A generalized linear time algorithm for an optimal k-distance dominating set of a weighted tree

Accepted Manuscript A generalized linear time algorithm for an optimal k-distance dominating set of a weighted tree

Sukhamay Kundu

PII: DOI: Reference:

S0020-0190(17)30187-4 https://doi.org/10.1016/j.ipl.2017.10.009 IPL 5599

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Information Processing Letters

Received date: Revised date: Accepted date:

9 January 2017 26 October 2017 28 October 2017

Please cite this article in press as: S. Kundu, A generalized linear time algorithm for an optimal k-distance dominating set of a weighted tree, Inf. Process. Lett. (2017), https://doi.org/10.1016/j.ipl.2017.10.009

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Highlights • It generalizes a previously known linear time algorithm for a k-hop optimal dominating set of an unweighted trees to the case of weighted trees. • This significantly increases the potential applications of dominating set in many real life problems. • The algorithm determines an optimal k-distance dominating set by selecting one node at a time in a bot- tom-up fashion using a rooted form of the tree.

Journal Logo Information Processing Letters (2017) 1–6

A Generalized Linear Time Algorithm for an Optimal k-distance Dominating Set of a Weighted Tree Sukhamay Kundu Dept of Computer Science and Engineering Louisiana State University, Baton Rouge, LA 70803, USA.

Abstract The new algorithm finds an optimal k-distance dominating set D in a weighted trees T = (V, E) with link-weights w(x, y) > 0 in time O(|V |). It generalizes a previous linear time algorithm for an optimal k-hop dominating set for an unweighted tree. This significantly increases the applications of dominating sets in trees. Keywords: k-distance dominating set, linear algorithm, weighted tree

1. Introduction Given an weighted (undirected) graph G = (V, E), with link weights w(x, y) > 0 for (x, y) ∈ E, and a constant k > 0, we call a subset of nodes D ⊆ V a k-distance dominating set, in short, a k-ddset if each node y in G has distance dist(x, y) ≤ k from some node x ∈ D. Here, dist(u, v) is the shortest length of a path in G connecting the nodes u and v (in short, an uv-path) based on the link lengths w(x, y). Clearly, D = V is a k-ddset for any G. We say D is an optimal k-ddset if |D| is the smallest among all k-ddsets. In [7], we presented a linear time algorithm for an optimal k-ddset for a tree T with each link-weight w(x, y) = 1. We generalize that algorithm for the case of an weighted tree without increasing the time complexity. If each w(x, y) = 1, the new algorithm creates the same optimal k-ddset as in [7]. We refer the reader to [7] for the literature review and the many applications of k-ddsets. y or, equivalently, y is dominated Let Δ(x) = {y : dist(x, y) ≤ k}; clearly, x ∈ Δ(x). We say x dominates  by x if y ∈ Δ(x). More generally, for S ⊆ V , we write Δ(S) = x∈S Δ(x) and for y ∈ Δ(S) we say S dominates y or y is S-dominated; for y ∈ / Δ(S), we say y is S-undominated. A set of nodes D is a k-ddset if and only if Δ(D) = V . If w(x, y) > k, then removing (x, y) from G does not alter Δ(u) for any u ∈ V and hence does not alter the k-ddsets or the optimal k-ddsets of G. We assume henceforth that each w(x, y) ≤ k. We sometimes write S + x to mean S ∪ {x}, when x ∈ / S. 2. Two Basic Theorems The results below are generalizations of similar results in [7] for weighted trees, and they play a critical role in the new algorithm presented in §3. Let T denote a rooted form of an unrooted tree T , with an Email address: [email protected] (Sukhamay Kundu)

