# On the graph inequality θ E (G) θ E (G m )

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2 H.H. Cho et al. / Discrete Mathematics 306 (2006) Then they found an infinite family of graphs that do not satisfy the inequality and brought up a question asking which graph G with the property θ E (G) < V (G) satisfies the inequality. As it is well-known that the edge clique cover number of a chordal graph is less than its vertex number (refer to [4]), it was natural to ask if chordal graphs satisfy the inequality (a graph G is chordal if every cycle in G of length greater than 3 possesses a chord, namely, an edge joining two nonconsecutive vertices on the cycle). Although they were able to show that interval graphs (a graph G is interval if it is an intersection graph of a family of intervals), dually chordal graphs (see [3] for definitions), and power-chordal graphs (a chordal graph G is power-chordal if G n is a chordal graph for any positive integer n) satisfy the inequality, the question about whether or not chordal graphs satisfy the inequality was left open. Section 2 answers this question by constructing chordal graphs that do not satisfy the inequality. In fact, those graphs even show that for any positive integer n, there exists a chordal graph G satisfying θ E (G 2 ) θ E (G) + n. Section 3 shows that θ E (G 2n+1 ) θ E (G) for any nonnegative integer n and constructs a graph G not satisfying θ E (G 2n ) θ E (G) for each positive integer n. Section 4 gives a formula for θ E (T n ) for any tree T. Section 5 presents some open problems. 2. Chordal graph G that does not satisfy the inequality θ E (G) θ E (G 2 ) We will construct chordal graphs that do not satisfy θ E (G) θ(g 2 ). A cycle C n = v 1 v 2...v n v 1 is a chordal cycle if the subgraph induced by {v 1,v 2,...,v n } is chordal. A subgraph S n of G is a sunflower if it consists of a chordal cycle C n = v 1 v 2...v n v 1 together with a set of n-independent vertices {u 1,u 2,...,u n } such that for each i, u i is adjacent to only v i and v j, where j i 1(mod n). A sunflower S n of G is called a suspended sunflower in G if there exists a vertex w/ S n such that w is adjacent to at least one pair of vertices u j and u k, where j / k ± 1(mod n). Laskar and Shier [8] showed that if a chordal graph contains a sunflower that is not suspended, then its square is not chordal. As it has been shown that power-chordal graphs satisfy the inequality, we consider the following chordal graphs containing a sunflower that is not suspended. Given a positive integer n, we construct a chordal graph W n by taking the union of complete graphs on the following vertex sets: K ={v 1,v 2,...,v 3n+12 }, X i ={v i,v i+1,u i,w i } (i = 1, 2,...,3n + 12; Identify 3n + 13 with 1), Y 1 ={v 1,v 2 } {u 3i+1 : 0 i n + 3}, Y 2 ={v 3i+1,v 3i+2,u 3i+1 : 1 i n + 3}. The subgraph of W 1 induced by the edges in cliques with vertex sets X i (i = 1, 2,...,15), Y 1, and Y 2 is drawn in Fig. 1. Now, we claim that W n is chordal. Before we do that, we need the following theorem which is one of the characterizations of the chordal graphs: If v is a vertex of a graph G, the neighborhood of v, denoted by N(v), is the set of vertices adjacent to v, while the closed neighborhood N[v]=N(v) {v}. Let a mapping σ :{1, 2,..., V (G) } V (G) be an ordering of the vertices in G. We say that σ is a perfect vertex elimination scheme if each set {σ(j) N(σ(i)) j>i} is a clique in G. Theorem 2 (Golumbic [4]). A graph G is chordal if and only if G has a perfect vertex elimination scheme. We will give a perfect vertex elimination scheme of the vertices of W n and it will follow from Theorem 2 that W n is a chordal graph. Order the vertices of W n as follows: w 1,...,w i,...,w 3n+12, u 2,u 3,...,u 3i 1,u 3i,...,u 3n+11,u 3n+12, u 1,u 4,...,u 3i 2,...,u 3n+10, v 1,v 2,...,v 3n+12. We denote the above ordering by σ. We can easily check that w i belong to exactly one clique X i from the construction of W n (i = 1,...,3n + 12), u 3i 1 belong to only X 3i 1, and u 3i belong to only X 3i (i = 1,...,n + 4). Thus, {σ(k) N(σ(j)) k>j} is a clique for j = 1,...,5n + 20.Nextisσ(5n + 21) = u 1, which belongs to exactly two

