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James Green-Armytage wrote:<br>
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cite="midfc.005b8ffc00f00cb23b9aca00fe2c6334.f00d37@mcgregor.edu">
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Am I correct in observing that Schwartz // SC-WMA is similar to completing
Condorcet with Schwartz and then Bucklin? How does the former differ from
the latter?
my best,
James
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CB: Bucklin hopelessly fails Clone Independence, and so (according to EMR)
is vulnerable to <a href="http://condorcet.org/emr/defn.shtml#crowding">
crowding</a> and <a href="http://condorcet.org/emr/defn.shtml#teaming">teaming</a>.<br>
I would never knowingly reccomend a method that does that.<br>
On the other hand Bucklin meets Mono-raise, and SC-WMA (I think fairly
benignly) doesn't.<br>
<br>
Here is a simple example from Marcus Schulze: <br>
2: A>B>C <br>
3: B>C>A <br>
4: C>A>B <br>
<br>
Bucklin and SC-WMA both elect C.<br>
Now we replace C with the clone set C1, C2, C3. <br>
<br>
2: A>B>C2>C1>C3 <br>
3: B>C3>C2>C1>A <br>
4: C1>C2>C3>A>B <br>
<br>
Now Bucklin elects B.<br>
<br>
SC-WMA "Weights": A:2, B:3, C1: 4, C2: 0, C3: 0.<br>
Each ballot approves from the top, until at least half (by "weight") of
the candidates have been approved. Thus:<br>
<br>
2:A,B<br>
3:B,C3,C2,C1<br>
4:C1,C2,C3,A<br>
<br>
Final "approval" scores: A:6, B:5, C1: 7, C2: 7, C3: 7.<br>
<br>
The three clones are tied for first. I would favour resolving this by electing
the tied candidate with the greatest "weight", C1.<br>
<br>
Chris Benham<br>
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