structured/0710.2139.structured.json

{"paper_meta":{"paper_id":"arxiv:0710.2139","title":"0710.2139","authors":[],"year":null,"published":null,"updated":null,"venue_or_source":"arxiv-papers-shard","primary_category":"cs","secondary_categories":[],"doi":null,"license":"unknown","source_type":"pdf+shard","source_url":null,"pdf_url":null},"abstract":{"raw":"","cleaned":"","offsets":[]},"full_paper_text":{"raw_ordered_text":"arXiv:0710.2139v1 [cs.CC] 10 Oct 2007\nApproximation algorithms and hardness for domination with\npropagation\nAshkan Aazami\naaazami@uwaterloo.ca\nDepartment of Combinatorics and Optimization\nUniversity of Waterloo\nMichael David Stilp\nmstilp3@gatech.edu\nSchool of Industrial and Systems Engineering\nGeorgia Institute of Technology\nNovember 2, 2018\nAbstract\nThe power dominating set (PDS) problem is the following extension of the well-known dom-\ninating set problem: find a smallest-size set of nodes S that power dominates all the nodes,\nwhere a node v is power dominated if (1) v is in S or v has a neighbor in S, or (2) v has a\nneighbor w such that w and all of its neighbors except v are power dominated. We show a\nhardness of approximation threshold of 2log1−ǫ n in contrast to the logarithmic hardness for the\ndominating set problem. We give an O(√n) approximation algorithm for planar graphs, and\nshow that our methods cannot improve on this approximation guarantee. Finally, we initiate\nthe study of PDS on directed graphs, and show the same hardness threshold of 2log1−ǫ n for\ndirected acyclic graphs. Also we show that the directed PDS problem can be solved optimally\nin linear time if the underlying undirected graph has bounded tree-width.\nKeywords: Approximation algorithms, Hardness of approximation, Dominating set, Power\ndominating set, Tree-width, Planar graphs, Greedy algorithms, PMU placement problem.\nAMS subject classifications: 68W25; 90C27\n1\nIntroduction\nA dominating set of an (undirected) graph G = (V, E) is a set of nodes S such that every node in\nthe graph is in S or has a neighbor in S. The problem of finding a dominating set of minimum size\nis an important problem that has been extensively studied, especially in the last 20 years, see the\nbooks by Haynes et al. [17, 18]. The problem is NP-hard [14], a simple greedy algorithm achieves\na logarithmic approximation guarantee∗[20], and, modulo the P ̸= NP conjecture, no polynomial\ntime algorithm gives a better approximation guarantee [27, 13].\nOur focus is on an extension called the Power Dominating Set (abbreviated as PDS) problem.\nPower domination is defined by two rules; the first rule is the same as the rule for the Dominating\nSet problem, but the second rule allows a type of indirect propagation. More precisely, given a set\nof nodes S, the set of nodes that are power dominated by S, denoted PS, is obtained as follows.\n∗An approximation algorithm for a (minimization) optimization problem means an algorithm that runs in poly-\nnomial time and computes a solution whose cost is within a guaranteed factor of the optimal cost; the approximation\nguarantee is the worst-case ratio, over all inputs of a given size, of the cost of the solution computed by the algorithm\nto the optimal cost.\n1\n\n(Rule 1) if node v is in S, then v and all of its neighbors are in PS;\n(Rule 2) (propagation) if node v is in PS, one of its neighbors w is not in PS, and all other neighbors\nof v are in PS, then w is inserted into PS.\nThe set PS is independent of the sequence in which nodes are inserted by Rule 2. Otherwise,\nthere is a minimal counter example with two maximal sequences of insertions and an “earliest”\nnode that occurs in one sequence but not the other; this is not possible. The PDS problem is\nto find a node-set S of minimum size that power dominates all nodes (i.e., find S ⊆V with |S|\nminimum such that PS = V ). We use Opt(G) to denote the size of an optimal solution for the PDS\nproblem for a graph G. Throughout, we use n to denote the number of nodes in the input graph.\nFor example, consider the planar graph in Figure 1; the graph has t disjoint triangles, and three\n(mutually disjoint) paths such that each path has exactly one node from each triangle; note that\n|V | = 3t. The minimum dominating set has size Θ(|V |), since the maximum degree is 4. The\nminimum power dominating set has size one – if S has any one node of the innermost (first)\ntriangle (like v), then PS = V †.\nv\nFigure 1: Illustrating those nodes power dominated by Rule 1 (denoted by a triangle) and Rule 2\n(denoted by a square); the picked node is shown by a circle.\nThe PDS problem arose in the context of electric power networks, where the aim is to monitor all\nof the network by placing a minimum-size set of very expensive devices called phase measurement\nunits; these units have the capability of monitoring remote elements via propagation (as in Rule 2);\nsee Brueni [6], Baldwin et al. [4], and Mili et al. [28]. In the engineering literature, the problem is\ncalled the PMU placement problem.\nOur motivation comes from the area of approximation algorithms and hardness results. The\nDominating Set problem is a so-called covering problem; we wish to cover all nodes of the graph\nby choosing as few node neighborhoods as possible. In fact, the Dominating Set problem is a\nspecial case of the well-known Set Covering‡ problem.\nSuch covering problems have been extensively investigated. One of the key positive results dates\nfrom the 1970’s, when Johnson [20], Lov ́asz [26] and later Chv ́atal [8] showed that the greedy method\nachieves an approximation guarantee of O(log |V |) where |V | denotes the size of the ground set, see\nalso [30]. Several negative results (on the hardness of approximation) have been discovered over\n†In more detail, we apply Rule 1 to see that all the nodes of the innermost (first) triangle and one node of the\nsecond triangle are in PS; then by two applications of Rule 2 (to each of the nodes in the first triangle not in S), we\nsee that the other two nodes of the second triangle are in PS; then by three applications of Rule 2 (to each of the\nnodes in the second triangle) we see that all three nodes of the third triangle are in PS; etc.\n‡Given a family of sets on a groundset, find the minimum number of sets whose union equals the groundset.\n2\n\nthe last few years: Lund and Yannakakis [27] showed that the Set Covering problem is hard to\napproximate within a ratio of Ω(logn) and later, Feige [13] showed that it is hard to approximate\nwithin a ratio of (1 −ǫ) ln n, modulo some variants of the P ̸= NP assumption.\nA natural question is what happens to covering problems (in the setting of approximation algo-\nrithms and hardness results) when we augment the covering rule with a propagation rule. PDS\nseems to be a key problem of this type, since it is obtained from the Dominating Set problem by\nadding a simple propagation rule.\n1.1\nPrevious literature\nApparently, the earliest publications on PDS are Brueni [6], Baldwin et al. [4], and Mili et al. [28].\nLater, Haynes et al. [16] showed that the problem is NP-complete even when the input graph is\nbipartite; they presented a linear-time algorithm to solve PDS optimally on trees. Kneis et al. [23]\ngeneralized this result to a linear-time algorithm that finds an optimal solution for graphs that have\nbounded tree-width, relying on earlier results of Courcelle et al. [9]. Kneis et al. [23] also showed\nthat PDS is a generalization of the Dominating Set problem as follows. Given a graph G we\ncan construct an augmented graph G′ such that S is an optimal solution for the Dominating Set\nproblem on G if and only if it is an optimal solution for PDS on G′; the graph G′ is obtained from\nG by adding a new node v′ for each node v in G and adding the edge vv′. Guo et al. [15] developed\na combinatorial algorithm based on dynamic-programming for optimally solving PDS on graphs of\ntree-width k. The running time of their algorithm is O(ck2 · n) where c is a constant. Guo et al.\nalso compared the tractability of the Dominating Set problem versus PDS on several classes of\ngraphs, that is, they study whether there are classes of graphs where the former problem is in P\nbut the latter one is NP-hard; but they have no result that “separates” the two problems. Even\nfor planar graphs, the Dominating Set problem is NP-hard [14], and the same holds for PDS\n[15]. Liao and Lee [25] proved that PDS on split graphs is NP-complete, and also they presented\na polynomial time algorithm for solving PDS optimally on interval graphs. Dorfling and Henning\ncomputed the power domination number, i.e. the size of optimal power dominating set, for n × m\ngrids [12].\nBrueni and Heath [7] have more results on PDS, especially the NP-completeness of\nPDS on planar bipartite graphs. To the best of our knowledge, no further results are known on\nsolving the PDS problem, either optimally or approximately. Some of the results in this paper have\nappeared in the thesis of the second author [31], and in the proceedings of a workshop [1].\n1.2\nOur contributions\nOur results substantially improve on the understanding of PDS in the context of approximation\nalgorithms. In particular, we show a substantial gap between the approximation guarantees for the\nDominating Set problem and PDS modulo a variant of the P ̸= NP conjecture. This seems to be\nthe first known “separation” result between the two problems, in any class of graphs.\n• We present a reduction from the MinRep problem to the PDS problem that shows that PDS\ncannot be approximated within a factor of 2log1−ǫn, unless NP ⊆DTIME(npolylog(n)).\n• For undirected graphs, we introduce the notion of strong regions and weak regions as a\nmeans of obtaining lower bounds on the size of an optimal solution for PDS. Based on this,\nwe develop an approximation algorithm for PDS that gives an approximation guarantee of\nO(k) for graphs that have tree-width k. The algorithm requires the tree decomposition as\n3\n\npart of the input, and runs in time O(n3) (independent of k). By slightly modifying this\nalgorithm we get an algorithm that solves PDS optimally on trees. Our algorithm provides\nan O(√n)-approximation algorithm for PDS on planar graphs because a tree decomposition\nof a planar graph with width O(√n) can be computed efficiently [2]. Moreover, we show\nthat our methods (specifically, the lower bounds used in our analysis) cannot improve on our\nO(√n) approximation guarantee.\n• We extend PDS in a natural way to directed graphs and prove that even for directed acyclic\ngraphs, PDS is hard to approximate within the same threshold as for undirected graphs\nmodulo the same complexity assumption.\n• We give a linear-time algorithm based on dynamic-programming for Directed PDS when\nthe underlying undirected graph has bounded tree-width. This builds on results and methods\nof Guo et al. [15].\n2\nPDS in Undirected Graphs\nIn this section we prove a result on the hardness of approximating PDS by a reduction from the\nMinRep problem. In Section 2.1 we define the MinRep problem, and then we give a gap preserving\nreduction from MinRep to PDS in Section 2.2.\n2.1\nThe MinRep problem\nIn the MinRep [24] problem we are given a bipartite graph G = (A, B, E) with a partition of A and\nB into equal-sized subsets. Let qA and qB denote the number of sets in the partition of A and B,\nrespectively. Let A = A1 ∪A2 ∪· · ·∪AqA denote the partition of A, and let B = B1 ∪B2 ∪· · ·∪BqB\ndenote the partition of B. This partition naturally defines a super bipartite graph H = (A, B, E).\nThe super nodes of H are A1, A2, . . . , AqA and B1, B2, . . . , BqB. There is a super edge between\nsuper nodes Ai and Bj if there exists some a ∈Ai and b ∈Bj such that ab is an edge in G. We\nsay that super edge AiBj is covered by nodes a, b if a ∈Ai, b ∈Bj, and there is an edge between\na and b in G. Given S ⊆A ∪B we say that the super edge AiBj is covered by S if there exists\na, b ∈S that covers AiBj. The goal in the MinRep problem is to pick a minimum-size set of nodes,\nA′ ∪B′ ⊆V (G), to cover all the super edges in H. Note that we need a pair of nodes to cover\na super edge, and the pair should induce an edge between the two super nodes of the super edge;\nmoreover, a node in A′ ∪B′ may be useful for covering more than one super edge. The following\nTheorem is from [24].\nTheorem 2.1 (Theorem 5.4 in [24]) The MinRep problem cannot be approximated within ra-\ntio 2log1−ǫn, for any fixed ǫ > 0, unless NP ⊆DTIME(npolylog(n)).\n2.2\nThe reduction to PDS\nTheorem 2.2 The PDS problem cannot be approximated within ratio 2log1−ǫn, for any fixed ǫ > 0,\nunless NP ⊆DTIME(npolylog(n)).\nThe reduction: Theorem 2.2 is proved by a reduction from the MinRep problem. We create an\ninstance G = (V , E) of the PDS problem from a given instance G = (A, B, E)(H = (A, B, E)) of\nthe MinRep problem. The idea is to replace each super edge with a “cover testing gadget”.\n4\n\n1. Start with a copy of each node in A ∪B in G. For convenience, we use the same notation for\nnodes (and set of nodes) in G and their copies in G.\n2. Add a new node w∗to the graph G, and connect w∗to all nodes in A ∪B. Also add new\nnodes w∗\n1, w∗\n2, w∗\n3 and connect them to w∗(the nodes w∗\n1, w∗\n2, w∗\n3 are added to force w∗to be\nin any optimal solution. See the proof of Lemma 2.3 for more details).\n3. ∀i ∈{1, . . . , qA} , j ∈{1, . . . , qB} if AiBj is a super edge, then do the following:\n(a) Let Eij denote the set of edges between Ai and Bj in G and let lij denote |Eij| (see\nFigure 3(a); for an example E11 has 3 edges, and E12 has 4 edges). We denote the edges\nin Eij by e1, e2, · · · , ek, · · ·.\n(b) Let Cij be a cycle of 3lij nodes. We sequentially label the nodes of Cij as u1, v1, w1,\nu2, v2, w2, · · · , uk, vk, wk, · · · (informally speaking, we associate each triple uk, vk, wk with\nan edge ek of Eij). Make λ = 4 new copies of the graph Cij (λ can be any constant greater\nthan 3; refer to the proof of Lemma 2.3 for more details). For each edge ek = akbk ∈Eij\nand for each of the 4 copies of Cij, we add an edge from ak to uk and an edge from bk\nto vk. See Figures 2(a), 2(b) for an illustration.\nCij\nv2\nu2\nu1\nv1\nvlij\nulij\nek\ne2\ne1\nelij\nvk\nuk\nw1\nw2\nwk\nwlij\n(a) The Cij graph\nCij\na1\nb1\na2\nb2\ne2\ne1\nv2\nu2\nu1\nv1\nvk\nuk\nek\nak\nbk\nvlij\nulij\nblij\nalij\nelij\n(b) Edges between Cij and Ai ∪Bj.\nFigure 2: The cover testing gadget.\n4. Let G = (V , E) be the obtained graph (see Figure 3 for an illustration).\nLet S be a feasible solution for the resulting PDS instance G, and suppose w∗∈S. Then all\nof the nodes in A ∪B are power dominated (by Rule 1 of PDS). Now consider a gadget Cij, and\nassume a node v of Cij is in S. By applying Rule 1 once and then repeatedly applying Rule 2 of\nPDS, the gadget Cij will be completely power dominated, that is, all nodes of the gadget will be\nin PS.\nThe next lemma shows that the size of an optimal solution in PDS is exactly one more than the\nsize of an optimal solution in MinRep. The number of nodes in the constructed graph is equal to\n V (G)\n = 4+ |V (G)| + 3λ |E(G)|. This will complete the proof of Theorem 2.2 by showing that the\nabove reduction is a gap preserving reduction from MinRep to PDS with the same gap (hardness\nratio) as the MinRep problem.\n5\n\nLemma 2.3 A∗∪B∗is an optimal solution to the instance G = (A, B, E) of the MinRep problem\nif and only if S∗= A∗∪B∗∪{w∗} ⊆V (G) is an optimal solution to the instance G of the PDS\nproblem.\nProof: First, we claim that w∗should be in any optimal solution of the PDS instance G. Suppose\nthat w∗is not in some optimal solutions. Then, in order to power dominate the nodes w∗, w∗\n1, w∗\n2, w∗\n3\nin G, the set S must contain at least two of the nodes (leaves) w∗\n1, w∗\n2, w∗\n3. This is a contradiction,\nsince we can replace these 2 nodes by w∗and obtain a smaller feasible solution.\nAssume that A∗∪B∗is a feasible solution for the MinRep instance G.\nWe will show that\nS = A∗∪B∗∪{w∗} is a feasible solution to the PDS instance G. Note that all nodes in A ∪B ∪\n{w∗, w∗\n1, w∗\n2, w∗\n3} are power dominated by applying Rule 1 on w∗. Now, we only need to show that\nall nodes in the gadgets Cij are power dominated. Consider any super edge AiBj of H. The set\nA∗∪B∗covers all the super edges in H. So there exists a pair of nodes ak ∈A∗∩Ai, bk ∈B∗∩Bj\nthat induces an edge of G. Since ak and bk are in S, their neighbors, uk and vk, in each of the\nλ = 4 copies of Cij in G, will be power dominated by applying Rule 1. Then the nodes uk and vk\nin each copy of Cij will power dominate the entire cycle by repeatedly applying Rule 2. To see this,\nnote that any node in Cij has exactly 2 neighbors in Cij and at most 1 neighbor not in Cij. The\nneighbors not in Cij are from Ai ∪Bj, and they are power dominated by w∗. Hence, if a node in Cij\nand one of its neighbors in Cij are power dominated, then by applying Rule 2 the other neighbor\nin Cij will be power dominated. Hence, by starting from vk and repeatedly applying Rule 2, we\ncan sequentially power dominate the nodes in Cij. This shows that S power dominates all nodes\nin G. Therefore, Opt(G) is at most |A∗∪B∗| + 1.\nLet S∗⊆V (G) be an optimal solution for PDS. By the above claim, w∗is in S∗. Now define\nA′ = A ∩S∗and B′ = B ∩S∗. First we prove that any optimal solution of PDS is contained in\nA ∪B ∪{w∗}, and then we show that A′ ∪B′ covers all super edges of the MinRep instance G.\nSuppose that S∗contains some nodes not in A ∪B ∪{w∗}. Hence, there are some gadgets that are\nnot completely power dominated by S∗∩(A ∪B ∪{w∗}). Let Cij be such a gadget. By symmetry\neach of the λ = 4 copies of Cij is not completely power dominated. Therefore, the optimal solution\nS∗needs to have at least 3 nodes from the 4 copies of Cij. By removing these 3 nodes from S∗and\nA1\nA2\nB1\nB2\n(a) MinRep Instance G\nw∗\n3\nw∗\n1\nw∗\n2\nw∗\nC2,2\nA2\nB2\nB1\nC1,1\nC1,2\nA1\n(b) PDS instance G: For each super edge AiBj we show\nonly 1 copy of Cij; in fact G has λ = 4 copies of Cij.\nFigure 3: The hardness construction\n6\n\nadding ak ∈Ai and bk ∈Bj to S∗for some arbitrary edge akbk ∈Eij, we can power dominate all\nof the 4 copies of Cij. This contradicts the minimality of S∗, and proves that S∗⊆A ∪B ∪{w∗}.\nTo see that A′ ∪B′ covers all super edges, note the following: suppose no node from any copy of\nCij is in the optimal solution; then any Cij can be power dominated only by taking a pair of nodes\na ∈Ai, b ∈Bj that induces an edge of G. This completes the proof of the lemma.\n□\n3\nApproximation Algorithms for Planar Graphs\nIn this section we describe an O(k)-approximation algorithm for PDS in graphs with tree-width\nk; the running time is O(n3), independent of k. This algorithm gives an O(√n)-approximation\nalgorithm for PDS in planar graphs, since the tree-width of a planar graph G with n nodes is\nO(√n) and in O(n\n3\n2) time we can find an O(√n) tree-width decomposition of the given planar\ngraph G [2]. Finally, we show that the analysis of our algorithm is tight on planar graphs. We use\nPlanar PDS to denote the special case of the PDS problem where the graph is planar.\nDefinition 3.1 [11] A tree decomposition of a graph G = (V, E) is a pair ⟨{Xi ⊆V |i ∈I} , T =\n(I, F)⟩such that T is a tree with V (T) = I, E(T) = F, and satisfying the following properties:\n(T1) S\ni∈I Xi = V , and every edge uv ∈E has both ends in some Xi,\n(T2) For all i, j, k ∈I if j is on the unique path from i to k in T then we have: Xi ∩Xk ⊆Xj,\nThe width of ⟨{Xi|i ∈I} , T⟩is the maxi∈I|Xi|−1. The tree-width of G is defined as the minimum\nwidth over all tree decompositions. The nodes of the tree are called T-nodes and the sets Xi are\ncalled bags.\nA nice tree decomposition is a tree decomposition ⟨{Xi ⊆V |i ∈I} , T = (I, F)⟩, where T is a\nrooted tree in which each node has at most 2 children.\nIf a node i ∈I has two children j, k\nthen Xi = Xj = Xk (i is called a Join node), and if i has one child j then either Xj ⊂Xi and\n|Xi \\ Xj| = 1 or Xi ⊂Xj and |Xj \\ Xi| = 1 (i is called an Insert or a Forget node, respectively).\nWe introduce the notion of a strong region before presenting our algorithm. Informally speaking,\na set of nodes R ⊆V is called strong if every feasible solution to the PDS problem has a node of R.\nFor a graph G = (V, E), the neighborhood of R ⊆V is nbr(R) = {v ∈V |∃uv ∈E, u ∈R, v /∈R},\nand the exterior of R is defined by ext(R) = nbr(V \\R), i.e., ext(R) consists of the nodes in R that\nare adjacent to a node in V \\ R.\nDefinition 3.2 Given a graph G = (V, E) and a set S ⊆V , the subset R ⊆V is called an S-strong\nregion if R ̸⊆PS∪nbr(R), otherwise, the set R is called an S-weak region. The region R is called\nminimal S-strong if it is an S-strong region and ∀r ∈R, R −r is an S-weak region.\nIt is easy to check from the definition that an S-strong region is also an ∅-strong (or shortly strong)\nregion. Any feasible solution to the PDS problem needs to have at least one node from every strong\nregion.\nLemma 3.3 A subset R ⊆V is an S-strong region if and only if for every feasible solution S ∪S∗\nof G, we have R ∩(S∗\\ S) ̸= ∅.\n7\n\nProof: It can be seen that the set S ∪(V \\ R) will power dominate the same set of nodes in R\nthat can be power dominated by S ∪nbr(R); this is valid for any subset R ⊆V .\nLet R be an S-strong region. By the definition of a strong region we have R ̸⊆PS∪nbr(R). Hence,\nby the above claim R ̸⊆PS∪(V \\R). This shows that every feasible solution S∗needs to have at least\none node from R that is not in S.\nNow assume that for every feasible solution S∗of G we have R ∩(S∗\\ S) ̸= ∅. Suppose that R\nis an S-weak region, so by the definition of a weak region we have R ⊆PS∪nbr(R). It follows that\nS∗= S ∪(V \\ R) is a feasible solution, but R has no intersection with S∗\\ S ⊆(V \\ R). This is a\ncontradiction, so R is an S-strong region.\n□\nOur algorithm makes one level-by-level and bottom-to-top pass over the tree T of the tree de-\ncomposition of G and constructs a solution S for PDS (initially, S = ∅). At each node rj of T we\ncheck whether the union of the bags in the subtree rooted at rj forms an S-strong region; if yes,\nthen the bag Xrj of rj is added to S, otherwise S is not updated. The key point in the analysis is\nto show that Opt(G) ≥m, where m is the number of nodes of T where we updated S.\nAlgorithm 1 O(k)-approximation Algorithm\n1: A tree decomposition ⟨{Xi|i ∈I} , T⟩of G is given, where T is rooted at r.\n2: Let Ilbe the set of T-nodes at distance lfrom the root, and let d be the maximum distance\nfrom r in T.\n3: S ←∅\n4: for i = d to 0 do\n5:\nLet Ii = {r1, . . . , rki} and denote by Trj the subtree in T rooted at rj.\n6:\nLet Yrj be the union of bags corresponding to the T-nodes in Trj.\n7:\nfor j = 1 to ki do\n8:\nif Yrj is an S-strong region then\n9:\nS ←S ∪Xrj; where Xrj is the bag corresponding to rj.\n10:\nend if\n11:\nend for\n12: end for\n13: Output So = S\n3.1\nAnalysis of the algorithm\nIn this subsection we show that our algorithm has an approximation guarantee of O(k).\nLet\nG = (V, E) denote the input graph, and let S ⊆V be any set of nodes.\nLemma 3.4 Suppose Z is an S-weak region such that ext(Z) ⊆S. Then we have Z ⊆PS.\nProof: Let Y = ext(Z), it is easy to check that nbr(Z \\ Y ) ⊆ext(Z). We claim that Z \\ Y is an\nS-weak region. Let S∗= V \\ (Z ∪S), it is easy to check that S ∪S∗is a feasible solution for the\ngraph G, but S∗∩(Z \\ Y ) = ∅. Hence, by Lemma 3.3, Z \\ Y is not an S-strong region, and so it\nis an S-weak region. Thus Z \\ Y ⊆PS∪nbr(Z\\Y ) ⊆PS∪ext(Z) = PS and this implies that Z ⊆PS as\nY = ext(Z) ⊆S.\n□\n8\n\nLemma 3.5 Let Z ⊆V be an S-strong region. Suppose that Y is a subset of V such that Y ⊆PS\nand ext(Y ) ⊆S. Then Z \\ Y is an S-strong region.\nProof: Assume for the sake of contradiction that Z \\Y is an S-weak region. Then by the definition\nof strong regions we have: Z \\Y ⊆PS∪nbr(Z\\Y ). It is easy to see that nbr(Z \\Y ) ⊆nbr(Z)∪ext(Y ).\nThis implies that Z \\Y ⊆PS∪nbr(Z\\Y ) ⊆PS∪nbr(Z)∪ext(Y ) = PS∪nbr(Z). The condition in the lemma\nstates that Y ⊆PS ⊆PS∪nbr(Z). Hence, we get Z = (Z \\ Y ) ∪(Z ∩Y ) ⊆PS∪nbr(Z), which means\nthat Z is an S-weak region. This is a contradiction, so the lemma is proved.\n□\nTheorem 3.6 Given a graph G = (V, E) and a tree decomposition of G of width k as input,\nAlgorithm 1 runs in time O(n · |E|), and achieves an approximation guarantee of (k + 1).\nProof: First, we show that the solution So found by the algorithm is feasible. Then we prove the\napproximation guarantee, and establish the running time.\nFor any node q of T, recall that Yq denotes the union of the bags corresponding to the T-nodes\nin the subtree rooted at q in T; let Gq denote the subgraph of G induced by Yq. We claim that\next(Yq) ⊆Xq.\nSuppose that q has m children in T, call them c1, . . . , cm.\nFor each edge qcj\n(j = 1, · · · , m), the set Xq ∩Xcj separates Ycj from the rest of the graph, that is, every path\nbetween a node in Ycj and a node in V \\Ycj contains a node of Xq ∩Xcj (see Lemma 12.3.1 in [11]).\nThus, ext(Ycj) ⊆Xq ∩Xcj ⊆Xq, and hence, for Yq = Xq ∪Yc1 ∪· · · ∪Ycm, we have ext(Yq) ⊆Xq.\nWe use induction on the height of the subtree of T rooted at q to prove the following: if Yq is\nS∗-strong, then Yq ⊆PS∗∪Xq, where S∗denotes the solution just before the algorithm examines\nYq. The statement clearly holds when q is a leaf of T (since Yq = Xq). Otherwise, let c1, . . . , cm\nbe the children of q in T. For each j = 1, . . . , m, when the algorithm examined Ycj, either Ycj was\nS-weak, in which case (by Lemma 3.4) we have Ycj ⊆PS∪ext(Ycj ) ⊆PS∪(Xcj ∩Xq) ⊆PS∗∪Xq or Ycj\nwas S-strong in which case Ycj ⊆PS∪Xcj by induction (note that S ∪Xcj ⊆S∗); we use S to denote\nthe solution just before the algorithm examines Ycj. Hence, Yq = Yc1 ∪· · · ∪Ycm ∪Xq ⊆PS∗∪Xq.\nThe above statement implies that V ⊆PSo because at the step when the algorithm examines the\nroot r of T either\n(i) Yr is S-strong, so So = S ∪Xr, and Yr ⊆PS∪Xr = PSo; or\n(ii) Yr is S-weak, and Yr ⊆PS∪ext(Yr) = PSo; since Yr = V (G) and ext(Yr) = ∅.\nTo show that the approximation guarantee is (k+1) we will construct a set ∆of pairwise disjoint\nstrong regions R1, R2, . . . , such that there is a strong region Rj corresponding to each step of the\nalgorithm that adds a non empty bag Xqj to S. Thus |So| ≤(k + 1) |∆| since each bag has ≤k + 1\nnodes, and Opt(G) ≥|∆| because every feasible solution has size ≥|∆|, by Lemma 3.3. Hence,\n|So| ≤(k + 1)Opt(G). We construct the sets R1, R2, . . . , during the execution of the algorithm\nas follows. Suppose the algorithm finds Yq to be S-strong while examining a node q of T. Let\nq1, . . . , ql−1 be the nodes of T where the algorithm updated the solution before examining q, and\nlet S be the solution just before the algorithm examines q. Then define Rl= Yql\\(Yq1 ∪· · ·∪Yql−1),\nwhere ql= q. We claim that Rlis an S-strong region. For each strong region Yqj (j = 1, . . . , l−1) we\nhave seen that ext(Yqj) ⊆Xqj ⊆S and Yqj ⊆PS; note that the algorithm added Xqj to the solution\nsince Yqj was a strong region. It follows that ext(Yq1 ∪· · · ∪Yql−1) ⊆S, and Yq1 ∪· · · ∪Yql−1 ⊆PS.\nHence, by Lemma 3.5, the set Rlis an S-strong region. Clearly, the sets R1, R2, . . . , are pairwise\ndisjoint. This completes the construction of ∆.\n9\n\nConsider the running time. Without loss of generality we can assume that the given tree decom-\nposition of width k has at most 4n bags (see Lemma 13.1.2 in [22]). Using standard algorithmic\ntechniques we can test in O(|E|) time whether a given set R ⊆V is an S-strong region (we compute\nPS∪nbr(R) and check if it contains R). Therefore, our algorithm has a running time of O(n · |E|). □\nIt is known that planar graphs have tree-width O(√n), and such a tree decomposition can be\nfound in O(n\n3\n2 ) time [2]. This fact together with the above theorem proves the following theorem.\nTheorem 3.7 Algorithm 1 achieves an approximation guarantee of O(√n) for the Planar PDS\nproblem.\nAs mentioned earlier, Haynes et al.[16] presented a linear-time algorithm for optimally solving\nPDS on trees. By modifying Algorithm 1, we can solve PDS optimally on trees. The resulting\nalgorithm differs from the algorithm of Haynes et al. since our algorithm uses strong regions.\nInformally, the algorithm makes a level-by-level and bottom-to-top pass over the tree G. At a node\nv of the tree G if the set of nodes in the subtree rooted at v forms a strong region, then we add v\nto the solution, otherwise we skip v. Formally, we define Xv = {v} for each v ∈V (G), and we run\nAlgorithm 1 on the tree G. Note that defining bags in this way does not give a tree decomposition\nof G.\nTheorem 3.8 A modification of Algorithm 1 runs in time O(n · |E|) and solves PDS optimally on\ntrees.\n3.2\nLower bounds via disjoint strong regions\nIn this part we show that any approximation algorithm for PDS that uses the number of disjoint\nstrong regions as a lower bound has an approximation guarantee of Ω(√n). In proposition 3.12\nwe give a lower bound on the optimal value for PDS on an l× m grid. Independently, [12] gave a\nstronger result for this.\nLemma 3.9 Any minimal S-strong region is connected.