Tiebreakers, Subcycle Rules
Mike Ositoff
ntk at netcom.com
Thu Aug 6 12:36:51 PDT 1998
I hadn't remembered Markus's tie-solution for Schulze. But
that reassures me that the situation I described hasn't been
ignored. But what if the Schwartz set is the whole set
of candidates, and you can't separate out a set that has
un-countered beat-paths to the rest? I don't know if my
indecisiveness example is such a case. I haven't gotten anywhere
with it yet.
Here's my indecisiveness example:
{A,B,C} is a subcycle, but not a clone set.
A>B 2
B>C 3
C>A 1
{A,B,C}>D 7
E>A 8
E>B 5
E>C 4
As I said, I don't yet know if Markus's tiebreaker
solves this, or if this set of candidates can be divided
into 1 subset that has un-countered beat-paths to all the
others.
But surely that can happen, can't it? If so, then, in that
instance, a further tiebreaker or subcycle rule is needed.
***
Trying those 4 methods on Markus's recent 4-candidate example,
I got a single winner with Schulze, but got a 2-candidate tie
with the other 3. So either Tideman isn't equivalent to Schulze,
or I don't yet know Tideman's own rule. David, would you post
the entire wording of Tideman's rule?
But though Petry's method does seem the same as the brief
wording of it that I suggested, it isn't the same as successively
dropping the smallest defeats till something is unbeaten. That
would be plain Condorcet(EM).
Try them on this:
{A,B,C} is a clone set subcycle.
A>B 20
B>C 30
C>A 40
E>{A,B,C} 1
{A,B,C}>D 3
D>E 2
***
Must quit before the keyboard decides to inexplicably quit.
Mike
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