NP-Completeness and APX-Completeness of Restrained Domination in Graphs

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Theoretical Computer Science 448 (2012) 1–8

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Theoretical Computer Science

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NP-completeness and APX-completeness of restrained domination in graphs

Lei Chen a, Weiming Zeng a, Changhong Lu b,∗

a College of Information Engineering, Shanghai Maritime University, Shanghai 201306, China b Department of Mathematics, East China Normal University, Shanghai 200241, China

article info a b s t r a c t

Article history: Let G = (V , E) be a simple graph. A vertex set S ⊆ V is a restrained dominating set Received 21 October 2011 if every vertex not in S is adjacent to a vertex in S and to a vertex in V − S. In this Received in revised form 18 February 2012 paper, we investigate the NP-completeness of the restrained domination problem in planar Accepted 3 May 2012 graphs and split graphs. Meanwhile, it is proved that the restrained domination problem is Communicated by D.-Z. Du APX-complete for bounded-degree graphs. Crown Copyright © 2012 Published by Elsevier B.V. All rights reserved. Keywords: NP-complete APX-complete Approximation ratio Domination problems Restrained domination

1. Introduction

Let G = (V , E) be a simple graph. For a vertex v ∈ V , the neighborhood of v in G is defined as NG(v) = {u ∈ V | uv ∈ E}. The degree of v in G, denoted by dG(v), is defined as |NG(v)|. We use N(v) for NG(v) and d(v) for dG(v) if there is no ambiguity. All graphs in this paper are connected. Some other notations and terminology not introduced here can be found in [28]. Domination and its variations in graphs have been extensively studied [3,17,18]. A vertex set S ⊆ V is a restrained dominating set if every vertex not in S is adjacent to a vertex in S and to a vertex in V − S. The restrained domination number of G, denoted by γr (G), is the minimum cardinality of a restrained dominating set. The restrained domination problem is to determine the restrained domination number of any graph. The concept of restrained domination was introduced by Telle and Proskurowski [27] in 1997, albeit indirectly, as a vertex partitioning problem. The restrained domination has been widely studied in the past ten years, please see [5,6,12–16,19,29]. It has been proved that the restrained domination problem is NP-complete even for bipartite graphs and chordal graphs [5]. In terms of the complexity of the restrained domination problem in graphs, a linear-time algorithm has been proposed for the restrained domination problem in trees [5]. In this paper, we continue studying the complexity of the restrained domination problem in some special graphs and extend these studies by investigating the approximation hardness of the restrained domination problem in graphs. An approximation algorithm is a polynomial time algorithm that outputs a solution whose cost can be compared to the optimal one. The ratio (more precisely, the worst-case ratio over all input instances) between these two costs is called the approximation ratio. General references on approximation algorithms can be found in [2,21]. The paper is organized as follows. In Section2, we prove that the restrained domination problem is NP-complete for planar graphs and split graphs. In Section3, a lower bound and an upper bound on the approximation ratio for the restrained

∗ Corresponding author. Tel.: +86 21 54863276. E-mail address: [email protected] (C. Lu).

a ✟a ❍ 1 ✟ ❍ 2 se e ✟❍ s ❍✟ e ❍ ✟ b s b✟❍❍✟ s 1 ✟ ❍ 2 sf f ✟❍ s ❍✟ f H⇒ ❍ ✟ c s c ✟❍❍✟ s 1 ✟ ❍ 2 sg g ✟❍ s ❍✟ g ❍ ✟ d s d❍✟ s s s Fig. 1. A reduction from the vertex cover problem to the restrained domination problem in planar graph. domination problem in general graphs are provided. In Section4, it is proved that the restrained domination problem is APX-complete for bounded-degree graphs.

