On the hardness of deciding the equality of the induced and the uniquely restricted matching number

TL;DR

This paper proves that deciding whether the maximum induced matching equals the maximum uniquely restricted matching in a graph is NP-hard, indicating the problem's computational difficulty and the unlikelihood of a simple characterization.

Contribution

It establishes the NP-hardness of the decision problem for equality of induced and uniquely restricted matching numbers in graphs.

Findings

Deciding equality of the two matching numbers is NP-hard.

The result suggests a likely absence of a simple characterization.

The problem's complexity impacts related graph matching research.

Abstract

If $G(M)$ denotes the subgraph of a graph $G$ induced by the set of vertices that are covered by some matching $M$ in $G$, then $M$ is an induced or a uniquely restricted matching if $G(M)$ is $1$-regular or if $M$ is the unique perfect matching of $G(M)$, respectively. Let $\nu_s(G)$ and $\nu_{ur}(G)$ denote the maximum cardinality of an induced and a uniquely restricted matching in $G$. Golumbic, Hirst, and Lewenstein (Uniquely restricted matchings, Algorithmica 31 (2001) 139-154) posed the problem to characterize the graphs $G$ with $\nu_{ur}(G) = \nu_{s}(G)$. We prove that the corresponding decision problem is NP-hard, which suggests that a good characterization is unlikely to be possible.

Equations

$$\nu (G)\ge {\nu }_{ur}(G)\ge {\nu }_{ac}(G)\ge {\nu }_s(G).$$

$$E_G(X,Y)=\{uv\in E(G):u\in X,v\in Y\},$$

$${\nu }_{ac}(G)$$

$$\le n-\frac{|M_3|}{2}+m-\frac{|M_3|}{2}+|M_3|$$

$$=n+m,$$

Full text

Institute of Optimization and Operations Research, Ulm University, Germany

maximilian.fuerst@uni-ulm.de

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1 Introduction

We consider only simple, finite, and undirected graphs, and use standard terminology. For a graph $G$, and a matching $M$ in $G$, let $V(M)$ be the set of vertices that are covered by $M$, and let $G(M)$ be the subgraph of $G$ induced by $V(M)$. A matching $M$ in $G$ is

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induced [1] if $G(M)$ is $1$-regular,

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acyclic [7] if $G(M)$ is a forest, or

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uniquely restricted [8] if $M$ is the unique perfect matching of $G(M)$.

The maximum cardinality of an ordinary, a uniquely restricted, an acyclic, and an induced matching is denoted by $\nu(G)$, $\nu_{ur}(G)$, $\nu_{ac}(G)$, and $\nu_{s}(G)$, respectively. While the ordinary matching number is tractable [3], the three remaining restricted matching numbers are NP-hard [7, 8, 14]. Golumbic et al. [8] observed that a matching $M$ in $G$ is uniquely restricted if and only if $G$ contains no $M$-alternating cycle, which implies that $\nu_{ur}(G)\geq\nu_{ac}(G)$. Since every induced matching is also acyclic, we obtain that

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In order to understand how those different restricted matching numbers relate to each other, it seems to be interesting to characterize the graphs achieving equality in one or more of the inequalities in (1). On the positive side, deciding whether a given graph $G$ satisfies $\nu(G)=\nu_{s}(G)$ or $\nu(G)=\nu_{ur}(G)$, and deciding whether a given subcubic graph $G$ satisfies $\nu_{ur}(G)=\nu_{s}(G)$ is tractable [2, 9, 10, 11, 12, 5]. On the negative side, the hardness of deciding $\nu(G)=\nu_{ac}(G)$ and $\nu_{ur}(G)=\nu_{ac}(G)$ was shown in [4]. In 2001, Golumbic et al. [8] posed the problem to characterize the graphs $G$ with $\nu_{ur}(G)=\nu_{s}(G)$. In this short note, we will prove the hardness of deciding $\nu_{ur}(G)=\nu_{ac}(G)$ and $\nu_{ur}(G)=\nu_{s}(G)$. This shows that a good characterization does not exist unless NP $=$ co-NP. Note that it is not obvious whether these two decision problems belong to NP.