\nProof: Assume that R is a minimal S-strong region that is not connected.\nLet C ⊂R be a\nconnected component of R. The set C is an S-weak region since R is a minimal S-strong region.\nBy the definition of a weak region we have C ⊆PS∪nbr(C). The set C is a connected component of\nR, so the neighborhood of C has no intersection with R \\ C. This implies that nbr(C) ⊆nbr(R),\nand consequently we have C ⊆PS∪nbr(C) ⊆PS∪nbr(R). The same argument as above shows that\nR \\ C ⊆PS∪nbr(R). Hence, R ⊆PS∪nbr(R) which is a contradiction.\n□\nLemma 3.10 The number of disjoint strong regions in an l× m grid is exactly one.\nProof: For the sake of contradiction, assume that the given grid has two disjoint strong regions.\nTake as few nodes as possible from these strong regions until we get minimal strong regions, say\nR1 and R2. It is easy to check that the set of nodes of any row or any column of the grid power\ndominates all nodes in the grid. By Lemma 3.3, R1 and R2 should have at least one node from\nevery feasible solution. In the other words, R1 and R2 must have at least one node from each row\nand also from each column. By lemma 3.9, we know that R1 and R2 induce connected subgraphs.\nHence, in R1 there is a path from a node in the top row to a node in the bottom row, and also in R2\n10\n\nthere is a path from a node in the rightmost column to a node in the leftmost column. Obviously\nthese two paths share a common node. This is a contradiction, since R1 and R2 are assumed to be\ndisjoint.\n□\nWe denote by Pi\nS a set of nodes that are power dominated after applying propagation rule, Rule\n2, for i number of times to P0\nS = S ∪nbr(S). Obviously this depends on the order of applying the\npropagation rule. We use the notation without specifying the order of applying the propagation\nrule.\nLemma 3.11 (Propagation lemma) Given an ordering of propagation rules applied to S ∪nbr(S)\nwith Pk\nS = PS we have:\n ext(Pj\nS)\n ≤\n ext(Pi\nS)\n , ∀0 ≤i < j ≤k.\nProof: We will prove that\n ext(Pi+1\nS\n)\n ≤\n ext(Pi\nS)\n , for all 0 ≤i ≤k −1. Consider the set Pi\nS\nand assume that in the (i + 1)-st step we apply Rule 2 to v ∈ext(Pi\nS) and power dominate u; i.e.\nu ∈Pi+1\nS\n, u /∈Pi\nS. To apply Rule 2 to v all neighbors of v except u should be power dominated, so\nwe have nbr(v)\\Pi\nS = {u}. Also since we power dominate u at step (i+1), we have nbr(v) ⊆Pi+1\nS\n.\nTherefore, v is not in ext(Pi+1\nS\n), but u may be in the exterior of Pi+1\nS\n. Hence, we have ext(Pi+1\nS\n) ⊆\n ext(Pi\nS) \\ {v}\n \n∪{u}. It follows that\n ext(Pi+1\nS\n)\n ≤\n ext(Pi\nS) \\ {v}\n + |{u}| =\n ext(Pi\nS)\n .\n□\nProposition 3.12 Let G be an l× m grid with l≤m, then Opt(G) = Θ(l).\nProof: First note that any row or any column of the grid power dominates all nodes. In the\nfollowing we prove that any feasible solution of PDS needs to have at least l−1\n5\nnodes. Assume that\nthere exists S ⊆V such that |S| < l−1\n5\nand PS = V (G). The maximum degree in G is 4, so we have\n ext(P0\nS)\n ≤|S ∪nbr(S)| < l−1\n5 · 5 = l −1. Therefore, P0\nS contains no full row or no full column of\nG. The set S power dominates G, so there is an i such that Pi\nS contains a full row or a full column.\nConsider the smallest i with this property. Hence, Pi−1\nS\nhas no full row or full column. But some\nrow or some column must have at least l −1 nodes in Pi−1\nS\nsince Pi\nS contains an entire row or an\nentire column. Without loss of generality assume that these l −1 nodes are from a column. By the\ndefinition of i, each of the l −1 nodes in this column are in a row which is not a subset of Pi−1\nS\n.\nTherefore, there are at least l −1 rows with at least one node in Pi−1\nS\nand at least one node not\nin Pi−1\nS\n. This implies that\n ext(Pi−1\nS\n)\n ≥l −1. Finally, by using Lemma 3.11 we get the following\ncontradiction: l −1 >\n P0\nS\n ≥\n ext(P0\nS)\n ≥\n ext(Pi−1\nS\n)\n ≥l −1.\n□\nConsider any approximation algorithm for PDS that uses only the number of disjoint strong\nregions as a lower bound on the size of an optimal solution. By Lemma 3.10, this algorithm finds\na lower bound of 1 on the size of an optimal solution on a grid. The √n × √n grid has an optimal\nsolution of size Θ(√n) by Proposition 3.12. This shows that the approximation guarantee of the\nalgorithm is Ω(√n), even on planar graphs.\nProposition 3.13 Consider any approximation algorithm for PDS that uses only the number of\ndisjoint strong regions as a lower bound on the optimal value. Then the approximation guarantee\nis Ω(√n).\n4\nPDS in Directed Graphs\nIn this section we extend the PDS problem to directed graphs to obtain the Directed Power\nDominating Set (Directed PDS) problem. Our motivation for studying the directed prob-\nlem comes from theoretical considerations.\nThe Dominating Set problem is studied on both\n11\n\nundirected and directed graphs, and there is extensive literature on the latter (see [17, 18]). The\nsimilarities between the Dominating Set problem and the PDS problem led us to define and study\nthe Directed PDS problem. We give a result on the hardness of approximation of Directed\nPDS. Then we reformulate the Directed PDS problem in terms of valid coloring of the edges.\nUsing this, we design an algorithm for solving Directed PDS in linear-time on a special class of\ndirected graphs.\nLet G = (V, E) be a directed graph.\nA node w is called an out-neighbor (in-neighbor) of a\nnode v if there is a directed edge from v to w (from w to v) in G. The number of out-neighbors\n(in-neighbors) of a node v is called the out-degree (in-degree) of v and is denoted by d+\nG(v) (or\nsimilarly d−\nG(v)). For a set of nodes X, the subgraph of G induced by X is denoted by G[X]. The\ndirected graphs that we consider here have no loops nor parallel edges, but may have two edges\nwith different directions on the same two end nodes (we call such edges antiparallel).\nGiven a\ndirected graph G by the underlying undirected graph we mean the undirected graph obtained from\nG, by removing the direction of edges and also removing any parallel edges that are introduced\nafter removing the directions.\nDefinition 4.1 (the Directed PDS problem) Let G be a directed graph. Given a set of nodes\nS ⊆V (G), the set of nodes that are power dominated by S, denoted by PS, is obtained as follows:\n(D1) if node v is in S, then v and all of its out-neighbors are in PS;\n(D2) (propagation) if node v is in PS, one of its out-neighbors w is not in PS, and all other\nout-neighbors of v are in PS, then w is inserted into PS.\nWe say that S power dominates G if PS = V (G). The Directed PDS problem is to find a node\nset S with minimum size that power dominates all the nodes in G.\nWe prove a threshold of 2log n1−ǫ for the hardness of approximation of Directed PDS modulo\nthe same complexity assumption as in Theorem 2.2. The proof uses a reduction from the MinRep\nproblem to the Directed PDS problem in a directed acyclic graph. This reduction is similar to\nthe reduction in Theorem 2.2; the main difference comes from the gadget for modeling the super\nedges.\nTheorem 4.2 The Directed PDS problem even when restricted to directed acyclic graphs cannot\nbe approximated within ratio 2log1−ǫ n, for any fixed ǫ > 0, unless NP ⊆DTIME(npolylog(n)).\nThe reduction: We create an instance G = (V , E) of the Directed PDS problem from a given\ninstance G = (A, B, E)(H = (A, B, E)) of the MinRep problem.\n1. Start with a copy of each node in A ∪B in G. For convenience, we use the same notation for\nnodes (and sets of nodes) in G and their copies in G.\n2. Add a new node w∗to the graph G, and add a directed edge from w∗to each node in A ∪B.\n3. ∀i ∈{1, . . . , qA} , j ∈{1, . . . , qB} if AiBj is a super edge, then do the following:\n(a) Let Eij be the set of edges between Ai and Bj in G, and let lij denote |Eij|. We denote\nthe edges in Eij by e1, e2, . . . , ek, . . . , elij.\n12\n\n(b) Let Dij be the graph on 6lij + 1 nodes as shown in Figure 4(a). In Dij there are 6\nnodes uk, vk, dk, αk, βk, γk associated with an edge ek of Eij. The part of Dij associated\nwith an edge ek is shown in Figure 4(b); note that all these parts share a common node,\ncalled the center node, in Dij. Make λ = 4 new copies of the graph Dij (λ can be any\nconstant greater than 3). For each edge ek = akbk ∈Eij and for each of the 4 copies of\nDij, we add a directed edge from ak to uk and a directed edge from bk to vk. In addition\nto these edges, there are directed edges from w∗to some nodes inside Dij; these directed\nedges are denoted by a dashed line in Figure 4(a).\nw∗\nv1\nu2\nvk\nu1\nv2\nd2\ndk\nd1\nulij\ndlij\nulij\nuk\n(a) The Dij graph\nbk\nak\nvk\nuk\ndk\nγk\nβk\nαk\ncenter node\nek\n(b) Part of gadget corre-\nsponding to an edge ek\nFigure 4: The cover testing gadget.\n4. Let G = (V , E) be the obtained graph.\nThe next lemma shows that the size of an optimal solution in Directed PDS is exactly one\nmore than the size of an optimal solution in the MinRep instance. The number of nodes in the\nconstructed graph is at most\n V (G)\n ≤1 + |V (G)| + 7λ |E(G)|. This will complete the proof of\nTheorem 4.2 by showing that the above reduction is a gap preserving reduction from MinRep to\nDirected PDS with the same gap (hardness ratio) as the MinRep problem.\nLemma 4.3 A∗∪B∗is an optimal solution to the instance G = (A, B, E) of the MinRep problem\nif and only if S∗= A∗∪B∗∪{w∗} ⊆V (G) is an optimal solution to the instance G of the Directed\nPDS problem.\nProof: First note that w∗should be in any feasible solution; because it has in-degree zero.\nAssume that A∗∪B∗is a feasible solution for the MinRep instance G.\nWe will show that\nS = A∗∪B∗∪{w∗} is a feasible solution to the Directed PDS instance G. Note that all nodes\nin A ∪B and some nodes inside the gadgets Dij are power dominated by applying rule (D1) on\nw∗. Now, we only need to show that all nodes in the gadgets Dij are power dominated. Consider\na super edge AiBj of H. The set A∗∪B∗covers all the super edges in H. Hence, there exists\na pair of nodes ak ∈A∗∩Ai, bk ∈B∗∩Bj that induces an edge of G.\nSince ak and bk are\nin S their out-neighbors uk and vk in each of the λ = 4 copies of the Dij graph will be power\ndominated by applying rule (D1). Now node dk, that is already power dominated by w∗, will power\ndominate the center node by an application of rule (D2). Now we claim that the center node will\npower dominate the remaining nodes in the gadget Dij. Consider the part of gadget (shown in\n13\n\nFigure 4(b)) corresponding to an edge er ∈Eij. Note that the nodes γr, αr, and dr are already\npower dominated by w∗. It is easy to check that the nodes βr, ur, vr will be power dominated by\nsequentially applying rule (D2) on γr, αr, and dr. This shows that S power dominates all nodes in\nG. Therefore, Opt(G) is at most |A∗∪B∗| + 1.\nLet S∗⊆V (G) be an optimal solution for Directed PDS. As we showed above w∗should be in\nany feasible solution for Directed PDS. Now define A′ = A∩S∗and B′ = B ∩S∗. First we prove\nthat any optimal solution of Directed PDS is contained in A ∪B ∪{w∗}, and then we show that\nA′ ∪B′ covers all the super edges of H. Suppose that S∗contains some nodes not in A ∪B ∪{w∗}.\nHence, there are some gadgets that are not completely power dominated by S∗∩(A ∪B ∪{w∗}).\nLet Dij be such a gadget. By symmetry each of the λ = 4 copies of Dij is not completely power\ndominated. Therefore, the optimal solution S∗needs to have at least 3 nodes from the 4 copies of\nDij. By removing these 3 nodes from S∗and adding ak ∈Ai and bk ∈Bj to S∗for some arbitrary\nedge ek = akbk ∈Eij, we can power dominate all 4 copies of Dij. This contradicts the optimality\nof S∗, and proves that S∗⊆A ∪B ∪{w∗}. To see that A′ ∪B′ covers all super edges, it is enough\nto note the following. Suppose no node from any copy of Dij is in the optimal solution, then any\nDij can be power dominated only by taking a pair of nodes a ∈Ai and b ∈Bj that induces an\nedge of G. Otherwise, any power dominated node in the gadget has at least 2 out-neighbors that\nare not power dominated, so rule (D2) cannot be applied. This completes the proof of the lemma.\n□\nThere are several notions for the tree-width of directed graphs such as DAG width [29], directed\ntree-width [21], and Kelly-width [19]. Directed acyclic graphs have width equal to zero for the first\ntwo notions [29], and have Kelly-width of 1. Hence, Theorem 4.2 gives a hardness threshold of\nO(2log n1−ǫ) even if the directed graph has width ≤1 according to any of the above three notions.\nWe reformulate Directed PDS in terms of valid colorings of the edges in order to develop\nan algorithm based on dynamic-programming for Directed PDS. Guo et al. [15] introduced the\nnotion of valid orientations to get a new formulation for PDS (in undirected graphs). They also\ndesigned a linear-time dynamic-programming algorithm based on valid orientations for optimally\nsolving PDS on graphs of bounded tree-width. Our method applies to directed graphs such that\nthe underlying undirected graph has bounded tree-width.\nDefinition 4.4 A coloring of a directed graph G = (V, E) is a partitioning of the edges in G into\nred and blue edges. We denote a coloring by C = (V, Er ∪Eb) where Er is the set of red edges and\nEb is the set of blue edges.\nWe reformulate the Directed PDS problem via a so-called valid coloring of directed graphs;\ninformally speaking, these colorings “model” the application of rules (D1) and (D2) of Directed\nPDS.\nDefinition 4.5 A valid coloring C = (V, Er ∪Eb) of a directed graph G = (V, E) is a coloring of\nG with the following properties:\n1. No two antiparallel edges can be colored red.\n2. The subgraph induced by the red edges, Gr = (V, Er), has the following properties:\n(a) ∀v ∈G : d−\nGr(v) ≤1, and\n(b) ∀v ∈G : d−\nGr(v) = 1 =⇒d+\nGr(v) ≤1.\n14\n\n3. G has no dependency cycle. A dependency cycle is a sequence of directed edges whose under-\nlying undirected graph forms a cycle such that all the red edges are in one direction, all the\nblue edges are in the other direction, and there are no two consecutive blue edges.\nWe call a node an origin of C if it has no incoming edges in Gr.\nOur dynamic-programming algorithm for Directed PDS is based on the following lemma.\nLemma 4.6 Given a directed graph G and S ⊆V (G), S power dominates G if and only if there\nis a valid coloring of G with S as the set of origins.\nProof: Suppose S ⊆V power dominates G. Then we give a valid coloring C with S as the set\nof origins by coloring the edges in G according to the way that S power dominates G. We color\nan edge (v, w) red if node w is power dominated by applying the power domination rules on v;\neither by the domination rule (D1) or by the propagation rule (D2). Note that when we apply\nthe propagation rule (D2), then we do not power dominate the previously power dominated nodes.\nAlso when we apply rule (D1) on v, then we power dominate all (not some subset of) neighbors\nof v that are not power dominated. We write v < u when a node u is power dominated after\nv. It is easy to check that with this coloring the degree requirements are satisfied; each node can\nbe power dominated only once, and if it is a power dominated node (not in S) , then it cannot\npower dominate more than one of its out-neighbors due to rule (D2). Now, we need to prove that\nthere is no dependency cycle.\nBy way of contradiction, suppose that C∗= u1, u2, . . . , um is a\ndependency cycle. Focus on the edges of C∗. Call the direction of the red edges forward, and call\nthe direction of the blue edges backward. Observe that a dependency cycle has ≥1 red edges, and\neach of its red edges corresponds to an application of rule (D2). Assume that all the edges in C∗\nare red. Then the red edges (ui, ui+1) imply that ui < ui+1 for all i = 1, 2, . . . , m −1; therefore\nu1 < u2 < · · · < um, but this is a contradiction since the last red edge from um back to u1 implies\nthat um < u1. Hence, there is no dependency cycle with all edges colored red. Now, assume that\nthe dependency cycle C∗has some blue edges. We show that a similar contradiction occurs when\nthere are no two consecutive blue edges. Consider a blue edge (u, v) of C∗and note that the other\nedge of C∗incident to u is a red edge, say (u, w). By rule (D2), we see that v should be power\ndominated before u can power dominate w; thus we have w > v. Repeating this argument, we\nget an ordering for the occurrences of power domination of some of the nodes in C∗that gives a\ncontradiction, e.g., if m is even and the edges of C∗are alternatively blue and red (starting with\nblue) we get u1 < u3 < u5 < · · · < um−1 < u1 (see Figure 5 for an example). Hence, G has a valid\ncoloring with S as the set of origins.\nB\nR\nR\nB\nR\nu1\nu2\nu3\nu4\nu5\nFigure 5:\nThe blue edge (u4, u3) means that u3 should be power dominated before we can power\ndominate u5 by applying rule (D2) on u4; thus u3 < u5. Again the blue edge from u1 to u5 implies\nthat u5 < u2. Finally , the red edge (u2, u3) shows that u2 < u3. Combining these dependencies\nwe get u3 < u5 < u2 < u3. This is a contradiction, and shows that there cannot be a dependency\ncycle in a coloring obtained from the application of rules (D1) and (D2) of Directed PDS.\nNow suppose that G has a valid coloring C = (V, Er ∪Eb) with S ⊆V (G) as the set of origins.\nThe nodes in S and all of their out-neighbors in Gr = (V, Er) are power dominated by applying\n15\n\nthe rule (D1). Now we prove that S will power dominate all nodes in G. Suppose that this does\nnot happen. Let X ⊂V be the maximal set of nodes that can be power dominated by S. We\nclaim that there is at least one red edge from X to V \\ X. Note that all of the origins are in X,\nso each of the nodes in V \\ X has in-degree 1 in Gr. Hence, if there is no red edge from X to\nV \\ X, then there should be a directed cycle of red edges in G[V \\ X]. This is not possible since\nthere are no dependency cycles. Therefore there is at least one red edge from X to V \\ X. Let\ne1 = (x1, y1), . . . , ek = (xk, yk) be all of the red edges from X to V \\X. If some xi has all of its out-\nneighbors in X except yi, then by applying rule (D2) on xi the node yi will be power dominated. By\nthe maximality assumption of X this cannot happen. Therefore, each xi has another out-neighbor,\nsay zi, in V \\X. Then (xi, zi) is a blue edge, otherwise, xi would be an origin and yi would be power\ndominated by applying rule (D1) on xi. Now, we construct a dependency cycle as follows: starting\nfrom x1, use a blue edge to move to a node z1 in V \\ X; then move in the reverse direction over a\nsequence of red edges (z2, z1), (z3, z2), · · · until we reach a red edge (zi, zi−1) with zi−1 ∈V \\ X and\nzi ∈X (such an edge exists since G[V \\X] has no directed cycle of red edges); note that (zi, zi−1) is\none of the red edges (x1, y1), · · · , (xk, yk). If zi = x1, then we have a dependency cycle; otherwise,\nwe again use a blue edge (zi, zi+1) to move to a node in V \\ X. By repeating these steps, we will\neventually find a dependency cycle. Note that all the blue edges are in one direction, and all the red\nedges are in the other direction. This is a contradiction, since C has no dependency cycle. Hence,\nwe have X = V , so S power dominates G.\n□\nTheorem 4.7 Given a directed graph G and a tree decomposition of width k of its underlying\nundirected graph, Directed PDS can be optimally solved in O(ck2 · n) time for a global constant\nc.\nA consequence of the above theorem is a linear-time algorithm for solving the Directed PDS\nproblem optimally on directed graphs, given a bounded tree-width decomposition of the underlying\nundirected graph. Also since the tree-width decomposition for graphs with bounded tree-width can\nbe computed in polynomial-time [5], there is a polynomial-time algorithm to solve Directed PDS\noptimally on the class of directed graphs such that the underlying undirected graph has bounded\ntree-width.\n5\nConclusions\nWe studied the PDS problem from the perspective of approximation algorithms. We introduced\na natural extension of the problem to directed graphs. We showed that both problems have a\nthreshold of O(2log n1−ǫ) for the hardness of approximation.\nWe presented an O(√n) approxi-\nmation algorithm for Planar PDS. We designed a dynamic-programming algorithm for solving\nthe Directed PDS problem optimally in linear-time for those directed graphs whose underlying\nundirected graph has bounded tree-width.\nHere, we describe an algorithm with an approximation guarantee of O(\nn\nlog n) for the PDS problem.\nThe algorithm works as follows. Partition the nodes of the graph G into log n equal-sized sets\nV1, V2, · · ·. Next, consider all possible ways of picking these sets (we pick all nodes in a set). Among\nall these different candidates, output the one that power dominates G and has the minimum\nnumber of nodes. Note that in the algorithm we only consider 2log n = n different candidates.\nClearly, the algorithm runs in polynomial time, since the feasibility of each candidate can be tested\nin polynomial time. Let S∗be an optimal solution.\nIt is easy to see that the set of Vi’s that\n16\n\nintersect S∗forms a feasible solution for the PDS problem in G; this solution has size at most\nn\nlog n · |S∗|. This establishes the approximation guarantee. The same algorithm and analysis applies\nto the Directed PDS problem.\nProposition 5.1 There is a polynomial time\nn\nlog n-approximation algorithm for both the PDS prob-\nlem and the Directed PDS problem.\nThere is a gap between our hardness threshold of O(2log n1−ǫ) and our approximation guarantee of\nO(\nn\nlog n), and narrowing this gap is an open question.\nA major open question in the area is whether there exists a PTAS (polynomial time approximation\nscheme) for Planar PDS. A first step may be to obtain an improvement on our approximation\nguarantee of O(√n). There has been a lot of research on designing PTASs for NP-hard problems on\nplanar graphs. Some of the most important developments are the outerplanar layering technique by\nBaker [3], and the bidimensionality theory by Demaine and Hajiaghayi [10]. Unfortunately, these\nmethods do not apply to Planar PDS.\nBaker [3] showed that the Dominating Set problem in planar graphs has a PTAS. In the Baker\nmethod we first partition the graph into smaller graphs. Then we solve the problem optimally on\neach subgraph, and finally we return the union of the solutions as a solution for the original graph.\nThe example in Figure 1 shows that this method does not apply to Planar PDS. The size of an\noptimal solution is 1, but if we apply the Baker method, then the size of the output solution will\nbe at least as large as the number of subgraphs in the partition which can be Θ(n).\nDemaine and Hajiaghayi [10] introduced the bidimensionality theory and used it to obtain PTASs\nfor several variants of the Dominating Set problem on planar graphs. An important property\nof bidimensionality is that when an edge is contracted the size of an optimal solution should not\nincrease. Consider the example in Figure 6. If we contract edges e1, e2, . . . , en in G, then we get\nthe graph G′. It can be checked that Opt(G) = 1 but Opt(G′) = Θ(n). Thus the bidimensionality\ntheory does not apply to Planar PDS, since the optimum value may increase when an edge is\ncontracted.\nv\n=⇒\nv\nu1\nun\nen\ne1\nG\nG′\nFigure 6: Optimal value of PDS increases when edges are contracted.\nLastly, we consider some variations of greedy algorithms for PDS and show that they perform very\npoorly even on planar graphs. In contrast, for other related problems such as Dominating Set and\nSet Covering, greedy algorithms perform well since they achieve a logarithmic approximation\nguarantee, and no substantial improvement is possible by any polynomial time algorithm, under\ncomplexity assumptions like P ̸= NP. The most natural greedy algorithm for PDS is the one that\nstarts with S = ∅, and in each step, adds a new node v to the current solution S such that v power\ndominates the maximum number of new nodes.\n17\n\nUnfortunately, this greedy algorithm may find a solution S such that |S| ≥Θ(n) · Opt(G). To see\nthis, consider a graph G that is obtained from a 9l× 9m grid by subdividing all row-edges, except\nwith minor changes in the four corners as shown in Figure 7(a). Partition the graph G into 9 × 9\ngrids (ignoring the nodes introduced by subdivision), see Figure 7(b). It is easy to check that any\nsingle node can power dominate at most 7 nodes, and the center node of any one of the 9 × 9 grids\nachieves this maximum. So the greedy algorithm at the first iteration may pick the center node\nof any one of the 9 × 9 grids. Assuming all nodes picked by the algorithm so far have been these\ncenter nodes, we see that picking another center node maximizes\n PS∪{v} \\ PS\n over all v ∈V . So\nthe greedy algorithm could continue picking center nodes, and after that possibly picking other\nnodes until it finds a feasible solution S. The size of the output S is at least m · l= Θ(n). By\n(a) Grid\n(b) 9 × 9 grid\nFigure 7: Bad example for the greedy algorithm\nProposition 3.12, we have Opt(G) = Θ(l). Now by fixing l= Θ(1) we can see that the size of the\noutput solution can be bigger than Opt(G) by a factor of Θ(n).\nProposition 5.2 The greedy algorithm for the PDS problem may find a solution S such that |S| ≥\nΘ(n) · Opt(G).\nWe will consider two other variations of the greedy algorithm, namely Proximity and Cleanup.\nWe have examples of Planar PDS showing that these variations of the greedy algorithm perform\npoorly.\nProximity: In each step of the Proximity algorithm we choose a node such that the set of all\npower dominated nodes induces a connected subgraph, and subject to this, the number of newly\npower dominated nodes is maximized. Informally, this is to escalate the use of the propagation\nrule.\nThe bad example for the proximity version of the greedy algorithm is obtained by modifying the\ncenter row of the h × (2m + 1) grid, as shown in Figure 8, by inserting lsubdividing nodes into\nthe edges in the middle row, and also subdividing all of the other row-edges except some of the\ncorner edges. Figure 8 illustrates an example of such a grid for l = 5 and h = 9 rows, but for a\nbad example for the proximity greedy algorithm we need h to be sufficiently large constant (h = 17\nsuffices). We use the figure for illustration, to show the working of the proximity greedy algorithm.\nIt is easy to check that by picking all nodes of the first column we can power dominate the entire\ngraph, so the optimal solution is Θ(1). The proximity greedy algorithm starts by picking a node\nthat power dominates maximum number of nodes (which is 17 = 2l+ 7); any white node satisfies\nthis requirement. Therefore the algorithm may pick for example the first white node (from the\n18\n\nleft). It is easy to check that in the next step the algorithm will pick the white node to the right of\nthe first one, since all of the power dominated region stays connected and also it power dominates\nmaximum number of new nodes (which is 16 = 2l+ 6). The algorithm continues picking all white\nnodes and at the end it will pick possibly more nodes to get a feasible solution.\n(The shaded\nregion shown in Figure 8 indicates the nodes that will be power dominated by picking all white\nnodes.) Therefore, the size of the solution found by the algorithm is at least m = Θ(n). Hence, the\nproximity greedy algorithm may find a feasible solution that is Θ(n) times worse than the optimal\nsolution.\nFigure 8: Bad example for the proximity greedy algorithm\nCleanup Step: Some of the recent approximation algorithms, especially some based on the primal-\ndual method, use a clean up step at the end: this step removes redundant elements from the solution\nin some sequential order. In the Cleanup algorithm, we first run the greedy algorithm to find a\nsolution (node set) S, then we repeatedly remove nodes from S, until S is an inclusionwise minimal\npower dominating set. Although a cleanup step may substantially improve on the solution found\nby the greedy algorithm on some examples, this does not hold for all examples.\nThe same bad example for the proximity version is also a bad example for the cleanup version of\nthe greedy algorithm. The cleanup greedy algorithm may again pick the first white node (from the\nleft), and after picking this white node, it may pick the third white node. Since both of them power\ndominate maximum number of new nodes (which is 17 = 2l+ 7). Note that in the original greedy\nalgorithm there is no need to have a connected subgraph induced on power dominated nodes. The\nalgorithm continues to pick all the odd indexed white nodes, and after that it will start picking the\neven indexed white nodes since any one of them power dominates maximum number of new nodes\n(which is at least 15 = 2l+ 5). At this stage of the algorithm the set of power dominated nodes\nare those in the shaded region of Figure 8. It is easy to check that we need to pick nodes from\nboth upper and lower parts in order to power dominate the entire graph. The greedy algorithm\nmay pick some nodes from the leftmost column in the top part and some nodes from the rightmost\ncolumn in the bottom part to power dominate the entire graph. Now we start doing the cleanup\nprocess. It can be checked that if we remove any two consecutive white nodes from the obtained\nsolution, the graph cannot be power dominated completely. So we need to keep at least half of the\nwhite nodes. Therefore, the size of the output solution at the end of cleanup process is at least\nm\n2 = Θ(n), but the optimal solution is just Θ(1) as before.