Fixed-Parameter Tractability Results for Full-Degree Spanning Tree and Its Dual

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1 Fixed-Parameter Tractability Results for Full-Degree Spanning Tree and Its Dual Jiong Guo, Rolf Niedermeier, and Sebastian Wernicke Institut für Informatik, Friedrich-Schiller-Uniersität Jena Ernst-Abbe-Platz 2, D Jena, Germany Abstract. We proide first-time fixed-parameter tractability results for the NP-complete problems Maximum Full-Degree Spanning Tree and Minimum-Vertex Feedback Edge Set. These problems are dual to each other: In Maximum Full-Degree Spanning Tree, the task is to find a spanning tree for a gien graph that maximizes the number of ertices that presere their degree. For Minimum-Vertex Feedback Edge Set the task is to minimize the number of ertices that end up ith a reduced degree. Parameterized by the solution size, e exhibit that Minimum-Vertex Feedback Edge Set is fixed-parameter tractable and has a problem kernel ith the number of ertices linearly depending on the parameter k. Our main contribution for Maximum Full-Degree Spanning Tree, hich is W[1]-hard, is a linear-size problem kernel hen restricted to planar graphs. Moreoer, e present subexponential-time algorithms in the case of planar graphs. 1 Introduction The NP-complete Maximum Full-Degree Spanning Tree (FDST) problem is defined as follos. Input: An undirected graph G = (V, E) and an integer k 0. Task: Find a spanning tree T of G (called solution tree) in hich at least k ertices hae the same degree as in G. Referring to ertices that maintain their degree as full-degree ertices and to the remaining ones as reduced-degree ertices, the task basically is to maximize the number of full-degree ertices. FDST is motiated by applications in ater distribution and electrical netorks [3, 4, 13]. It is a notoriously hard problem and, as such, not polynomial-time approximable ithin a factor of O(n 1/2 ɛ ) for any ɛ > 0 unless NP-complete problems hae randomized polynomial-time algorithms [3]. The approximability bound is almost tight in that Bhatia et al. [3] proide an algorithm ith an approximation ratio of Θ(n 1/2 ). FDST remains Supported by the Deutsche Forschungsgemeinschaft, Emmy Noether research group PIAF (fixed-parameter algorithms), NI 369/4. Supported by the Deutsche Telekom Stiftung.

2 NP-complete in planar graphs; hoeer, polynomial-time approximation schemes (PTAS) are knon here [3, 4]. Broersma et al. [4] present further tractability and intractability results for arious special graph classes. By ay of contrast, the parameterized complexity [7, 10, 14] of FDST has so far been unexplored. The complement (dual) problem of FDST, called Minimum-Vertex Feedback Edge Set (VFES), is to find a feedback edge set 1 in a gien graph such that this set is incident to as fe ertices as possible. In other ords, e ant to minimize the number of reduced-degree ertices. VFES is motiated by an application of placing pressure-meters in fluid netorks [12, 15, 16]. Khuller et al. [12] sho that VFES is APX-hard and present a polynomial-time approximation algorithm ith ratio (2 + ɛ) for any fixed ɛ > 0. Moreoer, they deelop a PTAS for VFES in planar graphs. As ith FDST, the parameterized complexity of VFES has so far not been inestigated. Parameterized by the respectie solution size k, this ork proides firsttime parameterized complexity results for FDST and its dual VFES. Somehat analogously to the study of approximability properties, e obsere that FDST seems to be the harder problem hen compared to its dual VFES: Whereas VFES is fixed-parameter tractable, FDST is W[1]-hard. 2 More specifically, our findings are as follos: VFES has a problem kernel ith less than 4k ertices and it can be soled in O(4 k k 2 + m) time for an m-edges graph. FDST becomes fixed-parameter tractable in the case of planar graphs. In particular, as the main technical contribution of this paper, e proe a linearsize problem kernel for FDST hen restricted to planar graphs. For planar graphs, both VFES and FDST are solable in subexponential time. More specifically, in n-ertices planar graphs VFES is solable in O(2 O( k log k) +k 5 +n) time and FDST in O(2 O( k log k) +k 5 +n 3 ) time. Herein, e make use of tree decomposition-based dynamic programming. We remark that, hen restricted to planar graphs, this ork amends the so far fe examples here both a problem and its dual possess linear-size problem kernels. Other examples e are aare of (again restricted to planar graphs) are Vertex Coer and its dual Independent Set and Dominating Set and its dual Nonblocker [1, 5, 6]. We omit most proofs in this extended abstract due to a lack of space. 2 Minimum-Vertex Feedback Edge Set Minimum-Vertex Feedback Edge Set (VFES), the dual of Maximum Full-Degree Spanning Tree (FDST), appears to be better tractable from a parameterized point of ie than FDST. We subsequently present a simple problem kernelization ith a kernel graph hose number of ertices linearly depends 1 A feedback edge set is a set of edges hose deletion destroys all cycles in a graph. 2 The reduction is from Independent Set and the same as the one in [3,12]. 2