Theorem 1**.**

Deciding whether a given graph $G$ of maximum degree $4$ satisfies $\nu_{ac}(G)=\nu_{s}(G)$ is NP-hard.

Theorem 2**.**

Deciding whether a given bipartite graph $G$ satisfies $\nu_{ur}(G)=\nu_{s}(G)$ is NP-hard.

The proofs of Theorem 1 and 2 are postponed to the following sections. We close the introduction with a few notations. For a graph $G$ and two disjoint sets $X$ and $Y$ of vertices of $G$, let

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and let $E_{G}(X)=E(G[X])$. For every positive integer $k$, let $[k]=\{1,\ldots,k\}$.

2 Proof of Theorem 1

We prove the statement by a reduction from Satisfiability that remains NP-complete (cf. e.g. [6]) for instances where every clause contains two or three literals, every positive literal appears in at most two different clauses, every negative literal appears in at most one clause, and no clause contains a literal and its negation. Let $\Gamma$ be such an instance of Satisfiability with variables $x_{1},\ldots,x_{n}$ and clauses $c_{1},\ldots,c_{m}$. For every $j\in[m]$, let $|c_{j}|$ be the number of literals that belong to the clause $c_{j}$.

Let $G$ be a graph that arises from the union of $n$ disjoint triangles with vertex sets $X_{1},\ldots,X_{n}$, and $m$ disjoint cliques with vertex sets $C_{1},\ldots,C_{m}$, where $|C_{j}|=|c_{j}|+1$ for every $j\in[m]$. For every $i\in[n]$, let $X_{i}=\{t_{i},f_{i},u_{i}\}$. For every $j\in[m]$, identify $|c_{j}|$ vertices in $C_{j}$ with the literals in the clause $c_{j}$, and let $v_{j}$ be the vertex in $C_{j}$ that is not identified with a literal from the clause $c_{j}$. Let $i$ be in $[n]$, let $w_{1}$ and $w_{2}$ be the vertices in $\bigcup_{j=1}^{m}C_{j}$ that are identified with the literal $x_{i}$, and let $w_{3}$ be the vertex in $\bigcup_{j=1}^{m}C_{j}$ that is identified with the literal $\bar{x}_{i}$. Now, add the edges $f_{i}w_{1}$, $f_{i}w_{2}$, and $t_{i}w_{3}$, see Figure 1 for an illustration.

Lemma 2.1**.**

$\nu_{ac}(G)=n+m$.

Proof.

Let $M$ arise from $\{u_{i}t_{i}:i\in[n]\}$ by adding, for every $j\in[m]$, an arbitrary edge of $E_{G}(C_{j})$. Suppose, for a contradiction, that $M$ is not an acyclic matching in $G$. Since $d_{G(M)}(u_{i})=1$ and $d_{G(M)}(t_{i})\leq 2$, neither $u_{i}$ nor $t_{i}$ are contained in any cycle of $G(M)$ for every $i\in[n]$. By construction, this implies a contradiction. Hence, $\nu_{ac}(G)\geq n+m$.

Let $M=M_{1}\cup M_{2}\cup M_{3}$ be some maximum acyclic matching with

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$M_{1}\subseteq\bigcup_{i=1}^{n}E_{G}(X_{i})$,

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$M_{2}\subseteq\bigcup_{j=1}^{m}E_{G}(C_{j})$, and

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M_{3}\subseteq E_{G}\big{(}\bigcup_{i=1}^{n}{X_{i}},V(G)\setminus\bigcup_{i=1}^{n}{X_{i}}\big{)}

