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<p>Forest,<br>
<br>
I was impressed by your "Max Covered Approval" method idea, which
seems to be similar to this.<br>
<br>
It differs by requiring any new addition to the chain to cover
(not just pairwise beat) the last one.<br>
<br>
So if the most approved candidate was uncovered it won, otherwise
the most most approved candidate<br>
that covered it if that candidate is uncovered, otherwise the most
approved candidate that covered it,<br>
and so on.<br>
<br>
This is one of my two or three favourite (distinctly different
from each other) Condorcet methods.<br>
<br>
Chris Benham<br>
<br>
</p>
<div class="moz-cite-prefix">On 3/03/2019 7:50 am, Forest Simmons
wrote:<br>
</div>
<blockquote type="cite"
cite="mid:CAP29one53hBxkhGX3oesSVgv63j-RN_kRDZpqNFiP8_6yFQcDw@mail.gmail.com">
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<div>A few years back Jobst suggested "chain climbing" as a
seamless, Condorcet compliant way of selecting an alternative
from a given ordered list.</div>
<div><br>
</div>
<div>For example electing a winner from a list of candidates c1,
c2, ... given in decreasing order of approval.</div>
<div><br>
</div>
<div>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.<br>
</div>
<div><br>
</div>
<div>The candidate who ends up at the top of the chain is
elected.</div>
<div><br>
</div>
<div>Since a beats all candidate will never be defeated, the
method is Condorcet compliant. It also turns out to be clone
resistant and monotonic. <br>
</div>
<div><br>
</div>
<div>Another nice property is that it always selects from the
Banks set, a nice game theoretic subset set of the set of
uncovered candidates. <br>
</div>
<div><br>
</div>
<div>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. <br>
</div>
<div><br>
</div>
<div>Here's my proposed improvement:</div>
<div><br>
</div>
<div>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.</div>
<div><br>
</div>
<div>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.</div>
<div><br>
</div>
<div>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..</div>
<div><br>
</div>
<div>For example in the A>B>C example above it elects A.<br>
</div>
<div><br>
</div>
<div>What do you think?</div>
<div><br>
</div>
<div>Forest<br>
</div>
<div><br>
</div>
<div><br>
</div>
<div><br>
</div>
</div>
<br>
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