On the Connectivity of a Graph and its Complement Graph

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1 On the Connectivity of a Graph and its Complement Graph Jia De Lin 1, Yue Li Wang 1, Jou Ming Chang 2 and Hung Chang Chan 3 1 Department of Information Management, National Taiwan University of Science and Technology, Taipei, Taiwan, R.O.C. 2 Department of Information Management, National Taipei College of Business, Taipei, Taiwan, R.O.C 3 Department of Computer Science and Information Engineering, National Taiwan University of Science and Technology, Taipei, Taiwan, R.O.C. Abstract In this paper, we study the connectivity of the complement of a graph. We first establish a simple formula to compute the connectivity of a graph that comes from the join of graphs. Base on this formula, we give an upper and a lower bound to the connectivity of the complement of a graph. Keyword: connectivity, complement, join, vertex of attachment. 1 Introduction All graphs considered in this paper are finite and simple (i.e., without loops and multiple edges). We denote the vertex set and edge set of a graph G by V (G) and E(G), respectively. The order of G is defined to be n(g) = V (G). The complement of G is the garph denoted by G such that V (G) = V (G) and E(G) = {uv : uv / E(G)}. For a subset S V (G), the subgraph of G induced by S is denoted by G[S]. Also, we use G S to denote the graph G[V \ S] where V \ S is the set {v V v / S}. Similarly, for F E(G) we denote the graph with vertex set V (G) and edge set E(G) \ F by G F. If S or F is a singleton x, we use the notation G x instead of G {x}. If H is a subgraph of G, then for simplicity, we sometimes write H to mean V (H). A graph G is connected if there is at least one path between any two vertices of G, otherwise G is disconnected. The vertex-connectivity (or connectivity for simplicity) of a graph G, denoted by κ(g), is the minimum number of vertices whose removal from G results in a disconnected or trivial graph, i.e., a graph has only one vertex. Thus, κ(k n ) = n 1, where K n is a complete graph with n vertices. A graph G is k-connected if κ(g) = k, or equivalently, it has at least k + 1 vertices and for every pair of nonadjacent vertices u and v there are at least k internally-vertex-disjoint paths joining u and v in G. In [1], Ando defined that a graph G is critical connected if κ(g v) = κ(g) 1 for each vertex v V (G) and (k, k )-critical connected if κ(g v) = k 1 or κ(g v) = k 1 for each vertex v V (G), where k = κ(g) and k = κ(g). As a result, Ando obtained an upper bound of the number of vertices for (k, k )- critical connected graphs using the terms k and k. Ando and Kaneko [2] defined that a graph G is bi-3-connected if both G and G are 3-connected. As a consequence, they proved that every bi-3- connected graph of order at least 22 has a twovertex set called bi-contractible pair whose contraction results still in a bi-3-connected graph. Recently, Ito and Yokoyama [4] showed that κ(g) can be computed in linear time if κ(g) 3. In this paper, we shall propose an upper bound and a lower bound of κ(g) related to κ(g). The rest of this paper is organized as follows. In Section 2, we give some definitions and notation. Furthermore, some prior results are introduced. In Section 3, we discuss the connectivity of a graph which is obtained from the join of graphs. In Section 4, we give an upper and a lower bound to the connectivity of the complement of a graph. Finally, some concluding remarks are given in Section

