[EM] Election-Methods Digest, Vol 118, Issue 8

[Forest Simmons]

Questions about compliances of Implicit Approval Chain Climbing




> ---------- Forwarded message ----------
> From: Steve Eppley <SEppley at alumni.caltech.edu>
> Date: Sun, Apr 20, 2014 at 7:48 PM
> Subject: Is Chain Climbing really monotonic??
> To: Michael Ossipoff <email9648742 at gmail.com>
>
>
> Begin with:
>      7: A B C
>      6: B C A
>      5: C A B
> First add C to S since C is bottom-ranked by the most (7).
> Then add B to S since only B is unbeaten pairwise by a candidate in S.
> Then elect B.
>
> Suppose two voters raise B from the bottom:
>      7: A B C
>      6: B C A
>      3: C A B
>      2: C B A
> First add A to S since A is bottom-ranked by the most (8).
> Now B can never be added to S since B is beaten pairwise by a candidate in
> S.
> So now B can't win.
>
> Have I made a mistake?
>

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.

>
>
> ------------------------------
>
> Message: 2
> Date: Mon, 21 Apr 2014 14:21:45 -0400
> From: Michael Ossipoff <email9648742 at gmail.com>
> To: "election-methods at electorama.com"
>         <election-methods at electorama.com>
> Subject: [EM] Fwd: Is Chain Climbing really independent of clones??
>
> ---------- Forwarded message ----------
> From: Steve Eppley <SEppley at alumni.caltech.edu>
> Date: Sun, Apr 20, 2014 at 7:22 PM
> Subject: Is Chain Climbing really independent of clones??
> To: Michael Ossipoff <email9648742 at gmail.com>
>
>
> The "least implicit approval" score used in Chain Climbing doesn't look
> cloneproof.
>
> Under Plurality Rule, two "top" clones can split their vote, allowing a
> third candidate to win by spoiling.  Chain Climbing seems very similar: Two
> "bottom" clones can split their "implicit disapproval" so that a third
> candidate will instead be added to S.
>
> Begin with the example from my recent email:
>      5: A B C
>      4: B C A
>      3: C A B
> First add C to S since C is bottom-ranked by the most (5).
> Then add B to S since only B is unbeaten pairwise by a candidate in S.
> Then elect B.
>
> Suppose we add a clone of C:
>      3: A B C C'
>      2: A B C' C
> 4: B C C' A
>      3: C C' A B
> First add A to S since A is bottom-ranked by the most (4).
> Now B can never be added to S since B is beaten pairwise by a candidate in
> S.
>
> Have I made a mistake?
>
> In the context of approval, two candidates that are ranked or rated  on
> opposite sides of the approval cutoff are not considered clones.



> 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.
> ------------------------------
>
> Message: 3
> Date: Mon, 21 Apr 2014 14:34:01 -0400
> From: Michael Ossipoff <email9648742 at gmail.com>
> To: "election-methods at electorama.com"
>         <election-methods at electorama.com>
> Subject: [EM] Fwd: Does Chain Climbing fail Resolvability?? (was Re:
>         Is Chain Climbing really monotonic??)
>
>
> Steve questioned TACC's compliance with Condocet, in situations where two
> candidates are unbeaten from S, and have equal implict-approval totals, if
> TACC adds them both simultaneously to set S. I replied that Jobst and
> Forest probably intended a random choice, by some means, between those two
> candidates, to determine which to first add to S. In that way, there
> doesn't remain any difficulty with Condorcet compliance.
>
> Below is a reply from Steve. Because he requested that I forward his other
> TACC comments, I assume that he'd like me to forward this one as well:
>
>
> ---------- Forwarded message ----------
> From: Steve Eppley <SEppley at alumni.caltech.edu>
> Date: Mon, Apr 21, 2014 at 12:38 PM
> Subject: Does Chain Climbing fail Resolvability?? (was Re: Is Chain
> Climbing really monotonic??)
> To: Michael Ossipoff <email9648742 at gmail.com>
>
> I wrote a few days ago about Chain Climbing's lack of decisiveness. In the
> same vein, I suspect Resolvability is failed by TACC with the random
> tiebreaker (assuming 'implicit disapproval' is defined by Bottom(A) rather
> than Bottom(Z)... 'absolute implicit disapproval' rather than 'relative
> implicit disapproval').  Suppose one terrible candidate is unanimously
> ranked bottom, and a bunch of other candidates cycle.  Call the cyclic
> candidates C.  Given the random tiebreaker to add one at a time to S, many
> candidates in C have a non-zero chance to win.  To try to make the winner
> deterministic, a vote added to the collection of votes can rank a subset of
> C at the bottom.  Call that subset Cb, and let Ct denote the rest of C.  If
> C contains many candidates, then at least one of Cb & Ct must contain two
> or more candidates.  If at some point TACC needs to add one of those
> "still-tied" candidates to S, the randomness of that pick may cause the
> winner to still be non-deterministic.
>
Steve,

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.

There are many deterministic ways of breaking ties in addition to the
random ones.

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.

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.

Forest
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