Characterization of total restrained domination edge critical unicyclic graphs

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1 AUSTRALASIAN JOURNAL OF COMBINATORICS Volume 47 (2010), Pages Characterization of total restrained domination edge critical unicyclic graphs Nader Jafari Rad School of Mathematics Institute for Research in Fundamental Sciences (IPM) P.O. Box , Tehran Iran Abstract A graph with no isolated vertices is edge critical with respect to total restrained domination if for any non-edge e of G, the total restrained domination number of G + e is less than the total restrained domination number of G. We call these graphs γ tr -edge critical. In this paper, we characterize all γ tr -edge critical unicyclic graphs. 1 Introduction A vertex in a graph Gdominatesitself and its neighbors. A set of vertices S in a graph G is a dominating set if each vertex not in S is dominated by some vertex of S. The domination number of G, denoted γ(g), is the minimum cardinality of a dominating set of G. A dominating set S is called a total dominating set if each vertex x is dominated by some vertex y of S with y x. The total domination number of G, denoted γ t (G), is the minimum cardinality of a total dominating set of G. For terminology and notation in general we follow [5]. A leaf in a graph G is a vertex of degree one, and a remote vertex is a vertex which is adjacent to a leaf. Let R(G), or just R, denote the set of remote vertices of G. Chen et al. [1] and Zelinka [6] introduced the study of total restrained domination, which was further studied by Cyman et al. [2] and Hattingh et al. [4]. S V (G) is a total restrained dominating set, denoted TRDS, if every vertex is adjacent to a vertex in S and every vertex in V (G)\S is also adjacent to a vertex in V (G)\S. The Also at Department of Mathematics, Shahrood University of Technology, Iran. This research was in part supported by a grant from IPM (No ).

2 78 NADER JAFARI RAD total restrained domination number of G, denoted γ tr (G), is the minimum cardinality of a total restrained dominating set of G. A TRDS of cardinality γ tr (G) is called a γ tr (G)-set. Gera et al. [3] introduced the study of total restrained domination edge critical graphs. A graph G is total restrained domination edge critical, or just γ tr -edge critical, if for any e E(G), we have γ tr (G + e) <γ tr (G). In this paper, we continue the study of total restrained domination edge critical graphs, and characterize all γ tr -edge critical unicyclic graphs. 2 Unicyclic graphs We begin with some known results. Theorem 1 ([1]) (1) For a path P n on n 2 vertices, γ tr (P n )=n 2 n 2 4, (2) For a cycle C n on n 3 vertices, γ tr (C n )=n 2 n 4. Proposition 2 ([3]) Assume G is a γ tr -edge critical graph. If R is the set of remote vertices, then R induces the complete graph K R. Proposition 3 ([3]) Assume G is a γ tr -edge critical graph. Let {r 1,...,r l } be the set of remote vertices of G, andletl i be the set of leaves adjacent to r i for i =1,...,l. If l 2, then L i =1for i =1,...,l. The following is easily verified. Observation 4 Let x, y be two non-adjacent vertices in a γ tr -edge critical graph G, and let S be a γ tr (G + xy)-set. (1) If {x, y} S, thenn G (x) S = or N G (y) S =. (2) If {x, y} S =, thenn G (x) S, orn G (y) S. (3) If x S, y S, thenn G (y) S =. Theorem 5 For n 3, the cycle C n is γ tr -edge critical if and only if n =5, 6, or n 3 (mod 4). Proof. Let G = C n and let V (G) ={v 1,v 2,..., v n }, where v i is adjacent to v i+1 for i =1, 2,..., n 1, and v n is also adjacent to v 1. We consider the following cases. Case 1. n 0(mod4). Let n =4k for some integer k. By Theorem 1, γ tr (G) =2k. If n {4, 8}, then γ tr (G) =γ tr (G + v 1 v 3 ), and so G is not γ tr -edge critical. Suppose that n 12. Let S be a γ tr (G+v 1 v 6 )-set. It is obvious that S {v 2,v 3,v 4,v 5 } 2. If {v 1,v 6 } S, then S γ tr (C n 4 )+2 = 2k, where C n 4 is the cycle obtained from G+v 1 v 6 by removing

3 TOTAL RESTRAINED DOMINATION 79 v 2,v 3,v 4,v 5. So assume v 6 S. We show that S 2k. If v 1 S, thenv 4 S. If v 3 S, then{v 2,v 3,v 4 } S, and we may assume that v 7 S. Then we consider the path P n 5 obtained by removing v 3,v 4,..., v 7. It follows that S 2+γ tr (P n 5 ) 2k. If v 3 S, thenv 5 S, andsov 7 S. This time we consider the path P n 6 obtained by removing v 2,v 3,..., v 7. It follows that S γ tr (P n 6 )+2 2k. It remains to suppose that v 1 S. If {v 7,v n } S, then we consider the path P n 6 obtained from G by removing v 1,v 2,..., v 6. So S γ tr (P n 6 )+2 2k. So without loss of generality assume that v 7 S. With a similar argument we obtain S 2k. Case 2. n 1(mod4). If n = 5, then we can easily see that G is γ tr -edge critical. So we let n 9. Let n = 4k + 1 for some integer k. By Theorem 1, γ tr (G) =2k +1. Let S be a γ tr (G + v 1 v 5 )- set. We show that S 2k +1. If{v 1,v 5 } S, then S 1+γ tr (C n 3 )=2k +1, where C n 3 is the cycle obtained from G + v 1 v 5 by removing v 2,v 3,v 4. So suppose that v 5 S. Ifv 1 S, then{v 2,v 3 } S, andso S 1+γ tr (P n 4 )=2k +1, where P n 4 is the path obtained from G by removing v 3,v 4,v 5,v 6. It remains to assume that v 1 S. This time with a similar argument we obtain S 2k +1. Case 3. n 2(mod4). It is a routine matter to see that C 6 is γ tr -edge critical. Also γ tr (C 10 )=γ tr (C 10 +v 1 v 6 ) which implies that C 10 is not γ tr -edge critical. So we let n 14. Let n =4k +2 for some integer k. By Theorem 1, γ tr (G) =2k +2. LetS be a γ tr (G + v 1 v 6 )-set. We show that S 2k +2. If{v 1,v 6 } S, then S γ tr (C n 4 )+2 = 2k +2, where C n 4 is the cycle obtained from G + v 1 v 6 by removing v 2,v 3,v 4,v 5.Wemayso assume that v 6 S. Assume v 1 S. We may assume that S {v 2,v 3,v 4,v 5 } =2; other possiblities are similarly verified. Let P n 6 be the path obtained from G by removing v 1,v 2,..., v 6. It follows that S 2+γ tr (P n 6 )=2k +2. Itremainsto assume that v 1 S. This time a similar argument gives that S 2k +2. Case 4. n 3(mod4). By definition C 3 is γ tr -edge critical. It is also straightforward to see that C 7 is γ tr - edge critical. So suppose that n 11. Let n =4k + 3 for some integer k. By Theorem 1, γ tr (G) =2k +3. Let S = {v 4i+3,v 4i+4 :0 i k 1} {v 4k+1,v 4k+2,v 4k+3 }. Then S is a γ tr (G)-set. Notice that v 4k+3 only dominates v 1. Since G is vertex transitive, we prove that for each j =2, 3,..., diam(g) =2k + 1, there is a vertex x S such that d(x, v 1 )=j. For any i>2k +2, we rename v i by w (4k+3) i+2. So V (G) =V 1 V 2, where V 1 = {v 1,v 2,..., v 2k+2 } and V 2 = {w 2,w 3,..., w 2k+2 }. Let S 1 = S V 1 and S 2 = S V 2. It follows that {d(v 1,v):v S 1 } = { 4i +2, 4i +3:0 i< } k, 2 while {d(v 1,v):v S 2 } = { 1, 2, 3, 4i, 4i +1:1 i } k. 2

