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

]]>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

]]>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!

]]>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.

]]>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.

]]>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?

]]>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).]]>