3 740 H.H. Cho et al. / Discrete Mathematics 306 (2006) w v 1 1 v 2 u 1 v 3 v 4 u 4 w 4 v 5 Fig. 1. The subgraph of W 1 induced by the edges in cliques with vertex sets X i (i = 1, 2,...,15), Y 1, and Y 2. cliques X 1 and Y 1. Since σ 1 (w 1 )<σ 1 (u 1 ) and X 1 \{w 1 } Y 1, {σ(k) N(σ(5n + 21)) k>5n + 21} is a clique. It is not hard to check that the subgraph induced by the remaining vertices {σ(j) j>5n + 21} is the union of a complete graph on K and a complete graph on Y 1 {u 1 }. Thus, σ is a perfect elimination scheme and W n is chordal. The following theorem claims that W n does not satisfy the inequality. In fact, it even shows that θ E (G 2 ) can be arbitrarily larger than θ E (G) for a chordal graph G. Theorem 3. For any positive integer n, θ E (W n ) + n θ E (W 2 n ). Proof. Recall that W n is the union of the complete graphs on the following vertex sets: K ={v 1,v 2,...,v 3n+12 }, X i ={v i,v i+1,u i,w i } (i = 1, 2,...,3n + 12, Identify 3n + 13 with 1), Y 1 ={v 1,v 2 } {u 3i+1 : 0 i n + 3}, Y 2 ={v 3i+1,v 3i+2,u 3i+1 : 1 i n + 3}. It immediately follows that θ E (W n ) 3n Also, it is not difficult to check that each edge v i w i is covered only by X i ; edge v 1 v 3 is covered only by K; edge u 1 u 4 is covered only by Y 1 ; edge v 4 u 7 is covered only by Y 2. Thus, θ E (W n ) 3n + 15 and therefore θ E (W n ) = 3n By the definition of W n, w i v i+1 w i+1 (i = 1,...,3n + 12), u 1 u 3j+1 w 3j+1 (j = 1,...,n+ 3) are paths of length 2inW n. Thus, the following are edges of W 2 n : w 1 w 2,...,w i w i+1,...,w 3n+12 w 1,u 1 w 4,...,u 1 w 3j+1,...,u 1 w 3n+10. We will show that no two of these edges are covered by one clique of W 2 n, which implies that θ E(W 2 n ) 4n It suffices if every two of these edges have ends at distance at least 3 in W n because these ends are not adjacent in W 2 n. For edges w iw i+1 and w j w j+1 (1 i<j 3n + 12), and edges u 1 w 3i+1 and u 1 w 3j+1 (1 i<j n + 3), it can easily be checked by the definition of W n that if k l / ±1(mod 3n + 12), then d Wn (w k,w l ) = 3. Now for edges w i w i+1 and u 1 w 3j+1 (1 i 3n + 12 and 1 j n + 3), note that d Wn (w i,w 3j+1 ) = 3ifi = 3j, 3j + 1, or 3j + 2; d Wn (w i+1,w 3j+1 ) = 3ifi = 3j + 2; d Wn (u 1,w i ) = 3ifi = 3j; d Wn (u 1,w i+1 ) = 3ifi = 3j + 1. Hence θ E (W 2 n ) 4n + 15.

4 3. Graph inequality θ E (G) θ E (G m ) H.H. Cho et al. / Discrete Mathematics 306 (2006) In this section, we first show that θ E (G m ) θ E (G) for each positive odd integer m. Then we show that this inequality does not hold for any positive even integer by constructing a graph G satisfying θ E (G 2n )>θ E (G) for each positive integer n. The clique graph K(G) of G has the set of maximal cliques of G as its vertex set and an edge between two vertices if and only if the corresponding maximal cliques have a nonempty intersection. In other words, the clique graph of G is the intersection graph of the maximal cliques of G. Let G be a graph and K(G) be the clique graph of G. Let H (G) be a subgraph of K(G) induced by the vertices whose corresponding maximal cliques form an edge clique cover of G. For each X V (H (G)), let X (n) denote the union of all maximal cliques of G corresponding to the vertices in N H (G) n[x]. Then the following are true: Lemma 4. For each X V (H (G)), X (n) forms a clique in G 2n+1. Proof. Take two vertices v and w in X (n). Then v X v and w X w for X v and X w in N H (G) n[x]. Since X v and X w are adjacent to X in H (G) n, there are paths X v Y 1 Y 2...Y k 1 X and XZ 1 Z 2...Z l 1 X w in H (G) for some k, l n. By the definition of H (G), there exist vertices v i Y i 1 Y i for i = 1,...,i = k and w i Z i 1 Z i for i =1,...,i=l. Since Y i and Z j are cliques for i =0,...,i=k and j =0,...,i=l, both sequences P =vv 1 v k 1 v k and Q = w 0 w 1 w l 1 w are walks in G. Now,PQ is a walk in G of length k + l + 1 which is less than or equal to 2n + 1. Thus, v and w are joined in G 2n+1. Lemma 5. The set {X (n) X V (H (G))} is