A *linear hypergraph* is a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(V,E)$ such that 


1. $|e|\geq 2$ for all $e\in E$,
2. $|e\_1\cap e\_2|\leq 1$ for all $e\_1, e\_2\in E$ with $e\_1\neq e\_2$.


We call a linear hypergraph *complete* if there is equality in statement 2 above, i.e. if $|e\_1\cap e\_2| = 1$ for all $e\_1, e\_2\in E$ with $e\_1\neq e\_2$.


Moreover, a linear hypergraph is *saturated* if for all $v\in V$ we have $|\{e\in E:v\in e\}| \geq 2.$


For $n>2$ set $\mathbb{N}\_n =\{1,\ldots,n\}$ and let $$\ell\_1(n)=\min\{|E|: E\subseteq{\cal P}(\mathbb{N}\_n) \text{ and }(\mathbb{N}\_n, E) \text{ is linear, saturated}\},$$ and 
$$\ell\_2(n)=\min\{|E|: E\subseteq{\cal P}(\mathbb{N}\_n) \text{ and }(\mathbb{N}\_n, E) \text{ is linear, saturated & complete}\}.$$


Obviously, we have $\ell\_1(n) \leq \ell\_2(n)$, but do we have equality?



Complete and saturated linear hypergraphs
You have equality, yes.


The same proof [I gave here](https://mathoverflow.net/questions/255230/minimal-number-of-edges-for-complete-linear-hypergraphs) applies equally well to this situation.


Here's a generalization of both (also implied by that proof).



> 
> Suppose $\mathcal{H}$ is a hypergraph on $[n]$ where any $r$ edges share at most $1$ point and each point is contained in at least $r$ edges. Then if $m$ is the number of edges in $\mathcal{H}$, we need $n \leq {m \choose r}$ and this is tight.
> 
> 
> 


(The modification of the proof is to consider $r$-element subsets of $V\_i$.)