3 ertices in V =1 ertices in V X =2 ertex in V G =2 ertices in V 3 Fig.1. Gien a spanning tree for a graph (bold edges), the proof of the linear kernel for VFES in Theorem 1 partitions them into four disjoint subsets as illustrated aboe (see proof for details). on the parameter (Sect. 2.1) and, based on this, an efficient fixed-parameter algorithm (Sect. 2.2) for VFES. 2.1 Data Reduction and Problem Kernel To reduce a gien instance of VFES to a problem kernel, e make use of to ery simple data reduction rules already used by Khuller et al. [12]. Rule 1. Remoe all degree-one ertices. Rule 2. For any to neighboring degree-to ertices that do not hae a common neighbor, contract the edge beteen them. The correctness and the linear running time of these to rules are easy to erify; e call an instance of VFES reduced if the reduction rules cannot be applied any more. Theorem 1. A reduced instance (G = (V, E), k) of VFES only has a solution tree if V < 4k. Proof. Assume that G has a solution tree T. Let X denote the set of the reduceddegree ertices in T. We partition the ertices in V into three disjoint subsets according to their degree in T, namely V =1 contains all degree-1 ertices, V =2 contains all degree-2 ertices, and V 3 contains all ertices of degree higher than 2. Furthermore, let V=2 X := V =2 X and V=2 G := V =2 \ V=2. X The partition is illustrated in Figure 1. Since G does not contain any degree-1 ertices by Rule 1, eery degree-1 ertex in T is a reduced-degree ertex. Hence, T can hae at most k leaes and V =1 X k. Since T is a tree, this directly implies V 3 k 2. As for V =2, the ertices in V =1 V 3 are either directly connected to each other or ia a path P consisting of ertices from V =2. Because Rule 2 contracts edges beteen to degree-2 ertices that hae no common neighbor in the input graph, at least one of eery to neighboring ertices of P has to be a reduceddegree ertex. Clearly, V=2 X V 1 X. Since T is a tree, it follos that V=2 G V =1 V 3 V=2 X 1 2k 3. Oerall, this shos that V = V =1 V=2 X V =2 G V 3 < 4k as claimed. 3

