/kloc-how to divide the 0/3 team into two-loss elimination systems?

All teams should be organized into a certain competition order, and two adjacent teams should compete. The winner will enter the next round, and the loser will compete with the neighboring losers. Lose two games and be eliminated. The winner is the one who loses only one game. The winner is the champion. This is the concept of double elimination.

According to the above, I think it should be 1 goodbye (counted as the winning round), 2-34-56-78-910-112-13, if The negative team is 3-4 6-911-12 (whoever loses will be eliminated). This round of 3, 6, 1 1 wins, 4, 9, 12 is out.

The following1-2; 3-4; 5-6; 7-8; The winner (if) 1, 4, 5, 8, 1 1 bye, and the negative team 2-3; 6-7; Win 2, 6, lose 3, 7, and then 2-6. Win 2,6,2-1 1 (Win 2,11), further down:1-4,5. (The winner is 1, 5) The loser is 4-8, and the winner on the 4th is eliminated 8), and then it is 2-4 (2 wins and 4 elimination); Finally, 1-5 (the winner's champion), the loser, competed with the remaining No.2 for the second place (the two runners who lost only one game). I think so. I didn't look closely either.