# How can 12 teams rotate through 6 games without overlaps?

Given the following:

• Six Number Teams (1 - 6)
• Six Letter Teams (A - F)
• Six Games (Basketball, Football, Baseball, Volleyball, Hockey, Rugby)
• Six Time Slots (1pm - 6pm)

Set up a game schedule that follows these rules:

• Each team must play each game once.
• Each letter team must play against each number team exactly once.
• Every game must be played once (by one letter team and one number team) during each time slot

Please provide either a solution or a mathematical prove of why a solution is impossible.

• For that last bullet point, do you mean that during each time slot, only 2 teams can play, total, or only 2 teams per game can play each time slot? – JonTheMon Mar 13 '15 at 13:14
• @JonTheMon Only 2 teams/game/slot. All 12 teams must play each time slot. – snumpy Mar 13 '15 at 13:30
• I have noted that this is simple with an odd number of everything. I've thrown together a solution using 5 letters, 5 numbers, 5 games, 5 time slots simply by rotating the letters in one direction and the numbers in the other. However, finding a solution with an even number of objects (even two) is baffling. – snumpy Mar 13 '15 at 17:00
• To do this, wouldn't we need mutually orthogonal latin squares of order 6, a known impossibility? cut-the-knot.org/arithmetic/latin3.shtml – Trenin Mar 13 '15 at 17:11
• @Trenin Yup, make it an answer. – Lopsy Mar 13 '15 at 17:49

In order to do this puzzle, you'd need to create Mutually Orthogonal Latin Squares of order 6.

For example, say that instead, you had 6 teams (1-3 and A-C), 3 sports (baseball, football, hockey), 3 timeslots. Then, you could make the following schedule:

    A   B   C
1  b@1 f@2 h@3
2  h@2 b@3 f@1
3  f@3 h@1 b@2


So in this example, team A plays baseball against team 1 at 1:00pm.

This uses 2 Mutually Orthogonal Latin Squares of order 3.

1 2 3    b f h
2 3 1    h b f
3 1 2    f h b


This allows us to conform to the following rules:

• Every lettered team plays every numbered team exactly once - simply by design of the table
• Every team plays every sport exactly once
• Every team plays in every timeslot exactly once

However, it is a known impossibility to create two MOLS of order 6, so the original question is not possible.

I have named the sports U, V, W, X, Y and Z for convenience

This can be solved simply by rotating through the teams and sports:

Timeslot 1
A vs 1 at U ; B vs 2 at W ; C vs 3 at Y ; D vs 4 at U ; E vs 5 at W ; F vs 6 at Y

Timeslot 2
A vs 2 at V ; B vs 3 at X ; C vs 4 at Z ; D vs 5 at V ; E vs 6 at X ; F vs 1 at Z

Timeslot 3
A vs 3 at W ; B vs 4 at Y ; C vs 5 at U ; D vs 6 at W ; E vs 1 at Y ; F vs 2 at U

Timeslot 4
A vs 4 at X ; B vs 5 at Z ; C vs 6 at V ; D vs 1 at X ; E vs 2 at Z ; F vs 3 at V

Timeslot 5
A vs 5 at Y ; B vs 6 at U ; C vs 1 at W ; D vs 2 at Y ; E vs 3 at U ; F vs 4 at W

Timeslot 6
A vs 6 at Z ; B vs 1 at V ; C vs 2 at X ; D vs 3 at Z ; E vs 4 at V ; F vs 5 at X

• If A and 1 are playing U during time slot 1, then D and 4 can't be playing U as well. Only two teams can play a game during each time slot (which means that every game will have to be played during each time slot). – snumpy Mar 13 '15 at 13:35
• The prmopt seemed a little confusing. It came across that each sport needed to be played twice at each time slot (which is impossible with 6 possible games and only 6 sports), but I can see where Andy got this. – Aggie Kidd Mar 13 '15 at 13:46
• @AggieKidd I'm not sure how "only two teams can play a game during each time slot" could be interpreted to mean "a game must be played twice (by four teams) during each time slot", but I've edited the question to clarify. – snumpy Mar 13 '15 at 13:52
• It's early in the morning. My brain can do anything, so I extend that same thought towards everyone else as well. Didn't say it made sense, just that is how my brain initially read it. – Aggie Kidd Mar 13 '15 at 13:55
• Yeah, I think it was having the word "two" in there that made me assume it could be played twice. I was surprised at how easy the puzzle was, so not surprised I got it wrong! I've edited the question to remove the word "two" - makes it really clear now. – AndyT Mar 13 '15 at 13:58

There doesn't seem to be a valid setup. Picking team 1, team A and game U at random, team 1 will plays game U against team A during the first time slot. That will leave five other teams left that team 1 will have to play against, each of those have played one of the remaining games. Team 1 must not play its first game type again, none of the other five teams is allowed to play the first game type they played again, after playing at most four additional games no further valid match is possible.

Simplifying things a bit, suppose there are just two time slots, two number teams (1, 2), two letter teams (A, B) and two games (U, V). Team 1 will play game U against team A in the first time slot. Team 1 can't play game U against Team B in the second time slot because team 1 already played game U. Team 1 also can't play game V against team B because team B already played game V during the first time slot.

• Although your conclusion may ultimately be correct, comparing your simplification to the actual problem is like comparing apples to oranges. I'll go on to state that I fairly easily found a solution where there are three in each category. – Warlord 099 Mar 13 '15 at 16:50
• Maybe I'm not reading the rules correctly, I don't really see how you could find a solution for three. It doesn't work for two, the final combination will always be invalid. Increasing the numbers won't change that, the last combination will always clash. You can't play n game types against n teams without duplicates, you'd either need n+1 opponents or n+1 game types. – Moghwyn Mar 13 '15 at 16:59
• @Moghwyn You may actually be on to something; however, note that it is possible with 3, 5, 7,... of each object (see my comment on the question). Having an even number of each object is either difficult or impossible - and that's what I want to know. – snumpy Mar 13 '15 at 17:04
• Ugh, right, it really seems to be possible with odd numbers. Tricky. – Moghwyn Mar 13 '15 at 17:13
• @snumpy It is actually possible for every number != 2/6. But that is a difficult thing to prove here. An example for 10 came out in the 1940s I think, and the others (14, 18, etc) can be built through complex constructions. – Trenin Mar 13 '15 at 18:04