4 2.2 A Fixed-Parameter Algorithm The problem kernel obtained in Theorem 1 suggests a simple fixed-parameter algorithm for Minimum-Vertex Feedback Edge Set: For i = 1,...,k, e consider all ( ) 4k i size-i subsets X of kernel ertices. For each of these subsets, all edges beteen the ertices in X are remoed from the input graph, that is, they become reduced-degree ertices. If the remaining graph is a forest, e hae found a solution. 3 The correctness of this algorithm is obious by its exhaustie nature, but on the running-time side it requires the consideration of k ( 4k ) i=1 i > 9.45 k ertex subsets. The next theorem shos that e can do better because an exhaustie approach does not need to consider all ertices of the kernel but only those ith degree at least three. Theorem 2. For an m-edge instance (G = (V, E), k) of VFES, it can be decided in O(4 k k 2 + m) time hether it has a solution tree. Proof. Gien an instance (G = (V, E), k) of VFES, e first perform the kernelization hich needs O(m) time. By Theorem 1, e kno that the remaining graph only has a solution tree if it contains feer than 4k ertices. We no partition the ertices in V according to their degree: ertices in V =2 hae degree to and V 3 contains all ertices ith degree at least three. For eery size-i subset X 3 V 3, 1 i k, the folloing to steps are performed: Step 1. Remoe all edges beteen ertices in X 3. Call the resulting graph G = (V, E ). Step 2. For each edge e E, assign it a eight (e) = m + 1 if it is incident to a ertex in V 3 \X 3 and a eight of 1, otherise. Then, find a maximum-eight spanning tree for eery connected component of G. If the total eight of edges that are not in a spanning tree is at most k i, then the original VFES instance (G = (V, E), k) has a solution tree and the algorithm terminates; otherise, the next subset X 3 is tried. To justify Step 2, obsere that the edges incident to ertices in V 3 \X 3 hae to be in the solution tree because these ertices presere their degree. Therefore, if a component in G contains cycles e can only destroy these by remoing edges beteen a ertex in X 3 and a ertex in V =2. Moreoer, remoing such an edge results in exactly one additional reduced-degree ertex. Thus, searching for a maximum-eight spanning tree in the second step leads to a solution tree (the components can easily be reconnected by edges beteen ertices in X 3 ). The running time of the algorithm is composed of the linear preprocessing time, the number of subsets that need to be considered, and the time needed for Steps 1 and 2. By Theorem 1, a reduced graph contains less than 4k ertices and, hence, O(k 2 ) edges, upper-bounding the time needed for Steps 1 and 2 by O(k 2 ). From the proof of Theorem 1, e kno that V 3 2k, and hence the oerall running time is bounded aboe by O(m + ) 1 i k k 2 ) = O(4 k k 2 + m). 3 Since the input graph must be connected, e can add some edges from G that are beteen the ertices in X in order to reconnect connected components ith each other and thus obtain a spanning tree. ( 2k i 4

5 3 Maximum Full-Degree Spanning Tree in Planar Graphs The reduction from Independent Set to Maximum Full-Degree Spanning Tree (FDST) that is gien in [3] already shos that FDST is W[1]-hard ith respect to the number k of full-degree ertices. 4 For planar graphs, hoeer, this does not hold; in this section, e sho that it is een possible to achiee a linear-size problem kernel for FDST in planar graphs. While the proof of the linear size upper bound is ery inoled and technical, the actual computation of the problem kernel is based on seeral straightforard data reduction rules on a gien instance of FDST. 5 Reduction Rules. Let N() denote the neighborhood of a ertex. For to ertices ith N() N() 3, perform the folloing reductions: 1. Let,, be three ertices in N() N(). 1.1 If N( ) = {, } and additionally either N( ) = {, } or N( ) = {,, }, then remoe. 1.2 If N( ) = {,, }, N( ) = {,,, }, and N( ) = {,, }, then remoe. 2. Let,,, u 4 be four ertices in N() N(). 2.1 If N( ) = {,, }, N( ) = {,, }, N( ) = {u 4,, }, and N(u 4 ) = {,, }, then remoe and u If N( ) = {,,, }, N( ) = {, u 4,, }, and there is no edge {, u 4 }, then remoe the edge {, }. 3. Let,,, u 4, u 5 be fie ertices in N() N(). If N( ) = {,,, }, N( ) = {,, }, N(u 4 ) = {u 5,, }, and N(u 5 ) = {u 4,, }, then remoe. The fie subcases are illustrated in Figure 2. Omitting a formal proof here, exhaustiely applying these rules yields a graph that has a spanning tree ith k full-degree ertices iff the original graph has a spanning tree ith k full-degree ertices, that is, the reduction rules are correct. As e shall sho in the remainder of this section, a planar graph that is reduced ith respect to the reduction rules is a linear-size problem kernel for Maximum Full-Degree Spanning Tree in planar graphs: Theorem 3. For a gien planar n-ertex graph G, let k be the maximum number of full-degree ertices in any spanning tree for G. If G is reduced ith respect to the gien reduction rules, then n = O(k), that is, e hae a linear-size problem kernel for FDST on G that can be computed in O(n 3 ) time. The proof of this theorem is quite inoled. We basically achiee it by contradiction, that is, e assume that e are gien an optimal solution tree to FDST 4 W[1]-hardness stands for (presumable) parameterized intractability. Refer to [7, 10, 14] for details. 5 Note that the reduction rule applies to an arbitrary graph but the linear-size kernel performance-guarantee is only shon for planar graphs. 5

