Chapter 4 SIGNED DOMINATION NUMBER OF A TREE.

Transcription

1 Chapter 4 SIGNED DOMINATION NUMBER OF A TREE. 74

2 4.1 Introduction: As we have already discussed about the signed domination number 7a(G) of a graph G and found some bounds on js(g) and also dealt with some elementary properties. In this Chapter, we are interested in signed domination number for trees. Prom the literature, [1] we kwow that only two bounds for signed domination number of tree T are known and nothing much is said about y3(g) for class of trees. In this Chapter, we find two more bounds on 7S(T) in terms of number of end vertices and order and also in terms of its order and diameter. In section two, we find exact value of %(T) when T is a binary tree and caterpillar and in section 3, we characterize the class of trees with small and large signed domination number. 4.2 Some existing results: We state some existing results for our reference and some are useful in dealing with new results. Proposition [1]: If g is a signed dominating function for a graph G and vertices u and v satisfy N(v) = {«}, then g(u) = g(v) = 1. or In other words, if v is a pendent vertex and u is a support adjacent to v, then g(u) = g(v) = 1. Proposition [1]: For any integer n,j,(pn) = n [^J 75

3 Proposition [1]: Let T be a tree of order n > 2, Then 7S(T) > with equahty if and only if T is a path on 3k + 2 vertices for some nonnegative integer k. Proposition [1]: For any tree T on n > 2 vertices, i(t) < 7S(T) 1; where i(t) denotes the cardinality of the smallest maximal independent set of vertices in T. 4.3 Bounds on 7S(T): The following bound on 7a(G) holds for any arbitrary graph G and hence holds for trees also. Proposition 4.3.1: For any graph G of order n,7a(g) > n 2r, where r denotes the number of vertices which are neither pendent vertices nor supports. Further, the equality holds if and only if for any minimum signed dominating function /, f(u) = 1, implies u is a pendent vertex or a support. Proof: Let G be a graph of order n and g be a minimum signed dominating function of G which partitions the vertex set V into a sets of vertices P and M that g assigns the value +1 and 1 respectively. Then, 7S(G) = w(g) = P M. By the Proposition 4-2.1, \P\>n r holds. Thus, 7,(G) = P - \M\ = P - (n - P ) = 2 P - n, since P + M = n. This implies that, 7a(G) >n 2r. Moreover, if 7S(G) = n 2r, then for every minimum signed dominating function f,w(f) n 2r. As above, let w(f) = P M, where P and M have 76

4 the usual meaning. Now, we claim that, if u P, then u is a support or a pendent vertex. If not, let P = P\ U P2 where Pi is the set of all pendent vertices and supports and P2 = P Pi; then P2 ^ (j> P > Pi + 1. Therefore, by assumption, we have n = P + \M\ > Pr \M\ =* \M\ < n - Pj - 1. therefore n 2r = js = P \M\ > Pi M > Pi (n - Pi - 1) = 2 Pi n. This implies that, 2n 2r > 2 Pi + 2 n r > Pi + 1 Therefore, r < n Pi. But n = r + P1 <n 1, a contradiction. Conversely, suppose that for any minimum signed dominating function /, f{u) = 1 implies that u is either a pendent vertex or a support. This implies that, every vertex v with f(v) = 1, is neither a support nor a pendent vertex. Thus, M = r. Hence 7S(<?) = w(f) = P \M\ = n 2\M\ = n 2r holds. THEOREM 4.3.1: Let T be a tree of order n with the vertex set V and edge set E and / be a minimum signed dominating function of T. If the vertices assigned with 1 by / induces a tree, then 7s(T) > 3n + 2e 4 _ where, e denotes the number of end vertices. Furthermore,the bound is sharp. Proof: Let T be a tree of order n with the vertex set V and edge set E and / be a minimum signed dominating function of T. Let P and M denote the sets of 77

5 positive vertices and negative vertices respectively. Further, let Pi be the set of end vertices and supports in T and P2 = P Pi. Now, let us label the vertices of M as vi,v2,..., vk and d* = deg^)vi, for i = 1,2,3, Then, by the property of signed dominating function, each vertex vt of M is adjacent to at least dj + 2 vertices in P2 and as T is a tree, no two vertices in P2 are adjacent, As 0(T) = n, so the number of edges in T is n - 1. Thus, n 1 = m(m)+m(p2)+m(pi)+ m(m, Pi)+m(M,P2)+ m(pi,p2) (4.1) where, m(s) denote the number of edges in an induced subgraph (S) and m(s, T) denote the number of edges joining a vertex of S with a vertex of S with a vertex of T. Clearly, m(p2) = m{p\) = 0 and m(m, Pi) = 0. Also, by the hypothesis, M induces a tree and hence m(m) = k 1. Thus, (4.1) reduces to k n 1 > k ~ 1 + ^2(di + 2) + e (4.2) i=l where e denotes the number of pendent vertices in T. Again, by the hypothk esis, Y/di = 2(k-l), so (4.2) becomes «=i n l>(k l) + 2k 2 + 2k + e =>n l>5k + e 3 (4.3) 78