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arbitrary node chosen as the root = root(T ). Because the xy-path for two nodes x and y = x is the same in T and in T , we have the distance dist(x, y), the k-ddsets, and the optimal k-ddsets are the same for T and T . As in [7], we work with a rooted form T of T to provide a ”direction” for efficient identification of successive nodes that can be safely included in an optimal k-ddset D, starting from D = ∅. In particular, we choose the successive nodes of an optimal k-ddset D in a bottom-up fashion, meaning that following the addition of a node z to D no other node in the subtree T z of T at z will be added to D. Theorem 1 below shows how to select a bottom-most node in D and Corollary 2 shows how to select all bottom-most nodes in D. For selecting the other nodes in D, we use a generalization of Theorem 1 given in Theorem 3. Let depth(T x ) = max {dist(x, y) : y ∈ T x }, parent(x) = the parent of x when x = root, and children(x) = {y : parent(y) = x}. Clearly, depth(T x ) = 0 if and only if x is a terminal node, and each node x = root is dominated by parent(x). If depth(T ) > k, then either some x ∈ children(root) has depth(T x ) > k or each x ∈ children(root) has depth(T x ) ≤ k and hence a child x of root such that depth(T x ) ≤ k < depth(T ) = depth(T x ) + w(parent(x), x). By applying the argument repeatedly, we see that depth(T ) > k implies there is a node x ∈ T − {root} such that depth(T x ) ≤ k < depth(T x ) + w(parent(x), x). By abuse of language, we sometimes use T x to denote the set of nodes in T x . Theorem 1. If depth(T ) ≤ k, then D = {root} is an optimal k-ddset of T ; otherwise, there is a node x ∈ T − {root} such that depth(T x ) ≤ k < depth(T x ) + w(parent(x), x) and, for each such node x, T has an optimal k-ddset D containing x. Proof. We only need to consider the case depth(T ) > k. Let y ∈ T x be a terminal node in T such that dist(x, y) = depth(T x ) ≤ k. If π(y) = yq , · · · , yp , · · · , y1 , y is the path from root(T ) = yq to y, where yp = x, then q > p because x = root; moreover, k < depth(T x ) + w(parent(x), x) implies dist(x, y) = depth(T x ) > 0 and hence p ≥ 1. See Fig. 1. Note that depth(T parent(x) ) may be larger than depth(T x ) + w(parent(x), x). / D, then there is a node z ∈ D ∩ T x , z = x, because parent(x) does Let D be an optimal k-ddset of T . If x ∈ not dominate y and hence no node outside T x dominates y. Also, because T x ⊆ Δ(x) and Δ(z) ∩ (T − T x ) ⊆ Δ(x) ∩ (T − T x ), we have Δ(z) ⊆ Δ(x). Thus, D = D − z + x is a k-ddset of T and hence an optimal k-ddset. (Note: the choice of x in D does not constrain the choice of the other nodes in D , if any.)

root=y q ⋅⋅⋅

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Figure 1. The path from root(T) to a terminal node y, where dist(x, y) = depth(T x ) ≤ k. The remark at the end of the proof of Theorem 1 immediately shows that if depth(T ) > k and B = {x : depth(T x ) ≤ k < depth(T x ) + w(parent(x), x)}, then T has an optimal k-ddset D ⊇ B. The optimal k-ddset computed by our algorithm will also contain B. The following definition is motivated by the fact that we build an optimal k-ddset D in a bottom-up fashion. Definition 2.1. A subset of nodes S ⊆ V is called descendant-dominating (in short, desc-dominating) if  {T z : z ∈ S} ⊆ Δ(S), i.e., each descendant of a node z ∈ S is dominated by some node in S. The empty-set is considered to be trivially desc-dominating. (Theorem 3 below reduces to Theorem 1 when S = ∅.) A set S = {z} is desc-dominating if and only if depth(T z ) ≤ k. On the other hand, if S is desc-dominating, z ∈ S, and depth(T z ) > k, then T z contains some other nodes of S. Clearly, each k-ddset D is desc-dominating. The following lemma is immediate from Def. 2.1 and is stated here without a proof. 2

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It shows that to expand a desc-dominating set S into an optimal k-ddset we would never add to S any descendant of S-nodes. Lemma 2. If S is a desc-dominating set, z ∈ S, and D ⊇ S is an optimal k-ddset then (D − S) ∩ T z = ∅. For a desc-dominating set S and an arbitrary node x, we write U (x, S) = T x − Δ(S ∩ T x ) ⊆ T x , the set of nodes in T x that are not dominated by the S-nodes in T x ; clearly, U (x, S) does not depend on S − T x , U (x, S) = U (x, S ∩ T x ), and U (x, ∅) = T x . In general, the nodes U (x, S) may not form a connected subset of nodes. Thus, we define U + (x, S) = {y  : y  ∈ xy-path for some y ∈ U (x, S)}, which is a connected subset of nodes in T x containing x if U (x, S) = ∅. Figs. 2(i)-(ii) show a tree with two different choices of the root. For S = {5} and k = 4, both Figs. 2(i)-(ii) give U (1, S) = {2, 14} and U + (1, S) = {1, 2, 13, 14}. However, if we let w(3, 4) = 3 ≤ k, then Fig. 2(i) gives U (1, S) = {2, 4, 14} and U + (1, S) = {1, 2, 3, 4, 13, 14} whereas Fig. 2(ii) gives the same U (1, S) and U + (1, S) as before.

