[EM] Improvement on Jobst's Chain climbing method

[Toby Pereira]

Is this (the objection) definitely correct? I might be being stupid or might have misunderstood the method, but the chain starts:
1. c1 (most approvals)2. c23. c3 (least approvals)
c3 then attempts to climb the chain, but is bumped of by their first opponent, c2, because c2 beats c3 in the head to head. Then c2 attempts to climb the chain but is bumped off by c1 who beats c2 in the head to head. c1 is then elected.
Toby

      From: Forest Simmons <fsimmons at pcc.edu>
 To: EM <election-methods at lists.electorama.com> 
 Sent: Saturday, 2 March 2019, 21:20
 Subject: [EM] Improvement on Jobst's Chain climbing method
   
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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