Maximum locally irregular induced subgraphs via minimum irregulators

Abstract

If a graph $G$ is such that no two of its adjacent vertices have the same degree, we say that $G$ is locally irregular. In this work we introduce and study the problem of finding a largest locally irregular induced subgraph of a given graph $G$. Equivalently, given a graph $G$, find a subset $S$ of $V\left(G\right)$ with minimum order, such that deleting the vertices of $S$ from $G$ results in a locally irregular graph; we denote with $I\left(G\right)$ the order of such a set $S$. We first examine some easy graph families, namely paths, cycles, complete bipartite and complete graphs. However, we show that the decision version of the introduced problem is $\mathrm{NP}$-Complete, even for restricted families of graphs, such as subcubic planar bipartite, or cubic bipartite graphs. We then show that we cannot even approximate an optimal solution within a ratio of $O\left(n^{1-\frac{1}{k}}\right)$, where $k\ge 1$ and $n$ is the order of the graph, unless $P$=$\mathrm{NP}$, even when the input graph is bipartite.

Then, looking for more positive results, we turn our attention towards computing $I\left(G\right)$ through the lens of parameterised complexity. In particular, we provide two algorithms that compute $I\left(G\right)$, each one considering different parameters. The first one considers the size of the solution $k$ and the maximum degree $\Delta$ of $G$ with running time ${\left(2\Delta \right)}^k{n}^{O\left(1\right)}$, while the second one considers the treewidth $tw$ and $\Delta$ of $G$, and has running time ${\Delta }^{3tw}{n}^{O\left(1\right)}$. Therefore, we show that the problem is in FPT by both $k$ and $tw$ if the graph has bounded maximum degree $\Delta$.

Since the algorithms we present are not in FPT for graphs with unbounded maximum degree (unless we consider $\Delta +k$ or $\Delta +tw$ as the parameter), it is natural to wonder if there exists an algorithm that does not include additional parameters (other than $k$ or $tw$) in its dependency. We answer negatively to this question. In particular, we prove that there is no algorithm that computes $I\left(G\right)$ with dependence $f\left(k\right)n^{o\left(k\right)}$ or $f\left(tw\right)n^{o\left(tw\right)}$, unless the ETH fails, showing that our algorithms are essentially optimal.