minimizing $|M_{3}|$. Let $j\in[m]$. If $E_{G}(C_{j},V(G)\setminus C_{j})\cap M_{3}\neq\emptyset$, then $E_{G}(C_{j})\cap M_{2}=\emptyset$. Since $M$ is acyclic, we have that $|E_{G}(C_{j},V(G)\setminus C_{j})\cap M_{3}|\leq 2$, and, if $E_{G}(C_{j},V(G)\setminus C_{j})\cap M_{3}$ contains exactly one edge $uv$ with $u\in C_{j}$, then the matching $(M\cup\{v_{j}u\})\setminus\{uv\}$ is acyclic, which is a contradiction to the minimality of $|M_{3}|$. Therefore, $|E_{G}(C_{j},V(G)\setminus C_{j})\cap M_{3}|\in\{0,2\}$. Let $i\in[n]$. If $E_{G}(X_{i},V(G)\setminus X_{i})\cap M_{3}\neq\emptyset$, then $E_{G}(X_{i})\cap M_{1}=\emptyset$. If $E_{G}(X_{i},V(G)\setminus X_{i})\cap M_{3}$ contains exactly one edge $uv$ with $u\in X_{i}$, then the matching $(M\cup\{u_{i}u\})\setminus\{uv\}$ is acyclic, which is a contradiction to the minimality of $|M_{3}|$. Therefore, $|E_{G}(X_{i},V(G)\setminus X_{i})\cap M_{3}|\in\{0,2\}$. Hence, we obtain that

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which completes the proof. ∎

Lemma 2.2**.**

$\Gamma$* is satisfiable if and only if $\nu_{ac}(G)=\nu_{s}(G)$.*

Proof.

Let $\Gamma$ be satisfiable, and let $t:\{x_{1},\ldots,x_{n}\}\rightarrow\{0,1\}$ be some satisfying truth assignment of $\Gamma$. Let $M$ arise from $\{u_{i}f_{i}:i\in[n]\text{ and }t(x_{i})=0\}\cup\{u_{i}t_{i}:i\in[n]\text{ and }t(x_{i})=1\}$ by adding, for every $j\in[m]$, some edge $v_{j}w$ where the literal that is identified with the vertex $w$ is true under $t$. Suppose that $M$ is not an induced matching, that is, there is some edge $e$ between two edges $e_{1}$ and $e_{2}$ in $M$. By construction, we may assume that $e_{1}=u_{i}y_{i}$ for $y_{i}\in\{f_{i},t_{i}\}$ and some $i\in[n]$, and $e_{2}=v_{j}w$ for some $j\in[m]$ where the literal that is identified with the vertex $w$ is true under $t$. Thus, $y_{i}$ and $w$ are adjacent. First, we assume that $y_{i}=f_{i}$. By construction, the literal that is identified with the vertex $w$ is $x_{i}$, which is a contradiction to the choice of $M$. Hence, we may assume that $y_{i}=t_{i}$. By construction, the literal that is identified with the vertex $w$ is $\bar{x}_{i}$, which is a contradiction to the choice of $M$. Since $M$ has size $n+m$, Lemma 2.1 implies that $\nu_{ac}(G)=\nu_{s}(G)$.

Let $\nu_{ac}(G)=\nu_{s}(G)$, which, by Lemma 2.1, implies that $\nu_{s}(G)=n+m$. Let $M$ be some maximum induced matching maximizing $|\{u_{i}:i\in[n]\}\cap V(M)|$. If there is some edge $uv$ with $u\in X_{i}$ and $v\in C_{j}$ for some $i\in[n]$ and $j\in[m]$, then the matching $M=(M\cup\{u_{i}u\})\setminus\{uv\}$ is induced, which is a contradiction to the maximality of $|\{u_{i}:i\in[n]\}\cap V(M)|$. Therefore, E_{G}\big{(}\bigcup_{i=1}^{n}{X_{i}},V(G)\setminus\bigcup_{i=1}^{n}{X_{i}}\big{)}=\emptyset. Moreover, if there is some edge $f_{i}t_{i}$ in $M$ for some $i\in[n]$, then the matching $(M\cup\{u_{i}f_{i}\})\setminus\{f_{i}t_{i}\}$ is also induced, which is a contradiction to the maximality of $|\{u_{i}:i\in[n]\}\cap V(M)|$. Since $\nu_{s}(G)=n+m$, this implies that either $u_{i}f_{i}$ or $u_{i}t_{i}$ belong to $M$ for every $i\in[n]$, and $E_{G}(C_{j})\cap M\neq\emptyset$ for every $j\in[m]$. Let $t:\{x_{1},\ldots,x_{n}\}\rightarrow\{0,1\}$ be defined as $t(x_{i})=0$ if $u_{i}f_{i}\in M$ and $t(x_{i})=1$ if $u_{i}t_{i}\in M$. Suppose, for a contradiction, that $\Gamma$ is not satisfied under $t$, that is, there is some clause $c_{j}$ such that no literal is true under $t$. By construction, there is some vertex $w\in V(M)\cap C_{j}$ where its corresponding literal $y$ is not true under $t$. By construction, $w$ is adjacent to $f_{i}$ or $t_{i}$ for some $i\in[n]$. If $w$ is adjacent to $f_{i}$, then $y=x_{i}$, and, since $M$ is induced, the edge $u_{i}t_{i}$ is in $M$, which implies that $t(x_{i})=1$, a contradiction. Hence, we may assume that $w$ is adjacent to $t_{i}$. This implies that $y=\bar{x}_{i}$ and that the edge $u_{i}f_{i}$ is in $M$, that is, $t(\bar{x}_{i})=1$, a contradiction. ∎

