[EM] Improvement on Jobst's Chain climbing method

[Richard Lung]

More than anyone could possibly want to know about the terminology in 
question is found in my (Smashwords or Amazon kdp) book -- FAB STV: Four 
Averages Binomial Single Transferable Vote.

from Richard Lung.

On 04/03/2019 21:56, Forest Simmons wrote:
>
> Greetings again EM list friends:
>
>
> I appreciate the response from Toby, P., Chris B., R. Lung, Kevin V. 
> and Jobst H.
>
>
> All of your observations are very close to my own thoughts, and I 
> heartily agree with them all, except perhaps from Richard Lung who 
> used some terminology with which I am unfamiliar. [I do not doubt its 
> value, but I am not qualified to judge.]
>
>
> Unfortunately, a rather subtle effect destroys mono-raise winner.There 
> is no problem if the only improvement in status of the winner is from 
> increased approval.But when the winner W adds another pairwise defeat 
> (say candidate W over candidate X) this X may newly qualify for a 
> position at the bottom of the chain, thus preventing some candidate Y 
> lower down the approval list from occupying that bottom chain position 
> any more, thus removing the only impassable obstacle from the rise of 
> (even lower approval) candidate Z to the very top of the chain, 
> thereby electing Z instead of W.
>
>
> Right now I do not see any way around this, so chain climbing (taught 
> to us by Jobst) is the only monotone Banks method that I know of.
>
> Sorry to get your hopes up in vain.For me trying to improve on chain 
> climbing is a kind of isometric exercise;by straining against a hard, 
> perhaps impossible problem, you get stronger (if it does not kill you).
>
>
> And Chris is right; the idea for the covering chain method that starts 
> at the top of the approval order and works its way down was inspired 
> by my attempts at finding a monotone Banks method.I do not have the 
> time here to tell you about some of the other spinoff from these attempts.
>
>
> Thanks Guys,
>
>
> Forest
>
>
> On Sat, Mar 2, 2019 at 1:20 PM Forest Simmons <fsimmons at pcc.edu 
> <mailto:fsimmons at pcc.edu>> 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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