Rest for three days before the next game

Question

$N$ ($N\gt 2$) teams compete in a single round-robin tournament, that is, every team must play every other team once. There will be one match every day, so the tournament lasts for $\frac {N(N-1)}{2}$ days. In order to allow teams to fully rest after each game, the organizers want to arrange the tournament in such a way that any team will rest for at least 3 days after participating in a game. What is the minimum number of $N$?

Answer 1

Lower bound:

With a minimum 3-day rest, teams playing on day one cannot play again until day five. The first four days must have 8 distinct teams, and we need a 9th team to play on day five against one of the teams from day one. This simple observation means $N\ge 9$

Proof positive:

  1 A B
  2 C D
  3 E F
  4 G H
  5 A I
  6 B C
  7 D E
  8 F G
  9 H I
 10 A C
 11 B E
 12 D G
 13 F H
 14 C I
 15 A E
 16 B G
 17 D H
 18 F I
 19 C E
 20 A G
 21 B H
 22 D F
 23 E I
 24 C G
 25 A H
 26 B F
 27 D I
 28 E G
 29 C H
 30 A F
 31 B D
 32 G I
 33 E H
 34 C F
 35 A D
 36 B I

Answer 2

It can't be 3,4,5,6,7,8 because of the three day rest. Meanwhile if it is 9 the teams would get a 3 day rest and the match and the number of matches would be of course 36 (just like the handshake problem). We get a perfect 36 days from equation, hence the answer is 9.

3,4,5,6,7,8 all have a problem:

• if 3 is the N: when after the first match 3 days will be lost because the other team would have no competitor... and putting 3 on equation we get 3 days so 3 is ruled out

• if 4 in N: After two matches 2 days will be lost only 6 days the matches can last

• if 5 is N: Here also after 2 matches 1 day will be lost

• if 6 is N: Also a day is lost

• if 7 is N: Also a day is lost

• if 8 is N: Also a day is lost because of the three day rest