Let $G=(V,E,w)$ be a signed directed graph where every vertex $v\in V$ has out-degree $2$, and every directed edge $e\in E$ has sign $w(e)\in\{-1, +1\}$.
A walk $z = (v_1, e_1, v_2, e_2, \ldots, v_{k-1}, e_{k-1}, v_k)$ in $G$ is an alternating sequence of vertices and directed edges where $e_i=(v_i, v_{i+1})$ for $i=1, \ldots, k-1$. Let us designate the walk $z$ as elementary if the following two conditions hold:
- $\sum_{i=1}^{k-1} w(e_i)=0$.
- For all $1 \leqslant k' < k-1$, the value of $\sum_{i=1}^{k'} w(e_i)$ is non-zero.
Let $W(G)$ be the set of all elementary walks in $G$.
I am looking to prove (or disprove by counterexample) the statement: If $G$ is finite, then $|W(G)|$ is finite.
If anyone has an intuition for this type of problem or has seen results that seem related to this, I would appreciate any pointers.
My example in comments is built on the following principle: if the graph contains a positive cycle and a negative cycle and a path between them, you can go around the positive cycle as many times as you like before walking over to the negative cycle and moving back to 0. The simplest (loopless) way to do that is with the edges $$ \{ a \overset{+}{\to} b, b \overset{+}{\to} a, a \overset{+}{\to} c, c \overset{-}{\to} d, d \overset{-}{\to} c\} \cup \{b \to ??, c \to ??, d \to ??\}, $$ where the last three are just there to make the degree condition work and their signs and termini don't matter.
In fact there is a simpler way to do it, namely, by having a cycle of net weight 0 with a positive path into it and a negative path out. This can be achieved with just three vertices: $$ \{ a \overset{+}{\to} b, \quad b \overset{+}{\to} c, c \overset{-}{\to} b, \quad b \overset{-}{\to} a\} \cup \{ a \to c, c \to a\}, $$ where again the last two edges are there just to make the degree conditions work.