In the Parts 1 and 2, the 'seeds' are ballots. The winners come/develop from these seeds.
The problem with this is that the methods I discussed do not guarantee n winners. So, I decided to take it from another direction. What if the "seeds" were the candidates themselves. If each candidate x were given the ballots that have x ranked first, when the ballots that candidate x has are tallied up into a pairwise matrix, candidate x would be the MMPO winner. (I chose MMPO, because it is a simple method. Also note that candidate x is the Condorcet winner in this circumstance.) But this is not fair as some candidates have more ballots than other candidates. What we need to get to is for n candidates to have N/n ballots each, where N is the total number of ballots (i.e. voters). However, those N/n ballots that each candidate x has must have x as the MMPO winner. So, what if a candidate x has more than N/n ballots? It could be deemed that the candidate has a sufficient number of ballots to be elected. That means that the candidate does not need to contest the election any further. So, the ballots that were against candidate x are moved to the the candidates that are ranked next on the ballots. I then realised that this was basically STV. If I hadn't have learnt about the basics of how STV worked so recently, may be I would have realised more quickly what was going on. This begs an interesting question. If there were to be an STV election, but you were allowed to choose a pairwise method to determine which candidate should be eliminated at each stage, what pairwise method would you choose? Thanks, Gervase. ---- Election-methods mailing list - see http://electorama.com/em for list info