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Figure 2. All link weights in T equal 1 unless indicated otherwise. Some of the optimal 4-ddsets are D = {1, 5}, {3, 5}, {1, 8}, and {5, 13}. Definition 2.2. If S is desc-dominating, then let mxdu(x, S) = max{dist(x, y) : y ∈ U (x, S)} when U (x, S) = ∅ and otherwise let mxdu(x, S) = −1. Clearly, mxdu(x, ∅) = depth(T x ), and mxdu(x, S) = 0 if and only if U (x, S) = {x}. The following theorem shows how to extend a desc-dominating set S that is a proper subset of an optimal k-ddset of T by adding a node x so that S +x is descended-dominating and is a subset of an optimal k-ddset. A repeated use of Theorem 3 clearly gives an optimal k-ddset. Because w(u, v) ≤ k for each link (u, v), the condition k < mxdu(x, S) + w(parent(x), x) in Theorem 3 implies mxdu(x, S) > 0 and hence U (x, S) = ∅. Likewise, the condition 0 ≤ mxdu(root(T ), S) ≤ k and S ⊂ D implies root(T ) ∈ / S. Theorem 3. Suppose S is desc-dominating and a proper subset of an optimal k-ddset of T . Then, T has an optimal k-ddset containing S  = S + x, if mxdu(x, S) ≤ k < mxdu(x, S) + w(parent(x), x). Moreover, if 0 ≤ mxdu(root(T ), S) ≤ k, then S  = S + root(T ) is an optimal k-ddset. In each case, S  is desc-dominating. Proof. Consider the case mxdu(x, S) ≤ k < mxdu(x, S) + w(parent(x), x) and hence x = root and mxdu(x, S) > 0. Let y ∈ U (x, S) ⊆ T x such that dist(x, y) = mxdu(x, S) and therefore k < dist(x, y) + w(parent(x), x). Thus, y is not dominated by parent(x) and hence not dominated by any node not in T x ; moreover, y is not dominated by S ∩ T x . This means an optimal k-ddset D ⊃ S must contain a node z ∈ T x . If z = x, then as in the proof of Theorem 1 the set D − z + x is also an optimal k-ddset containing S + x. Also, mxdu(x, S) ≤ k implies S + x is desc-dominating. For 0 ≤ mxdu(root(T ), S) ≤ k, S+root(T ) is trivially an optimal k-ddset and hence is desc-dominating. Example 1. Consider the rooted tree T in Fig. 2(i) and let k = 4. The only node that satisfies Theorem 1 is x = 5. For S = {5}, the only nodes not in Δ(S) are {2, 14} = U (1, S); on the other hand, U (3, S) = ∅ 3

