Algorithmic and complexity aspects of problems related to total restrained domination for graphs

Domination is a well-researched topic in graph theory, the literature on this subject is well described in two books by Haynes et al. (1998a, 1998b). And numerous researchers have thoroughly investigated the variations of domination in graphs.

Let G be a graph with vertex set V and edge set E. For vertex \(v\in V\), the open neighborhood \(N_{G}(v)\) of v is the set \(\{u\in V| uv\in E\}\), its cardinality \(|N_{G}(v)|\) is called the degree of v, and the closed neighborhood of v is the set \(N_{G}[v]=N(v)\cup \{v\}\). Moreover, when no confusion can arise, \(N_{G}(v)\) and \(N_{G}[v]\) are simplified by N(v) and N[v], respectively. The maximum degree of G will be denoted by \(\Delta (G)\). A leaf of G is a vertex of degree 1, a support vertex is a vertex adjacent to a leaf. For a subset \(D\subseteq V\). D is a total dominating set (TDS), introduced by Cockayne et al. (1980), of G if every vertex in V is adjacent to a vertex in D. D is a restrained dominating set (RDS), introduced by Domke et al. (1999), of G if every vertex in \(V\setminus D\) is adjacent to a vertex in D and another vertex in \(V\setminus D\). D is a total restrained dominating set (TRDS), introduced by Ma et al. (2005), of G if every vertex in V is adjacent to a vertex in D and every vertex in \(V\setminus D\) is adjacent to a vertex in \(V\setminus D\). Obviously, a TRDS is exactly both a TDS and a RDS. By the definitions we easily get two observations:

Observation 1.1

Each \(\textrm{RDS}\) in graph G includes all its leaf vertices.

Observation 1.2

Each \(\textrm{TRDS}\) in graph G contains all its leaf vertices and support vertices.

The minimum cardinality of total dominating sets (resp. restrained dominating sets, total restrained dominating sets) of G is called the total domination number (resp. restrained domination number, total restrained domination number) of G, denoted by \(\gamma _{t}(G)\) (resp. \(\gamma _{r}(G)\), \(\gamma _{tr}(G)\)). The MINIMUM TOTAL RESTRAINED DOMINATION (MTRD) problem (resp. the MINIMUM TOTAL DOMINATION (MTD) problem) for a graph G is to find a TRDS (resp. a TDS) of minimum cardinality of G. Given a positive integer k and a graph G, the TOTAL RESTRAINED DOMINATION DECISION (TRDD) problem (resp. the TOTAL DOMINATION DECISION (TDD) problem) is to decide whether G has a TRDS (resp. a TDS) of cardinality at most k. For detail research on total domination, see a survey (Henning 2009).

Obviously, for a graph G with no isolated vertices, \(\gamma _{tr}(G)\ge \gamma _{t}(G)\) and \(\gamma _{tr}(G)\ge \gamma _{r}(G)\). Raczek and Cyman (2008) characterized the trees T with \(\gamma _{tr}(T)=\gamma _{t}(T)\) and Raczek (2007) characterized the trees T with \(\gamma _{tr}(T)=\gamma _{r}(T)\). Ma et al. (2005) proved that for bipartite graphs and chordal graphs, the TRDD problem is NP-complete. Chen et al. (2012) also proved that the TRDD problem is NP-complete for planar graphs and split graphs, respectively.

G is called an undirected path graph (circle graph) if there exists direct mapping between V and a family of paths of a tree (a set of chords of a circle) such that two vertices are adjacent if and only if the corresponding paths (chords) intersect. It is well known that undirected path graph is a subclass of chordal graphs.

A comb is a tree obtained from a path P (called backbone) by joining a pendant leaf (called a tooth) to each vertex of P. A star is defined as a tree with at most one non-leaf. A bipartite graph G with bipartition (X, Y) (roles of X and Y can be exchanged) is called tree-convex, if there exists a tree T with vertex set X such that for every vertex \(y\in Y\), the neighborhood of y induces a subtree of T. When T is a comb (resp. a star), then G is called comb-convex (resp. star-convex) bipartite graph. We abbreviate the two bipartite graphs above as C-CB graph and S-CB graph, respectively. Bao and Zhang (2012) demonstrated that the construction of the associated tree T can be completed in linear time, and tree-convex bipartite graphs can be efficiently identified in linear time.

As mentioned in Ma et al. (2005), an application of total restrained domination is that of guards and inmates. Each vertex in the TRDS represents a guard and each vertex not in the TRDS represents a inmate. For supervise inmates and ensure the safety of guards, each prisoner and each guard must be seen by some guards; However, to safeguard the rights of inmates, it is necessary to ensure that each innate is observed by another inmate; To ensure the lowest possible cost, it is preferable to deploy the fewest number of guards feasible.

The present paper is organized as follows. In Sect. 2, we prove that the TRDD problem is NP-complete for undirected path graphs, circle graphs, S-CB graphs and C-CB graphs, respectively, and that, for a S-CB graph or C-CB graph with n vertices, the MTRD problem cannot be approximated within a factor of \((1-\epsilon )\textrm{ln} n\) for any \(\epsilon >0\) unless \(NP\subseteq DTIME(n^{O(\textrm{loglog} n)})\). Next, for a graph G, we prove that the problem of deciding whether \(\gamma _{r}(G) =\gamma _{tr}(G)\) is NP-hard even when G is restricted to planar graphs with maximum degree at most 4, and that the problem of deciding whether \(\gamma _{t}(G) =\gamma _{tr}(G)\) is NP-hard even when G is restricted to planar bipartite graphs with maximum degree at most 5. Finally, we show that the MTRD problem is APX-complete for bipartite graphs with maximum degree at most 4. In Sect. 3, we design a linear-time algorithm for solving this problem for generalized series–parallel graphs.

In this section, we firstly show that the TRDD problem is NP-complete for undirected path graphs, circle graphs, S-CB graphs and C-CB graphs, respectively. Secondly, we discuss the lower bounds of approximation ratio for the MTRD problem in S-CB graphs and C-CB graphs, and the upper bound of approximation ratio for the MTRD problem in general graphs. Thirdly, we show that for a planar graph G with \(\Delta (G)\le 4\), it is NP-hard to decide whether \(\gamma _{r}(G)=\gamma _{tr}(G)\), and that it is NP-hard to decide whether \(\gamma _{t}(G)=\gamma _{tr}(G)\), even if G is a planar bipartite graph with \(\Delta (G)\le 5\). Finally, we show that the MTRD problem is APX-complete for bipartite graphs with maximum degree at most 4.

2.1 NP-completeness for the TRDD problem

To establish the NP-completeness of a problem P, it is enough to show that P belongs to the class NP and that a known NP-complete problem is reducible to P in polynomial time. Here, We use the TDD problem, which is already known to be NP-complete for undirected path graphs (Laskar et al. 1984), circle graphs (Keil 1993), respectively.

Now we present a reduction from the TDD problem to the TRDD problem as follows. Let \(G=(V,E)\) be a graph with \(V=\{v_{1}, v_{2}, \ldots , v_{n}\}\), without isolated vertices, we construct a new graph \(G_{1}=(V_{1},E_{1})\), where \(V_{1}=V\cup \{a_{i},b_{i},c_{i},d_{i},e_{i},f_{i}|1\le i\le n\}\) and \(E_{1}=E\cup \{v_{i}a_{i},a_{i}b_{i},b_{i}c_{i},c_{i}d_{i},a_{i}e_{i},e_{i}f_{i}|\) \(1\le i\le n\}\) (see Fig. 1).

Theorem 2.1

Let G and \(G_1\) be graphs defined as above. Then G has a \(\textrm{TDS}\) of size at most k if and only if \(G_{1}\) has a \(\textrm{TRDS}\) of size at most \(k+4n\).

Proof

Sufficiency. Let D be a TDS of G with \(|D|\le k\). It is obvious that \(D_{1}=D\cup \{c_{i},d_{i},f_{i},e_{i}|1\le i\le n\}\) is a TRDS of \(G_1\) with \(|D_{1}|\le k+4n\).

Necessity. Let \(D'\) be a TRDS with \(|D'|\le k+4n\) in \(G_{1}\) such that \(|D'\cap \{a_{i},b_{i}|1\le i\le n\}|\) is as small as possible. For \(1\le i\le n\), since \(d_{i}\), \(f_{i}\) are leaves of \(G_{1}\), \(\{c_{i}, d_{i}, f_{i}, e_{i}\}\subseteq D'\) by Observation 1.2. By the choice and definition of \(D'\), either \(\{a_i, b_i\} \subseteq D'\) or \(\{a_i, b_i\} \cap D'=\emptyset \). But we have

Claim. \(D'\cap \{a_{i},b_{i}\}=\emptyset \) for \(1\le i\le n\).