2. NP-completeness of restrained domination problem

A graph is called a planar graph if it can be embedded in the plane, i.e., it can be drawn on the plane in such a way that its edges intersect only at their vertices. A graph is chordal if every cycle of length greater than three has a chord, i.e. an edge jointing two nonconsecutive vertices in the cycle. A graph is split if its vertex set can be partitioned into a stable set and a clique. It is easy to know that split graphs is a subclass of chordal graphs. It has been proved that the restrained domination problem is NP-complete even for bipartite graphs and chordal graphs [5]. We continue investigating its NP-completeness in other special graphs. In detail, we prove that it is also NP-complete for planar graphs and split graphs, which in part extend the results in [5]. We need to employ a well-known NP-completeness result, called the vertex cover problem, which is defined as follows. The vertex cover problem is for a given nontrivial graph and a positive integer k to answer if there is a vertex set of size at most k such that each edge of the graph has at least one end vertex in this set. Theorem 1 ([8,23]). The vertex cover problem is NP-complete even for planar graphs. Theorem 2. The restrained domination problem is NP-complete for planar graphs. Proof. For a planar graph G = (V , E), a positive integer k, and an arbitrary subset S ⊆ V with |S| ≤ k, it is easy to verify in polynomial time whether S is a restrained dominating set of G. Hence, the restrained domination problem in planar graphs is in NP. We construct a reduction from the vertex cover problem to the restrained domination problem. Given a nontrivial planar = = { } = { } = { i i i } = graph G (V , E), where V v1, v2, . . . , vn and E e1, e2,..., em , let Ei e1, e2,..., em (i 1, 2). Construct a ′ = ′ ′ ′ = ∪ ∪ ′ = ∪ { i | = graph G (V , E ) with vertex set V V E1 E2, and edge set E E vjek i 1, 2 and vj is incident to ek in } ∪ { 1 2 | = } ′ G ej ej j 1, 2,..., m . Since G is a planar graph, it is easy to verify that G is also a planar graph. Fig.1 shows an example of the reduction. Next, we will show that G has a vertex cover of size at most k if and only if G′ has a restrained dominating set of size at most k. Let VC = {vi , vi , . . . , vi } be a vertex cover of G. Then it is obvious that VC is a restrained dominating set of ′ 1 2 k ′ G and its size is k. For the converse, let RD be a restrained dominating set of G with |RD| ≤ k. Suppose RD1 = RD ∩ V and E∗ = {e ∈ E | e1 ∈ RD or e2 ∈ RD}. Take one end vertex of each edge in E∗ to compose a vertex set V ∗. Obviously, ∗ ∗ ∗ |V | ≤ |E | ≤ |RD − RD1|. Let VC = RD1 ∪ V , then it is easy to verify that VC is a vertex cover of G. Furthermore, ∗ |VC| ≤ |RD1| + |V | ≤ |RD1| + |RD − RD1| ≤ |RD| ≤ k. Finally, one can construct G′ from G in polynomial time. By Theorem1, we conclude that the restrained domination problem is NP-complete for planar graphs. Theorem 3. The restrained domination problem is NP-complete for split graphs. Proof. We still construct a reduction from the vertex cover problem to the restrained domination problem. Given a = = { } = { } = { i i i } = nontrivial graph G (V , E), where V v1, v2, . . . , vn and E e1, e2,..., em , let Ei e1, e2,..., em (i 1, 2). ′ = ′ ′ ′ = ∪ ∪ ′ = { 1| Construct a graph G (V , E ) with vertex set V V E1 E2, and edge set E vjek vj is incident to ek in } ∪ { 1 2 | = } ∪ { | ∈ ∪ } ′ ∪ G ej ej j 1, 2,..., m xy x, y V E2 . Then G is a split graph, where the vertices in V E2 form a clique and vertices in E1 form an independent set. Fig.2 shows an example of the reduction. With the similar argument of Theorem2, G has a vertex cover of size at most k if and only if G′ has a restrained dominating set of size at most k. On the other hand, one can construct G′ from G in polynomial time. Therefore, we conclude that the restrained domination problem is NP-complete for split graphs by Theorem1. Remark. A particular case of the restrained domination, say total restrained domination, was proposed in 2005 by Ma and Chen [24]. A set S ⊆ V is a total restrained dominating set if every vertex is adjacent to a vertex in S and every vertex L. Chen et al. / Theoretical Computer Science 448 (2012) 1–8 3

a ✟a ❍ 1 ✟ ▲❏ ❍ 2 se e ❍✟ s▲❏ ✟❍ e ❍ ✟ ✡☞ b s b✟❍❍✟▲ ❏✡☞ s 1 ✟ ❏ ❍▲✡❏☞ 2 sf f ❍✟ s ❏✡ ▲☞✟❍❏ f H⇒ ❍ ✟ ✡ c s c ✟❍❍✟✡ ☞❏▲✡ s 1 ✟ ☞❍✡❏▲ 2 sg g ❍✟ s☞✡ ✟❍❏▲ g ❍ ✟ d s d❍✟☞✡ s s s Fig. 2. A reduction from the vertex cover problem to the restrained domination problem in split graph.