\n19\n\nReferences\n[1] A. Aazami and M. D. Stilp. Approximation algorithms and hardness for domination with\npropagation. In Proceedings of the 10th International Workshop on Approximation Algorithms\nfor Combinatorial Optimization Problems, volume 4627 of LNCS, pages 1–15. Springer, 2007.\n[2] J. Alber, H. L. Bodlaender, H. Fernau, T. Kloks, and R. Niedermeier.\nFixed parameter\nalgorithms for dominating set and related problems on planar graphs. Algorithmica, 33(4):461–\n493, 2002.\n[3] B. S. Baker. Approximation algorithms for NP-complete problems on planar graphs. J. ACM,\n41(1):153–180, 1994.\n[4] T. L. Baldwin, L. Mili, M. B. Boisen, and R. Adapa. Power system observability with minimal\nphasor measurement placement. IEEE Transactions on Power Systems, 8(2):707–715, 1993.\n[5] H. L. Bodlaender. A linear-time algorithm for finding tree-decompositions of small treewidth.\nSIAM J. Comput., 25(6):1305–1317, 1996.\n[6] D. J. Brueni. Minimal PMU placement for graph observability, a decomposition approach.\nMaster’s thesis, Virginia Polytechnic Institute and State University, Blacksburg, VA, 1993.\n[7] D. J. Brueni and L. S. Heath.\nThe PMU placement problem.\nSIAM J. Discret. Math.,\n19(3):744–761, 2005.\n[8] V. Chvatal. A greedy heuristic for the set covering problem. Math. Oper. Res., 4:233–235,\n1979.\n[9] B. Courcelle, J. A. Makowsky, and U. Rotics. Linear time solvable optimization problems\non graphs of bounded clique width. In Proceedings of the 24th International Workshop on\nGraph-Theoretic Concepts in Computer Science, volume 1517 of LNCS, pages 1–16. Springer,\n1998.\n[10] E. D. Demaine and M. T. Hajiaghayi. Bidimensionality: new connections between FPT al-\ngorithms and PTASs. In Proceedings of the 16th Annual ACM-SIAM Symposium on Discrete\nAlgorithms, pages 590–601, 2005.\n[11] R. Diestel. Graph Theory. Springer-Verlag, New York, 2nd edition, 2000.\n[12] M. Dorfling and M. A. Henning. A note on power domination in grid graphs. Discrete Applied\nMathematics, 154(6):1023–1027, 2006.\n[13] U. Feige. A threshold of ln n for approximating set cover. J. ACM, 45(4):634–652, 1998.\n[14] M. R. Garey and D. S. Johnson. Computers and Intractability: A Guide to the Theory of\nNP-Completeness. W. H. Freeman and Co., New York, NY, USA, 1979.\n[15] J. Guo, R. Niedermeier, and D. Raible. Improved algorithms and complexity results for power\ndomination in graphs. In Proceedings of the 15th International Symposium on Fundamentals\nof Computation Theory, volume 3623 of LNCS, pages 172–184. Springer, 2005 (to appear in\nAlgorithmica).\n20\n\n[16] T. W. Haynes, S. M. Hedetniemi, S. T. Hedetniemi, and M. A. Henning. Domination in graphs\napplied to electric power networks. SIAM J. Discrete Math., 15(4):519–529, 2002.\n[17] T. W. Haynes, S. T. Hedetniemi, and P. J. Slater. Domination in Graphs: Advanced Topics.\nMarcel Dekker, New York, 1998.\n[18] T. W. Haynes, S. T. Hedetniemi, and P. J. Slater. Fundamentals of Domination in Graphs.\nMarcel Dekker, New York, 1998.\n[19] P. Hunter and S. Kreutzer. Digraph measures: Kelly decompositions, games, and orderings.\nIn Proceedings of the 18th Annual ACM Symposium on Discrete Algorithms, pages 637–644,\nPhiladelphia, PA, USA, 2007.\n[20] D. S. Johnson. Approximation algorithms for combinatorial problems. J. Comput. Syst. Sci.,\n9(3):256–278, 1974.\n[21] T. Johnson, N. Robertson, P. D. Seymour, and R. Thomas. Directed tree-width. J. Comb.\nTheory, Ser. B, 82(1):138–154, 2001.\n[22] T. Kloks. Treewidth, Computations and Approximations, volume 842 of LNCS. Springer, 1994.\n[23] J. Kneis, D. M ̈olle, S. Richter, and P. Rossmanith. Parameterized power domination complex-\nity. Inf. Process. Lett., 98(4):145–149, 2006.\n[24] G. Kortsarz. On the hardness of approximating spanners. Algorithmica, 30(3):432–450, 2001.\n[25] C. S. Liao and D. T. Lee. Power domination problem in graphs. In Proceedings of the 11th\nInternational Computing and Combinatorics Conference, volume 3595 of LNCS, pages 818–\n828. Springer, 2005.\n[26] L. Lov ́asz.\nOn the ratio of optimal integral and fractional covers.\nDiscrete Mathematics,\n13:383–390, 1975.\n[27] C. Lund and M. Yannakakis. On the hardness of approximating minimization problems. J.\nACM, 41(5):960–981, 1994.\n[28] L. Mili, T.L. Baldwin, and A.G. Phadke.\nPhasor measurements for voltage and transient\nstability monitoring and control. In Proceedings of the EPRI-NSF Workshop on Application\nof Advanced Mathematics to Power Systems, 1991.\n[29] J. Obdrz ́alek. Dag-width: connectivity measure for directed graphs. In Proceedings of the 17th\nAnnual ACM Symposium on Discrete Algorithms, pages 814–821. ACM Press, 2006.\n[30] P. Slav ́ık. A tight analysis of the greedy algorithm for set cover. In Proceedings of the 28th\nAnnual ACM Symposium on Theory of Computing, pages 435–441, New York, NY, USA, 1996.\nACM Press.\n[31] M. D. Stilp. On power dominating sets. Master’s thesis, Combinatorics and Optimization,\nUniversity of Waterloo, Ontario, Canada, 2006.\n21\n\nA\nDynamic Programming\nIn this section we describe our dynamic-programming algorithm for the Directed PDS problem.\nThis algorithm is similar to the dynamic-programming algorithm designed by Guo et al.[15] to\noptimally solve PDS for undirected graphs with bounded tree-width. It is known that any tree\ndecomposition of width-k can be transformed to a nice tree decomposition with width k in linear-\ntime [22] (Lemma 13.1.3). So we can assume that we are given a nice tree decomposition of the\nunderlying undirected graph of G call it ⟨{Xi|i ∈I} , T⟩. Let Ti denote the subtree of T rooted\nat T-node i, and Yi =\n S\nj∈V (Ti) Xj\n \n\\ Xi. Also let Gi be the subgraph induced by Yi ∪Xi, i.e.\nGi = G [Yi ∪Xi], and let G′i = G[Xi].\nConsider a valid coloring C of the graph G. We store the color of the edges in each bag by\nassigning a state to that bag (the formal definition of a state will follow). We can reconstruct the\ncoloring C from the states of all bags in the tree decomposition of G; so there is no need to store\nthe coloring C in the dynamic-programming.\nThe state of a bag: Given a coloring C, the state of a bag Xi describes the coloring of the edges\nin G′\ni. In order to detect the dependency cycles in the coloring C without reconstructing the whole\ncoloring, we need to store some more information in a state. This extra information enables us to\ndetect a dependency cycle in Gi which goes through Xi, by considering only the state of the bag\nXi. A bag state s contains the following: state of each edge, state of each node, and state of each\npair of nodes in G′\ni = G[Xi].\n• State of an edge: The state of an edge e ∈E(G[Xi]) denoted by s(e) is the color that is\nassigned to e in the coloring C; s(e) ∈{R, B}.\n• State of a node: The state of a node v ∈Xi denoted by s(v) shows the number of red\nedges between v and Yi.\n– s(v) = 1: There is exactly one red edge from a node in Yi to v and no red edge from v\nto Yi,\n– s(v) = 2: There is exactly one red edge from a node in Yi to v and exactly one red edge\nfrom v to Yi,\n– s(v) = 3: There is no red edge between Yi and v,\n– s(v) = 4: There are at least two red edges from v to Yi and no red edge from Yi to v,\n– s(v) = 5: There is exactly one red edge from v to Yi and no red edge from Yi to v.\n• State of a pair of nodes: A dependency path from u to v is a path P where all red edges\nin P are directed from u to v and all blue edges are directed from v to u. We categorize\ndependency paths according to the color of their first and last edges. There are 4 possible\ntypes RR, RB, BR, BB; for example a path of type RB is a path with the first edge colored\nred and the last edge colored blue. For a pair (u, v) ∈Xi × Xi (u ̸= v) the state of (u, v)\ndenoted by s(u, v) shows the type of dependency paths from u to v in G[Yi ∪{u, v}]; that is,\ns(u, v) ⊆{RR, RB, BR, BB}. Note that there are 24 = 16 different states for each pair of\nnodes.\nDetecting dependency cycles:: An important part of the dynamic-programming algorithm is\nto detect dependency cycles in the coloring C. Assume we are at bag Xi and we are given the bag\nstate s corresponding to the coloring C. We can detect the dependency cycles in G′\ni = G[Xi] by\n22\n\nenumerating all possible cycles; note that the coloring of edges in G′\ni is given in the state s. The\ndependency cycles in Gi can be detected by considering the state of each pair of nodes in Xi. For\nexample assume that RB ∈s(u, v) and RR ∈s(v, u). Then, by combining a dependency path of\ntype RB from u to v and a dependency path of type RR from v to u we obtain a dependency cycle\ngoing through u and v in Gi.\nLet us denote by Λi the set of all possible states for the bag Xi. The dynamic-programming will\ncompute a mapping Ai : Λi →N ∪{+∞}. For a bag state s ∈Λi the value Ai(s) is the minimum\nnumber of origins in an optimal valid coloring C of Gi under the restriction that the state of nodes,\nedges, and pairs of nodes in Xi is given by s. Now, we describe how our dynamic-programming\nworks.\nStep 1: (Initialization): In this step for each leaf node i of T, we initialize the mapping Ai as\nfollows. For a given state s, we define Ai(s) as +∞if s has a dependency cycle, a node v with\ns(v) ̸= 3, or a pair of nodes u and v such that s(u, v) ̸= ∅. Otherwise, we define Ai(s) as the\nnumber of nodes with no in-coming red edges in the coloring defined by s.\nStep 2: (Bottom-Up Computation): After initialization, we visit the nodes in T in a bottom-\nup fashion and at each bag Xi we compute the mapping Ai corresponding to Xi. The update\nprocess depends on the type of T-nodes that we are considering. Here, we only consider the update\nprocess at an Insert Node. The other cases are similar to this one.\nInsert Node: Suppose i is an insert node with the child j, and assume that Xi = Xj ∪{x}. For\neach bag state s ∈Λi do the following:\n1. Check whether the coloring given by s forms a valid coloring of G′\ni; if not, define Ai(s) = +∞.\n2. Compute the set Λj(s) containing bag states of j that are “compatible” with the bag state s.\n3. For each s′ ∈Λj(s), check if a valid coloring of Gj “compatible” with s′ can be extended to\na valid coloring of Gi “compatible” with s.\n4. Compute Ai based on the mapping Aj.\nCompatible bag state (Step 2): A bag state s′ ∈Λj is said to be compatible with the bag state\ns ∈Λi if the state of each node, each edge, and each pair of nodes in V (G′\nj) in the bag state s′ is the\nsame as the corresponding state in the bag state s. If s(x) ̸= 3, or ∃v ∈Xj : s(x, v) ̸= ∅∨s(v, x) ̸= ∅\nthen we define Λj(s) = ∅.\nDetecting dependency cycles (Step 3): The conditions of a valid coloring can be violated due\nto degree constraints on the new node x, or by the existence of a dependency cycle going through\nx. Both these cases can be tested by considering the bag states s and s′.\nComputing Ai (Step 4): The addition of x may change the number of origins in Gi. The node x\nwill be an origin if it has at least one outgoing red edge in s. But an origin node v ∈Xj (in s′) that\nhas an incoming red edge from x, is no longer an origin. So by considering the red edges going out\nof x we can update the number of origins and compute the mapping Ai. If the coloring compatible\nwith s is not a valid coloring, then we define Ai(s) to be +∞.\nStep 3: (At root r): Finally, we compute the number of origins in an optimal valid coloring of\nG by finding the minimum of Ar(s) over all possible s ∈Λr. It is easy to see that each bag Xi has\nat most 16(k+1)2 · 5k+1 · 2(k+1)2 states; note that |Xi| ≤(k + 1). It can be checked that the total\nrunning time of our algorithm is O(ck2 · n), for some global constant c.\n23","paragraphs":[{"paragraph_id":"p1","order":1,"text":"arXiv:0710.2139v1 [cs.CC] 10 Oct 2007\nApproximation algorithms and hardness for domination with\npropagation\nAshkan Aazami\naaazami@uwaterloo.ca\nDepartment of Combinatorics and Optimization\nUniversity of Waterloo\nMichael David Stilp\nmstilp3@gatech.edu\nSchool of Industrial and Systems Engineering\nGeorgia Institute of Technology\nNovember 2, 2018\nAbstract\nThe power dominating set (PDS) problem is the following extension of the well-known dom-\ninating set problem: find a smallest-size set of nodes S that power dominates all the nodes,\nwhere a node v is power dominated if (1) v is in S or v has a neighbor in S, or (2) v has a\nneighbor w such that w and all of its neighbors except v are power dominated. We show a\nhardness of approximation threshold of 2log1−ǫ n in contrast to the logarithmic hardness for the\ndominating set problem. We give an O(√n) approximation algorithm for planar graphs, and\nshow that our methods cannot improve on this approximation guarantee. Finally, we initiate\nthe study of PDS on directed graphs, and show the same hardness threshold of 2log1−ǫ n for\ndirected acyclic graphs. Also we show that the directed PDS problem can be solved optimally\nin linear time if the underlying undirected graph has bounded tree-width.\nKeywords: Approximation algorithms, Hardness of approximation, Dominating set, Power\ndominating set, Tree-width, Planar graphs, Greedy algorithms, PMU placement problem.\nAMS subject classifications: 68W25; 90C27\n1\nIntroduction\nA dominating set of an (undirected) graph G = (V, E) is a set of nodes S such that every node in\nthe graph is in S or has a neighbor in S. The problem of finding a dominating set of minimum size\nis an important problem that has been extensively studied, especially in the last 20 years, see the\nbooks by Haynes et al. [17, 18]. The problem is NP-hard [14], a simple greedy algorithm achieves\na logarithmic approximation guarantee∗[20], and, modulo the P ̸= NP conjecture, no polynomial\ntime algorithm gives a better approximation guarantee [27, 13].\nOur focus is on an extension called the Power Dominating Set (abbreviated as PDS) problem.\nPower domination is defined by two rules; the first rule is the same as the rule for the Dominating\nSet problem, but the second rule allows a type of indirect propagation. More precisely, given a set\nof nodes S, the set of nodes that are power dominated by S, denoted PS, is obtained as follows.\n∗An approximation algorithm for a (minimization) optimization problem means an algorithm that runs in poly-\nnomial time and computes a solution whose cost is within a guaranteed factor of the optimal cost; the approximation\nguarantee is the worst-case ratio, over all inputs of a given size, of the cost of the solution computed by the algorithm\nto the optimal cost.\n1"},{"paragraph_id":"p2","order":2,"text":"(Rule 1) if node v is in S, then v and all of its neighbors are in PS;\n(Rule 2) (propagation) if node v is in PS, one of its neighbors w is not in PS, and all other neighbors\nof v are in PS, then w is inserted into PS.\nThe set PS is independent of the sequence in which nodes are inserted by Rule 2. Otherwise,\nthere is a minimal counter example with two maximal sequences of insertions and an “earliest”\nnode that occurs in one sequence but not the other; this is not possible. The PDS problem is\nto find a node-set S of minimum size that power dominates all nodes (i.e., find S ⊆V with |S|\nminimum such that PS = V ). We use Opt(G) to denote the size of an optimal solution for the PDS\nproblem for a graph G. Throughout, we use n to denote the number of nodes in the input graph.\nFor example, consider the planar graph in Figure 1; the graph has t disjoint triangles, and three\n(mutually disjoint) paths such that each path has exactly one node from each triangle; note that\n|V | = 3t. The minimum dominating set has size Θ(|V |), since the maximum degree is 4. The\nminimum power dominating set has size one – if S has any one node of the innermost (first)\ntriangle (like v), then PS = V †.\nv\nFigure 1: Illustrating those nodes power dominated by Rule 1 (denoted by a triangle) and Rule 2\n(denoted by a square); the picked node is shown by a circle.\nThe PDS problem arose in the context of electric power networks, where the aim is to monitor all\nof the network by placing a minimum-size set of very expensive devices called phase measurement\nunits; these units have the capability of monitoring remote elements via propagation (as in Rule 2);\nsee Brueni [6], Baldwin et al. [4], and Mili et al. [28]. In the engineering literature, the problem is\ncalled the PMU placement problem.\nOur motivation comes from the area of approximation algorithms and hardness results. The\nDominating Set problem is a so-called covering problem; we wish to cover all nodes of the graph\nby choosing as few node neighborhoods as possible. In fact, the Dominating Set problem is a\nspecial case of the well-known Set Covering‡ problem.\nSuch covering problems have been extensively investigated. One of the key positive results dates\nfrom the 1970’s, when Johnson [20], Lov ́asz [26] and later Chv ́atal [8] showed that the greedy method\nachieves an approximation guarantee of O(log |V |) where |V | denotes the size of the ground set, see\nalso [30]. Several negative results (on the hardness of approximation) have been discovered over\n†In more detail, we apply Rule 1 to see that all the nodes of the innermost (first) triangle and one node of the\nsecond triangle are in PS; then by two applications of Rule 2 (to each of the nodes in the first triangle not in S), we\nsee that the other two nodes of the second triangle are in PS; then by three applications of Rule 2 (to each of the\nnodes in the second triangle) we see that all three nodes of the third triangle are in PS; etc.\n‡Given a family of sets on a groundset, find the minimum number of sets whose union equals the groundset.\n2"},{"paragraph_id":"p3","order":3,"text":"the last few years: Lund and Yannakakis [27] showed that the Set Covering problem is hard to\napproximate within a ratio of Ω(logn) and later, Feige [13] showed that it is hard to approximate\nwithin a ratio of (1 −ǫ) ln n, modulo some variants of the P ̸= NP assumption.\nA natural question is what happens to covering problems (in the setting of approximation algo-\nrithms and hardness results) when we augment the covering rule with a propagation rule. PDS\nseems to be a key problem of this type, since it is obtained from the Dominating Set problem by\nadding a simple propagation rule.\n1.1\nPrevious literature\nApparently, the earliest publications on PDS are Brueni [6], Baldwin et al. [4], and Mili et al. [28].\nLater, Haynes et al. [16] showed that the problem is NP-complete even when the input graph is\nbipartite; they presented a linear-time algorithm to solve PDS optimally on trees. Kneis et al. [23]\ngeneralized this result to a linear-time algorithm that finds an optimal solution for graphs that have\nbounded tree-width, relying on earlier results of Courcelle et al. [9]. Kneis et al. [23] also showed\nthat PDS is a generalization of the Dominating Set problem as follows. Given a graph G we\ncan construct an augmented graph G′ such that S is an optimal solution for the Dominating Set\nproblem on G if and only if it is an optimal solution for PDS on G′; the graph G′ is obtained from\nG by adding a new node v′ for each node v in G and adding the edge vv′. Guo et al. [15] developed\na combinatorial algorithm based on dynamic-programming for optimally solving PDS on graphs of\ntree-width k. The running time of their algorithm is O(ck2 · n) where c is a constant. Guo et al.\nalso compared the tractability of the Dominating Set problem versus PDS on several classes of\ngraphs, that is, they study whether there are classes of graphs where the former problem is in P\nbut the latter one is NP-hard; but they have no result that “separates” the two problems. Even\nfor planar graphs, the Dominating Set problem is NP-hard [14], and the same holds for PDS\n[15]. Liao and Lee [25] proved that PDS on split graphs is NP-complete, and also they presented\na polynomial time algorithm for solving PDS optimally on interval graphs. Dorfling and Henning\ncomputed the power domination number, i.e. the size of optimal power dominating set, for n × m\ngrids [12].\nBrueni and Heath [7] have more results on PDS, especially the NP-completeness of\nPDS on planar bipartite graphs. To the best of our knowledge, no further results are known on\nsolving the PDS problem, either optimally or approximately. Some of the results in this paper have\nappeared in the thesis of the second author [31], and in the proceedings of a workshop [1].\n1.2\nOur contributions\nOur results substantially improve on the understanding of PDS in the context of approximation\nalgorithms. In particular, we show a substantial gap between the approximation guarantees for the\nDominating Set problem and PDS modulo a variant of the P ̸= NP conjecture. This seems to be\nthe first known “separation” result between the two problems, in any class of graphs.\n• We present a reduction from the MinRep problem to the PDS problem that shows that PDS\ncannot be approximated within a factor of 2log1−ǫn, unless NP ⊆DTIME(npolylog(n)).\n• For undirected graphs, we introduce the notion of strong regions and weak regions as a\nmeans of obtaining lower bounds on the size of an optimal solution for PDS. Based on this,\nwe develop an approximation algorithm for PDS that gives an approximation guarantee of\nO(k) for graphs that have tree-width k. The algorithm requires the tree decomposition as\n3"},{"paragraph_id":"p4","order":4,"text":"part of the input, and runs in time O(n3) (independent of k). By slightly modifying this\nalgorithm we get an algorithm that solves PDS optimally on trees. Our algorithm provides\nan O(√n)-approximation algorithm for PDS on planar graphs because a tree decomposition\nof a planar graph with width O(√n) can be computed efficiently [2]. Moreover, we show\nthat our methods (specifically, the lower bounds used in our analysis) cannot improve on our\nO(√n) approximation guarantee.\n• We extend PDS in a natural way to directed graphs and prove that even for directed acyclic\ngraphs, PDS is hard to approximate within the same threshold as for undirected graphs\nmodulo the same complexity assumption.\n• We give a linear-time algorithm based on dynamic-programming for Directed PDS when\nthe underlying undirected graph has bounded tree-width. This builds on results and methods\nof Guo et al. [15].\n2\nPDS in Undirected Graphs\nIn this section we prove a result on the hardness of approximating PDS by a reduction from the\nMinRep problem. In Section 2.1 we define the MinRep problem, and then we give a gap preserving\nreduction from MinRep to PDS in Section 2.2.\n2.1\nThe MinRep problem\nIn the MinRep [24] problem we are given a bipartite graph G = (A, B, E) with a partition of A and\nB into equal-sized subsets. Let qA and qB denote the number of sets in the partition of A and B,\nrespectively. Let A = A1 ∪A2 ∪· · ·∪AqA denote the partition of A, and let B = B1 ∪B2 ∪· · ·∪BqB\ndenote the partition of B. This partition naturally defines a super bipartite graph H = (A, B, E).\nThe super nodes of H are A1, A2, . . . , AqA and B1, B2, . . . , BqB. There is a super edge between\nsuper nodes Ai and Bj if there exists some a ∈Ai and b ∈Bj such that ab is an edge in G. We\nsay that super edge AiBj is covered by nodes a, b if a ∈Ai, b ∈Bj, and there is an edge between\na and b in G. Given S ⊆A ∪B we say that the super edge AiBj is covered by S if there exists\na, b ∈S that covers AiBj. The goal in the MinRep problem is to pick a minimum-size set of nodes,\nA′ ∪B′ ⊆V (G), to cover all the super edges in H. Note that we need a pair of nodes to cover\na super edge, and the pair should induce an edge between the two super nodes of the super edge;\nmoreover, a node in A′ ∪B′ may be useful for covering more than one super edge. The following\nTheorem is from [24].\nTheorem 2.1 (Theorem 5.4 in [24]) The MinRep problem cannot be approximated within ra-\ntio 2log1−ǫn, for any fixed ǫ > 0, unless NP ⊆DTIME(npolylog(n)).\n2.2\nThe reduction to PDS\nTheorem 2.2 The PDS problem cannot be approximated within ratio 2log1−ǫn, for any fixed ǫ > 0,\nunless NP ⊆DTIME(npolylog(n)).\nThe reduction: Theorem 2.2 is proved by a reduction from the MinRep problem. We create an\ninstance G = (V , E) of the PDS problem from a given instance G = (A, B, E)(H = (A, B, E)) of\nthe MinRep problem. The idea is to replace each super edge with a “cover testing gadget”.\n4"},{"paragraph_id":"p5","order":5,"text":"1. Start with a copy of each node in A ∪B in G. For convenience, we use the same notation for\nnodes (and set of nodes) in G and their copies in G.\n2. Add a new node w∗to the graph G, and connect w∗to all nodes in A ∪B. Also add new\nnodes w∗\n1, w∗\n2, w∗\n3 and connect them to w∗(the nodes w∗\n1, w∗\n2, w∗\n3 are added to force w∗to be\nin any optimal solution. See the proof of Lemma 2.3 for more details).\n3. ∀i ∈{1, . . . , qA} , j ∈{1, . . . , qB} if AiBj is a super edge, then do the following:\n(a) Let Eij denote the set of edges between Ai and Bj in G and let lij denote |Eij| (see\nFigure 3(a); for an example E11 has 3 edges, and E12 has 4 edges). We denote the edges\nin Eij by e1, e2, · · · , ek, · · ·.\n(b) Let Cij be a cycle of 3lij nodes. We sequentially label the nodes of Cij as u1, v1, w1,\nu2, v2, w2, · · · , uk, vk, wk, · · · (informally speaking, we associate each triple uk, vk, wk with\nan edge ek of Eij). Make λ = 4 new copies of the graph Cij (λ can be any constant greater\nthan 3; refer to the proof of Lemma 2.3 for more details). For each edge ek = akbk ∈Eij\nand for each of the 4 copies of Cij, we add an edge from ak to uk and an edge from bk\nto vk. See Figures 2(a), 2(b) for an illustration.\nCij\nv2\nu2\nu1\nv1\nvlij\nulij\nek\ne2\ne1\nelij\nvk\nuk\nw1\nw2\nwk\nwlij\n(a) The Cij graph\nCij\na1\nb1\na2\nb2\ne2\ne1\nv2\nu2\nu1\nv1\nvk\nuk\nek\nak\nbk\nvlij\nulij\nblij\nalij\nelij\n(b) Edges between Cij and Ai ∪Bj.\nFigure 2: The cover testing gadget.\n4. Let G = (V , E) be the obtained graph (see Figure 3 for an illustration).\nLet S be a feasible solution for the resulting PDS instance G, and suppose w∗∈S. Then all\nof the nodes in A ∪B are power dominated (by Rule 1 of PDS). Now consider a gadget Cij, and\nassume a node v of Cij is in S. By applying Rule 1 once and then repeatedly applying Rule 2 of\nPDS, the gadget Cij will be completely power dominated, that is, all nodes of the gadget will be\nin PS.\nThe next lemma shows that the size of an optimal solution in PDS is exactly one more than the\nsize of an optimal solution in MinRep. The number of nodes in the constructed graph is equal to\n V (G)\n = 4+ |V (G)| + 3λ |E(G)|. This will complete the proof of Theorem 2.2 by showing that the\nabove reduction is a gap preserving reduction from MinRep to PDS with the same gap (hardness\nratio) as the MinRep problem.\n5"},{"paragraph_id":"p6","order":6,"text":"Lemma 2.3 A∗∪B∗is an optimal solution to the instance G = (A, B, E) of the MinRep problem\nif and only if S∗= A∗∪B∗∪{w∗} ⊆V (G) is an optimal solution to the instance G of the PDS\nproblem.\nProof: First, we claim that w∗should be in any optimal solution of the PDS instance G. Suppose\nthat w∗is not in some optimal solutions. Then, in order to power dominate the nodes w∗, w∗\n1, w∗\n2, w∗\n3\nin G, the set S must contain at least two of the nodes (leaves) w∗\n1, w∗\n2, w∗\n3. This is a contradiction,\nsince we can replace these 2 nodes by w∗and obtain a smaller feasible solution.\nAssume that A∗∪B∗is a feasible solution for the MinRep instance G.\nWe will show that\nS = A∗∪B∗∪{w∗} is a feasible solution to the PDS instance G. Note that all nodes in A ∪B ∪\n{w∗, w∗\n1, w∗\n2, w∗\n3} are power dominated by applying Rule 1 on w∗. Now, we only need to show that\nall nodes in the gadgets Cij are power dominated. Consider any super edge AiBj of H. The set\nA∗∪B∗covers all the super edges in H. So there exists a pair of nodes ak ∈A∗∩Ai, bk ∈B∗∩Bj\nthat induces an edge of G. Since ak and bk are in S, their neighbors, uk and vk, in each of the\nλ = 4 copies of Cij in G, will be power dominated by applying Rule 1. Then the nodes uk and vk\nin each copy of Cij will power dominate the entire cycle by repeatedly applying Rule 2. To see this,\nnote that any node in Cij has exactly 2 neighbors in Cij and at most 1 neighbor not in Cij. The\nneighbors not in Cij are from Ai ∪Bj, and they are power dominated by w∗. Hence, if a node in Cij\nand one of its neighbors in Cij are power dominated, then by applying Rule 2 the other neighbor\nin Cij will be power dominated. Hence, by starting from vk and repeatedly applying Rule 2, we\ncan sequentially power dominate the nodes in Cij. This shows that S power dominates all nodes\nin G. Therefore, Opt(G) is at most |A∗∪B∗| + 1.