Let \(I=\{i|a_i,b_i\in D'\}\) and \(D''=D'\cup \bigcup _{i\in I}\{v_i,v_{i'}\}{\setminus }\bigcup _{i\in I}\{a_i,b_i\}\), where \(v_{i'}\in N_{G_1}(v_i)\cap V\). Now we show that \(D''\) is a TRDS of \(G_1\). For any \(1\le i\le n\), \(a_i\) is adjacent to \(e_i\in D''\) and \(b_i\) is adjacent to \(c_i\in D''\). Consider any \(v_i\in V\). If \(i\in I\), then there is \(v_{i'}\in N_{G_1}(v_i)\cap V\) such that \(v_{i'}\in D''\). Otherwise \(i\notin I\), then there is \(v_j\in N_{G_1}(v_i)\cap V\cap D'\). By the construction of \(D''\), \(v_j\in D''\). Any vertex \(v_i\in V{\setminus } D''\) is adjacent to \(a_i\notin D''\). Thus \(D''\) is a TRDS of \(G_1\) with \(|D''|\le |D'|\) and \(D''\cap \{a_i,b_i|1\le i\le n\}=\emptyset \), a contradiction with the choice of \(D'\).

From Claim and \(\{c_{i}, d_{i}, f_{i}, e_{i}\}\subseteq D'\) for \(1\le i\le n\), it follows immediately that \(D'\cap V\) is a TDS with \(|D'\cap V|\le k\) in G. \(\square \)

From definitions of undirected path graphs and circle graphs, the following lemma emerges.

Lemma 2.1

Assume that \(G_1\) is constructed from a graph G as above, shown in Fig. 1. If G is an undirected path graph or circle graph, then so is \(G_1\).

Clearly, the TRDD problem is in NP for undirected path graphs and circle graphs, respectively. Combined with Theorem 2.1 and Lemma 2.1, the following theorem emerges, the following theorem emerges.

Theorem 2.2

The \(\textrm{TRDD}\) problem is NP-complete for undirected path graphs and circle graphs, respectively.

2.2 Lower bounds on the approximation ratio

To obtain a lower bound, we give an approximation preserving reduction from the MINIMUM SET COVER problem.

MINIMUM SET COVER (MSC) problem: Let U be any non-empty set and \(\mathcal {S}\) be a collection of (non-empty) subsets of U such that \(\cup _{S\in \mathcal {S}} S=U\) (and \((U,\mathcal {S})\) is called a set system). A subcollection \(\mathcal {S}'\subseteq \mathcal {S}\) is called a set cover (SC) of U, if every element of U belongs to at least one element of \(\mathcal {S}'\), i.e., \(\cup _{S\in \mathcal {S}'} S=U\). The MSC problem is finding the smallest SC for a given set system \((U, \mathcal {S})\). The SET COVER DECISION problem is to decide whether U has a SC of cardinality at most k, for a given positive integer k and a set system \((U,\mathcal {S})\).

A lower bound of the approximation ratio for the MSC problem has been investigated in Feige (1998).

Theorem 2.3

(Feige 1998) The \(\textrm{MSC}\) problem for the input instance \((U,\mathcal {S})\) does not admit a \((1-\epsilon )\textrm{ln}|U|\)-approximation algorithm for any \(\epsilon >0\) unless \(NP\subseteq DTIME(|U|^{O(\textrm{loglog} |U|)})\). Furthermore, this inapproximability result holds for the case when the size of the input collection \(\mathcal {S}\) is no more than the size of the base set U.

Now, we show that a hardness result for the MTRD problem in S-CB graphs. First, the characterization of S-CB graphs was given as follows.

Lemma 2.2

(Pandey and Panda 2019) Let \(G= (X, Y, E)\) be a bipartite graph. G is a S-CB graph if, and only if, there is a vertex v in X, and every vertex in Y is either a pendant leaf or a neighbor of v.

Theorem 2.4

The \(\textrm{MTRD}\) problem for a S-CB graph G with n vertices does not admit a \((1-\epsilon )\textrm{ln}n\)-approximation algorithm for any \(\epsilon >0\) unless \(NP\subseteq DTIME(n^{O(\textrm{loglog} n)})\).

Proof

It is sufficient to reduce the MSC problem to the MTRD problem. Let the set system \((U,\mathcal {S})\), where \(U=\{u_{1}, u_{2},\ldots ,u_{p}\}\) and \(\mathcal {S}=\{S_{1},S_{2},\ldots ,S_{m}\}\), \(m\le p\) be an instance of the MSC problem. We construct the bipartite graph \(G = (X, Y, E)\) as follows.

1. 1.

   For \(u_{i}\in U\), add a vertex \(x_{i}\) in X; add an additional vertex \(x_{p + 1}\) in X; for \(S_{j}\in \mathcal {S}\), add two vertices \(y_{j,1}\), \(y_{j,2}\) in Y.

2. 2.

   For \(u_i\in S_j\), add edges \(x_{i}y_{j,1}, x_{i}y_{j,2}\) in E; add \(\{x_{p + 1}y_{j,1}, x_{p + 1}y_{j,2}|1\le j\le m \}\) in E.

Clearly, in polynomial time, G can be constructed. By the construction of G, \(x_{ p + 1}\) is a neighbor of every vertex in Y and further G is a S-CB graph with \(n = p + 2m + 1\) vertices by Lemma 2.2. See Fig. 2a for an example. Then we have

Construction \(\star \): Let D be a TRDS of G defined as above. Construct a SC \(\mathcal {S}'\) of U from D, where \(\mathcal {S}'=\{S_j|y_{j, 1} \text{ or } y_{j, 2} \in D\cap Y, 1\le j\le m\}\) with \(|\mathcal {S}'|\le |U|-1\).

First, we show that \(\mathcal {S}'\) is a SC, i.e., for any \(1\le i\le p\), there exists \(S_j\in \mathcal {S}'\) such that \(u_i\in S_j\). By the definition of D, there exists a neighbor y of \(x_i\) such that \(y\in D\). Since G is bipartite, \(y=y_{j, 1}\) or \(y_{j, 2}\) for some \(1\le j\le m\), further \(u_i\in S_j\) by the definition of G and \(S_j\in \mathcal {S}'\) by the definition of \(\mathcal {S}'\). Therefore, \(\mathcal {S}'\) is a SC of U. Note that \(|\mathcal {S}'|\le |D\cap Y|\). Then we show that \(|\mathcal {S}'|\le |D|-1\). Since D is a TRDS of bipartite graph \(G=(X, Y, E)\), \(D\cap X\ne \emptyset \). Thus \(|\mathcal {S}'|\le |D\cap Y|\le |D\cap X|+|D\cap Y|-1=|D|-1\).

Claim. For a positive integer k, U has a SC of size at most k if and only if G has a TRDS of size at most \(k + 1\).

Let \(\mathcal {S}'\) be a SC of U and \(|\mathcal {S}'|\le k\). Obviously, the set \(D = \{x_{ p + 1 }, y_{j, 1}|S_{j}\in \mathcal {S}'\}\) is a TRDS of G and \(|D|\le k+1\). Conversely, it can be obtained directly from Construction \(\star \).

By Claim, if \(OPT_{sc}\) is a MSC of U and \(OPT_{trd}\) is a minimum TRDS of G, then \(|OPT_{trd}| = |OPT_{sc}| + 1\). Suppose to the contrary that there exists a polynomial time algorithm, say Algorithm A, to approximate the MTRD problem within a ratio of \((1-\epsilon )\textrm{ln} n\) for S-CB graphs with n vertices. Then Algorithm 1 (defined as follows) constructs a solution for the MSC problem.

Since each step of Algorithm 1 can be accomplished within polynomial time, Algorithm 1 can be computed in polynomial time. Let l be a positive integer such that \(\frac{1}{l}< \epsilon \). If \(|OPT_{sc}|\le l\), then \(OPT_{sc}\) can be computed in polynomial time. Next we assume that \(|OPT_{sc}|>l\). If \(\mathcal {S}'\) is a SC of U computed by Algorithm 1, then \(|\mathcal {S}'| < |\)D\(|\le ((1-\epsilon )\textrm{ln} n)|OPT_{trd}|= ((1-\epsilon )\textrm{ln} n) (|OPT_{sc}|+1)= ((1-\epsilon )\textrm{ln} n) (1 + \frac{1}{|OPT_{sc}|})|OPT_{sc}|< ((1-\epsilon )\textrm{ln} n)(1 + \frac{1}{l} )|OPT_{sc}|<((1- \epsilon ^{2}) \textrm{ln}n)|OPT_{sc}|\). Since \(m\le p\), we have \(n=p+2m+1\le 4p\). Then

For sufficiently large \(p(=|U|)\), the term \(1+\frac{\textrm{ln}4}{\textrm{ln}p}\) can be bounded by \(1+\epsilon ^2\). Now we have

where \(\epsilon '=\epsilon ^4\).

