Dear Daniel, > Step 1: Sort candidates according to your favourite rule. > Step 2: Pick the bottom two candidates. Remove the pairwise loser. > Step 3: Repeat until only 1 candidate is left. > > Every method in this category is Smith-efficient, so it automatically > meets many important rules like Condorcet loser, Mutual Majority, > and ISDA. This proposal doesn't automatically meet ISDA. Markus Schulze

Dear Markus, On Tue, Jun 29, 2021 at 7:16 AM Markus Schulze <markus.schulze8@gmail.com> wrote: > Dear Daniel, > > > Step 1: Sort candidates according to your favourite rule. > > Step 2: Pick the bottom two candidates. Remove the pairwise loser. > > Step 3: Repeat until only 1 candidate is left. > > > > Every method in this category is Smith-efficient, so it automatically > > meets many important rules like Condorcet loser, Mutual Majority, > > and ISDA. > > This proposal doesn't automatically meet ISDA. > Hmm... yes, I see that. My bad. I think that if you change Step 2 so that the bottom candidate is compared against every other candidate *AND* you pick a sorting rule that puts the Smith set on top, then I think that would meet IDSA. For example: Step 1: Rank candidates by number of pairwise wins (Copeland sorting). Use first-place votes to break ties. Step 2: Pick the bottom candidate and compare him against every other candidate. If he loses any match, kick him out. Step 3: Repeat until you find a Condorcet winner or only 1 candidate is left. I think *this* version would meet ISDA. Am I right? Someone on Reddit has made the case to try to make the system resistant to DH3. The only way I can think of to avoid DH3 is to sort by first-place votes, but then I suspect that that wouldn't meet ISDA. Cheers, Daniel

 Dear Daniel,

   > Step 1: Sort candidates according to your favourite rule.
   > Step 2: Pick the bottom two candidates. Remove the pairwise loser.
   > Step 3: Repeat until only 1 candidate is left.
   >
   > Every method in this category is Smith-efficient, so it
 automatically
   > meets many important rules like Condorcet loser, Mutual Majority,
   > and ISDA.

 This proposal doesn't automatically meet ISDA.

On 6/29/21 11:18 PM, Daniel Carrera wrote: > Dear Markus, > > On Tue, Jun 29, 2021 at 7:16 AM Markus Schulze > <markus.schulze8@gmail.com <mailto:markus.schulze8@gmail.com>> wrote: > > Dear Daniel, > >  > Step 1: Sort candidates according to your favourite rule. >  > Step 2: Pick the bottom two candidates. Remove the pairwise loser. >  > Step 3: Repeat until only 1 candidate is left. >  > >  > Every method in this category is Smith-efficient, so it > automatically >  > meets many important rules like Condorcet loser, Mutual Majority, >  > and ISDA. > > This proposal doesn't automatically meet ISDA. > > > Hmm... yes, I see that. My bad. I think that if you change Step 2 so > that the bottom candidate is compared against every other candidate > *AND* you pick a sorting rule that puts the Smith set on top, then I > think that would meet IDSA. For example: I don't think that works. Suppose we have an election where A, B, and C are in the Smith set (in an ABCA cycle), and D and E are not, but the base method fails ISDA. Suppose that the base method outcome for the election is A>B>C>D>E with all candidates present, but if D and E are eliminated first, then it's B>C>A (hence ISDA failure). Then in the case where all candidates remain, the list removal method will go like this: - E is kicked off (not in Smith set) - D is kicked off (ditto) - C is kicked off (beaten by B) - B is kicked off (beaten by A) A wins. If everybody outside the Smith set were eliminated first, the list removal method would go: - A is kicked off (beaten by C) - C is kicked off (beaten by B) and B wins. > Step 1: Rank candidates by number of pairwise wins (Copeland sorting). > Use first-place votes to break ties. > Step 2: Pick the bottom candidate and compare him against every other > candidate. If he loses any match, kick him out. > Step 3: Repeat until you find a Condorcet winner or only 1 candidate is > left. > > I think *this* version would meet ISDA. Am I right? I would have to fire up my linear programming solver to craft a scenario for that particular method, but I think the general sketch above should work. -km

On 6/30/21 1:13 AM, Kristofer Munsterhjelm wrote: > I would have to fire up my linear programming solver to craft a scenario > for that particular method, but I think the general sketch above should > work. ... by which I mean the general disproof sketch, not that the method would work :-) Just to clear up any ambiguity. -km