6 Since 7»(T) = n 2k and hence _ n - 7,(T) (4.4) Therefore, from (4.3) and (4.4), we have n_!>5(ni^s2)+e_3 By simplifying, we get the desired result. Next, We show that the bound is sharp. Consider the tree Ta = S(Kit2k+2) with a central vertex u as labelled in the Figure 4-1, where S(H) denote the subdivision of a graph H. FIGURE-4.1 and Ti = S(Kit3) with central vertex v< for i = 1,2,...,& as labelled in the Figure 4A 79

7 FIGURE-4.2 Let T be a tree obtained from Ta and T^ by adding edges uu,- for i = 1,2,A: as in the Figure 4-3, T FIGURE

8 Assigning the value 1 to a vertex u and the vertices V\,V2,..., Vk and rest of the vertices by +1, produces a signed dominating function / on T with weight w(f) = 7k+3. Thus, 73(T) < 7k 4-3. On the other hand, T is a tree of order 9k + 5 with e = 4fc + 2 end vertices and hence by the bound above, 7* CO > 3(9fc + 5) + 2(4fc + 2) = 7k + 3. which establishes the sharpness of the bound. diameter. Next, theorem gives the upper bound on ja(t) in terms of order and THEOREM 4.3.2: For any tree T of order n with diameter d,.. d 1 7s (O < n - 2, where zj denotes the largest integer not larger than x. Proof: Let T be a tree of order n and diam(t) = d. Let u = Uq,Ui,U2,U3,..., Ud-i,Ud = t be an u v diameteral path in T. Define a mapping / : V(0 * { 1.1} by /(u3i_i) = -1, for a vertex u3i_i on u v path, for % > 1 and /(x) = 1, for every x ^ u3i_x e V(T). Clearly, / is a signed dominating function and the number of vertices assigned with 1 by / in T is [~J. Thus, %(T) < n 2 holds. The following theorem proves the sharpness of the upper bound. 81

9 THEOREM 4.3.3: For any integer n > 2, and 1 < d < n 1, there exists a tree T of order n, diameter d, satisfying 7,(T) = n 2 Proof: Let n > 2 and 1 < d < n 1. Let T be a tree obtained from a path P :u0, tii, «2> -., Ud-1, Ud and n d 1 pendent edges of the form Ui«j; i d +1, d + 2,n. The tree T so obtained is as in the Figuer 4-4> FIGURE-4.4 Any minimum signed dominating function / on T, assigns 4-1 to the vertices,«n and only vertices which are assigned with 1 by / must be on a path P : ito,ui,u2,ti3,...,ud-i,ud. By the Proposition 4-2.2, the number of vertices on P assigned with 1 by / is [^J. Thus, 7.(T) = n-2 ^p holds. 82

10 4.4 Signed Domination number of Some Special classes of trees: In this section, we find the exact value of 7»(T), when T is a caterpiller and complete binary tree Caterpiller: A caterpiller is a tree T whose derived tree T' obtained by deleting all its end vertices is a path P* : «i,u2,u3,thus, each vertex of T is either on P* or is adjacent to a vertex on P*. Every caterpiller T has an associated sequence of non-negative integers d = d\dt...d*, where <!k is the number of end vertices to which Ui is adjacent and di;d» > 0 when n > 1. Obviously, this sequence uniquely determines T. For example, Figure 4.5 shows a caterpiller with the sequence d = 3,2,1,4,0,0,0,2,0,0,1,1,2. T: FIGURE