a a ✟✟❍❍ 1 ✟ ❍ 2 se e ❍ s ✟ e ❍❍✟✟ b s b✟❍ s 1 ✟ ❍ 2 sf f ✟ s ❍ f H⇒ ❍ ✟ ′ ❍❍✟✟ w w c s c ✟❍ s s s 1 ✟ ❍ 2 sg g ✟ s ❍ g ❍❍ ✟✟ d s d❍✟ s s s Fig. 3. A reduction from the vertex cover problem to the total restrained domination problem in planar graph.

a ✟a ❍ 1 ✟ ▲❏ ❍ 2 se e ✟ s ❍ e ❍❍ ▲❏✟✟✡☞ b b❍✟▲ ❏✡☞ s ✟❍ s 2 1 ✟ ❏ ❍▲✡❏☞ f sf f ✟❍ s ❏✡ ▲☞❍✟❏ H⇒ ❍ ✟ ✡ w w′ c c ❍✟✡ ☞❏▲✡ s ✟❍ s 2 s s 1 ✟ ☞❍✡❏▲ g sg g ✟ s ❍❏▲ ❍❍ ☞✡✟✟ d s d❍✟☞✡ s s s Fig. 4. A reduction from the vertex cover problem to the total restrained domination problem in split graph.

in V − S is adjacent to a vertex in V − S. The total restrained domination number of G, denoted by γtr (G), is the minimum cardinality of a total restrained dominating set. Recently, the theoretical properties of total restrained domination have been well studied [11,12,20,22,26,29]. For its complexity, it has been proved to be NP-complete for bipartite graphs and chordal graphs [24]. In here, we point out that the total restrained domination problem is also NP-complete for planar graphs and split graphs with the similar reduction in Theorems1 and2. The reduction is simply described as follows (the detail of the proofs is left to readers). (1) Reduction for planar graph. Given a nontrivial graph G = (V , E), where V = {v1, v2, . . . , vn} and E = {e1, e2,..., em}, = { i i i } = ′ = ′ ′ ′ = ∪ ∪ ∪ { ′} let Ei e1, e2,..., em (i 1, 2). Construct a graph G (V , E ) with vertex set V V E1 E2 w, w , and edge set ′ = { i | = } ∪ { 1 2 | = } ∪ { | ∈ } ∪ { ′} E vjek i 1, 2 and vj is incident to ek in G ej ej j 1, 2,..., m wv v V ww . Fig.3 shows an example of the reduction. (2) Reduction for split graph. Given a nontrivial graph G = (V , E), where V = {v1, v2, . . . , vn} and E = {e1, e2,..., em}, let = { i i i } = ′ = ′ ′ ′ = ∪ ∪ ∪ { ′} Ei e1, e2,..., em (i 1, 2). Construct a graph G (V , E ) with vertex set V V E1 E2 w, w , and edge set ′ = { 1| } ∪ { 1 2 | = } ∪ { | ∈ ∪ ∪ { }} ∪ { ′} E vjek vj is incident to ek in G ej ej j 1, 2,..., m xy x, y V E2 w ww . Fig.4 shows an example of the reduction.

3. Approximation ratio of restrained domination problem in general graphs

In this section, we investigate the approximation ratio of the restrained domination problem in general graphs. In detail, we give a lower bound on the ratio of any approximation algorithm for the restrained domination problem in Section 3.1, and an approximation algorithm and its approximation ratio are provided in Section 3.2. 4 L. Chen et al. / Theoretical Computer Science 448 (2012) 1–8

3.1. Lower bounds on approximation ratio

To obtain the lower bound, we need another problem, say the set cover problem. We formalize the considered problems in this section as follows.