By Theorem 2.3, if the MSC problem can be approximated within \((1- \epsilon ')\textrm{ln}|U|\), then \(NP\subseteq DTIME(|U|^{O(\textrm{loglog} |U|)})\). It follows that if the MTRD problem can be approximated within\((1- \epsilon )\textrm{ln}n\) for any \(\epsilon >0\), then \(NP\subseteq DTIME (n^{O(\textrm{loglog} n)})\). Hence, the MTRD problem cannot be approximated within \((1- \epsilon )\textrm{ln}n\) unless \(NP\subseteq DTIME(n^{O(\textrm{loglog} n)})\). \(\square \)

Theorem 2.5

(Karp 1972). The \(\mathrm {S{ET}}\) \(\mathrm {C{OVER}}\) \(\mathrm {D{ECISION}}\) problem is NP-complete.

Clearly, the TRDD problem is in NP for S-CB graphs. Combined Theorem 2.5 with Claim in Theorem 2.4, we get

Theorem 2.6

The \(\textrm{TRDD}\) problem is NP-complete for S-CB graphs.

Next, we show that the hardness result for the MTRD problem in C-CB graphs by giving the similar proof as in Theorem 2.4, but the construction of the bipartite graph is different.

Given a set system \((U,\mathcal {S})\), where \(U=\{u_{1}, u_{2}, \ldots , u_{p}\}\) and \(\mathcal {S}=\{S_{1}, S_{2}, \ldots , S_{m}\}\), \(m\le p\), we construct a C-CB graph \(H = (X, Y, E )\) in the following way:

1. 1.

   For \(u_{i}\in U\), add two vertices \(x_{i, 1}\) and \(x_{i, 2}\) in X and a vertex \(y_{i}'\) in Y; for \(S_{j}\in \mathcal {S}\), add a vertex \(y_{j}\) in Y; Denote by \(X_{k}=\{x_{i,k}|1\le i\le p\}\), \(k=1,2\), \(Y'=\{y_{i}'|1\le i\le p\}\).

2. 2.

   Add the edge \(x_{i,1}y_{i}'\), \(1\le i\le p\);

3. 3.

   If \(u_{i}\in S_{j}\), then we add the edge \(x_{i,1}y_{j}\), \(1\le j\le m\);

4. 4.

   Form a complete bipartite graph between \(X_2\) and Y. Note that H can be constructed in polynomial time. Figure 2b shows the construction of the C-CB graph H from the set system \((U,\mathcal {S})\), where \(U= \{u_{1}, u_{2}, u_{3}, u_{4}\}\), and \(\mathcal {S}=\{\{u_{1},u_{2}\},\{u_{2},u_{3}\}, \{u_{3},u_{4}\}\}\). It is easy to see that H is a C-CB graph if \(X_{2}\) is taken as the backbone and \(X_{1}\) is taken as the teeth of the comb. Next, we give a construction. Construction \(\star \star \): Let D be a TRDS of H defined as above, f a mapping from \(D\cap Y'\) to \(\mathcal {S}\) such that, for \(y_i'\in D\cap Y'\) with \(1\le i\le p\), \(u_i\in f(y_i')\). Construct a SC \(|\mathcal {S}'|\le |U|-1\) of U from D by \(\mathcal {S}'=\{S_j|y_j\in D, 1\le j\le m\}\cup \{f(y_i')\in \mathcal {S}|y_i'\in D, 1\le i\le p\}\). Firstly, we show that \(\mathcal {S}'\) is a SC, i.e., for any \(1\le i\le p\), there exists \(S_j\in \mathcal {S}'\) such that \(u_i\in S_j\). By the definition of D, there exists a neighbor y of \(x_{i,1}\) such that \(y\in D\). Since H is bipartite, either \(y=y_{j}\) for some \(1\le j\le m\), or \(y=y'_{i}\). If \(y=y_{j}\), then \(u_i\in S_j\) by the definition of H and \((u_i\in )S_j\in \mathcal {S}'\) by the definition of \(\mathcal {S}'\). Otherwise, \(y=y'_{i}\). Then \(u_i\in f(y_i')\) by the definition of f and \((u_i\in )f(y_i')\in \mathcal {S}'\) by the definition of \(\mathcal {S}'\). Therefore, \(\mathcal {S}'\) is a SC of U. Secondly, we show that \(|\mathcal {S}'|\le |D|-1\). Since D is a TRDS of bipartite graph \(H=(X, Y, E)\), \(D\cap X\ne \emptyset \). Thus \(|\mathcal {S}'|\le |D\cap Y|\le |D\cap X|+|D\cap Y|-1=|D|-1\).

Lemma 2.3

Let \((U,\mathcal {S})\) be a set system and H the bipartite graph associated with it, defined as above. Then \((U,\mathcal {S})\) has a \(\textrm{SC}\) of size at most k if and only if H has a \(\textrm{TRDS}\) of size at most \(k+1\). In particular, the size of a \(\textrm{MSC}\) of U equals the size of a minimum \(\textrm{TRDS}\) of H minus 1.

Proof

Let \(\mathcal {S}'\) be a SC of U with \(|\mathcal {S}'|=k\). Evidently, the set \(D=\{x_{1,2}\}\cup \{y_j|S_j\in \mathcal {S}'\}\) is a TRDS of H with \(|D|=k+1\). Conversely, it can be obtained directly from Construction \(\star \star \). \(\square \)

Clearly, the TRDD problem is in NP for C-CB graphs. Combined Theorem 2.5 with Lemma 2.3, we get

Theorem 2.7

The \(\textrm{TRDD}\) problem is NP-complete for C-CB graphs.

Similar to Theorem 2.4 and its proof, we have

Theorem 2.8

The \(\textrm{MTRD}\) problem for a C-CB graph G with n vertices does not admit a \((1-\epsilon )\) \(\textrm{ln}n\)-approximation algorithm for any \(\epsilon >0\) unless \(NP\subseteq DTIME(n^{O(\textrm{loglog} n)})\).

2.3 Upper bound on the approximation ratio

Let \(G= (V,E)\) be a graph of maximum degree \(\Delta \) with no isolated vertices, D be a minimum TRDS of G. Since for any \(v\in D\) has a neighbor in D by the definition of TRDS, and \(V = \cup _{v\in D}N_{G}(v)\), we have \(|V|=| \cup _{v\in D} N_{G}(v)|\le \sum _{v\in D}|N_{G}(v)| = \sum _{v\in D}d_{G}(v) \le \sum _{v\in D}\Delta = \Delta \cdot |D|\). So \( |V|\le \Delta \cdot |D|\). It follows that (take the full vertex set as a TRDS).

Theorem 2.9

For any graph \(G = (V,E)\) with \(\Delta (G)=\Delta \), the minimum TRDS can be approximated within an approximation ratio of \(\Delta \).

2.4 Hardness results on \(\gamma _{r}(G)=\gamma _{tr}(G)\) (resp. \(\gamma _{t}(G)=\gamma _{tr}(G)\)) problems for a graph G

It is evident that, for a graph G without isolated vertices, \(\gamma _{r}(G)\le \gamma _{tr}(G)\) and \(\gamma _{t}(G)\le \gamma _{tr}(G)\). A natural question arises: for which graphs does the respective equality hold? Raczek (2007) characterized the trees T with \(\gamma _{r}(T)=\gamma _{tr}(T)\) and Raczek and Cyman (2008) characterized the trees T with \(\gamma _{t}(T)=\gamma _{tr}(T)\), respectively. In this section, we show that it is NP-hard to decide whether \(\gamma _{r}(G)=\gamma _{tr}(G)\) holds for planar graphs G with \(\Delta (G)\le 4\) and whether \(\gamma _{t}(G)=\gamma _{tr}(G)\) holds for planar bipartite graphs G with \(\Delta (G)\le 5\), respectively. Here we use the following NP-complete problem, due to Dahlhaus et al. (1994), as the reduction.