11 We call this sequence d = d^d2...dn assiciated with the caterpiller T, a code of T. Further, we reduce this sequence d = d\d2...dn to a sequenceof subsequences aubi,a2,b2,a3,...bk-i,ak', where a, dj,dj+i,...,d/, such that di+i = 0 and bt = dr,dr+1,...jdm, such that d3 = 0 for r < s < m, dr_i ^ 0 and dm+1 ^ 0. Farther, note that the length of 6* is the number of 0'B between a* and Oj+i- For example, the sequence d = 3,2,1,4,0,0,0,3,2,0,0,1,1,2 of the caterpiller T of the Figure 4-5, has a reduced code d 01,61,02,62)03 where ai=3,2,l,4; 61 = 0,0,0; 62 = 0,0 and 03 = 1,1,2. Let l(oj) and l(bi) denote the lengths of the subsequences Oj and 6, respectively. Thus, from the Figure 4-5, l(oi) = 4,l(bi) = 3,l(a2) = 2,1(62) = 2 and l(a3) = 3. THEOREM : If T is a caterpillar of order n with code d = 01,61,02,62,03,...6fc_i,Ofc, then 7,(T) = n-2(g l(bi + 2) Proof: If T is a caterpillar of order n with code d = 01,61,02,62,03,...6fc_i,a*, and / be a minimum signed dominating function on T. By the properties of signed dominating function /, assigns +1 to the pendent vertices and supports in T. The vertices assigned with 1 by / will occur for the vertices with codes bts in T. For each 6j, there associates a path Pi on Z(6f) + 4 vertices in which vertices with two pendent vertices. Thus, by Proposition 4-2-2, there are exactly 84

12 2j negative vertices under any minimum signed dominating func- I HK 2) I tion. Hence, the total number of negative vertices by / are 2J j ^ Thus, 7.(T) = w(/) = (n-2)g[mj. The following example explain the proof of the above Theorem. Consider a Caterpiller T with code d = 2,3,1,0,0,0,0,1,2,0,3,1,0,0,0,2,2 = oi, 61,02,621 «3i h, <* FIGURB-4.6 Here, 0(T) = n = 34, ((61)=4,/(W = 1,((6S) = 3. Thus, = 34-2( ) = 34-2 x 4 = 34-8 =

13 4.4.2 Complete binary trees: A tree T consisting of a vertex of degree two and remaining vertices are of degree three or one is called a binary tree. The vertex of degree two is called a root. In the laungage of computer science, the root is also called as parent and the two neighbors are called as children of level T'. These vertices act as parent if they have two more neighbors each of these vertices are called as children of level,2,.) and so on. Therefore, both of these vertices acts as parents as wel as children. The vertices which do not have children are called siblings. In other words, the pendent vertices of a binary tree are called siblings. Consider a binary tree with 'V levels that is, 0th, 1st, 2nd,..., (I l)t/l levels. If there ends with 2Z_1 siblings in the (l l)th level of a binary tree, then the tree is called as Complete binary tree. In other words, a binary tree T is called a complete binary tree iff T contains exactly 2,_1 siblings. PROPOSITION : For any complete binary tree T with l levels, 7,(T) > 2I_1 + 1, for l > 2. Further, the equality holds if and only if l = 3. Proof: Let T be a complete binary tree with l levels and L0, L\, L2,..., L*_i be the i-i levels of T. Clearly, L* contains 2 vertices, and hence 0(T) = ]T = 2'- 1. In i= 1 T, the vertices at level (l 1) are pendent vertices and at level (l 2) are supports. Thus, any minimum signed dominating function / assigns +1 to all the 1-3 vertices in levels L/_i and Lj_2. But all the remaining vertices ^ 2i = 2i-2-1 are»=i non-supports and non-pendent vertices. Hence, by the Proposition 4-3.1, we have 86

14 7,(T)>0(T)-2r = 2* 1 2 (2l~2-1) = 2l - 2l~x + 1 = 2l~l + 1. LEMMA : If / is a minimum signed dominating function on a complete binary tree T with l levels, then the maximum number of vertices in ith level L. (2 < i < l 3) which receive negative signs is 2t_1. Proof: If possible assume that the level Li for 2 < i < l 1, contains vertices of negative sign assigned by a minimum signed dominating function /. Then by the Pigeonhole principle, there exists at least one parent in the level L,_i, which as both of its children receive negative sign. All the vertices under consideration are of degree three, should have at most one negative neighbor, which is a contradiction. Hence the lemma. THEOREM 4.4.3: The signed domination number of a complete binary tree of level l is given by f(8fc-l), if l=3k-l 7s(T)= < f (8fc 1) 1, if 1 3k f (8fc - 1) + 1, if l=3k+l. 87

