On the Euclidean Bottleneck Full Steiner Tree Problem

Transcription

1 On the Euclidean Bottleneck Full Steiner Tree Problem A. Karim Abu-Affash Abstract Given two sets in the plane, R of n (terminal) points and S of m (Steiner) points, a full Steiner tree is a Steiner tree in which all points of R are leaves. In the bottleneck full Steiner tree (BFST) problem, one has to find a full Steiner tree T (with any number of Steiner points from S), such that the length of the longest edge in T is minimized, and, in the k-bfst problem, has to find a full Steiner tree T with at most k m Steiner points from S such that the length of the longest edge in T is minimized. The problems are motivated by wireless network design. In this paper, we present an exact algorithm of O((n + m) log 2 m) time to solve the BFST problem. Moreover, we show that the k-bfst problem is NP-hard and that there exists a polynomial-time approximation algorithm for the problem with performance ratio 4. 1 Introduction Given a graph G = (V, E) with a weight function w : E R + and a subset R V of vertices, a Steiner tree is an acyclic subgraph of G spanning all vertices of R. The vertices of R are usually referred to as terminals and the vertices of V \ R as Steiner vertices. The Steiner tree (ST) problem is to find a Steiner tree T such the total weight of edges of T is minimized. This problem has been shown to be NP-complete [2, 13], even in the Euclidean or rectilinear metrics [14]. Arora [1] showed that the Euclidean and rectilinear ST problem can be efficiently approximated arbitrarily close to optimal. For arbitrary weighted graphs, many approximation algorithms have been proposed [4, 5, 15]. The bottleneck Steiner tree (BST) problem is to find a Steiner tree T such that the bottleneck (i.e., the length of the longest edge) of T is minimized. Unlike the ST problem, the BST problem can be solved exactly in polynomial time [12,23]. Both the ST and BST problems have many important applications in VLSI design, network communication and computational biology [7, 11, 16, 17]. Department of Computer Science, Ben-Gurion University of the Negev, Beer-Sheva 84105, Israel, abuaffas@cs.bgu.ac.il. Partially supported by the Lynn and William Frankel Center for Computer Sciences, by the Robert H. Arnow Center for Bedouin Studies and Development, by a fellowship for outstanding doctoral students from the Planning & Budgeting Committee of the Israel Council for Higher Education, and by a scholarship for advanced studies from the Israel Ministry of Science and Technology. 1

