[EM] Three rounds

[Kristofer Munsterhjelm]

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.

> 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".

> 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?