15 Proof: By the property of signed dominating function /, we have /(«) > i, us/v[u] for any vertex v in G and in particular for a tree T. Keeping this in mind, let T be a complete binary tree with l levels L0, L\, L2,Lj_x. By the Proposition 4-4-%-h the vertices of levels L(_i and L;_2 must receive +1 and assign -t-1 and 1 to the children of level L;_3 to each single parent of level L;_4 and assign +1 to all the vertices of levels Lj_4 and L(_5 and repeat the process to the vertices of level Lj_6 and continue untill we reach the level L0. The vertex in level L0 receive either +1 or -1 depending on the nature of l. For this, we consider the following three cases: Case-1: Let l = 3A: 1; for some integer k > 1. Let / be a signed dominating function which assigns +1 and -1 to the vertices of a complete binary tree T as described above. That is, the vertices in the levels L$k-2, Lzk-s receive +1 and half of the vertices in Lzk-i receive -1 and other half receive +1 as described above. Continuing this process, the vertices in L0 and L\ receive +1. Thus, the weight w(f) = fe"3 + 2?k~2 Thus, = ( fc~3) + ( fc~2) 2 (23fc 1) 21(23fc 1) 3. 23_i CT) < w(f) =?(8-1). (4.5) On the other hand, by the Lemma 4-4-S-2, the maximum number of vertices in Li which receive 1 is 2-1 by any signed dominating function. As l = 3fc 1 the only vertices which receive 1 are in the levels L2, L5, Lg,L3k-i- Thus, if M is 88

16 the set of vertices in T which receive -1 by a minimum signed dominating function /, then \M\ < *~2 2(23*-3 1) 23fc 3 1 = - = -. Thus, /o3fc-2 _ o\ 7s(T) < w(f) = 0(T) - 2 M > 23fc_ f j 7 x 23* fc_ x 23fc_1-3 _ - = f (2* - 1) =?(8k - 1). (4.6) Case-2: Let l = 3k. As in the case 1, the levels L3Ifc_2, L^-i receive +1 and half of the vertices in L^-z receive 1 and finally the levels Lo, L\ and L2 receive +1. The negative vertices occur in the levels Lz,LstLs,..., L3k-z- In a similar way as in case 1, we can prove that 7a(T) = f (8* - 1) - 1. Case-3: Let l = 3k+ 1. By choosing a suitable minimum signed dominating function / as in case 1, we deduce that 7S(T) = y(8fc - 1)

17 4.5 Trees with Small and Large Signed domination number: For any arbitrary graph G, the signed domination number "ys(g), as we know, may be negative, or positive or zero. But, signed domination number of a tree T is always positive, which can be seen by the following arguments. Let T be a tree of order n and e be the number of pendent vertices in T. Let / : V(T) { 1,1} be a minimum signed dominating function of T, So that js(t) = w(f). Then, / forms a partition of a vertex set V into two sets M and P, where M consists of all vertices assigned assigned with -1 and P consists of all vertices with +1. If \M\ = k, then 7S(T) = n 2k. Further, we can partition the set P into Pi and P2. where Pi is the set of all pendent vertices and P2 = P Pi. Relabel the sets M, Pi, P2 as Vi, V3, V2 respectively, then we have 3 n - 1 = X>(V5) + m(vhv2) + m(v1, V3) + m(v2, V3) (4.7) i=l where m(vi) denotes the number of edges in an induced subgraph (V^) and m(vi,vj) denote the number of edges joining a vertex of V* to the vertex of V). By the property of signed dominating function /, each vertex of V\ is adjacent to at least two vertices in V2 and as V3 consists of only pendent vertices, and hence fn(vi,v2) > 2k and 771(1^, V3) = e and m(v3) = 0 and m(v\,v3) = 0. Thus, (4.7) reduces to 90

18 n 1 = m(vi) + m{\2) + 2 A; + e (4.8) But m(\1) > 0 and m(v2) > 0 and thus, (4.8) reduces to n 1 > 2k + e => n 2fc > e + 1 Thus, 7«(r) > e+ 1 This inequality and the bound 7S(T) > ^ given in Theorem 4.2.3, we characterize the class of trees with small and large signed domination number. By the inequality (4.9), we have ja{t) > 3, as e > 2 and 7S(T) < n. In this section, we consider the trees T with 7S(T) = 3 and 4 and on the other side 7S(T) = n and n 2. PROPOSITION 4.5.1: For any tree T of order n, 7a(T) = 3 if and only if T = P3 or P5. Proof: Suppose 7a(T) = 3. Then, by the above equation (3), we have e < 2, which implies that T Pn. But, we know that 3 = 7a(T) > which gives n < 5. As 7a(T) = 3 and hence 3 < n < 5, which gives us that T = P3 or P4 or P5. But T can not be P4, otherwise 7a(T) = 4. Thus, T must be either P3 or P5. Converse is obvious. 91

19 PROPOSITION 4.5.2: Let T be a tree of order n, %(T) = 4 if and only if T is one of the following graphs shown in Figure 4-7: Tl=P4: T2=P6:n FIGURE-4.7 Proof: Let T be a tree of order n with j,(t) 4. Then by the inequality (4.5.3), we have e < 3 and by the Theorem 4.2.3, we have 4 = 7,(T) > which gives us that n < 8. To prove the result, we consider two cases: Case-1: Let e = 2. 92

