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Motivation. In my town, every student spends the school year with the same set of students; that set is referred to as a "school class". My eldest son is in 6th grade, and that grade consists of $3$ school classes, each consisting of the same number of students. This week, school management announced they will rearrange the three school classes with the aim of maximizing the number of new students that each students gets to be with in the new school class. This gives rise to a bunch of mathematical questions, one of which I want to formulate here.

Formalization. Given a positive integer $n$, let $[n] = \{1,\ldots, n\}$. Let $k, n$ be integers greater than $1$. We call a partition $P$ of $[kn]$ a $(k,n)$-partition if $|P| = k$ and every member of $P$ has $n$ elements. (We can imagine $k$ to be the number of classes, and $n$ to be the number of students in each class.) The collection of $(k,n)$-partitions on the set $[kn]$ is denoted by $\newcommand{\Part}{\text{Part}}\Part(k,n)$. If $s\in [kn]$, let $[s]_P$ denote the unique member of $P$ containing $s$.

Given $s\in [kn]$ and $P,Q\in \Part(k,n)$, let $$\text{diff}_{P, Q}(s) = \frac{\text{card}\big([s]_Q \setminus [s]_P\big)}{n-1}.$$ The intuition behind this is: if we are the given school class assignment (= partition) $P$, how many new school class mates does student $s$ get under assignment $Q$? (We divide by $n-1$ because $s$ can get at most $n-1$ new class mates.) The quality of the assignment $Q$ with respect to $P$ is defined by $\text{qual}(P,Q) = \min\{\text{diff}_{P,Q}(s):s\in [kn]\}$.

As always in life, we want to maximize quality and set $\text{maxqual}(k,n) = \max\{\text{qual}(P,Q): P,Q\in\Part(k,n)\}$.

Question. If we fix $k\geq 2$, what is the value of $\liminf_{n\to\infty}\text{maxqual}(k,n)$?

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1 Answer 1

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$\operatorname{maxquel}(k, n) = \frac{\lfloor \frac{k-1}k n\rfloor}{n-1}$. In particular, the limit as $n \to \infty$ for a fixed $k$ is $\frac{k-1}k$.

For a lower bound, we can arbitrarily order the partitions and elements in each partition, and assign the $i$-th element in the $j$-th partition to the $i + j \bmod k$-th partition. By symmetry the parts in the resulting partition will all have the same size, which must be $n$. Every person remains together with at most $\lceil\frac{n}{k}\rceil$ members from $P$, so it has at least $n - \lceil\frac{n}{k}\rceil = \lfloor\frac{k-1}k n\rfloor$ new members, so the quality is $\frac{\lfloor\frac{k-1}k n\rfloor}{n-1}$.

For an upper bound, choose some part $A$ in $P$. Because $\sum_{B \in Q} |A \cap B| = |A|$ there must be some $B$ with $\lceil\frac nk\rceil \le |A\cap B|$. Choosing some $s \in A \cap B$, we get $\operatorname{diff}_{P,Q}(s) = \frac{|A \setminus B|}{n-1} = \frac{n - |A \cap B|}{n-1} \le \frac{n - \lceil\frac nk\rceil}{n-1} = \frac{\lfloor\frac{k-1}k n\rfloor}{n-1}$.

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