PLANAR EXACTLY 3-BOUNDED 3-SAT problem: A formula \(\Phi \) with variable set X and clause set C such that, firstly, each variable has exactly three literals with one of them occurring in two clauses and the other in one, and each clause in the disjunction of either two or three literals; secondly, the bipartite graph B with bipartition (X, C), obtained by connecting an edge between \(x \in X\) and \(c \in C\) if c contains one of the literals x and \(\overline{x}\), is planar. Notice that \(\Delta (B)\le 3\). The PLANAR EXACTLY 3-BOUNDED 3-SAT is the problem of deciding whether a given Boolean formula satisfied the above conditions is satisfiable.

Theorem 2.10

For a given planar graph G with \(\Delta (G)\le 4\), it is NP-hard to decide whether \(\gamma _{tr}(G)=\gamma _{r}(G)\).

Proof

We reduce from the PLANAR EXACTLY 3-BOUNDED 3-SAT problem. Consider a set of clauses \(C=\{c_{1}, c_{2}, \ldots , c_{p}\}\) with variables \(X=\{x_{1}, x_{2}, \ldots , x_{n}\}\), as the input for the PLANAR EXACTLY 3-BOUNDED 3-SAT problem, say, B the planar bipartite graph associated with it, now we construct a graph \(G=(V,E)\). For any variable \(x_{i}\), we introduce a triangle \(G_{x_{i}}\) with two distinguished literal vertices \(x_{i}\) and \(\overline{x_{i}}\), for any clause \(c_{j}\in C\), we denoted by \(G_{c_{j}}\) the graph described in Fig. 3 (left). Finally, for \(x_{i}\in c_j\) (resp. \(\overline{x_{i}}\in c_j\)), we add the edge \(x_{i}a_{j}\) (resp. \(\overline{x_{i}}a_{j}\)). See Fig. 3 (right) for an example. In fact, G is obtained from planar B mainly by splitting each of vertices of degree 3 corresponding to variables into two vertices connecting an edge. Clearly, G is also planar and \(\Delta (G)\le 4\).

Claim. C is satisfiable if and only if \(\gamma _{tr}(G)=\gamma _{r}(G)\).

First we show that \(\gamma _{r}(G)=n+5p\). Let D be a RDS of G. Then \(|D\cap V(G_{x_{i}})|\ge 1\) for \(1\le i\le n\). For \(1\le j\le p\), \(\{e_{j}, k_{j}\}\subseteq D\) by Observation 1.1, \(|D\cap N[m_{j}]|\ge 1\), \(|D\cap N[g_{j}]|\ge 1\) and \(|D\cap N[b_{j}]|\ge 1\) by the definition of D. Note that \(N[m_{j}], N[g_{j}]\) and \(N[b_{j}]\) are pairwise disjoint. Thus \(|D\cap V(G_{c_{j}})|\ge 5\). Therefore, \(\gamma _{r}(G)\ge n+5p\). Moreover, \(\{x_{i}, a_{j}, d_{j}, e_{j}, k_{j}, l_{j}|1\le i\le n, 1\le j\le p\}\) is a RDS of G. Therefore, \(\gamma _{r}(G)= n+5p\).

Now, assuming that \(\Phi \) is satisfiable. Given a satisfying truth assignment of \(\Phi \), next we will prove there exist a TRDS D of G with \(|D|=n+5p\). For \(1\le i\le n\), we add \(x_{i}\) to D, if \(x_{i}\) evaluate to true, or \(\overline{x_{i}}\), otherwise. For \(1\le j \le p\), we add \(\{a_{j}, d_{j}, e_{j}, k_{j}, l_{j}\}\) to D. Clearly, \(|D|=n+5p\). Next we show that D is a TRDS of G. By the construction of D, it is obvious that D is a dominating set and for any \(v\in V\setminus D\), v has a neighbor in \(V{\setminus } D\). It is sufficient to show that for any \(v\in D\), v has a neighbor in D. Obviously, \(d_{j}\) is adjacent to \(e_{j}\), \(l_{j}\) is adjacent to \(k_{j}\) for \(1\le j\le p\) and each literal vertex whose corresponding literal evaluates to true is adjacent to one of \(a_{j}\) for \(1\le j\le p\). Since \(\Phi \) is satisfiable, each \(a_{j}\) has a neighbor that is a literal vertex whose corresponding literal is true, thus \(a_{j}\) is adjacent to a vertex in D.

Conversely, let’s assume that \(\gamma _{tr}(G)=\gamma _{r}(G)(=n+5p)\) and let T be a TRDS of size \(n+5p\) in G. Similar to the proof above of \(\gamma _r(G)=n+5p\), we have \(|G_{x_{i}}\cap T|=1\) and \(|T\cap G_{c_{j}}|=5\). By Observation 1.2, \(\{d_{j}, e_{j}, l_{j},k_{j}\}\subseteq T\). If \(b_{j}\in T\), then \(|T\cap N[b_{j}]|\ge 2\), a contradiction with \(|G_{c_{j}}\cap T|=5\). If \(b_{j}\notin T\), \(f_{j}\in T\), then \(h_{j}\in T\), otherwise \(h_{j}\) is an isolated vertex in \(V\setminus T\). Thus \(|G_{c_{j}}\cap T|\ge 6\), a contradiction with \(|G_{c_{j}}\cap T|=5\). So \(a_j\in T\). Thus \(T\cap V(G_{c_{j}})=\{a_{j}, d_{j}, e_{j}, l_{j}, k_{j}\}\).

For \(1\le i\le n\), obviously, T contains exactly one of \(x_i\) and \(\overline{x_i}\) in \(V(G_{x_{i}})\). We establish a truth assignment for \(\Phi \) in the following manner: set \(x_{i}\) to true if \(x_{i}\in T\) and to false if \(\overline{x_{i}}\in T\). Since each \(a_{j}\) belongs to T and \(b_{j}\notin T\), by the definition of T, \(a_{j}\) has a neighbor that is a literal vertex in T, and the corresponding literal is true. Consequently, \(\Phi \) satisfies all clauses. \(\square \)

Theorem 2.11

For a given planar bipartite graph G with \(\Delta (G)\le 5\), it is NP-hard to decide whether \(\gamma _{tr}(G)=\gamma _{t}(G)\).

Proof

We reduce from the PLANAR EXACTLY 3-BOUNDED 3-SAT problem. Consider a set of clauses \(C=\{c_{1}, c_{2}, \ldots , c_{p}\}\) with variables \(X=\{x_{1}, x_{2},\ldots , x_{n}\}\), as the input for the PLANAR EXACTLY 3-BOUNDED 3-SAT problem, now we construct a graph \(G=(V,E)\). For any variable \(x_{i}\), we introduce a \(G_{x_{i}}\) with two special vertices \(x_{i}\) and \(\overline{x_{i}}\), see Fig. 4 (left). For any clause \(c_{j}\in C\), we introduce a \(G_{c_{j}}\) described in Fig. 4 (right). If literal \(x_{i}\) (resp. \(\overline{x_{i}}\)) in \(c_{j}\), then we add the edge \(x_{i}u^{1}_{j}\) (resp. \(\overline{x_{i}}u^{1}_{j}\)) in G. By simple analysis, we get G is planar bipartite graph and \(\Delta (G)\le 5\).

Claim 1. \(\gamma _{t}(G)=4n+5p\).

Let D be a TDS of G. By simple analysis, for each \(G_{c_{j}}\), we have the subgraph induced by \(\{v_j^1, v_j^2, \ldots , v_j^9\}\) (resp. by \(\{u_j^1, u_j^2, \ldots , u_j^7\}\)) contains at least three (resp. two) vertices in D, thus \(|D\cap V(G_{c_j})|\ge 5\); for each \(G_{x_i}\), we have the subgraphs induced by \(\{r_i, x_i, q_i, z_i\}\) and by \(\{\overline{x_i}, t_i, s_i, w_i\}\) both contain at least two vertices in D, thus \(|D\cap V(G_{x_i})|\ge 4\). Then, \(\gamma _{t}(G)\ge 4n+5p\). Moreover, \(\{x_{i}, z_{i}, w_{i}, \overline{x_{i}}, u^{2}_{j}, u^{3}_{j}, v^{1}_{j},v^{5}_{j},v^{9}_{j}|1\le i\le n, 1\le j\le p\}\) of size \(4n+5p\) is a TDS of G. Therefore \(\gamma _{t}(G)\le 4n+5p\), then \(\gamma _{t}(G)= 4n+5p\).

Next, we show

Claim 2. \(\Phi \) is satisfiable if and only if \(\gamma _{tr}(G)=\gamma _{t}(G)\).

