[EM] Improvement on Jobst's Chain climbing method

[Kevin Venzke]

 Hi Forest, I agree your revision seems better. I also don't like the potential for C3 to change the outcome like that.
It seems like the revised method would usually elect C1, with the most common exception being when C2 pairwise defeats C1. As you proceed down the list it starts to become hard to imagine how the candidate could do well enough in pairwise contests (to be elected) while doing so poorly in approval.
So this seems to me like a clone-independent way of fixing an overly simple method like "elect the approval winner unless second place beats him pairwise."
Kevin


 Le samedi 2 mars 2019 à 15:20:39 UTC−6, Forest Simmons <fsimmons at pcc.edu> a écrit : 

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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