A [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(V,E)$ is *sane* if $V$ is finite, $E \neq \emptyset$, and $\emptyset \notin E$, and $e\not\subseteq e'$ whenever $e\neq e' \in E$. Moreover, we call $H$ *summable* if there is a map $f:V\to \mathbb{Z}\_{\geq 0}$ such that for all $e, e' \in E$ we have $$\sum\_{v\in e}f(v) = \sum\_{w\in e'}f(w) > 0.$$
(The $>0$ requirement is there to exclude $f$ being the constant $0$-map.)


**Question.** Is every sane hypergraph summable?



Summable hypergraphs
No, not every sane hypergraph is summable. To see this, let $G$ be a star with five leaves $\ell\_1, \dots, \ell\_5$ all adjacent to a vertex $u$. Then turn $G$ into a sane hypergraph $H$ by adding the hyperedges $\{\ell\_1,\ell\_2\}$ and $\{\ell\_3, \ell\_4, \ell\_5\}$. Towards a contradiction, suppose that $f: V(H) \to \mathbb{Z}\_{\geq 0}$ is a function which shows that $H$ is summable. Since $\{u,\ell\_i\}$ is an edge of $H$ for all $i \in [5]$, we must have $f(\ell\_i)=f(\ell\_j)$ for all $i,j \in [5]$. But now the hyperedges $\{\ell\_1, \ell\_2\}$ and $\{\ell\_3, \ell\_4, \ell\_5\}$ have different sums.