MIN SET COVER Instance: Set system G = (U, S), where U is a universe and S is a collection of (nonempty) subsets of U such that ∪S = U. Solution: A set cover of G, i.e., a subcollection S′ ⊆ S such that ∪S′ = U. Measure: Cardinality of the set cover, i.e., |S′|.

MIN RE-DOM SET Instance: A simple graph G = (V , E). Solution: A restrained dominating set of G, i.e., a subset V ′ ⊆ V such that every vertex in V − V ′ is adjacent to a vertex in V ′ and to a vertex in V − V ′. Measure: Cardinality of the restrained dominating set, i.e., |V ′|.

The lower bound of the approximation ratio for MIN SET COVER has been investigated in [7]. Theorem 4 ([7]). If there is some ϵ > 0 such that a polynomial time algorithm can approximate MIN SET COVER within (1 − ϵ)lnn, then NP ⊆ DTIME(nO(loglog n)). Theorem 5. If there is some ϵ > 0 such that a polynomial time algorithm can approximate MIN RE-DOM SET within (1 − ϵ)lnn, then NP ⊆ DTIME(nO(loglog n)). Proof. Construct an approximation preserving reduction from MIN SET COVER to MIN RE-DOM SET as follows. Let G = (U, S), where U = {u1, u2,..., un} and S = {S1, S2,..., Sm}, be an instance of MIN SET COVER. Define Ui = { i i i } = = = ∪ ∪ = u1, u2,..., un (i 1, 2). Construct a instance of MIN RE-DOM SET G (V , E), where V U1 U2 S and E { k | ∈ = } ∪ { | ≤ ̸= ≤ } ∪ { 1 2 | = } ui Sj ui Sj, k 1, 2 SiSj 1 i j m ui ui i 1, 2,..., n . ′ = { } = ′ | | ≤ | ′| Let S Si1 , Si2 ,..., Sil be a set cover of G, then RD S is a restrained dominating set in G of size RD S . In particular, |RD∗| ≤ |S∗|, where RD∗ and S∗ are the optimal restrained dominating set of G and the optimal set cover of G, = ∩ ∗ = { ∈ | 1 ∈ respectively. For the converse, let RD be a restrained dominating set of G. Suppose RD1 RD S and U ui U ui RD 2 ∈ } ∈ ∗ ∈ ∈ ∗ ′ = ∪ ∗ or ui RD . For each uj U , take exactly one set Sk S such that uj Sk to compose a collection S . Let S RD1 S . Then S′ is a set cover of G and |S′| ≤ |RD|. Now, we assume that MIN RE-DOM SET can be approximated with ratio α by using an polynomial time algorithm A. We construct an algorithm A′ as follows:

Algorithm A′. Input: An instance of MIN SET COVER G = (U, S). 1. Construct a graph G = (V , E) as described in the first paragraph of the proof. 2. Compute a restrained dominating set RD of G using algorithm A. 3. Construct a subcollection S′ of G as described in the second paragraph of the proof. 4. Output S′.

Obviously, algorithm A′ runs in polynomial time as A is a polynomial time algorithm. Let RD∗ and S∗ be the optimal restrained dominating set of G and the optimal set cover of G, respectively. Algorithm A′ can compute a set cover of size |S′| ≤ |RD| ≤ α|RD∗| ≤ α|S∗|. Thus, algorithm A′ can approximate MIN SET COVER within a ratio α. If α ≤ (1 − ε) ln n, then O(loglog n) NP ⊆ DTIME(n ) according to Theorem4.

3.2. Upper bounds on approximation ratio

In this section, a greedy algorithm for finding a restrained dominating set in general graphs is provided and we prove the upper bound on the approximation ratio of this algorithm. Firstly, the greedy algorithm is described as follows.