\nLet S∗⊆V (G) be an optimal solution for PDS. By the above claim, w∗is in S∗. Now define\nA′ = A ∩S∗and B′ = B ∩S∗. First we prove that any optimal solution of PDS is contained in\nA ∪B ∪{w∗}, and then we show that A′ ∪B′ covers all super edges of the MinRep instance G.\nSuppose that S∗contains some nodes not in A ∪B ∪{w∗}. Hence, there are some gadgets that are\nnot completely power dominated by S∗∩(A ∪B ∪{w∗}). Let Cij be such a gadget. By symmetry\neach of the λ = 4 copies of Cij is not completely power dominated. Therefore, the optimal solution\nS∗needs to have at least 3 nodes from the 4 copies of Cij. By removing these 3 nodes from S∗and\nA1\nA2\nB1\nB2\n(a) MinRep Instance G\nw∗\n3\nw∗\n1\nw∗\n2\nw∗\nC2,2\nA2\nB2\nB1\nC1,1\nC1,2\nA1\n(b) PDS instance G: For each super edge AiBj we show\nonly 1 copy of Cij; in fact G has λ = 4 copies of Cij.\nFigure 3: The hardness construction\n6"},{"paragraph_id":"p7","order":7,"text":"adding ak ∈Ai and bk ∈Bj to S∗for some arbitrary edge akbk ∈Eij, we can power dominate all\nof the 4 copies of Cij. This contradicts the minimality of S∗, and proves that S∗⊆A ∪B ∪{w∗}.\nTo see that A′ ∪B′ covers all super edges, note the following: suppose no node from any copy of\nCij is in the optimal solution; then any Cij can be power dominated only by taking a pair of nodes\na ∈Ai, b ∈Bj that induces an edge of G. This completes the proof of the lemma.\n□\n3\nApproximation Algorithms for Planar Graphs\nIn this section we describe an O(k)-approximation algorithm for PDS in graphs with tree-width\nk; the running time is O(n3), independent of k. This algorithm gives an O(√n)-approximation\nalgorithm for PDS in planar graphs, since the tree-width of a planar graph G with n nodes is\nO(√n) and in O(n\n3\n2) time we can find an O(√n) tree-width decomposition of the given planar\ngraph G [2]. Finally, we show that the analysis of our algorithm is tight on planar graphs. We use\nPlanar PDS to denote the special case of the PDS problem where the graph is planar.\nDefinition 3.1 [11] A tree decomposition of a graph G = (V, E) is a pair ⟨{Xi ⊆V |i ∈I} , T =\n(I, F)⟩such that T is a tree with V (T) = I, E(T) = F, and satisfying the following properties:\n(T1) S\ni∈I Xi = V , and every edge uv ∈E has both ends in some Xi,\n(T2) For all i, j, k ∈I if j is on the unique path from i to k in T then we have: Xi ∩Xk ⊆Xj,\nThe width of ⟨{Xi|i ∈I} , T⟩is the maxi∈I|Xi|−1. The tree-width of G is defined as the minimum\nwidth over all tree decompositions. The nodes of the tree are called T-nodes and the sets Xi are\ncalled bags.\nA nice tree decomposition is a tree decomposition ⟨{Xi ⊆V |i ∈I} , T = (I, F)⟩, where T is a\nrooted tree in which each node has at most 2 children.\nIf a node i ∈I has two children j, k\nthen Xi = Xj = Xk (i is called a Join node), and if i has one child j then either Xj ⊂Xi and\n|Xi \\ Xj| = 1 or Xi ⊂Xj and |Xj \\ Xi| = 1 (i is called an Insert or a Forget node, respectively).\nWe introduce the notion of a strong region before presenting our algorithm. Informally speaking,\na set of nodes R ⊆V is called strong if every feasible solution to the PDS problem has a node of R.\nFor a graph G = (V, E), the neighborhood of R ⊆V is nbr(R) = {v ∈V |∃uv ∈E, u ∈R, v /∈R},\nand the exterior of R is defined by ext(R) = nbr(V \\R), i.e., ext(R) consists of the nodes in R that\nare adjacent to a node in V \\ R.\nDefinition 3.2 Given a graph G = (V, E) and a set S ⊆V , the subset R ⊆V is called an S-strong\nregion if R ̸⊆PS∪nbr(R), otherwise, the set R is called an S-weak region. The region R is called\nminimal S-strong if it is an S-strong region and ∀r ∈R, R −r is an S-weak region.\nIt is easy to check from the definition that an S-strong region is also an ∅-strong (or shortly strong)\nregion. Any feasible solution to the PDS problem needs to have at least one node from every strong\nregion.\nLemma 3.3 A subset R ⊆V is an S-strong region if and only if for every feasible solution S ∪S∗\nof G, we have R ∩(S∗\\ S) ̸= ∅.\n7"},{"paragraph_id":"p8","order":8,"text":"Proof: It can be seen that the set S ∪(V \\ R) will power dominate the same set of nodes in R\nthat can be power dominated by S ∪nbr(R); this is valid for any subset R ⊆V .\nLet R be an S-strong region. By the definition of a strong region we have R ̸⊆PS∪nbr(R). Hence,\nby the above claim R ̸⊆PS∪(V \\R). This shows that every feasible solution S∗needs to have at least\none node from R that is not in S.\nNow assume that for every feasible solution S∗of G we have R ∩(S∗\\ S) ̸= ∅. Suppose that R\nis an S-weak region, so by the definition of a weak region we have R ⊆PS∪nbr(R). It follows that\nS∗= S ∪(V \\ R) is a feasible solution, but R has no intersection with S∗\\ S ⊆(V \\ R). This is a\ncontradiction, so R is an S-strong region.\n□\nOur algorithm makes one level-by-level and bottom-to-top pass over the tree T of the tree de-\ncomposition of G and constructs a solution S for PDS (initially, S = ∅). At each node rj of T we\ncheck whether the union of the bags in the subtree rooted at rj forms an S-strong region; if yes,\nthen the bag Xrj of rj is added to S, otherwise S is not updated. The key point in the analysis is\nto show that Opt(G) ≥m, where m is the number of nodes of T where we updated S.\nAlgorithm 1 O(k)-approximation Algorithm\n1: A tree decomposition ⟨{Xi|i ∈I} , T⟩of G is given, where T is rooted at r.\n2: Let Ilbe the set of T-nodes at distance lfrom the root, and let d be the maximum distance\nfrom r in T.\n3: S ←∅\n4: for i = d to 0 do\n5:\nLet Ii = {r1, . . . , rki} and denote by Trj the subtree in T rooted at rj.\n6:\nLet Yrj be the union of bags corresponding to the T-nodes in Trj.\n7:\nfor j = 1 to ki do\n8:\nif Yrj is an S-strong region then\n9:\nS ←S ∪Xrj; where Xrj is the bag corresponding to rj.\n10:\nend if\n11:\nend for\n12: end for\n13: Output So = S\n3.1\nAnalysis of the algorithm\nIn this subsection we show that our algorithm has an approximation guarantee of O(k).\nLet\nG = (V, E) denote the input graph, and let S ⊆V be any set of nodes.\nLemma 3.4 Suppose Z is an S-weak region such that ext(Z) ⊆S. Then we have Z ⊆PS.\nProof: Let Y = ext(Z), it is easy to check that nbr(Z \\ Y ) ⊆ext(Z). We claim that Z \\ Y is an\nS-weak region. Let S∗= V \\ (Z ∪S), it is easy to check that S ∪S∗is a feasible solution for the\ngraph G, but S∗∩(Z \\ Y ) = ∅. Hence, by Lemma 3.3, Z \\ Y is not an S-strong region, and so it\nis an S-weak region. Thus Z \\ Y ⊆PS∪nbr(Z\\Y ) ⊆PS∪ext(Z) = PS and this implies that Z ⊆PS as\nY = ext(Z) ⊆S.\n□\n8"},{"paragraph_id":"p9","order":9,"text":"Lemma 3.5 Let Z ⊆V be an S-strong region. Suppose that Y is a subset of V such that Y ⊆PS\nand ext(Y ) ⊆S. Then Z \\ Y is an S-strong region.\nProof: Assume for the sake of contradiction that Z \\Y is an S-weak region. Then by the definition\nof strong regions we have: Z \\Y ⊆PS∪nbr(Z\\Y ). It is easy to see that nbr(Z \\Y ) ⊆nbr(Z)∪ext(Y ).\nThis implies that Z \\Y ⊆PS∪nbr(Z\\Y ) ⊆PS∪nbr(Z)∪ext(Y ) = PS∪nbr(Z). The condition in the lemma\nstates that Y ⊆PS ⊆PS∪nbr(Z). Hence, we get Z = (Z \\ Y ) ∪(Z ∩Y ) ⊆PS∪nbr(Z), which means\nthat Z is an S-weak region. This is a contradiction, so the lemma is proved.\n□\nTheorem 3.6 Given a graph G = (V, E) and a tree decomposition of G of width k as input,\nAlgorithm 1 runs in time O(n · |E|), and achieves an approximation guarantee of (k + 1).\nProof: First, we show that the solution So found by the algorithm is feasible. Then we prove the\napproximation guarantee, and establish the running time.\nFor any node q of T, recall that Yq denotes the union of the bags corresponding to the T-nodes\nin the subtree rooted at q in T; let Gq denote the subgraph of G induced by Yq. We claim that\next(Yq) ⊆Xq.\nSuppose that q has m children in T, call them c1, . . . , cm.\nFor each edge qcj\n(j = 1, · · · , m), the set Xq ∩Xcj separates Ycj from the rest of the graph, that is, every path\nbetween a node in Ycj and a node in V \\Ycj contains a node of Xq ∩Xcj (see Lemma 12.3.1 in [11]).\nThus, ext(Ycj) ⊆Xq ∩Xcj ⊆Xq, and hence, for Yq = Xq ∪Yc1 ∪· · · ∪Ycm, we have ext(Yq) ⊆Xq.\nWe use induction on the height of the subtree of T rooted at q to prove the following: if Yq is\nS∗-strong, then Yq ⊆PS∗∪Xq, where S∗denotes the solution just before the algorithm examines\nYq. The statement clearly holds when q is a leaf of T (since Yq = Xq). Otherwise, let c1, . . . , cm\nbe the children of q in T. For each j = 1, . . . , m, when the algorithm examined Ycj, either Ycj was\nS-weak, in which case (by Lemma 3.4) we have Ycj ⊆PS∪ext(Ycj ) ⊆PS∪(Xcj ∩Xq) ⊆PS∗∪Xq or Ycj\nwas S-strong in which case Ycj ⊆PS∪Xcj by induction (note that S ∪Xcj ⊆S∗); we use S to denote\nthe solution just before the algorithm examines Ycj. Hence, Yq = Yc1 ∪· · · ∪Ycm ∪Xq ⊆PS∗∪Xq.\nThe above statement implies that V ⊆PSo because at the step when the algorithm examines the\nroot r of T either\n(i) Yr is S-strong, so So = S ∪Xr, and Yr ⊆PS∪Xr = PSo; or\n(ii) Yr is S-weak, and Yr ⊆PS∪ext(Yr) = PSo; since Yr = V (G) and ext(Yr) = ∅.\nTo show that the approximation guarantee is (k+1) we will construct a set ∆of pairwise disjoint\nstrong regions R1, R2, . . . , such that there is a strong region Rj corresponding to each step of the\nalgorithm that adds a non empty bag Xqj to S. Thus |So| ≤(k + 1) |∆| since each bag has ≤k + 1\nnodes, and Opt(G) ≥|∆| because every feasible solution has size ≥|∆|, by Lemma 3.3. Hence,\n|So| ≤(k + 1)Opt(G). We construct the sets R1, R2, . . . , during the execution of the algorithm\nas follows. Suppose the algorithm finds Yq to be S-strong while examining a node q of T. Let\nq1, . . . , ql−1 be the nodes of T where the algorithm updated the solution before examining q, and\nlet S be the solution just before the algorithm examines q. Then define Rl= Yql\\(Yq1 ∪· · ·∪Yql−1),\nwhere ql= q. We claim that Rlis an S-strong region. For each strong region Yqj (j = 1, . . . , l−1) we\nhave seen that ext(Yqj) ⊆Xqj ⊆S and Yqj ⊆PS; note that the algorithm added Xqj to the solution\nsince Yqj was a strong region. It follows that ext(Yq1 ∪· · · ∪Yql−1) ⊆S, and Yq1 ∪· · · ∪Yql−1 ⊆PS.\nHence, by Lemma 3.5, the set Rlis an S-strong region. Clearly, the sets R1, R2, . . . , are pairwise\ndisjoint. This completes the construction of ∆.\n9"},{"paragraph_id":"p10","order":10,"text":"Consider the running time. Without loss of generality we can assume that the given tree decom-\nposition of width k has at most 4n bags (see Lemma 13.1.2 in [22]). Using standard algorithmic\ntechniques we can test in O(|E|) time whether a given set R ⊆V is an S-strong region (we compute\nPS∪nbr(R) and check if it contains R). Therefore, our algorithm has a running time of O(n · |E|). □\nIt is known that planar graphs have tree-width O(√n), and such a tree decomposition can be\nfound in O(n\n3\n2 ) time [2]. This fact together with the above theorem proves the following theorem.\nTheorem 3.7 Algorithm 1 achieves an approximation guarantee of O(√n) for the Planar PDS\nproblem.\nAs mentioned earlier, Haynes et al.[16] presented a linear-time algorithm for optimally solving\nPDS on trees. By modifying Algorithm 1, we can solve PDS optimally on trees. The resulting\nalgorithm differs from the algorithm of Haynes et al. since our algorithm uses strong regions.\nInformally, the algorithm makes a level-by-level and bottom-to-top pass over the tree G. At a node\nv of the tree G if the set of nodes in the subtree rooted at v forms a strong region, then we add v\nto the solution, otherwise we skip v. Formally, we define Xv = {v} for each v ∈V (G), and we run\nAlgorithm 1 on the tree G. Note that defining bags in this way does not give a tree decomposition\nof G.\nTheorem 3.8 A modification of Algorithm 1 runs in time O(n · |E|) and solves PDS optimally on\ntrees.\n3.2\nLower bounds via disjoint strong regions\nIn this part we show that any approximation algorithm for PDS that uses the number of disjoint\nstrong regions as a lower bound has an approximation guarantee of Ω(√n). In proposition 3.12\nwe give a lower bound on the optimal value for PDS on an l× m grid. Independently, [12] gave a\nstronger result for this.\nLemma 3.9 Any minimal S-strong region is connected.\nProof: Assume that R is a minimal S-strong region that is not connected.\nLet C ⊂R be a\nconnected component of R. The set C is an S-weak region since R is a minimal S-strong region.\nBy the definition of a weak region we have C ⊆PS∪nbr(C). The set C is a connected component of\nR, so the neighborhood of C has no intersection with R \\ C. This implies that nbr(C) ⊆nbr(R),\nand consequently we have C ⊆PS∪nbr(C) ⊆PS∪nbr(R). The same argument as above shows that\nR \\ C ⊆PS∪nbr(R). Hence, R ⊆PS∪nbr(R) which is a contradiction.\n□\nLemma 3.10 The number of disjoint strong regions in an l× m grid is exactly one.\nProof: For the sake of contradiction, assume that the given grid has two disjoint strong regions.\nTake as few nodes as possible from these strong regions until we get minimal strong regions, say\nR1 and R2. It is easy to check that the set of nodes of any row or any column of the grid power\ndominates all nodes in the grid. By Lemma 3.3, R1 and R2 should have at least one node from\nevery feasible solution. In the other words, R1 and R2 must have at least one node from each row\nand also from each column. By lemma 3.9, we know that R1 and R2 induce connected subgraphs.\nHence, in R1 there is a path from a node in the top row to a node in the bottom row, and also in R2\n10"},{"paragraph_id":"p11","order":11,"text":"there is a path from a node in the rightmost column to a node in the leftmost column. Obviously\nthese two paths share a common node. This is a contradiction, since R1 and R2 are assumed to be\ndisjoint.\n□\nWe denote by Pi\nS a set of nodes that are power dominated after applying propagation rule, Rule\n2, for i number of times to P0\nS = S ∪nbr(S). Obviously this depends on the order of applying the\npropagation rule. We use the notation without specifying the order of applying the propagation\nrule.\nLemma 3.11 (Propagation lemma) Given an ordering of propagation rules applied to S ∪nbr(S)\nwith Pk\nS = PS we have:\n ext(Pj\nS)\n ≤\n ext(Pi\nS)\n , ∀0 ≤i < j ≤k.\nProof: We will prove that\n ext(Pi+1\nS\n)\n ≤\n ext(Pi\nS)\n , for all 0 ≤i ≤k −1. Consider the set Pi\nS\nand assume that in the (i + 1)-st step we apply Rule 2 to v ∈ext(Pi\nS) and power dominate u; i.e.\nu ∈Pi+1\nS\n, u /∈Pi\nS. To apply Rule 2 to v all neighbors of v except u should be power dominated, so\nwe have nbr(v)\\Pi\nS = {u}. Also since we power dominate u at step (i+1), we have nbr(v) ⊆Pi+1\nS\n.\nTherefore, v is not in ext(Pi+1\nS\n), but u may be in the exterior of Pi+1\nS\n. Hence, we have ext(Pi+1\nS\n) ⊆\n ext(Pi\nS) \\ {v}"},{"paragraph_id":"p12","order":12,"text":"∪{u}. It follows that\n ext(Pi+1\nS\n)\n ≤\n ext(Pi\nS) \\ {v}\n + |{u}| =\n ext(Pi\nS)\n .\n□\nProposition 3.12 Let G be an l× m grid with l≤m, then Opt(G) = Θ(l).\nProof: First note that any row or any column of the grid power dominates all nodes. In the\nfollowing we prove that any feasible solution of PDS needs to have at least l−1\n5\nnodes. Assume that\nthere exists S ⊆V such that |S| < l−1\n5\nand PS = V (G). The maximum degree in G is 4, so we have\n ext(P0\nS)\n ≤|S ∪nbr(S)| < l−1\n5 · 5 = l −1. Therefore, P0\nS contains no full row or no full column of\nG. The set S power dominates G, so there is an i such that Pi\nS contains a full row or a full column.\nConsider the smallest i with this property. Hence, Pi−1\nS\nhas no full row or full column. But some\nrow or some column must have at least l −1 nodes in Pi−1\nS\nsince Pi\nS contains an entire row or an\nentire column. Without loss of generality assume that these l −1 nodes are from a column. By the\ndefinition of i, each of the l −1 nodes in this column are in a row which is not a subset of Pi−1\nS\n.\nTherefore, there are at least l −1 rows with at least one node in Pi−1\nS\nand at least one node not\nin Pi−1\nS\n. This implies that\n ext(Pi−1\nS\n)\n ≥l −1. Finally, by using Lemma 3.11 we get the following\ncontradiction: l −1 >\n P0\nS\n ≥\n ext(P0\nS)\n ≥\n ext(Pi−1\nS\n)\n ≥l −1.\n□\nConsider any approximation algorithm for PDS that uses only the number of disjoint strong\nregions as a lower bound on the size of an optimal solution. By Lemma 3.10, this algorithm finds\na lower bound of 1 on the size of an optimal solution on a grid. The √n × √n grid has an optimal\nsolution of size Θ(√n) by Proposition 3.12. This shows that the approximation guarantee of the\nalgorithm is Ω(√n), even on planar graphs.\nProposition 3.13 Consider any approximation algorithm for PDS that uses only the number of\ndisjoint strong regions as a lower bound on the optimal value. Then the approximation guarantee\nis Ω(√n).\n4\nPDS in Directed Graphs\nIn this section we extend the PDS problem to directed graphs to obtain the Directed Power\nDominating Set (Directed PDS) problem. Our motivation for studying the directed prob-\nlem comes from theoretical considerations.\nThe Dominating Set problem is studied on both\n11"},{"paragraph_id":"p13","order":13,"text":"undirected and directed graphs, and there is extensive literature on the latter (see [17, 18]). The\nsimilarities between the Dominating Set problem and the PDS problem led us to define and study\nthe Directed PDS problem. We give a result on the hardness of approximation of Directed\nPDS. Then we reformulate the Directed PDS problem in terms of valid coloring of the edges.\nUsing this, we design an algorithm for solving Directed PDS in linear-time on a special class of\ndirected graphs.\nLet G = (V, E) be a directed graph.\nA node w is called an out-neighbor (in-neighbor) of a\nnode v if there is a directed edge from v to w (from w to v) in G. The number of out-neighbors\n(in-neighbors) of a node v is called the out-degree (in-degree) of v and is denoted by d+\nG(v) (or\nsimilarly d−\nG(v)). For a set of nodes X, the subgraph of G induced by X is denoted by G[X]. The\ndirected graphs that we consider here have no loops nor parallel edges, but may have two edges\nwith different directions on the same two end nodes (we call such edges antiparallel).\nGiven a\ndirected graph G by the underlying undirected graph we mean the undirected graph obtained from\nG, by removing the direction of edges and also removing any parallel edges that are introduced\nafter removing the directions.\nDefinition 4.1 (the Directed PDS problem) Let G be a directed graph. Given a set of nodes\nS ⊆V (G), the set of nodes that are power dominated by S, denoted by PS, is obtained as follows:\n(D1) if node v is in S, then v and all of its out-neighbors are in PS;\n(D2) (propagation) if node v is in PS, one of its out-neighbors w is not in PS, and all other\nout-neighbors of v are in PS, then w is inserted into PS.\nWe say that S power dominates G if PS = V (G). The Directed PDS problem is to find a node\nset S with minimum size that power dominates all the nodes in G.\nWe prove a threshold of 2log n1−ǫ for the hardness of approximation of Directed PDS modulo\nthe same complexity assumption as in Theorem 2.2. The proof uses a reduction from the MinRep\nproblem to the Directed PDS problem in a directed acyclic graph. This reduction is similar to\nthe reduction in Theorem 2.2; the main difference comes from the gadget for modeling the super\nedges.\nTheorem 4.2 The Directed PDS problem even when restricted to directed acyclic graphs cannot\nbe approximated within ratio 2log1−ǫ n, for any fixed ǫ > 0, unless NP ⊆DTIME(npolylog(n)).\nThe reduction: We create an instance G = (V , E) of the Directed PDS problem from a given\ninstance G = (A, B, E)(H = (A, B, E)) of the MinRep problem.\n1. Start with a copy of each node in A ∪B in G. For convenience, we use the same notation for\nnodes (and sets of nodes) in G and their copies in G.\n2. Add a new node w∗to the graph G, and add a directed edge from w∗to each node in A ∪B.\n3. ∀i ∈{1, . . . , qA} , j ∈{1, . . . , qB} if AiBj is a super edge, then do the following:\n(a) Let Eij be the set of edges between Ai and Bj in G, and let lij denote |Eij|. We denote\nthe edges in Eij by e1, e2, . . . , ek, . . . , elij.\n12"},{"paragraph_id":"p14","order":14,"text":"(b) Let Dij be the graph on 6lij + 1 nodes as shown in Figure 4(a). In Dij there are 6\nnodes uk, vk, dk, αk, βk, γk associated with an edge ek of Eij. The part of Dij associated\nwith an edge ek is shown in Figure 4(b); note that all these parts share a common node,\ncalled the center node, in Dij. Make λ = 4 new copies of the graph Dij (λ can be any\nconstant greater than 3). For each edge ek = akbk ∈Eij and for each of the 4 copies of\nDij, we add a directed edge from ak to uk and a directed edge from bk to vk. In addition\nto these edges, there are directed edges from w∗to some nodes inside Dij; these directed\nedges are denoted by a dashed line in Figure 4(a).\nw∗\nv1\nu2\nvk\nu1\nv2\nd2\ndk\nd1\nulij\ndlij\nulij\nuk\n(a) The Dij graph\nbk\nak\nvk\nuk\ndk\nγk\nβk\nαk\ncenter node\nek\n(b) Part of gadget corre-\nsponding to an edge ek\nFigure 4: The cover testing gadget.\n4. Let G = (V , E) be the obtained graph.\nThe next lemma shows that the size of an optimal solution in Directed PDS is exactly one\nmore than the size of an optimal solution in the MinRep instance. The number of nodes in the\nconstructed graph is at most\n V (G)\n ≤1 + |V (G)| + 7λ |E(G)|. This will complete the proof of\nTheorem 4.2 by showing that the above reduction is a gap preserving reduction from MinRep to\nDirected PDS with the same gap (hardness ratio) as the MinRep problem.\nLemma 4.3 A∗∪B∗is an optimal solution to the instance G = (A, B, E) of the MinRep problem\nif and only if S∗= A∗∪B∗∪{w∗} ⊆V (G) is an optimal solution to the instance G of the Directed\nPDS problem.\nProof: First note that w∗should be in any feasible solution; because it has in-degree zero.\nAssume that A∗∪B∗is a feasible solution for the MinRep instance G.\nWe will show that\nS = A∗∪B∗∪{w∗} is a feasible solution to the Directed PDS instance G. Note that all nodes\nin A ∪B and some nodes inside the gadgets Dij are power dominated by applying rule (D1) on\nw∗. Now, we only need to show that all nodes in the gadgets Dij are power dominated. Consider\na super edge AiBj of H. The set A∗∪B∗covers all the super edges in H. Hence, there exists\na pair of nodes ak ∈A∗∩Ai, bk ∈B∗∩Bj that induces an edge of G.\nSince ak and bk are\nin S their out-neighbors uk and vk in each of the λ = 4 copies of the Dij graph will be power\ndominated by applying rule (D1). Now node dk, that is already power dominated by w∗, will power\ndominate the center node by an application of rule (D2). Now we claim that the center node will\npower dominate the remaining nodes in the gadget Dij. Consider the part of gadget (shown in\n13"},{"paragraph_id":"p15","order":15,"text":"Figure 4(b)) corresponding to an edge er ∈Eij. Note that the nodes γr, αr, and dr are already\npower dominated by w∗. It is easy to check that the nodes βr, ur, vr will be power dominated by\nsequentially applying rule (D2) on γr, αr, and dr. This shows that S power dominates all nodes in\nG. Therefore, Opt(G) is at most |A∗∪B∗| + 1.\nLet S∗⊆V (G) be an optimal solution for Directed PDS. As we showed above w∗should be in\nany feasible solution for Directed PDS. Now define A′ = A∩S∗and B′ = B ∩S∗. First we prove\nthat any optimal solution of Directed PDS is contained in A ∪B ∪{w∗}, and then we show that\nA′ ∪B′ covers all the super edges of H. Suppose that S∗contains some nodes not in A ∪B ∪{w∗}.\nHence, there are some gadgets that are not completely power dominated by S∗∩(A ∪B ∪{w∗}).\nLet Dij be such a gadget. By symmetry each of the λ = 4 copies of Dij is not completely power\ndominated. Therefore, the optimal solution S∗needs to have at least 3 nodes from the 4 copies of\nDij. By removing these 3 nodes from S∗and adding ak ∈Ai and bk ∈Bj to S∗for some arbitrary\nedge ek = akbk ∈Eij, we can power dominate all 4 copies of Dij. This contradicts the optimality\nof S∗, and proves that S∗⊆A ∪B ∪{w∗}. To see that A′ ∪B′ covers all super edges, it is enough\nto note the following. Suppose no node from any copy of Dij is in the optimal solution, then any\nDij can be power dominated only by taking a pair of nodes a ∈Ai and b ∈Bj that induces an\nedge of G. Otherwise, any power dominated node in the gadget has at least 2 out-neighbors that\nare not power dominated, so rule (D2) cannot be applied. This completes the proof of the lemma.\n□\nThere are several notions for the tree-width of directed graphs such as DAG width [29], directed\ntree-width [21], and Kelly-width [19]. Directed acyclic graphs have width equal to zero for the first\ntwo notions [29], and have Kelly-width of 1. Hence, Theorem 4.2 gives a hardness threshold of\nO(2log n1−ǫ) even if the directed graph has width ≤1 according to any of the above three notions.\nWe reformulate Directed PDS in terms of valid colorings of the edges in order to develop\nan algorithm based on dynamic-programming for Directed PDS. Guo et al. [15] introduced the\nnotion of valid orientations to get a new formulation for PDS (in undirected graphs). They also\ndesigned a linear-time dynamic-programming algorithm based on valid orientations for optimally\nsolving PDS on graphs of bounded tree-width. Our method applies to directed graphs such that\nthe underlying undirected graph has bounded tree-width.\nDefinition 4.4 A coloring of a directed graph G = (V, E) is a partitioning of the edges in G into\nred and blue edges. We denote a coloring by C = (V, Er ∪Eb) where Er is the set of red edges and\nEb is the set of blue edges.\nWe reformulate the Directed PDS problem via a so-called valid coloring of directed graphs;\ninformally speaking, these colorings “model” the application of rules (D1) and (D2) of Directed\nPDS.\nDefinition 4.5 A valid coloring C = (V, Er ∪Eb) of a directed graph G = (V, E) is a coloring of\nG with the following properties:\n1. No two antiparallel edges can be colored red.\n2. The subgraph induced by the red edges, Gr = (V, Er), has the following properties:\n(a) ∀v ∈G : d−\nGr(v) ≤1, and\n(b) ∀v ∈G : d−\nGr(v) = 1 =⇒d+\nGr(v) ≤1.\n14"},{"paragraph_id":"p16","order":16,"text":"3. G has no dependency cycle. A dependency cycle is a sequence of directed edges whose under-\nlying undirected graph forms a cycle such that all the red edges are in one direction, all the\nblue edges are in the other direction, and there are no two consecutive blue edges.\nWe call a node an origin of C if it has no incoming edges in Gr.\nOur dynamic-programming algorithm for Directed PDS is based on the following lemma.\nLemma 4.6 Given a directed graph G and S ⊆V (G), S power dominates G if and only if there\nis a valid coloring of G with S as the set of origins.\nProof: Suppose S ⊆V power dominates G. Then we give a valid coloring C with S as the set\nof origins by coloring the edges in G according to the way that S power dominates G. We color\nan edge (v, w) red if node w is power dominated by applying the power domination rules on v;\neither by the domination rule (D1) or by the propagation rule (D2). Note that when we apply\nthe propagation rule (D2), then we do not power dominate the previously power dominated nodes.\nAlso when we apply rule (D1) on v, then we power dominate all (not some subset of) neighbors\nof v that are not power dominated. We write v < u when a node u is power dominated after\nv. It is easy to check that with this coloring the degree requirements are satisfied; each node can\nbe power dominated only once, and if it is a power dominated node (not in S) , then it cannot\npower dominate more than one of its out-neighbors due to rule (D2). Now, we need to prove that\nthere is no dependency cycle.\nBy way of contradiction, suppose that C∗= u1, u2, . . . , um is a\ndependency cycle. Focus on the edges of C∗. Call the direction of the red edges forward, and call\nthe direction of the blue edges backward. Observe that a dependency cycle has ≥1 red edges, and\neach of its red edges corresponds to an application of rule (D2). Assume that all the edges in C∗\nare red. Then the red edges (ui, ui+1) imply that ui < ui+1 for all i = 1, 2, . . . , m −1; therefore\nu1 < u2 < · · · < um, but this is a contradiction since the last red edge from um back to u1 implies\nthat um < u1. Hence, there is no dependency cycle with all edges colored red. Now, assume that\nthe dependency cycle C∗has some blue edges. We show that a similar contradiction occurs when\nthere are no two consecutive blue edges. Consider a blue edge (u, v) of C∗and note that the other\nedge of C∗incident to u is a red edge, say (u, w). By rule (D2), we see that v should be power\ndominated before u can power dominate w; thus we have w > v. Repeating this argument, we\nget an ordering for the occurrences of power domination of some of the nodes in C∗that gives a\ncontradiction, e.g., if m is even and the edges of C∗are alternatively blue and red (starting with\nblue) we get u1 < u3 < u5 < · · · < um−1 < u1 (see Figure 5 for an example). Hence, G has a valid\ncoloring with S as the set of origins.