Social Golfer Problem - Quintets I wrote an article on the Social Golfer Problem, which has questions like: 
Each day, 16 people play Munchkin in foursomes simultaneously. How many days can they play with no two people playing with each other twice? tournament
I compiled best-known solutions for pairs, triplets, and quadruplets at the Social Golfer Problem Demonstration. To compile this data, I had to read through a few hundred books and papers. Very often, an answer would be given in new-to-me difficult notation, or it would reference a previous paper, or it would just be "obvious".  
Now that I am considered an "expert", I've been asked about quintets and sextets, and for larger number of groups. I don't recall seeing solutions for any of the below problems in the papers I perused, but it's possible I missed them.


*

*30 play in groups of 3 each day. No two play together more than once. How many days?

*33 play in groups of 3 each day. No two play together more than once. How many days?

*36 play in groups of 3 each day. No two play together more than once. How many days?

*40 play in groups of 4 each day. No two play together more than once. How many days?

*44 play in groups of 4 each day. No two play together more than once. How many days?

*48 play in groups of 4 each day. No two play together more than once. How many days?

*52 play in groups of 4 each day. No two play together more than once. How many days?

*25 play in groups of 5 each day. No two play together more than once. How many days?

*30 play in groups of 5 each day. No two play together more than once. How many days?

*35 play in groups of 5 each day. No two play together more than once. How many days?

*40 play in groups of 5 each day. No two play together more than once. How many days?

*45 play in groups of 5 each day. No two play together more than once. How many days?

*50 play in groups of 5 each day. No two play together more than once. How many days?

*55 play in groups of 5 each day. No two play together more than once. How many days?

*60 play in groups of 5 each day. No two play together more than once. How many days?

*36 play in groups of 6 each day. No two play together more than once. How many days?

*42 play in groups of 6 each day. No two play together more than once. How many days?

*48 play in groups of 6 each day. No two play together more than once. How many days?

*54 play in groups of 6 each day. No two play together more than once. How many days?

*60 play in groups of 6 each day. No two play together more than once. How many days?

*66 play in groups of 6 each day. No two play together more than once. How many days?

*72 play in groups of 6 each day. No two play together more than once. How many days?


Does anyone have any solutions to any of these?
 A: Have you seen Ivan Dotu and Pascal Van Hentenryck, Scheduling social golfers locally? They claim  
45 golfers in groups of 5 can play 7 days;  
54 golfers in groups of 6 can play 6 days;  
50 golfers in groups of 5 can play 8 days;  
and some other results that lie outside the bounds of your question. But they don't actually present any instances of the solutions, just their algorithm and the amount of time it took to find the instances. 
A: I don't have a lot of advice on quintets in particular, but perhaps the information below will be useful nevertheless.
Some of the solutions you are looking for correspond to the necessary conditions for a resolvable balanced incomplete block designs and, if such a RBIBD exists, then this will lead immediately to a solution with the maximal number of days.  The items in this category are:


*

*33 play in groups of 3 for 16 days - the much studied Kirkman Triple System

*40 play in groups of 4 for 13 days - see Kageyama (1972)   

*52 play in groups of 4 for 17 days - see Lam & Miao (1999) Lemma 5.3  

*25 play in groups of 5 for 6 days - see comment above

*45 play in groups of 5 for <=11 days - RBIBD is an open problem

*36 play in groups of 6 for <7 days - RBIBD does not exist

*66 play in groups of 6 for <=13 days - RBIBD is an open problem


In addition to the above, I can offer examples of the following:


*

*30 play in groups of 3 for 14 days [maximal]

*36 play in groups of 3 for 17 days [maximal]

*44 play in groups of 4 for 13 days

*48 play in groups of 4 for 14 days [see Ge and Lam(2003) Lemma 3.1]

*45 play in groups of 5 for 9 days.


I have also been confused by the large literature on this subject.  I feel that 44/4/14 days, and 48/4/15 days should both be possible, but I have not come across a construction yet.
References:
Ge & Lam (2003) "Resolvable group divisible designs with block size four and
 group size six"; Discrete Mathematics, 268, 139 – 151.
Kageyama (1972) "A survey of resolvable solutions of balanced incomplete block designs"; Rev. Inst. Internat. Statist., 40, 269–273.
Lam & Miao (1999) "On Cyclically Resolvable Cyclic Steiner 2-Designs", Journal of Combinatorial Theory, Series A 85, 194-207.
A: The Social Golfer Problem with triples and an odd number of groups is studied using
Kirkman Triple Systems. This page claims that $6n+3$ golfers playing in triplets can play for $3n+1$ days (for all nonnegative $n$). This is proven in the following paper (though I haven't checked this myself).


*

*Solution of Kirkman's schoolgirl problem, Ray-Chaudhuri and Wilson, 1971. In Proc. of Symp. in Pure Math, Vol 19.


EDIT:
I did some more research. If $6n \ge 18$ players are playing in triplets, they can play for $3n-1$ days. This follows from the study of Nearly Kirkman Triple Systems. The proof is spread out over multiple papers, the following paper cites them:


*

*Nearly Kirkman triple systems of order 18 and Hanani triple systems of order 19, Charles J.Colbourn et al. Discrete Mathematics 2011 311:827-834.


Furthermore, if $12n+4$ players play in groups of 4, they can play for $4n+1$ days. This is proven in


*

*On resolvable designs, Haim Hanani, D.K. Ray-Chaudhuri, Richard M. Wilson Discrete Mathematics, 1972, 3:343-357.


Furthermore, Asymptotic Existence of Nearly Kirkman Systems, Hao Chen, Wen-Song Chu makes the following claims (citing other papers for the first two claims and proving the third)


*

*If $12n\ge24$ players play in groups of 4, they can play for $4n-1$ days, except possibly for $12n\in\{84, 132, 264, 372, 456, 552, 660, 804, 852\}$.

*For a fixed group size $k$ for all but finitely many $n \equiv k \mod k(k-1)$ we have that $n$ golfers can play the optimal $\frac{n-1}{k-1}$ number of days.

*For a fixed group size $k$ for all but finitely many $n \equiv 0 \mod k(k-1)$ we have that $n$ golfers can play the optimal $\frac{n}{k-1}-1$ number of days.


This answers the following questions:


*

*30 play in groups of 3 for 14 days.

*33 play in groups of 3 for 16 days.

*36 play in groups of 3 for 17 days.

*40 play in groups of 4 for 13 days.

*48 play in groups of 4 for 15 days.

*52 play in groups of 4 for 17 days.

A: A new (Dec 2020) paper on this topic uses symmetry to approach the problem, and publishes solutions for groups up to 50 participants: Breakout Group Allocation Schedules and the Social Golfer Problem with Adjacent Group Sizes
