0
$\begingroup$

For any set $X$ and any cardinal $\kappa$, let $[X]^\kappa$ denote the subsets of $X$ having cardinality $\kappa$.

A linear hypergraph is a hypergraph such that for all $e\neq e_1 \in E$ we have $|e\cap e_1|\leq 1$. For any positive integer $n$ let $[n] = \{1,\ldots,n\}$. We say that $([n^2], E)$ is a square hypergraph if it is linear and every element of $E$ has $n$ elements.

For any integer $n>1$ let $m(n)$ denote the maximum cardinality of $E$ where $([n^2], E)$ is a square hypergraph. For instance we have $m(2) = 6$. What is the value of $$\lim (\sup)_{n\geq 2}\frac{m(n)}{n^2}\;?$$ (Note: I added $(\sup)$ in case the limit doesn't exist.)

$\endgroup$
4
$\begingroup$

In a linear hypergraph, any pair of vertices is contained in at most one hyperedge. Since any hyperedge contains $n$ vertices, it contains ${n \choose 2}$ pairs. Double counting gives $$m(n) \leq \frac{n^2 \choose 2}{n \choose 2} = n (n+1).$$

This bound is sharp when $n$ is a prime power (take an affine plane of order $n$), so $$ \limsup \frac {m(n)}{n^2} = 1. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.