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by kroeckx
1887 days ago
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Not 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. |
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