 Solution

The printed formula for the edge label has mismatched variable names. The type-correct reading is $x_i=\sigma (v_{i-1},v_i)$. Interpret the requested route as a directed walk of consecutive labelled edges, allowing repeated vertices; the distinction from a vertex-simple directed path is discussed below.

Let $R_t$ be the set of vertices reachable from the specified start after reading exactly the first $t$ labels. Initialize $R_0=\{v_0\}$ and updateAfter $k$ stages, answer yes exactly when $R_k\ne \varnothing$. A directed walk realizing a prefix decomposes into a walk realizing the shorter prefix and one correctly labelled final edge; conversely every such extension is valid. This proves the dynamic programming invariant by induction on $t$. Store reachable vertices in Boolean arrays and scan the edges per stage. The running time is $O(k(|V|+|E|))$ with $O(|V|)$ rolling-table space complexity, in addition to graph storage. The empty label sequence is realized by staying at $v_0$.

If “path” is instead required to mean vertex-simple, augment the state to $(S,v)$, where $S$ is the set of visited vertices. Start with $R_0(\{v_0\},v_0)$ and extend a matching edge to $w$ only when $w\notin S$, replacing $S$ by $S\cup \{w\}$. This gives an exact dynamic programming algorithm for that interpretation too; reject immediately when $k\ge |V|$. A straightforward bound is $O(k{2}^{|V|}(|V|+|E|))$ running time with $O(2^{|V|}|V|)$ rolling-table storage. The exponential distinction is substantive: all labels equal and $k=|V|-1$ include the Hamiltonian path problem from a fixed start. The efficient layered recurrence answers the consecutive-edge interpretation without pretending to enforce distinct vertices.