Clinton Mead wrote:
What would be the simplest paper count that will produce a winner in the
smith set?
I'm not completely sure how your method works, but how about this?
Count the number of ballots on which each candidate is ranked (in any
position). Call each candidate's count his "approval score". Then while
there is more than one candidate left, eliminate the pairwise loser of
the two remaining candidates with the least approval score.
Eliminating losers won't turn a ranked candidate into a non-ranked
candidate (or vice versa), so unlike IRV, you don't have to do a recount
for every round.
It should also pick a winner from the Smith set. Say all but one of the
Smith set candidates have been eliminated. Then the remaining Smith set
candidate will, by definition of the Smith set, beat everybody else
pairwise, and so nobody will be able to eliminate him. Therefore, that
remaining Smith set member will be elected. QED.
The counting burden isn't that hard, either: you need one pass to
calculate the approval scores, and then (n-1) (for n candidates)
pairwise counts. I don't think you can do better than (n-1) pairwise
counts and still be sure to elect someone in the Smith set.
You could replace the approval count with a Plurality count to get
something simpler, but that could also be unfair to Smith set candidates
that have few first place votes, and it wouldn't be consistent:
eliminating candidates from a Plurality count would mean other
candidates could be exposed, and so you'd have to recount as in IRV.
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