2 2 Preliminaries A subgraph H of a graph G is called a component if H is a maximal connected subgraph of G. A cutset of G is a subset of vertices whose removal either increases the number of components or increases the number of isolated vertices in G. A cutset with the minimum size is called a minimum cutset. Using this modification of the standard definition permits one to compute κ(g) by finding the size of a minimum cutset for any connected graph G. Specifically, a cutset with only one vertex is called a cut vertex. Let G 1 and G 2 be two vertex-disjoint graphs. Then the join, G 1 G 2, of G 1 and G 2 is the graph with vertex set V (G 1 ) V (G 2 ) and edge set E(G 1 ) E(G 2 ) {uv : u V (G 1 ) and v V (G 2 )}. Obviously, it follows from the definition that G 1 G 2 is connected. For convenience, we write k G i = G 1 G 2 G k. We use an example to depict the concept of the join of two graphs. Figure 1(a) shows that two separated graphs G 1 and G 2, and Figure 1(b) shows the graph G 1 G 2. Note that both G 1 and G 2 are subgraphs of G 1 G 2 induced by V (G 1 ) and V (G 2 ), respectively. a variant pertaining to the connectivity of graphs. Let G be a graph and v V (G). The vertex i- connectivity of v, denoted by κ i (v, G), is the cardinality of a minimum vertex set T V (G) whose removal results in a subgraph containing v as a cut vertex. In particular, κ i (v, G) = 0 if v is a cut vertex of G, and κ i (v, G) = n(g) 1 if T does not exist. As a result, Boland et al. provided a way to find the vertex i-connectivity of the join of two graphs. Theorem 1 [3] Let G and H be two arbitrary connected graphs and v V (G). Then κ i (v, G H) = min{κ i (v, G)+n(H), κ(h)+n(g) 1}. For example, Figure 2 shows a graph C 4 K 2, where C 4 contains a vertex v. From Theorem 1, we are easy to verify that κ i (v, C 4 ) = 1, n(k 2 ) = 2, κ(k 2 ) = 1, and n(c 4 ) = 4. Thus κ i (v, C 4 K 2 ) = 3. In this figure, the part of shadowed vertices is the minimum vertex set T such that v becomes a cut vertex if T is removed. C 4 K 2 G 1 v G 2 G 1 G 2 Figure 2: κ i (v, C 4 K 2 ) = 3 where v V (G). (a) (b) Figure 1: Two graphs and their join. For a subgrarph H of G, the attachment of H in G, denoted by A G (H), is the maximal subset of V (H) such that every vertex in the set is adjacent to at least one vertex of G H. Thus, A G (G) = A G ( ) = 0. For two nonempty subsets S 1, S 2 V (G), the set containing all edges of G with one end vertex in S 1 and the other end vertex in S 2 is denoted by E G [S 1, S 2 ]. When no ambiguity arises, we drop the index G in the above notations. Boland et al. in [3] studied the inclusive connectivity (i-connectivity for short) of graphs, which is 3 The connectivity of the join of graphs In this section, we discuss the connectivity of the join of graphs. The following lemma is similar to Theorem 1 that gives a formula to compute the connectivity of the join of two graphs. Lemma 2 Let G 1 and G 2 be two arbitrary graphs that each has a minimum cut set C i for i = 1, 2, respectively. Then κ(g 1 G 2 ) = min{n(g 1 )+κ(g 2 ), n(g 2 )+κ(g 1 )}. 166

3 In particular, if κ(g 1 G 2 ) = n(g i )+κ(g j ) where i, j {1, 2} and i j, then V (G i ) C j is a minimum cut set of G 1 G 2. Proof. If both G 1 and G 2 are complete graphs, then so is G 1 G 2. In this case, it is clear that κ(g 1 G 2 ) = n(g 1 ) + n(g 2 ) 1 and the lemma holds. We now suppose that at least one of G 1 and G 2 is not a complete graph and let C be a minimum cut set in G 1 G 2. Then κ(g 1 G 2 ) = C. Since in G 1 G 2 every vertex of G 1 is adjacent to every vertex of G 2, if two vertices u and v are separated in (G 1 G 2 ) C, then both u and v must come from G 1 or G 2. Thus, either V ((G 1 G 2 ) C) V (G 1 ) or V ((G 1 G 2 ) C) V (G 2 ), which is dependent on the size of C. Without loss of generality, we assume V ((G 1 G 2 ) C) V (G 1 ). Clearly, V (G 2 ) C. Since G 1 G 2 is not a complete graph, (G 1 G 2 ) C has at least two components that each contains only the vertices from G 1. This implies that C must contain a cut set of G 1. Thus, C n(g 2 ) + κ(g 1 ). Since C is a minimum cut set in G 1 G 2, it must consist of exactly V (G 2 ) and a minimum cut set of G 1. In particular, we