How about this modification of this variant?
Start with the pairwise matrix P in which the number in the row
corresponding to candidate X and the column corresponding to candidate Y
is the number of ballots ranking (or rating or grading) X strictly above
Y.
Order the candidates according to the maximum entry in their columns, i.e.
their maximum pairwise oppositions. [Of the two possible orders, choose
the one that puts the max max opposition candidate at the bottom of the
list and the min max opposition candidate at the top.]
Bubble sort this order to get its local Kemenization.
This is done by starting at the top and working your way down through the
order, letting each candidate percolate up as far as possible by
transpositions of adjacent candidates whenever dictated by the pairwise
win matrix. [This pairwise win matrix is obtained from the pairwise matrix
P by subtracting its transpose from it, replacing all resulting positive
numbers by ones, and zeroing out the rest of the matrix.]
I like this method because it comes close to satisfying the FBC.
Why would you rank Compromise over Favorite? Only if you thought that
Favorite would get in the way of Compromise during the percolation
process, i.e. only if ...
(1) Favorite can beat Compromise head-to-head, and (2) Favorite's max
opposition is smaller than Compromise's max opposition, AND (3) Compromise
can beat all the candidates who at worst give up fewer points than
Favorite, including at least one that Favorite cannot beat.
You would have to be pretty fickle to abandon Favorite on account
of some poll pretending to be accurate enough to predict all of that.
Forest
Post by Steve EppleyPost by Forest SimmonsHas anybody ever proposed minimizing the maximum opposition
rather than minimizing the maximum defeat?
It's a reasonably good method, although it doesn't satisfy some
criteria I consider important such as Minimal Defense (which is
similar to Mike Ossipoff's Strong Defensive Strategy Criterion) and
Clone Independence. (That's why I think the best method is a
variation of Ranked Pairs which I call Maximize Affirmed Majorities,
or MAM. See the web pages at www.alumni.caltech.edu/~seppley for
more info and rigorous proofs. The website is still under
construction, not yet friendly for people who aren't social
scientists, and most of the web pages requires a web browser that
supports HTML 4.0 and Microsoft's "symbol" font to be viewed
properly.)
Minimax(pairwise opposition) even satisfies a criterion promoted by
Let w denote the winning alternative given some set
of ballots. If one or more ballots that had only w
higher than bottom are changed so some other
"compromise" alternative x is raised to second
place (still below w but raised over all the other
alternatives) then w must still win.
The proof that Minimax(pairwise opposition) satisfies Uncompromising
is simple: Raising x increases the pairwise opposition for all
candidates except w and x, and does not decrease the pairwise
opposition for any candidate, so w must still have the smallest
maximum pairwise opposition.
Changing pairwise indifferences to strict preferences in ballots that
ranked w top cannot increase w's pairwise opposition or decrease any
other alternatives' pairwise opposition.
Post by Forest SimmonsI know that theoretically this could elect the Condorcet Loser, but it
seems very unlikely that it would do so.
-snip-
Minimax(pairwise defeat) can also elect a Condorcet Loser. Suppose
there are 4 alternatives, 3 of them in the top cycle but involved in
a "vicious" cycle. If the pairwise defeats of the 4th are slim
majorities, the 4th wins.
As for the likelihood, this may depend on how you model voters'
preferences. In a spatial model with sincere voting, I think you're
right. But what if supporters of the Condorcet Loser vote
strategically to create a vicious cycle?
Minimax(pairwise opposition) can also defeat a "weak" Condorcet
Winner, one that wins all its pairings but does not have more than
half the votes in each of its pairings. The CW may defeat candidate
x pairwise by a plurality such as 48% (less than half the votes) and
that might be x's largest opposition, whereas the CW may have a
larger opposition, such as 49%, in some pairing.
If there is a "strong" Condorcet Winner (an alternative
that wins each of its pairings by more than half of the
votes) then it must be elected.
Thus anyone who claims no Condorcet-consistent method satisfies
Uncompromising is incorrect.
-- Steve Eppley
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