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and thus mxdu(3, S) = −1. Because 0 < mxdu(1, S) = 3 ≤ k, D = {5, 1} is an optimal desc-dominating 4-ddset by Theorem 3. Consider now the rooted tree in Fig. 2(i) and let k = 3. The only node satisfying Theorem 1 is x = 8. For S = {8}, Δ(S) = {3, 5, 6, 8, 9, 10, 11, 12} and because of node 7 ∈ / Δ(S) the only node which satisfies Theorem 3 is x = 5, with mxdu(5, S) = 3 ≤ 3 < mxdu(5, S) + w(parent(5), 5) = 3+1. This means there is an optimal 3-ddset of T containing the desc-dominating set S = {5, 8}. Now, the only nodes not in Δ(S) are {2, 13, 14} = U (1, S) and only the root = 1 satisfies Theorem 3. This gives us an optimal desc-dominating 3-ddset D = {8, 5, 1}. (Note that for S = {8}, U (6, S) = {6, 7} and mxdu(6, S) = 2 ≤ 3 ≮ 2 + w(parent(6), 6) = 3, and therefore Theorem 3 does not apply to expand S = {8} to S + 6 = {8, 6}. However, {8, 6, 1} is an optimal 3-ddset containing {8, 6}. Thus, the condition mxdu(x, S) ≤ k < mxdu(x, S) + w(parent(x), x) in Theorem 3 is not a necessary condition for the existence of an optimal k-ddset containing S + x.) If we remove node 14 from T in Fig. 2(i), we would still get the optimal 3-ddset D = {8, 5, 1} as before. However, if we remove node 7 in Fig. 2(i), then we would expand S = {8} to an optimal 3-ddset D = {8, 1}. 3. Algorithm for Optimal k-ddset The algorithm OptKddset(x) given below determines an optimal k-ddset D of T when called with x = root(T ). It also assigns for each node y ∈ T a node dom(y) ∈ D such that for each y ∈ D we have dom(y) = y and the set domBy(y) = {y  : dom(y  ) = y} ⊆ Δ(y) is a connected subset of nodes in T . As we indicated earlier, nodes are added to D in a bottom-up fashion, and at any intermediate stage D is desc-dominating. Each time we add a node x to the current D we keep x as high (close to root) as possible. We add x to the current D if and only if the following conditions hold: (1) D ∩ T x does not dominate T x , (2) Either x has no ancestor or none of them dominates the nodes U (x, D) ⊆ T x , and (3) U (x, D) ⊆ Δ(x). If x is added to D, then all nodes y ∈ U + (x, D), including x, are assigned dom(y) = x. We introduce the following concepts for efficient verification of the conditions (1)-(3). Definition 3.1. Let nrnd(x, S) be an S-node in T x which is nearest to x with distance distNrnd(x, S) = dist(x, nrnd(x, S)) ≤ k from x, if any; x ∈ S if and only if nrnd(x, S) = x, i.e., distNrnd(x, S) = 0. Henceforth, we use the short notations mxdu(x) for mxdu(x, S), nrnd(x) for nrnd(x, S), and distNrnd(x) for distNrnd(x, S)), when S is the current set of nodes added to the optimal k-ddset D being constructed. We assume that the nodes are numbered 0 ≤ x < n, where n = |V | and we write nrnd(x) = dom(x) = −1 to indicate they are undefined, as is the case at the start of our OptKddset-algorithm; likewise, we write mxdu(x) = −1 to mean U (x, D) = ∅, i.e, all nodes y ∈ T x is assigned a valid dom(y) ≥ 0. In particular, nrnd(x) ≥ 0 implies 0 ≤ distNrnd(x) ≤ k but it does not imply mxdu(x) = −1 because there can be nodes y ∈ U (x, D), y = x. We use an auxiliary function AssignDom(x, y) to assign dom(x ) = y to each node in x ∈ U + (x, D), i.e., the nodes x ∈ T x with dom(x ) = −1 in a bottom-up fashion. That is, if dom(x) = −1, then AssignDom(x, y) calls AssignDom(x , y) for each child x of x with dom(x ) = −1 and on return from those calls AssignDom(x, y) lets dom(x) = y and mxdu(x) = −1. On return from OptKddset(x), i.e., after completion of processing a node x (which includes the computation of nrnd(x), distNrnd(x), mxdu(x), etc based on those of the children of x), one of the following holds: (1) Each node y ∈ T x , including x, is assigned a valid dom(y) ∈ D ∩ T x and mxdu(y) is reset to −1, if needed. (2) The node x (and perhaps some other nodes in T x ) is not assigned a valid dom(x) ≥ 0 because no D-node in T x (and, in particular, the node nrnd(x), if any) dominates all of U + (x, D) and there is a possibility that some ancestor of x can dominate U + (x, D), i.e., mxdu(x) + w(parent(x), x) ≤ k. Algorithm OptKddset(x): //x = root in initial call Input: A constant k > 0 and a rooted form T of a tree T with ≥ 2 nodes, an arbitrary root, parent(x) = parent of x (parent(root) = root), and link-weights 0 < w(x, y) < k. Output: An optimum k-ddset D of T and an assignment dom(x) ∈ D for nodes in T such that z ∈ domBy(z) = {x : dom(x) = z} ⊆ Δ(z) is a connected set of nodes in T for each z ∈ D. 1. If (x = root, i.e., parent(x) = x), then initialize D = ∅, let dom(y) = nrnd(y) = mxdu(y) = −1 and distNrnd(y) = 1 + k (indicating they are undefined) for each node y ∈ T root . 4

