Cyman, Joanna

Lemańska, Magdalena

Raczek, Joanna

Czech digital mathematics library

Lower bound on the distance $k$-domination number of a tree

Institute of Mathematics AS CR, v. v. i.

Mathematica SlovacaJoanna Raczek; Magdalena Lemańska; Joanna CymanLower bound on the distance k-domination number of a treeMathematica Slovaca, Vol. 56 (2006), No. 2, 235--243Persistent URL: http://dml.cz/dmlcz/130763Terms of use:© Mathematical Institute of the Slovak Academy of Sciences, 2006Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access todigitized documents strictly for personal use. Each copy of any part of this document must containthese Terms of use.This paper has been digitized, optimized for electronic delivery and stampedwith digital signature within the project DML-CZ: The Czech Digital MathematicsLibrary http://project.dml.czMathematica Slovaca ©2006 -. .-.- //-.>-.«̂ \ ... /-. «.-.-- «..« Mathematical Institute Math. SlOVaCa, 5 6 ( 2 0 0 6 ) , NO. 2, 2 3 5 - 2 4 3 Slovák Academy of Sciences LOWER BOUND ON THE DISTANCE fc-DOMINATION NUMBER OF A TREE J O A N N A R A C Z E K — M A G D A L E N A L E M A Ň S K A — J O A N N A C Y M A N (Communicated by Martin Skoviera ) ABSTRACT. A subset D of vertices of a graph G = (V, E) is said to be a distance k -dominating set of G if every vertex of V — D is at distance at most k from some vertex of D. The minimum size of a distance k-dominating set of G is called the distance k-domination number of G. We prove that for each tree T of order n with nx end-vertices, the distance k-domination number is bounded below by (n + 2k — k-n1)/(2k-fl) and we characterize the corresponding extremal trees. 1. Introduction Let G be a finite simple graph and let k > 1 be an integer. A set D of vertices of G is said to be distance k-dominating if any vertex not in D is within distance k from some vertex of D. The distance k-domination number lk(G) of G is the smallest number of vertices of a k-dominating set in G. Note that the distance 1-domination number is the domination number j(G). This kind of domination was defined by H e n n i n g , O e l l e r m a n n and S w a r t [1], As an illustration, let G be a graph associated with the road grid of a city, the vertices of G corresponding to the street intersections. Two vertices of G are adjacent if and only if the corresponding street intersections are adjacent (i.e. block apart). Then 7(G) is the smallest number of policemen who may be placed at intersections so that every intersection is at most one block away from a policeman. If the prescribed distance of each intersection (or intersection and policeman) from a policeman is changed from at most one block to at most k — 1 blocks, k > 2, then the minimum number of policemen required is 7*.(G). In this paper we consider the distance domination number of non-trivial trees. Let n = n(T) be the order of T and let nx = nx(T) denote the number of end-vertices of T. 2000 M a t h e m a t i c s S u b j e c t C l a s s i f i c a t i o n : Primary 05C05, 05C69. K e y w o r d s : distance k-domination number, tree. 235 JOANNA RACZEK — MAGDALENA LEMANSKA — JOANNA CYMAN L e m a n s k a [2] proved that each non-trivial tree T satisfies the inequality 7(-0 > (n + 2 — n 1 ) / 3 and characterized the extremal trees. The purpose of this paper is to generalize this result to the distance version of the domination number. In particular, we prove that for each tree T of order n the distance fc-domination number is bounded below by (n + 2fc — fc • n1)/(2fc + 1) and we characterize the corresponding extremal trees. 