\nB\nR\nR\nB\nR\nu1\nu2\nu3\nu4\nu5\nFigure 5:\nThe blue edge (u4, u3) means that u3 should be power dominated before we can power\ndominate u5 by applying rule (D2) on u4; thus u3 < u5. Again the blue edge from u1 to u5 implies\nthat u5 < u2. Finally , the red edge (u2, u3) shows that u2 < u3. Combining these dependencies\nwe get u3 < u5 < u2 < u3. This is a contradiction, and shows that there cannot be a dependency\ncycle in a coloring obtained from the application of rules (D1) and (D2) of Directed PDS.\nNow suppose that G has a valid coloring C = (V, Er ∪Eb) with S ⊆V (G) as the set of origins.\nThe nodes in S and all of their out-neighbors in Gr = (V, Er) are power dominated by applying\n15"},{"paragraph_id":"p17","order":17,"text":"the rule (D1). Now we prove that S will power dominate all nodes in G. Suppose that this does\nnot happen. Let X ⊂V be the maximal set of nodes that can be power dominated by S. We\nclaim that there is at least one red edge from X to V \\ X. Note that all of the origins are in X,\nso each of the nodes in V \\ X has in-degree 1 in Gr. Hence, if there is no red edge from X to\nV \\ X, then there should be a directed cycle of red edges in G[V \\ X]. This is not possible since\nthere are no dependency cycles. Therefore there is at least one red edge from X to V \\ X. Let\ne1 = (x1, y1), . . . , ek = (xk, yk) be all of the red edges from X to V \\X. If some xi has all of its out-\nneighbors in X except yi, then by applying rule (D2) on xi the node yi will be power dominated. By\nthe maximality assumption of X this cannot happen. Therefore, each xi has another out-neighbor,\nsay zi, in V \\X. Then (xi, zi) is a blue edge, otherwise, xi would be an origin and yi would be power\ndominated by applying rule (D1) on xi. Now, we construct a dependency cycle as follows: starting\nfrom x1, use a blue edge to move to a node z1 in V \\ X; then move in the reverse direction over a\nsequence of red edges (z2, z1), (z3, z2), · · · until we reach a red edge (zi, zi−1) with zi−1 ∈V \\ X and\nzi ∈X (such an edge exists since G[V \\X] has no directed cycle of red edges); note that (zi, zi−1) is\none of the red edges (x1, y1), · · · , (xk, yk). If zi = x1, then we have a dependency cycle; otherwise,\nwe again use a blue edge (zi, zi+1) to move to a node in V \\ X. By repeating these steps, we will\neventually find a dependency cycle. Note that all the blue edges are in one direction, and all the red\nedges are in the other direction. This is a contradiction, since C has no dependency cycle. Hence,\nwe have X = V , so S power dominates G.\n□\nTheorem 4.7 Given a directed graph G and a tree decomposition of width k of its underlying\nundirected graph, Directed PDS can be optimally solved in O(ck2 · n) time for a global constant\nc.\nA consequence of the above theorem is a linear-time algorithm for solving the Directed PDS\nproblem optimally on directed graphs, given a bounded tree-width decomposition of the underlying\nundirected graph. Also since the tree-width decomposition for graphs with bounded tree-width can\nbe computed in polynomial-time [5], there is a polynomial-time algorithm to solve Directed PDS\noptimally on the class of directed graphs such that the underlying undirected graph has bounded\ntree-width.\n5\nConclusions\nWe studied the PDS problem from the perspective of approximation algorithms. We introduced\na natural extension of the problem to directed graphs. We showed that both problems have a\nthreshold of O(2log n1−ǫ) for the hardness of approximation.\nWe presented an O(√n) approxi-\nmation algorithm for Planar PDS. We designed a dynamic-programming algorithm for solving\nthe Directed PDS problem optimally in linear-time for those directed graphs whose underlying\nundirected graph has bounded tree-width.\nHere, we describe an algorithm with an approximation guarantee of O(\nn\nlog n) for the PDS problem.\nThe algorithm works as follows. Partition the nodes of the graph G into log n equal-sized sets\nV1, V2, · · ·. Next, consider all possible ways of picking these sets (we pick all nodes in a set). Among\nall these different candidates, output the one that power dominates G and has the minimum\nnumber of nodes. Note that in the algorithm we only consider 2log n = n different candidates.\nClearly, the algorithm runs in polynomial time, since the feasibility of each candidate can be tested\nin polynomial time. Let S∗be an optimal solution.\nIt is easy to see that the set of Vi’s that\n16"},{"paragraph_id":"p18","order":18,"text":"intersect S∗forms a feasible solution for the PDS problem in G; this solution has size at most\nn\nlog n · |S∗|. This establishes the approximation guarantee. The same algorithm and analysis applies\nto the Directed PDS problem.\nProposition 5.1 There is a polynomial time\nn\nlog n-approximation algorithm for both the PDS prob-\nlem and the Directed PDS problem.\nThere is a gap between our hardness threshold of O(2log n1−ǫ) and our approximation guarantee of\nO(\nn\nlog n), and narrowing this gap is an open question.\nA major open question in the area is whether there exists a PTAS (polynomial time approximation\nscheme) for Planar PDS. A first step may be to obtain an improvement on our approximation\nguarantee of O(√n). There has been a lot of research on designing PTASs for NP-hard problems on\nplanar graphs. Some of the most important developments are the outerplanar layering technique by\nBaker [3], and the bidimensionality theory by Demaine and Hajiaghayi [10]. Unfortunately, these\nmethods do not apply to Planar PDS.\nBaker [3] showed that the Dominating Set problem in planar graphs has a PTAS. In the Baker\nmethod we first partition the graph into smaller graphs. Then we solve the problem optimally on\neach subgraph, and finally we return the union of the solutions as a solution for the original graph.\nThe example in Figure 1 shows that this method does not apply to Planar PDS. The size of an\noptimal solution is 1, but if we apply the Baker method, then the size of the output solution will\nbe at least as large as the number of subgraphs in the partition which can be Θ(n).\nDemaine and Hajiaghayi [10] introduced the bidimensionality theory and used it to obtain PTASs\nfor several variants of the Dominating Set problem on planar graphs. An important property\nof bidimensionality is that when an edge is contracted the size of an optimal solution should not\nincrease. Consider the example in Figure 6. If we contract edges e1, e2, . . . , en in G, then we get\nthe graph G′. It can be checked that Opt(G) = 1 but Opt(G′) = Θ(n). Thus the bidimensionality\ntheory does not apply to Planar PDS, since the optimum value may increase when an edge is\ncontracted.\nv\n=⇒\nv\nu1\nun\nen\ne1\nG\nG′\nFigure 6: Optimal value of PDS increases when edges are contracted.\nLastly, we consider some variations of greedy algorithms for PDS and show that they perform very\npoorly even on planar graphs. In contrast, for other related problems such as Dominating Set and\nSet Covering, greedy algorithms perform well since they achieve a logarithmic approximation\nguarantee, and no substantial improvement is possible by any polynomial time algorithm, under\ncomplexity assumptions like P ̸= NP. The most natural greedy algorithm for PDS is the one that\nstarts with S = ∅, and in each step, adds a new node v to the current solution S such that v power\ndominates the maximum number of new nodes.\n17"},{"paragraph_id":"p19","order":19,"text":"Unfortunately, this greedy algorithm may find a solution S such that |S| ≥Θ(n) · Opt(G). To see\nthis, consider a graph G that is obtained from a 9l× 9m grid by subdividing all row-edges, except\nwith minor changes in the four corners as shown in Figure 7(a). Partition the graph G into 9 × 9\ngrids (ignoring the nodes introduced by subdivision), see Figure 7(b). It is easy to check that any\nsingle node can power dominate at most 7 nodes, and the center node of any one of the 9 × 9 grids\nachieves this maximum. So the greedy algorithm at the first iteration may pick the center node\nof any one of the 9 × 9 grids. Assuming all nodes picked by the algorithm so far have been these\ncenter nodes, we see that picking another center node maximizes\n PS∪{v} \\ PS\n over all v ∈V . So\nthe greedy algorithm could continue picking center nodes, and after that possibly picking other\nnodes until it finds a feasible solution S. The size of the output S is at least m · l= Θ(n). By\n(a) Grid\n(b) 9 × 9 grid\nFigure 7: Bad example for the greedy algorithm\nProposition 3.12, we have Opt(G) = Θ(l). Now by fixing l= Θ(1) we can see that the size of the\noutput solution can be bigger than Opt(G) by a factor of Θ(n).\nProposition 5.2 The greedy algorithm for the PDS problem may find a solution S such that |S| ≥\nΘ(n) · Opt(G).\nWe will consider two other variations of the greedy algorithm, namely Proximity and Cleanup.\nWe have examples of Planar PDS showing that these variations of the greedy algorithm perform\npoorly.\nProximity: In each step of the Proximity algorithm we choose a node such that the set of all\npower dominated nodes induces a connected subgraph, and subject to this, the number of newly\npower dominated nodes is maximized. Informally, this is to escalate the use of the propagation\nrule.\nThe bad example for the proximity version of the greedy algorithm is obtained by modifying the\ncenter row of the h × (2m + 1) grid, as shown in Figure 8, by inserting lsubdividing nodes into\nthe edges in the middle row, and also subdividing all of the other row-edges except some of the\ncorner edges. Figure 8 illustrates an example of such a grid for l = 5 and h = 9 rows, but for a\nbad example for the proximity greedy algorithm we need h to be sufficiently large constant (h = 17\nsuffices). We use the figure for illustration, to show the working of the proximity greedy algorithm.\nIt is easy to check that by picking all nodes of the first column we can power dominate the entire\ngraph, so the optimal solution is Θ(1). The proximity greedy algorithm starts by picking a node\nthat power dominates maximum number of nodes (which is 17 = 2l+ 7); any white node satisfies\nthis requirement. Therefore the algorithm may pick for example the first white node (from the\n18"},{"paragraph_id":"p20","order":20,"text":"left). It is easy to check that in the next step the algorithm will pick the white node to the right of\nthe first one, since all of the power dominated region stays connected and also it power dominates\nmaximum number of new nodes (which is 16 = 2l+ 6). The algorithm continues picking all white\nnodes and at the end it will pick possibly more nodes to get a feasible solution.\n(The shaded\nregion shown in Figure 8 indicates the nodes that will be power dominated by picking all white\nnodes.) Therefore, the size of the solution found by the algorithm is at least m = Θ(n). Hence, the\nproximity greedy algorithm may find a feasible solution that is Θ(n) times worse than the optimal\nsolution.\nFigure 8: Bad example for the proximity greedy algorithm\nCleanup Step: Some of the recent approximation algorithms, especially some based on the primal-\ndual method, use a clean up step at the end: this step removes redundant elements from the solution\nin some sequential order. In the Cleanup algorithm, we first run the greedy algorithm to find a\nsolution (node set) S, then we repeatedly remove nodes from S, until S is an inclusionwise minimal\npower dominating set. Although a cleanup step may substantially improve on the solution found\nby the greedy algorithm on some examples, this does not hold for all examples.\nThe same bad example for the proximity version is also a bad example for the cleanup version of\nthe greedy algorithm. The cleanup greedy algorithm may again pick the first white node (from the\nleft), and after picking this white node, it may pick the third white node. Since both of them power\ndominate maximum number of new nodes (which is 17 = 2l+ 7). Note that in the original greedy\nalgorithm there is no need to have a connected subgraph induced on power dominated nodes. The\nalgorithm continues to pick all the odd indexed white nodes, and after that it will start picking the\neven indexed white nodes since any one of them power dominates maximum number of new nodes\n(which is at least 15 = 2l+ 5). At this stage of the algorithm the set of power dominated nodes\nare those in the shaded region of Figure 8. It is easy to check that we need to pick nodes from\nboth upper and lower parts in order to power dominate the entire graph. The greedy algorithm\nmay pick some nodes from the leftmost column in the top part and some nodes from the rightmost\ncolumn in the bottom part to power dominate the entire graph. Now we start doing the cleanup\nprocess. It can be checked that if we remove any two consecutive white nodes from the obtained\nsolution, the graph cannot be power dominated completely. So we need to keep at least half of the\nwhite nodes. Therefore, the size of the output solution at the end of cleanup process is at least\nm\n2 = Θ(n), but the optimal solution is just Θ(1) as before.\n19"},{"paragraph_id":"p21","order":21,"text":"References\n[1] A. Aazami and M. D. Stilp. Approximation algorithms and hardness for domination with\npropagation. In Proceedings of the 10th International Workshop on Approximation Algorithms\nfor Combinatorial Optimization Problems, volume 4627 of LNCS, pages 1–15. Springer, 2007.\n[2] J. Alber, H. L. Bodlaender, H. Fernau, T. Kloks, and R. Niedermeier.\nFixed parameter\nalgorithms for dominating set and related problems on planar graphs. Algorithmica, 33(4):461–\n493, 2002.\n[3] B. S. Baker. Approximation algorithms for NP-complete problems on planar graphs. J. ACM,\n41(1):153–180, 1994.\n[4] T. L. Baldwin, L. Mili, M. B. Boisen, and R. Adapa. Power system observability with minimal\nphasor measurement placement. IEEE Transactions on Power Systems, 8(2):707–715, 1993.\n[5] H. L. Bodlaender. A linear-time algorithm for finding tree-decompositions of small treewidth.\nSIAM J. Comput., 25(6):1305–1317, 1996.\n[6] D. J. Brueni. Minimal PMU placement for graph observability, a decomposition approach.\nMaster’s thesis, Virginia Polytechnic Institute and State University, Blacksburg, VA, 1993.\n[7] D. J. Brueni and L. S. Heath.\nThe PMU placement problem.\nSIAM J. Discret. Math.,\n19(3):744–761, 2005.\n[8] V. Chvatal. A greedy heuristic for the set covering problem. Math. Oper. Res., 4:233–235,\n1979.\n[9] B. Courcelle, J. A. Makowsky, and U. Rotics. Linear time solvable optimization problems\non graphs of bounded clique width. In Proceedings of the 24th International Workshop on\nGraph-Theoretic Concepts in Computer Science, volume 1517 of LNCS, pages 1–16. Springer,\n1998.\n[10] E. D. Demaine and M. T. Hajiaghayi. Bidimensionality: new connections between FPT al-\ngorithms and PTASs. In Proceedings of the 16th Annual ACM-SIAM Symposium on Discrete\nAlgorithms, pages 590–601, 2005.\n[11] R. Diestel. Graph Theory. Springer-Verlag, New York, 2nd edition, 2000.\n[12] M. Dorfling and M. A. Henning. A note on power domination in grid graphs. Discrete Applied\nMathematics, 154(6):1023–1027, 2006.\n[13] U. Feige. A threshold of ln n for approximating set cover. J. ACM, 45(4):634–652, 1998.\n[14] M. R. Garey and D. S. Johnson. Computers and Intractability: A Guide to the Theory of\nNP-Completeness. W. H. Freeman and Co., New York, NY, USA, 1979.\n[15] J. Guo, R. Niedermeier, and D. Raible. Improved algorithms and complexity results for power\ndomination in graphs. In Proceedings of the 15th International Symposium on Fundamentals\nof Computation Theory, volume 3623 of LNCS, pages 172–184. Springer, 2005 (to appear in\nAlgorithmica).\n20"},{"paragraph_id":"p22","order":22,"text":"[16] T. W. Haynes, S. M. Hedetniemi, S. T. Hedetniemi, and M. A. Henning. Domination in graphs\napplied to electric power networks. SIAM J. Discrete Math., 15(4):519–529, 2002.\n[17] T. W. Haynes, S. T. Hedetniemi, and P. J. Slater. Domination in Graphs: Advanced Topics.\nMarcel Dekker, New York, 1998.\n[18] T. W. Haynes, S. T. Hedetniemi, and P. J. Slater. Fundamentals of Domination in Graphs.\nMarcel Dekker, New York, 1998.\n[19] P. Hunter and S. Kreutzer. Digraph measures: Kelly decompositions, games, and orderings.\nIn Proceedings of the 18th Annual ACM Symposium on Discrete Algorithms, pages 637–644,\nPhiladelphia, PA, USA, 2007.\n[20] D. S. Johnson. Approximation algorithms for combinatorial problems. J. Comput. Syst. Sci.,\n9(3):256–278, 1974.\n[21] T. Johnson, N. Robertson, P. D. Seymour, and R. Thomas. Directed tree-width. J. Comb.\nTheory, Ser. B, 82(1):138–154, 2001.\n[22] T. Kloks. Treewidth, Computations and Approximations, volume 842 of LNCS. Springer, 1994.\n[23] J. Kneis, D. M ̈olle, S. Richter, and P. Rossmanith. Parameterized power domination complex-\nity. Inf. Process. Lett., 98(4):145–149, 2006.\n[24] G. Kortsarz. On the hardness of approximating spanners. Algorithmica, 30(3):432–450, 2001.\n[25] C. S. Liao and D. T. Lee. Power domination problem in graphs. In Proceedings of the 11th\nInternational Computing and Combinatorics Conference, volume 3595 of LNCS, pages 818–\n828. Springer, 2005.\n[26] L. Lov ́asz.\nOn the ratio of optimal integral and fractional covers.\nDiscrete Mathematics,\n13:383–390, 1975.\n[27] C. Lund and M. Yannakakis. On the hardness of approximating minimization problems. J.\nACM, 41(5):960–981, 1994.\n[28] L. Mili, T.L. Baldwin, and A.G. Phadke.\nPhasor measurements for voltage and transient\nstability monitoring and control. In Proceedings of the EPRI-NSF Workshop on Application\nof Advanced Mathematics to Power Systems, 1991.\n[29] J. Obdrz ́alek. Dag-width: connectivity measure for directed graphs. In Proceedings of the 17th\nAnnual ACM Symposium on Discrete Algorithms, pages 814–821. ACM Press, 2006.\n[30] P. Slav ́ık. A tight analysis of the greedy algorithm for set cover. In Proceedings of the 28th\nAnnual ACM Symposium on Theory of Computing, pages 435–441, New York, NY, USA, 1996.\nACM Press.\n[31] M. D. Stilp. On power dominating sets. Master’s thesis, Combinatorics and Optimization,\nUniversity of Waterloo, Ontario, Canada, 2006.\n21"},{"paragraph_id":"p23","order":23,"text":"A\nDynamic Programming\nIn this section we describe our dynamic-programming algorithm for the Directed PDS problem.\nThis algorithm is similar to the dynamic-programming algorithm designed by Guo et al.[15] to\noptimally solve PDS for undirected graphs with bounded tree-width. It is known that any tree\ndecomposition of width-k can be transformed to a nice tree decomposition with width k in linear-\ntime [22] (Lemma 13.1.3). So we can assume that we are given a nice tree decomposition of the\nunderlying undirected graph of G call it ⟨{Xi|i ∈I} , T⟩. Let Ti denote the subtree of T rooted\nat T-node i, and Yi =\n S\nj∈V (Ti) Xj"},{"paragraph_id":"p24","order":24,"text":"\\ Xi. Also let Gi be the subgraph induced by Yi ∪Xi, i.e.\nGi = G [Yi ∪Xi], and let G′i = G[Xi].\nConsider a valid coloring C of the graph G. We store the color of the edges in each bag by\nassigning a state to that bag (the formal definition of a state will follow). We can reconstruct the\ncoloring C from the states of all bags in the tree decomposition of G; so there is no need to store\nthe coloring C in the dynamic-programming.\nThe state of a bag: Given a coloring C, the state of a bag Xi describes the coloring of the edges\nin G′\ni. In order to detect the dependency cycles in the coloring C without reconstructing the whole\ncoloring, we need to store some more information in a state. This extra information enables us to\ndetect a dependency cycle in Gi which goes through Xi, by considering only the state of the bag\nXi. A bag state s contains the following: state of each edge, state of each node, and state of each\npair of nodes in G′\ni = G[Xi].\n• State of an edge: The state of an edge e ∈E(G[Xi]) denoted by s(e) is the color that is\nassigned to e in the coloring C; s(e) ∈{R, B}.\n• State of a node: The state of a node v ∈Xi denoted by s(v) shows the number of red\nedges between v and Yi.\n– s(v) = 1: There is exactly one red edge from a node in Yi to v and no red edge from v\nto Yi,\n– s(v) = 2: There is exactly one red edge from a node in Yi to v and exactly one red edge\nfrom v to Yi,\n– s(v) = 3: There is no red edge between Yi and v,\n– s(v) = 4: There are at least two red edges from v to Yi and no red edge from Yi to v,\n– s(v) = 5: There is exactly one red edge from v to Yi and no red edge from Yi to v.\n• State of a pair of nodes: A dependency path from u to v is a path P where all red edges\nin P are directed from u to v and all blue edges are directed from v to u. We categorize\ndependency paths according to the color of their first and last edges. There are 4 possible\ntypes RR, RB, BR, BB; for example a path of type RB is a path with the first edge colored\nred and the last edge colored blue. For a pair (u, v) ∈Xi × Xi (u ̸= v) the state of (u, v)\ndenoted by s(u, v) shows the type of dependency paths from u to v in G[Yi ∪{u, v}]; that is,\ns(u, v) ⊆{RR, RB, BR, BB}. Note that there are 24 = 16 different states for each pair of\nnodes.\nDetecting dependency cycles:: An important part of the dynamic-programming algorithm is\nto detect dependency cycles in the coloring C. Assume we are at bag Xi and we are given the bag\nstate s corresponding to the coloring C. We can detect the dependency cycles in G′\ni = G[Xi] by\n22"},{"paragraph_id":"p25","order":25,"text":"enumerating all possible cycles; note that the coloring of edges in G′\ni is given in the state s. The\ndependency cycles in Gi can be detected by considering the state of each pair of nodes in Xi. For\nexample assume that RB ∈s(u, v) and RR ∈s(v, u). Then, by combining a dependency path of\ntype RB from u to v and a dependency path of type RR from v to u we obtain a dependency cycle\ngoing through u and v in Gi.\nLet us denote by Λi the set of all possible states for the bag Xi. The dynamic-programming will\ncompute a mapping Ai : Λi →N ∪{+∞}. For a bag state s ∈Λi the value Ai(s) is the minimum\nnumber of origins in an optimal valid coloring C of Gi under the restriction that the state of nodes,\nedges, and pairs of nodes in Xi is given by s. Now, we describe how our dynamic-programming\nworks.\nStep 1: (Initialization): In this step for each leaf node i of T, we initialize the mapping Ai as\nfollows. For a given state s, we define Ai(s) as +∞if s has a dependency cycle, a node v with\ns(v) ̸= 3, or a pair of nodes u and v such that s(u, v) ̸= ∅. Otherwise, we define Ai(s) as the\nnumber of nodes with no in-coming red edges in the coloring defined by s.\nStep 2: (Bottom-Up Computation): After initialization, we visit the nodes in T in a bottom-\nup fashion and at each bag Xi we compute the mapping Ai corresponding to Xi. The update\nprocess depends on the type of T-nodes that we are considering. Here, we only consider the update\nprocess at an Insert Node. The other cases are similar to this one.\nInsert Node: Suppose i is an insert node with the child j, and assume that Xi = Xj ∪{x}. For\neach bag state s ∈Λi do the following:\n1. Check whether the coloring given by s forms a valid coloring of G′\ni; if not, define Ai(s) = +∞.\n2. Compute the set Λj(s) containing bag states of j that are “compatible” with the bag state s.\n3. For each s′ ∈Λj(s), check if a valid coloring of Gj “compatible” with s′ can be extended to\na valid coloring of Gi “compatible” with s.\n4. Compute Ai based on the mapping Aj.\nCompatible bag state (Step 2): A bag state s′ ∈Λj is said to be compatible with the bag state\ns ∈Λi if the state of each node, each edge, and each pair of nodes in V (G′\nj) in the bag state s′ is the\nsame as the corresponding state in the bag state s. If s(x) ̸= 3, or ∃v ∈Xj : s(x, v) ̸= ∅∨s(v, x) ̸= ∅\nthen we define Λj(s) = ∅.\nDetecting dependency cycles (Step 3): The conditions of a valid coloring can be violated due\nto degree constraints on the new node x, or by the existence of a dependency cycle going through\nx. Both these cases can be tested by considering the bag states s and s′.\nComputing Ai (Step 4): The addition of x may change the number of origins in Gi. The node x\nwill be an origin if it has at least one outgoing red edge in s. But an origin node v ∈Xj (in s′) that\nhas an incoming red edge from x, is no longer an origin. So by considering the red edges going out\nof x we can update the number of origins and compute the mapping Ai. If the coloring compatible\nwith s is not a valid coloring, then we define Ai(s) to be +∞.\nStep 3: (At root r): Finally, we compute the number of origins in an optimal valid coloring of\nG by finding the minimum of Ar(s) over all possible s ∈Λr. It is easy to see that each bag Xi has\nat most 16(k+1)2 · 5k+1 · 2(k+1)2 states; note that |Xi| ≤(k + 1). It can be checked that the total\nrunning time of our algorithm is O(ck2 · n), for some global constant c.\n23"}],"pages":[{"page":1,"text":"arXiv:0710.2139v1 [cs.CC] 10 Oct 2007\nApproximation algorithms and hardness for domination with\npropagation\nAshkan Aazami\naaazami@uwaterloo.ca\nDepartment of Combinatorics and Optimization\nUniversity of Waterloo\nMichael David Stilp\nmstilp3@gatech.edu\nSchool of Industrial and Systems Engineering\nGeorgia Institute of Technology\nNovember 2, 2018\nAbstract\nThe power dominating set (PDS) problem is the following extension of the well-known dom-\ninating set problem: find a smallest-size set of nodes S that power dominates all the nodes,\nwhere a node v is power dominated if (1) v is in S or v has a neighbor in S, or (2) v has a\nneighbor w such that w and all of its neighbors except v are power dominated. We show a\nhardness of approximation threshold of 2log1−ǫ n in contrast to the logarithmic hardness for the\ndominating set problem. We give an O(√n) approximation algorithm for planar graphs, and\nshow that our methods cannot improve on this approximation guarantee. Finally, we initiate\nthe study of PDS on directed graphs, and show the same hardness threshold of 2log1−ǫ n for\ndirected acyclic graphs. Also we show that the directed PDS problem can be solved optimally\nin linear time if the underlying undirected graph has bounded tree-width.\nKeywords: Approximation algorithms, Hardness of approximation, Dominating set, Power\ndominating set, Tree-width, Planar graphs, Greedy algorithms, PMU placement problem.\nAMS subject classifications: 68W25; 90C27\n1\nIntroduction\nA dominating set of an (undirected) graph G = (V, E) is a set of nodes S such that every node in\nthe graph is in S or has a neighbor in S. The problem of finding a dominating set of minimum size\nis an important problem that has been extensively studied, especially in the last 20 years, see the\nbooks by Haynes et al. [17, 18]. The problem is NP-hard [14], a simple greedy algorithm achieves\na logarithmic approximation guarantee∗[20], and, modulo the P ̸= NP conjecture, no polynomial\ntime algorithm gives a better approximation guarantee [27, 13].\nOur focus is on an extension called the Power Dominating Set (abbreviated as PDS) problem.\nPower domination is defined by two rules; the first rule is the same as the rule for the Dominating\nSet problem, but the second rule allows a type of indirect propagation. More precisely, given a set\nof nodes S, the set of nodes that are power dominated by S, denoted PS, is obtained as follows.\n∗An approximation algorithm for a (minimization) optimization problem means an algorithm that runs in poly-\nnomial time and computes a solution whose cost is within a guaranteed factor of the optimal cost; the approximation\nguarantee is the worst-case ratio, over all inputs of a given size, of the cost of the solution computed by the algorithm\nto the optimal cost.\n1"},{"page":2,"text":"(Rule 1) if node v is in S, then v and all of its neighbors are in PS;\n(Rule 2) (propagation) if node v is in PS, one of its neighbors w is not in PS, and all other neighbors\nof v are in PS, then w is inserted into PS.\nThe set PS is independent of the sequence in which nodes are inserted by Rule 2. Otherwise,\nthere is a minimal counter example with two maximal sequences of insertions and an “earliest”\nnode that occurs in one sequence but not the other; this is not possible. The PDS problem is\nto find a node-set S of minimum size that power dominates all nodes (i.e., find S ⊆V with |S|\nminimum such that PS = V ). We use Opt(G) to denote the size of an optimal solution for the PDS\nproblem for a graph G. Throughout, we use n to denote the number of nodes in the input graph.\nFor example, consider the planar graph in Figure 1; the graph has t disjoint triangles, and three\n(mutually disjoint) paths such that each path has exactly one node from each triangle; note that\n|V | = 3t. The minimum dominating set has size Θ(|V |), since the maximum degree is 4. The\nminimum power dominating set has size one – if S has any one node of the innermost (first)\ntriangle (like v), then PS = V †.\nv\nFigure 1: Illustrating those nodes power dominated by Rule 1 (denoted by a triangle) and Rule 2\n(denoted by a square); the picked node is shown by a circle.