[EM] Improvement on Jobst's Chain climbing method

[Forest Simmons]

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> 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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