, > > This method is a little tricky to double-check so I have a little less > confidence than usual in offering scenarios... > > Hmm, but try this one: > > 0.345: C>D>B>A > 0.320: A>C>D>B > 0.243: B>A>C>D > 0.090: D>B>C>A  -->  D>C>B>A > > It seems to me this moves the win from C to A. > > Defeats are A>C>B>A, A>D, C>D, D>B. They're unchanged by the vote. > Changing B>C to C>B seems to help A by reducing B's effective first pref count > for the three-set {A,B,C}. Yeah, it checks out. The example can be reduced to: 7: C>D>B>A 6: A>C>D>B 5: B>A>C>D 2: D>B>C>A so it shows a problem with my monotonicity checker. Since my method is pretty similar to IFPP, it may be useful to look at the latter to get an indication of whether the "wrong" winner is D before the raising or A after. IFPP elects C in both cases. OTOH, the set narrowing method I mentioned earlier (where I was having problems with cycles) gives the outcome as a tie between A, C, and D before; and as A uniquely after. So who knows? >> Can anyone find a DMTBR counterexample for minx(fpA-fpC)? I haven't been >> able to for either four or five candidates. Since DMTBR plus Condorcet >> and either monotonicity or summability is a high bar in itself, I want >> to be *very* sure before proclaiming that I've accomplished something >> spectacular. > > How about this: > > 0.389: D>A>C>B > 0.334: A>D>B>C  -->  A>B>C>D > 0.162: B>A>D>C > 0.114: C>B>D>A > > Initially D is the CW and has over a third of first prefs. I think this vote > change elects A. > > Lowering D creates a B>D win. The other wins are A>B, A>C, C>B, D>A, D>C. > D's worst score becomes -.107 (from {B,C,D}) while A's worst is -.055 (from > {A,B,D}). That seems right too. I managed to round it off to: 9: D>A>C>B 8: A>D>B>C 3: B>A>D>C 3: C>B>D>A Again, my Monte-Carlo program doesn't seem to recognize the "before burial" election to be vulnerable to burial. (The exact manipulability results for 5 voters should still hold, though; that's a different calculation and it doesn't check specifically for burial.) Thanks! -km