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Title: Ping Pong tournament Post by Altamira_64 on Jun 7th, 2012, 12:30pm 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). |
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Title: Re: Ping Pong tournament Post by Altamira_64 on Jun 24th, 2012, 3:54pm Well, I have spent many days on this, but obviously it needs a program and I have no clue about programming whatsoever :) So, any volunteers, please help!! |
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Title: Re: Ping Pong tournament Post by atyq on Jul 30th, 2012, 2:38pm Here is the answer of all games: [hideb]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[/hideb] But: [hideb]39 1 58 2 46 7 [/hideb] is the only correct answer :). It is the is the only time I had to break the second rule, player 1 and 2 only played in one game before arbitrating again. |
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Title: Re: Ping Pong tournament Post by Altamira_64 on Sep 16th, 2012, 4:30am Thanks for the great solution, atyq. The publisher of the riddle claims that there is a solution with only ONE breaking of the second condition (yours has two, because both players 1 and 2 only played in one game before arbitrating again). Any ideas? I tried to do some shuffling in the order of the rows in your array, but still got 2 or more breakings of the 2nd condition. |
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Title: Re: Ping Pong tournament Post by pex on Sep 16th, 2012, 6:09am on 09/16/12 at 04:30:08, Altamira_64 wrote:
Ermm... this can be fixed by switching the last two rounds, can't it? |
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