[EM] A family of easy-to-explain Condorcet methods

Kristofer Munsterhjelm km_elmet at t-online.de
Tue Jun 29 16:13:23 PDT 2021

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 at gmail.com <mailto:markus.schulze8 at 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


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 


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