Jonathan - right! Defining \\(\mathcal{X}(x,y)\\) in terms of a _meet_ is also useful in the case when there are _infinitely many_ paths from \\(x\\) to \\(y\\). As long as our symmetric monoidal preorder has all meets, we're fine. And a "quantale" has all meets.

Here's how I'd give the formula. In category theory, a **graph** is a set \\(V\\) of **vertices** and a set \\(E\\) of **edges**, together with maps \\(s,t : E \to V\\) giving the source and target (starting-point and ending-point) of each edge. Given a graph, a **path** from a vertex \\(x\\) to a vertex \\(y\\) is a finite sequence of edges \\((e_1, \dots, e_n)\\) with

\[ s(e_1) = x, \quad t(e_1) = s(e_2), \quad \dots, \quad t(e_{n-1}) = s(e_n), \quad t(e_n) = y. \]

For any symmetric monoidal poset \\(\mathcal{V}\\), a **\\(\mathcal{V}\\)-weighted graph** is a graph together with a map

\[ \ell: E \to \mathcal{V} \]

assigning each edge an element of \\(\mathcal{V}\\). We define the **length** of a path \\(p = (e_1, \dots, e_n)\\) in a \\(\mathcal{V}\\)-weighted graph to be

\[ L(p) = \ell(e_1) \otimes \cdots \otimes \ell(e_n) .\]

Then we define

\[ \mathcal{X}(x,y) = \bigwedge \\{ L(p) : \; p \textrm{ is a path from } x \textrm{ to } y \\} .\]

This will exist if \\(\mathcal{V}\\) has all meets.

**Puzzle 97.** Under what conditions on \\(\mathcal{V}\\) will this construction of \\(\mathcal{X}(x,y) \\) give a \\(\mathcal{V}\\)-category?

A unital commutative quantale will do the job, but it's more fun to check the two \\(\mathcal{V}\\)-category axioms and see what will make them work.