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Looks like two more votes for option 4 over option 7 would have made a Condorcet cycle at the top, and they'd have had to break out the fun beatpath stuff.
by mjw1007 6y ago
Looks like two more votes for option 4 over option 7 would have made a Condorcet cycle at the top, and they'd have had to break out the fun beatpath stuff.
- kroeckx 6y agoThat is not enough to create a cycle. Option 7 would still beat option 4 via option 3 with 8 and via option 2 with 4 votes. Note that there is already a cycle between option 1, 3, and 4. You're right that it doesn't take that many votes to create more cycles and complicate the process.
- mjw1007 6y agoIf option 4 had beat option 7 pairwise but lost transitively by some longer path, isn't that what we mean by the term "cycle"?
- kroeckx 6y agoNot as far as I know. If an option X beats an option Y, and Y beats option Z, and Z beats option X, you have a cycle. This is the case in the GR for option 1, 3, and 4. If there are no other options beating X, Y or Z, they all 3 end up in the Schwartz set. We then have to determine which is the weakest defeat in the Schwartz set and remove it, and then determine the Schwartz set again. Edit: Clearly I was confused, and changing it creates additional cycles.