[EM] Improvement on Jobst's Chain climbing method

[Chris Benham]

Forest,

I was impressed by your  "Max Covered Approval" method idea, which seems 
to be similar to this.

It differs by requiring any new addition to the chain to cover (not just 
pairwise beat) the last one.

So if the most approved candidate was uncovered it won, otherwise the 
most most approved candidate
that covered it if that candidate is uncovered, otherwise the most 
approved candidate that covered it,
and so on.

This is one of my two or three favourite (distinctly different from each 
other) Condorcet methods.

Chris Benham

On 3/03/2019 7:50 am, 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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