\nThe PDS problem arose in the context of electric power networks, where the aim is to monitor all\nof the network by placing a minimum-size set of very expensive devices called phase measurement\nunits; these units have the capability of monitoring remote elements via propagation (as in Rule 2);\nsee Brueni [6], Baldwin et al. [4], and Mili et al. [28]. In the engineering literature, the problem is\ncalled the PMU placement problem.\nOur motivation comes from the area of approximation algorithms and hardness results. The\nDominating Set problem is a so-called covering problem; we wish to cover all nodes of the graph\nby choosing as few node neighborhoods as possible. In fact, the Dominating Set problem is a\nspecial case of the well-known Set Covering‡ problem.\nSuch covering problems have been extensively investigated. One of the key positive results dates\nfrom the 1970’s, when Johnson [20], Lov ́asz [26] and later Chv ́atal [8] showed that the greedy method\nachieves an approximation guarantee of O(log |V |) where |V | denotes the size of the ground set, see\nalso [30]. Several negative results (on the hardness of approximation) have been discovered over\n†In more detail, we apply Rule 1 to see that all the nodes of the innermost (first) triangle and one node of the\nsecond triangle are in PS; then by two applications of Rule 2 (to each of the nodes in the first triangle not in S), we\nsee that the other two nodes of the second triangle are in PS; then by three applications of Rule 2 (to each of the\nnodes in the second triangle) we see that all three nodes of the third triangle are in PS; etc.\n‡Given a family of sets on a groundset, find the minimum number of sets whose union equals the groundset.\n2"},{"page":3,"text":"the last few years: Lund and Yannakakis [27] showed that the Set Covering problem is hard to\napproximate within a ratio of Ω(logn) and later, Feige [13] showed that it is hard to approximate\nwithin a ratio of (1 −ǫ) ln n, modulo some variants of the P ̸= NP assumption.\nA natural question is what happens to covering problems (in the setting of approximation algo-\nrithms and hardness results) when we augment the covering rule with a propagation rule. PDS\nseems to be a key problem of this type, since it is obtained from the Dominating Set problem by\nadding a simple propagation rule.\n1.1\nPrevious literature\nApparently, the earliest publications on PDS are Brueni [6], Baldwin et al. [4], and Mili et al. [28].\nLater, Haynes et al. [16] showed that the problem is NP-complete even when the input graph is\nbipartite; they presented a linear-time algorithm to solve PDS optimally on trees. Kneis et al. [23]\ngeneralized this result to a linear-time algorithm that finds an optimal solution for graphs that have\nbounded tree-width, relying on earlier results of Courcelle et al. [9]. Kneis et al. [23] also showed\nthat PDS is a generalization of the Dominating Set problem as follows. Given a graph G we\ncan construct an augmented graph G′ such that S is an optimal solution for the Dominating Set\nproblem on G if and only if it is an optimal solution for PDS on G′; the graph G′ is obtained from\nG by adding a new node v′ for each node v in G and adding the edge vv′. Guo et al. [15] developed\na combinatorial algorithm based on dynamic-programming for optimally solving PDS on graphs of\ntree-width k. The running time of their algorithm is O(ck2 · n) where c is a constant. Guo et al.\nalso compared the tractability of the Dominating Set problem versus PDS on several classes of\ngraphs, that is, they study whether there are classes of graphs where the former problem is in P\nbut the latter one is NP-hard; but they have no result that “separates” the two problems. Even\nfor planar graphs, the Dominating Set problem is NP-hard [14], and the same holds for PDS\n[15]. Liao and Lee [25] proved that PDS on split graphs is NP-complete, and also they presented\na polynomial time algorithm for solving PDS optimally on interval graphs. Dorfling and Henning\ncomputed the power domination number, i.e. the size of optimal power dominating set, for n × m\ngrids [12].\nBrueni and Heath [7] have more results on PDS, especially the NP-completeness of\nPDS on planar bipartite graphs. To the best of our knowledge, no further results are known on\nsolving the PDS problem, either optimally or approximately. Some of the results in this paper have\nappeared in the thesis of the second author [31], and in the proceedings of a workshop [1].\n1.2\nOur contributions\nOur results substantially improve on the understanding of PDS in the context of approximation\nalgorithms. In particular, we show a substantial gap between the approximation guarantees for the\nDominating Set problem and PDS modulo a variant of the P ̸= NP conjecture. This seems to be\nthe first known “separation” result between the two problems, in any class of graphs.\n• We present a reduction from the MinRep problem to the PDS problem that shows that PDS\ncannot be approximated within a factor of 2log1−ǫn, unless NP ⊆DTIME(npolylog(n)).\n• For undirected graphs, we introduce the notion of strong regions and weak regions as a\nmeans of obtaining lower bounds on the size of an optimal solution for PDS. Based on this,\nwe develop an approximation algorithm for PDS that gives an approximation guarantee of\nO(k) for graphs that have tree-width k. The algorithm requires the tree decomposition as\n3"},{"page":4,"text":"part of the input, and runs in time O(n3) (independent of k). By slightly modifying this\nalgorithm we get an algorithm that solves PDS optimally on trees. Our algorithm provides\nan O(√n)-approximation algorithm for PDS on planar graphs because a tree decomposition\nof a planar graph with width O(√n) can be computed efficiently [2]. Moreover, we show\nthat our methods (specifically, the lower bounds used in our analysis) cannot improve on our\nO(√n) approximation guarantee.\n• We extend PDS in a natural way to directed graphs and prove that even for directed acyclic\ngraphs, PDS is hard to approximate within the same threshold as for undirected graphs\nmodulo the same complexity assumption.\n• We give a linear-time algorithm based on dynamic-programming for Directed PDS when\nthe underlying undirected graph has bounded tree-width. This builds on results and methods\nof Guo et al. [15].\n2\nPDS in Undirected Graphs\nIn this section we prove a result on the hardness of approximating PDS by a reduction from the\nMinRep problem. In Section 2.1 we define the MinRep problem, and then we give a gap preserving\nreduction from MinRep to PDS in Section 2.2.\n2.1\nThe MinRep problem\nIn the MinRep [24] problem we are given a bipartite graph G = (A, B, E) with a partition of A and\nB into equal-sized subsets. Let qA and qB denote the number of sets in the partition of A and B,\nrespectively. Let A = A1 ∪A2 ∪· · ·∪AqA denote the partition of A, and let B = B1 ∪B2 ∪· · ·∪BqB\ndenote the partition of B. This partition naturally defines a super bipartite graph H = (A, B, E).\nThe super nodes of H are A1, A2, . . . , AqA and B1, B2, . . . , BqB. There is a super edge between\nsuper nodes Ai and Bj if there exists some a ∈Ai and b ∈Bj such that ab is an edge in G. We\nsay that super edge AiBj is covered by nodes a, b if a ∈Ai, b ∈Bj, and there is an edge between\na and b in G. Given S ⊆A ∪B we say that the super edge AiBj is covered by S if there exists\na, b ∈S that covers AiBj. The goal in the MinRep problem is to pick a minimum-size set of nodes,\nA′ ∪B′ ⊆V (G), to cover all the super edges in H. Note that we need a pair of nodes to cover\na super edge, and the pair should induce an edge between the two super nodes of the super edge;\nmoreover, a node in A′ ∪B′ may be useful for covering more than one super edge. The following\nTheorem is from [24].\nTheorem 2.1 (Theorem 5.4 in [24]) The MinRep problem cannot be approximated within ra-\ntio 2log1−ǫn, for any fixed ǫ > 0, unless NP ⊆DTIME(npolylog(n)).\n2.2\nThe reduction to PDS\nTheorem 2.2 The PDS problem cannot be approximated within ratio 2log1−ǫn, for any fixed ǫ > 0,\nunless NP ⊆DTIME(npolylog(n)).\nThe reduction: Theorem 2.2 is proved by a reduction from the MinRep problem. We create an\ninstance G = (V , E) of the PDS problem from a given instance G = (A, B, E)(H = (A, B, E)) of\nthe MinRep problem. The idea is to replace each super edge with a “cover testing gadget”.\n4"},{"page":5,"text":"1. Start with a copy of each node in A ∪B in G. For convenience, we use the same notation for\nnodes (and set of nodes) in G and their copies in G.\n2. Add a new node w∗to the graph G, and connect w∗to all nodes in A ∪B. Also add new\nnodes w∗\n1, w∗\n2, w∗\n3 and connect them to w∗(the nodes w∗\n1, w∗\n2, w∗\n3 are added to force w∗to be\nin any optimal solution. See the proof of Lemma 2.3 for more details).\n3. ∀i ∈{1, . . . , qA} , j ∈{1, . . . , qB} if AiBj is a super edge, then do the following:\n(a) Let Eij denote the set of edges between Ai and Bj in G and let lij denote |Eij| (see\nFigure 3(a); for an example E11 has 3 edges, and E12 has 4 edges). We denote the edges\nin Eij by e1, e2, · · · , ek, · · ·.\n(b) Let Cij be a cycle of 3lij nodes. We sequentially label the nodes of Cij as u1, v1, w1,\nu2, v2, w2, · · · , uk, vk, wk, · · · (informally speaking, we associate each triple uk, vk, wk with\nan edge ek of Eij). Make λ = 4 new copies of the graph Cij (λ can be any constant greater\nthan 3; refer to the proof of Lemma 2.3 for more details). For each edge ek = akbk ∈Eij\nand for each of the 4 copies of Cij, we add an edge from ak to uk and an edge from bk\nto vk. See Figures 2(a), 2(b) for an illustration.\nCij\nv2\nu2\nu1\nv1\nvlij\nulij\nek\ne2\ne1\nelij\nvk\nuk\nw1\nw2\nwk\nwlij\n(a) The Cij graph\nCij\na1\nb1\na2\nb2\ne2\ne1\nv2\nu2\nu1\nv1\nvk\nuk\nek\nak\nbk\nvlij\nulij\nblij\nalij\nelij\n(b) Edges between Cij and Ai ∪Bj.\nFigure 2: The cover testing gadget.\n4. Let G = (V , E) be the obtained graph (see Figure 3 for an illustration).\nLet S be a feasible solution for the resulting PDS instance G, and suppose w∗∈S. Then all\nof the nodes in A ∪B are power dominated (by Rule 1 of PDS). Now consider a gadget Cij, and\nassume a node v of Cij is in S. By applying Rule 1 once and then repeatedly applying Rule 2 of\nPDS, the gadget Cij will be completely power dominated, that is, all nodes of the gadget will be\nin PS.\nThe next lemma shows that the size of an optimal solution in PDS is exactly one more than the\nsize of an optimal solution in MinRep. The number of nodes in the constructed graph is equal to\n V (G)\n = 4+ |V (G)| + 3λ |E(G)|. This will complete the proof of Theorem 2.2 by showing that the\nabove reduction is a gap preserving reduction from MinRep to PDS with the same gap (hardness\nratio) as the MinRep problem.\n5"},{"page":6,"text":"Lemma 2.3 A∗∪B∗is an optimal solution to the instance G = (A, B, E) of the MinRep problem\nif and only if S∗= A∗∪B∗∪{w∗} ⊆V (G) is an optimal solution to the instance G of the PDS\nproblem.\nProof: First, we claim that w∗should be in any optimal solution of the PDS instance G. Suppose\nthat w∗is not in some optimal solutions. Then, in order to power dominate the nodes w∗, w∗\n1, w∗\n2, w∗\n3\nin G, the set S must contain at least two of the nodes (leaves) w∗\n1, w∗\n2, w∗\n3. This is a contradiction,\nsince we can replace these 2 nodes by w∗and obtain a smaller feasible solution.\nAssume that A∗∪B∗is a feasible solution for the MinRep instance G.\nWe will show that\nS = A∗∪B∗∪{w∗} is a feasible solution to the PDS instance G. Note that all nodes in A ∪B ∪\n{w∗, w∗\n1, w∗\n2, w∗\n3} are power dominated by applying Rule 1 on w∗. Now, we only need to show that\nall nodes in the gadgets Cij are power dominated. Consider any super edge AiBj of H. The set\nA∗∪B∗covers all the super edges in H. So there exists a pair of nodes ak ∈A∗∩Ai, bk ∈B∗∩Bj\nthat induces an edge of G. Since ak and bk are in S, their neighbors, uk and vk, in each of the\nλ = 4 copies of Cij in G, will be power dominated by applying Rule 1. Then the nodes uk and vk\nin each copy of Cij will power dominate the entire cycle by repeatedly applying Rule 2. To see this,\nnote that any node in Cij has exactly 2 neighbors in Cij and at most 1 neighbor not in Cij. The\nneighbors not in Cij are from Ai ∪Bj, and they are power dominated by w∗. Hence, if a node in Cij\nand one of its neighbors in Cij are power dominated, then by applying Rule 2 the other neighbor\nin Cij will be power dominated. Hence, by starting from vk and repeatedly applying Rule 2, we\ncan sequentially power dominate the nodes in Cij. This shows that S power dominates all nodes\nin G. Therefore, Opt(G) is at most |A∗∪B∗| + 1.\nLet S∗⊆V (G) be an optimal solution for PDS. By the above claim, w∗is in S∗. Now define\nA′ = A ∩S∗and B′ = B ∩S∗. First we prove that any optimal solution of PDS is contained in\nA ∪B ∪{w∗}, and then we show that A′ ∪B′ covers all super edges of the MinRep instance G.\nSuppose that S∗contains some nodes not in A ∪B ∪{w∗}. Hence, there are some gadgets that are\nnot completely power dominated by S∗∩(A ∪B ∪{w∗}). Let Cij be such a gadget. By symmetry\neach of the λ = 4 copies of Cij is not completely power dominated. Therefore, the optimal solution\nS∗needs to have at least 3 nodes from the 4 copies of Cij. By removing these 3 nodes from S∗and\nA1\nA2\nB1\nB2\n(a) MinRep Instance G\nw∗\n3\nw∗\n1\nw∗\n2\nw∗\nC2,2\nA2\nB2\nB1\nC1,1\nC1,2\nA1\n(b) PDS instance G: For each super edge AiBj we show\nonly 1 copy of Cij; in fact G has λ = 4 copies of Cij.\nFigure 3: The hardness construction\n6"},{"page":7,"text":"adding ak ∈Ai and bk ∈Bj to S∗for some arbitrary edge akbk ∈Eij, we can power dominate all\nof the 4 copies of Cij. This contradicts the minimality of S∗, and proves that S∗⊆A ∪B ∪{w∗}.\nTo see that A′ ∪B′ covers all super edges, note the following: suppose no node from any copy of\nCij is in the optimal solution; then any Cij can be power dominated only by taking a pair of nodes\na ∈Ai, b ∈Bj that induces an edge of G. This completes the proof of the lemma.\n□\n3\nApproximation Algorithms for Planar Graphs\nIn this section we describe an O(k)-approximation algorithm for PDS in graphs with tree-width\nk; the running time is O(n3), independent of k. This algorithm gives an O(√n)-approximation\nalgorithm for PDS in planar graphs, since the tree-width of a planar graph G with n nodes is\nO(√n) and in O(n\n3\n2) time we can find an O(√n) tree-width decomposition of the given planar\ngraph G [2]. Finally, we show that the analysis of our algorithm is tight on planar graphs. We use\nPlanar PDS to denote the special case of the PDS problem where the graph is planar.\nDefinition 3.1 [11] A tree decomposition of a graph G = (V, E) is a pair ⟨{Xi ⊆V |i ∈I} , T =\n(I, F)⟩such that T is a tree with V (T) = I, E(T) = F, and satisfying the following properties:\n(T1) S\ni∈I Xi = V , and every edge uv ∈E has both ends in some Xi,\n(T2) For all i, j, k ∈I if j is on the unique path from i to k in T then we have: Xi ∩Xk ⊆Xj,\nThe width of ⟨{Xi|i ∈I} , T⟩is the maxi∈I|Xi|−1. The tree-width of G is defined as the minimum\nwidth over all tree decompositions. The nodes of the tree are called T-nodes and the sets Xi are\ncalled bags.\nA nice tree decomposition is a tree decomposition ⟨{Xi ⊆V |i ∈I} , T = (I, F)⟩, where T is a\nrooted tree in which each node has at most 2 children.\nIf a node i ∈I has two children j, k\nthen Xi = Xj = Xk (i is called a Join node), and if i has one child j then either Xj ⊂Xi and\n|Xi \\ Xj| = 1 or Xi ⊂Xj and |Xj \\ Xi| = 1 (i is called an Insert or a Forget node, respectively).\nWe introduce the notion of a strong region before presenting our algorithm. Informally speaking,\na set of nodes R ⊆V is called strong if every feasible solution to the PDS problem has a node of R.\nFor a graph G = (V, E), the neighborhood of R ⊆V is nbr(R) = {v ∈V |∃uv ∈E, u ∈R, v /∈R},\nand the exterior of R is defined by ext(R) = nbr(V \\R), i.e., ext(R) consists of the nodes in R that\nare adjacent to a node in V \\ R.\nDefinition 3.2 Given a graph G = (V, E) and a set S ⊆V , the subset R ⊆V is called an S-strong\nregion if R ̸⊆PS∪nbr(R), otherwise, the set R is called an S-weak region. The region R is called\nminimal S-strong if it is an S-strong region and ∀r ∈R, R −r is an S-weak region.\nIt is easy to check from the definition that an S-strong region is also an ∅-strong (or shortly strong)\nregion. Any feasible solution to the PDS problem needs to have at least one node from every strong\nregion.\nLemma 3.3 A subset R ⊆V is an S-strong region if and only if for every feasible solution S ∪S∗\nof G, we have R ∩(S∗\\ S) ̸= ∅.\n7"},{"page":8,"text":"Proof: It can be seen that the set S ∪(V \\ R) will power dominate the same set of nodes in R\nthat can be power dominated by S ∪nbr(R); this is valid for any subset R ⊆V .\nLet R be an S-strong region. By the definition of a strong region we have R ̸⊆PS∪nbr(R). Hence,\nby the above claim R ̸⊆PS∪(V \\R). This shows that every feasible solution S∗needs to have at least\none node from R that is not in S.\nNow assume that for every feasible solution S∗of G we have R ∩(S∗\\ S) ̸= ∅. Suppose that R\nis an S-weak region, so by the definition of a weak region we have R ⊆PS∪nbr(R). It follows that\nS∗= S ∪(V \\ R) is a feasible solution, but R has no intersection with S∗\\ S ⊆(V \\ R). This is a\ncontradiction, so R is an S-strong region.\n□\nOur algorithm makes one level-by-level and bottom-to-top pass over the tree T of the tree de-\ncomposition of G and constructs a solution S for PDS (initially, S = ∅). At each node rj of T we\ncheck whether the union of the bags in the subtree rooted at rj forms an S-strong region; if yes,\nthen the bag Xrj of rj is added to S, otherwise S is not updated. The key point in the analysis is\nto show that Opt(G) ≥m, where m is the number of nodes of T where we updated S.\nAlgorithm 1 O(k)-approximation Algorithm\n1: A tree decomposition ⟨{Xi|i ∈I} , T⟩of G is given, where T is rooted at r.\n2: Let Ilbe the set of T-nodes at distance lfrom the root, and let d be the maximum distance\nfrom r in T.\n3: S ←∅\n4: for i = d to 0 do\n5:\nLet Ii = {r1, . . . , rki} and denote by Trj the subtree in T rooted at rj.\n6:\nLet Yrj be the union of bags corresponding to the T-nodes in Trj.\n7:\nfor j = 1 to ki do\n8:\nif Yrj is an S-strong region then\n9:\nS ←S ∪Xrj; where Xrj is the bag corresponding to rj.\n10:\nend if\n11:\nend for\n12: end for\n13: Output So = S\n3.1\nAnalysis of the algorithm\nIn this subsection we show that our algorithm has an approximation guarantee of O(k).\nLet\nG = (V, E) denote the input graph, and let S ⊆V be any set of nodes.\nLemma 3.4 Suppose Z is an S-weak region such that ext(Z) ⊆S. Then we have Z ⊆PS.\nProof: Let Y = ext(Z), it is easy to check that nbr(Z \\ Y ) ⊆ext(Z). We claim that Z \\ Y is an\nS-weak region. Let S∗= V \\ (Z ∪S), it is easy to check that S ∪S∗is a feasible solution for the\ngraph G, but S∗∩(Z \\ Y ) = ∅. Hence, by Lemma 3.3, Z \\ Y is not an S-strong region, and so it\nis an S-weak region. Thus Z \\ Y ⊆PS∪nbr(Z\\Y ) ⊆PS∪ext(Z) = PS and this implies that Z ⊆PS as\nY = ext(Z) ⊆S.\n□\n8"},{"page":9,"text":"Lemma 3.5 Let Z ⊆V be an S-strong region. Suppose that Y is a subset of V such that Y ⊆PS\nand ext(Y ) ⊆S. Then Z \\ Y is an S-strong region.\nProof: Assume for the sake of contradiction that Z \\Y is an S-weak region. Then by the definition\nof strong regions we have: Z \\Y ⊆PS∪nbr(Z\\Y ). It is easy to see that nbr(Z \\Y ) ⊆nbr(Z)∪ext(Y ).\nThis implies that Z \\Y ⊆PS∪nbr(Z\\Y ) ⊆PS∪nbr(Z)∪ext(Y ) = PS∪nbr(Z). The condition in the lemma\nstates that Y ⊆PS ⊆PS∪nbr(Z). Hence, we get Z = (Z \\ Y ) ∪(Z ∩Y ) ⊆PS∪nbr(Z), which means\nthat Z is an S-weak region. This is a contradiction, so the lemma is proved.\n□\nTheorem 3.6 Given a graph G = (V, E) and a tree decomposition of G of width k as input,\nAlgorithm 1 runs in time O(n · |E|), and achieves an approximation guarantee of (k + 1).\nProof: First, we show that the solution So found by the algorithm is feasible. Then we prove the\napproximation guarantee, and establish the running time.\nFor any node q of T, recall that Yq denotes the union of the bags corresponding to the T-nodes\nin the subtree rooted at q in T; let Gq denote the subgraph of G induced by Yq. We claim that\next(Yq) ⊆Xq.\nSuppose that q has m children in T, call them c1, . . . , cm.\nFor each edge qcj\n(j = 1, · · · , m), the set Xq ∩Xcj separates Ycj from the rest of the graph, that is, every path\nbetween a node in Ycj and a node in V \\Ycj contains a node of Xq ∩Xcj (see Lemma 12.3.1 in [11]).\nThus, ext(Ycj) ⊆Xq ∩Xcj ⊆Xq, and hence, for Yq = Xq ∪Yc1 ∪· · · ∪Ycm, we have ext(Yq) ⊆Xq.\nWe use induction on the height of the subtree of T rooted at q to prove the following: if Yq is\nS∗-strong, then Yq ⊆PS∗∪Xq, where S∗denotes the solution just before the algorithm examines\nYq. The statement clearly holds when q is a leaf of T (since Yq = Xq). Otherwise, let c1, . . . , cm\nbe the children of q in T. For each j = 1, . . . , m, when the algorithm examined Ycj, either Ycj was\nS-weak, in which case (by Lemma 3.4) we have Ycj ⊆PS∪ext(Ycj ) ⊆PS∪(Xcj ∩Xq) ⊆PS∗∪Xq or Ycj\nwas S-strong in which case Ycj ⊆PS∪Xcj by induction (note that S ∪Xcj ⊆S∗); we use S to denote\nthe solution just before the algorithm examines Ycj. Hence, Yq = Yc1 ∪· · · ∪Ycm ∪Xq ⊆PS∗∪Xq.\nThe above statement implies that V ⊆PSo because at the step when the algorithm examines the\nroot r of T either\n(i) Yr is S-strong, so So = S ∪Xr, and Yr ⊆PS∪Xr = PSo; or\n(ii) Yr is S-weak, and Yr ⊆PS∪ext(Yr) = PSo; since Yr = V (G) and ext(Yr) = ∅.\nTo show that the approximation guarantee is (k+1) we will construct a set ∆of pairwise disjoint\nstrong regions R1, R2, . . . , such that there is a strong region Rj corresponding to each step of the\nalgorithm that adds a non empty bag Xqj to S. Thus |So| ≤(k + 1) |∆| since each bag has ≤k + 1\nnodes, and Opt(G) ≥|∆| because every feasible solution has size ≥|∆|, by Lemma 3.3. Hence,\n|So| ≤(k + 1)Opt(G). We construct the sets R1, R2, . . . , during the execution of the algorithm\nas follows. Suppose the algorithm finds Yq to be S-strong while examining a node q of T. Let\nq1, . . . , ql−1 be the nodes of T where the algorithm updated the solution before examining q, and\nlet S be the solution just before the algorithm examines q. Then define Rl= Yql\\(Yq1 ∪· · ·∪Yql−1),\nwhere ql= q. We claim that Rlis an S-strong region. For each strong region Yqj (j = 1, . . . , l−1) we\nhave seen that ext(Yqj) ⊆Xqj ⊆S and Yqj ⊆PS; note that the algorithm added Xqj to the solution\nsince Yqj was a strong region. It follows that ext(Yq1 ∪· · · ∪Yql−1) ⊆S, and Yq1 ∪· · · ∪Yql−1 ⊆PS.\nHence, by Lemma 3.5, the set Rlis an S-strong region. Clearly, the sets R1, R2, . . . , are pairwise\ndisjoint. This completes the construction of ∆.\n9"},{"page":10,"text":"Consider the running time. Without loss of generality we can assume that the given tree decom-\nposition of width k has at most 4n bags (see Lemma 13.1.2 in [22]). Using standard algorithmic\ntechniques we can test in O(|E|) time whether a given set R ⊆V is an S-strong region (we compute\nPS∪nbr(R) and check if it contains R). Therefore, our algorithm has a running time of O(n · |E|). □\nIt is known that planar graphs have tree-width O(√n), and such a tree decomposition can be\nfound in O(n\n3\n2 ) time [2]. This fact together with the above theorem proves the following theorem.\nTheorem 3.7 Algorithm 1 achieves an approximation guarantee of O(√n) for the Planar PDS\nproblem.\nAs mentioned earlier, Haynes et al.[16] presented a linear-time algorithm for optimally solving\nPDS on trees. By modifying Algorithm 1, we can solve PDS optimally on trees. The resulting\nalgorithm differs from the algorithm of Haynes et al. since our algorithm uses strong regions.\nInformally, the algorithm makes a level-by-level and bottom-to-top pass over the tree G. At a node\nv of the tree G if the set of nodes in the subtree rooted at v forms a strong region, then we add v\nto the solution, otherwise we skip v. Formally, we define Xv = {v} for each v ∈V (G), and we run\nAlgorithm 1 on the tree G. Note that defining bags in this way does not give a tree decomposition\nof G.\nTheorem 3.8 A modification of Algorithm 1 runs in time O(n · |E|) and solves PDS optimally on\ntrees.\n3.2\nLower bounds via disjoint strong regions\nIn this part we show that any approximation algorithm for PDS that uses the number of disjoint\nstrong regions as a lower bound has an approximation guarantee of Ω(√n). In proposition 3.12\nwe give a lower bound on the optimal value for PDS on an l× m grid. Independently, [12] gave a\nstronger result for this.\nLemma 3.9 Any minimal S-strong region is connected.\nProof: Assume that R is a minimal S-strong region that is not connected.\nLet C ⊂R be a\nconnected component of R. The set C is an S-weak region since R is a minimal S-strong region.\nBy the definition of a weak region we have C ⊆PS∪nbr(C). The set C is a connected component of\nR, so the neighborhood of C has no intersection with R \\ C. This implies that nbr(C) ⊆nbr(R),\nand consequently we have C ⊆PS∪nbr(C) ⊆PS∪nbr(R). The same argument as above shows that\nR \\ C ⊆PS∪nbr(R). Hence, R ⊆PS∪nbr(R) which is a contradiction.\n□\nLemma 3.10 The number of disjoint strong regions in an l× m grid is exactly one.\nProof: For the sake of contradiction, assume that the given grid has two disjoint strong regions.\nTake as few nodes as possible from these strong regions until we get minimal strong regions, say\nR1 and R2. It is easy to check that the set of nodes of any row or any column of the grid power\ndominates all nodes in the grid. By Lemma 3.3, R1 and R2 should have at least one node from\nevery feasible solution. In the other words, R1 and R2 must have at least one node from each row\nand also from each column. By lemma 3.9, we know that R1 and R2 induce connected subgraphs.\nHence, in R1 there is a path from a node in the top row to a node in the bottom row, and also in R2\n10"},{"page":11,"text":"there is a path from a node in the rightmost column to a node in the leftmost column. Obviously\nthese two paths share a common node. This is a contradiction, since R1 and R2 are assumed to be\ndisjoint.\n□\nWe denote by Pi\nS a set of nodes that are power dominated after applying propagation rule, Rule\n2, for i number of times to P0\nS = S ∪nbr(S). Obviously this depends on the order of applying the\npropagation rule. We use the notation without specifying the order of applying the propagation\nrule.\nLemma 3.11 (Propagation lemma) Given an ordering of propagation rules applied to S ∪nbr(S)\nwith Pk\nS = PS we have:\n ext(Pj\nS)\n ≤\n ext(Pi\nS)\n , ∀0 ≤i < j ≤k.\nProof: We will prove that\n ext(Pi+1\nS\n)\n ≤\n ext(Pi\nS)\n , for all 0 ≤i ≤k −1. Consider the set Pi\nS\nand assume that in the (i + 1)-st step we apply Rule 2 to v ∈ext(Pi\nS) and power dominate u; i.e.\nu ∈Pi+1\nS\n, u /∈Pi\nS. To apply Rule 2 to v all neighbors of v except u should be power dominated, so\nwe have nbr(v)\\Pi\nS = {u}. Also since we power dominate u at step (i+1), we have nbr(v) ⊆Pi+1\nS\n.\nTherefore, v is not in ext(Pi+1\nS\n), but u may be in the exterior of Pi+1\nS\n. Hence, we have ext(Pi+1\nS\n) ⊆\n ext(Pi\nS) \\ {v}\n \n∪{u}. It follows that\n ext(Pi+1\nS\n)\n ≤\n ext(Pi\nS) \\ {v}\n + |{u}| =\n ext(Pi\nS)\n .\n□\nProposition 3.12 Let G be an l× m grid with l≤m, then Opt(G) = Θ(l).\nProof: First note that any row or any column of the grid power dominates all nodes. In the\nfollowing we prove that any feasible solution of PDS needs to have at least l−1\n5\nnodes. Assume that\nthere exists S ⊆V such that |S| < l−1\n5\nand PS = V (G). The maximum degree in G is 4, so we have\n ext(P0\nS)\n ≤|S ∪nbr(S)| < l−1\n5 · 5 = l −1. Therefore, P0\nS contains no full row or no full column of\nG. The set S power dominates G, so there is an i such that Pi\nS contains a full row or a full column.\nConsider the smallest i with this property. Hence, Pi−1\nS\nhas no full row or full column. But some\nrow or some column must have at least l −1 nodes in Pi−1\nS\nsince Pi\nS contains an entire row or an\nentire column. Without loss of generality assume that these l −1 nodes are from a column. By the\ndefinition of i, each of the l −1 nodes in this column are in a row which is not a subset of Pi−1\nS\n.\nTherefore, there are at least l −1 rows with at least one node in Pi−1\nS\nand at least one node not\nin Pi−1\nS\n. This implies that\n ext(Pi−1\nS\n)\n ≥l −1. Finally, by using Lemma 3.11 we get the following\ncontradiction: l −1 >\n P0\nS\n ≥\n ext(P0\nS)\n ≥\n ext(Pi−1\nS\n)\n ≥l −1.\n□\nConsider any approximation algorithm for PDS that uses only the number of disjoint strong\nregions as a lower bound on the size of an optimal solution. By Lemma 3.10, this algorithm finds\na lower bound of 1 on the size of an optimal solution on a grid. The √n × √n grid has an optimal\nsolution of size Θ(√n) by Proposition 3.12. This shows that the approximation guarantee of the\nalgorithm is Ω(√n), even on planar graphs.\nProposition 3.13 Consider any approximation algorithm for PDS that uses only the number of\ndisjoint strong regions as a lower bound on the optimal value. Then the approximation guarantee\nis Ω(√n).