Dominating Sets for Outerplanar Graphs

VAL PINCIU Department of Mathematics Southern Connecticut State University 501 Crescent, New Haven, CT 06515 UNITED STATES

Abstract: - We provide lower and upper bounds for the domination numbers and the connected domination numbers for outerplanar graphs. We also provide a recursive algorithm that finds a connected domination set for an outerplanar graph. Finally, we show that for outerplanar graphs where all bounded faces are 3-cycles, the problem of determining the connected domination number is equivalent to an art gallery problem, which is known to be NP-hard.

Key-Words: - dominating sets, star forests, outerplanar graphs, art gallery

1 Introduction numbers and the connected domination numbers In this paper all graphs are assumed to be have been found for special classes of graphs [2]. undirected and simple. A graph is outerplanar if it Planar graphs have been studied in [5] for example. can be embedded in a plane such that all vertices lie In this paper we will consider domination problems on the outer face. Given a graph G=(V,E), and v in for outerplanar graphs. Studies on outerplanar V, the open neighborhood of v, denoted by N(v), is graphs usually focus on simple graphs that are the set of all vertices adjacent to v, and the closed nonseparable (removing one vertex will leave a neighborhood of v, denoted by N[v], is the union connected graph). An outerplanar graph is between N(v) and {v}. A set of vertices D is a nonseparable if and only if they have a plane dominating set, if for every u in V, there exits v in D representation with a hamiltonian face. For such that u is in N[v]. The domination number of a nonseparable outerplanar graphs the hamiltonian graph G, denoted by (G), is the minimum size of a face is unique [7]. Such graphs have no degree 1 dominating set of vertices in G. vertices. A set of vertices D is a connecting dominating set if D is a dominating set, and the subgraph induced by D on G is connected. The connected domination number of a graph G, denoted by c(G), is the minimum size of a dominating set of vertices in G. Obviously, (G) c(G). A graph is a star if it isomorphic to K1,n for some nonnegative integer n. The center of a star is the unique vertex of maximum degree. (if n = 1 either vertex can be its center) A star forest of a Fig. 1: A graph with = c = 1. graph G, is a spanning subgraph of G in which each component is a star. The size of a star forest is the size of its edge set, which is the number of vertices in G minus the number of components in the star forest. The largest size of a star forest is denoted by (G). The set of the centers of a star forest is a dominating set. Also, for every dominating set D, one can construct a star forest by joining each vertex v in D, with the vertices in N(v). (see [1]) Therefore: (G) = |V| − (G) (1) The problem of determining the domination number for a graph is a well known NP-hard problem. Over the years, bounds for the domination Fig. 2: A graph with = c = |V| / 2. If the graphs have vertices of degree 1, the bounds exists a nonseparable outerplanar graph with n one can obtain for outerplanar graphs are no better vertices and m edges such that: than the bounds we can obtain for all graphs. The »2V − E ÿ γ = graphs like the one in Fig. 1 can have = c = 1 and (G) … Ÿ . 3 the graphs like the one in Fig 2 can have = c = … Ÿ |V| / 2 respectively. To construct such a graph, start with an n-cycle, and Therefore, throughout this paper we will assume connect the first vertex with vertices 3, 4, … , and that G is a nonseparable outerplanar graph. In m − n + 2. sections 2 and 3 we give bounds for the domination numbers and connected domination numbers, respectively. Bounds for outerplanar graphs with 3 Connected Dominating Sets small diameter can be found in [5], but no general As in the previous section we consider a bounds for outerplanar graphs have been found nonseparable outerplanar graph G = (V,E). We also before. In section 4 we provide a recursive algorithm consider its unique planar representation that has a that finds a connected domination set for a given hamiltonian face. The hamiltonian face will be the outerplanar graph. The algorithm can be adapted in outer or unbounded face, while the other faces are the case the graph has degree 1 vertices. Finally, in the inner or bounded faces. We start with two easy section 5, we show that in the case of outerplanar to prove lemmas and a proposition that will help us graphs with a maximum number of edges (every obtain bounds for the connected domination bounded face is a triangle), the problem of numbers, as well obtain the algorithm described in determining a connected domination set is section 4. equivalent to an art gallery problem, which is known Lemma 1 The complement of any connected to be NP-hard. domination set in an n-cycle contains no more than 2 vertices. If there are 2 vertices in the complement, then they must be adjacent. 