2 replies to this topic

[bmx]

9 ping pong players will participate in a tournament. There are only 3 tables where 3 games can be played simultaneously. Two players will be playing in each game, while a third will be acting as the arbitrator. For example, the first round would be 12 3 45 6 78 9 with 3, 6 and 9 being the arbitrators and 12 45 78 playing against each other.
There are two rules for the tournament:
1. It must be completed in 12 rounds of 3 simultaneous games, where each player will play against each of the other 8 only once, and will be arbitrating exactly 4 games.
2. After each player arbitrates one game, he must play at least 2 times against another athlete before being allowed to arbitrate again.
You will realize that it is impossible to have all two conditions met together. Can you write a schedule that would meet the first condition and would break the second condition for a minimum number of times? The answer must be 12 rows of 9 digits each, where the 3rd, 6th and 9th digit of each row will be the arbitrator, while all the others will be the players playing against each other, e.g. 12 3 45 6 78 9 for the first round (1 is playing against 2 and 3 arbitrates, 4 against 5 etc).

[MikeD]

Spoiler for a first try

12 3 45 6 78 9
36 5 91 7 24 8
35 1 67 2 89 4
14 3 28 6 75 9
13 5 26 7 49 8
37 1 59 2 86 4
15 3 27 6 48 9
16 5 29 7 43 8
38 1 56 2 79 4
18 3 25 6 47 9
39 1 58 2 46 7 <--
17 5 23 8 69 4

At the arrow is the only time I had to break the second rule, player 1 and 2 only played in one game before arbitrating again.

[bonanova]

Spoiler for a first try

12 3 45 6 78 9
36 5 91 7 24 8
35 1 67 2 89 4
14 3 28 6 75 9
13 5 26 7 49 8
37 1 59 2 86 4
15 3 27 6 48 9
16 5 29 7 43 8
38 1 56 2 79 4
18 3 25 6 47 9
39 1 58 2 46 7 <--
17 5 23 8 69 4

At the arrow is the only time I had to break the second rule, player 1 and 2 only played in one game before arbitrating again.

I'm thinking it would be hard to improve on that. Nice.