Hacker News new | past | comments | ask | show | jobs | submit login

Shouldn't it be possible to remove the Schwartz Set candidates and then re-run the count to find out the remaining relative rankings?

That's probably a nice way of doing it - iteratively work out all schwartz sets (if there are multiple roots, there can be multiple schwartz sets) and remove them until there are none left to work out the rankings.

Now I have to go back to the code!

I did something semi-related with beatpaths.com when it was active. I created a cyclic graph of NFL team wins. There would obviously be loops (such as two teams in a division splitting the series). So I'd simply remove the loops from the graph; smallest loops first. Eventually you'd end up with a DAG (perhaps with multiple roots), and I'd run a modified tsort (with various tiebreakers) to come up with a power ranking. It was kind of a silly website but I wanted to see how accurate NFL picks could be using only wins and losses. (Answer: somewhat competitive with other methods, but not very - but still, fun, and with pretty pictures!)

Anyway, yes - you can come up with a full ranking of candidates in a Condorcet method by finding the condorcet winner (or schwartz set), removing, and running again, and keep going until you've fully ranked all candidates.

No. Assuming no changes in preferences among the members of the Schwartz set, re-running an election among the Schwarz set won't change anything.

I meant remove the Schwartz set candidates from the ballots, and then recount the ballots to find the next ranked candidate.

Guidelines | FAQ | Support | API | Security | Lists | Bookmarklet | Legal | Apply to YC | Contact