2. Proof of the bound Our aim in this section is to present and prove a lower bound on the distance fc-domination number of a tree. LEMMA 1. If T is a tree with >yk(T) = 1, then k • nx(T) > n(T) - 1. P r o o f . We proceed by induction on the number of end-vertices of a tree T . If nx(T) = 2, then T is a path P. As jk(T) = 1, it follows that the path has at most 2fc + 1 vertices. Thus fc • n x (P) = fc • 2 = (2fc + 1) - 1 > n(P) - 1. Assume now that the result is true for all trees T' with n1(T') = 2 , . . . , j and 7A.(T/) = 1. Let T be a tree with ^k(T) = 1 and nx(T) = j + 1. Let p be the smallest integer such that ( x 0 , . . . ,xp) is a path in T with dT(x0) = 1 and dT(xp) > 2. Observe that 1 < p < fc, as diam(T) < 2fc, and T is not a path. Let V1 = {x0,xv... , x p _ 1 } . Obviously \VX\ < fc. Let us remove from T all vertices of V1. By induction we have k-n^T-VJ ^ ( T - ^ j - l . Since nx(T) - l = n1(T-V1) and n(T - VJ > n(T) - fc, it follows that fc- ( n x ( T ) - l ) >n(T)-k-l. Thus k>nx(T) > n ( T ) - l , which completes the induction step. • The open k-neighbourhood of a vertex x G V(G), denoted NQ(X) , is the set {v e V(G) : 0 < dG(v,x) < fc}. The set N%[x] = N%(x) U {x} is called the closed k-neighbourhood of v in G. Let us define PNQ[X,D] = NQ[X] — NQ[D — {x}] to be the private distance k-neighbourhood of a vertex x, with respect to a set D. If y G PNQ[X,D], we say that y is a private distance k-neighbour of x. The set of end-vertices of G is denoted by Q(G). We continue with a basic property of minimal fc-dominating sets, due to H e n n i n g , O e l l e r m a n and S w a r t [1]. 236 LOWER BOUND ON THE DISTANCE k-DOMINATION NUMBER OF A TREE PROPOSITION 2. Let D be a distance k-dominating set of a graph G for some k > 1. Then D is a minimal distance k-dominating set if and only if each vertex u G D satisfies at least one of the following conditions: 1. there exists a vertex v G V(G) — D for which NG(v) HD = {u}; 2. dG(u, w) > k for every vertex w G D — {u}. For a given tree T , let S = (s0 , s l 5 . . . , st) be a longest path in T . Assume I > 2k. Let P0,P1,... ,Pt be a partition of V(T) such that p. =- {y G V(T) : dT(v, 3i) = dT (v, V(S))} . Observe that dT(s{, x) < min{z, l—i} only if x G P{, as otherwise there would be a path longer than 5 in T . Let D be a minimum distance ^-dominating set in T . We say that D has a property T if sk G D and ^ dT(v,V(S)) is minimum. veD LEMMA 3. If D has the property T, then sk distance k-dominates all vertices mP0U.--UPk. P r o o f . As 5 is a longest path in T and sk G V(S), the result is straight-forward. D LEMMA 4. If D has the property T, then (P0 U • • • U Pk) n D = {sk} . P r o o f . Suppose that the result is not true, i.e. let x G (P0 U • • • U Pk) n D and x ^ sk.By Lemma 3, sk distance k-dominates x. Thus, by Proposition 2, x has a private distance k-neighbour, say y. As y is not distance k-dominated by sk, we have y ^ P0 U • • • U Pk and dT(sk,y) > k. It follows that dT(x, y) = dT(x,sk) + dT(sk,y) > k, which is a contradiction with the fact, that y is a private