[EM] A method that excludes unfortunates for three candidates

[Kristofer Munsterhjelm]

This method should work to exclude an unfortunate candidate - one who 
can't win an n-1 of n Droop proportional election because other 
candidates are forced to win by the Droop constraints - for three seats. 
It's also decisive for more ambiguous elections with no single 
Droop-consistent outcome.

Do Bucklin. As soon as everybody but one candidate has more than 1/3 of 
total support, elect the candidates who have.

The proof idea is like this: suppose that C is unfortunate. Then either 
fpA and fpB are both over 1/3 relative support, or C's last preference 
count is > 2/3 of the total.

In the former case, A and B will immediately reach the 1/3 target when 
only the first rank has been counted.

In the latter case, because C's last preference count is greater than 
2/3, the other two candidates must have less than 1/3 last prefs each. 
Because the sum of all ranked preferences for a single candidate is 
equal to 100% support, it follows that both A and B must reach 1/3 
before last rank. Hence they're elected and C is not.

This all suggests the following more general pseudo(?)proportional method:

For k seats, n candidates:
	1. Add in ranks until everybody but one candidate has support
		exceeding a Droop quota for n-1 seats, i.e. |V|/n.
	2a. If k == n-1, elect these candidates.
	2b. Otherwise, eliminate the remaining candidate, reducing n by one, 
and loop to 1.

Interestingly, this method has no reweighting of votes. I don't know 
whether it's only partially proportional or fully so, but I suspect the 
former. It's also not summable and probably not monotone. But it's novel.

(Unfortunately it also has center squeeze because it's possible to 
arrange LCR so that Droop forces L and R to be elected for two seats. 
Hence C gets eliminated as part of the 2 of 3 election.)

-km