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On Euler Paths and the Maximum Degree Growth of Iterated ...
Aryan Sanghi, Anubhav Dhar, Sudeshna Kolay · 2026-02-10 · via math.CO updates on arXiv.org

Given a simple graph $G$, its line graph, denoted by $L(G)$, is obtained by representing each edge of $G$ as a vertex, with two vertices in $L(G)$ adjacent whenever the corresponding edges in $G$ share a common endpoint. By applying the line graph operation repeatedly, we obtain higher order line graphs, denoted by $L^{r}(G)$. In other words, $L^{0}(G) = G$, and for any integer $r \ge 1$, $L^{r}(G) = L(L^{r-1}(G))$. Given a graph $G$ on $n$ vertices, we wish to efficiently find out (i) if $L^k(G)$ has an Euler path, (ii) the value of $Δ(L^k(G))$. Note that the size of a higher order line graph could be much larger than that of $G$. For the first question, we show that for a graph $G$ with $n$ vertices and $m$ edges the largest $k$ where $L^k(G)$ has an Euler path satisfies $k = \mathcal O(nm)$. We also design an $\mathcal{O}(n^2m)$-time algorithm to output all $k$ such that $L^k(G)$ has an Euler path. For the second question, we study the growth of maximum degree of $L^k(G)$, $k \ge 0$. It is easy to calculate $Δ(L^k(G))$ when $G$ is a path, cycle or a claw. Any other connected graph is called a prolific graph and we denote the set of all prolific graphs by $\mathcal G$. We extend the works of Hartke and Higgins to show that for any prolific graph $G$, there exists a constant rational number $dgc(G)$ and an integer $k_0$ such that for all $k \ge k_0$, $Δ(L^k(G)) = dgc(G) \cdot 2^{k-4} + 2$. We show that $\{dgc(G) \mid G \in \mathcal G\}$ has first, second, third, fourth and fifth minimums, namely, $c_1 = 3$, $c_2 = 4$, $c_3 = 5.5$, $c_4 = 6$ and $c_5=7$; the third minimum stands out surprisingly from the other four. Moreover, for $i \in \{1, 2, 3, 4\}$, we provide a complete characterization of $\mathcal G_i = \{dgc(G) = c_i \mid G \in \mathcal G \}$. Apart from this, we show that the set $\{dgc(G) \mid G \in \mathcal G, 7 < dgc(G) < 8\}$ is countably infinite.