3 Proof of Theorem 2

Given a boolean formula in conjunctive normal form, Exact Satisfiability is the problem is to decide whether there is a truth assignment of the variables so that every clause contains exactly one true literal. If there is such a truth assignment, then the instance is exact satisfiable.

Lemma 3.1**.**

Exact Satisfiability* remains NP-complete when restricted to instances where the literals occur only positively, every literal occurs at most three times, every clause has size exactly three, and no literal appears twice in one clause.*

Proof.

It was proved recently [13] that Exact Satisfiability remains NP-complete when restricted to instances where the literals occur only positively, every literal occurs exactly three times, and every clause has size exactly three. Let $\Gamma$ be such an instance of Exact Satisfiability.

Suppose that some literal $x$ appears twice in some clause $c$ of $\Gamma$. This implies that $x$ must be false and that the other literal in $c$ must be true. Therefore, some variables already have a unique truth value, which might result in a contradiction in which case $\Gamma$ is not exact satisfiable. If not, then we delete all variables with a unique truth value. If we apply this process iteratively, then we obtain an equivalent instance $\Gamma^{\prime}$ of Exact Satisfiability where the literals occur only positively, every literal appears exactly three times, every clause has size two or three, and no literal appears twice in one clause.

We construct a new instance that is equivalent to $\Gamma^{\prime}$ where the literals occur only positively, every literal appears at most three times, every clause has size exactly three, and no literal appears twice in one clause. Suppose that there is a clause of size two with literals $x$ and $y$ in $\Gamma^{\prime}$. Since no literal appears twice in one clause, $x\neq y$. We delete the clause $x\lor y$ and we add four new clauses $x\lor y\lor a_{1}$, $a_{1}\lor a_{2}\lor a_{3}$, $a_{1}\lor a_{2}\lor a_{4}$, and $a_{2}\lor a_{3}\lor a_{4}$. It is easy to see that this instance of Exact Satisfiability satisfies all desired constraints. Since the only possible solution of the above four clauses is obtained by assigning $x$ or $y$ to $1$, $a_{2}$ to $1$, and $a_{1}$, $a_{3}$, and $a_{4}$ to [math], the newly constructed instance of Exact Satisfiability is equivalent to $\Gamma^{\prime}$. ∎

Let $\Gamma$ be an instance of Exact Satisfiability as in Lemma 3.1 with variables $x_{1},\ldots,x_{n}$ and clauses $c_{1},\ldots,c_{m}$. Let $G$ be the graph that arises from the union of $n$ disjoint copies of a $K_{1,2}$ with vertex sets $X_{1},\ldots,X_{n}$, and $m$ disjoint copies of a $K_{1,3}$ with vertex sets $C_{1},\ldots,C_{m}$. For every $i\in[n]$, let $f_{i}$ and $t_{i}$ be the leaves of $X_{i}$, and let $u_{i}$ be the vertex of degree two in $X_{i}$. For every $j\in[m]$, identify the three leaves of $C_{j}$ with the literals that belong to the clause $c_{j}$, and let $v_{j}$ be the vertex of degree three in $C_{j}$. Let $i$ be in $[n]$, let $W$ be the set of vertices in $\bigcup_{j=1}^{m}C_{j}$ that are identified with the literal $x_{i}$, and let $J=\{j\in[m]:W\cap C_{j}\neq\emptyset\}$. Furthermore, let W^{\prime}=\big{(}\bigcup_{j\in J}{C_{j}\setminus\{v_{j}\}}\big{)}\setminus W. Now, add the edges $f_{i}w$ for every $w$ in $W$, and the edges $t_{i}w^{\prime}$ for every $w^{\prime}$ in $W^{\prime}$, see Figure 2 for an illustration.