can choose C = V (G 2 ) C 1. We now use an example to illustrate Lemma 2. Two graphs G 1 and G 2 are shown in Figure 2(a), and G 1 G 2 is shown in Figure 2(b). Since n(g 1 ) = 3, κ(g 1 ) = 1, n(g 2 ) = 5, and κ(g 2 ) = 2, by Lemma 2, κ(g 1 G 2 ) = min{5 + 3, 3 + 2} = 5. The shadowed part in each graph indicates the minimum cutset of G 1, G 2, and G 1 G 2, respectively. According to Lemma 2, the following extension can easily be proved by induction. Theorem 3 Let G 1, G 2,..., G n be distinct graphs that each has a minimum cut set C i for i = 1, 2,..., n. Then n κ( G i ) = min 1 j n {κ(g j) + 1 i n n(g i )} In particular, if κ( n G i)=κ(g j )+ 1 i n n(g i ) for some j {1,..., n}, then ( 1 i n V (G i )) C j is a minimum cut set of n G i. Proof. We prove this theorem by induction on n. For n = 2, the result is shown in Lemma 2. Assume that this theorem holds for any k graphs G 1, G 2,..., G k, where k 2. We now consider G 1 G 2 (a) G 1 G 2 (b) Figure 3: The minimum cutsets of G 1, G 2 and G 1 G 2. k + 1 graphs G 1, G 2,..., G as follows. _ G i) κ(! k_ = κ ( G i) G k_ k_ = min{n( G i) + κ(g ), n(g ) + κ( G i)} = min{ k n(g i) + κ(g ), n(g ) + min 1 j k {κ(gj) + 1 i k = min{ n(g i) + κ(g ) n(g ), = n(g i)}} n(g i) + min {κ(gj) n(gj)}} 1 j k n(g i) + min{κ(g ) n(g ), min {κ(gj) n(gj)}} 1 j k 167

4 = n(g i) + min 1 j = min 1 j {κ(gj) + {κ(gj) n(gj)} 1 i n(g i)} This shows that the connectivity of n G i is κ(g j ) + 1 i n n(g i ) for some j {1,..., n}. In this case, to find out a minimum cutset C of G n, we can choose C = ( 1 i n V (G i )) C j. This completes the proof. D 1 D 2 H 1 H 2 A 1 A 2 C A 3 H 3 D 3 Corollary 4 If G is a disconnected graph with k components, then κ(g) k 1. Proof. Let H 1, H 2,..., H k be the k components of G. Clearly, G = k H i is connected. By Theorem 3 and the fact that κ(h i ) 0 and n(h i ) 1 for every i = 1,..., k, we have κ(g) = κ( k H i) = min 1 j k {κ(h j ) + 1 i k n(h i )} k 1. (a) G D 1 D 2 H 1 H 2 A 1 A 2 C 4 The connectivity of the complement of a Graph H 3 A 3 Throughout the rest, we consider that G is a nontrival, noncomplete, connected graph with a cutset C, and suppose that there are k 2 components in G C, say H i for i = 1,..., k. Clearly, G C = H 1 H 2 H k and the attachment A G (H i ), i = 1,..., k, are nonempty sets. For simplicity, we write A i to mean A G (H i ) and let D i = V (H i A i ). With respect to the cutset C of G, we first define a special graph G C. Then, we will show that κ(g C ) is a lower bound of the connectivity of G and it can be computed by a simple way. Formally, we define as follows: G C = G E G [C, k[ k[ A i] ( E G [A i, D i]) (1) Note that G C is a spanning subgraph of G. For example, a visual illustration of a graph G with a cutset C is shown in Figure 4(a), where G C contains three components H 1, H 2, H 3 and the edge sets E[C, 3 A i] and E[A i, D i ] for 1 i 3 are drawn by dotted lines. According to the definition (1), another visual illustration of the graph G C is shown in Figure 4(b), where the edge sets E GC [H i, H j ] = {xy : x V (H i ), y V (H j )} for D 3 (b) G C Figure 4: (a) A visualization of a graph G with a cutset C and the three components of G C; (b) The graph G C. i j and E GC [C, D i ] = {cd : c C, d D i } for 1 i 3 are drawn by solid lines. Lemmas 5 and 6 are used to achieve κ(g C ). Lemma 5 Let G be a graph and H, K be two disjoint connected induced subgraphs of G. Suppose that K has a minimum cutset C K. Then, the following statements are true: (1) If κ(g[k H]) > κ(k), then E[S, H], where S is any component of K C K. (2) If κ(g[k H]) κ(k), then a minimum custset of G[K H] is either C K or a subset of A G[K H] (K) V (H). 168

