[EM] A few more Bucklin variants, because why not?

[Kristofer Munsterhjelm]

On 2024-09-21 13:53, Etjon Basha wrote:
> Dear gentlemen,
> 
> 
> A while ago I did write here about the Iterated Bucklin 
> <https://electowiki.org/wiki/Iterated_Bucklin> method on which I’ve 
> recently had a chance to think and generalize about a bit more. Maybe 
> some of the below could be novel or otherwise of interest.
> 
> 
> First, and for our purposes today, let's define the *Serious Candidates 
> Set* in the context of a ranked ballot, to include those candidates who 
> would win an approval count if they served as the approval cutoff across 
> all ballots.
> 
> 
> In the [2:A>B, 3:C>A, 4:A>B] election as an example, the Set would 
> include A and B only, as applying the cutoff at C would still elect B.
> 
> 
> I’ve been checking some random simulations from Kevin Venzke’s 
> votingmethods.net <http://votingmethods.net>, and here are some 
> properties of this Set that I *suspect*:
> 
> 1.If there is a Condorcet Winner, this Set should always include them.

I think that should hold, at least with strict ranks. Suppose A beats B 
pairwise. Then if we set the approval cutoff to A inclusive, every 
ballot that has B>A will give a point to B and A, whereas every ballot 
that has A>B will give a point only to A. Since A beats B pairwise, the 
number of A>B ballots exceed the number of B>A ballots. Thus A must have 
a higher approval score than B.

Since A is the CW, this reasoning holds for all other B. A must obtain a 
higher approval score than B for any other B. Hence A is the winner and 
belongs to the SCS.

> 2.Otherwise, this Set should always partially overlap with the Smith Set.

The reasoning above gets us part of the way. Suppose A is in the Smith 
set. For any non-Smith member B, A must obtain more points than B for 
the same reason as above. So the non-Smith members can't win.

What we have to show is that it's impossible for e.g. A to be the winner 
when the approval cutoff is set at B, B be the winner when it's set at 
C, and C be the winner when it's set at A. I'm not sure how to prove 
that, though.

-km