[EM] Three rounds

[Juho Laatu]

--- On Wed, 12/11/08, Kristofer Munsterhjelm <km-elmet at broadpark.no> wrote:

> From: Kristofer Munsterhjelm <km-elmet at broadpark.no>
> Subject: Re: [EM] Three rounds
> To: juho4880 at yahoo.co.uk
> Cc: election-methods at lists.electorama.com
> Date: Wednesday, 12 November, 2008, 3:01 AM
> Juho Laatu wrote:
> > Is Schwartz set specifically "Condorcetian"?
> Also methods like minmax
> > could be said to be strongly "Condorcetian"
> (although they do not
> > necessarily elect from the Schwartz/Smith sets).
> 
> Schwartz is Condorcet-like because a CW will always be in
> the Schwartz set, and Smith (and Schwartz) is a reasonable
> extension of the Condorcet criterion (from "a candidate
> who is preferred to all others should win" to "of
> a group where the group is preferred to all outside the
> group, a group member should win"). Minmax is Condorcet
> yet not Schwartz, but anything that's Schwartz is also
> Condorcet.


Clones of members of the Smith set (I'll ignore ties in this mail) are in the Smith set but Smith set members need not be clones. The Smith set is thus not a (unified) group of clones (but just a group of candidates that happen to best all the others in a pairwise comparison).

There are also other potential "reasonable extensions" of the Condorcet criterion. One interesting question is if it is more important for the elected candidate to have weak opposition or to have a narrow opposition.

(Beatpaths may be considered a "winner evaluation criterion" too - although their meaning in real life situations is not clear - maybe they are used just to identify clones in some approximate way rather than describe the value of the to be winner.)

One basic example.
17: A>B>D>C
16: A>D>B>C
17: B>C>D>A
16: B>D>C>A
17: C>A>D>B
16: C>D>A>B
A, B and C are in a strong loop. A, B and C form the Smith set but they are not clones. Each of them is beaten badly by another member of the loop. D loses to all the Smith set members, but only with a very narrow margin.

One could say that D is the most acceptable choice, and that electing the candidate with weakest opposition (against any single one of the other candidates) is a natural extension of the Condorcet criterion.

(D is the Condorcet loser but it is also very close to being the Condorcet winner. The visual impression of "being below the top three" positions D somewhere deep down at the bottom of the picture and at the end of the preference list, but obviously such 2D visualization does not describe the cyclic relations in the best way.)


> 
> > Also party lists could be used. One approach to
> breaking cycles and
> > identifying clone sets would be to not to count the
> votes of fellow
> > party members against the candidates (not before all
> candidates of
> > the competing parties/branches have been eliminated).
> How did you use
> > D'Hondt without parties?
> 
> I used the Condorcet idea. First iteration: elect as for a
> single winner (this includes the legitimate single winner so
> that a runoff never hurts him). Second iteration: count
> ballots as usual, but all preferences below the winner of
> the first iteration count half. So, for instance,
> 
> A > B > C > D
> with B as first iteration winner
> counts C > D as 0.5 victory of C over D.
> 
> Then generate a social ordering based on the new matrix,
> eliminate the winner of the first iteration from that
> ordering, and pick whoever won as the winner of the second
> iteration. The winners from each iteration go to the runoff.
> 
> The full method is given somewhere in the archive, just
> search for "D'Hondt without lists"; but for
> runoffs, there are just two iterations, so it's as
> simple as this (I think; it's late and so I may have
> made an error).
> 
> If you're going to use party list, I don't see much
> point in a runoff. Either it'll be multiwinner, in which
> case a runoff doesn't make much sense, or it'll be
> single-winner, in which case you can just use the equivalent
> single-winner method. For ordinary party list, that method
> would be Plurality; instead of voting for a party, vote for
> the party's designated "appointee".


I was thinking of single winner elections only.

The party lists could be more interesting when breaking Condorcet cycles. But in a runoff one could first vote between parties and only then between candidates of the winning party. I'm not sure that this is very useful, but this way one could e.g. reduce the risk of the best compromise candidate of a party being eliminated too early.

For example
40: A1>A2>B>C
08: A2>A1>B>C
07: A2>B>A1>C
25: B>A2>C>A1
20: C>B>A2>A1
A2 would be eliminated first in IRV but here A1 and A2 form a party (with 55 first preference votes) and therefore C will be eliminated first, B next, and then A2 will win.


> 
> > Sorry about making only "opposite proposals"
> :-). Schwartz and
> > partyless approaches may be ok too. And use of some
> sequential
> > approach to break a Condorcet cycle as well.
> 
> What kind of sequential approaches were you thinking of?


Just some quite traditional ones like IRV.

Juho