Golumbic et al. [8] showed that, if $G$ is a bipartite graph and $M$ is some uniquely restricted matching in $G$, then $G(M)$ has a vertex of degree $1$ in $G(M)$. We shall use this within the proof of the following lemma.

Lemma 3.2**.**

$\nu_{ur}(G)=n+m$.

Proof.

The matching $M$ that arises from $\{u_{i}t_{i}:i\in[n]\}$ by adding, for each $j\in[m]$, an edge between $v_{j}$ and one of its neighbors is uniquely restricted, because all edges in $M$ are pendant edges in $G(M)$. Suppose, for a contradiction, that $\nu_{ur}(G)>n+m$, and let $M=M_{1}\cup M_{2}\cup M_{3}$ be some maximum uniquely restricted matching with

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$M_{1}\subseteq\bigcup_{i=1}^{n}E_{G}(X_{i})$,

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$M_{2}\subseteq\bigcup_{j=1}^{m}E_{G}(C_{j})$, and

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M_{3}\subseteq E_{G}\big{(}\bigcup_{i=1}^{n}X_{i},V(G)\setminus\bigcup_{i=1}^{n}X_{i}\big{)}

minimizing $|M_{3}|$. Since $|M|>n+m$, it follows that $M_{3}\neq\emptyset$.

Claim 1**.**

If $i\in[n]$ is such that $M_{3}\cap E_{G}(X_{i},V(G)\setminus X_{i})\neq\emptyset$, then $M_{1}\cap E_{G}(X_{i})=\emptyset$.

Proof.

Suppose, for a contradiction, that $M_{1}\cap E_{G}(X_{i})\neq\emptyset$. Let $j\in[m]$ be such that $M_{3}\cap E_{G}(X_{i},C_{j})\neq\emptyset$, and let $w_{1},w_{2}$, and $w_{3}$ be the vertices in $C_{j}$ distinct from $v_{j}$ so that $w_{1}$ is adjacent to $f_{i}$, and $w_{2}$ and $w_{3}$ are both adjacent to $t_{i}$.

First, we assume that $f_{i}w_{1}$ and $u_{i}t_{i}$ belong to $M$. Since $v_{j}$ is not covered by $M$, the set $M^{\prime}=\left(M\cup\{v_{j}w_{1}\}\right)\setminus\{f_{i}w_{1}\}$ is a matching in $G$, which, by the minimality of $|M_{3}|$, is not uniquely restricted. Since $f_{i}$ is not covered by $M^{\prime}$, there is an $M^{\prime}$-alternating cycle in $G$ disjoint from $X_{i}$, which, by symmetry, can be written as $v_{j}w_{1}Pw_{2}v_{j}$ for some $M^{\prime}$-alternating path $P$ in $G$. Since $P$ is also $M$-alternating, the cycle $w_{1}Pw_{2}t_{i}u_{i}f_{i}w_{1}$ is $M$-alternating in $G$, which is a contradiction.

