James Green-Armytage wrote:
CB: Bucklin hopelessly fails Clone Independence, and so (according to EMR) is vulnerable to crowding and teaming.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
I would never knowingly reccomend a method that does that.
On the other hand Bucklin meets Mono-raise, and SC-WMA (I think fairly benignly) doesn't.
Here is a simple example from Marcus Schulze:
2: A>B>C
3: B>C>A
4: C>A>B
Bucklin and SC-WMA both elect C.
Now we replace C with the clone set C1, C2, C3.
2: A>B>C2>C1>C3
3: B>C3>C2>C1>A
4: C1>C2>C3>A>B
Now Bucklin elects B.
SC-WMA "Weights": A:2, B:3, C1: 4, C2: 0, C3: 0.
Each ballot approves from the top, until at least half (by "weight") of the candidates have been approved. Thus:
2:A,B
3:B,C3,C2,C1
4:C1,C2,C3,A
Final "approval" scores: A:6, B:5, C1: 7, C2: 7, C3: 7.
The three clones are tied for first. I would favour resolving this by electing the tied candidate with the greatest "weight", C1.
Chris Benham
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