Suppose that \(\Phi \) is satisfiable. Given a satisfying truth assignment of \(\Phi \), next we construct a TRDS D of size \(4n+5p\) of G. In each \(G_{x_i}\), we add \(\{\overline{x_{i}}, w_{i},q_{i},r_{i}\}\) to D, if \(x_{i}\) evaluates to true, and add \(\{x_{i}, z_{i},s_{i},t_{i}\}\) to D, otherwise. In each \(G_{c_{j}}\), we add \(\{u^{2}_{j}, u^{3}_{j}, v^{1}_{j},v^{5}_{j},v^{9}_{j}\}\) to D. Clearly, \(|D|=4n+5p\). Now we show that D is a TRDS of G. Obviously, any \(v\in V\) has a neighbor in D, every vertex in \(V{\setminus } (D\cup \{u^{1}_{j}|1\le j\le p\})\) has a neighbor in \(V\setminus D\). It is sufficient to prove that \(u^{1}_{j}\), \(1\le j\le p\), is adjacent to a vertex in \(V\setminus D\). Since \(\Phi \) is satisfiable, each \(u^{1}_{j}\) has a neighbor that is a literal vertex whose corresponding literal is true, which is not in D by the definition of D, thus \(u^{1}_{j}\) is adjacent to a vertex in \(V\setminus D\).

Conversely, let’s assume that \(\gamma _{tr}(G)=\gamma _{t}(G)\) and let T be a TRDS of size \(4n+5p\) in G. By the similar analysis as above, we have \(|T\cap V(G_{x_{i}})|= 4\) and \(|T\cap V(G_{c_{j}})|= 5\).

Fact 1. In each \(G_{x_{i}}\), T contains exactly one of \(x_{i}\) and \(\overline{x_{i}}\).

In \(G_{x_{i}}\), since \(y_{i}\) is totally dominated by T, either \(x_{i}\in T\) or \(\overline{x_{i}}\in T\). If both of them are in T, then \(y_{i}\in T\). Thus the remained one vertex in \(T\cap V(G_{x_i})\) can not dominate \(q_i\) and \(s_i\) simultaneously, a contradiction.

Fact 2. In each \(G_{c_{j}}\), \(u_j^1\not \in T\), \(u_j^2\in T\).

Similar to the proof of Claim 1, \(|T\cap \{u_j^1, u_j^2, \ldots , u_j^7\}|=2\). Then \(u_j^1\not \in T\) by the definition of T. If \(u_j^2\not \in T\), then, by the definition of T, \(T\cap \{u_j^1, u_j^2, \ldots , u_j^7\}=\{u_j^3, u_j^4\}\) or \(\{u_j^3, u_j^6\}\). Assume that \(T\cap \{u_j^1, u_j^2, \ldots , u_j^7\}=\{u_j^3, u_j^4\}\). To dominate \(u_j^7\), then \(v_j^5\in T\). By simple analysis, we have \(T\cap \{v_j^1, v_j^2, \ldots , v_j^9\}=\{v_j^1, v_j^5, v_j^9\}\). Then \(u_j^5\) is isolated in \(V\setminus T\), a contradiction with the definition of T. Therefore \(u_j^2\in T\).

We establish a truth assignment for \(\Phi \) by Fact 1 in the following manner: set \(x_i\) to be true if \(x_i\not \in T\), to be false otherwise. By Fact 2 and the definition of T, \(u^{1}_{j}\) is adjacent to a literal vertex which is not in T such that the corresponding literal is true. Consequently, \(\Phi \) satisfies all clauses. \(\square \)

2.5 APX-completeness

In this section we will prove the APX-completeness of the MTRD problem P for bipartite graphs with maximum degree at most 4. We need to prove that both P is in APX, that is, P can be approximated within a constant ratio, and P is APX-hard. To show the problem P is APX-hard, it is sufficient to show that there is an L-reduction from an APX-complete problem to P. We first recall the notation of L-reduction as follows.

L-reduction ( Papadimitriou and Yannakakis 1991): Consider two NP-optimization problems F and G, along with a polynomial-time transformation f which maps instances from F and G. We can categorize f as an L-reduction if there are positive constants \(\alpha \) and \(\beta \), such that the following condition holds for every instance x of F,

1. 1.

   \(opt_{G}(f(x)) \le \alpha \cdot opt_{F}(x)\),

2. 2.

   For every feasible solution y of f(x) with objective value \(c_{2}\), in polynomial time one can find a solution \(y'\) of x with objective value \(c_{1}\) such that \(|opt_{F}(x)-c_{1}| \le \beta \cdot |opt_{G}( f (x))-c_{2}|\). Next we define the MINIMUM VERTEX COVER (MVC) problem. Given a graph \(G= (V,E)\), a set \(S \subseteq V\) of G is called a vertex cover (VC) of G if for every edge \(uv \in E\), either \(u \in S\) or \(v \in S\). The MVC problem is to find a VC of G with minimum cardinality.

Theorem 2.12

The \(\textrm{MTRD}\) problem is APX-complete for bipartite graphs with maximum degree at most 4.

Proof

By Theorem 2.9, the MTRD problem is in APX for bipartite graphs with maximum degree 4. The MVC problem for graphs with maximum degree at most 3 is APX-complete (Papadimitriou and Yannakakis 1991). Next, we use the MVC problem for graphs with maximum degree at most 3 to L-reduction the MTRD problem for bipartite graphs with maximum degree at most 4. Given a graph \(G= (V, E)\) with \(\Delta (G)\le 3\), where \(V =\{v_{1},v_{2},\ldots ,v_{n}\}\) and \(E =\{e_{1}, e_{2}, \ldots , e_{m}\}\), next we construct a graph \(G^{*} = (V', E')\) (see Fig. 5 for an example):

1. 1.

   For each edge \(e_{i}=uv\) in G, we first subdivide the edge by introducing an additional vertex \(q_{i}\) (called subdividing vertex corresponding to \(e_i\)), connecting \(q_i\) and u, v, respectively; then we introduce three additional vertices \(r_{i}\), \(s_{i}\), \(t_{i}\) and add the edges \(q_{i}r_{i}\), \(r_{i}s_{i}\), \(s_{i}t_{i}\).

2. 2.

   For each vertex \(v_{i}\) in G, take two additional vertices \(o_{i}\), \(p_{i}\) and add the edges \(v_{i}o_{i}\), \(o_{i}p_{i}\). Obviously, \(G^*\) is bipartite and \(\Delta (G^*)\le 4\). For the sake of simplicity, we denote by \(O=\{o_{i} |1\le i \le n\}\), \(P =\{p_{i} |1\le i \le n\}\), \(R =\{r_{i} |1\le i\le m\}\) and \(Q =\{q_{i} |1\le i\le m\}\), \(T =\{t_{i} |1\le i \le m\}\), \(S=\{s_{i} |1\le i \le m\}\).

Claim. Let D be any TRDS of \(G^*\). There exists a TRDS \(D'\) with \(|D'|\le |D|\) such that \((Q\cup R)\cap D'=\emptyset \). Furthermore, if \(C^*\) is a MVC of G and \(D^*\) is a minimum TRDS of \(G^{*}\), then \(|D^*|=|C^*|+2n+2\,m\).

First, \(O\cup P\cup S\cup T\subseteq D\) by Observation 1.2. If \((Q\cup R)\cap D=\emptyset \), then we can choose \(D'=D\), we have finished. Otherwise we can construct another TRDS \(D''\) from D such that \(|(Q\cup R)\cap D''|<|(Q\cup R)\cap D|\). Then we can continue to carry out the process repeatedly until we find \(D'\).

Suppose there exists some \(1\le i\le m\) such that \(\{q_i, r_i\}\cap D\ne \emptyset \) with \(q_{i}\in Q\) corresponding to the edge \(v_jv_k\) in G. If \(q_i\in D\), then \(r_i\in D\). Otherwise \(r_i\) is isolated in \(V(G^*){\setminus } D\), a contradiction with the definition of D. Next, we will break down the discussion into two cases.

Case 1. \(\{q_i, r_i\} \subseteq D\).

If either \(v_j\) or \(v_k\) is in D, then, by the fact that \(\{o_j, o_k\}\subseteq D\), \(D''=D{\setminus } \{q_i, r_i\}\) is as desired. Otherwise \(D''=D\setminus \{q_i\}\) is as desired.

Case 2. \(q_i\not \in D\) and \(r_i\in D\).

If either \(v_j\) or \(v_k\) is in D, then \(D''=D\setminus \{r_i\}\) is as desired. Suppose \(\{v_j, v_k\}\cap D=\emptyset \) in the following. Let \(L=\{l|q_l\in N_{G^*}(v_j), q_l\not \in D, r_l\in D\}\). Note that \(i\in L\). Then \(D''=D\cup \{v_j\}{\setminus } \{r_l|l\in L\}\) is as desired (see Fig. 6 for the illustration).

