[EM] Simplest paper count to produce a winner in the smith set.

[Kristofer Munsterhjelm]

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.