Algorithm GREEDY-RE-DOM Input: A graph G = (V , E) 1. RD := ∅; := ′ := ∅ 2. i 0, Si ; − ′ ∪ · · · ∪ ′ ̸= ∅ 3. While (V (S0 Si ) ) do 4. i := i + 1; 5. For each vertex u ∈ V − RD, calculate N1(u) = {v ̸∈ RD | u is the only neighbor of v such that u ̸∈ RD} = [ ] − ′ ∪ ′ ∪ · · · ∪ ′ N2(u) N u (S0 S1 Si−1) | | D(u) = N2(u) |N1(u)|+1 L. Chen et al. / Theoretical Computer Science 448 (2012) 1–8 5

6. Choose a vertex w ∈ V − RD such that D(w) = maxu̸∈RD D(u) 7. Si := N1(w) ∪ {w}; ′ := [ ] − ′ ∪ · · · ∪ ′ 8. Si N w (S0 Si−1); 9. RD := RD ∪ Si; 10. Output RD. Lemma 6. Algorithm GREEDY-RE-DOM can provide a restrained dominating set in polynomial time. Proof. Since a restrained dominating set requires that a vertex not in the restrained dominating set must be adjacent to another vertex not in the restrained dominating set, we put the selected vertex and some of its neighbors such that their only neighbor not in RD is the selected vertex at each loop. In other words, vertices not in RD must have a neighbor such that it is not in RD after each loop. On the other hand, it is clear that RD is a dominating set. Therefore, the output set RD is a restrained dominating set. The number of loops in the algorithm must be bounded by |V |, which is polynomial. Furthermore, each loop is also polynomial. The lemma follows. ∈ ′ Lemma 7. For each vertex u V , there is exactly one set Si containing u. ′ Proof. By Lemma6, RD is a restrained dominating set. Thus u is contained in at least one set Si . Furthermore, when u is first ′ ̸∈ ′ ∈ ′ ∪ ′ ∪ · · · ∪ ′ contained in set Sk, then u Sj for j > k due to u S0 S1 Sj−1. Suppose that the number of loops in the algorithm GREEDY-RE-DOM is k for some input G = (V , E). By Lemma7, there | | ∈ { } ∈ ′ ∈ = Si is only one index i 1, 2,..., k such that u Si for each u V . Let du | ′| . Si Theorem 8. The MIN RE-DOM SET in any graph G = (V , E) of maximum degree ∆ can be approximated with an approximation ratio of (∆ + 1)(ln(∆ + 1) + 1).  a = Proof. Let k be the number of loops in the algorithm GREEDY-RE-DOM for some input. By the fact that x∈Y |Y | a and Lemma7, we have

k k    |Si|  |RD| = |Si| = = dw. |S′| i=1 i=1 ∈ ′ i w∈V w Si Let RD∗ be a minimum restrained dominating set of G. Then each vertex w ∈ V has at least a neighbor in RD∗ or is contained in RD∗, i.e. there is at least one vertex u ∈ RD∗ such that w ∈ N[u]. Hence, we obtain    dw ≤ dw. w∈V u∈RD∗ w∈N[u] ∈ ∗ = | [ ] − ′ ∪ ′ ∪ · · · ∪ ′ | = ≥ = Let u RD and zi N u (S0 S1 Si ) for i 1, 2,..., k. Clearly, zi−1 zi for i 1, 2,..., k. Suppose that l = ′ − [ ] is the smallest index such that zl 0. At the i-th loop of the algorithm, Si contains zi−1 zi vertices in N u . Thus, l   |Si| dw = (zi−1 − zi) . |S′| w∈N[u] i=1 i ′ |S | Since we select a vertex w such that D(w) = i is maximum at the i-th loop of the algorithm, we have |Si| ′ ′ ′ N2(u) |N[u] − (S ∪ S ∪ · · · ∪ S − )| zi−1 D(w) ≥ = 0 1 i 1 ≥ . |N1(u)| + 1 |N1(u)| + 1 ∆ + 1 It follows that  l ∆ + 1 dw ≤ (zi−1 − zi) . z − w∈N[u] i=1 i 1 − ≥ b−a ≥ = p 1 = By the fact that H(b) H(a) b for all integer b a, where H(p) i=1 i and H(0) 0, we have l l   zi−1 − zi  dw ≤ (∆ + 1) ≤ (∆ + 1) (H(zi−1) − H(zi)) = (∆ + 1)H(|N[u]|). z − w∈N[u] i=1 i 1 i=1 Since the maximum degree of G is ∆, |N[u]| ≤ ∆ + 1. It follows that      |RD| = dw ≤ dw ≤ (∆ + 1)H(|N[u]|) ≤ (∆ + 1) H(∆ + 1) w∈V u∈RD∗ w∈N[u] u∈RD∗ u∈RD∗ ∗ ∗ = (∆ + 1)|RD |H(∆ + 1) ≤ |RD |(∆ + 1)(ln(∆ + 1) + 1).

