2016 MacBook Pro shutdowns when the lid is close with cause 0

To answer Jon Noel's question in the comments, there is no such example for finite hypergraphs.

Claim. Let $H=(V,E)$ be a finite hypergraph such that $|e| > 1$ for all $e \in E$ and for all distinct $e_1, e_2 \in E$, $|e_1 \cap e_2| \neq 1$. Then $H$ is $2$-chromatic.

Proof. We proceed by induction on $|E|$. The base case is clear. Now arbitrarily choose $e \in E$ and let $G=(V, E \setminus e)$. By induction, $G$ has a $2$-coloring $c$. If two vertices in $e$ receive distinct colors from $c$, then $c$ is a $2$-colouring of $H$ and we are done. So we may assume that all vertices in $e$ are red. We now choose a vertex $x \in e$ and recolor $x$ blue. This is a valid $2$-colouring of $H$ unless there is an edge $f \in E$ such that $x \in f$ and all other vertices in $f$ are blue. But this implies that $|e \cap f|=1$, which is a contradiction.

Try this

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