Good seating arrangements for sequence of meals and tables of size k for a group of people Given a set $S$ of people I'd like to sit them for a sequence of meals at tables of size $k$. (Of course, there are enough tables to sit all $|S|$ for each meal.) I'd like to arrange this such that nobody shares a table with the same person twice. Typical values are $|S|=45$ and $k=5$ and 6 to 10 meals.

Put in a more abstract way, I'd like to find a sequence of partitions of $S$ such that each partition consists of pairwise disjoint subsets of cardinality $k$ and the added global property that any intersection between two such subsets contains no more than one element. I suspect this can be formulated as a graph theoretical or combinatorical problem. 

I'd be grateful for a better formulation of the problem and pointers to relevant literature as it is outside of my domain.

The background: this could be used for seating arrangements at Schloss Dagstuhl where many computer scientists come to discuss their research over the course of a week. Currently seating is done randomly and unsurprisingly some people find themselves sitting with the same people twice (or more often) over the course of a week. Also unsurprisingly, we receive some complaints about this and vague suggestions how to improve this. I'd like to understand this better. A stronger formulation of the problem involves optimizing who is sitting next to each other but I believe this is not relevant for tables of size 5. 

Outside of the application I think the interesting question is for the maximum number of meals that can be served for a given $S$ and $k$, i.e., how many such partitions exist.
Here's a variant of the original answer (below) that gives the desired setting: tables of size 5, 45 people, and 10 meals, except that one meal has a few tables of size 4.

Let $F$ be the field of size 9.  Pick 4 vertical, degenerate lines  $\{(b,x) | x \in F\}$ for every $b = 0,1,2,3$ and declare their people "empty."  We are left with 81 - 9x4 = 45 people.

9 meals are given by slopes $a = 0,1,\ldots,8$.  The intersections with the 4 empty degenerate lines reduce the table size to 9-4=5.

An additional meal is given by the remaining degenerate lines $\{(b,x) | x \in F\}$ for every $b = 4,5,6,7,8$.  Here the table size is 9.  However (in any solution) we can break a table of size 9 into a table of size 5 and one of size 4.

If there are a few more people one can use the field of size 11.



First let us handle $k^2$ people and $k$ meals.

Pick a finite field $F$ of size $k$ and identify people with $F \times F$.  To each meal there corresponds a slope, to a table a line parallel to that slope.

Specifically, meal $a$ has $k$ tables $\{(x,ax+b) | x \in F\}$ for every $b \in F$.

The intersection property you want is the fact that lines with distinct slopes intersect in exactly one point.



To handle $2k^2$ people, divide them up in two groups of $k^2$ each, and apply the construction above to each group.  To handle $2k^2 - k = 45$, label (in the first group) a fixed line such as $\{(x,x)|x\in F\}$ as "empty."  You may have a few tables with $k-1$ people.

For more meals one could e.g. pick a different partition in two groups at the beginning of the 6th meal. (Say you interleave the original partition, to ensure that the two groups "mix.") Though of course this may result in some intersections.