Dan Oneata wrote:

> Does anyone know how to solve the second part of the puzzle, that is, how to change the monoidal poset to count paths of length \\(n\\) separately for each \\(n\\)?

Perhaps the right thing would be the monoidal poset with object set \\(\mathbb{N}^\mathbb{N}\\), infinite sequences of natural numbers, with the product partial order, and with monoidal operation given by convolution?

Edit: Actually, I think there's a problem with this if the graph contains cycles. Then if \\(p: x\to y\\), \\(q: y\to y\\), and \\(r: y\to z\\) are paths, we'd be double-counting the path \\(p;q;r : x\to z\\) both as the concatenation of \\(p;q\\) with \\(r\\) and as the concatenation of \\(p\\) with \\(q; r\\). This problem doesn't arise in the "all paths" version Dan describes, since the concatenation function \\(\mathrm{Path}(x,y)\times\mathrm{Path}(y,z) \to \mathrm{Path}(x,z)\\) is either injective or both sides are infinite.

> Does anyone know how to solve the second part of the puzzle, that is, how to change the monoidal poset to count paths of length \\(n\\) separately for each \\(n\\)?

Perhaps the right thing would be the monoidal poset with object set \\(\mathbb{N}^\mathbb{N}\\), infinite sequences of natural numbers, with the product partial order, and with monoidal operation given by convolution?

Edit: Actually, I think there's a problem with this if the graph contains cycles. Then if \\(p: x\to y\\), \\(q: y\to y\\), and \\(r: y\to z\\) are paths, we'd be double-counting the path \\(p;q;r : x\to z\\) both as the concatenation of \\(p;q\\) with \\(r\\) and as the concatenation of \\(p\\) with \\(q; r\\). This problem doesn't arise in the "all paths" version Dan describes, since the concatenation function \\(\mathrm{Path}(x,y)\times\mathrm{Path}(y,z) \to \mathrm{Path}(x,z)\\) is either injective or both sides are infinite.