The last inequality is obtained from the fact H(p) ≤ ln(p) + 1 for any positive integer p. 6 L. Chen et al. / Theoretical Computer Science 448 (2012) 1–8

yv,2

y y y y y sv,1 sv,3 sv,4 sv,5 sv,6 ∗ Fig. 5. Fv in reduction from MIN DOM SET-3 to MIN RE-DOM SET-3.

4. APX-completeness of restrained domination problem in bounded-degree graphs

By Theorem8, the restrained domination problem can be approximated within a constant ratio if the degree of the graph is bounded by a constant. Thus the restrained domination problem in bounded-degree graphs is in APX. Furthermore, we show that the restrained domination problem is APX-complete, i.e., there is no PATS, even if the degree of the graph is bounded by three. Three is a critical number for the degree, because several NP-hard problems become polynomial time solvable for graphs with maximum degree two [9,10]; however, they are still NP-hard even if restricted to the graphs with degree at most three. Some variation domination problems have been proved that they are APX-complete for graphs with maximum degree at most three [1,4]. We first recall the notation of L-reduction [2,25]. Given two NP optimization problems F and G and a polynomial time transformation f from instances of F to instances of G, we say that f is an L-reduction if there are positive constants α and β such that for every instance x of F

1. optG(f (x)) ≤ α · optF (x); ′ 2. for every feasible solution y of f (x) with objective value mG(f (x), y) = c2 we can in polynomial time find a solution y of ′ x with mF (x, y ) = c1 such that |optF (x) − c1| ≤ β · |optG(f (x)) − c2|. To show that a problem P ∈ APX is APX-complete, it is enough to show that there is an L-reduction from some APX- complete problem to P . Many problems in graph theory have been proved to be APX-complete even if the degree of the graph is at most three, see [1,4]. We formalize the considered problems in this section as follows.

MIN DOM SET-B Instance: A graph G = (V , E) with degree at most B. Solution: A dominating set of G, i.e., a subset V ′ ⊂ V such that each vertex u ∈ V − V ′ has at least one neighbor in V ′. Measure: Cardinality of the dominating set, i.e., |V ′|.

MIN RE-DOM SET-B Instance: A graph G = (V , E) with degree at most B. Solution: A restrained dominating set of G, i.e., a subset V ′ ⊂ V such that every vertex in V − V ′ is adjacent to one vertex in V ′ and to a vertex in V − V ′. Measure: Cardinality of the restrained dominating set, i.e., |V ′|. Theorem 9 ([1]). MIN DOM SET-3 is APX-complete. Theorem 10. MIN RE-DOM SET-3 is APX-complete. Proof. By Theorem9, MIN DOM SET-3 is APX-complete. It is enough to construct an L-reduction f from MIN DOM SET-3 to MIN RE-DOM SET-3. Given a graph G = (V , E) with degree at most 3, construct a graph G′ with degree at most 3 in the ≤ ∗ following manner. For each vertex v in G with dG(v) 2, attach a Fv in Fig.5 at v by adding an edge vyv,1 . For each vertex v in G with dG(v) = 3, each vertex is split and transformed as shown in Fig.6. Let D be a dominating set of G. A vertex set RD can be constructed by the following manner: ∈ ≤ 1. If v D and dG(v) 2, then v, yv,3 , yv,6 are put into RD; ̸∈ ≤ 2. If v D and dG(v) 2, then yv,3 , yv,6 are put into RD; ∈ = 3. If v D and dG(v) 3, then v1, v2, yv,3 , yv,7 , yv,10 , yv,12 are put into RD; 4. If v ̸∈ D, then one of a, b, c must be in D. ̸∈ 4.1. If at least one of a, b is in D and c D, then yv,3 , yv,5 , yv,7 , yv,10 , yv,12 are put into RD; ̸∈ ∈ 4.2. If a, b D and c D, then yv,1 , yv,3 , yv,7 , yv,10 , yv,12 are put into RD; ∈ 4.3. If exactly one of a, b is in D and c D, then yv,1 , yv,3 , yv,7 , yv,10 , yv,12 are put into RD; ∈ 4.4. If a, b, c D, then yv,3 , yv,5 , yv,7 , yv,10 , yv,12 are put into RD. It is easy to check that RD is a restrained dominating set of G′ and |RD| = |D| + 2(|V (G)| − s) + 5s, where s is the number of vertices in G with degree three. In particular, |RD∗| ≤ |D∗| + 2(|V (G)| − s) + 5s, where RD∗ is a minimum restrained dominating set of G′ and D∗ is a minimum dominating set of G. Since G is a graph with degree at most three, |D∗| ≥ |V (G)|/4. Hence, |RD∗| ≤ |D∗| + 2|V (G)| + 3s ≤ |D∗| + 5|V (G)| ≤ |D∗| + 5 × 4|D∗| = 21|D∗|. L. Chen et al. / Theoretical Computer Science 448 (2012) 1–8 7 c

