[EM] RCV Challenge
forest.simmons21 at gmail.com
Fri Dec 24 20:00:27 PST 2021
I'm not sure if there is a strong convention on the precise definition of
implicit approval, but (without any fanfare) I took the liberty of defining
it for my purposes in such a way that equal bottom rank is equivalent to
truncation ... so by my (uncoventional?) definition the IA winner W is the
candidate with the most ballots on which it is ranked strictly above at
least one other candidate... so in this version just being explicitly
ranked last on every ballot is not enough for 100 percent IA ...in fact, in
my version, (given its bottom ranking though not necessarily bottom status)
the only ballots that would contribute anything other than zero to its IA
would be the ballots on which some other extant candidate was totally
unranked ... not even ranked bottom ... i.e. truncated... bottom ranked
candidates are considered to be ranked above unranked candidates... but we
cannot say unranked candidates are ranked below ranked candidates, because
that would be an oxymoron.
Perhaps some article about Coombs method would reveal the standard
definitions of "ranked last", "lowest ranking", "unranked", "truncated",
In a three candidate Coombs race do the ballots A>B>C count the same as
I prefer to distinguish the two ballots in terminology, but not
instrumentally. C is ranked last or bottom on the ballot A>B>C but
truncated or unranked on A>B ... while still enjoying bottom status,
assuming a three candidate race.
In general, all unranked/truncated candidates share (equal) last/bottom
status on a given ballot.
My remark about IA being determined *before* any eliminations is also
germane to this conversation. Suppose at some point after some elimination
steps a ballot looks like A>B. Does or did this ballot contribute implicit
approval to B? Answer: only if there exists a candidate C that B is now or
once was ranked ahead of. To say that B is ranked ahead of C on the ballot
A>B implies that the unranked C has not yet been eliminated from the race.
As you know, the IA's have to be determined at the beginning without
"renormalizations," in order for the method to be monotone.
So many details to watch out for!
El vie., 24 de dic. de 2021 3:27 p. m., Kristofer Munsterhjelm <
km_elmet at t-online.de> escribió:
> On 25.12.2021 00:18, Forest Simmons wrote:
> > My proposal, Implicit Approval Chain Climbing (IACC), needs a tie
> > breaker. For that purpose we require each candidate to submit a
> > recommended tie breaking order. The order submitted by the highest
> > implicit approval candidate will be used.
> > In the rare event that the tie is for highest implicit approval, break
> > this tie by applying the IACC method recursively to a copy of the ballot
> > set restricted to the candidates tied for highest IA.
> > Does that "tie up" all of the loose ends?
> I'd probably just use random ballot; seems simpler... although there is
> the possible summability problem.
> However, the method has (as all implicit approval methods have) the
> somewhat undesirable property that if everybody fully ranks the
> candidates, then it's completely indecisive.
-------------- next part --------------
An HTML attachment was scrubbed...
More information about the Election-Methods