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A linear hypergraph is a pair $\pi=(X, L)$ where $X\neq \emptyset$ is a set and $L\subseteq {\cal P}(X)$ has the following properties:

  1. for $e\in L$ we have $|e|\geq 2$;
  2. if $e_1\neq e_2 \in L$ then $|e_1\cap e_2|\leq 1$.

We set $X(\pi)=X$ and $L(\pi)=L$. The graph $G_\pi$ associated to a linear hypergraph $\pi$ is given by $G=(V,E)$ where $V = L$ and $E = \{\{e_1, e_2\} \subseteq L: e_1\neq e_2\text{ and } e_1\cap e_2\neq \emptyset\}$. It turns out that for any graph $G$ there is a linear hypergraph $\pi$ such that $G\cong G_\pi$. For any graph $G$ the we set $$\ell(G) := \text{min}\{|X(\pi)|:\pi \text{ is a linear hypergraph such that } G_{\pi} \cong G\}$$ and call this the linear intersection number of $G$. (For infinite graphs, this concept is boring: $\ell(G) = |V(G)|$ for infinite graphs.)

For any graph $G$ let $\tau(G)$ be the minimum cardinality of a vertex cover of $G$.

Does $\tau(G) \leq \ell(G)$ hold for all graphs $G$?

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    $\begingroup$ $V$ is used for too many things. $\endgroup$ – domotorp Jan 28 '15 at 14:24
  • $\begingroup$ That's right, thanks for pointing it out - I amended it.. $\endgroup$ – Dominic van der Zypen Jan 28 '15 at 15:35
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Let $H$ be a graph on $n$ vertices, thought of as the hypergraph in the question. Then a vertex cover in $G_H$ is precisely a subgraph of $H$ that contains all but at most one edge at each vertex of $H$. Such a subgraph is obtained from $H$ by deleting a subgraph of maximum degree at most $1$, which contains at most $n/2$ edges. So any graph $H$ with more than $3n/2$ edges will provide a counterexample.

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No, it might not hold. Let $\pi=K_{4,4}$, so $|X|=8$ and each $e\in L$ contains exactly two vertices. This implies that $\ell(G_\pi)\le 8$. But for each $x\in X$ we must select at least three of the four edges meeting there in any vertex cover of $G_\pi$, thus $\tau(G_\pi)\ge 3\cdot 8/2=12$.

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  • $\begingroup$ Very nicely written - I would like to accept both Ben's and your answer! $\endgroup$ – Dominic van der Zypen Jan 29 '15 at 12:41
  • $\begingroup$ @Dominic: I guess he deserves it more as he spotted the mistake in my earlier attempt. $\endgroup$ – domotorp Jan 29 '15 at 15:45
  • $\begingroup$ OK - then I hope you don't mind if I accepted his answer (as I can only accept one answer). $\endgroup$ – Dominic van der Zypen Jan 29 '15 at 15:47
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This answer is incorrect! As pointed out by Ben, I confused dominating set and vertex cover.

Yes, it is true. For every vertex $v$ of the hypergraph $\pi$, consider the edges of the graph $G$ that run between hyperedges that intersect in $v$. This will be a clique, which we can denote by $K_v$. We can select any vertex of the clique for each $v$, this gives a vertex cover of size at most "$\ell$".

The linearity seems not needed. It would imply, btw, that any two such cliques are edge-disjoint, that is, $K_v\cap K_u\cap E=\emptyset$. Therefore in this case the edgeset of the graph, $E$, is an edge-disjoint union of cliques.

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    $\begingroup$ This is certainly a dominating set (if you pick one hyperedge containing each hypervertex then every hyperedge meets some hyperedge in the set), but is it a vertex cover? For that you would need that for every intersecting pair of hyperedges, one of the pair is in the cover. Or have I misunderstood? $\endgroup$ – Ben Barber Jan 29 '15 at 11:27
  • $\begingroup$ Oops - I fell into the same trap... Thanks for the clarification! $\endgroup$ – Dominic van der Zypen Jan 29 '15 at 12:35

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