[EM] hey Markus, can you confirm something?...

[Warren Smith]

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 is because if there is a directed N-cycle
for some N>3, you can always "draw a chord" in the "circle"
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... do Schulze & Ranked Pairs always return the same winner? No.
If the Smith set has 3 or fewer members? Yes.

-- 
Warren D. Smith
http://RangeVoting.org  <-- add your endorsement (by clicking
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and
math.temple.edu/~wds/homepage/works.html