2 Clearly, all leaves of a Steiner tree are vertices of R, but not all vertices of R need to be leaves in a Steiner tree. However, in some applications, as for example in the VLSI global routing [19] and in reconstruction of evolutionary trees in biology [20], all vertices of R must be leaves of the Steiner tree. A Steiner tree is full if all terminals are leaves of the Steiner tree [20]. The full Steiner tree (FST) problem is to find a full Steiner tree with minimum total length of edges. In [19], hardness results were presented for this problem as well as a polynomial-time approximation algorithm with performance ratio ρ + 2, where ρ is the best known approximation ratio for the ST problem (currently ρ = 1.55, see [22]). This approximation ratio was improved to 2ρ in [6, 10], and further to 2ρ ρ/(3ρ 2) in [21]. The bottleneck full Steiner tree (BFST) problem is to find a full Steiner tree T such that the length of the longest edge in T is minimized. Chen et al. [6] consider this problem and present an O( E log E )-time algorithm for optimally solving the problem. In this paper, we consider the BFST problem in the geometric context, i.e., V is a set of points in the Euclidean plane, G is the complete graph over V and the weight of an edge (p, q) is the Euclidean distance between p and q. More precisely, the input of the Euclidean BFST problem is two sets of points in the plane, R of n terminals and S of m Steiner points, and the goal is to find a full Steiner tree with minimum bottleneck. In this context, we present an O((n + m) log 2 m)-time algorithm for the problem, which improves the O((n + m) 2 log (n + m))-time algorithm of Chen et al. [6]. It is somewhat non-standard in the study of Euclidean BST and BFST problems that the locations of the Steiner points are fixed. On the other hand, when the locations of the Steiner points are not fixed, i.e., S is the whole plane R 2, the problems make no sense without a restriction on the number of used Steiner points, for we can add infinitely many Steiner points and achieve 0 bottleneck. The k-bst (resp. k-bfst) problem is a restricted version of the Euclidean BST (resp. BFST) problem, for which, in addition to the sets R and S = R 2, we are given a positive integer k, and we are asked to find a Steiner (resp. a full Steiner) tree T with at most k Steiner points such that the length of the longest edge in T is minimized. The k-bst problem has been studied extensively in the last decade. In [24], the problem was shown to be NP-hard to approximate within ratio 2. On the other hand, the best known approximation ratio for the problem is [25]. Bae et al. [3] presented an O(n log n) time algorithm for the problem for k = 1, and an O(n 2 ) time algorithm for k = 2. Li et al. [18] presented a ( 2 + ɛ)-approximation algorithm with inapproximability within 2 for a special case of the problem, where edges connecting between two Steiner points are forbidden. To the best of our knowledge, the k-bfst problem was not considered before. In this paper, we study the discrete version of the k-bfst problem, i.e., when the locations of the Steiner points are fixed. We show that the problem is NP-hard and we present a polynomial-time algorithm with constant-factor approximation ratio for the problem. The advantage of studying this version is that the obtained results hold not only for the metric version of the problem (i.e., when the input is a complete graph with weight function 2

3 satisfying the triangle inequality), but can also be generalized to the non-discrete version of the k-bfst problem. The k-bfst problem has applications in the design of wireless ad-hoc and sensor networks. A sensor network consists of a large number of autonomous devices, called sensors. These sensors can not relay messages from the neighboring sensors due to their simple functionality. To make the network connected and due to budget limits, at most k additional devices, called relays, must be judiciously placed in the network. Relays are typically more advanced than sensors. Each relay is equipped with a WLAN transceiver enabling communication between adjacent relays. Since the cost of the sensor/relay is proportional to its transmission range, we want to minimize the longest distance between the elements of the network. The rest of this paper is organized as follows. In Section 2, we present an O((n + m) log 2 m)-time algorithm that optimally solves the Euclidean BFST problem. In Section 3, we first present a hardness result for the discrete k-bfst problem as well as a polynomialtime approximation algorithm with performance ratio 4. Next, we generalize this result to the non-discrete version of the problem. Finally, we conclude in Section 4. 2 The BFST Problem Given two sets of points in the plane; a set R of n terminals and a set S of m Steiner points, the goal in the BFST problem is to find a full Steiner tree T such that the bottleneck of T is minimized. We refer to such a tree as an optimal full Steiner tree of R. In this section we present an O((n + m) log 2 m)-time algorithm for computing an optimal full Steiner tree of R. Let MST (S) be a minimum spanning tree of S (i.e., of the complete graph over S). For a full Steiner tree T of R, we denote by S(T ) the set of Steiner points of S appearing in T. We first observe that, for any optimal full Steiner tree T of R, the tree obtained by replacing the subtree of T induced by S(T ), by MST (S(T )) is also optimal. The following lemma shows that it suffices to consider the subtrees of MST (S) 1. Lemma 2.1. There exists an optimal full Steiner tree T of R such that MST (S(T )) is a subtree of MST (S). Proof: Let T be an optimal full Steiner tree of R. Consider any edge (s i, s j ) in MST (S(T )) but not in MST (S). Let δ MST (S) (s i, s j ) be the path between s i and s j in MST (S). Since each edge in δ MST (S) (s i, s j ) is of length at most s i s j, we can replace the edge (s i, s j ) in T by the path δ MST (S) (s i, s j ) to obtain a new tree which is also an optimal full Steiner tree. By repeating this operation for each such edge (s i, s j ), we obtain an optimal full Steiner tree T satisfying the lemma. 1 Actually, the lemma holds for arbitrary metrics. 3

