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That 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 b
by kroeckx 5y ago
That 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 5y 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 5y 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.