[EM] UncAAO, DMC, ASM, and TACC
chrisjbenham at optusnet.com.au
Sun Mar 11 06:24:18 PDT 2007
>Forest W Simmons wrote (March 11, 2007):
>Another nice method, Jobst's TACC (Total Approval Chain Climbing), also
>gives b as winner.
>Recall that Jobst initializes a "chain" with the lowest approval
>candidate, and (moving up the approval list) adds only those candidates
>to the chain that pairwise defeat each of the other candidates
>currently in the chain.
> From Jobst Heitzig (March 4, 2005):
>TACC (Total Approval Chain Climbing):
>1. Sort the candidates by increasing total approval.
>2. Starting with an empty "chain of candidates", consider each
>candidate in the above order. When the
>candidate defeats all candidates already in the chain, add her at the
>top of the chain. The last added
Chris Benham wrote:
>TACC seems to have the curious property that if there are three
>candidates in a top cycle, the most approved can't win unless it pairwise
>loses to the second-most approved candidate. This doesn't seem to cause any
>monotonicity problem, but it does seem to be much more vulnerable to
>Burial than the other candidate methods:
>44: B>>C (sincere is B or B>>A)
>A>B>C>A. Approvals: A46, B44, C10.
>TACC elects the buriers' candidate B (chain: B>C) while all our other
>candidate methods (UncAOO, ASM,DMC,AWP) elect A.
TACC having that curious property and so electing B here shows that it
spectacularly fails the
Definite Majority criterion. Maybe that is forgivable for a FBC method
like MAMPO, but not for a
Condorcet method that bases its result on nothing but pairwise and
>A>B>C>A Approvals: A35, B32, C33.
TACC makes the chain B>C and so elects B, but A is more approved than B
and also pairwise beats B
so this is also an example of Definite Majority failure.
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