6 Case 1.1 u 4 Case 1.2 Case 2.1 u 4 u 4 u 5 Case 2.2 Case 3 u 4 u 4 u 5 Fig.2. The fie subcases of the reduction rule for planar FDST that yields a linear-size problem kernel. on G and sho that either V = O(k) must hold or this optimality is contradicted. 6 Throughout this section, e denote the set of full-degree ertices in the optimal solution tree by F (note that F = k). To this end, the folloing lemma is ery helpful. Lemma 4. Eery ertex in G has distance at most 2 to a ertex in F. Using this lemma, our strategy to proe Theorem 3 is for eery ertex in F to partition the set of ertices that hae distance at most 2 to it into to sets and separately upper-bound their size. More specifically, Section 3.1 introduces the concept of region decompositions hich use the set F to diide the input graph into small areas. 7 In Section 3.2, it is shon that the number of regions in a region decomposition is bounded aboe by O( F ) (Lemma 9) and that in the reduced graph, eery region contains only a constant number of ertices (Lemma 13). Oerall, this shos that there are at most O( F ) ertices lying inside regions (Proposition 14). Essentially folloing the same strategy in Section 3.3, e upper-bound the number of ertices that do not lie inside regions by O( F ) (Proposition 18). The linear kernel claimed in Theorem 3 directly follos from the O( F ) upper bounds on the number of ertices in regions and outside regions. 3.1 Neighborhood Partition and Region Decomposition This section prepares the proof of the size-o( F ) upper bound of a reduced graph by introducing to partitions of the ertices in V \ F. One partition is 6 This sort of proof strategy has first been used in ork dealing ith the Max Leaf Spanning Tree problem [8, 9]. 7 The proof concept is, in this sense, similar to the one for the linear-size problem kernel for Dominating Set in planar graphs due to Alber et al. [1]. 6

7 local in that for eery ertex in F, it concerns ertices ithin distance at most 2 to it; e partition this 2-neighborhood into so-called exit ertices and prison ertices. The other partition, called region decomposition, is somehat more global in that it concerns the union of 2-neighborhoods for some pairs of ertices in F. These partitions are subsequently used in Sections 3.2 and 3.3 to sho the desired linear size of a reduced planar graph. As to the notation used, for a set V V the subgraph of G induced by V is denoted by G[V ] = (V, E ). For a ertex V, e denote the ertices haing distance 2 to by N 2 (). By N 1,2 (), e denote N() N 2 (). Furthermore, e set N 1 () := N() \ F and let N 2 () denote all ertices that hae distance exactly 2 to in the induced graph G[(V \F) {}]. Finally, N 1,2 () := N 1 () N 2 (). This section and the folloing alays consider some fixed embedding of G in the plane (hence, e call G plane instead of planar). As already mentioned, Lemma 4 shos that eery ertex in G has distance at most 2 to a full-degree ertex. Thus, if e can upper-bound the number of ertices in F N1,2 () by O( F ), the linear problem kernel follos as claimed. In order to do this, e partition the ertices in N 1,2 () into to subsets, separately upper-bounding their sizes in Sections 3.2 and 3.3: N 1,2 exit () := {u N1,2 () u N() for a ertex / N 1,2 ()}, N 1,2 prison () := N1,2 () \ N 1,2 exit (). For a subset V V, e use N 1,2 exit (V ) to denote V N1,2 exit (). The intuition of the partition is that the exit ertices hae edges to ertices that lie outside of N 1,2 () hereas the prison ertices hae no edge to a ertex different from N 1,2 (). 8 As an example, the folloing illustration shos the partition of a neighborhood N 1,2 () into N 1,2 exit () and N1,2 prison (): F N 1,2 exit N 1,2 prison As it turns out, eery ertex in N 1,2 exit () is caught beteen to ertices in F, that is, for these ertices a stronger ariant of Lemma 4 holds. Lemma 5. Gien a ertex F, eery ertex u N 1,2 exit () lies on a length-atmost-fie path beteen and an other ertex F (here ). Lemma 5 can be used to eentually upper-bound the size of N 1,2 exit (F) because the length-at-most-fie paths form bounded areas beteen ertices from F called regions. 8 See [1] for a similar notion in the context of Dominating Set. 7

