2-subcoloring is NP-complete for planar comparability graphs

TL;DR

This paper proves that the problem of 2-subcoloring remains NP-complete even when restricted to planar comparability graphs with maximum degree 4, extending the known complexity results to this class.

Contribution

It establishes NP-completeness of 2-subcoloring for planar comparability graphs with maximum degree 4, a previously unknown complexity result for this class.

Findings

2-subcoloring is NP-complete for planar comparability graphs with maximum degree 4.

Extends NP-completeness results to a new subclass of graphs.

Shows the problem remains hard under planar and degree constraints.

Abstract

A $k$-subcoloring of a graph is a partition of the vertex set into at most $k$ cluster graphs, that is, graphs with no induced $P_3$. 2-subcoloring is known to be NP-complete for comparability graphs and three subclasses of planar graphs, namely triangle-free planar graphs with maximum degree 4, planar perfect graphs with maximum degree 4, and planar graphs with girth 5. We show that 2-subcoloring is also NP-complete for planar comparability graphs with maximum degree 4.

Full text

1 Introduction

A $k$-subcoloring of a graph is a partition of the vertex set into at most $k$ cluster graphs, that is, graphs with no induced $P_{3}$. Unlike $k$-coloring, $k$-subcoloring is already NP-complete for $k=2$:

Theorem 1**.**

$2$-subcoloring is NP-complete for the following classes:

1. (1)

($K_{4}$, bull, house, butterfly, gem, odd-hole)-free graphs with maximum degree 5 **[1]**, 2. (2)

triangle-free planar graphs with maximum degree 4 **[2, 3]**, 3. (3)

($K_{1,3},K_{4},K_{4}^{-},C_{4}$, odd-hole) planar graphs **[4]**, 4. (4)

planar graphs with girth 5 **[7]**.

A graph $G$ is $(d_{1},\ldots,d_{k})$-colorable if the vertex set of $G$ can be partitioned into subsets $V_{1},\ldots,V_{l}$ such that the graph induced by the vertices of $V_{i}$ has maximum degree at most $d_{i}$ for every $1\leqslant i\leqslant k$. Notice that every $(1,1)$-colorable graph is $2$-subcolorable. Moreover, on triangle-free graphs, $(1,1)$-colorable is equivalent to $2$-subcolorable. As it is well known, for every $a,b\geqslant 0$, every graph with maximum degree $a+b+1$ is $(a,b)$-colorable [6]. Thus, every graph with maximum degree 3 is $2$-subcolorable, so that the degree bound of 4 in Theorems 1.(2) and 1.(3) is best possible.

Notice that the graphs in Theorem 1.(1) are comparability graphs since they are (bull, house, odd-hole)-free [5]. Our main result restricts the class in Theorem 1.(1) to planar graphs and lowers the maximum degree from 5 to 4.

Theorem 2**.**

Let $\mathcal{G}$ denote the class of ($K_{4}$, bull, house, butterfly, gem, odd-hole)-free planar graphs with maximum degree 4. $2$-subcoloring is NP-complete for $\mathcal{G}$.

2 Main result

The reduction is from $2$-subcoloring (or equivalently $(1,1)$-coloring) on triangle-free planar graphs with maximum degree 4, which is NP-complete by Theorem 1.(2). From an instance graph $G$ of this problem, we construct a graph $G^{\prime}$ in $\mathcal{G}$. Every vertex $v$ of $G$ is replaced by a copy $H_{v}$ of the vertex gadget $H$ depicted in Figure 1. For every edge $uv$ of $G$, we use two copies of the edge gadget $E$ depicted in Figure 2 to connect $H_{u}$ and $H_{v}$ as follows:

• •

We identify the vertex $x_{1}$ of the first (resp. second) edge gadget with a vertex $a_{2p}$ (resp. $a_{2p+1}$) of $H_{u}$, with $0\leqslant p\leqslant 3$.

• •

We identify the vertex $x_{2}$ of the first (resp. second) edge gadget with a vertex $a_{i}$ (resp. $a_{j}$) of $H_{v}$ such that $\min\left(i,j\right)\equiv 0\pmod{2}$, $\left|j-i\right|=1$, and no edge crossing is created.

It is easy to check that $G^{\prime}$ can be made planar and with maximum degree 4. Moreover, $G^{\prime}$ is ($K_{4}$, bull, house, gem, butterfly)-free. By removing the vertices whose neighborhood induces a $P_{3}$, we obtain a bipartite graph. This shows that $G^{\prime}$ is odd-hole free. Thus $G^{\prime}$ belongs to $\mathcal{G}$.

Let us show that $G^{\prime}$ is $2$-subcolorable if and only if $G$ is $2$-subcolorable. Given a $2$-subcoloring of a graph, we say that a vertex $p$ is saturated if there exists a monochromatic edge $pq$ and is unsaturated otherwise. We will need the following properties of $E$:

In every $2$-subcoloring of $E\setminus\left\{x_{1},x_{2},y_{1},y_{2}\right\}$, the vertices $z_{1}$ and $z_{2}$ get distinct colors and are saturated. 2. 2.

In every $2$-subcoloring of $E\setminus\left\{x_{1},x_{2}\right\}$, the vertices $y_{1}$ and $y_{2}$ get distinct colors and are not saturated. 3. 3.

There exists a $2$-subcoloring of $E$ such that the vertices $x_{1}$ and $x_{2}$ get distinct colors and are not saturated. 4. 4.

In every $2$-subcoloring of $E$ such that the vertices $x_{1}$ and $x_{2}$ get the same color, exactly one vertex in $\left\{x_{1},x_{2}\right\}$ is saturated.

The six vertices labeled $a_{i}$ in $H$ are called ports. Using properties (1) and (2), we obtain that every $2$-subcoloring of $H$ (in colors red and blue) forces the color of many vertices (see Figure 1). In particular, all the ports in $H$ get the same color. This common color is said to be the color of $H_{v}$ corresponds to the color of $v$ in a $2$-subcoloring of $G$. We also check that in every $2$-subcoloring of $H$, at most one of the ports is not saturated.

Suppose that $uv$ is an edge in $G$. Consider the $2$-subcolorings of the subgraph of $G^{\prime}$ induced by $H_{u}$, $H_{v}$, and the two edge gadgets for the edge $uv$. If distinct colors are given to $H_{u}$ and $H_{v}$, then this $2$-subcoloring can be extended to the edge gadgets using property (3). Since this extension does not saturate any of the considered ports of $H_{u}$ and $H_{v}$, $H_{u}$ can be connected to any number of vertex gadgets with the color distinct from the color of $H_{u}$. If the same color is given to $H_{u}$ and $H_{v}$, then this $2$-subcoloring can be extended using property (4). However, this coloring extension saturates the unique unsaturated port in both $H_{u}$ and $H_{v}$. Thus, $H_{u}$ can be connected to at most one vertex gadget with the same color as $H_{u}$.

This shows that $G^{\prime}$ is $2$-subcolorable if and only if $G$ is $2$-subcolorable.