\n4\nPDS in Directed Graphs\nIn this section we extend the PDS problem to directed graphs to obtain the Directed Power\nDominating Set (Directed PDS) problem. Our motivation for studying the directed prob-\nlem comes from theoretical considerations.\nThe Dominating Set problem is studied on both\n11"},{"page":12,"text":"undirected and directed graphs, and there is extensive literature on the latter (see [17, 18]). The\nsimilarities between the Dominating Set problem and the PDS problem led us to define and study\nthe Directed PDS problem. We give a result on the hardness of approximation of Directed\nPDS. Then we reformulate the Directed PDS problem in terms of valid coloring of the edges.\nUsing this, we design an algorithm for solving Directed PDS in linear-time on a special class of\ndirected graphs.\nLet G = (V, E) be a directed graph.\nA node w is called an out-neighbor (in-neighbor) of a\nnode v if there is a directed edge from v to w (from w to v) in G. The number of out-neighbors\n(in-neighbors) of a node v is called the out-degree (in-degree) of v and is denoted by d+\nG(v) (or\nsimilarly d−\nG(v)). For a set of nodes X, the subgraph of G induced by X is denoted by G[X]. The\ndirected graphs that we consider here have no loops nor parallel edges, but may have two edges\nwith different directions on the same two end nodes (we call such edges antiparallel).\nGiven a\ndirected graph G by the underlying undirected graph we mean the undirected graph obtained from\nG, by removing the direction of edges and also removing any parallel edges that are introduced\nafter removing the directions.\nDefinition 4.1 (the Directed PDS problem) Let G be a directed graph. Given a set of nodes\nS ⊆V (G), the set of nodes that are power dominated by S, denoted by PS, is obtained as follows:\n(D1) if node v is in S, then v and all of its out-neighbors are in PS;\n(D2) (propagation) if node v is in PS, one of its out-neighbors w is not in PS, and all other\nout-neighbors of v are in PS, then w is inserted into PS.\nWe say that S power dominates G if PS = V (G). The Directed PDS problem is to find a node\nset S with minimum size that power dominates all the nodes in G.\nWe prove a threshold of 2log n1−ǫ for the hardness of approximation of Directed PDS modulo\nthe same complexity assumption as in Theorem 2.2. The proof uses a reduction from the MinRep\nproblem to the Directed PDS problem in a directed acyclic graph. This reduction is similar to\nthe reduction in Theorem 2.2; the main difference comes from the gadget for modeling the super\nedges.\nTheorem 4.2 The Directed PDS problem even when restricted to directed acyclic graphs cannot\nbe approximated within ratio 2log1−ǫ n, for any fixed ǫ > 0, unless NP ⊆DTIME(npolylog(n)).\nThe reduction: We create an instance G = (V , E) of the Directed PDS problem from a given\ninstance G = (A, B, E)(H = (A, B, E)) of the MinRep problem.\n1. Start with a copy of each node in A ∪B in G. For convenience, we use the same notation for\nnodes (and sets of nodes) in G and their copies in G.\n2. Add a new node w∗to the graph G, and add a directed edge from w∗to each node in A ∪B.\n3. ∀i ∈{1, . . . , qA} , j ∈{1, . . . , qB} if AiBj is a super edge, then do the following:\n(a) Let Eij be the set of edges between Ai and Bj in G, and let lij denote |Eij|. We denote\nthe edges in Eij by e1, e2, . . . , ek, . . . , elij.\n12"},{"page":13,"text":"(b) Let Dij be the graph on 6lij + 1 nodes as shown in Figure 4(a). In Dij there are 6\nnodes uk, vk, dk, αk, βk, γk associated with an edge ek of Eij. The part of Dij associated\nwith an edge ek is shown in Figure 4(b); note that all these parts share a common node,\ncalled the center node, in Dij. Make λ = 4 new copies of the graph Dij (λ can be any\nconstant greater than 3). For each edge ek = akbk ∈Eij and for each of the 4 copies of\nDij, we add a directed edge from ak to uk and a directed edge from bk to vk. In addition\nto these edges, there are directed edges from w∗to some nodes inside Dij; these directed\nedges are denoted by a dashed line in Figure 4(a).\nw∗\nv1\nu2\nvk\nu1\nv2\nd2\ndk\nd1\nulij\ndlij\nulij\nuk\n(a) The Dij graph\nbk\nak\nvk\nuk\ndk\nγk\nβk\nαk\ncenter node\nek\n(b) Part of gadget corre-\nsponding to an edge ek\nFigure 4: The cover testing gadget.\n4. Let G = (V , E) be the obtained graph.\nThe next lemma shows that the size of an optimal solution in Directed PDS is exactly one\nmore than the size of an optimal solution in the MinRep instance. The number of nodes in the\nconstructed graph is at most\n V (G)\n ≤1 + |V (G)| + 7λ |E(G)|. This will complete the proof of\nTheorem 4.2 by showing that the above reduction is a gap preserving reduction from MinRep to\nDirected PDS with the same gap (hardness ratio) as the MinRep problem.\nLemma 4.3 A∗∪B∗is an optimal solution to the instance G = (A, B, E) of the MinRep problem\nif and only if S∗= A∗∪B∗∪{w∗} ⊆V (G) is an optimal solution to the instance G of the Directed\nPDS problem.\nProof: First note that w∗should be in any feasible solution; because it has in-degree zero.\nAssume that A∗∪B∗is a feasible solution for the MinRep instance G.\nWe will show that\nS = A∗∪B∗∪{w∗} is a feasible solution to the Directed PDS instance G. Note that all nodes\nin A ∪B and some nodes inside the gadgets Dij are power dominated by applying rule (D1) on\nw∗. Now, we only need to show that all nodes in the gadgets Dij are power dominated. Consider\na super edge AiBj of H. The set A∗∪B∗covers all the super edges in H. Hence, there exists\na pair of nodes ak ∈A∗∩Ai, bk ∈B∗∩Bj that induces an edge of G.\nSince ak and bk are\nin S their out-neighbors uk and vk in each of the λ = 4 copies of the Dij graph will be power\ndominated by applying rule (D1). Now node dk, that is already power dominated by w∗, will power\ndominate the center node by an application of rule (D2). Now we claim that the center node will\npower dominate the remaining nodes in the gadget Dij. Consider the part of gadget (shown in\n13"},{"page":14,"text":"Figure 4(b)) corresponding to an edge er ∈Eij. Note that the nodes γr, αr, and dr are already\npower dominated by w∗. It is easy to check that the nodes βr, ur, vr will be power dominated by\nsequentially applying rule (D2) on γr, αr, and dr. This shows that S power dominates all nodes in\nG. Therefore, Opt(G) is at most |A∗∪B∗| + 1.\nLet S∗⊆V (G) be an optimal solution for Directed PDS. As we showed above w∗should be in\nany feasible solution for Directed PDS. Now define A′ = A∩S∗and B′ = B ∩S∗. First we prove\nthat any optimal solution of Directed PDS is contained in A ∪B ∪{w∗}, and then we show that\nA′ ∪B′ covers all the super edges of H. Suppose that S∗contains some nodes not in A ∪B ∪{w∗}.\nHence, there are some gadgets that are not completely power dominated by S∗∩(A ∪B ∪{w∗}).\nLet Dij be such a gadget. By symmetry each of the λ = 4 copies of Dij is not completely power\ndominated. Therefore, the optimal solution S∗needs to have at least 3 nodes from the 4 copies of\nDij. By removing these 3 nodes from S∗and adding ak ∈Ai and bk ∈Bj to S∗for some arbitrary\nedge ek = akbk ∈Eij, we can power dominate all 4 copies of Dij. This contradicts the optimality\nof S∗, and proves that S∗⊆A ∪B ∪{w∗}. To see that A′ ∪B′ covers all super edges, it is enough\nto note the following. Suppose no node from any copy of Dij is in the optimal solution, then any\nDij can be power dominated only by taking a pair of nodes a ∈Ai and b ∈Bj that induces an\nedge of G. Otherwise, any power dominated node in the gadget has at least 2 out-neighbors that\nare not power dominated, so rule (D2) cannot be applied. This completes the proof of the lemma.\n□\nThere are several notions for the tree-width of directed graphs such as DAG width [29], directed\ntree-width [21], and Kelly-width [19]. Directed acyclic graphs have width equal to zero for the first\ntwo notions [29], and have Kelly-width of 1. Hence, Theorem 4.2 gives a hardness threshold of\nO(2log n1−ǫ) even if the directed graph has width ≤1 according to any of the above three notions.\nWe reformulate Directed PDS in terms of valid colorings of the edges in order to develop\nan algorithm based on dynamic-programming for Directed PDS. Guo et al. [15] introduced the\nnotion of valid orientations to get a new formulation for PDS (in undirected graphs). They also\ndesigned a linear-time dynamic-programming algorithm based on valid orientations for optimally\nsolving PDS on graphs of bounded tree-width. Our method applies to directed graphs such that\nthe underlying undirected graph has bounded tree-width.\nDefinition 4.4 A coloring of a directed graph G = (V, E) is a partitioning of the edges in G into\nred and blue edges. We denote a coloring by C = (V, Er ∪Eb) where Er is the set of red edges and\nEb is the set of blue edges.\nWe reformulate the Directed PDS problem via a so-called valid coloring of directed graphs;\ninformally speaking, these colorings “model” the application of rules (D1) and (D2) of Directed\nPDS.\nDefinition 4.5 A valid coloring C = (V, Er ∪Eb) of a directed graph G = (V, E) is a coloring of\nG with the following properties:\n1. No two antiparallel edges can be colored red.\n2. The subgraph induced by the red edges, Gr = (V, Er), has the following properties:\n(a) ∀v ∈G : d−\nGr(v) ≤1, and\n(b) ∀v ∈G : d−\nGr(v) = 1 =⇒d+\nGr(v) ≤1.\n14"},{"page":15,"text":"3. G has no dependency cycle. A dependency cycle is a sequence of directed edges whose under-\nlying undirected graph forms a cycle such that all the red edges are in one direction, all the\nblue edges are in the other direction, and there are no two consecutive blue edges.\nWe call a node an origin of C if it has no incoming edges in Gr.\nOur dynamic-programming algorithm for Directed PDS is based on the following lemma.\nLemma 4.6 Given a directed graph G and S ⊆V (G), S power dominates G if and only if there\nis a valid coloring of G with S as the set of origins.\nProof: Suppose S ⊆V power dominates G. Then we give a valid coloring C with S as the set\nof origins by coloring the edges in G according to the way that S power dominates G. We color\nan edge (v, w) red if node w is power dominated by applying the power domination rules on v;\neither by the domination rule (D1) or by the propagation rule (D2). Note that when we apply\nthe propagation rule (D2), then we do not power dominate the previously power dominated nodes.\nAlso when we apply rule (D1) on v, then we power dominate all (not some subset of) neighbors\nof v that are not power dominated. We write v < u when a node u is power dominated after\nv. It is easy to check that with this coloring the degree requirements are satisfied; each node can\nbe power dominated only once, and if it is a power dominated node (not in S) , then it cannot\npower dominate more than one of its out-neighbors due to rule (D2). Now, we need to prove that\nthere is no dependency cycle.\nBy way of contradiction, suppose that C∗= u1, u2, . . . , um is a\ndependency cycle. Focus on the edges of C∗. Call the direction of the red edges forward, and call\nthe direction of the blue edges backward. Observe that a dependency cycle has ≥1 red edges, and\neach of its red edges corresponds to an application of rule (D2). Assume that all the edges in C∗\nare red. Then the red edges (ui, ui+1) imply that ui < ui+1 for all i = 1, 2, . . . , m −1; therefore\nu1 < u2 < · · · < um, but this is a contradiction since the last red edge from um back to u1 implies\nthat um < u1. Hence, there is no dependency cycle with all edges colored red. Now, assume that\nthe dependency cycle C∗has some blue edges. We show that a similar contradiction occurs when\nthere are no two consecutive blue edges. Consider a blue edge (u, v) of C∗and note that the other\nedge of C∗incident to u is a red edge, say (u, w). By rule (D2), we see that v should be power\ndominated before u can power dominate w; thus we have w > v. Repeating this argument, we\nget an ordering for the occurrences of power domination of some of the nodes in C∗that gives a\ncontradiction, e.g., if m is even and the edges of C∗are alternatively blue and red (starting with\nblue) we get u1 < u3 < u5 < · · · < um−1 < u1 (see Figure 5 for an example). Hence, G has a valid\ncoloring with S as the set of origins.\nB\nR\nR\nB\nR\nu1\nu2\nu3\nu4\nu5\nFigure 5:\nThe blue edge (u4, u3) means that u3 should be power dominated before we can power\ndominate u5 by applying rule (D2) on u4; thus u3 < u5. Again the blue edge from u1 to u5 implies\nthat u5 < u2. Finally , the red edge (u2, u3) shows that u2 < u3. Combining these dependencies\nwe get u3 < u5 < u2 < u3. This is a contradiction, and shows that there cannot be a dependency\ncycle in a coloring obtained from the application of rules (D1) and (D2) of Directed PDS.\nNow suppose that G has a valid coloring C = (V, Er ∪Eb) with S ⊆V (G) as the set of origins.\nThe nodes in S and all of their out-neighbors in Gr = (V, Er) are power dominated by applying\n15"},{"page":16,"text":"the rule (D1). Now we prove that S will power dominate all nodes in G. Suppose that this does\nnot happen. Let X ⊂V be the maximal set of nodes that can be power dominated by S. We\nclaim that there is at least one red edge from X to V \\ X. Note that all of the origins are in X,\nso each of the nodes in V \\ X has in-degree 1 in Gr. Hence, if there is no red edge from X to\nV \\ X, then there should be a directed cycle of red edges in G[V \\ X]. This is not possible since\nthere are no dependency cycles. Therefore there is at least one red edge from X to V \\ X. Let\ne1 = (x1, y1), . . . , ek = (xk, yk) be all of the red edges from X to V \\X. If some xi has all of its out-\nneighbors in X except yi, then by applying rule (D2) on xi the node yi will be power dominated. By\nthe maximality assumption of X this cannot happen. Therefore, each xi has another out-neighbor,\nsay zi, in V \\X. Then (xi, zi) is a blue edge, otherwise, xi would be an origin and yi would be power\ndominated by applying rule (D1) on xi. Now, we construct a dependency cycle as follows: starting\nfrom x1, use a blue edge to move to a node z1 in V \\ X; then move in the reverse direction over a\nsequence of red edges (z2, z1), (z3, z2), · · · until we reach a red edge (zi, zi−1) with zi−1 ∈V \\ X and\nzi ∈X (such an edge exists since G[V \\X] has no directed cycle of red edges); note that (zi, zi−1) is\none of the red edges (x1, y1), · · · , (xk, yk). If zi = x1, then we have a dependency cycle; otherwise,\nwe again use a blue edge (zi, zi+1) to move to a node in V \\ X. By repeating these steps, we will\neventually find a dependency cycle. Note that all the blue edges are in one direction, and all the red\nedges are in the other direction. This is a contradiction, since C has no dependency cycle. Hence,\nwe have X = V , so S power dominates G.\n□\nTheorem 4.7 Given a directed graph G and a tree decomposition of width k of its underlying\nundirected graph, Directed PDS can be optimally solved in O(ck2 · n) time for a global constant\nc.\nA consequence of the above theorem is a linear-time algorithm for solving the Directed PDS\nproblem optimally on directed graphs, given a bounded tree-width decomposition of the underlying\nundirected graph. Also since the tree-width decomposition for graphs with bounded tree-width can\nbe computed in polynomial-time [5], there is a polynomial-time algorithm to solve Directed PDS\noptimally on the class of directed graphs such that the underlying undirected graph has bounded\ntree-width.\n5\nConclusions\nWe studied the PDS problem from the perspective of approximation algorithms. We introduced\na natural extension of the problem to directed graphs. We showed that both problems have a\nthreshold of O(2log n1−ǫ) for the hardness of approximation.\nWe presented an O(√n) approxi-\nmation algorithm for Planar PDS. We designed a dynamic-programming algorithm for solving\nthe Directed PDS problem optimally in linear-time for those directed graphs whose underlying\nundirected graph has bounded tree-width.\nHere, we describe an algorithm with an approximation guarantee of O(\nn\nlog n) for the PDS problem.\nThe algorithm works as follows. Partition the nodes of the graph G into log n equal-sized sets\nV1, V2, · · ·. Next, consider all possible ways of picking these sets (we pick all nodes in a set). Among\nall these different candidates, output the one that power dominates G and has the minimum\nnumber of nodes. Note that in the algorithm we only consider 2log n = n different candidates.\nClearly, the algorithm runs in polynomial time, since the feasibility of each candidate can be tested\nin polynomial time. Let S∗be an optimal solution.\nIt is easy to see that the set of Vi’s that\n16"},{"page":17,"text":"intersect S∗forms a feasible solution for the PDS problem in G; this solution has size at most\nn\nlog n · |S∗|. This establishes the approximation guarantee. The same algorithm and analysis applies\nto the Directed PDS problem.\nProposition 5.1 There is a polynomial time\nn\nlog n-approximation algorithm for both the PDS prob-\nlem and the Directed PDS problem.\nThere is a gap between our hardness threshold of O(2log n1−ǫ) and our approximation guarantee of\nO(\nn\nlog n), and narrowing this gap is an open question.\nA major open question in the area is whether there exists a PTAS (polynomial time approximation\nscheme) for Planar PDS. A first step may be to obtain an improvement on our approximation\nguarantee of O(√n). There has been a lot of research on designing PTASs for NP-hard problems on\nplanar graphs. Some of the most important developments are the outerplanar layering technique by\nBaker [3], and the bidimensionality theory by Demaine and Hajiaghayi [10]. Unfortunately, these\nmethods do not apply to Planar PDS.\nBaker [3] showed that the Dominating Set problem in planar graphs has a PTAS. In the Baker\nmethod we first partition the graph into smaller graphs. Then we solve the problem optimally on\neach subgraph, and finally we return the union of the solutions as a solution for the original graph.\nThe example in Figure 1 shows that this method does not apply to Planar PDS. The size of an\noptimal solution is 1, but if we apply the Baker method, then the size of the output solution will\nbe at least as large as the number of subgraphs in the partition which can be Θ(n).\nDemaine and Hajiaghayi [10] introduced the bidimensionality theory and used it to obtain PTASs\nfor several variants of the Dominating Set problem on planar graphs. An important property\nof bidimensionality is that when an edge is contracted the size of an optimal solution should not\nincrease. Consider the example in Figure 6. If we contract edges e1, e2, . . . , en in G, then we get\nthe graph G′. It can be checked that Opt(G) = 1 but Opt(G′) = Θ(n). Thus the bidimensionality\ntheory does not apply to Planar PDS, since the optimum value may increase when an edge is\ncontracted.\nv\n=⇒\nv\nu1\nun\nen\ne1\nG\nG′\nFigure 6: Optimal value of PDS increases when edges are contracted.\nLastly, we consider some variations of greedy algorithms for PDS and show that they perform very\npoorly even on planar graphs. In contrast, for other related problems such as Dominating Set and\nSet Covering, greedy algorithms perform well since they achieve a logarithmic approximation\nguarantee, and no substantial improvement is possible by any polynomial time algorithm, under\ncomplexity assumptions like P ̸= NP. The most natural greedy algorithm for PDS is the one that\nstarts with S = ∅, and in each step, adds a new node v to the current solution S such that v power\ndominates the maximum number of new nodes.\n17"},{"page":18,"text":"Unfortunately, this greedy algorithm may find a solution S such that |S| ≥Θ(n) · Opt(G). To see\nthis, consider a graph G that is obtained from a 9l× 9m grid by subdividing all row-edges, except\nwith minor changes in the four corners as shown in Figure 7(a). Partition the graph G into 9 × 9\ngrids (ignoring the nodes introduced by subdivision), see Figure 7(b). It is easy to check that any\nsingle node can power dominate at most 7 nodes, and the center node of any one of the 9 × 9 grids\nachieves this maximum. So the greedy algorithm at the first iteration may pick the center node\nof any one of the 9 × 9 grids. Assuming all nodes picked by the algorithm so far have been these\ncenter nodes, we see that picking another center node maximizes\n PS∪{v} \\ PS\n over all v ∈V . So\nthe greedy algorithm could continue picking center nodes, and after that possibly picking other\nnodes until it finds a feasible solution S. The size of the output S is at least m · l= Θ(n). By\n(a) Grid\n(b) 9 × 9 grid\nFigure 7: Bad example for the greedy algorithm\nProposition 3.12, we have Opt(G) = Θ(l). Now by fixing l= Θ(1) we can see that the size of the\noutput solution can be bigger than Opt(G) by a factor of Θ(n).\nProposition 5.2 The greedy algorithm for the PDS problem may find a solution S such that |S| ≥\nΘ(n) · Opt(G).\nWe will consider two other variations of the greedy algorithm, namely Proximity and Cleanup.\nWe have examples of Planar PDS showing that these variations of the greedy algorithm perform\npoorly.\nProximity: In each step of the Proximity algorithm we choose a node such that the set of all\npower dominated nodes induces a connected subgraph, and subject to this, the number of newly\npower dominated nodes is maximized. Informally, this is to escalate the use of the propagation\nrule.\nThe bad example for the proximity version of the greedy algorithm is obtained by modifying the\ncenter row of the h × (2m + 1) grid, as shown in Figure 8, by inserting lsubdividing nodes into\nthe edges in the middle row, and also subdividing all of the other row-edges except some of the\ncorner edges. Figure 8 illustrates an example of such a grid for l = 5 and h = 9 rows, but for a\nbad example for the proximity greedy algorithm we need h to be sufficiently large constant (h = 17\nsuffices). We use the figure for illustration, to show the working of the proximity greedy algorithm.\nIt is easy to check that by picking all nodes of the first column we can power dominate the entire\ngraph, so the optimal solution is Θ(1). The proximity greedy algorithm starts by picking a node\nthat power dominates maximum number of nodes (which is 17 = 2l+ 7); any white node satisfies\nthis requirement. Therefore the algorithm may pick for example the first white node (from the\n18"},{"page":19,"text":"left). It is easy to check that in the next step the algorithm will pick the white node to the right of\nthe first one, since all of the power dominated region stays connected and also it power dominates\nmaximum number of new nodes (which is 16 = 2l+ 6). The algorithm continues picking all white\nnodes and at the end it will pick possibly more nodes to get a feasible solution.\n(The shaded\nregion shown in Figure 8 indicates the nodes that will be power dominated by picking all white\nnodes.) Therefore, the size of the solution found by the algorithm is at least m = Θ(n). Hence, the\nproximity greedy algorithm may find a feasible solution that is Θ(n) times worse than the optimal\nsolution.\nFigure 8: Bad example for the proximity greedy algorithm\nCleanup Step: Some of the recent approximation algorithms, especially some based on the primal-\ndual method, use a clean up step at the end: this step removes redundant elements from the solution\nin some sequential order. In the Cleanup algorithm, we first run the greedy algorithm to find a\nsolution (node set) S, then we repeatedly remove nodes from S, until S is an inclusionwise minimal\npower dominating set. Although a cleanup step may substantially improve on the solution found\nby the greedy algorithm on some examples, this does not hold for all examples.\nThe same bad example for the proximity version is also a bad example for the cleanup version of\nthe greedy algorithm. The cleanup greedy algorithm may again pick the first white node (from the\nleft), and after picking this white node, it may pick the third white node. Since both of them power\ndominate maximum number of new nodes (which is 17 = 2l+ 7). Note that in the original greedy\nalgorithm there is no need to have a connected subgraph induced on power dominated nodes. The\nalgorithm continues to pick all the odd indexed white nodes, and after that it will start picking the\neven indexed white nodes since any one of them power dominates maximum number of new nodes\n(which is at least 15 = 2l+ 5). At this stage of the algorithm the set of power dominated nodes\nare those in the shaded region of Figure 8. It is easy to check that we need to pick nodes from\nboth upper and lower parts in order to power dominate the entire graph. The greedy algorithm\nmay pick some nodes from the leftmost column in the top part and some nodes from the rightmost\ncolumn in the bottom part to power dominate the entire graph. Now we start doing the cleanup\nprocess. It can be checked that if we remove any two consecutive white nodes from the obtained\nsolution, the graph cannot be power dominated completely. So we need to keep at least half of the\nwhite nodes. Therefore, the size of the output solution at the end of cleanup process is at least\nm\n2 = Θ(n), but the optimal solution is just Θ(1) as before.\n19"},{"page":20,"text":"References\n[1] A. Aazami and M. D. Stilp. Approximation algorithms and hardness for domination with\npropagation. In Proceedings of the 10th International Workshop on Approximation Algorithms\nfor Combinatorial Optimization Problems, volume 4627 of LNCS, pages 1–15. Springer, 2007.\n[2] J. Alber, H. L. Bodlaender, H. Fernau, T. Kloks, and R. Niedermeier.\nFixed parameter\nalgorithms for dominating set and related problems on planar graphs. Algorithmica, 33(4):461–\n493, 2002.\n[3] B. S. Baker. Approximation algorithms for NP-complete problems on planar graphs. J. ACM,\n41(1):153–180, 1994.\n[4] T. L. Baldwin, L. Mili, M. B. Boisen, and R. Adapa. Power system observability with minimal\nphasor measurement placement. IEEE Transactions on Power Systems, 8(2):707–715, 1993.\n[5] H. L. Bodlaender. A linear-time algorithm for finding tree-decompositions of small treewidth.\nSIAM J. Comput., 25(6):1305–1317, 1996.\n[6] D. J. Brueni. Minimal PMU placement for graph observability, a decomposition approach.\nMaster’s thesis, Virginia Polytechnic Institute and State University, Blacksburg, VA, 1993.\n[7] D. J. Brueni and L. S. Heath.\nThe PMU placement problem.\nSIAM J. Discret. Math.,\n19(3):744–761, 2005.\n[8] V. Chvatal. A greedy heuristic for the set covering problem. Math. Oper. Res., 4:233–235,\n1979.\n[9] B. Courcelle, J. A. Makowsky, and U. Rotics. Linear time solvable optimization problems\non graphs of bounded clique width. In Proceedings of the 24th International Workshop on\nGraph-Theoretic Concepts in Computer Science, volume 1517 of LNCS, pages 1–16. Springer,\n1998.\n[10] E. D. Demaine and M. T. Hajiaghayi. Bidimensionality: new connections between FPT al-\ngorithms and PTASs. In Proceedings of the 16th Annual ACM-SIAM Symposium on Discrete\nAlgorithms, pages 590–601, 2005.\n[11] R. Diestel. Graph Theory. Springer-Verlag, New York, 2nd edition, 2000.\n[12] M. Dorfling and M. A. Henning. A note on power domination in grid graphs. Discrete Applied\nMathematics, 154(6):1023–1027, 2006.\n[13] U. Feige. A threshold of ln n for approximating set cover. J. ACM, 45(4):634–652, 1998.\n[14] M. R. Garey and D. S. Johnson. Computers and Intractability: A Guide to the Theory of\nNP-Completeness. W. H. Freeman and Co., New York, NY, USA, 1979.\n[15] J. Guo, R. Niedermeier, and D. Raible. Improved algorithms and complexity results for power\ndomination in graphs. In Proceedings of the 15th International Symposium on Fundamentals\nof Computation Theory, volume 3623 of LNCS, pages 172–184. Springer, 2005 (to appear in\nAlgorithmica).\n20"},{"page":21,"text":"[16] T. W. Haynes, S. M. Hedetniemi, S. T. Hedetniemi, and M. A. Henning. Domination in graphs\napplied to electric power networks. SIAM J. Discrete Math., 15(4):519–529, 2002.\n[17] T. W. Haynes, S. T. Hedetniemi, and P. J. Slater. Domination in Graphs: Advanced Topics.\nMarcel Dekker, New York, 1998.\n[18] T. W. Haynes, S. T. Hedetniemi, and P. J. Slater. Fundamentals of Domination in Graphs.\nMarcel Dekker, New York, 1998.\n[19] P. Hunter and S. Kreutzer. Digraph measures: Kelly decompositions, games, and orderings.\nIn Proceedings of the 18th Annual ACM Symposium on Discrete Algorithms, pages 637–644,\nPhiladelphia, PA, USA, 2007.\n[20] D. S. Johnson. Approximation algorithms for combinatorial problems. J. Comput. Syst. Sci.,\n9(3):256–278, 1974.\n[21] T. Johnson, N. Robertson, P. D. Seymour, and R. Thomas. Directed tree-width. J. Comb.\nTheory, Ser. B, 82(1):138–154, 2001.\n[22] T. Kloks. Treewidth, Computations and Approximations, volume 842 of LNCS. Springer, 1994.\n[23] J. Kneis, D. M ̈olle, S. Richter, and P. Rossmanith. Parameterized power domination complex-\nity. Inf. Process. Lett., 98(4):145–149, 2006.\n[24] G. Kortsarz. On the hardness of approximating spanners. Algorithmica, 30(3):432–450, 2001.\n[25] C. S. Liao and D. T. Lee. Power domination problem in graphs. In Proceedings of the 11th\nInternational Computing and Combinatorics Conference, volume 3595 of LNCS, pages 818–\n828. Springer, 2005.