8 Definition 6. A region R(, ) beteen to ertices, F is a closed subset of the plane ith the folloing properties: 1. The boundary of R(, ) is formed by to paths beteen and ; the length of each path is at most fie. 2. All ertices hich lie strictly inside of R(, ) that is, they do not lie on the boundary are from N 1,2 () N 1,2 (). (Obsere that no ertex from F lies strictly inside of a region.) To denote the ertices that lie in R(, ), e use V (R(, )), that is, V (R(, )) is the union of the boundary ertices and the ertices lying strictly inside of R(, ). Using this definition, a plane graph can be partitioned into a number of regions by a so-called region decomposition. Definition 7. An F-region decomposition of G is a set R of regions R(, ) ith, F such that the folloing holds: 1. Except for and, no ertex from V (R(, )) belongs to F. 2. There is no ertex that lies strictly inside of more than one region from R. (The boundaries of regions may touch each other.) For an F-region decomposition R, e let V (R) := R R V (R). An F-region decomposition R is called maximal if there is no region R / R such that R := R {R} is an F-region decomposition ith V (R) V (R ). An example of a (maximal) F-region decomposition is the folloing (the fulldegree ertices in F are colored black, the shaded areas are the regions of the decomposition): Our notions of region and region decomposition are similar to the technique used by Alber et al. [1] for proing a linear-size kernel for Dominating Set in planar graphs. Hoeer, the problem structure of FDST makes the proof somehat more inoled: Whereas Alber et al. ere able to bound their regions by length-3 paths, the FDST problem appears to require longer bounds (that is, length-5 paths in our proofs). The reason for this is that in Dominating Set, a ertex affects only its direct neighborhood hereas the full-degree property affects ertices in the 2-neighborhood. The folloing lemma shos that it is possible to construct a maximal F- region decomposition R satisfying N 1,2 exit (F) V (R). By subsequently bounding 8

9 the number of ertices that lie in regions in Section 3.2, this allos us to upperbound the number of ertices in N 1,2 exit (F) so that e only hae to deal ith ertices from N 1,2 prison (F) in Section 3.3. Lemma 8. For a plane graph G = (V, E), there exists a maximal F-region decomposition R of G such that N 1,2 exit (F) V (R). 3.2 Bounding the Number of Vertices in Regions In this section, e sho that the F-region decomposition R from Lemma 8 has O( F ) ertices lying inside of regions (Proposition 14). The proof of this is achieed in seeral steps: First, Lemma 9 shos that the total number of regions is O( F ). Then, Proposition 12 upper-bounds the number of length-2 paths that can occur beteen to ertices in the reduced graph; this proposition is heaily used to proe that eery region contains at most O(1) ertices in Lemma 13. Finally, Proposition 14 follos from Lemmas 9 and 13. Lemma 9. For F 3, the maximal F-region decomposition R in Lemma 8 consists of at most 6 F 12 regions. To sho the constant number of ertices in each region, e make use of the folloing structure: Definition 10. Let and be to distinct ertices in a plane graph G. A diamond D(, ) is a closed area of the plane that is bounded by to length-2 paths beteen and such that eery ertex that lies inside of the closed area is a neighbor of both and. If i ertices lie strictly inside of a diamond, then the diamond is said to hae (i + 1) facets (a facet is an area enclosed by to length-2 paths). Lemma 11. A reduced plane graph does not contain a diamond ith more than 5 facets. The only to possible 5-facet diamonds (the orst-case diamonds, so to say) that might remain after the data reduction are the folloing: The absence of diamonds ith too many facets is central to shoing the linear size of the problem kernel for FDST: At most 2 ertices of any diamond can be full-degree, and hence the possibility of arbitrarily large diamonds ould prohibit a proable upper bound of the reduced graph. For the same reason, it is important that there cannot be arbitrarily many length-2 paths beteen to ertices of the input graph. Therefore, Lemma 11 is generalized in order to obtain a upper bound on the number of length-2 paths beteen to ertices. 9

