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Topic: Ping Pong tournament (Read 7853 times) |
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Altamira_64
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Ping Pong tournament
« on: Jun 7th, 2012, 12:30pm » |
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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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Altamira_64
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Re: Ping Pong tournament
« Reply #1 on: Jun 24th, 2012, 3:54pm » |
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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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Technologeek
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Re: Ping Pong tournament
« Reply #2 on: Jul 30th, 2012, 2:38pm » |
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Here is the answer of all games:
| hidden: | 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 |
But: 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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Against Wikipedia totalitarism - Proofs Wiki: The Mean value theorem proof and
Fundamental theorem of calculus
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Altamira_64
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Re: Ping Pong tournament
« Reply #3 on: Sep 16th, 2012, 4:30am » |
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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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pex
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Re: Ping Pong tournament
« Reply #4 on: Sep 16th, 2012, 6:09am » |
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on Sep 16th, 2012, 4:30am, Altamira_64 wrote: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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Ermm... this can be fixed by switching the last two rounds, can't it?
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