Is is possible to have 0% risk of jam in swiss system?
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I'm trying to implement my algorithm for pairing using the Swiss system and I wonder if it's possible to reduce the risk to 0 of jamming pairing, for example, we have 12 participants. In 8 rounds we have 3 players left who cannot play with each other (because they played before). Is it possible to eliminate such a case or is it not possible and can be solved only manually?