10 Proposition 12. Let and be to ertices in a reduced plane graph G such that an area A(, ) of the plane is enclosed by to length-2 paths beteen and. If neither the middle ertices of the enclosing paths nor any ertex inside of the area are contained in F, then the folloing holds: 1. If, F, at most eight length-2 paths from to lie inside of A(, ). 2. If F or F, at most sixteen length-2 paths from to lie inside of A(, ). Using the upper bound on the number of length-2 paths beteen to ertices that Proposition 12 establishes, it is possible to bound aboe the number of ertices that can lie inside of a region. Lemma 13. Eery region R(, ) R contains O(1) ertices. In conjunction ith the O( F ) upper bound on the number of regions that as established in Lemma 9, this directly allos us to bound aboe the number of ertices that lie inside of regions. Proposition 14. The total number of ertices lying inside of regions of R is O( F ). 3.3 Bounding the Number of Vertices Outside of Regions In the last section, e hae bounded aboe the number of ertices that lie inside of regions of a maximal F-region decomposition R. It remains to bound aboe the ertices that do not lie in regions. By Lemma 8, eery one of these remaining ertices is in N 1,2 prison () for some ertex F. To bound the number of these ertices by O( F ), e essentially follo the same strategy as in the last section: We sho that each ertex in N 1,2 prison () lies in one of O( F ) so-called prison areas and that each such prison area contains a constant number of ertices. Definition 15. Gien a maximal F-region decomposition, a prison area for a ertex F is a closed area of the plane ith the folloing properties: 1. All ertices that lie strictly inside of the area are from N 1,2 prison (). 2. The area cannot be extended to include any ertex from N 1,2 prison () ithout iolating the first condition. Analogously to the preceding section, e first sho that the number of prison areas is upper-bounded by O( F ) and then sho that each area contains a constant number of ertices. Lemma 16. Gien a maximal F-region decomposition R, the ertices that do not lie inside of any region of R form at most 12 F 24 prison areas (again, e assume F 3). Lemma 17. Eery prison area contains O(1) ertices. Proposition 18. The number of ertices lying outside of regions of R is bounded aboe by O( F ). 10

11 4 A Tree Decomposition-Based Algorithm There exists a tree decomposition-based algorithm that soles both Minimum- Vertex Feedback Edge Set and Maximum Full-Degree Spanning Tree in O((2ω) 3ω ω n) time on an n-ertex graph ith a gien tree decomposition of idth ω. 9 We omit its description here due to lack of space. Theorem 19. For an n-ertex graph G ith a gien idth-ω tree decomposition (T, X), both Minimum-Vertex Feedback Edge Set and Maximum Full-Degree Spanning Tree can be soled in O((2ω) 3ω ω n) time. Using the linear problem kernels for VFES and FDST that ere established in Theorems 1 and 3, Theorem 19 implies subexponential-time algorithms for these problems hen restricted to planar graphs: Theorem 20. In n-ertex planar graphs, Minimum-Vertex Feedback Edge Set is solable in O(2 O( k log k) + k 5 + n) time and Maximum Full-Degree Spanning Tree in O(2 O( k log k) + k 5 + n 3 ) time, here k is the number of degree-reduced ertices or full-degree ertices, respectiely. Proof. Both problems hae linear-size kernels in planar graphs, that is, the reduced graphs hae at most O(k) ertices after performing the respectie kernelization. The claim follos from Theorem 19 and the facts that planar O(k)-ertex graphs hae treeidth ω = O( k) and that a corresponding tree decomposition of idth 3ω/2 can be computed in O(k 5 ) time [17]. 5 Conclusion Our main technical contribution is the proof of the linear-size problem kernel for FDST in planar graphs. It is easily conceiable that there is room for significantly improing the inoled orst-case constant factors the situation is comparable (but perhaps een more technical) ith analogous results for Dominating Set in planar graphs [1, 5]. Haing obtained linear problem kernel sizes for a problem and its dual, the ay for applying the loer bound technique for kernel size due to Chen et al. [5] no seems open. Another interesting line of future research might be to inestigate hether our problem kernel result for planar graphs can be lifted to superclasses of planar graphs corresponding results for Dominating Set in this direction are reported in [11]. Perhaps een more importantly, it ould be interesting to pursue experimental studies (similar to Bhatia et al. [3]) ith real-orld data in order to explore the practical usefulness of our algorithms and problem kernelizations. 9 Broersma et al. [4] hae already shon that both VFES and FDST are solable in linear time for graphs ith bounded treeidth by expressing the problems in monadic second-order logic. The linear-time solability follos then from a result of Arnborg et al. [2]. Hoeer, this approach does not proide an exact dependency of the running time on the treeidth. Therefore, their result cannot imply a subexponential-time algorithm. 11

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