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.