20 Then T is a path Pn. By the Proposition 4.2.2, we have 4 = 7.(*n) = n- 2. That 2 [a=2j = n - 4. The solutions of the above equation are n = 4,6 and 8. Thus, T is either P4 or P6 or P8. Case-2: Let e = 3. From the above inequality (3), and the facts that n < 8 and e = 3, we have ro(vi) + m(v2) + 2k<4 (4.10) Now, claim that k = 1. For if k > 2, then n > 9, a contradiction. Therefore, the inequality (4) turns out to be, 1 + m(v2) + 2 < 4 = m(v2) ^ 1. Therefore, m(v2) = 0. This gives us that T = T4 or T5 as given in the Figure 4-8, 93

21 FIGURE-4.8 The converse is obvious. As in [1], Dunbar et. al. proved that %{T) = n if and only if every vertex of T is either a support or a pendent vertex. Here, we consider the characterization of of trees with 7t(T) n 2 in the following proposition. PROPOSITION 4.5.3: Let T be a tree of order n, 7,(T) = n 2 if and only if there exists v e T such that T v is a forest in which is either a support or a pendent vertex. Proof: Suppose v T such that T v is a forest in which every vertex is a support or a pendent vertex. Let Ti,r2>...,T* be the component of T-v. Define /: V(Ti) * {-1,1} by f(v) = -1 and /(«) = 1, Vu 6 V(T{), i = 1,2,3,... k. => f is a signed dominating function = 7a(T) < w(f) = n - 2. Now, we claim that 7,(T) n 2 94

22 For if 7a(T) < n 4, then there exist at least two vertices it, v and g be a signed dominating function defined by g(u) = g(v) = -1. Then, T v contains a component containing u which is neither a support nor a pendent vertex., a contradiction. Conversely, Suppose 7S{T) n 2. Then, for any minimum signed dominating function /, there is exactly one vertex v such that f(v) = 1. Now, consider T v and let 7\, T2,..., 7* be the components of T v. We have to prove that every vertex in Tj for i = 1,2,3,...k is either a pendent vertex or a support. If this is not true, then there is at least one Tt in which, there is a vertex u which is neither a pendent vertex nor a support. With out loss of generality, let Ti be such a component. By the Proposition 4-3.1, 7a(Xi) < 0(T{) 2. Now,let us define a function / : V(T) > { 1,1} by f(v) 1 and f(u) = -1,f(w) = 1 for all w ^ u,v. Clearly / is a signed dominating function Therefore, n 2 = 7S(T) < w(f) k < (Ti) ^ 0(Ti) i=2 Hence the proof. = E (T<)-3»=i n 1 3 = n 4, a contradiction. 95

23 REFERENCES [1] J.Dunbar, S.T.Hedetniemi, M.A.Henning and P.J.Slater: Signed domination in graphs, Graph Theory, Combinatorics and Applications, Proceedings 7th Internat. Conf. Combinatorics, Graph Theory, Applications, Vol. 1 (Y.Alavi, A.J.Schwenk, eds.) John Willey and Sons, Inc., (1995) [2] O.Favaron: Signed domination in regular graphs, Discrete Math. 158(1995) [3] F.Harary: Graph Theory, Narosa Publishing House, New-Delhi 1988 [4] Haynes T.W, Hedetniemi S.T and Slater P.J(edj): Fundamentals of Domination in Graphs, New York; Marcel Dekker 1998 [5] J.Hattingh, M.A.Henning and P.J.Slater: Three-Valued k-neighbourhood domination in graphs, Australas. J. Combin. 9(1994), [7] M.A.Henning and P.J.Slater: Inequalities related domination parameters in cubic graphs, Discrete Math. 158(1996) [8] R.Hass and T.B.Wexler: Signed domination number of a graphs and its complement, Preprint submitted to Elsevier Preprint, on 7 September (2001). [9] Jiri Martousek: On the signed domination in graphs, Combinatorica, 20(2000), [10] Jaeun Lee and Xin-zhong Lu: Signed domination numbers of graphs, Mathematics, Yeungnam University, Kyongsan, Korea. [11] Savita B.Basapur: Domination theory in graphs and related topics, Ph.D. thesis, Kamatak University, Dharawad (1993). [12] Z.Zhang, B.Xu, Y.Li and Liu: A note on the lower bounds of signed domination number of a graph, Discrete Math. 195(1999)

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