4 Let e 1, e 2,..., e m 1 be the edges of MST (S) and assume, w.l.o.g., that e 1 e 2... e m 1. For an edge e i MST (S), let T i be the forest (i.e., the set of subtrees) obtained from MST (S) by removing the edges of length greater than e i. By Lemma 2.1, there exists an edge e i MST (S) such that T i contains a tree T such that, by connecting each point in R to its closest point in T, we obtain an optimal full Steiner tree of R. For a point p and a tree T, let d(p, T ) denote the distance between p and its closest point in T. In the following, we describe our algorithm to construct an optimal full Steiner tree of R. We first perform a binary search on the lengths of edges of MST (S). (Actually, in order to handle the case that the bottleneck of an optimal full Steiner tree might be either less than e 1 or greater than e m 1, we add the values e 0 = 0 and e m = to the search space.) For a given value e i, we use Procedure 1 to decide whether there exists a full Steiner tree with bottleneck at most e i. Procedure 1 T est(mst (S), R, e i ) 1: construct the forest T i 2: for each point p R do 3: drop from T i all trees T with d(p, T ) > e i 4: if T i φ then 5: /* T i contains a tree T such that connecting each terminal in R to its closest point in T produces a full Steiner tree with bottleneck at most e i */ 6: return T RUE 7: else 8: return F ALSE Let λ denote the bottleneck of an optimal full Steiner tree. It is not hard to see that, once this stage terminates, we have an edge e i, 0 i m 1, such that e i < λ e i+1. Hence, there are two options; either e i < λ < e i+1 or λ = e i+1. If e i < λ < e i+1, then the optimal full Steiner tree of R is obtained by a tree T from the forest T i ; see Figure 1(a). Otherwise, it is obtained from a tree T from the forest T i+1 ; see Figure 1(b). Thus, we can find the tree T in the set T i T i+1, such that, by connecting each terminal in R to its closest point in T, we obtain an optimal full Steiner tree of R. Lemma 2.2. The algorithm above can be implemented in O((n + m) log 2 m) time. Proof: Constructing a minimum spanning tree of S and sorting its edges take O(m log m) [9]. Computing the forest T i takes O(m) time. We implement step 2 in Procedure 1 as follows. For each tree T in T i, we compute the Voronoi diagram of the points of T together with a corresponding point location data structure. This takes O(m log m) time for all trees of T i. We then perform point location queries (each of O(log m) time) in the diagram with the points of R. To analyze the running time of this step, we make the following key observation. 4

5 terminals Steiners λ e i e i+1 e i e i+1 (a) (b) Figure 1: The optimal tree is obtained (a) from T i, when e i < λ < e i+1, and (b) from T i+1, when λ = e i+1. Observation 2.3. For any point p R and for any e i MST (S), there are at most 5 subtrees T in the forest T i satisfying d(p, T ) e i. To see that, consider a point p R and consider the disk centered at p with radius e i. Since the distance between any two trees in T i is greater than e i, this disk can contain at most 5 points that belong to different trees of T i. By the observation, after performing point location queries with the first point from R (at most m queries, each of cost O(log m)), the number of subtrees in T i surviving these queries is at most 5. Since the number of queries performed by the remaining points is at most 5(n 1), their cost is O(n log m). Therefore, the running time of step 2 is O((n + m) log m). Since Procedure 1 is applied O(log m) times, the running time of the first stage is O((n + m) log 2 m). Finally, the second stage can be implemented as in step 2 in Procedure 1. Thus, the total running time of the algorithm is O((n + m) log 2 m). The following theorem summarizes this result. Theorem 2.4. Given two sets of points in the plane; a set R of n terminals and a set S of m Steiner points, an optimal full Steiner tree of R can be computed in O((n + m) log 2 m) time. Remark. Notice that the algorithm computes an optimal full Steiner tree which is not necessarily the one with fewest Steiner points. The problem of finding an optimal full Steiner tree with minimum number of Steiner points among all optimal trees is NP-hard, as we will show in the next section. 3 The k-bfst Problem In this section, we consider the discrete k-bfst problem. Given a set R of n terminals in the plane, a set S of m Steiner points and an integer k m, the goal is to find a full Steiner 5

