Let's get more specific now.

Let \\(t\\) be a transition, and Stoich(\\(t\\)) be its stoichiometric vector.

For a species \\(z\\), let InpDegree(\\(t,z\\)) be the number of places that \\(t\\) inputs from \\(z\\), and OutDegree(\\(t,z\\)) be the number of places \\(t\\) outputs to \\(z\\).

Let \\(x \in \mathbb{N}^S\\) be a definite state, and let \\(x' = x\\) + Stoich(\\(t\\)).

If \\(x'\\) contains any negative components, then it is outside of \\(\mathbb{N}^S\\), and it is not a valid state. That indicates that the transition \\(t\\) doesn't have a sufficient number of input tokens in \\(x\\) to fire.

If however \\(x'\\) does belong to \\(\mathbb{N}^S\\), then we add an edge \\(e(t,x,x')\\) from \\(x\\) to \\(x'\\) in the graph.

Let \\(t\\) be a transition, and Stoich(\\(t\\)) be its stoichiometric vector.

For a species \\(z\\), let InpDegree(\\(t,z\\)) be the number of places that \\(t\\) inputs from \\(z\\), and OutDegree(\\(t,z\\)) be the number of places \\(t\\) outputs to \\(z\\).

Let \\(x \in \mathbb{N}^S\\) be a definite state, and let \\(x' = x\\) + Stoich(\\(t\\)).

If \\(x'\\) contains any negative components, then it is outside of \\(\mathbb{N}^S\\), and it is not a valid state. That indicates that the transition \\(t\\) doesn't have a sufficient number of input tokens in \\(x\\) to fire.

If however \\(x'\\) does belong to \\(\mathbb{N}^S\\), then we add an edge \\(e(t,x,x')\\) from \\(x\\) to \\(x'\\) in the graph.