5 Proof. To prove statement (1), we let S be any component of K C K. Since κ(g[k H]) > κ(k), G[K H] C K is connected and it contains S as an induced subgraph. In this case, since there exist no edges joining vertices between any two components of K C K, S must be directly connected to H. Thus E[S, H]. We prove statement (2) by contradiction. Suppose to the contrary that κ(g[k H]) κ(k) and there is a minimum cutset C of G[K H] which is not C K or a subset of A G[K H] (K) V (H). Let and = C ( A G[K H] (K) V (H) ) Y = C. Clearly, and Y is not a cutset of G[K H]. Thus G[K H] Y is connected. Let Z = A G[K H] (K) Y. Since A G[K H] (K) is a cutset of G[K H], Y cannot contains all the vertices of A G[K H] (K). This implies that Z is nonempty. By the fact C = Y, is a cutset of G[K H] Y. Moreover, we can find that the vertices of (H Y ) Z must be contained in one of the components of G[K H] Y. Therefore, is also a cutset of K Y. Now, if K[C] is a cutset of K and K[C] C K, then n(k[c]) > κ(k). However, by assumption κ(k) κ(g[k H]) = n(c) n(k[c]). It is a contradiction. Lemma 6 For any graph G with a cutset C, κ(g C ) = min{ k n(d i ), min 1 i k {n(g C H i)}} Proof. Let H = G C = k H i and C H be a minimum cutset of H. By Theorem 3, all components of H C H must be contained in H j for a certain j {1,..., k}. In particular, there is a component of H C H, say S, that contains only the vertices of A j since by definition κ(h j ) = 0 and E GC [A j, D j ] =. Obviously, G C = G C [V (H) C]. We now claim that κ(g C ) κ(h). Suppose not, by Lemma 5 we know that every component of H C H is directly connected to C. Thus E GC [S, C]. However, this is impossible since S contains only the vertices of A j and by definition there exist no edges between A j and C in G C. Since κ(g C ) κ(h), again by Lemma 5, a minimum cutset of G C is either C H or a subset of A GC (H) C. Thus we consider the following two cases. Case 1: C H is a minimum cutset of G C. In this case κ(g C ) = κ(h) = C H. By Theorem 3, we can compute the connectivity of G C as follows. κ(h) = κ( k H i) = min {κ(hj) + n(h i)} 1 j k 1 i k = min {n(g C Hi)} (2) 1 i k The last equality holds due to the fact that κ(h j ) = 0 and n(h) n(h i ) = n(g C H i ). Case 2: There is a subset of A GC (H) C, say T, which forms a minimum cutset of G C. Since each vertex of C is adjacent to each vertex of D i in G C for all i = 1,..., k, either C T or ( k D i) T. For the former case, since T C is a cutset of H, it follows from Case 1 that T C +κ(h) = C C H > κ(g C ), a contradiction. Contrastively, since E GC [ k A i, C] = and T is a minimum cutset of G C, the later case implies that T = k D i and κ(g C ) = T = k n(d i). As a consequence, we have the following result. Theorem 7 Let G be an arbitrary graph with a minimum cutset C and G C be defined as in (1). Then where, κ(g C ) κ(g) κ(g C) + κ(g) κ(g C) = min 1 i k {n(g C H i)}. Proof. The lower bound of κ(g) follows from Lemma 6 and the fact that G C is a spanning subgraph of G. Similarly, since G is a spanning subgraph of G C C, κ(g) κ(g C C). By Theorem 3, κ(g) min{κ(g C) + n(c), κ(c) + n(g C)} κ(g C) + n(c) = κ(g C) + κ(g). Moreover, κ(g C) can be computed as in (2). 169

6 5 Concluding remarks In this paper, we derive an upper bound and a lower bound of the connectivity of the complement of a graph G, i.e., κ(g C ) κ(g) κ(g C) + κ(g), where κ(g C ) and κ(g C) can be computed by a simple way. We are now trying to find the class of graphs so that the equality holds in the above inequality. References [1] K. Ando, An upper bound for orders of certain (k, k)-connected graphs, Discrete Mathematics 135 (1994) [2] K. Ando and A. Kaneko, A remark on the connectivity of the complement of a 3- connected graph, Discrete Mathematics 151 (1996) [3] J.W. Boland, L.M. Lawson and R.D. Ringeisen, i-connectivity of the join of two Graphs, Congressus Numerantium 109 (1995) [4] H. Ito and M. Yokoyama, Linear time algorithms for graph search and connectivity determination on complement graphs, Information Processing Letters 66 (1998)