Chris Benham 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
>candidate wins.

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:
>46: A>>B
>44: B>>C  (sincere is B or B>>A)
>10: C
>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 
approval information.

>31: A>>B
>04: A>>C
>32: B>>C
>33: C>>A
>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.

Chris Benham

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