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Given $g \ge 2$, $k \ge 1$ and a population of $p = kg$ workers, I'm trying to figure out the longest series of work shifts such that:

  • during each shift, all workers work in $k$ teams of g people;
  • any two workers work together at most once.

In more formal terms: I'd like to build the largest set of partitions of $1,\dots,p$ in sets of $g$ items so that any two sets taken from any two distinct partitions have at most one element in common.

I have a simple enough solution for $g = 2$, that achieves the theoretical maximum of $\lfloor {p - 1 \over g - 1} \rfloor$ shifts (obtained by considering that worker #1, for instance, must be associated with $g - 1$ unique other workers during each shift). That raw upper bound cannot apparently be reached for $g > 2$ in general. Even with $g = 2$, I don't know how to enumerate all solutions efficiently.

Has this problem been studied before? Any pointers would be much appreciated.

Peter Taylor
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Yann David
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    I just found out that this is known as "the social golfer problem". It is in general an unresolved problem. – Yann David Dec 29 '14 at 20:02
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    For what it's worth, my solution assigns the pair of workers / players (i, j) with 0 ≤ i < j ≤ p - 1 (0..p-1 is actually more convenient than 1..p) to the shift / tournament (i + j) mod (p - 1) if j < p - 1, 2i mod (p - 1) if j = p - 1 (with shits numbered from 0 to p - 2). – Yann David Jan 03 '15 at 00:18
  • In this answer I showed that if $k=g$ is a power of a prime number than the largest set contains $k+1$ partitions. – Alex Ravsky Aug 13 '17 at 07:46

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