4 80 NADER JAFARI RAD Hence the result follows. We are now ready to give the main result of this paper. Let E be the class of graphs G such that G belongs to E if and only if one of the following holds. (1) G is obtained from K 1,n for a positive integer n 5 by joining two leaves, (2) G is obtained from the corona of C 3 by removing at most one leaf, (3) G {C 5,C 6,C n : n 3(mod4)}. Theorem 6 A unicyclic graph G is γ tr -edge critical if and only if G E. Proof. It is a routine matter to see that any graph in E is a γ tr -edge critical unicyclic graph. Let G be a γ tr -edge critical unicyclic graph, and let C be the unique cycle in G. If G = C, then by Theorem 5, G E. Suppose that G C. Let v 1 L(G) such that the distance between v 1 and C is maximum, and let P : v 1 v 2...v k be the shortest path from v 1 to C. Note that V (C) V (P )={v k }. Let V (C) ={u 1,u 2,..., u t }, where u 1 = v k,ande(c) ={u i u i+1 :1 i t 1} {u 1 u t }. So N G [v k ] V (C) ={u 1,u 2,u t }. By Propositions 2 and 3, if k>3, then deg(v i )=2 for 3 i k 1. Moreover, if k > 2, then deg(u 1 )=3anddeg(u i )=2for 2 i t. We first show that k 6. Suppose to the contrary that k 7. Let S 1 be a γ tr (G + v 2 v 7 )-set. If v 7 S 1,thenN G (v 7 ) S 1 =, since S 1 is not a TRDS for G. Since v 5 is dominated by S 1,wehave{v 3,v 4 } S 1.Butthen(S 1 \{v 4 }) {v 6 } is a TRDS for G, a contradiction. So we assume that v 7 S 1. Again we have N G (v 7 ) S 1 =, since S 1 is not a TRDS for G. Since v 6 is dominated by S 1,we have {v 3,v 4,v 5 } S 1.Now(S 1 \{v 3,v 4 }) {v 6 } is a TRDS for G, a contradiction. Thus, k 6. Suppose now that k = 6. Let S 2 be a γ tr (G + v 2 u 2 )-set. If u 2 S 2, then N G (u 2 ) S 2 =. If u t S 2 then {v 3,v 4 } S 2,andso(S 2 \{v 4 }) {v 6 } is a TRDS for G, a contradiction. So u t S 2,andso{v 3,v 4,v 5 } S 2. It follows that (S 2 \{v 3,v 4 }) {v 6,u t } is a TRDS for G, a contradiction. We thus assume that u 2 S 2. Since S 2 is not a TRDS for G, N G (u 2 ) S 2 =. If u t S 2,then {v 3,v 4 } S 2,andso(S 2 \{v 3,v 4 }) {v 5,v 6 } is a TRDS for G. This contradiction yields that u t S 2. But v 5 is dominated by S 2. So {v 3,v 4 } S 2. This time {u i 1 (mod t): u i S 2 } {v 1,v 2,v 5 } is a TRDS for G, a contradiction. We deduce that k 5. Next suppose that k =5. LetS 3 be a γ tr (G + v 4 u 2 )-set. If {v 4,u 2 } S 3 =, then either N G (u 2 ) S 3 or N G (v 4 ) S 3.IfN G (v 4 ) S 3,thenS 3 \{v 3 } is a TRDS for G, while if N G (u 2 ) S 3 then ({u i 1 (mod t): u i S 3 }) (S 3 {v 1,v 2,v 3 })isa TRDS for G. Both are contradiction. So {v 4,u 2 } S 3. If {v 4,u 2 } S 3,then v 5 S 3 and (S 3 \{v 3,v 4 }) {v 5,v t } is a TRDS for G, which again is a contradiction. We deduce that {v 4,u 2 } S 3 =1. Ifv 4 S 3,thenN G [u 2 ] S 3 =, andso (S 3 \{v 3,v 4 }) {u t,v 5 } is a TRDS for G, a contradiction. It remains to assume that u 2 S 3. This time {u i 1 (mod t): u i S 3 } {v 1,v 2 } is a TRDS for G, a contradiction. Thus, k 4.