2 Dominating Sets Lemma 2 If D is a connected domination set for In this section we consider a nonseparable an outerplanar graph G, then every edge of G that outerplanar graph G = (V,E). Since G is outerplanar, separates two inner faces in the planar representation it contains a simple cycle of length |V|, so (G) of G must have at least one endpoint in D. (|V| + 2)/ 3. This bound is tight, since n-cycles are Proposition 1 A set of vertices D is a connected outerplanar graphs. domination set for a nonseparable outerplanar graph The following theorem provides also a lower G if and only if the following two conditions are bound for (G). true: Theorem 1 If G = (V,E) is a nonseparable i. For every bounded face F of the graph G, outerplanar graph, then: D P V(F) is a domination set for the cycle that »2V − E ÿ » V ÿ bounds F. … Ÿ ≤ γ (G) ≤ … Ÿ (2) ii. Every edge of G that separates two inner … 3 Ÿ … 3 Ÿ faces in the planar representation of G has at Proof: The upper bound follows from the previous least one endpoint in D. remark. For the lower bound, let's consider a star The proof of Proposition 1 follows easily from the forest of maximum size that spans G. The star two lemmas. This proposition is the key ingredient forests has (G) connected components, and (G) in the algorithm described in section 4. edges. If di represent the number of edges in each Next, we will obtain bounds for the connected connected component of the star forest, then at least domination numbers, as we did in the previous di - 2 of them separate bounded faces in the unique section for the domination numbers. Since G is planar representation of G that has a hamiltonian outerplanar, it contains a simple cycle of length |V|, face. Since there are |V| edges on the hamiltonian so c(G) |V| − 2. This bound is tight, since n-cycles face, we obtain ( di − 2) |E| − |V|, or (G) − 2 (G) are outerplanar graphs. The following theorem |E| − |V|. By combining this inequality with (1), we provides also a lower bound for c(G). obtain the lower bound in (2). Theorem 2 If G = (V,E) is a nonseparable outerplanar graph then: Remark: The lower bound provided in the previous 3V − 2 E − 2 ≤ γ (G) ≤ V − 2 (3) theorem is tight. Indeed, for any integer n 3, and c for any integer m, such that n m 2n − 3, there Proof: The upper bound follows from the previous remark. In order to prove the lower bound, first let's note that if e is an edge of G that does not bound the bound from (3) cannot be tight. The graph in Figure outer face of G, then it naturally splits the graph G 3 is the largest graph on 10 vertices that has a into two outerplanar subgraphs H and K. These two domination set with of size 3|V| −2|E| - 2. subgraphs share two vertices and one edge. (the endpoints of e and e respectively). Since a connected domination set for G induces connected domination 4 An Algorithm sets for H and K, and H and K share only two In this section we provide a recursive algorithm vertices, we obtain: that finds minimum size dominating sets that contain γ ≥ γ + γ − c (G) c (H ) c (K) 2 (4) a given subset of vertices for a connected Since the connected domination number of a 3-cycle outerplanar graph with all vertices of degree 2 or is 1, if either H or K is a 3-cycle, inequality (4) can higher. Later we will show how one can use this be replaced by: algorithm to find a connecting dominating set for γ (G) ≥ γ (H ) + γ (K) −1 (5) any connected outerplanar graph, including those c c c that have degree 1 vertices. Using all edges of G that do not bound the outer face Here is the algorithm: of G, we can split G into k cycles. Let k be the n Algorithm number of n-cycles in this natural splitting of G. Input: a connected outerplanar graph G = (V,E), Using inequalities (4) or (5) k − 1 times we obtain: with all vertices of degree 2 or more, and S, a subset γ ≥ γ − − + c (G) ƒ kn c (Cn ) k3 2ƒ kn 2 (6) of V. n≥3 n≥4 Output: a minimum size dominating set D = D(G,S) which can be simplified as: for the graph G that contains S. γ ≥ + + + + c (G) 2 k5 2k6 3k7 .... (7) a. If G has only one bounded face, then G must be a Since every n-cycle has n edges, and every edge cycle of length n 3. If S = V, then let D = V. If S bounds two inner faces, or an inner face and the V, then let D be the complement of the set of two outer face, we obtain: adjacent vertices that are not in S, if such vertices V + 3k + 4k + 5k + ... = 2 E (8) exist. Otherwise, let D be the complement of the set 3 4 5 of one vertex that is not in S. Using Euler's Formula for planar graphs we obtain: b. Find a bounded face F of the graph G that − + + + + + = V E 1 k3 k4 k5 ... 