an edge clique cover of G 2n+1. Proof. By Lemma 4, X (n) is a clique in G 2n+1 for each X V (H (G)). Let vw be an edge in G 2n+1. Then there exists a (v, w)-walk P = v 0 v 1 v 2 v l 1 v l (v = v 0 and w = v l )ingfor some l,1 l 2n + 1. The ends of each edge on P belong to the same clique that is a vertex in H (G). Let K i be a clique in V (H (G)) containing v i 1 and v i for i = 1,...,l. Since v i K i K i+1, K i and K i+1 are adjacent in H (G).Now,K 1 K 2 K l is a walk of length at most 2n, and so K 1 and K l both are adjacent to K l/2 in H (G) n. Hence, vw K (n) l/2 and the lemma follows. Theorem 6. θ E (G 2n+1 ) θ E (G) for any positive integer n. Proof. By Lemma 5, θ E (G 2n+1 ) {X (n) X V (H (G))}. Now {X (n) X V (H (G))} = V (H (G)) =θ E (G) and the theorem follows. The following theorem tells us that for every even power, there exists a graph not satisfying the inequality θ E (G m ) θ E (G). Theorem 7. For every positive integer n, there exist graphs G satisfying the inequality θ E (G) < θ E (G 2n ). Proof. Given a positive integer n, we construct a graph G as follows: let C 8n be a cycle x 1 x 2 x 8n x 1 of length 8n and P 4n a path y 1 y 2 y 4n of length 4n 1. In addition, let K and L be complete graphs with vertex set {x 1,x 8n,y 1 } and {x 4n,x 4n+1,y 4n }, respectively. Let G = C 8n P 4n 1 K L (see Fig. 2 for the case n = 2). It can easily be checked that θ E (G) = 12n 1. Consider the following edges of G 2n (where all the subscripts are reduced modulo 8n, with 0 identified with 8n): (I) x 1 x 2n+1,x 2 x 2n+2,...,x i x 2n+i,...,x 8n x 2n ; (II) x 1 y 2n,x 2 y 2n 1,...,x i y 2n i+1,...,x n y n+1 ; (III) x 4n+1 y 2n+1,x 4n+2 y 2n+2,...,x 4n+i y 2n+i,...,x 5n y 3n ; (IV) y 1 y 2n+1,y 2 y 2n+2,...,y 2n 1 y 4n 1,y 2n y 4n.

5 742 H.H. Cho et al. / Discrete Mathematics 306 (2006) x 16 x 1 y 1 y 8 x9 G x 8 Fig. 2. θ E (G) = 23 and θ E (G 4 ) 24. Note that the ends of each of these edges is at distance 2n in G. To show that no two of the above edges belong to the same clique of G 2n, we will check that every two of these edges have ends at distance at least 2n + 1inG. Consider any pair of edges where one edge is type (I) and the other edge is type (II), (III), or (IV). Then one edge has an end in {x n+1,...,x 3n } or {x 5n+1,...,x 7n } and the other has an end in {y n+1,...,y 3n }. Since these ends have distance at least 2n + 1inG, the pair of edges are not in any clique of G 2n. Since cycles and C 8n,D= x 1 x 2 x 4n y 4n y 4n 1 y 1 x 1, E = x 4n+1 x 4n+2 x 8n y 1 y 2 y 4n x 4n+1 have length 8n and do not have a chord, any two distinct paths of length 2n that lie in C 8n, D, ore have endpoints at distance at least 2n + 1. Now, the specified edges of G 2n cannot be in one clique unless one is type (II) and the other is type (III). In this case, one edge has an end in {x 1,...,x n } and the other has an end in {x 4n+1,...,x 5n }. Since the distance between such vertices is at least 2n + 1, these edges are not in a clique, either. Since there are 12n edges no two of which are in the same clique of G 2n, it is true that θ E (G 2n ) 12n. 4. The edge clique cover number of T m Kang et al. [7] gave a formula for θ E (T 2 ) for a tree T with at least three vertices by showing that θ E (T 2 )= V(T) t, where t is the number of pendent vertices in T. In the rest of this section, we will derive a formula for θ E (T m ) for any tree T and positive integer m, and we will give a very simple formula for the case m = 3. A center of a graph G is a vertex u such that max v V (G) d(u, v) is as small as possible, where d(u, v) denotes the distance between u and v. It is a well-known theorem of Jordan [6] that a tree has either one center or two adjacent centers. We start with the following lemma: Lemma 8. Let P i be the (s i,t i )-path of length at least m in a tree T for i = 1, 2. Suppose that there is a center c i of P i that is not a center of the other path (a center of a path P means a center of the path graph P as a subgraph) for i = 1, 2. Then one of s 1, t 1 and one of s 2, t 2 are at distance at least m + 1.

