Ted Stern tedstern-at-mailinator.com |EMlist| wrote:
Thanks to both of your responses, I have an idea now that I think will work, and it should have (my) desired quality of encouraging generous approval cutoff and ranking of candidates below the cutoff.
Basically, the idea is simply Beatpath: Break each cycle at the weakest link.
But what should be the weakest link? Why not call it the defeat made by the
candidate with lowest approval? We could call this Total Approval Beatpath (TAB), but suggest a better name if you want.
Forest's idea can proceed almost identically as before:
(1) Sort candidates by approval, from highest to lowest, into a list {A_1, A_2,..., A_n}.
(2) Initialize a linked list (or chain, if you prefer) of candidates B with Approval Winner A_1. The beginning of the list is the winning end, and the end is the losing end.
(3) For each A_i, i=2:n,
B node points to losing end candidate.
While A_i defeats B-node, follow B-list backward to next stronger candidate, set B-node to that candidate. End while
Insert A_i before B-node. End For
The winner is the candidate at the left most end of the list. The list gives you the social ordering of the candidates.
This is just like Forest's proposal but doesn't eliminate candidates from the B list.
Thanks for explaining your method. It is interesting, but I'm not sure I understand it completely.
In step 3, I presume that by "defeats" you mean pairwise (as opposed to Approval-wise). Suppose an A_i candidate does not pairwise defeat any of the other candidates already on the chain. Is A_i then dropped or added to the end of the chain?
If the latter is true, then by the time the least-approved candidate comes along all the other candidates will be on the chain. Hence, the least approved candidate cannot win unless he is the Condorcet winner. Is it possible for the least-approved candidate to be the Condorcet winner?
--Russ ---- Election-methods mailing list - see http://electorama.com/em for list info