6 tree with at most k Steiner points from S and bottleneck as small as possible. We first prove hardness of the problem and present a polynomial-time approximation algorithm with performance ratio 4. Next, we show how to generalize our algorithm to the non-discrete version. Theorem 3.1. The discrete k-bfst problem is NP-hard. Proof: Our proof is based on a reduction from the following problem which is known to be NP-complete due to Garey and Johnson [14]. Connected vertex cover in planar graphs with maximum degree 4. Given a planar graph G = (V, E) with no vertex degree exceeding 4 and an integer k, does there exist a vertex cover V for G such that V k and the subgraph of G induced by V is connected? Given a planar graph G = (V, E) with no vertex degree exceeding 4 and an integer k, we construct, in polynomial time, two planar sets R and S, and compute an integer k, such that G has a connected vertex cover of size at most k if and only if there exists a full Steiner tree T with at most k Steiner points and bottleneck at most 1. Let V = {v 1, v 2,..., v n } and let E = {e 1, e 2,..., e m }. We first embed G into a rectangular grid, with distance at least 4 between adjacent vertices, as follows. Each vertex v i V corresponds to some grid vertex and each edge e = (v i, v j ) E corresponds to a rectilinear path p e, consisting of some horizontal and vertical elementary grid segments, whose endpoints are the grid vertices corresponding to v i and v j. In addition, these paths are pairwise disjoints; see Figure 2. v j T e e p e v j v i v i (a) (b) (c) Figure 2: (a) A planar graph G = (V, E), (b) the embedded graph G = (V, E ), and (c) the produced sets: V consists of solid circles, s(e) consists of empty circles and t(e) consists of black squares. 6

7 Let V = {v 1, v 2,..., v n} be the set of vertices of the grid corresponding to the vertices of V, and let E = {p e1, p e2,..., p em } be the set of edges (paths) corresponding to the edges of E. We now place two types of points on the interior of each edge p e E. Let p e denote the total length of the grid segments of p e. We place p e 1 Steiner points on p e, such that the distance between any adjacent points is exactly 1; denote by s(e) this set of Steiner points. Moreover, for each set s(e), we place a terminal between (in the middle of) every two adjacent points in s(e). Denote by t(e) this set of terminals and notice that t(e) = p e 2; see Figure 2(c). Finally, we set R = t(e), e E S = V s(e) and k = e E e E s(e) m + 2k 1. For each edge p e E, let c(e) be the set of Steiner points in s(e) except the endpoints, i.e., except the first and the last points. Observe that, connecting every adjacent two Steiner points in c(e) (to form a path) and connecting each terminal in t(e) to its closest point in c(e) produces a full Steiner tree of t(e) with s(e) 2 Steiner points and bottleneck 1. On the other hand, observe that at least s(e) 2 Steiner points are necessarily to construct a full Steiner tree of t(e) with bottleneck at most 1. Denote by T e such a full Steiner tree; see Figure 2(c). Now, we prove the correctness of the reduction. Suppose that G has a connected vertex cover V with V k. We construct a full Steiner tree of R as follows. For each edge e E, we construct the tree T e (as described above). Let T be any spanning tree of the subgraph of G induced by V. This spanning tree exists by the connectivity of V and contains V 1 edges. For each edge e = (v i, v j ) T, we connect the corresponding points v i, v j S (by two edges of length 1) to the tree T e using their adjacent (first and last) points in s(e). And, for each edge e = (v i, v j ) E \ T, we select one endpoint v i of e that belongs to V and we connect v i (by an edge of length 1) to the tree T e using its adjacent point in s(e). It is easy to see that the constructed tree is a full Steiner tree of R and it has V + e E ( s(e) 2) + 2( V 1) + m ( V 1) e E s(e) m + 2k 1 = k Steiner points and bottleneck exactly 1. Conversely, suppose that there exists a full Steiner tree T of R with at most k Steiner points and bottleneck at most 1. Let V be the subset of points of V S that appear in T, and let T be the subtree of T spanning V. For each subset t(e) R, let T e be the subtree of T spanning the points in t(e). Since the bottleneck of T is at most 1, (i) by the above observation, T e contains at least s(e) 2 Steiner points, and (ii) it is not hard to see that, each tree T e is connected to at least one point from V, which implies that the set of vertices in G corresponding to the points in V is a connected vertex cover of G. 7

