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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

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: 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. Moreover, 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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,956

Hi anna_gg;

It must be completed in 12 rounds of 3 simultaneous games

where each player will play against each of the other 8 only once

be arbitrating exactly 4 games

Moreover, after each player arbitrates one game, he must play at least 2 times against another athlete before being allowed to arbitrate again.

There are 4 conditions here. Which of these are not to broken and which can be?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

Well, you are right; actually I was considering the first 3 conditions as one These 3 can't be broken.

The 4th condition, which is "after each player arbitrates one game, he must play at least 2 times against another athlete before being allowed to arbitrate again", can be broken, but we request that this happens the least number of times.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,956

Hi anna_gg;

Okay, thank you. You do understand that this problem is somewhat more difficult than a progressive dinner or social golfer problem and that most of them do not have solutions.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

Hi Bobbym,

I didn't say it is easy After all, we are not here for the easy ones

This one, however, does have a solution because it was published it a riddles site.

Have a nice weekend!

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

Let's start with

123456789 where 3, 6 and 9 are the arbitrators.

Then 132465798

231564897 where 1 4 and 7 have played 2 games, in order for them to be allowed to arbitrate.

That was the easy part, working on the next steps

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

143286759

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,956

Hi anna_gg;

My feeling is that a program will be necessary. So far none have worked.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

You are absolutely right, but I don't have any experience in programming

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 98,956

I have lots and it is not helping.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **Thinking is cheating.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

Am sure someone will show up with a brilliant idea

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