Assuming no drawn games, determining a clear winner (and, incidentally, a clear loser) would require the same number of rounds as that of a knockout tournament, which is the binary logarithm of the number of players rounded up. Thus three rounds can handle eight players, four rounds can handle sixteen players and so on. If fewer than this minimum number of rounds are played, two or more players could finish the tournament with a perfect score, having won all their games but never having faced each other. Due to the fact that players should meet each other at most once and pairings are chosen dependent on the results, there is a natural upper bound on the number of rounds of a Swiss-system tournament, which is equal to half of the number of players rounded up. If more than this number of rounds should be played, the tournament might run into the situation that there is either no feasible round, or some players have to play each other a second time.