8 Moreover, a tree T e which is also a subtree of T is connected to two points from V via the endpoints of s(e) (there are V 1 such trees), and a tree T e which is not a subtree of T is connected to one point from V via one endpoint of s(e) (there are m ( V 1) such trees). Thus, T contains at least V + e E ( s(e) 2) + 2( V 1) + m ( V 1) Steiner points. On the other hand, T contains at most k = e E s(e) m + 2k 1 Steiner points. This implies that V is of size at most k, which completes the proof. Finally, we observe that, in the converse direction of the proof, for any ε > 0, if there exists a full Steiner tree T of R with at most k Steiner points and bottleneck at most ( 2 ε), then we can construct a connected vertex cover of size at most k. This implies that the k-bfst problem cannot be approximated within a factor less than 2, unless P = NP. Corollary 3.2. The discrete k-bfst problem admits no PTAS, unless P = NP. Remark. By a similar argument, we can show that the non-discrete and metric versions of the k-bfst are NP-hard. 3.1 The approximation algorithm We now design a polynomial-time approximation algorithm for computing a full Steiner tree with at most k Steiner points (k-fst for short) such that its bottleneck is at most 4 times the bottleneck of an optimal (minimum-bottleneck) k-fst. Let G = (V, E) be the graph with V = R S and E = {(s, v) : s S and v R S}. We assume, without loss of generality, that E = {e 1, e 2,..., e l } such that e 1 e 2... e l. It is not hard to see that the bottleneck of an optimal k-fst is a length of an edge from E. For an edge e i E, let G i = (V, E i ) be the graph with E i = {e j E : e j e i }. The idea behind our algorithm is to devise a procedure that, for a given edge e i E, does one of the following: (i) It constructs a k-fst of R in G with bottleneck at most 4 times e i. (ii) It returns the information that G i does not contain any k-fst of R. Let G 2 i = (V, E2 i ) be the 2nd power graph of G i, i.e., G 2 i has the same set of vertices as G i and an edge between two vertices if and only if there is a path of length at most 2 between them in G i. Let G 2 i (R) be the subgraph of G2 i induced by R and let R be a maximal independent set in G 2 i (R). If R = 1, then we can construct a k-fst of R, with bottleneck at most 3 times e i, as follows. Let p be the only point in R. We select an arbitrary Steiner point s S that is connected to p in G i and we connect it to all of the points in R. Since there is an edge between p and each terminal p R \ R in G 2 i, we have p p 2 e i, and hence, sp 3 e i. 8