Hence, by symmetry, we may assume that $t_{i}w_{2}$ and $u_{i}f_{i}$ belong to $M$. Since $v_{j}$ is not covered by $M$, the set $M^{\prime}=\left(M\cup\{v_{j}w_{2}\}\right)\setminus\{t_{i}w_{2}\}$ is a matching in $G$, which, by the minimality of $|M_{3}|$, is not uniquely restricted. Therefore, there is an $M^{\prime}$-alternating cycle $C$ that contains the edge $v_{j}w_{2}$. Since $t_{i}$ is not covered by $M^{\prime}$, we have that $X_{i}\cap V(C)=\emptyset$. If $C$ can be written as $w_{2}v_{j}w_{3}Pw_{2}$ for some $M^{\prime}$-alternating path $P$, then the cycle $w_{2}t_{i}w_{3}Pw_{2}$ is $M$-alternating, which is a contradiction. Hence, we may assume that $C$ can be written as $w_{2}v_{j}w_{1}Pw_{2}$ for some $M^{\prime}$-alternating path $P$, which implies that the cycle $w_{2}t_{i}u_{i}f_{i}w_{1}Pw_{2}$ is $M$-alternating in $G$, which is a contradiction. ∎

Claim 2**.**

If $j\in[m]$ is such that $M_{3}\cap E_{G}(C_{j},V(G)\setminus C_{j})\neq\emptyset$, then $M_{2}\cap E_{G}(C_{j})=\emptyset$.

Proof.

Suppose, for a contradiction, that $M_{2}\cap E_{G}(C_{j})\neq\emptyset$. Let $i\in[n]$ be such that $M_{3}\cap E_{G}(X_{i},C_{j})\neq\emptyset$, and let $w_{1},w_{2}$, and $w_{3}$ be the vertices in $C_{j}$ distinct from $v_{j}$ such that $w_{1}$ is adjacent to $f_{i}$, and $w_{2}$ and $w_{3}$ are both adjacent to $t_{i}$.

First, we assume that $f_{i}w_{1}$ and $v_{j}w_{3}$ belong to $M$. Since $u_{i}$ is not covered by $M$, the set $M^{\prime}=\left(M\cup\{u_{i}f_{i}\}\right)\setminus\{f_{i}w_{1}\}$ is a matching, which, as before, implies that there is an $M^{\prime}$-alternating cycle $C$ that contains the edge $u_{i}f_{i}$. If $v_{j}w_{3}$ is not contained in $E(C)$, then the vertices $w_{1}$, $v_{j}$, and $w_{3}$ are not contained in $V(C)$, which implies that $C$ can be written as $f_{i}u_{i}t_{i}Pf_{i}$ for some $M^{\prime}$-alternating path $P$ in $G$. Since $P$ is also $M$-alternating, the cycle $f_{i}w_{1}v_{j}w_{3}t_{i}Pf_{i}$ is $M$-alternating, which is a contradiction. Hence, we may assume that $v_{j}w_{3}$ is contained in $E(C)$. In this case, the cycle can either be written as $f_{i}Pw_{2}v_{j}w_{3}P^{\prime}t_{i}u_{i}f_{i}$ or as $f_{i}Qw_{3}v_{j}w_{2}Q^{\prime}t_{i}u_{i}f_{i}$. In the first case, the cycle $f_{i}w_{1}v_{j}w_{3}P^{\prime}t_{i}w_{2}Pf_{i}$ is $M$-alternating in $G$, while in the second case the cycle $w_{2}Q^{\prime}t_{i}w_{2}$ is $M$-alternating, which, in both cases, is a contradiction.

Hence, by symmetry, we may assume that $t_{i}w_{2}$ and $v_{j}w_{1}$ belong to $M$. Since $u_{i}$ is not covered by $M$, the set $M^{\prime}=\left(M\cup\{u_{i}t_{i},v_{j}w_{2}\}\right)\setminus\{t_{i}w_{2},v_{j}w_{1}\}$ is a matching, which, as before, implies that there is an $M^{\prime}$-alternating cycle $C$. If $C$ contains exactly one of the edges $u_{i}t_{i}$ or $v_{j}w_{2}$, then $C$ can be written as $f_{i}Pt_{i}u_{i}f_{i}$ or as $v_{j}w_{3}Qw_{2}v_{j}$. In the first case, the cycle $f_{i}Pt_{i}w_{2}v_{j}w_{1}f_{i}$ is $M$-alternating, while in the second case the cycle $w_{3}Qw_{2}t_{i}w_{3}$ is $M$-alternating, which, in both cases, is a contradiction. Hence, we may assume that both $u_{i}t_{i}$ and $v_{j}w_{2}$ are contained in $E(C)$. In this case, the cycle can either be written as $f_{i}Pw_{2}v_{j}w_{3}P^{\prime}t_{i}u_{i}f_{i}$ or as $f_{i}Qw_{3}v_{j}w_{2}t_{i}u_{i}f_{i}$. In the first case, the cycle $w_{1}f_{i}Pw_{2}t_{i}P^{\prime}w_{3}v_{j}w_{1}$ is $M$-alternating in $G$, while in the second case the cycle $f_{i}Qw_{3}v_{j}w_{1}f_{i}$ is $M$-alternating in $G$, which, in both cases, is a contradiction. ∎