distance k-neighbour of x. Thus such a vertex x does not exist, as claimed. D LEMMA 5. If D has the property T and x G D C\Pi, then dT(si,x) <i — k. P r o o f . Suppose to the contrary that D is a minimum distance k-domi-nating set having the property T in T and let x G Dr\Pi such that dT(s{, x) > i — k. Thus i — k < dT(s^x) < i. It follows that there exists a vertex y such that dT(s{1x) = dT(si,y) + dT(y,x) and dT(si,y) = i — k. Hence dT(y,x) = dT(svx) -dT(s^y) < k. We claim that NT[x] C NT[y]. Suppose, contrary to our claim, that there exists a vertex z belonging to NT[x] — NT[y]. Then from dT(z,y) > k and dT(si,y) = i — k we find that either dT(si,z) > i or dT(svz) < i — 2k. The inequality dT(si, z) > i implies that there is a longer path than S in T , which is a contradiction. If dT(si, z) < i — 2k, then dT(x, z) > dT(x, sz) — dT(s^ z) > i — k — (i — 2k) = k, which contradicts z G N^[x]. 237 JOANNA RACZEK — MAGDALENA LEMANSKA — JOANNA CYMAN Consequently, the set D' = D -{x} U {y} is a minimum distance fc-dominating set in T, where J ] dT(v,V(S)) < ^ dT(v, V(S)), again a contradiction. • veD' veD LEMMA 6. If PN*[sk,D] n P. ^ 0 for i G {fc+1, . . . , 2fc}, then D n (Pk+l U . - . U P . ) = 0. P r o o f . For a contradiction let us suppose that x G PNT[sk,D]nPi for i G {fc+1, . . . , 2fc} and let there exist a vertex z G D n P r , where r G {fc+1, . . . , i}. By Lemma 5, dT(z,sr) < r — fc. As sk distance fc-dominates x, we have dT(sk, x) = dT(sk, sr) + dT(sr, x) = r - fc + dT(sr, x) < k. Hence, dT(sr,x) < 2fc — r . On the other hand dT(z, x) = dT(z, sr) + dT(sr, x) < (r - fc) + (2fc - r) = fc , which means that z distance fc-dominates x, a contradiction, as x is a private distance fc-neighbour of sk. • By Lemmas 4 and 6 we find the following corollary. COROLLARY 7. If PNk[sk, D] n P - / 0 for i G {fc+1,. . . ,2fc}, tfien fln(P0U •-UPt) = {sk}. LEMMA 8. I/PjY|[5fc,I9] n P - T-- 0 /Or i G {fc+1, . . . , 2fc}. £fterc 5^ distance k-dominates all vertices in PkJrl U • • • U P^. P r o o f . Suppose that the result is not true, i.e. suppose that x G PNT[sk,D] n P% for i G {fc+1, . . . , 2fc} and sk does not distance fc-dominate a vertex y G P- , where / G {fc+1, . . . , i}. Then there exists a vertex z G H f l P r such that z distance fc-dominates y and z does not distance fc-dominate x. Corollary 7 implies that i < r. We need to consider the inequality chain: k < j < i < r. As z distance fc-dominates y, we have dT(z,y) = dT(z,sr) + dT(sr,Sj) + dT(sj,y) = dT(z,sr) +r - j + dT(s^y) < fc. As z does not distance fc-dominate x we can write that dT(z, x) = dT(z, sr) + dT(sr, 5-) + dT(sv x) = dT(z, sr)+r - i + dT(s{, x) > k . Combining these inequalities we obtain dT(spy) -j< dT(st1x) - i . As sk distance fc-dominates x, we have dT(sk,x) < k and thus dT(si,x) < 2fc — i. Similarly, as sk does not distance fc-dominate T/, we have dT(s-,y) > 2fc - j . Thus -2j + 2fc < dT(Sj,y) - j < dT(s{, x) - i < 2fc - 2i , which gives i < j , which is impossible. • From Lemma 3 and Lemma 8 we have: 238 LOWER BOUND ON THE DISTANCE k-DOMINATION NUMBER OF A TREE .» COROLLARY 9. If PN*[sk,D] n Pi ^ 0 for i G { k + 1 , . . . , 2 k } , then sk dis-tance k-dominates all vertices in P0 U • • • U Pi. Here and subsequently, for a tree T and an edge uv, let Tx and T2 be the components of T — uv to which vertices