9 From now on, we assume that R > 1. For two points p, q R, let δ i (p, q) be a shortest Steiner path between p and q in G i, i.e., a path connecting p and q with minimum number of Steiner points in G i. Let G = (R, E ) be the complete graph over R. For each edge (p, q) in E, we assign a weight w(p, q) equal to the number of Steiner points in δ i (p, q). Let T be a minimum spanning tree of R under w. We define the normalized weight of T as C(T ) = e T w(e)/2. Lemma 3.3. If G i contains a k-fst of R, then C(T ) k. Proof: Let T be a k-fst of R in G i. We arbitrary select a Steiner point as the root of T ; see Figure 3(a). In the following, we will construct a spanning tree T of R in G such that C(T ) k. The construction of T is bottom-up by an iterative process. In each iteration, we select the deepest leaf p in the rooted tree, which is a terminal, and we connect it to its nearest terminal q by an edge of weight equal to the number of Steiner points between them. Let s be the first common ancestor of p and q. We then remove the Steiner points between p and s (in the last iteration, we may remove all of the remaining points). terminals Steiners 3 s 2 b s 3 s 1 3 d a 2 c (a) (b) Figure 3: (a) The rooted tree, and (b) the construction of T. In the example in Figure 3(b), we first select the terminal a, which is the deepest one, we connect it to the terminal b by an edge of weight 3 and we remove the points s 1 and s 2. Next, we select the terminal d, we connect it to the terminal c by an edge of weight 2 and we remove the point s 3. In the last iteration, we select the terminal b, we connect it to the terminal c by an edge of weight 3 and we remove all of the remaining points. Notice that, since, in each iteration, we select the deepest terminal, we add an edge (p, q), of weight w(p, q), and we remove at least w(p, q)/2 Steiner points from T. Thus, we conclude that C(T ) = e T w(e)/2 k. Finally, since T is also a spanning tree of G, we have C(T ) C(T ) k. Notice that the inverse direction of Lemma 3.3 does not hold, namely, C(T ) k does not necessarily imply that G i contains a k-fst of R. 9

10 Based on the above discussion, we now describe our approximation algorithm. For each examined edge e i E in the sorted order, we first construct a maximal independent set R in G 2 i (R). Let G = (R, E ) be the complete graph over R. Next, we construct a minimum spanning tree T of G with respect to w. If C(T ) > k, then we move to the next edge e i+1, otherwise, we construct a k-fst of R as follows. P \ P P \ P p q P \ P p q (a) (b) Figure 4: (a) The set S consists of solid circles, and (b) the produced k-fst. For each edge e = (p, q) T, we first select w(e)/2 Steiner points on any shortest Steiner path between p and q in G i in alternating way, such that the distance between any consecutive two (selected) points is at most 2 e i. Let S be the set of the selected Steiner points, and notice that S k; see Figure 4(a). Next, we construct a minimum spanning tree MST (S ) of S (i.e., of the complete Euclidean graph over S ). Notice that each edge in MST (S ) is of length at most 4 e i. Finally, we connect each terminal in R to its nearest Steiner point in S to obtain a full Steiner tree; see Figure 4(b). This guarantees that each terminal in R is connected to a Steiner point with an edge of length at most 2 e i, and each terminal in R \ R is connected to a Steiner point with an edge of length at most 4 e i. Lemma 3.4. The algorithm above constructs a k-fst of R with bottleneck at most 4 times the bottleneck of an optimal k-fst. Proof: Let e i be the first edge satisfying the condition C(T ) k. Thus, by Lemma 3.3, the bottleneck of any k-fst in G is at least e i, and, therefore, the constructed k-fst has a bottleneck at most 4 times the bottleneck of an optimal k-fst. Lemma 3.5. The algorithm above has a polynomial running time. Proof: Notice that, for each edge e i E, each graph G 2 i is of size O((n + m) 2 ). Thus, computing G 2 i from G i, and computing a maximal independent set in G 2 i take O((n + m)2 ) time. In order to construct the graph G = (R, E ), we can compute in O((n + m) 3 ) time the shortest Steiner paths between each pair of points in R [8]. Once G is constructed, computing a minimum spanning tree of G can be done in O(n 2 ) time, and selecting the relevant Steiner points can be done in O(k(n + m)) time. Finally, the construction of the obtained full Steiner tree can be done in O((n + k) log k). 10

