Generalized domination structure in cubic graphs

TL;DR

This paper introduces a novel polynomial-time method for determining the minimum dominating set in cubic graphs by leveraging structural properties and labeling schemes, improving upon the APX-complete complexity result.

Contribution

It presents a new approach that solves the minimum dominating set problem in cubic graphs efficiently, which was previously known to be APX-complete.

Findings

Polynomial-time algorithm for minimum dominating set in cubic graphs

Structural characterization of dominating sets using labeling schemes

Reduction of problem complexity from APX-complete to polynomial time

Abstract

The minimum dominating set problem asks for a dominating set with minimum size. First, we determine some vertices contained in the minimum dominating set of a graph. By applying a particular scheme, we ensure that the resulting graph is 2-connected and the length of each formed induced cycle is 0 mod 3. We label every three vertices in the induced cycles of length 0 mod 3. Then there is a way of labeling in which the set of all labeled vertices is the minimum dominating set of the resulting graph, and is contained in the minimum dominating set of the original graph. We also consider the remaining vertices of the minimum dominating set of the original graph and determine all vertices contained in the minimum dominating set of a graph with maximum degree 3. The complexity of the minimum dominating set problem for cubic graphs was shown to be APX-complete in 2000 and this problem is solved by our arguments in polynomial time.

Full text

MSC : 05C69

1 Notation

In this paper, a graph $G$ is finite, undirected, and simple with the vertex set $V$ and edge set $E$. We follow [1] for basic notation. For a vertex $v\in V(G)$, the open neighborhood, denoted by $N_{G}(v)$, is $\{u\in V(G)\colon\ uv\in E(G)\}$, and the closed neighborhood, denoted by $N_{G}[v]$, is $N_{G}(v)\cup\{v\}$, also for a set $W\subseteq V(G)$, let $N_{G}(W)=\bigcup_{v\in W}N_{G}(v)$ and $N_{G}[W]=N_{G}(W)\cup W$. A dominating set $X\subseteq V(G)$ is such that $N_{G}[X]=V(G)$. For a set $S\subseteq V(G)$, as is clear from the context, $S$ denotes $G[S]$. A minimum dominating set, called a d-set, is a dominating set of minimum cardinality. Two cycles $C_{1}$ and $C_{2}$ are said to be connecting without seams if $C_{1}\cap C_{2}$ is one path. For a graph $G$, structure $H$ is the union of maximal number of cycles of length 0 mod 3 in $G$ where each cycle of length 0 mod 3 is connecting without seams for some other cycle of length 0 mod 3, or one cycle of length 0 mod 3 in $H$, in addition, if $V(G-H)=\emptyset$, we call this structure domination structure. Let $\mathcal{F}(G)$ be the set of all structures in a graph $G$.

2 Generalized domination structure in cubic graphs

We consider a connected graph $G$, otherwise consider each component one by one. We introduce the construction scheme ${\bf K}$ as follows.

${\bf K}$: Input a connected graph $G$.

(1) Let $G_{0}=G$ and $k=0$.

(2) Let $v$ be a cut vertex of $G_{k}$. For every pair of components $C_{1}$ and $C_{2}$ of $G_{k}-v$ and for every pair of vertices $v_{1}\in C_{1}\cap N_{G_{k}}(v)$ and $v_{2}\in C_{2}\cap N_{G_{k}}(v)$, add an edge $v_{1}v_{2}$. Increment $k$.

(3) Let $D_{2}$ be an induced cycle of length 2 mod 3 in $G_{k}$. Take a vertex $w\in V(D_{2})$, and set $N_{D_{2}}(w)=\{w_{1},w_{2}\}$. Now, add an edge $w_{1}w_{2}$. Set $w_{1}ww_{2}\alpha\subseteq D_{2}$. Now, add two edges $w_{1}\alpha$ and $w\alpha$. Increment $k$.

(4) Let $D_{1}$ be an induced cycle of length 1 mod 3 in $G_{k}$. Take a vertex $x\in V(D_{1})$, and set $N_{D_{1}}(x)=\{x_{1},x_{2}\}$. Now, add an edge $x_{1}x_{2}$. Set $x_{1}xx_{2}\alpha\subseteq D_{1}$. Now, add an edge $x\alpha$. Increment $k$.

(5) Repeat (2)-(4).

(6) Return the resulting graph $G_{k}$.

The next remark is a basic concept of the following proofs.

Remark 2.1**.**

Let $X$ be a dominating set of $G$. Every subset $D\subseteq X$ is a d-set of $N_{G}[D]$ if and only if $X$ is a d-set of $G$.

Let ${\bf K}(G)$ be a graph constructed by applying ${\bf K}$ to $G$. Note that ${\bf K}(G)$ is not unique and constructed from $G$ arbitrarily.

Proposition 2.1**.**

${\bf K}(G)$* is domination structure. Moreover, $|\mathcal{F}({\bf K}(G))|=1$.*

Proof.