u and v belong, respectively. LEMMA 10. Ifjk(T) > 1, then there exists an edge uv in T such that 7^(--ri) + %(T2) = %(T). P r o o f . Let S = (s0, s1,..., st) be a longest path in T. As jk(T) > 1, we have I > 2k. Let D be a minimum distance k-dominating set having property J7 in T. According to Lemma 4, we conclude that s0 G PNj,[sk,D]. Let i = max{j : P- n PN^[sk, D] ^ 0} . We need to consider two cases: Case 1: i < k. If i < k, then we remove the edge sksk+1 and we obtain two trees: Tx = (P0U- • -UP,) and T2 = (P*+1U- • -UPZ). By Lemma 4, (P0U- -UPk)nD = {sk} and by Lemma 3, sk distance k-dominates all vertices in Tx. Moreover, sk has no private distance k-neighbour among vertices of PA,+1 U- ••UP/ i n T . Therefore 7,(r1) = iand7 ifc(r2) = 7 f c ( r ) - i . Case 2: k <i < 2k. If k < i < 2k, then we remove the edge sisi+1 and we obtain two trees: Tx = (P0 U • • • U P.) and T2 = (P . + 1 U - - U P , ) . According to Corollary 7, (P0 U • • • UP{) nD = {sk} and by Corollary 9, sk distance k-dominates all vertices in Tx and sk has no private distance k-neighbour among vertices of P i + 1 U • • • UPt in T. Therefore 7fc (Tx) = 1 and 7* (T2) = 7fc (T) - 1. • We are now in position to prove the main result of this paper. THEOREM 11 . IfT is a tree, then k • nx(T) > n(T) + 2k- (2k + l)ik(T). P r o o f . We proceed by induction on 7^ (T). If ^k (T) = 1, the result follows by Lemma 1. We assume now that the result is true for all trees T' with ik(T') = 1,.. .,j. Let T be a tree with jk(T) = j + 1. Let S = (s0,s1:... ,st) be a longest path in T. Note that / > 2k, as jk(T) > 1. Let D be a minimum distance k-dominating set in T such that sk e D and X) ^ T C ^ ^ ( ^ ) ) i s minimum. v<ED From Lemma 10 there exists an edge uv of T such that for the two components Tx, T2 of T - uv, we have jk(T) = 7*(r i ) + 7/c(T2) • % induction we have the following inequalities for Tx and T2 : 239 JOANNA RACZEK — MAGDALENA LEMANSKA — JOANNA CYMAN fc • njCTi) > n(Tt) + 2k- (2k + l)7fc(-"1) and k • nr(T2) > n(T2) + 2k- (2k + l)ik(T2). Summing those inequalities we obtain k • (n^TJ + nx(T2)) -2k> n(T) + 2k - (2k + l)-yk(T). Observe that nx(T) > nx(Tx) + nx(T2) - 2, so k • na (T) > n(T) + 2k - (2k + l)lk (T), which completes the proof of the inequality. • 3. Characterization of the extremal trees We are now able to provide a characterization of all the trees for which k • nx(T) = n(T) + 2k - (2k + 1)%(T). For this purpose, we define a family 11 to be a family of all trees for which dT(u, v) = 2k mod (2k + 1) for each two end-vertices u, v, where u ^ v. For a given integer j > 2, a k-spider is a graph obtained by attaching j disjoint paths of length k to a single vertex of Kx. Let T be a tree belonging to the family 1Z and let S = (s0 , 5 1 ? . . . , st) be a longest path in T . Since s0, st are end-vertices of T, we conclude that / = /(2k + 1) + 2k for some non-negative integer t. Let I = {0, /} U {j(2k + 1) + k : j = 0 , 1 , . . . , t} be a set of indexes. LEMMA 12. If T is a tree belonging to the family 1Z and S = (s0 , s 1 ? . . . , st) is a longest path in T, then Pi D fi(T) = 0 for i £ I. P r o o f . Suppose that x is an end-vertex belonging to Pi and i £ I. As T G K, we conclude that dT(sQjx) = tx(2k + 1) + 2k for some non-negative integer t1 and thus dT(svx) = tx(2k + 1) + 2k — i. On the other