If \(C^*\) is a MVC of G, then \(C^*\cup O \cup P\cup S \cup T\) is a TRDS of \(G^*\). Hence \(|D^*|\le |C^*|+2n+2m\). Conversely, by the former part of Claim, we can assume that \(D^*\) is a minimum TRDS of \(G^*\) such that \(D^*\cap (Q\cup R)=\emptyset \). Then for any vertex \(q_i\in Q\), it is dominated by some \(v_j\) in \(D^*\), which is an end of \(e_i\) in G by the definition of \(G^*\). Thus \(D^*{\setminus } (O \cup P\cup S \cup T)\) is a VC of G. Therefore, \(|C^*|\le |D^*|-2n-2\,m\).

Finally, we prove that the previously introduced construction leads to an L-reduction. Let \(D^*\) be a minimum TRDS of \(G^{*}\) and \(C^*\) be a VC of G with the smallest cardinality. Since \(\Delta (G)\le 3\), \(m\le \frac{3}{2}n\) and \(n\le 4|C|\) for any VC C. Hence \(|D^*|=|C^*|+2n+2m\le 21|C^*|\). Moreover, given a TRDS D of \(G^*\), there exists a TRDS \(D'\) in \(G^*\) with \(|D'|\le |D|\) such that \(D'\cap (Q\cup R)=\emptyset \) by Claim. Then \(C=D'{\setminus } (O \cup P\cup S \cup T)\) is a VC of G by the same discuss above. Thus, \(|C|-|C^*|= (|D'|-2n-2\,m)-|C^*|\le |D|-(2n+2\,m+|C^*|)=|D|-|D^*|\). Therefore f is an L-reduction with \(\alpha =21\), \(\beta =1\). \(\square \)

In this section we study generalized series–parallel graphs, including outerplanar graphs and trees (Korneyenko 1994).

Generalized series–parallel (GSP) graphs with two distinct vertices u and v, called its left and right terminals respectively, and denoted by G(u; v), are defined recursively as follows.

(1) A graph with only two vertices, denoted as u and v, connected by an edge represents a fundamental GSP graph with u and v as its terminals, and

(2) For two GSP graphs \(G_{1}(u_{1}; v_{1})\) and \(G_{2}(u_{2}; v_{2})\), performing any of the three operations below results in a GSP graph:

Series-1 (\(S_{1}\)) composition: sticking \(v_{1}\) and \(u_{2}\) while designating \(u_{1}\) and \(v_{2}\) as the left and right terminals, respectively, of the newly generated graph;

Series-2 (\(S_{2}\)) composition: sticking \(v_{1}\) and \(u_{2}\) (still named as \(v_{1}\)) while designating \(u_{1}\) and \(v_{1}\) as the left and right terminals, respectively, of the newly generated graph;

Parallel (P) composition: sticking \(u_{1}\) and \(u_{2}\) (denoted by \(u_1\)), \(v_{1}\) and \(v_{2}\) (denoted by \(v_1\)), respectively, while designating \(u_{1}\) and \(v_{1}\) as the left and right terminals, respectively, of the newly generated graph. We assume that this composition will not result in multiple edges.

For any GSP graph G, we introduce the definition of a parse tree PT(G). A parse tree PT(G) is a rooted binary tree where every non-leaf vertex serves as a representation of a subgraph within G:

1. (1)

   Each leaf of PT(G) exactly represents an edge in G;

2. (2)

   Each internal node u has a label l(u) either \(S_{1}\), \(S_{2}\) or P, meaning that the subgraph represented by the node is constructed from two subgraphs represented by its two children by l(u) composition;

3. (3)

   The root represents G itself.

The structure of any GSP graph G can be succinctly represented by its PT(G). There are algorithms, with linear time complexity, to determine if a graph G is GSP and to construct a PT(G) (Eppstein 1992; Ho et al. 1999). And note that parse trees of a GSP graph are possibly not unique. An example for a GSP graph and one of its parse tree is shown in Fig. 7. In what follows, by using the structure of a parse tree, we will design an algorithm (linear-time) to solve the MTRD problem on GSP graphs.

First, for a GSP graph \(G=G(u;v)\) and \(i, j\in \{1,2,\ldots , 5\}\), we define a subset, denoted by \(T_{ij}(G)\), of vertex set V(G) such that:

1. (1)

   \(N(w)\cap T_{ij}(G)\ne \emptyset \) for every \(w\in V(G){\setminus }\{u,v\}\).

2. (2)

   \(N(w)\cap (V(G)\setminus T_{ij}(G))\ne \emptyset \) for every \(w\in V(G)\setminus (\{u,v\}\cup T_{ij}(G))\).

3. (3)

   \(u\in T_{1j}\) and \(N(u)\cap T_{1j}\ne \emptyset \). (resp. \(v\in T_{i1}\) and \(N(v)\cap T_{i1}\ne \emptyset \)).

4. (4)

   \(u\in T_{2j}\) and \(N(u)\cap T_{2j}=\emptyset \). (resp. \(v\in T_{i2}\) and \(N(v)\cap T_{i2}=\emptyset \)).

5. (5)

   \(u\notin T_{3j}\), \(N(u)\cap T_{3j}\ne \emptyset \) and \(N(u)\cap (V(G)\setminus T_{3j})\ne \emptyset \). (resp. \(v\notin T_{i3}\), \(N(v)\cap T_{i3}\ne \emptyset \) and \(N(v)\cap (V(G){\setminus } T_{i3})\ne \emptyset \)).

6. (6)

   \(u\notin T_{4j}\), \(N(u)\cap T_{4j}\ne \emptyset \) and \(N(u)\cap (V(G)\setminus T_{4j})=\emptyset \). (resp. \(v\notin T_{i4}\), \(N(v)\cap T_{i4}\ne \emptyset \) and \(N(v)\cap (V(G){\setminus } T_{i4})=\emptyset \)).

7. (7)

   \(u\notin T_{5j}\), \(N(u)\cap T_{5j}=\emptyset \) and \(N(u)\cap (V(G)\setminus T_{5j})\ne \emptyset \). (resp. \(v\notin T_{i5}\), \(N(v)\cap T_{i5}=\emptyset \) and \(N(v)\cap (V(G){\setminus } T_{i3})\ne \emptyset \)). Let \(MT_{ij}(G)\) be a minimum cardinality \(T_{ij}(G)\) and \(m_{ij}(G)=|MT_{ij}(G)|\). Note that \(T_{ij}(G)\) may not exist for some i, j. If \(T_{ij}(G)\) does not exist, denoted by \(\Omega \), where we let \(S \cup \Omega =\Omega \) for any set S, then \(m_{ij}(G)\) is infinity by convention. For example, \(m_{11}(K_{2})=2\), \(m_{24}(K_{2})=1\), \(m_{42}(K_{2})=1\), \(m_{55}(K_{2})=0\), \(m_{ij}(K_{2})=\infty \) otherwise.

Let \(F_1,\ldots ,F_s\) be subsets of V(G). Define \(\min \{F_t|1\le t\le s\}\) be a set which has minimum cardinality among all \(\{F_t|1\le t\le s\}\), that is, \(|\min \{F_t|1\le t\le s\}|=\min _{1\le t\le s}|F_t|\). Obviously, \(\min \{MT_{ij}(G)|i, j\in \{1,3\}\}\) is a minimum TRDS of G.

Lemma 3.1

Let \(G(u_{1}; v_{2})\) be a GSP graph obtained by applying a \(S_1\) composition to GSP graphs \(G_{1}(u_{1}; v_{1})\) and \(G_{2}(u_{2}; v_{2})\). Then, for \(i, j \in \{1, 2, 3, 4, 5\}\), the set

is a \(MT_{ij}(G)\). Here \(v_1\) in \(G_1\) and \(u_2\) in \(G_2\) are considered as the same vertex.

Proof

In the following, we just show the statement holds for the case of \(i=1\), \(j=1\), the other cases can be proven in similar ways. It is obvious that \(S_{11}\) defined in Lemma 3.1 is a \(T_{11}(G)\). So \(|S_{11}|\ge m_{11}(G)\). All that needs to be proven is that, for any chosen \(MT_{11}(G)\), \(|S_{11}|\le |MT_{11}(G)|\). We divide \(MT_{11}(G)\) into two cases to discuss.

Case 1. \(v_{1}\in MT_{11}(G)\).