6 H.H. Cho et al. / Discrete Mathematics 306 (2006) Proof. Let P be the minimal path between centers of P 1 and P 2. The intersection of P and P 1 must be contained in a subpath P 1 from one of its endpoints t to the closest center to t, because T is acyclic. The distance from P to the other endpoint of P 1 is at least m/2, with equality only if m is even and P 1 has length m. Similarly, P 2 has an endpoint at distance at least m/2 from P, with equality only if m is even and P 2 has length m. These endpoints of P 1 and P 2 are joined by a unique path in T, so this path must consist of P and portions of P 1 and P 2. It must have length greater than 2 (m/2) unless m is even, P 1 and P 2 each have length m, and P is trivial. But then P 1 and P 2 would have a single center which must be the same vertex, a contradiction. Theorem 9. Given a tree T and a positive integer m, { V(T m/2 θ E (T m ) if m is even or T m/2 is trivial, ) = V(T m/2 ) 1 if m is odd and T m/2 is not trivial, where for i = 1,..., m/2, T i is the tree obtained from T i 1 by deleting all pendent vertices of T i 1 if T i 1 is not K 2 and by deleting one end of K 2 if T i 1 is K 2 (and T 0 = T ). Proof. If T m/2 is a trivial tree (that is, V(T m/2 ) =1), then the diameter of T is less than or equal to m and T m is a complete graph. Thus θ E (T m ) = 1. Now, suppose that T m/2 is not trivial. For each vertex v in T m/2, v is not a pendent vertex in T m/2 1 (for otherwise it is deleted when T m/2 is obtained). Let u 1 and w 1 be two distinct vertices adjacent to v in T m/2 1.If m/2 2, then u 1 and w 1 are not pendent vertices in T m/2 2 and so there are vertices u 2 and w 2, both distinct from v, that are adjacent to u 1 and w 1 in T, respectively. By continuing in this way, we find a path P =w m/2 w m/2 1 w 1 vu 1 u m/2 1 u m/2 of length 2 m/2 in T such that v is a center of P. We first consider the case where m is even. Then for each v V(T m/2 ), let K v be the set of vertices at distance at most m/2 from v in T. We claim that C ={K v : v V(T m/2 )} is an edge clique cover of T m of the smallest size. Since any two vertices in K v are at distance at most m, K v is a clique of T m. Take an edge e of T m. Let x and y be the ends of e. Then there exists an (x, y)-path of length at most m in T. Let z be a center of this path and u be the vertex that is the closest to z among the vertices in V(T m/2 ).Ifu is at distance at least m/2 + 1 from x, then the vertex on the (x, u)-path in T that is at distance m/2 from x is in V(T m/2 ) and closer to z, which contradicts the choice of u. Thus, u is at distance at most m/2 from x. For the same reason, u is at distance at most m/2 from y. Hence, e is covered by K u and so C is an edge clique cover of T m. Now, we show that C is an edge clique cover of the smallest size. Take two distinct vertices v and w in T m/2. Then there are paths P v and P w of length m such that v and w are centers of P v and P w, respectively, by the argument above. Since m is even, v and w are only centers of P v and P w, respectively. Let s v and t v (resp., s w and t w ) be the endpoints of P v (resp., P w ). Now, P v and P w satisfy the condition given in Lemma 8 and so one of s v, t v and one of s w, t w are at distance at least m + 1 and so edges s v t v and s w t w of T m cannot be in the same clique. Hence, at least V(T m/2 ) cliques are needed to cover edges s v t v for v V(T m/2 ) and so θ E (T m ) V(T m/2 ). Since C is an edge clique cover of T m with cardinality V(T m/2 ), it is an edge clique cover of the smallest size. Now, suppose that m is odd. Then for each e E(T m/2 ), let K e be the set of vertices at distance at most (m 1)/2 from (the closer end of) e in T. We claim that D ={K e : e E(T m/2 )} is an edge clique cover of T m of the smallest size. Since any two vertices in K e are at distance at most m, K e is a clique of T m. Take an edge e of T m. Let x and y be the ends of e. Then there exists an (x, y)-path of length at most m in T. Let f