2 (9) corresponds to a leaf in weakly dual tree of G. Since Combining equations (7), (8), and (9), we obtain: the graph is outerplanar, the weakly dual graph of G γ (G) ≥ 2 + k + 2k + 3k + .... = is a tree, so it has at least two leaves. Let uv be the c 5 6 7 edge that separates this bounded face from another + + + + + − + + = 2 k3 (3k3 4k4 5k5 ...) 4(k3 k4 ...) bounded face of G. (uv is unique since the bounded 2 + k + 2 E − V − 4( E − V + 1) = face corresponds to a leaf) 3 c. Let H be the sugraph of G induced by erasing all − − + 3V 2 E 2 k3 the vertices of F except u and v. d. If F is a triangle and either u or v are in S, then let which concludes our proof since k3 is nonnegative. D be the output of the D(H, S P V(H)) S S . e. If F is a triangle, but neither u, nor v are in S, then D is either D(H, (S P V(H)) S {u}) S S or D(H, (S P V(H)) S {v}) S S whichever has a smaller size. Remark: It is easy to see that if u has degree 3 for example, then |D(H, (S P V(H)) S {v})| |D(H, (S P V(H)) S {u})| , so we don't need to test the algorithm for both u and v when one of them has degree 3. f. If F contains two adjacent vertices w1 and w2 that Fig 3: A graph with |V| = 10, |E| = 12, and c = 4. are not in S S {u, v}, then let replace S with S S (V(F) − {w1, w2}) and go to step l. Remark: It can be shown that the lower bound in (3) g. If vertices w1, u, v and w2 occur in this order in the is tight for outerplanar graphs with a small number cycle F, and none of them is in S, then let S1 = S S of edges relative to the number of vertices. (V(F) − {u, w1}), and let S2 = S S (V(F) − {v, w2}) Obviously, for a large number of edges, there must and go to step m. be 3-cycles in the natural splitting of G, so the lower h. If there exist a vertex w in V − (S S {u, v}), then let replace S with S S (V(F) − {w}) and go to step l. vertices has a connected vertex guard set of size no i. If u is in S, but v is not, or v is in S, and u is not in more than (n - 2)/2. S, then let D be D(H, S P V(H)) S S. Proposition 2 Let T be a triangulation graph. j. If u and v are both in S, then go to step l. Then a set G of vertices is a connected vertex guard k. Let S1 = S S (V(F) − {u}), S2 = S S (V(F) − {v}) set for T, if and only if G is a connected domination and go to step m. set for T. l. Let D be D(H, S P V(H)) S S. Proof: G is a vertex guard set for T, if and only if it m. Let D be either D(H, S1 P V(H)) S S1 or D(H, S2 satisfies condition i from Proposition 1. Since it is a P V(H)) S S2, whichever has a smaller size. connected guard set, it should also satisfy condition Remark: It is easy to see that if u has degree 3 for ii from Proposition 1. Otherwise, if there was an example, then |D(H, S2 P V(H))| |D(H, S1 P V(H))|, edge e with no endpoints in G, vertices u and v in so we don't need to test the algorithm for both u and the two triangles bounded by e would have to be in v when one of them has degree 3. G, since G is a guard set, but would not be We can use the algorithm described above to connected. So G has to be a connected domination find a connected dominating set for a connected set for T. The converse follows easily from outerplanar graph with no degree 1 vertices by Proposition 1. initializing S as the empty set. If G has vertices of degree 1, for each vertex v of degree 1, let v* the Corollary 1 If G is a triangulation graph of a unique vertex in the longest simple cycle of G that is polygon with n vertices, then: closest to v. If V be the set of all degree 1 vertices, − 1 γ ≤ …n 2Ÿ and V the set vertices that are between v and v* for c (G) (10) 2 … 2 ⁄Ÿ some vertex v of degree 1, apply the algorithm described above for the subgraph H of G induced on Corollary 2 The problem of finding a connected domination set for an outerplanar graph is V(G) − (V1 S V2). For the initial set S use {v* : vE NP-hard. V1}. Then D(H,S) S V2 is a minimum size connected domination set for G. This result is stronger than a previously known result, that the problem is NP-hard for planar graphs.

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