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