From the rules, ${\bf K}(G)$ is 2-connected. Hence ${\bf K}(G)$ has an ear decomposition. Since all induced cycles are of length 0 mod 3, the domination structure $H$ is obtained by finding 0 mod 3 induced cycles connecting without seams one by one, so that $H={\bf K}(G)$. We had the claim. ∎

Suppose that for all $v\in V(G)$, $d_{G}(v)\geq 3$.

Fact 2.1**.**

For domination structure $H={\bf K}(G)$, label every three vertices on the induced cycles that constitute $H$ in order of connecting without seams. Note that a certain induced cycle of $H$ is not counted for the labeling, and may have no labels, where the vertices of the induced cycle are all in other induced cycles of $H$. There exist at most $|V(G)|$ cases of labeling. (i) For every labeling, the set of all labeled vertices is a dominating set of ${\bf K}(G)$. (ii) For at least one labeling, the set of all labeled vertices is a d-set of ${\bf K}(G)$.

Proof.

By the first vertex choice for the labeling, all labeled vertices are uniquely determined in $V(G)$ since for all $v\in V(G)$, $d_{G}(v)\geq 3$, and so there exist at most $|V(G)|$ cases of labeling. The statement (i) is obvious. The statement (ii) follows from Remark 2.1. ∎

Let $Y$ be a d-set of ${\bf K}(G)$ that is obtained by applying Fact 2.1. Let $\mathcal{Y}$ be the set of all $Y$. Let $X$ be a d-set of $G$. Let $\mathcal{X}$ be the set of all $X$.

Proposition 2.2**.**

For some $X\in\mathcal{X}$, and some $Y\in\mathcal{Y}$, $Y\subseteq X$.

Proof.

For some $Y\in\mathcal{Y}$, if $Y$ is a dominating set of $G$, then for some $X\in\mathcal{X}$, $X=Y$. Otherwise, for all $Y\in\mathcal{Y}$, $Y$ is not a dominating set of $G$. Now, for some $v\in Y$, and some $w\in V(G)\setminus Y$, $vw$ is an added edge for ${\bf K}(G)$ and $N_{G}(w)\subseteq V(G)\setminus Y$. Now, we consider such $v$ and $w$. Let $Z^{\prime}=\{w\in V(G)\setminus Y\colon\ v\in Y,vw\in E({\bf K}(G))\setminus E(G),N_{G}(w)\subseteq V(G)\setminus Y\}$ and $E^{\prime}=\{vw\in E({\bf K}(G))\setminus E(G)\colon\ w\in V(G)\setminus Y,v\in Y,N_{G}(w)\subseteq V(G)\setminus Y\}$. Let $E^{\prime\prime}(vw)$ be an induced cycle of length 0 mod 3 in ${\bf K}(G)$ such that for $vw\in E^{\prime}$, $vw\in E^{\prime\prime}(vw)$ holds. Let $\mathcal{E}(vw)$ be the set of all $E^{\prime\prime}(vw)$ for $vw\in E^{\prime}$ and let $\mathcal{E}=\bigcup_{vw\in E^{\prime}}\mathcal{E}(vw)$. Let $J$ be the union of all induced cycles of length 0 mod 3 in ${\bf K}(G)$ other than the induced cycles in $\mathcal{E}$. By the definition of $Y$ and Remark 2.1, $J\cap Y$ is a d-set of $N_{G}[J\cap Y]$, and $Y$ is a d-set of $G-Z^{\prime}=N_{G}[Y]$. Let $W$ be a subset of $V(G)\setminus Y$ of minimum cardinality such that $Y\cup W$ is a dominating set of $G$. By the definition of $W$, $W$ is a d-set of $N_{G}[W]$. Since for all $x\in V(G)$, $d_{G}(x)\geq 3$, and by Fact 2.1, $E(G[Y])=\emptyset$, and $Y\cup W$ is a minimal dominating set of $G$. Therefore, by Remark 2.1, for some $X\in\mathcal{X}$, $X=Y\cup W$. ∎

Suppose that $Y$ is a d-set of ${\bf K}(G)$ that satisfies Proposition 2.2.

Fact 2.2**.**

Let $G^{\prime}$ be constructed by deleting $Y$, and for every pair $w_{1},w_{2}\in\bigcup_{y\in Y}N_{G}(y)$, adding an edge $w_{1}w_{2}$ to $G$. Let $Z_{1}$ be a d-set of $G^{\prime}$. Let $G^{\prime\prime}=G-N_{G}[Y]$ and $Z_{2}$ be a d-set of $G^{\prime\prime}$. If $|Z_{1}|<|Z_{2}|$, then $Y\cup Z_{1}$ is a d-set of $G$, and $Z_{1}\cap\bigcup_{y\in Y}N_{G}(y)\neq\emptyset$. If $|Z_{1}|\geq|Z_{2}|$, then $Y\cup Z_{2}$ is a d-set of $G$.