yv,5 v2 b b s ❅ ❅ ❅ s ❅ c H⇒ s ❅ s s ❅ v v1 yv,1❅ yv,4 yv,11 yv,12 s s s s ❅ s s s a a yv,2 yv,3❅ s s s s ❅ yv,6 yv,8 yv,9 yv,10 s s ss yv,7 s

Fig. 6. The transformation of a vertex with degree 3 in reduction from MIN DOM SET-3 to MIN RE-DOM SET-3.

′ ′ ′ = { Let RD be a restrained dominating set of G . We construct a vertex set D of G as follows. Let T (v) v, yv,1 , yv,2 , yv,3 , } ≤ = { } = = | ∩ ′| yv,4 , yv,5 , yv,6 for dG(v) 2 and T (v) v1, v2, yv,1 ,..., yv,12 for dG(v) 3. Calculate k(v) T (v) RD for each vertex v ∈ V (G). If dG(v) ≤ 2, v is put into D if and only if k(v) ≥ 3. If dG(v) = 3, v is put into D if and only if k(v) ≥ 6. We claim that D′ is a dominating set of G. To prove this claim, we give two obvious facts as follows.

Fact (1) If dG(v) ≤ 2 and v ∈ RD, then k(v) ≥ 3; (2) If dG(v) = 3 and one of v1 and v2 is in RD, then k(v) ≥ 6.

′ ′ Now we prove that D is a dominating set of G. Suppose that there is a vertex w ∈ V (G) such that NG[w] ∩ D = ∅. If ′ dG(w) ≤ 2, then k(w) ≤ 2. It is easy to obtain that w in G is not dominated by any vertex in T (w). Thus w is dominated by ∈ ′ ′ ≥ ′ ≤ ′ ≥ ′ = a neighbor other than yw,1 , say u, and let u T (u ). By the above fact, k(u ) 3 if dG(u ) 2 or k(u ) 6 if dG(u ) 3. By the rules of choosing vertices, u′ ∈ D′, which is a contradiction. With a similar argument, we can also obtain a contradiction ′ if dG(w) = 3. Thus D is a dominating set of G. On the other hand, it is easy to obtain that k(v) ≥ 2 if dG(v) ≤ 3 and ′ ′ ∗ ∗ k(v) ≥ 5 if dG(v) = 3. Thus |D | ≤ |RD | − 2(|V (G)| − s) − 5s. In particular, |D | ≤ |RD | − 2(|V (G)| − s) − 5s. Therefore, |D∗| = |RD∗| − 2(|V (G)| − s) − 5s. In addition, |D′| − |D∗| ≤ |RD′| − |RD∗|. As a result, f is an L-reduction with α = 21 and β = 1.

Acknowledgments

The first and the third authors are supported in part by the National Natural Science Foundation of China (No. 10971248) and the Fundamental Research Funds for the Central Universities; the second author is supported in part by the National Natural Science Foundation of China (No. 31170952) and the Program for Professor of Special Appointment (Eastern Scholar) at Shanghai Institutions of Higher Learning, Innovation Program of Shanghai Municipal Education Commission (No. 11ZZ143, No. 12YZ120). The first author is also supported in part by Science & Technology Program of Shanghai Maritime University (No. 20120105). The third author is also supported in part by the National Natural Science Foundation of China (No. 61073198) and the Program for New Century Excellent Talents in Universities.

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