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<div class="moz-cite-prefix">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.</div>
<div class="moz-cite-prefix"><br>
</div>
<div class="moz-cite-prefix">from Richard Lung.<br>
</div>
<div class="moz-cite-prefix"><br>
</div>
<div class="moz-cite-prefix">On 04/03/2019 21:56, Forest Simmons
wrote:<br>
</div>
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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" moz-do-not-send="true">fsimmons@pcc.edu</a>>
wrote:<br>
</div>
<blockquote class="gmail_quote" style="margin:0px 0px 0px
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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>
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