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<p class="MsoNormal"><span>Greetings again
EM list friends:</span></p><p class="MsoNormal"><span><br></span></p>
<p class="MsoNormal"><span>I appreciate the
response from Toby, P., Chris B., R. Lung, Kevin V. and Jobst H.</span></p><p class="MsoNormal"><span><br></span></p>
<p class="MsoNormal"><span>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.]</span></p>
<p class="MsoNormal"><span><br></span></p><p class="MsoNormal"><span>Unfortunately, a
rather subtle effect destroys mono-raise winner.<span> </span>There is no problem if the only improvement
in status of the winner is from increased approval.<span> </span>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.</span></p><p class="MsoNormal"><span><br></span></p>
<p class="MsoNormal"><span>Right now I do
not see any way around this, so <span> </span>chain
climbing (taught to us by Jobst) is the only monotone Banks method that I know
of.</span></p>
<p class="MsoNormal"><span>Sorry to get your
hopes up in vain.<span> </span>For me trying to
improve on chain climbing is a kind of isometric exercise;<span> </span>by straining against a hard, perhaps
impossible problem, you get stronger (if it does not kill you).</span></p>
<p class="MsoNormal"><span><br></span></p><p class="MsoNormal"><span>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.<span> </span>I do not have the
time here to tell you about some of the other spinoff from these attempts.</span></p><p class="MsoNormal"><span><br></span></p><p class="MsoNormal"><span>Thanks Guys,</span></p><p class="MsoNormal"><span><br></span></p><p class="MsoNormal"><span>Forest<br></span></p>
</div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Sat, Mar 2, 2019 at 1:20 PM Forest Simmons <<a href="mailto:fsimmons@pcc.edu">fsimmons@pcc.edu</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div dir="ltr"><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>
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