By Claim 1 and 2, all edges in $M_{1}\cup M_{2}$ are pendant edges in $G(M)$. Let $H=G(M_{3})$. Since $H$ is bipartite, there is a vertex $u$ in $H$ of degree $1$. Let $uv\in M_{3}$. First, we assume that $u\in X_{i}$ for some $i\in[n]$. By Claim 1, the vertex $u_{i}$ is not covered by $M$. Hence, the set $M^{\prime}=(M\cup\{u_{i}u\})\setminus\{uv\}$ is a matching. Since $u$ is only adjacent to vertices in $G(M)$ that are covered by edges in $M_{2}$, the matching $M^{\prime}$ is uniquely restricted, which is a contradiction to the minimality of $|M_{3}|$. Hence, we may assume that $u\in C_{j}$ for some $j\in[m]$. By Claim 2, the vertex $v_{j}$ is not covered by $M$. Hence, the set $M^{\prime}=(M\cup\{v_{j}u\})\setminus\{uv\}$ is a matching. Since $u$ is only adjacent to vertices in $G(M)$ that are covered by edges in $M_{1}$, the matching $M^{\prime}$ is uniquely restricted, which is a contradiction to the minimality of $|M_{3}|$. ∎

Lemma 3.3**.**

$\Gamma$* is exact satisfiable if and only if $\nu_{ur}(G)=\nu_{s}(G)$.*

Proof.

Let $\Gamma$ be exact satisfiable, and let $t:\{x_{1},\ldots,x_{n}\}\rightarrow\{0,1\}$ be some satisfying truth assignment of $\Gamma$ such that each clause contains exactly one literal that is true under $t$. Let $M$ arise from $\{u_{i}f_{i}:i\in[n]\text{ and }t(x_{i})=0\}\cup\{u_{i}t_{i}:i\in[n]\text{ and }t(x_{i})=1\}$ by adding, for each $j\in[m]$, the edge $v_{j}v$ where $v$ is identified with a literal in $c_{j}$ that is true under $t$. Suppose, for a contradiction, that $M$ is not an induced matching, that is, there is some edge $e$ between two edges $e_{1}$ and $e_{2}$ in $M$. By construction, we may assume that $e_{1}=u_{i}y_{i}$ for $y_{i}\in\{f_{i},t_{i}\}$ and $i\in[n]$, and $e_{2}=v_{j}z_{j}$, where $z_{j}$ is the vertex in $C_{j}$ that is identified with the unique literal in $c_{j}$ that is true under $t$, for some $j\in[m]$. Therefore, $y_{i}$ and $z_{j}$ are adjacent. If $y_{i}=f_{i}$, then, by construction, $z_{j}$ is identified with the literal $x_{i}$, which is a contradiction to the choice of $M$. Hence, we may assume that $y_{i}=t_{i}$, which implies that $t(x_{i})=1$. Furthermore, by construction, $z_{j}$ is identified with a literal from the clause $c_{j}$ distinct from $x_{i}$, which, by the choice of $M$, is also true under $t$, a contradiction. Hence, $M$ is an induced matching of size $n+m$, which, by Lemma 3.2, implies that $\nu_{ur}(G)=\nu_{s}(G)$.

