[EM] My Favorite Deterministic Condorcet Efficient Method: TACC

[C.Benham]

31: A>B
32: B>C
37: C>A

Approvals:  B63,  A68,  C69.   A>B>C>A.

TACC elects A, but  C is positionally the dominant candidate and 
pairwise beats A.

For a Condorcet method with pretension to mathematical elegance, I don't 
see how that
can be justified.

Chris Benham

PS:  Could someone please refresh our memories: What is the "Banks Set"?


 From Jobst Heitzig (March 2005):

> ROACC (Random Order Acrobatic Chain Climbing):
> --------------------------------------------------------------
> 1. Sort the candidates into a random order.
> 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.
>
> The good thing about ROACC is that it is both
> - monotonic and
> - the winner is in the Banks Set,
> in particular, the winner is uncovered and thus the method is Smith-, 
> Pareto-, and Condorcet-efficient.
>
> Until yesterday ROACC was the only way I knew of to choose an 
> uncovered candidate in a monotonic way. But Forest's idea of needles 
> tells us that it can be done also in another way.
> The only difference is that in step 1 we use approval scores instead 
> of a random process:
>
> TACC (Total Approval Chain Climbing):
> ------------------------------------------------
> 1. Sort the candidates by increasing total approval.
> 2. Exactly as above.
>
> The main differences in properties are: TACC is deterministic where 
> ROACC was randomized, and TACC respects approval information where 
> ROACC only uses the defeat information.
> And, most important: TACC is clone-proof where ROACC was not! That was 
> something Forest and I tried to fix without violating monotonicity but 
> failed. More precisely, ROACC was
> only weakly clone-proof in the sense that cloning cannot change the 
> set of possible winners but can change the actual probabilites of 
> winning. With TACC, this makes no difference since it
> is deterministic and so the set of possible winners consists of only 
> one candidate anyway.
>