Proof.

Obviously, $Y\cup Z_{2}$ is a d-set of $G$ if and only if $|Z_{1}|\geq|Z_{2}|$. For a set $A\subseteq V(G)$, suppose that $Y\cup A$ is a d-set of $G$. Let $A^{\prime}=A\setminus\bigcup_{y\in Y}N_{G}(y)$. By Remark 2.1, $A^{\prime}$ is a d-set of $N_{G}[A^{\prime}]$. Suppose that $|Z_{1}|<|Z_{2}|$, then $A^{\prime}$ is not a dominating set of $G^{\prime\prime}$, and $A\cap\bigcup_{y\in Y}N_{G}(y)\neq\emptyset$. Now, $A$ is a minimal dominating set of $G^{\prime}$. By the definition of $G^{\prime}$, $Y\cup Z_{1}$ is a minimal dominating set of $G$, and so it suffices that $A=Z_{1}$. Thus $Z_{1}\cap\bigcup_{y\in Y}N_{G}(y)\neq\emptyset$. ∎

Suppose that $G$ is cubic.

Theorem 2.1**.**

For some $X\in\mathcal{X}$, $X$ is determined in polynomial time.

Proof.

By Proposition 2.2, $Y\subseteq X$ for some $X\in\mathcal{X}$. Let $G_{0}=G$. Let $G_{1}$ be constructed by deleting $Y$, and for every pair $w_{1},w_{2}\in\bigcup_{y\in Y}N_{G}(y)$, adding an edge $w_{1}w_{2}$ to $G_{0}$. Let $G_{2}=G_{0}-N_{G_{0}}[Y]$. Let $W_{1}$ be a d-set of $G_{1}$ and $W_{2}$ be a d-set of $G_{2}$. Since $G_{0}$ is cubic and by the definition of $G_{2}$, each component of $G_{2}$ is path or cycle. Thus $W_{2}$ is determined in polynomial time. Suppose that $|W_{1}|<|W_{2}|$. Let $Y_{1}$ be a d-set of ${\bf K}(G_{1})$ that satisfies Proposition 2.2. By the definition of $Y_{1}$, it suffices that $Y_{1}\cap\bigcup_{y\in Y}N_{G_{0}}(y)\neq\emptyset$. Let $G_{3}$ be constructed by deleting $Y_{1}$, and for every pair $w_{1},w_{2}\in\bigcup_{y\in Y_{1}}N_{G_{1}}(y)$, adding an edge $w_{1}w_{2}$ to $G_{1}$. Let $G_{4}=G_{1}-N_{G_{1}}[Y_{1}]$. Let $W_{3}$ be a d-set of $G_{3}$ and $W_{4}$ be a d-set of $G_{4}$. By the definition of $G_{4}$, each component of $G_{4}$ is path. Thus $W_{4}$ is determined in polynomial time. Suppose that $|W_{3}|<|W_{4}|$. Let $Y_{2}$ be a d-set of ${\bf K}(G_{3})$ that satisfies Proposition 2.2. By the definition of $Y_{2}$, it suffices that $Y_{2}\cap\bigcup_{y\in Y_{1}}N_{G_{1}}(y)\neq\emptyset$. Let $G_{5}$ be constructed by deleting $Y_{2}$, and for every pair $w_{1},w_{2}\in\bigcup_{y\in Y_{2}}N_{G_{3}}(y)$, adding an edge $w_{1}w_{2}$ to $G_{3}$. Let $G_{6}=G_{3}-N_{G_{3}}[Y_{2}]$. Let $W_{5}$ be a d-set of $G_{5}$ and $W_{6}$ be a d-set of $G_{6}$. By the definition of $G_{6}$, $G_{6}$ is independent. Thus $W_{6}=V(G_{6})$. Suppose that $|W_{5}|<|W_{6}|$. Let $Y_{3}$ be a d-set of ${\bf K}(G_{5})$ that satisfies Proposition 2.2. By the definition of $Y_{3}$, it suffices that $Y_{3}\cap\bigcup_{y\in Y_{2}}N_{G_{3}}(y)\neq\emptyset$. Now, $Y_{3}$ is a dominating set of $G_{5}$ and so it suffices that $W_{5}=Y_{3}$. By Fact 2.2, if $|W_{5}|<|W_{6}|$, then it suffices that $W_{3}=Y_{2}\cup W_{5}$, otherwise, it suffices that $W_{3}=Y_{2}\cup W_{6}$. By Fact 2.2, if $|W_{3}|<|W_{4}|$, then it suffices that $W_{1}=Y_{1}\cup W_{3}$, otherwise, it suffices that $W_{1}=Y_{1}\cup W_{4}$. By Fact 2.2, if $|W_{1}|<|W_{2}|$, then it suffices that $X=Y\cup W_{1}$, otherwise, it suffices that $X=Y\cup W_{2}$. The proof is complete. ∎