[EM] Improvement on Jobst's Chain climbing method

[Richard Lung]

The American context of voting theory, in a country where single-winner 
elections predominate, is minimally democratic (maiorocracy, the tyranny 
of the majority: JS Mill) so unlikely to provide general conclusions.
And Condorcet is not an actual method, but a cross-checking of the 
result for any given method. Anyway, I suspect that this approach has 
turned into a search for the best way to eliminate candidates.
 From my point of view, it is an unnecessary task, because the 
exclusion  of unprefered candidates can be conducted in a same accurate, 
rational manner (using Meek keep values) as the election of prefered 
candidates: Binomial STV, actually FAB STV. This does away with 
discarding the votes of candidates before the whole count has been 
completed. And thereby does away with non-monotonicity (preferential 
illogic in the count).

from Richard Lung.



On 02/03/2019 21:20, Forest Simmons wrote:
> A few years back Jobst suggested "chain climbing" as a seamless, 
> Condorcet compliant way of selecting an alternative from a given 
> ordered list.
>
> For example electing a winner from a list of candidates c1, c2, ... 
> given in decreasing order of approval.
>
> Chain Climbing initializes a chain of candidates with the last (least 
> approved in this case) candidate in the list. Then moving up the list 
> each successive candidate "climbs the chain" as far it can before 
> being bumped off by a chain member that defeats it. If it makes it all 
> of the way to the top, it is added to the top of the chain.
>
> The candidate who ends up at the top of the chain is elected.
>
> Since a beats all candidate will never be defeated, the method is 
> Condorcet compliant. It also turns out to be clone resistant and 
> monotonic.
>
> Another nice property is that it always selects from the Banks set, a 
> nice game theoretic subset set of the set of uncovered candidates.
>
> The biggest objection to this method is that when applied to a list a 
> list where c1 beats c2 beats c3, and c3 beats c1, it elects c2.
>
> Here's my proposed improvement:
>
> Initialize the chain with c1.  Move down the list instead of up.  For 
> each successive candidate x (as we move down the list) if possible, 
> insert that candidate into the chain at a point where it is beaten by 
> every candidate above it and is not defeated by any candidate below 
> it.  If not possible, discard it.
>
> After going through the entire list (top to bottom) inserting new 
> candidates where possible into the totally ordered chain, we end up 
> with a maximal totally ordered chain of candidates (ordered by 
> pairwise defeat) The candidate at the top fo the completed chain (the 
> one who is not defeated by any of the others) is elected.
>
> It is easy to show that this method has all of the nice properties of 
> chain climbing, but retains more of the spirit of the original list..
>
> For example in the A>B>C example above it elects A.
>
> What do you think?
>
> Forest
>
>
>
>
> ----
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