Let $\nu_{ur}(G)=\nu_{s}(G)$, which, by Lemma 3.2, implies that $\nu_{s}(G)=n+m$. Let $M=M_{1}\cup M_{2}\cup M_{3}$ be some maximum induced matching in $G$ with

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$M_{1}\subseteq\bigcup_{i=1}^{n}E_{G}(X_{i})$,

• •

$M_{2}\subseteq\bigcup_{j=1}^{m}E_{G}(C_{j})$, and

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M_{3}\subseteq E_{G}\big{(}\bigcup_{i=1}^{n}{X_{i}},V(G)\setminus\bigcup_{i=1}^{n}{X_{i}}\big{)}

minimizing $|M_{3}|$. Suppose, for a contradiction, that $M_{3}$ is non-empty. Let $i\in[n]$ be such that $E_{G}(X_{i},V(G)\setminus X_{i})\cap M_{3}$ is non-empty. If $E_{G}(X_{i},V(G)\setminus X_{i})\cap M_{3}=\{uv\}$ where $u\in X_{i}$, then the matching $(M\cup\{u_{i}u\})\setminus\{uv\}$ is induced, which is a contradiction to the minimality of $|M_{3}|$. Hence, we may assume that $E_{G}(X_{i},V(G)\setminus X_{i})\cap M_{3}=\{t_{i}v,f_{i}w\}$. First, we assume that $v,w\in C_{j}$ for some $j\in[m]$. This implies that $v$ and $w$ are the only vertices in $C_{j}$ that are covered by $M$. Hence, the matching $(M\cup\{u_{i}f_{i},v_{j}v\})\setminus\{t_{i}v,f_{i}w\}$ is induced, which is a contradiction to the minimality of $|M_{3}|$. Hence, we may assume that $v$ belongs to $C_{j}$ for some $j\in[m]$, and that $w$ belongs to $C_{j^{\prime}}$ for some $j^{\prime}\in[m]\setminus\{j\}$. Again by construction, $v$ is the only vertex in $C_{j}$ that is covered by $M$. Hence, the matching $(M\cup\{u_{i}f_{i},v_{j}v\})\setminus\{t_{i}v,f_{i}w\}$ is induced, which is a contradiction to the minimality of $|M_{3}|$. Hence, we may assume that $M_{3}=\emptyset$. Since $|M|=n+m$, this implies that $M\cap E_{G}(X_{i})$ and $M\cap E_{G}(C_{j})$ are all non-empty for every $i\in[n]$ and $j\in[m]$. Let $t:\{x_{1},\ldots,x_{n}\}\rightarrow\{0,1\}$ be defined as $t(x_{i})=0$ if $u_{i}f_{i}\in M$ and $t(x_{i})=1$ if $u_{i}t_{i}\in M$, and suppose, for a contradiction, that $\Gamma$ is not exact satisfied under $t$, that is, there is some clause $c_{\ell}=x_{i}\lor x_{j}\lor x_{k}$ such that not exactly one literal is true under $t$. First, we assume that no literal in $c_{\ell}$ is true under $t$. This implies that $u_{i}f_{i},u_{j}f_{j}$, and $u_{k}f_{k}$ belong to $M$, which implies that no vertex in $C_{\ell}$ is covered by $M$, a contradiction. Hence, we may assume that at least two literals in $c_{\ell}$ are true under $t$, which, by symmetry, implies that $u_{i}t_{i}$ and $u_{j}t_{j}$ both belong to $M$. Again by construction, this implies that no vertex in $C_{\ell}$ is covered by $M$, a contradiction. ∎

The graphs constructed in the proof of Theorem 2 have maximum degree at most $7$. Replacing $X_{1},\ldots,X_{n}$ by $K_{3,3}$’s where the edges of some maximum matching are subdivided once, yields the hardness for graphs of maximum degree $5$. The proof of it proceeds along the lines of the proof of Theorem 2. However, the lemma corresponding to Lemma 3.2 becomes quite technical, and so the proof is omitted. Therefore, in view of [5], for restrictions imposed on the maximum degree, the only case left are the graphs with maximum degree $4$.

Acknowledgement

I would like to thank Dieter Rautenbach for valuable suggestions that helped to improve the presentation of this note.