On Feb 5, 2010, at 10:24 PM, Warren Smith wrote:
robert bristow-johnson:
In a Condorcet election in which a cycle occurs, if there are only
three
candidates in the Smith set, or even if there are more but the defeat
path is a simple single loop, is the outcome of the election any
different if decided by Schulze rules than if decided by Tideman
ruules
(Ranked Pairs)?
Of course, this question is open for anyone to answer.
Kristofer Munsterhjelm:
AFAIK, if there are three candidates in a cycle, RP and Schulze
returns
the same: they break the cycle by its weakest defeat. This is obvious
for RP, because the weakest defeat (least victory) gets sorted
last. For
Schulze, consider the Schwartz set implementation (CSSD) as
described on
http://en.wikipedia.org/wiki/
Schulze_method#The_Schwartz_set_heuristic .
Since there's a cycle and all three candidates are involved in it,
steps
one and two can't be done, so 3 is done, which also discards the
weakest
defeat.
Warren D Smith:
To complete KM's answer, the case where
"there are more [than three in the Smith set]
but the defeat path is a simple single loop"
does not exist.
that, i didn't expect.
That is because if there is a directed N-cycle
for some N>3, you can always "draw a chord" in the "circle"
<smacking forehead>
and no matter which way the arrow on that chord points, the result
is always
at least one shorter directed cycle, i.e. with smaller N.
Consequently, no
matter what the value of N>3, there is always a 3-cycle.
so everyone in the smith set is related to everyone else via a 3-
cycle? or is that also not general?
So... do Schulze & Ranked Pairs always return the same winner? No.
If the Smith set has 3 or fewer members? Yes.
i suspected these two answers but don't mind getting misconceptions
enlightened.
thanks, Warren.
--
r b-j [email protected]
"Imagination is more important than knowledge."
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