hand we have dT(svx) = 0JT(5/,5.) + rfT(5.,x) = £(2k + l) + 2 k - i + £1(2k+ l) + 2fc-z = t2(2k + l ) + 4 k - 2 i , where t2 = t+tl. As T G 7^, it follows that dT(st,x) = 2k mod ( 2 k + l ) . Hence, it is required to be 4k — 2i = t3(2k + 1) + 2k, where t3 is an integer and thus i = k — ^f-(2k + 1). Since i is a positive integer, t3 must be 0 or a negative even number. For t3 = 0, - 2 , - 4 , . . . , we obtain i = k, (2k+l)+k , 2 (2k+l)+k , respectively, which are elements of 7, and it is a contradiction with our assump-tions. • 240 LOWER BOUND ON THE DISTANCE k-DOMINATION NUMBER OF A TREE COROLLARY 13. If T € n, then for all vertices of a longest path S = (s0,..., st) there holds dT(si) = 2 only if i £ I. LEMMA 14. If Ten and ~fk(T) = 1, then k-n1(T) = n(T)-l. P r o o f . Let S = (s0,s1,...,sl) be a longest path in a tree T belonging to the family n. Since jk(T) = 1, we have dT(s0,st) < 2k. As T e n, we conclude that dT(s0, sz) = 2k and thus I = 2k. Moreover, dT(s0) = dT(s2k) = 1 and, by Corollary 13, dT(si) = 2 for i <£ {0,fc,2fc}. Hence, if dT(sk) = 2, then T is a path P on 2fc + 1 vertices. In this case k-n1(T) = k-2 = (2k + l)-l = n(T) - 1, so the equality holds. Otherwise, dT(sk) = j > 2 and Ten imply that T is a k-spider. In this case we have fc • nx(T) = k • j = (k • j + 1) - 1 - n(T) - 1, and the equality holds as well. • LEMMA 15. If T en, then k • nx(T) = n(T) + 2k- (2k + 1)7*CO • P r o o f . We proceed by induction on 7fc (T). If ^k (T) = 1, then by Lemma 14 the equality holds. Assume now that the result is true for all tress V belonging to the family n with 7fc(T') = 1 , . . . , j . Let T e n be a tree with 7 J T ) = j + 1 and let S = (s0,s1,...,sl) be a longest path in T . Let D be a minimum distance k-dominating set with property T in T . Lemma 12 implies that each vertex s{ where i £ I has degree 2. Thus, without loss of generality, we may assume that {sk+1,..., ssk}DD = 0 and ssk+1 e D. Now we remove the edge s2ks2kJrl from T to obtain trees Tx = (P0 U • • • U P2k) and T2 = (P2k+1 U • • • U Pt) • It is clear that 7fe(^1) = 1 as sk distance k-dominates all vertices in Tx. Moreover, ~fk(T2) = 7/c(T) —1 as sk has no private distance k-neighbour in P2k+1U- • -\JPt. Furthermore, as dT(s2k) = dT(s2k+1) = 2 and dTi(s2k) = dT2(s2k+1) = 1, we conclude that nt(Tx) + nx(T2) =nx(T) + 2. We claim that Tx e n. Indeed, dT(s0,s2k) = 2k. Moreover, as T G n, we have dT(s0, x) = 2k only if x e Vt(T) n Pk. Thus dT(x, s2k) = 2k. In this case Tx is a k-spider. If Vt(T) n Pk = 0, then T is a path on 2k + 1 vertices. Hence Tx en. We also claim that T2 e n. From T e n it follows that dT(s0 ,Sj) = 2k mod (2k+ 1). Thus dT(s2 f c + 1 ,x) = dT(s0,x) - (2k + 1) = 2k mod (2k + 1) for each x e fi(T2). Hence T2en. 241 JOANNA RACZEK — MAGDALENA LEMANSKA — JOANNA CYMAN By induction, for Tx and T2 we have equalities k • ni(Tx) = n(Tx) + 2k- (2k + lhfcCT.) and k • nx(T2) = n(T2) + 2k- (2k + l ) 7 f e(T 2) . Summing these equalities, we obtain k.