Then by the definition of \(MT_{11}(G)\), either \(N_{G_1}(v_1)\cap MT_{11}(G)\ne \emptyset \) or \(N_{G_2}(u_{2})\cap MT_{11}(G)\ne \emptyset \), if, \(N_{G_1}(v_1)\cap MT_{11}(G)\ne \emptyset \). Obviously, \(G_{1}\cap MT_{11}(G)\) is a \(T_{11}(G_{1})\), further \(|MT_{11}(G_1)|\le |G_1\cap MT_{11}(G)|\). If \(N_{G_2}(u_{2})\cap MT_{11}(G)\ne \emptyset \), then \(G_{2}\cap MT_{11}(G)\) is a \(T_{11}(G_{2})\), further \(|MT_{11}(G_2)|\le |G_2\cap MT_{11}(G)|\). Thus \(|MT_{11}(G_{1})\cup MT_{11}(G_{2})|\le |(G_1\cap MT_{11}(G))\cup (G_2\cap MT_{11}(G))|=|MT_{11}(G)|\). Note that \(MT_{11}(G_{1})\cap MT_{11}(G_{2})=(G_1\cap MT_{11}(G))\cap (G_2\cap MT_{11}(G))\). If \(N_{G_2}(u_{2})\cap MT_{11}(G)=\emptyset \), then \(G_{2}\cap MT_{11}(G)\) is a \(T_{21}(G_{2})\) by the definition. Similarly, we have \(|MT_{11}(G_{1})\cup MT_{21}(G_{2})|\le |MT_{11}(G)|\). If \(N_{G_1}(v_1)\cap MT_{11}(G)=\emptyset \), then \(N_{G_2}(u_{2})\cap MT_{11}(G)\ne \emptyset \), we can get \(|MT_{12}(G_{1})\cup MT_{11}(G_{2})|\le |MT_{11}(G)|\).

Case 2. \(v_{1}\notin MT_{11}(G)\).

Subcase 2.1. \(N_{G_1}(v_1)\cap MT_{11}(G)\ne \emptyset \) and \(N_{G_1}(v_1)\setminus MT_{11}(G)\ne \emptyset \).

Obviously, \(G_{1}\cap MT_{11}(G)\) is a \(T_{13}(G_{1})\). If \(N_{G_2}(u_2)\cap MT_{11}(G)=\emptyset \), then \(G_{2}\cap MT_{11}(G)\) is a \(T_{51}(G_{2})\). Thus, \(|MT_{13}(G_{1})\cup MT_{51}(G_{2})|\le |MT_{11}(G)|\). Note that \(T_{13}(G_1)\cap T_{51}(G_2)=(G_{1}\cap MT_{11}(G))\cap (G_{2}\cap MT_{11}(G))=\emptyset \). If \(N_{G_2}(u_2)\subseteq MT_{11}(G)\), then \(G_{2}\cap MT_{11}(G)\) is a \(T_{41}(G_{2})\). Thus, \(|MT_{13}(G_{1})\cup MT_{41}(G_{2})|\le |MT_{13}(G)|\). Otherwise, \(G_{2}\cap MT_{11}(G)\) is a \(T_{31}(G_{2})\). Thus, \(|MT_{13}(G_{1})\cup MT_{31}(G_{2})|\le |MT_{13}(G)|\).

Subcase 2.2. \(N_{G_1}(v_1)\subseteq MT_{11}(G)\).

Obviously, \(G_{1}\cap MT_{11}(G)\) is a \(T_{14}(G_{1})\) and \(N_{G_2}(u_2)\setminus MT_{11}(G)\ne \emptyset \). If \(N_{G_2}(u_2)\cap MT_{11}(G)=\emptyset \), then \(G_{2}\cap MT_{11}(G)\) is a \(T_{51}(G_{2})\). Thus \(|MT_{14}(G_{1})\cup MT_{51}(G_{2})|\le |MT_{11}(G)|\). Otherwise, \(G_{2}\cap MT_{11}(G)\) is a \(T_{31}(G_{2})\). Thus \(|MT_{14}(G_{1})\cup MT_{31}(G_{2})|\le |MT_{11}(G)|\).

Subcase 2.3. \(N_{G_1}(v_1)\cap MT_{11}(G)=\emptyset \).

Obviously, \(G_{1}\cap MT_{11}(G)\) is a \(T_{15}(G_{1})\) and \(N_{G_2}(u_2)\cap MT_{11}(G)\ne \emptyset \). If \(N_{G_2}(u_2)\subseteq MT_{11}(G)\), then \(G_{2}\cap MT_{11}(G)\) is a \(T_{41}(G_{2})\). Thus \(|MT_{15}(G_{1})\cup MT_{41}(G_{2})|\le |MT_{11}(G)|\). Otherwise, \(G_{2}\cap MT_{11}(G)\) is a \(T_{31}(G_{2})\). Thus \(|MT_{15}(G_{1})\cup MT_{31}(G_{2})|\le |MT_{11}(G)|\). \(\square \)

Lemma 3.2

Let \(G=G(u_{1}; v_{1})\) be a GSP graph obtained by applying a \(S_2\) composition to GSP graphs \(G_{1}(u_{1}; v_{1})\) and \(G_{2}(u_{2}; v_{2})\). Then, for any \(i \in \{1, 2, 3, 4, 5\}\), \(t\in \{1, 3\}\), \(q\in \{2,4,5\}\), the sets

         \(S_{iq} = \min \{MT_{iq}(G_{1}) \cup MT_{qt}(G_{2})\};\)

are \(MT_{i1}(G)\), \(MT_{iq}(G)\), \(MT_{i3}(G)\), respectively. Here \(v_1\) in \(G_1\) and \(u_2\) in \(G_2\) are considered as the same vertex.

Proof

In the following, we just show the statement holds for the case of \(i=1\), \(j=1\), and \(i=1\), \(j=3\), the other cases can be proven in similar ways. It is obvious that \(S_{11}\) and \(S_{13}\) defined in Lemma 3.2 is a \(T_{11}(G)\), \(T_{13}(G)\), respectively. So \(|S_{11}|\ge m_{11}(G)\), \(|S_{13}|\ge m_{13}(G)\). All that needs to be proven is that, for any chosen \(MT_{11}(G)\), \(|S_{11}|\le |MT_{11}(G)|\), for any chosen \(MT_{13}(G)\), \(|S_{13}|\le |MT_{13}(G)|\). First we show that \(|S_{11}|\le |MT_{11}(G)|\). By the definition of \(MT_{11}(G)\), \(v_{1}\in MT_{11}(G)\), and either \(N_{G_1}(v_{1})\cap MT_{11}(G)\ne \emptyset \) or \(N_{G_2}(u_2)\cap MT_{11}(G)\ne \emptyset \). We divide \(MT_{11}(G)\) into two cases to discuss.

Case 1. \(v_{2}\in MT_{11}(G)\).

Subcase 1.1. \(N_{G_1}(v_{1})\cap MT_{11}(G)\ne \emptyset \).

If \(N_{G_2}(u_2)\cap MT_{11}(G)\ne \emptyset \), obviously, \(G_{1}\cap MT_{11}(G)\) is a \(T_{11}(G_{1})\), \(G_{2}\cap MT_{11}(G)\) is a \(T_{11}(G_{2})\). Thus \(|MT_{11}(G_{1})\cup MT_{11}(G_{2})|\le |MT_{11}(G)|\). Otherwise, \(G_{1}\cap MT_{11}(G)\) is a \(T_{11}(G_{1})\), \(G_{2}\cap MT_{11}(G)\) is a \(T_{21}(G_{2})\). Thus \(|MT_{11}(G_{1})\cup MT_{21}(G_{2})|\le |MT_{11}(G)|\).

Subcase 1.2. \(N_{G_1}(v_{1})\cap MT_{11}(G)=\emptyset \).

By the definition of \(MT_{11}(G)\), \(N_{G_2}(u_2)\cap MT_{11}(G)\ne \emptyset \), then \(G_{1}\cap MT_{11}(G)\) is a \(T_{12}(G_{1})\), \(G_{2}\cap MT_{11}(G)\) is a \(T_{11}(G_{2})\). Thus \(|MT_{12}(G_{1})\cup MT_{11}(G_{2})|\le |MT_{11}(G)|\).

Case 2. \(v_{2}\notin MT_{11}(G)\).

The proof is similar to Case 1, thus we have \(|MT_{11}(G_{1})\cup MT_{13}(G_{2})|\le |MT_{11}(G)|\), \(|MT_{11}(G_{1})\cup MT_{23}(G_{2})|\le |MT_{11}(G)|\), \(|MT_{12}(G_{1})\cup MT_{13}(G_{2})|\le |MT_{11}(G)|\).

