This question came from a question asked earlier today linked here
The question implicitly asked how to make a schedule with his/her class of 24 students such that: 1) Everyday will consist of the 24 students split into 6 groups of 4 students. 2) Over the course of several days, every pair of students will have been in a group with one another.
This seems like it might be a Steiner system, but i'm not exactly sure. Steiner system $S(2,4,24)$. I've found that a necessary condition for this system existing is the existence of $S(1,3,23)$. Now here's where I'm not sure if I'm following the definition of Steiner system exactly. This system, $S(1,3,23)$ would be a $K_3$ decomposition of $K_{23}$, would it not? Perhaps this is not exactly a Steiner system, this is part of my question. Help me to clear this up
How many days would be required to schedule this?
Well, we need to partition the edges of $K_{24}$ which has $\binom{24}{2}$ edges. Each $K_4$ has 6 edges. So on a single day we would have 36 edges while we need a total of 276 edges. However, $\frac{276}{36}$ is not an integer, so this actually tells us that we can't do this if we require no edges to be repeated. So, some pairs of students will be in the same group with each-other more than once.
The Question: What is the minimum number of days required to schedule this? Will $\lceil\frac{276}{36}\rceil = 8$ days be sufficient? What exactly am I dealing with here? Some sort of design? A Steiner system? What?
This is known as the Social Golfer Problem. Best you can do without repetition is 7 days. But you can likely find an 8th day that fills in all the gaps.
day 1: ABKU IJSE QRCM DGFX HLNO PTVW
day 2: ACLV IKTF QSDN EHGR BMOP JUWX
day 3: ADMW ILUG QTEO FBHS CJNP KRVX
day 4: AENX IMVH QUFP GCBT DJKO LRSW
day 5: AFOR INWB QVGJ HDCU EKLP MSTX
day 6: AGPS IOXC QWHK BEDV FJLM NRTU
day 7: AHJT IPRD QXBL CFEW GKMN OSUV