[EM] SSD, RP wrap-up

Blake Cretney bcretney at postmark.net
Mon Apr 23 13:05:00 PDT 2001

Dear Markus,
> You wrote (19 Apr 2001):
> > RP can choose outside the initial Schwartz set. SSD & Cloneproof
> > will never do that. Choosing outside the initial Schwartz set
> > serious. I'm not aware of it causing a strategy problem, but that
> > doesn't mean I'm saying it doesn't. But it's an embarrassment,
> > as I said before, an aesthetic gaffe. And an avoidable one, since
> > we've got excellent methods that don't have that failing.
> In so far as your claim that Ranked Pairs could choose outside the
> Schwartz set seems to be your main argument against Ranked Pairs,
> it would be very helpful if you could post an explicit example where
> Ranked Pairs actually chooses a candidate outside the Schwartz set.
> Otherwise your claim that Ranked Pairs could choose outside the
> Schwartz set seems to be not justifiable.

The question was asked of Mike, but I'll respond anyway.  He must be
refering to an example like this:

A>B 40
B>C 39
C>A 38
A ties with D
D>B 5
D>C 4


Now, I don't take the Schwartz criterion particularly seriously, but
just for fun, here's another criterion to counter it.

Let's call it the Pairwise Dominance criterion.  It goes like this.  

p(X,Y) is the number of voters who rank X over Y.  
For all X, if exists Y s.t. (for all Z, p(Y,Z)>=p(X,Z) and
p(Z,Y)<=p(Z,X)) and p(Y,X)>p(X,Y)
then  X must not win

In other words, a candidate shouldn't win, if there is another
candidate that not only beats it head-to-head, but also does at least
as well in all the other contests, whether measured in terms of votes
for or against the candidate.

If you use the tiebreaker that Zavist and Tideman specify for Ranked
Pairs, it passes this criterion.  Initially, I misuderstood their
tiebreaker, and then viewed it as unnecessarily complicated, but I
recently realized that it passes a criterion of this form, and has
certain other subtle advantages.  They may have considered this, but
they don't mention it.

For those wanting a definition of that tie-breaker, here it goes. 
Let's say you have two different victories, like
A vs. B and C vs. D
this involves the candidates A, B, C, and D.  One of these candidates
is ranked highest on the tiebreaker.  Whichever victory involves that
candidate is considered higher after the tiebreaker.  The same
candidate can appear in both victories, as long as it isn't highest. 
For example, given the tiebreaker A>B>C>D>E, C>B would beat D>E, since
B is ranked highest on the tiebreaker.

But it might be, that you have the same candidate repeated, and this
candidate is the highest.  Not surprisingly, you look at the
candidates the victories don't share, and see which is higher of
those.  That victory is considered higher after the tiebreaker.  For
example, with the tiebreaker A>B>C>D>E, C>D would beat E>C, since D is
higher than E on the tiebreaker.

Next, you consider pairwise ties.  Score those as a victory for the
higher ranked over the lower ranked candidate.  So, with tiebreaker
A>B>C>D>E, B=E would cause B to be ranked over E.  Once there are no
pairwise ties, and no victories are considered equal, the normal
definitions give a result.


An example where Schulze/SSD fails PDC is the following,

1 ballot of each permutation of A B C D
2 A B C D
2 C D A B
2 D A B C

B shouldn't win by PDC, but with Schulze/SSD, it can (if you draw a
B-1st ballot).  It appears that there is little justification for
allowing B to win, since A does better than B by any reasonable
measure.  On the other hand, since this (and Schwartz) only show up
when there is a tie, this is really an argument about tiebreakers. 
You could make Schulze pass PDC, or RP pass Schwartz, by adding an
extra step to avoid electing the candidates that are declared
ineligible.  Of course, this provides another complication.

Blake Cretney

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