Next we show that \(|S_{13}|\le |MT_{13}(G)|\). By the definition of \(MT_{13}(G)\), \(v_{1}\notin MT_{13}(G)\), and either \(N_{G_1}(v_{1})\cap MT_{13}(G)\ne \emptyset \) or \(N_{G_2}(u_2)\cap MT_{13}(G)\ne \emptyset \). Next, we divide \(MT_{13}(G)\) into two cases to discuss.

Case 1. \(v_{2}\in MT_{13}(G)\).

Subcase 1.1. \(N_{G_1}(v_{1})\cap MT_{13}(G)\ne \emptyset \).

Subsubcase 1.1.1. \(N_{G_2}(u_2)\cap MT_{13}(G)\ne \emptyset \).

If \(N_{G_1}(v_{1})\setminus MT_{13}(G)\ne \emptyset \) and \(N_{G_2}(u_2)\setminus MT_{13}(G)\ne \emptyset \). Obviously, \(G_{1}\cap MT_{13}(G)\) is a \(T_{13}(G_{1})\), \(G_{2}\cap MT_{13}(G)\) is a \(T_{31}(G_{2})\). Thus \(|MT_{13}(G_{1})\cup MT_{31}(G_{2})|\le |MT_{13}(G)|\). If \(N_{G_1}(v_{1}){\setminus } MT_{13}(G)\ne \emptyset \) and \(N_{G_2}(u_2)\setminus MT_{13}(G)=\emptyset \). Obviously, \(G_{1}\cap MT_{13}(G)\) is a \(T_{13}(G_{1})\), \(G_{2}\cap MT_{13}(G)\) is a \(T_{41}(G_{2})\). Thus \(|MT_{13}(G_{1})\cup MT_{41}(G_{2})|\le |MT_{13}(G)|\). If \(N_{G_1}(v_{1}){\setminus } MT_{13}(G)=\emptyset \) and \(N_{G_2}(u_2)\setminus MT_{13}(G)\ne \emptyset \). Obviously, \(G_{1}\cap MT_{13}(G)\) is a \(T_{14}(G_{1})\), \(G_{2}\cap MT_{13}(G)\) is a \(T_{31}(G_{2})\). Thus \(|MT_{14}(G_{1})\cup MT_{31}(G_{2})|\le |MT_{13}(G)|\).

Subsubcase 1.1.2. \(N_{G_2}(u_2)\cap MT_{13}(G)=\emptyset \).

If \(N_{G_1}(v_{1})\setminus MT_{13}(G)\ne \emptyset \), Obviously, \(G_{1}\cap MT_{13}(G)\) is a \(T_{13}(G_{1})\), \(G_{2}\cap MT_{13}(G)\) is a \(T_{51}(G_{2})\). Thus \(|MT_{13}(G_{1})\cup MT_{51}(G_{2})|\le |MT_{13}(G)|\). Otherwise, \(G_{1}\cap MT_{13}(G)\) is a \(T_{14}(G_{1})\), \(G_{2}\cap MT_{13}(G)\) is a \(T_{51}(G_{2})\). Thus \(|MT_{14}(G_{1})\cup MT_{51}(G_{2})|\le |MT_{13}(G)|\).

Subcase 1.2. \(N_{G_1}(v_{1})\cap MT_{13}(G)=\emptyset \).

By the definition of \(MT_{13}(G)\), \(N_{G_2}(u_2)\cap MT_{13}(G)\ne \emptyset \). If \(N_{G_2}(u_2){\setminus } MT_{13}(G)\ne \emptyset \). Obviously, \(G_{1}\cap MT_{13}(G)\) is a \(T_{15}(G_{1})\), \(G_{2}\cap MT_{13}(G)\) is a \(T_{31}(G_{2})\). Thus \(|MT_{15}(G_{1})\cup MT_{31}(G_{2})|\le |MT_{13}(G)|\). If \(N_{G_2}(u_2)\setminus MT_{13}(G)=\emptyset \). Obviously, \(G_{1}\cap MT_{13}(G)\) is a \(T_{15}(G_{1})\), \(G_{2}\cap MT_{13}(G)\) is a \(T_{41}(G_{2})\). Thus \(|MT_{15}(G_{1})\cup MT_{41}(G_{2})|\le |MT_{13}(G)|\).

Case 2. \(v_{2}\notin MT_{13}(G)\).

The proof is similar to Case 1, thus we have \(|MT_{13}(G_{1})\cup MT_{33}(G_{2})|\) \(\le |MT_{13}(G)|\), \(|MT_{13}(G_{1})\) \(\cup MT_{43}(G_{2})|\le |MT_{13}(G)|\), \(|MT_{14}(G_{1})\cup MT_{33}(G_{2})|\le |MT_{13}(G)|\), \(|MT_{13}(G_{1})\cup MT_{53}(G_{2})|\le |MT_{13}(G)|\), \(|MT_{14}(G_{1})\cup MT_{53}(G_{2})|\le |MT_{13}(G)|\), \(|MT_{15}(G_{1})\cup MT_{33}(G_{2})|\le |MT_{13}(G)|\), \(|MT_{15}(G_{1})\cup MT_{43}(G_{2})|\le |MT_{13}(G)|\). \(\square \)

Lemma 3.3

Let \(G(u_{1}; v_{2})\) be a GSP graph obtained by applying a P composition to GSP graphs \(G_{1}(u_{1}; v_{1})\) and \(G_{2}(u_{2}; v_{2})\). Then, for any \(i, j \in \{2, 4, 5\}\), the sets

\(S_{11} =\min \{MT_{sx}(G_{1}) \cup MT_{ty}(G_{2})| s, t,x,y\in \{1, 2\}, st\ne 4,xy\ne 4\};\)

\(S_{1j} =\min \{MT_{sj}(G_{1}) \cup MT_{tj}(G_{2})| s, t\in \{1, 2\}, st\ne 4\};\)

\(S_{13} =\min \{MT_{sp}(G_{1}) \cup MT_{tq}(G_{2})|s, t\in \{1,2\}, p, q\in \{3,4,5\}, st\ne 4, pq\ne 16, 25\};\)

\(S_{i1}=\min \{MT_{is}(G_{1}) \cup MT_{it}(G_{2})| s, t\in \{1, 2\}, st\ne 4\};\)

\(S_{ij} =MT_{ij}(G_{1}) \cup MT_{ij}(G_{2});\)

\(S_{i3} =\min \{MT_{ip}(G_{1}) \cup MT_{iq}(G_{2})| p, q\in \{3,4,5\}, pq\ne 16, 25\};\)

\(S_{31} =\min \{MT_{ps}(G_{1}) \cup MT_{qt}(G_{2})|s, t\in \{1,2\}, p, q\in \{3,4,5\}, st\ne 4, pq\ne 16, 25\};\)

\(S_{3j} =\min \{MT_{pj}(G_{1}) \cup MT_{qj}(G_{2})| p, q\in \{3,4,5\}, pq\ne 16, 25\};\)

\(S_{33} =\min \{MT_{pm}(G_{1}) \cup MT_{qn}(G_{2})|p,q,m,n\in \{3,4,5\}, pq, mn\ne 16, 25\}.\)

are \(MT_{11}(G)\), \(MT_{1j}(G)\), \(MT_{13}(G)\), \(MT_{i1}(G)\), \(MT_{ij}(G)\),\( MT_{i3}(G)\), \(MT_{31}(G)\), \(MT_{3j}(G)\), \(MT_{33}(G)\), respectively. Here \(u_1\) in \(G_1\) and \(u_2\) in \(G_2\) are considered as the same vertex, \(v_1\) in \(G_1\) and \(v_2\) in \(G_2\) are considered as the same vertex.

Based on Lemmas 3.1-3.3, an algorithm for finding a minimum TRDS of GSP graphs is designed by using dynamic programming technique.

Theorem 3.1

Algorithm 2 produces a minimum \(\textrm{TRDS}\) of a GSP graph in linear-time.

Proof

The correctness of Algorithm 2 comes directly from Lemmas 3.1–3.3. Next, we analyze the time complexity. PT(G) can be constructed in linear-time in Eppstein (1992); Ho et al. (1999); Hopcroft and Tarjan (1973). And since PT(G) has O(|E|) leaves, the initialization of lines 1–3 of Algorithm 2 takes O(|E|) time. The iteration in lines 4–15 of Algorithm 2 is O(|E|), because PT(G) is a binary tree with O(|E|) nodes and the calculation of each iteration in lines 4–15 requires a constant time. Therefore, the time complexity of Algorithm 2 is linear-time. \(\square \)