(n1(T1)+n1(T2))-2fc = n(T) + 2 k - ( 2 k + l)7 f c(T) and from nx(Tx) + nx(T2) -2 = nx(T), k • nx(T) = n(T) + 2k - (2k + l)7fc(T) and the induction is completed. D LEMMA 16. If 7fc(T) = 1 and k- nx(T) = n(T) - 1, then T eK. P r o o f . We use induction on the number of end-vertices of a tree T . If nx(T) = 2 and k • 2 = n(T) — 1, T is a path P on 2k + 1 vertices. In this case dT(u,v) = 2k for the two end-vertices i n P , s o T G K . Assume now that the result is true for all trees T' with nx (T') = 2,..., j . Let T be a tree with nx(T) = j + 1, 7fe(T) = 1 and k-nx(T) = n(T)-l. Let p be the smallest integer such that (x0,..., x ) is a path in T , where dT(x0) = 1 and dT(xp) > 2. Observe that 1 < p < k, as diam(T) < 2k and T is not a path. Let ^i = { x 0 ' x i ' • • • > xp - i}- Obviously (VJ < k. Let T' be a tree obtained from T by removing of all vertices belonging to Vx. According to Theorem 11, we have inequality k • nx(T') > n(T') + 2k - (2k + l)7 f c(T') . As nx(T') =nx(T)-l, %(T') = 1 and |Vi| < k, we obtain k • nx(T) -k = k- nx(T') > n(T') - 1 > n(T) - k - 1 . Since k • n±(T) = n(T) - 1, it follows that k-nx(T)-k = k- n^T') = n(Tf) - 1 = n(T) - k - 1 . Thus \Vx\ = k and k-nx(T') = n(T') -1. By induction we find that T' e K. As Jk(T) = 7/e(T/) = 1 and |V̂ _| = k, we conclude that T is a k-spider and thus TeK. D LEMMA 17. If k • nx(T) = n(T) + 2k - (2k + l)jk(T), then T eK. P r o o f . We proceed by induction on 7 fc(T). If jk(T) = 1, then by Lemma 16 the equality holds. 242 LOWER BOUND ON THE DISTANCE k-DOMINATION NUMBER OF A TREE Assume now that the result is true for all trees T' with 7/,(T') = 1,..., j . Let T b e a t r e e w i t h 7 f c ( T ) =j + l and let k-nx(T) = n(T) + 2k-(2k + l)lk(T). By Lemma 10, there exists an edge uv G E(T) such that T — uv has two components Tx and T2 and lk(Tx) + *yk(T2) = 1k(T). Theorem 11 implies that k - n^TJ > n(Tx) + 2k- (2k + l)lk(Tx) (1) and k • n, (T2) > n(T2) + 2k - (2k + 1 ) 7 , (T 2 ) . (2) By summing (1) and (2) and applying n^T) > n1(T1) + n1(T2) — 2 we obtain k-nx(T)>k- (nx(TJ + nx(T2)) - 2k > n(T) + 2k - (2k + l)-yk(T). As k • nx(T) = n(T) + 2k- (2k + 1)7*.(T), we conclude that k-nx(T) = k- (ni(Tt) + nx(T2)) -2k = n(T) + 2k - (2k + l)-yk(T), which implies that in (1) and (2) we have equalities and nx(T) = n±(T^ + nx(T2) — 2. Thus, by induction, TX,T2 £ 1Z. Moreover, if e = uv was the edge we removed from T to obtain Tx and T2 , then dT (u) = dT (v) = 1. It follows that dTi(u,x) = 2k mod (2k + 1) for any x e f ^ T j and dT (v,y) = 2k mod (2k + 1) for any y G H(T2). Hence dT(x,y) = 2k mod (2k + 1) for all x,y e n(T) and thus T eK. • By Lemmas 15 and 17 we have the following: THEOREM 18. If T is a tree, then k • nx(T) = n(T) + 2k - (2k + l)jk(T) if and only if T belongs to the family 1Z. REFERENCES [1] HENNING, M. A.—OELLERMANN, O. R .—SWART, H. C : Bounds on distance domi-nation parameters, J. Combin. Inform. Sуstem Sci. 16 (1991), 11-18. [2] LEMANSKA, M . : Lower bound on the domination number of a tree, Discuss. Math. Graph Theorу 24 (2004), 165-169. Received September 21, 2004 Department of Discrete Mathematics Revised December 20, 2004 Faculty of Applied Physics and Mathematics Gdańsk University of Technology Narutowicza 11/12 PL-80-952 Gdańsk POLAND E-mail: gardenia@mif.pg-gda.pl  magda@mif.pg-gda.pl  joana@mif.pg.gda.pl 243 

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Lower bound on the distance $k$ -domination number of a tree

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Lower bound on the distance kkk-domination number of a tree

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Lower bound on the distance $k$ -domination number of a tree