\n[26] L. Lov ́asz.\nOn the ratio of optimal integral and fractional covers.\nDiscrete Mathematics,\n13:383–390, 1975.\n[27] C. Lund and M. Yannakakis. On the hardness of approximating minimization problems. J.\nACM, 41(5):960–981, 1994.\n[28] L. Mili, T.L. Baldwin, and A.G. Phadke.\nPhasor measurements for voltage and transient\nstability monitoring and control. In Proceedings of the EPRI-NSF Workshop on Application\nof Advanced Mathematics to Power Systems, 1991.\n[29] J. Obdrz ́alek. Dag-width: connectivity measure for directed graphs. In Proceedings of the 17th\nAnnual ACM Symposium on Discrete Algorithms, pages 814–821. ACM Press, 2006.\n[30] P. Slav ́ık. A tight analysis of the greedy algorithm for set cover. In Proceedings of the 28th\nAnnual ACM Symposium on Theory of Computing, pages 435–441, New York, NY, USA, 1996.\nACM Press.\n[31] M. D. Stilp. On power dominating sets. Master’s thesis, Combinatorics and Optimization,\nUniversity of Waterloo, Ontario, Canada, 2006.\n21"},{"page":22,"text":"A\nDynamic Programming\nIn this section we describe our dynamic-programming algorithm for the Directed PDS problem.\nThis algorithm is similar to the dynamic-programming algorithm designed by Guo et al.[15] to\noptimally solve PDS for undirected graphs with bounded tree-width. It is known that any tree\ndecomposition of width-k can be transformed to a nice tree decomposition with width k in linear-\ntime [22] (Lemma 13.1.3). So we can assume that we are given a nice tree decomposition of the\nunderlying undirected graph of G call it ⟨{Xi|i ∈I} , T⟩. Let Ti denote the subtree of T rooted\nat T-node i, and Yi =\n S\nj∈V (Ti) Xj\n \n\\ Xi. Also let Gi be the subgraph induced by Yi ∪Xi, i.e.\nGi = G [Yi ∪Xi], and let G′i = G[Xi].\nConsider a valid coloring C of the graph G. We store the color of the edges in each bag by\nassigning a state to that bag (the formal definition of a state will follow). We can reconstruct the\ncoloring C from the states of all bags in the tree decomposition of G; so there is no need to store\nthe coloring C in the dynamic-programming.\nThe state of a bag: Given a coloring C, the state of a bag Xi describes the coloring of the edges\nin G′\ni. In order to detect the dependency cycles in the coloring C without reconstructing the whole\ncoloring, we need to store some more information in a state. This extra information enables us to\ndetect a dependency cycle in Gi which goes through Xi, by considering only the state of the bag\nXi. A bag state s contains the following: state of each edge, state of each node, and state of each\npair of nodes in G′\ni = G[Xi].\n• State of an edge: The state of an edge e ∈E(G[Xi]) denoted by s(e) is the color that is\nassigned to e in the coloring C; s(e) ∈{R, B}.\n• State of a node: The state of a node v ∈Xi denoted by s(v) shows the number of red\nedges between v and Yi.\n– s(v) = 1: There is exactly one red edge from a node in Yi to v and no red edge from v\nto Yi,\n– s(v) = 2: There is exactly one red edge from a node in Yi to v and exactly one red edge\nfrom v to Yi,\n– s(v) = 3: There is no red edge between Yi and v,\n– s(v) = 4: There are at least two red edges from v to Yi and no red edge from Yi to v,\n– s(v) = 5: There is exactly one red edge from v to Yi and no red edge from Yi to v.\n• State of a pair of nodes: A dependency path from u to v is a path P where all red edges\nin P are directed from u to v and all blue edges are directed from v to u. We categorize\ndependency paths according to the color of their first and last edges. There are 4 possible\ntypes RR, RB, BR, BB; for example a path of type RB is a path with the first edge colored\nred and the last edge colored blue. For a pair (u, v) ∈Xi × Xi (u ̸= v) the state of (u, v)\ndenoted by s(u, v) shows the type of dependency paths from u to v in G[Yi ∪{u, v}]; that is,\ns(u, v) ⊆{RR, RB, BR, BB}. Note that there are 24 = 16 different states for each pair of\nnodes.\nDetecting dependency cycles:: An important part of the dynamic-programming algorithm is\nto detect dependency cycles in the coloring C. Assume we are at bag Xi and we are given the bag\nstate s corresponding to the coloring C. We can detect the dependency cycles in G′\ni = G[Xi] by\n22"},{"page":23,"text":"enumerating all possible cycles; note that the coloring of edges in G′\ni is given in the state s. The\ndependency cycles in Gi can be detected by considering the state of each pair of nodes in Xi. For\nexample assume that RB ∈s(u, v) and RR ∈s(v, u). Then, by combining a dependency path of\ntype RB from u to v and a dependency path of type RR from v to u we obtain a dependency cycle\ngoing through u and v in Gi.\nLet us denote by Λi the set of all possible states for the bag Xi. The dynamic-programming will\ncompute a mapping Ai : Λi →N ∪{+∞}. For a bag state s ∈Λi the value Ai(s) is the minimum\nnumber of origins in an optimal valid coloring C of Gi under the restriction that the state of nodes,\nedges, and pairs of nodes in Xi is given by s. Now, we describe how our dynamic-programming\nworks.\nStep 1: (Initialization): In this step for each leaf node i of T, we initialize the mapping Ai as\nfollows. For a given state s, we define Ai(s) as +∞if s has a dependency cycle, a node v with\ns(v) ̸= 3, or a pair of nodes u and v such that s(u, v) ̸= ∅. Otherwise, we define Ai(s) as the\nnumber of nodes with no in-coming red edges in the coloring defined by s.\nStep 2: (Bottom-Up Computation): After initialization, we visit the nodes in T in a bottom-\nup fashion and at each bag Xi we compute the mapping Ai corresponding to Xi. The update\nprocess depends on the type of T-nodes that we are considering. Here, we only consider the update\nprocess at an Insert Node. The other cases are similar to this one.\nInsert Node: Suppose i is an insert node with the child j, and assume that Xi = Xj ∪{x}. For\neach bag state s ∈Λi do the following:\n1. Check whether the coloring given by s forms a valid coloring of G′\ni; if not, define Ai(s) = +∞.\n2. Compute the set Λj(s) containing bag states of j that are “compatible” with the bag state s.\n3. For each s′ ∈Λj(s), check if a valid coloring of Gj “compatible” with s′ can be extended to\na valid coloring of Gi “compatible” with s.\n4. Compute Ai based on the mapping Aj.\nCompatible bag state (Step 2): A bag state s′ ∈Λj is said to be compatible with the bag state\ns ∈Λi if the state of each node, each edge, and each pair of nodes in V (G′\nj) in the bag state s′ is the\nsame as the corresponding state in the bag state s. If s(x) ̸= 3, or ∃v ∈Xj : s(x, v) ̸= ∅∨s(v, x) ̸= ∅\nthen we define Λj(s) = ∅.\nDetecting dependency cycles (Step 3): The conditions of a valid coloring can be violated due\nto degree constraints on the new node x, or by the existence of a dependency cycle going through\nx. Both these cases can be tested by considering the bag states s and s′.\nComputing Ai (Step 4): The addition of x may change the number of origins in Gi. The node x\nwill be an origin if it has at least one outgoing red edge in s. But an origin node v ∈Xj (in s′) that\nhas an incoming red edge from x, is no longer an origin. So by considering the red edges going out\nof x we can update the number of origins and compute the mapping Ai. If the coloring compatible\nwith s is not a valid coloring, then we define Ai(s) to be +∞.\nStep 3: (At root r): Finally, we compute the number of origins in an optimal valid coloring of\nG by finding the minimum of Ar(s) over all possible s ∈Λr. It is easy to see that each bag Xi has\nat most 16(k+1)2 · 5k+1 · 2(k+1)2 states; note that |Xi| ≤(k + 1). It can be checked that the total\nrunning time of our algorithm is O(ck2 · n), for some global constant c.\n23"}]},"section_tree":[],"assets":{"figures":[],"tables":[],"images":[]},"math_expressions":{"inline":[],"block":[{"equation_id":"eq1","equation_number":null,"raw_text":"A dominating set of an (undirected) graph G = (V, E) is a set of nodes S such that every node in","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq2","equation_number":null,"raw_text":"a logarithmic approximation guarantee∗[20], and, modulo the P ̸= NP conjecture, no polynomial","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq3","equation_number":null,"raw_text":"minimum such that PS = V ). We use Opt(G) to denote the size of an optimal solution for the PDS","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq4","equation_number":null,"raw_text":"|V | = 3t. The minimum dominating set has size Θ(|V |), since the maximum degree is 4. The","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq5","equation_number":null,"raw_text":"triangle (like v), then PS = V †.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq6","equation_number":null,"raw_text":"within a ratio of (1 −ǫ) ln n, modulo some variants of the P ̸= NP assumption.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq7","equation_number":null,"raw_text":"Dominating Set problem and PDS modulo a variant of the P ̸= NP conjecture. This seems to be","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq8","equation_number":null,"raw_text":"In the MinRep [24] problem we are given a bipartite graph G = (A, B, E) with a partition of A and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq9","equation_number":null,"raw_text":"respectively. Let A = A1 ∪A2 ∪· · ·∪AqA denote the partition of A, and let B = B1 ∪B2 ∪· · ·∪BqB","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq10","equation_number":null,"raw_text":"denote the partition of B. This partition naturally defines a super bipartite graph H = (A, B, E).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq11","equation_number":null,"raw_text":"instance G = (V , E) of the PDS problem from a given instance G = (A, B, E)(H = (A, B, E)) of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq12","equation_number":null,"raw_text":"an edge ek of Eij). Make λ = 4 new copies of the graph Cij (λ can be any constant greater","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq13","equation_number":null,"raw_text":"than 3; refer to the proof of Lemma 2.3 for more details). For each edge ek = akbk ∈Eij","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq14","equation_number":null,"raw_text":"4. Let G = (V , E) be the obtained graph (see Figure 3 for an illustration).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq15","equation_number":null,"raw_text":"= 4+ |V (G)| + 3λ |E(G)|. This will complete the proof of Theorem 2.2 by showing that the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq16","equation_number":null,"raw_text":"Lemma 2.3 A∗∪B∗is an optimal solution to the instance G = (A, B, E) of the MinRep problem","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq17","equation_number":null,"raw_text":"if and only if S∗= A∗∪B∗∪{w∗} ⊆V (G) is an optimal solution to the instance G of the PDS","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq18","equation_number":null,"raw_text":"S = A∗∪B∗∪{w∗} is a feasible solution to the PDS instance G. Note that all nodes in A ∪B ∪","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq19","equation_number":null,"raw_text":"λ = 4 copies of Cij in G, will be power dominated by applying Rule 1. Then the nodes uk and vk","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq20","equation_number":null,"raw_text":"A′ = A ∩S∗and B′ = B ∩S∗. First we prove that any optimal solution of PDS is contained in","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq21","equation_number":null,"raw_text":"each of the λ = 4 copies of Cij is not completely power dominated. Therefore, the optimal solution","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq22","equation_number":null,"raw_text":"only 1 copy of Cij; in fact G has λ = 4 copies of Cij.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq23","equation_number":null,"raw_text":"Definition 3.1 [11] A tree decomposition of a graph G = (V, E) is a pair ⟨{Xi ⊆V |i ∈I} , T =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq24","equation_number":null,"raw_text":"(I, F)⟩such that T is a tree with V (T) = I, E(T) = F, and satisfying the following properties:","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq25","equation_number":null,"raw_text":"i∈I Xi = V , and every edge uv ∈E has both ends in some Xi,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq26","equation_number":null,"raw_text":"A nice tree decomposition is a tree decomposition ⟨{Xi ⊆V |i ∈I} , T = (I, F)⟩, where T is a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq27","equation_number":null,"raw_text":"then Xi = Xj = Xk (i is called a Join node), and if i has one child j then either Xj ⊂Xi and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq28","equation_number":null,"raw_text":"|Xi \\ Xj| = 1 or Xi ⊂Xj and |Xj \\ Xi| = 1 (i is called an Insert or a Forget node, respectively).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq29","equation_number":null,"raw_text":"For a graph G = (V, E), the neighborhood of R ⊆V is nbr(R) = {v ∈V |∃uv ∈E, u ∈R, v /∈R},","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq30","equation_number":null,"raw_text":"and the exterior of R is defined by ext(R) = nbr(V \\R), i.e., ext(R) consists of the nodes in R that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq31","equation_number":null,"raw_text":"Definition 3.2 Given a graph G = (V, E) and a set S ⊆V , the subset R ⊆V is called an S-strong","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq32","equation_number":null,"raw_text":"of G, we have R ∩(S∗\\ S) ̸= ∅.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq33","equation_number":null,"raw_text":"Now assume that for every feasible solution S∗of G we have R ∩(S∗\\ S) ̸= ∅. Suppose that R","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq34","equation_number":null,"raw_text":"S∗= S ∪(V \\ R) is a feasible solution, but R has no intersection with S∗\\ S ⊆(V \\ R). This is a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq35","equation_number":null,"raw_text":"composition of G and constructs a solution S for PDS (initially, S = ∅). At each node rj of T we","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq36","equation_number":null,"raw_text":"4: for i = d to 0 do","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq37","equation_number":null,"raw_text":"Let Ii = {r1, . . . , rki} and denote by Trj the subtree in T rooted at rj.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq38","equation_number":null,"raw_text":"for j = 1 to ki do","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq39","equation_number":null,"raw_text":"13: Output So = S","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq40","equation_number":null,"raw_text":"G = (V, E) denote the input graph, and let S ⊆V be any set of nodes.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq41","equation_number":null,"raw_text":"Proof: Let Y = ext(Z), it is easy to check that nbr(Z \\ Y ) ⊆ext(Z). We claim that Z \\ Y is an","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq42","equation_number":null,"raw_text":"S-weak region. Let S∗= V \\ (Z ∪S), it is easy to check that S ∪S∗is a feasible solution for the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq43","equation_number":null,"raw_text":"graph G, but S∗∩(Z \\ Y ) = ∅. Hence, by Lemma 3.3, Z \\ Y is not an S-strong region, and so it","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq44","equation_number":null,"raw_text":"is an S-weak region. Thus Z \\ Y ⊆PS∪nbr(Z\\Y ) ⊆PS∪ext(Z) = PS and this implies that Z ⊆PS as","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq45","equation_number":null,"raw_text":"Y = ext(Z) ⊆S.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq46","equation_number":null,"raw_text":"This implies that Z \\Y ⊆PS∪nbr(Z\\Y ) ⊆PS∪nbr(Z)∪ext(Y ) = PS∪nbr(Z). The condition in the lemma","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq47","equation_number":null,"raw_text":"states that Y ⊆PS ⊆PS∪nbr(Z). Hence, we get Z = (Z \\ Y ) ∪(Z ∩Y ) ⊆PS∪nbr(Z), which means","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq48","equation_number":null,"raw_text":"Theorem 3.6 Given a graph G = (V, E) and a tree decomposition of G of width k as input,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq49","equation_number":null,"raw_text":"(j = 1, · · · , m), the set Xq ∩Xcj separates Ycj from the rest of the graph, that is, every path","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq50","equation_number":null,"raw_text":"Thus, ext(Ycj) ⊆Xq ∩Xcj ⊆Xq, and hence, for Yq = Xq ∪Yc1 ∪· · · ∪Ycm, we have ext(Yq) ⊆Xq.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq51","equation_number":null,"raw_text":"Yq. The statement clearly holds when q is a leaf of T (since Yq = Xq). Otherwise, let c1, . . . , cm","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq52","equation_number":null,"raw_text":"be the children of q in T. For each j = 1, . . . , m, when the algorithm examined Ycj, either Ycj was","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq53","equation_number":null,"raw_text":"the solution just before the algorithm examines Ycj. Hence, Yq = Yc1 ∪· · · ∪Ycm ∪Xq ⊆PS∗∪Xq.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq54","equation_number":null,"raw_text":"(i) Yr is S-strong, so So = S ∪Xr, and Yr ⊆PS∪Xr = PSo; or","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq55","equation_number":null,"raw_text":"(ii) Yr is S-weak, and Yr ⊆PS∪ext(Yr) = PSo; since Yr = V (G) and ext(Yr) = ∅.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq56","equation_number":null,"raw_text":"let S be the solution just before the algorithm examines q. Then define Rl= Yql\\(Yq1 ∪· · ·∪Yql−1),","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq57","equation_number":null,"raw_text":"where ql= q. We claim that Rlis an S-strong region. For each strong region Yqj (j = 1, . . . , l−1) we","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq58","equation_number":null,"raw_text":"to the solution, otherwise we skip v. Formally, we define Xv = {v} for each v ∈V (G), and we run","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq59","equation_number":null,"raw_text":"S = S ∪nbr(S). Obviously this depends on the order of applying the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq60","equation_number":null,"raw_text":"S = PS we have:","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq61","equation_number":null,"raw_text":"S = {u}. Also since we power dominate u at step (i+1), we have nbr(v) ⊆Pi+1","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq62","equation_number":null,"raw_text":"+ |{u}| =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq63","equation_number":null,"raw_text":"Proposition 3.12 Let G be an l× m grid with l≤m, then Opt(G) = Θ(l).","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq64","equation_number":null,"raw_text":"and PS = V (G). The maximum degree in G is 4, so we have","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq65","equation_number":null,"raw_text":"5 · 5 = l −1. Therefore, P0","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq66","equation_number":null,"raw_text":"Let G = (V, E) be a directed graph.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq67","equation_number":null,"raw_text":"We say that S power dominates G if PS = V (G). The Directed PDS problem is to find a node","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq68","equation_number":null,"raw_text":"The reduction: We create an instance G = (V , E) of the Directed PDS problem from a given","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq69","equation_number":null,"raw_text":"instance G = (A, B, E)(H = (A, B, E)) of the MinRep problem.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq70","equation_number":null,"raw_text":"called the center node, in Dij. Make λ = 4 new copies of the graph Dij (λ can be any","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq71","equation_number":null,"raw_text":"constant greater than 3). For each edge ek = akbk ∈Eij and for each of the 4 copies of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq72","equation_number":null,"raw_text":"4. Let G = (V , E) be the obtained graph.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq73","equation_number":null,"raw_text":"Lemma 4.3 A∗∪B∗is an optimal solution to the instance G = (A, B, E) of the MinRep problem","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq74","equation_number":null,"raw_text":"if and only if S∗= A∗∪B∗∪{w∗} ⊆V (G) is an optimal solution to the instance G of the Directed","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq75","equation_number":null,"raw_text":"S = A∗∪B∗∪{w∗} is a feasible solution to the Directed PDS instance G. Note that all nodes","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq76","equation_number":null,"raw_text":"in S their out-neighbors uk and vk in each of the λ = 4 copies of the Dij graph will be power","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq77","equation_number":null,"raw_text":"any feasible solution for Directed PDS. Now define A′ = A∩S∗and B′ = B ∩S∗. First we prove","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq78","equation_number":null,"raw_text":"Let Dij be such a gadget. By symmetry each of the λ = 4 copies of Dij is not completely power","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq79","equation_number":null,"raw_text":"edge ek = akbk ∈Eij, we can power dominate all 4 copies of Dij. This contradicts the optimality","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq80","equation_number":null,"raw_text":"Definition 4.4 A coloring of a directed graph G = (V, E) is a partitioning of the edges in G into","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq81","equation_number":null,"raw_text":"red and blue edges. We denote a coloring by C = (V, Er ∪Eb) where Er is the set of red edges and","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq82","equation_number":null,"raw_text":"Definition 4.5 A valid coloring C = (V, Er ∪Eb) of a directed graph G = (V, E) is a coloring of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq83","equation_number":null,"raw_text":"2. The subgraph induced by the red edges, Gr = (V, Er), has the following properties:","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq84","equation_number":null,"raw_text":"Gr(v) = 1 =⇒d+","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq85","equation_number":null,"raw_text":"By way of contradiction, suppose that C∗= u1, u2, . . . , um is a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq86","equation_number":null,"raw_text":"are red. Then the red edges (ui, ui+1) imply that ui < ui+1 for all i = 1, 2, . . . , m −1; therefore","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq87","equation_number":null,"raw_text":"Now suppose that G has a valid coloring C = (V, Er ∪Eb) with S ⊆V (G) as the set of origins.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq88","equation_number":null,"raw_text":"The nodes in S and all of their out-neighbors in Gr = (V, Er) are power dominated by applying","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq89","equation_number":null,"raw_text":"e1 = (x1, y1), . . . , ek = (xk, yk) be all of the red edges from X to V \\X. If some xi has all of its out-","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq90","equation_number":null,"raw_text":"one of the red edges (x1, y1), · · · , (xk, yk). If zi = x1, then we have a dependency cycle; otherwise,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq91","equation_number":null,"raw_text":"we have X = V , so S power dominates G.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq92","equation_number":null,"raw_text":"number of nodes. Note that in the algorithm we only consider 2log n = n different candidates.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq93","equation_number":null,"raw_text":"the graph G′. It can be checked that Opt(G) = 1 but Opt(G′) = Θ(n). Thus the bidimensionality","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq94","equation_number":null,"raw_text":"complexity assumptions like P ̸= NP. The most natural greedy algorithm for PDS is the one that","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq95","equation_number":null,"raw_text":"starts with S = ∅, and in each step, adds a new node v to the current solution S such that v power","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq96","equation_number":null,"raw_text":"nodes until it finds a feasible solution S. The size of the output S is at least m · l= Θ(n). By","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq97","equation_number":null,"raw_text":"Proposition 3.12, we have Opt(G) = Θ(l). Now by fixing l= Θ(1) we can see that the size of the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq98","equation_number":null,"raw_text":"corner edges. Figure 8 illustrates an example of such a grid for l = 5 and h = 9 rows, but for a","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq99","equation_number":null,"raw_text":"bad example for the proximity greedy algorithm we need h to be sufficiently large constant (h = 17","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq100","equation_number":null,"raw_text":"that power dominates maximum number of nodes (which is 17 = 2l+ 7); any white node satisfies","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq101","equation_number":null,"raw_text":"maximum number of new nodes (which is 16 = 2l+ 6). The algorithm continues picking all white","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq102","equation_number":null,"raw_text":"nodes.) Therefore, the size of the solution found by the algorithm is at least m = Θ(n). Hence, the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq103","equation_number":null,"raw_text":"dominate maximum number of new nodes (which is 17 = 2l+ 7). Note that in the original greedy","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq104","equation_number":null,"raw_text":"(which is at least 15 = 2l+ 5). At this stage of the algorithm the set of power dominated nodes","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq105","equation_number":null,"raw_text":"2 = Θ(n), but the optimal solution is just Θ(1) as before.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq106","equation_number":null,"raw_text":"at T-node i, and Yi =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq107","equation_number":null,"raw_text":"Gi = G [Yi ∪Xi], and let G′i = G[Xi].","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq108","equation_number":null,"raw_text":"i = G[Xi].","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq109","equation_number":null,"raw_text":"– s(v) = 1: There is exactly one red edge from a node in Yi to v and no red edge from v","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq110","equation_number":null,"raw_text":"– s(v) = 2: There is exactly one red edge from a node in Yi to v and exactly one red edge","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq111","equation_number":null,"raw_text":"– s(v) = 3: There is no red edge between Yi and v,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq112","equation_number":null,"raw_text":"– s(v) = 4: There are at least two red edges from v to Yi and no red edge from Yi to v,","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq113","equation_number":null,"raw_text":"– s(v) = 5: There is exactly one red edge from v to Yi and no red edge from Yi to v.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq114","equation_number":null,"raw_text":"red and the last edge colored blue. For a pair (u, v) ∈Xi × Xi (u ̸= v) the state of (u, v)","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq115","equation_number":null,"raw_text":"s(u, v) ⊆{RR, RB, BR, BB}. Note that there are 24 = 16 different states for each pair of","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq116","equation_number":null,"raw_text":"i = G[Xi] by","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq117","equation_number":null,"raw_text":"s(v) ̸= 3, or a pair of nodes u and v such that s(u, v) ̸= ∅. Otherwise, we define Ai(s) as the","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq118","equation_number":null,"raw_text":"Insert Node: Suppose i is an insert node with the child j, and assume that Xi = Xj ∪{x}. For","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq119","equation_number":null,"raw_text":"i; if not, define Ai(s) = +∞.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq120","equation_number":null,"raw_text":"same as the corresponding state in the bag state s. If s(x) ̸= 3, or ∃v ∈Xj : s(x, v) ̸= ∅∨s(v, x) ̸= ∅","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq121","equation_number":null,"raw_text":"then we define Λj(s) = ∅.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"}],"equation_refs":[]},"symbol_table":[],"references":{"in_text_markers":[],"bibliography":[]},"claims":[],"flow_graph":{"nodes":[],"edges":[]},"quality":{"text_char_count":69272,"parse_confidence":0.5,"equation_parse_rate_proxy":1.0,"citation_resolution_rate_proxy":0.0,"structure_coverage_proxy":0.0,"asset_coverage_proxy":0.0,"accepted_for_training":true}}