5 TOTAL RESTRAINED DOMINATION 81 Suppose now that k = 4. It is straightforward to see that for t {3, 4, 5}, γ tr (G) =γ tr (G+v 2 v 4 ), and for t =6,γ tr (G) =γ tr (G+v 3 u 3 ). So we assume that t 7. Let S 4 be a γ tr (G + u 4 u t 1 )-set. We first show that {u 4,u t 1 } S 4. Suppose that {u 4,u t 1 } S 4 =. Then N G (u 4 ) S 4 or N G (u t 1 ) S 4.IfN G (u 4 ) N G (u t 1 ) S 4, then {v 3,v 4,u 2,u 3,u t } S 4,and(S 4 \{v 3,v 4 }) {u 4,u t 1 } is a TRDS for G which is a contradiction. So either N G (u 4 ) S 4 or N G (u t 1 ) S 4. Assume that N G (u 4 ) S 4, and so N G (u t 1 ) S 4. It follows that {v 3,v 4 } S 4. Then S 4 \{u 3 } is a TRDS for G, a contradiction. Similarly, N G (u t 1 ) S 4 which produces a contradiction. We deduce that {u 4,u t 1 } S 4. If u 4 S 4,thenu t 1 S 4 and N G [u 4 ] S 4 =. Since u 3 is dominated by S 4, {v 3,v 4,u 2,u t } S 4. This time (S 4 \{v 3,v 4 }) {u 3 } is atrdsforg, a contradiction. So u 4 S 4, and by a similar argument u t 1 S 4. It follows that either N G (u 4 ) S 4 =, orn G (u t 1 ) S 4 =. If N G (u 4 ) S 4 =, then {v 3,v 4,u t } S 4,and(S 4 \{v 4 }) {u 3 } is a TRDS for G, a contradiction. So N G (u 4 ) S 4, and by symmetry N G (u t 1 ) S 4. ButthenS 4 is a TRDS for G, a contradiction. Thus, k 3. Suppose that k = 3. It is straightforward to see that for t = 3, γ tr (G) = γ tr (G + v 2 u 2 ), and for t =4,γ tr (G) =γ tr (G + v 2 u 2 ). So we suppose that t 5. Let S 5 be a γ tr (G + v 2 u 5 )-set. If u 5 S 5,thenN G (u 5 ) S 5 =. It follows that {u 2,v 3 } S 5,and{u i 4 (mod t): u i S 5 } {v 1,v 2 } is a TRDS for G, a contradiction. So u 5 S 5. But then {u 2,u 3 } S 5. Again {u i 4 (mod t): u i S 5 } {v 1,v 2 } is a TRDS for G, a contradiction. We conclude that k =2. For t = 3, it is straightforward to see that either G is obtained from the corona of C 3 by removing at most one pendant edge, or G is obtained from K 1,n foraninteger n 5 by joining two leaves. In both cases G E. So assume that t 4. We show that v 2 is the unique remote vertex of G. Suppose that v 2 is not the unique remote vertex of G. By Propositions 2 we may assume that u 2 is also a remote vertex. By Proposition 3, deg(v 2 )=deg(u 2 )=3. Letx be a leaf adjacent to u 2 and let S 6 be a γ tr (G + u 2 u t )-set. It follows that S 6 is a TRDS for G, a contradiction. We deduce that v 2 is the unique remote vertex of G. Fort {4, 5, 6} it is a routine matter to see that γ tr (G) =γ tr (G+u 2 u t ), and so G is not γ tr -edge critical. Let t 7, and let S 7 be a γ tr (G+v 2 u 6 )-set. If u 6 S 7,thenN G (u 6 ) S 7 =. This implies that {u 2,u 3 } S 7 and (S 7 \{u 2,u 3 }) {u 4,u 5 } is a TRDS for G, a contradiction. So u 6 S 7. It follows that N G [u 6 ] S 7 =. But then {u 2,u 3,u 4 } S 7,and(S 7 \{u 2,u 3 }) {u 5 } is a TRDS for G, a contradiction. Acknowledgment The author wishes to thank the referee for his/her remarks and suggestions that helped improve the manuscript.

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