A tight upper bound on the (2,1)-total labeling number of outerplanar graphs

TL;DR

This paper proves a conjecture that the (2,1)-total labeling number of outerplanar graphs is at most their maximum degree plus two, covering all cases including graphs with maximum degree up to four.

Contribution

The paper confirms Chen and Wang's conjecture for all outerplanar graphs, including those with maximum degree less than five, completing the proof.

Findings

The (2,1)-total labeling number is at most maximum degree plus two for all outerplanar graphs.

The conjecture holds true even for graphs with maximum degree less than five.

The result extends previous partial proofs to all outerplanar graphs.

Abstract

A $(2,1)$-total labeling of a graph $G$ is an assignment $f$ from the vertex set $V(G)$ and the edge set $E(G)$ to the set $\{0,1,...,k\}$ of nonnegative integers such that $|f(x)-f(y)|\ge 2$ if $x$ is a vertex and $y$ is an edge incident to $x$, and $|f(x)-f(y)|\ge 1$ if $x$ and $y$ are a pair of adjacent vertices or a pair of adjacent edges, for all $x$ and $y$ in $V(G)\cup E(G)$. The $(2,1)$-total labeling number $\lambda^T_2(G)$ of a graph $G$ is defined as the minimum $k$ among all possible assignments. In [D. Chen and W. Wang. (2,1)-Total labelling of outerplanar graphs. Discr. Appl. Math. 155, 2585--2593 (2007)], Chen and Wang conjectured that all outerplanar graphs $G$ satisfy $\lambda^T_2(G) \leq \Delta(G)+2$, where $\Delta(G)$ is the maximum degree of $G$, while they also showed that it is true for $G$ with $\Delta(G)\geq 5$. In this paper, we solve their conjecture completely, by proving that $\lambda^T_2(G) \leq \Delta(G)+2$ even in the case of $\Delta(G)\leq 4 $.