ntk at netcom.com
Thu Jul 30 22:37:51 PDT 1998
Norm asked how Tideman is inferior to Smith//Condorcet(EM), and
what worse violation by Tideman's method I was referring to.
Answering the 2nd question answers the first.
I'll give 2 examples, but first let me state the problem briefly:
According to Tideman's rules, as I interpreted them, the least
beaten alternative in a cycle of heavy defeats gets its defeat
skipped by Tideman. The defeats in that cycle are locked-in or
skipped before the others are gotten to, because all the defeats
in that cycle are greater than the other defeats.
So then we start locking in the other defeats. It happens that
there's another cycle containing the alternative whose defeat
was skipped. This other cycle has extremely light defeats. But
they all get locked-in, because none of them lock-in a cycle,
because of that first skipped defeat, which is common to both
I'll give an example now. Bruce Anderson proved that, for any
set of pairwise vote totals, there's a set of rankings that is
consistent with those pairwise vote totals, and which will
yield those pairwise vote totals. That means that, for methods
based on pairwise vote totals, there's no need for an example
to actually specify all the rankings, just the pairwise vote totals.
And the only important part of the pairwise vote totals, when we're
going by votes-against, is the votes-against total in a defeat.
I'll specify those. For the opposite pairwise vote total in each
pairing, you can of course fill in any number between zero &
1 less than the votes-against figure. For emphasis, I've sometimes
used votes-against totals of just 1. Of course for that to be
a defeat, it must be that if A beats B with a votes-against total
of 1, then zero people have ranked B over A.
A beats D beats B beats A.
D beats B beats C beats D.
D beats B.
A beats C.
The numbers written after the defeats are the votes-against totals:
A beats D 12
D beats B 10
B beats A 11
B beats C 1
C beats D 2
C beats A 5 (this defeat isn't really relevant to the problem)
First we lock-in the biggest defeat, A>D 12. Then B>A 11.
then we skip D>B 10, because that would lock-in a cycle.
Then we lock-in C>A 5, then C>D 2, then B>C 1. That doesn't lock
in a cycle, because D>B 10 has been skipped.
C, the least beaten alternative, which would win in Schulze's
method or Condorcet(EM) or Smith//Condorcet(EM) loses, and B,
with 10 votes against it in a defeat, wins.
I first noticed that problem in an example with a subcycle,
but, as the above example shows, it doesn't require a subcycle.
Sure, you could add a rule that a skipped defeat will be
"unskipped" before other defeats in a cycle involving that
skipped defeat, other than the cycle that resulted in the
skipping, are locked-in.
Good luck. That makes an already-messy procedure messier still,
More about the intuitive justification for Schulze:
If you feel that A is better than B, and that B is better than
C, then usually that means that A is better than C, and usually
If public preferences are cyclic, then there are mutually-
contradictory sequences of preferences.
If, as indicated by pairwise vote-counts, the public prefers
A to B, B to C, and C to A, then they're saying they like A
better than B, but they're also saying that B is better than C
and that C is better than A.
If someone says that A>B is so conclusive that B>C>A doesn't
count, remind them that everything is beaten, and so that
assumption gets us nowhere.
So if the defeat A>B is weaker than B>C and also weaker than
C>A, then, though the public is contradicting itself, the
implied preference for B>A is based on more nearly unanimous
vote totals than is A>B.
As convincing as the A>B defeat might seem, for comparing A
& B, every alternative is a loser if we go by that, in a
cycle. So, for comparing 2 alternatives, it makes sense to
compare those mutually-contradictory sequences of preferences.
So Schulze's method seems intuitive & natural for comparing
pairs of alternatives.
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