[EM] Method X

[Kristofer Munsterhjelm]

On 8/3/23 21:20, Kristofer Munsterhjelm wrote:

> I'm thinking that a more consistent or elegant tiebreaker is to use 
> leximax: find the next highest score obtainable, then the third highest, 
> etc. But it's a real pain to implement (particularly to pass through 
> vectors of scores recursively), so I haven't tried yet.

That might be much harder to do than I thought because you would want to 
avoid the Kemeny-esque clone problems that could result - e.g. adding a 
bunch of clones could give one candidate more high scores than another 
candidate due to there being more ways to eliminate the clones than before.

So perhaps it's better to consider the first preference scores 
throughout the optimal elimination order, e.g.

if the elimination order is "eliminate D, then C, then score is A>B"
then the extended score vector could be
	[A>B,  (first preference of A when D but not C is eliminated),
		A's first preferences]

though that could still produce clone problems, possibly... it's hard to 
reason about what would.

A fourth option is to relax the quota, e.g. make it 1/3 instead of 1/n 
and see which candidate gets the highest score, and use that as a 
tiebreaker. But it feels ad hoc. Quotas higher than 1/n still *work*, 
but used on their own (i.e. as methods by themselves, not as 
tiebreakers), the method X analogs are much more susceptible to strategy.

-km