be an edge of E(T m/2 ) that is closest to a center of this path; if the path has two centers that are both in T m/2, then let f be the edge between the centers. By applying a similar argument for the case where m is even, we can show that x and y are at distance at most (m 1)/2 from an end of f. Thus, e is covered by K f and D is an edge clique cover of T m. Now, we show that D is an edge clique cover of the smallest size. Take an edge e in T m/2. We contract e to obtain T e from T (that is resulting from deleting e and identifying the ends of e). Then it is easy to check that (T e) i = T i e. We denote the identified vertex by v. By the argument at the beginning of the proof, there is a path of length m 1 whose center is v in T e. Such a path corresponds to a path of length m whose centers are the ends of e in T. We denote one of such paths by P e. Note that the ends of e are the only centers of P e. For the same reason, for an edge f distinct from e, there exists a path P f of length m whose centers are the ends of f in T. Denote the endpoints of P e (resp., P f )bys e and t e (resp., s f and t f ). For distinct edges e and f of T m/2, P e and P f satisfy the condition

7 744 H.H. Cho et al. / Discrete Mathematics 306 (2006) in Lemma 8 and so one of s e, t e and one of s f, t f are at distance at least m + 1 and so edges s e t e and s f t f of T m cannot be in the same clique. Hence, at least E(T m/2 ) cliques are needed to cover edges s e t e for e E(T m/2 ) and so θ E (T m ) E(T m/2 ) = V(T m/2 ) 1. Since the size of D is V(T m/2 ) 1, it is an edge clique cover of the smallest size. By Theorem 9, θ E (T 2 ) = V(T 1 ).IfT has at least three vertices, then V(T 1 ) = V(T) t, where t is the number of pendent vertices in T. Thus, we immediately obtain the following corollary shown by Kang et al. [7]: Corollary 10. Let T be a tree with at least three vertices, and t be the number of the pendent vertices in T. Then θ E (T 2 ) = V(T) t. Again, by Theorem 9, θ E (T 3 ) = V(T 1 ) 1ifT 1 is not trivial. Now the following corollary holds: Corollary 11. Let T be a tree. If the diameter of T is greater than or equal to 3, θ E (T 3 ) = V(T) t 1, where t is the number of pendent vertices of T. 5. Open questions In this paper, for a positive integer n, we found chordal graphs G that satisfy θ E (G) + n θ E (G 2 ). We ask whether or not there exists a graph G with less vertices that still satisfies the inequality. We also propose the problem of characterizing the chordal graphs that satisfy θ E (G) θ E (G 2 ). Characterizing graphs satisfying θ E (G) θ E (G 2 ) still remains wide open. Acknowledgments We wish to acknowledge the anonymous referees for suggestions leading to improvements in the presentation of the results. We especially thank an anonymous referee for suggesting the statement of Theorem 9 and the proof idea. The author thanks KOSEF for its support under Grant MOST/KOSEF-R (SRC/ERC program). References [1] R. Balakrishnan, P. Paulraja, Graphs whose squares are chordal, Indian J. Pure Appl. Algebra 12 (1981) [2] R. Balakrishnan, P. Paulraja, Powers of chordal graphs, Austral. J. Math Ser. A 35 (1983) [3] A. Brandstädt, F. Dragan, V. Chepoi, V. Voloshin, Dually chordal graphs, SIAM J. Discrete Math. 11 (1998) [4] M.C. Golumbic, Algorithmic Graph Theory and Perfect Graphs, Academic Press, New York, [6] C. Jordan, Sur les assemblages de lignes, J. Reine Agnew. Math. 70 (1869) [7] I. Kang, S.-R. Kim, Y. Nam, Y. Shin, Graph satisfying inequality θ(g 2 ) θ(g), Discrete Math. 250 (2002) [8] R. Laskar, D. Shier, On powers and centers of chordal graphs, Discrete Appl. Math. 6 (1983) [9] J.R. Lundgren, S.K. Merz, C.W. Rasmussen, A characterization of graphs with interval squares, Congr. Numer. 98 (1993) [10] A. Raychaudhuri, On powers of strongly chordal and circular arc graphs, Ars Combin. 34 (1992) [12] F.S. Roberts, Applications of edge coverings by cliques, Discrete Appl. Math. 10 (1985)

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