3

Okay here is my problem. I am not sure if this is possible or not as the 8 hours i have spent on this havn't been the best.

There are 4 groups of 9 teams. So we have teams A1 to A9, B1 to B9, C1 to C9, D1 to D9. The are 9 rounds of 9 different events. Each event can hold 1 team from each group. How do i make it so that each team faces every other team that isn't in their group exactly once and no team twice. And every team goes to each event once and no event twice.

Some background, this is for a school event where our school has 4 different houses. I have been put in charge of setting together the timetable for the day and the way the teams mvoe around the events. There is no way i am able to just have 2 houses face each other at an event as 4 teams from different houses has to compete at each event. Principle wants 9 teams for each house to compete.

Is this possible? If this isnt possible, would it be possible if the number of teams went down to 8 with the number of events go down to 8 and rounds go down to 8 also?

Thanks

hmakholm left over Monica
  • 286,031
user126911
  • 31

2 Answers2

2

Consider a $9\times9$ array, where the row represents the event and the column represents the round. For each house, arrange the $9$ copies numbers from $1$ to $9$ in the array, so that each number appears once in each row and each column. That is, make a $9\times9$ Latin square. Call these arrays $(a_{ij}),(b_{ij}),(c_{ij}),(d_{ij})$ for groups A,B,C,D, respectively.

If $a_ij=m$ for example, it means that team $A_m$ will play event $i$ in round $j$. Now suppose that arrays $(a_{ij}),(b_{ij})$ are orthogonal, that is if $$a_{ij}=a_{mn}=x$$ and $$b_{ij}=b_{mn}=y$$ then $$x=y.$$ Then $A_x$ meets team $B_y$ at most once, and since $A_x$ must meets $9$ different teams, it meets team $B_y$ exactly once.

We see that the conditions can be fulfilled if and only if there exist $4$ pairwise orthogonal Latin squares. Since $9=3^2$ is a prime power, it is known that there exist $8$ pairwise orthogonal Latin squares, and the answer is "yes."

The construction is described here.

EDIT

On reflection, I think I should have included a specific construction. Here is a set of $8$ mutually orthogonal $9\times9$ Latin squares, the maximum possible.

$$\begin{array}{|c|c c c|c c c|c c c|}\hline&1&2&3&4&5&6&7&8&9\\ \hline 1 &11111111&56489723&98765432&69358247&72634859&24973568&85296374&37542986&43827695\\ 2 &22222222&64597831&79846513&47169358&83415967&35781649&96374185&18653794&51938476\\ 3 &33333333&45678912&87954621&58247169&91526748&16892457&74185296&29461875&62719584\\ \hline4 &44444444&89723156&32198765&93682571&15967283&57316892&28539617&61875329&76251938\\ 5 &55555555&97831264&13279846&71493682&26748391&68124973&39617428&42986137&84362719\\ 6 &66666666&78912345&21387954&82571493&34859172&49235781&17428539&53794218&95143827\\ \hline7 &77777777&23156489&65432198&36925814&48391526&81649235&52863941&94218653&19584362\\ 8 &88888888&31264597&46513279&14736925&59172634&92457316&63941752&75329461&27695143\\ 9 &99999999&12345678&54621387&25814736&67283415&73568124&41752863&86137542&38476251\\ \hline\end{array}$$

The rows represent rounds, and the columns represent events. The table is for $8$ houses, say $A,B,C,D,E,F,G,H.$ The sequence indicates which teams from each house will compete in a particular event in a given round. For example, at row $3$, column $4$ we find $58247169$. The interpretation of this is that in round $3$, the teams that compete in event $4$ are $$A_5,B_8,C_2,D_4,E_7,F_1,G_6,\text{ and }H_9$$

In the instant case, where there are only $4$ houses, the last $4$ digits or each table entry can simply be ignored.

saulspatz
  • 53,131
-1

Initially, there are $9^4$ to organize an event. But every time you do it, you're eliminating not only one possibility out of the $9^4$. Indeed, if I picked up $\{A_1,B_1,C_1,D_1\}$ for the first event, I could not pick later any order in which $\{A_1, B_1\}$ appeared, which is $9^2-1$. On top of that, there are $\binom 4 2=6$ ways of choosing a pair out of the first event, so the previous number gets multiplied by $6$.

So you organized your first event and now, out of the $9^4$ options, there are only $9^4-6\times(9^2-1)$ left. For how many events can you do that? Simply calculate $\frac {9^4}{6\times(9^2-1)}$ and take the integer part. This actually already answers your second question.

Leo Azevedo
  • 385