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2. If (x is a terminal node), then let mxdu(x) = 0. //0 ≤ mxdu(x) + w(parent(x), x) ≤ k 3. Otherwise, do (a), (b), and one of (c) and (d). (a) Call OptKddset(y) for each y ∈ children(x). (b) If (some y ∈ children(x) has mxdu(y) ≥ 0), then let tempMxdu = max{mxdu(y) + w(x, y) : y ∈ children(x) and mxdu(y) ≥ 0}. //here, 0 < tempMxdu ≤ k else tempMxdu = 0. (c) If (each y ∈ children(x) has mxdu(y) ≥ 0), then i. If ((x = root(T )) and (tempMxdu + w(parent(x), x) ≤ k)), then let mxdu(x) = tempMxdu. //x not added to D and 0 < mxdu(x) + w(parent(x), x) ≤ k ii. Otherwise, add x to D, let nrnd(x) = x, distNrnd(x) = 0, and call AssignDom(x, x). //dom(x) = x and now mxdu(x) = −1 (d) Otherwise, i.e., if (some y ∈ children(x) has mxdu(y) = −1 and hence dom(y) ≥ 0), then i. Let tempDistNrnd = min{distNrnd(y) + w(x, y) : y ∈ children(x) and mxdu(y) = −1} and let bestChild be an y for which the minimum is attained. //here, each distNrnd(y) ≤ k ii. [tempMxdu ≥ 0] If (tempDistNrnd + tempMxdu ≤ k) then let nrnd(x) = nrnd(bestChild), distNrnd(x) = tempDistNrnd, and call AssignDom(x, nrnd(bestChild)). //dom(x) = nrnd(bestChild) and now mxdu(x) = −1; x not added to D iii. Otherwise, if ((x = root) and (tempMxdu + w(parent(x), x) ≤ k)), then let mxdu(x) = tempMxdu. // nodes in U + (x, D) is dominated by parent(x), x not added to D iv. Otherwise, add x to D, let nrnd(x) = x and distNrnd(x) = 0, and call Assign-Dom(x, x). //dom(x) = x and now mxdu(x) = −1 Example 2. Fig. 3 shows the results applying OptKddset-algorithm for the tree in Fig. 2(i) and k = 4; the optimal 4-ddset obtained is D = {1, 5}. On completion of the call OptKddset(5) in step 3(c)(ii), we add the first node 5 to D, we set nrnd(5) = 5 and distNrnd(5) = 0, and the call AssignDom(5, 5) assigns dom(y) = 5 (and resets mxdu(y) = −1) for all y ∈ U + (5, ∅) = T 5 . On completion of the call OptKddset(3) in step 3(d)(ii), we get tempDistNrnd = 3 (bestChild = 5), we set nrnd(3) = 5 and distNrnd(3) = 2, and the call AssignDom(3, 5) assigns dom(y) = 5 for all y ∈ U + (3, {5}) = {3, 4}. Finally, on completion of the call OptKddset(1) in step 3(d)(iv), we get tempDistNrnd = 3 (bestChild = 3), tempMxdu = 3, and tempDistNrnd + tempMxdu = 3+3 = 6 > k, we add 1 to D, we set nrnd(1) = 1 and distNrnd(1) = 0, and the call AssignDom(1, 1) assigns dom(y) = 1 for all y ∈ U + (1, {5}) = {1, 2, 13, 14}. Now consider the tree in Fig. 2(ii), which is the same tree in Fig. 2(i) but with root = 3. The optimal 4-ddset obtained is now D = {3, 5}. We first add node 5 to D as before on completion of the call OptKddset(5) in step 3(c)(ii). Then, on completion of the call OptKddset(3) in step 3(d)(iv), we add 3 to D. Theorem 4. OptKddset(root) correctly computes an optimal k-ddset of T in O(|V |) time. Proof. The correctness of OptKddset(root) follows directly from Theorem 3. For the complexity, the function AssignDom(x, y) visits the set of nodes U + (x, S) by recursively calling AssignDom(x , y) for {x ∈ children(x) : dom(x ) = −1}, and then it assigns dom(x) = y and resets mxdu(x) = −1 to satisfy the property that mxdu(x) = −1 if and only if dom(x) ≥ 0. Thus, all calls of AssignDom(x, y) taken together visits a node z at most twice, once arriving at z for the first time and assigning dom(z) ≥ 0 and later we may arrive at z from parent(z) once more to find that dom(z) ≥ 0 and thus do not go further down. The total time for all calls of AssignDom(·, ·) is therefore O(|V |) time. Steps (1) and (2) for all nodes takes at most O(|V |) time. For each node x, we look at the links (x, y), y ∈ children(x), at most once in each of steps 3(b)-(d) to compute one or more of tempMxdu, tempDistNrnd, and bestChild, using a constant time per link. Also, the parent-link (parent(x), x) is looked at most once in each of steps 3(c)-(d). This gives the total computation time O(|V |) for OptKddset(root).

