<div dir="ltr">Questions about compliances of Implicit Approval Chain Climbing<br><div><div class="gmail_extra"><br><br><div class="gmail_quote"><br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
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From: Steve Eppley <<a href="mailto:SEppley@alumni.caltech.edu" target="_blank">SEppley@alumni.caltech.edu</a>><br>
Date: Sun, Apr 20, 2014 at 7:48 PM<br>
Subject: Is Chain Climbing really monotonic??<br>
To: Michael Ossipoff <<a href="mailto:email9648742@gmail.com" target="_blank">email9648742@gmail.com</a>><br>
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<br>
Begin with:<br>
7: A B C<br>
6: B C A<br>
5: C A B<br>
First add C to S since C is bottom-ranked by the most (7).<br>
Then add B to S since only B is unbeaten pairwise by a candidate in S.<br>
Then elect B.<br>
<br>
Suppose two voters raise B from the bottom:<br>
7: A B C<br>
6: B C A<br>
3: C A B<br>
2: C B A<br>
First add A to S since A is bottom-ranked by the most (8).<br>
Now B can never be added to S since B is beaten pairwise by a candidate in<br>
S.<br>
So now B can't win.<br>
<br>
Have I made a mistake?<br></blockquote><div><br></div><div>No, but since the approval order changed from B>A>C to B>C>A this is not a test of mono raise winner; i.e. since C was also raised (in the approval order) we shouldn't be surprised that C became the new winner. <br>
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Message: 2<br>
Date: Mon, 21 Apr 2014 14:21:45 -0400<br>
From: Michael Ossipoff <<a href="mailto:email9648742@gmail.com" target="_blank">email9648742@gmail.com</a>><br>
To: "<a href="mailto:election-methods@electorama.com" target="_blank">election-methods@electorama.com</a>"<br>
<<a href="mailto:election-methods@electorama.com" target="_blank">election-methods@electorama.com</a>><br>
Subject: [EM] Fwd: Is Chain Climbing really independent of clones??<br>
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From: Steve Eppley <<a href="mailto:SEppley@alumni.caltech.edu" target="_blank">SEppley@alumni.caltech.edu</a>><br>
Date: Sun, Apr 20, 2014 at 7:22 PM<br>
Subject: Is Chain Climbing really independent of clones??<br>
To: Michael Ossipoff <<a href="mailto:email9648742@gmail.com" target="_blank">email9648742@gmail.com</a>><br>
<br>
<br>
The "least implicit approval" score used in Chain Climbing doesn't look<br>
cloneproof.<br>
<br>
Under Plurality Rule, two "top" clones can split their vote, allowing a<br>
third candidate to win by spoiling. Chain Climbing seems very similar: Two<br>
"bottom" clones can split their "implicit disapproval" so that a third<br>
candidate will instead be added to S.<br>
<br>
Begin with the example from my recent email:<br>
5: A B C<br>
4: B C A<br>
3: C A B<br>
First add C to S since C is bottom-ranked by the most (5).<br>
Then add B to S since only B is unbeaten pairwise by a candidate in S.<br>
Then elect B.<br>
<br>
Suppose we add a clone of C:<br>
3: A B C C'<br>
2: A B C' C<br>
4: B C C' A<br>
3: C C' A B<br>
First add A to S since A is bottom-ranked by the most (4).<br>
Now B can never be added to S since B is beaten pairwise by a candidate in<br>
S.<br>
<br>
Have I made a mistake?<br>
<br></div>
In the context of approval, two candidates that are ranked or rated on opposite sides of the approval cutoff are not considered clones. </blockquote><div> </div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
Some Range devotees go even further and require true clones to have equal scores on all ballots. The idea is that they could differ infinitesimally, but then they would necessarily round to the same standard score when the set of allowable ratings is a standard finite set, as in any standard public election.<br>
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Message: 3<br>
Date: Mon, 21 Apr 2014 14:34:01 -0400<br>
From: Michael Ossipoff <<a href="mailto:email9648742@gmail.com" target="_blank">email9648742@gmail.com</a>><br>
To: "<a href="mailto:election-methods@electorama.com" target="_blank">election-methods@electorama.com</a>"<br>
<<a href="mailto:election-methods@electorama.com" target="_blank">election-methods@electorama.com</a>><br>
Subject: [EM] Fwd: Does Chain Climbing fail Resolvability?? (was Re:<br>
Is Chain Climbing really monotonic??)<br>
<br>
<br>
Steve questioned TACC's compliance with Condocet, in situations where two<br>
candidates are unbeaten from S, and have equal implict-approval totals, if<br>
TACC adds them both simultaneously to set S. I replied that Jobst and<br>
Forest probably intended a random choice, by some means, between those two<br>
candidates, to determine which to first add to S. In that way, there<br>
doesn't remain any difficulty with Condorcet compliance.<br>
<br>
Below is a reply from Steve. Because he requested that I forward his other<br>
TACC comments, I assume that he'd like me to forward this one as well:<br>
<br>
<br>
---------- Forwarded message ----------<br>
From: Steve Eppley <<a href="mailto:SEppley@alumni.caltech.edu" target="_blank">SEppley@alumni.caltech.edu</a>><br>
Date: Mon, Apr 21, 2014 at 12:38 PM<br>
Subject: Does Chain Climbing fail Resolvability?? (was Re: Is Chain<br>
Climbing really monotonic??)<br>
To: Michael Ossipoff <<a href="mailto:email9648742@gmail.com" target="_blank">email9648742@gmail.com</a>><br>
<br>
I wrote a few days ago about Chain Climbing's lack of decisiveness. In the<br>
same vein, I suspect Resolvability is failed by TACC with the random<br>
tiebreaker (assuming 'implicit disapproval' is defined by Bottom(A) rather<br>
than Bottom(Z)... 'absolute implicit disapproval' rather than 'relative<br>
implicit disapproval'). Suppose one terrible candidate is unanimously<br>
ranked bottom, and a bunch of other candidates cycle. Call the cyclic<br>
candidates C. Given the random tiebreaker to add one at a time to S, many<br>
candidates in C have a non-zero chance to win. To try to make the winner<br>
deterministic, a vote added to the collection of votes can rank a subset of<br>
C at the bottom. Call that subset Cb, and let Ct denote the rest of C. If<br>
C contains many candidates, then at least one of Cb & Ct must contain two<br>
or more candidates. If at some point TACC needs to add one of those<br>
"still-tied" candidates to S, the randomness of that pick may cause the<br>
winner to still be non-deterministic.<br></div></blockquote></div>Steve,<br><br></div><div class="gmail_extra">you seem to overlook the possibility of equal rankings and truncations. I think that most voters would truncate at least one of the three cyclic candidates along with the despised candidate. Since the despised candidate is covered there is no possibility of his election.<br>
<br></div><div class="gmail_extra">There are many deterministic ways of breaking ties in addition to the random ones. <br><br></div><div class="gmail_extra">For example if two candidates are in bottom position on the same number of ballots, the one ranked equal top on the most ballots comes out ahead of the other one.<br>
<br></div><div class="gmail_extra">As I mentioned to Michael Ossipoff, all of these worries disappear if we stick to score ballots, for example ballots with allowable scores 0, 1, ... 10. If two candidates have the same number of zeroes, the one with the fewest ones comes out ahead in the approval order. If that doesn't resolve it, then the one with the fewest number of twos, etc. In the very rare case that two candidates have identical distributions of scores, random ballot can be used.<br>
</div><div class="gmail_extra"><br></div><div class="gmail_extra">Forest<br></div><div class="gmail_extra"><br></div></div></div>