[EM] Fwd: Does Chain Climbing fail Resolvability?? (was Re: Is Chain Climbing really monotonic??)

[Michael Ossipoff]

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.
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