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Node temp- temp- bestdistmxdu(x) nrnd(x) dom(x) x Mxdu DistNrnd Child Nrnd(x) 2 − − − 0 −1 5 −1 4 − − − 0 −1 5 −1 7 − − − 0 −1 5 −1 6 2 − − 2 −1 5 −1 10 − − − 0 −1 5 −1 9 1 − − 1 −1 5 −1 11 − − − 0 −1 5 −1 8 3 − − 3 −1 5 −1 12 − − − 0 −1 5 −1 5 4 − − −1 5 0 5 3 1 2 5 −1 5 2 5 14 − − − 0 −1 5 −1 13 2 − − 2 −1 5 −1 1 3 3 3 −1 1 0 1

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Figure 3. Illustration of algorithm OptKddset for k = 4 and the tree in Fig. 2(i); nodes x are shown in the order of completion of the call OptKddset(x). 4. Conclusion We have given here a linear time O(|V |) algorithm to determine an optimal k-distance dominating set D of a weighted tree T . We use a rooted form of T , with an arbitrary choice of the root as in [7], to direct a bottom-up search for the nodes included in D, one at a time. Unlike [7], the decision to include a node x in D is now dependent on both the subtree T x and the weight of the link (parent(x), x), if exists. Finding all optimal k-distance dominating sets of a (weighted) tree T efficiently remains a challenging problem. Acknowledgement I thank the anonymous referee for several helpful suggestions to improve the presentation. [1] E. Cockayne, S. Goodman, and S. Hedetniemi. A Linear Algorithm for the Domination Number of a Tree, Information Processing Letters, 41–44, 1975. [2] E.D. Demaine, F.V. Fomin, M. Hajiaghayi, and D.M. Thilikos. Fixed-parameter Algorithms for (k, r)-center in Planar Graphs and Map Graphs, Transactions on Algorithms, Vol. 1, No. 1, 1–16, 2005. [3] M.R. Garey and D.S. Johnson. Computers and Intractability: A Guide to the Theory of NP-Completeness, W.H.Freeman & Co., 190, 1979. [4] F. Harary and T.W. Haynes. Double Domination in Graphs, Ars Combinatoria, 55:201–213, 2000. [5] T.W. Haynes, S.T. Hedetniemi, and P.J. Slater. Fundamentals of Domination in Graphs, Marcel Dekker, Inc., 1998. [6] T.W. Haynes, S.T. Hedetniemi, and P.J. Slater (Eds.). Domination in Graphs: Advanced Topics, Marcel Dekker, Inc., 1998. [7] S. Kundu and S. Majumder. A Linear Time Algorithm for Optimal k-hop Dominating Set of a Tree, Info. Processing Letters, 116:197–202, 2016. [8] J.K. Lan and G.J. Chang. Algorithmic Aspects of the k-domination Problem in Graphs, Discrete Applied Math., 161:10–11, 2013. [9] P.J. Slater. R-domination in Graphs, Assoc. Comput. Mach., 23:446–450, 1976.

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