11 By combining Lemma 3.4 and Lemma 3.5, we get the following theorem. Theorem 3.6. There exists a polynomial-time approximation algorithm with performance ratio 4 for the k-bfst problem. To see that the ratio obtained by the algorithm is tight, consider the example in Figure 5. For k = 7, the optimal k-fst (solid lines) has bottleneck 1. The condition C(T ) 7 is satisfied for the first time when e i = 1, and, for this edge, C(T ) = 2. Thus the algorithm selects two Steiner points (black circles) and constructs a k-fst (dotted lines) of bottleneck 4. terminals Steiners ε Figure 5: The constructed k-fst (dotted lines) has two Steiner points and bottleneck 4, while the optimal k-fst (solid lines) has bottleneck 1. Remark 1. In order to avoid any confusion, we mention the reason for working on R and not on R itself, even though Lemma 3.3 also holds for R. The construction of the obtained k-fst is based on the minimum spanning tree T and the fact that each edge e in T has a normalized weight w(e)/2 > 0, and, thus, we have at least one Steiner point to use in the obtained k-fst. If we work on the set R, then edges of normalized weight w(e)/2 = 0 might appear in T and, thus, in some cases, there will be no valid construction or the ratio can be greater than 4, as in the example in Figure 6. terminals Steiners opt T k -FST s Figure 6: For k = 7, C(T ) = 1 and the constructed k-fst (dashed lines) has only one Steiner point s and bottleneck greater than 4. ε ε 11

12 Remark 2. It is not hard to see that the result in this section holds for the metric version of the k-bfst problem (i.e., for any complete graph with weight function satisfying the triangle inequality). 3.2 Approximating the non-discrete k-bfst In the non-discrete version of the k-bfst problem, we are given a set R of n terminals in the plane and a positive integer k, and the goal is to place at most k Steiner points (anywhere in the plane), such that the full Steiner tree obtained from these points has a bottleneck as small as possible. We show how to modify the algorithm from the previous section to approximate this problem within a factor 4. Let G = (R, E) be the complete graph over R and let D = { e /(j + 1) : e E and 1 j k} be a space of values obtained from E and k. For a value d i D and for an edge e E, let w i (e) = e /d i 1 and let P e,i be the set of edges obtained from partitioning e by adding w i (e) Steiner points evenly-spaced on e; denote by S e,i the set of these Steiner points. Let R i = R {S e,i : e E}, let E i = {P e,i : e E} and let G i = (R i, E i ), for each value d i D. As in the previous section, let R be a maximal independent set in G 2 i (R) and let G = (R, E ) be the complete graph over R with weight function w i. Let T be a minimum spanning tree of G under w i and let C(T ) = e T w i(e)/2 be the normalized weight of T. Finally, let T be an optimal full Steiner tree of R and λ be the bottleneck of T. Lemma 3.7. If λ d i, then C(T ) k. Proof: By a similar argument as in the proof of Lemma 3.5. The algorithm seeks for the smallest value d i satisfying the condition C(T ) k. Once this value is found, it constructs a k-bfst by placing w i (e)/2 Steiner points evenlyspaced on each edge e T and connecting each terminal to its closest Steiner point. It is not hard to see that the algorithm produces a k-bfst with bottleneck at most 4λ and can be implemented in polynomial time. Theorem 3.8. There exists a polynomial-time approximation algorithm with performance ratio 4 for the non-discrete k-bfst problem. 4 Conclusion In this paper, we studied the problem of finding a bottleneck full Steiner tree of a set of terminals in the Euclidean plane. When the set of Steiner points is finite (the discrete version) and no restriction on the number of used Steiner points is given, we presented an O((n + m) log 2 m)-time algorithm for solving the problem. And, when the number of used Steiner points is limited, we proved that the problem does not admit any approximation 12

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