An Exponential Time 2-Approximation Algorithm for Bandwidth

Transcription

1 An Exponentia Time 2-Approximation Agorithm for Bandwidth Martin Fürer 1, Serge Gaspers 2, Shiva Prasad Kasiviswanathan 3 1 Computer Science and Engineering, Pennsyvania State University, furer@cse.psu.edu 2 CMM, Universidad de Chie, sgaspers@dim.uchie.c 3 Los Aamos Nationa Laboratory, kasivisw@gmai.com Abstract. The bandwidth of a graph G on n vertices is the minimum b such that the vertices of G can be abeed from 1 to n such that the abes of every pair of adjacent vertices differ by at most b. In this paper, we present a 2-approximation agorithm for the Bandwidth probem that takes worst-case O( n = O( n time and uses poynomia space. This improves both the previous best 2- and 3-approximation agorithms of Cygan et a. which have an O (3 n and O (2 n worst-case time bounds, respectivey. Our agorithm is based on constructing bucket decompositions of the input graph. A bucket decomposition partitions the vertex set of a graph into ordered sets (caed buckets of (amost equa sizes such that a edges are either incident on vertices in the same bucket or on vertices in two consecutive buckets. The idea is to find the smaest bucket size for which there exists a bucket decomposition. The agorithm uses a simpe divide-and-conquer strategy aong with dynamic programming to achieve this improved time bound. 1 Introduction Let G = (V, E be a graph on n vertices and b be an integer. The Bandwidth probem asks whether the vertices of G can be abeed from 1 to n such that the abes of every pair of adjacent vertices differ by at most b. The Bandwidth probem is a specia case of the Subgraph Isomorphism probem, as it can be formuated as foows: Is G isomorphic to a subgraph of P b n? Here, P b n denotes the graph obtained from P n (the path on n vertices by adding an edge between every pair of vertices at distance at most b in P n. A typica scenario in which the Bandwidth probem arises is that of minimizing the bandwidth of a symmetric matrix M to aow for more efficient storing and manipuating procedures [11]. The bandwidth of M is b if a its non-zero entries are at a distance of at most b from the diagona. Appying permutations on the rows and coumns to reduce the bandwidth of M corresponds then to reordering the vertices of a graph whose adjacency matrix corresponds to M by repacing a non-zero entries by 1. Visiting EPFL Lausanne and Universität Zürich. Research supported in part by NSF Grant CCF Partiay supported by the Research Counci of Norway (NFR and by the GRAAL project ANR-06-BLAN-0148 of the French Nationa Research Agency (ANR.

2 The Bandwidth probem is NP-hard [19], even for trees of maximum degree at most three [14] and caterpiars with hair ength at most three [17]. Even worse, approximating the bandwidth within a constant factor is NP-hard, even for caterpiars of degree three [21]. Further, it is known that the probem is hard for every fixed eve of the W-hierarchy [3] and unikey to be sovabe in f(bn o(b time [4]. Faced with this immense intractabiity, severa approaches have been proposed in the iterature for the Bandwidth probem. The first (poynomia time approximation agorithm with a poyogarithmic approximation factor was provided by Feige [10]. Later, Dunagan and Vempaa gave an O(og 3 n og og n-approximation agorithm. The current best approximation agorithm achieves an O(og 3 n(og og n 1/4 -approximation factor [16]. For arge b, the best approximation agorithm is the probabiistic agorithm of Bum et a. [2] which has an O( n/b og n-approximation factor. Super-poynomia time approximation agorithms for the Bandwidth probem have aso been widey investigated [5,8,9,12]. Feige and Tawar [12], and Cygan and Piipczuk [8] provided subexponentia time approximation schemes for approximating the bandwidth of graphs with sma treewidth. For genera graphs, a 2-approximation agorithm with a running time of O (3 n 4 is easiy obtained by combining ideas from [11] and [12] (as noted in [5]. Further, Cygan et a. [5] provide a 3-approximation agorithm with a running time of O (2 n, which they generaize to a (4r 1-approximation agorithm (for any positive integer r with a running time of O (2 n/r. Concerning exact exponentia time agorithms, the fastest poynomia space agorithm is sti the eegant O (10 n time agorithm of Feige [11]. When aowing exponentia space, this bound is improved in a sequence of agorithms by Cygan and Piipczuk; their O (5 n time agorithm uses O (2 n space [6], their O(4.83 n time agorithm uses O (4 n space [7], and their O(4.473 n time agorithm uses O(4.473 n space [8]. The most practica of these agorithms is probaby the O (5 n time agorithm as the space requirements of the other ones seems forbiddingy arge for practica appications. The Bandwidth probem can aso be soved exacty in O(n b time using dynamic programming [20,18]. Another recent approach to cope with the intractabiity of Bandwidth is through the concept of hybrid agorithms, introduced by Vassievska et a. [22]. They gave an agorithm that after a poynomia time test, either computes the minimum bandwidth of a graph in O (4 n+o(n time, or provides a poyogarithmic approximation ratio in poynomia time. This resut was recenty improved by Amini et a. [1] who give an agorithm which, after a poynomia time test, either computes the minimum bandwidth of a graph in O (4 n time, or provides an O(og 3/2 n-approximation in poynomia time. Our Resuts. Our main resut is a 2-approximation agorithm for the Bandwidth probem that takes worst-case O( n time (Theorem 1. This improves the O (3 n time bound achieved by Cygan et a. [5] for the same approximation ratio. Aso, the previous best 3-approximation agorithm of Cygan and Piipczuk [8] has an O (2 n time bound. Therefore, our 2-approximation agorithm is aso faster than the previous best 3-approximation agorithm. 4 O (f(n denotes n O(1 O(f(n.

3 Our agorithm is based on constructing bucket decompositions of the input graph. A bucket decomposition partitions the vertex set of a graph into ordered sets (caed buckets of (amost equa sizes such that a edges are either incident on vertices in the same bucket or on vertices in two consecutive buckets. The idea is to find the smaest bucket size for which there exists a bucket decomposition. This gives a 2-approximation for the Bandwidth probem (Lemmas 2 and 1. The agorithm uses a simpe divideand-conquer strategy aong with dynamic programming to achieve this improved time bound. 2 Preiminaries Let G = (V, E be a graph on n vertices. A inear arrangement of G is a bijective function L : V [n] = {1,..., n}, that is a numbering of its vertices from 1 to n. The stretch of an edge (u, v is the absoute difference between the numbers assigned to its endpoints L(u L(v. The bandwidth of a inear arrangement is the maximum stretch over a the edges of G and the bandwidth of a graph is the minimum bandwidth over a inear arrangements of G. A bucket arrangement of G is a pacement of its vertices into buckets such that for each edge, its endpoints are either in the same bucket or in two consecutive buckets [12]. The buckets are ineary ordered and numbered from eft to right. A capacity vector C is a vector of positive integers. The ength of a capacity vector C = (C[1],..., C[k] is k and its size is k i=1 C[i]. Given a capacity vector C of size n, a C-bucket arrangement of G is a bucket arrangement in which exacty C[i] vertices are paced in bucket i, for each i. For integers n and with < n/2, an (n, -capacity vector is a capacity vector (a,,,...,, b }{{} n 2 times of size n such that a, b. We say that an (n, -capacity vector is eft-packed if a = and baanced if a b = 1. Let X V be a subset of the vertices of G. We denote by G[X] the subgraph of G induced on X, and by G \ X the subgraph of G induced on V \ X. The open neighborhood of a vertex v is denoted by N G (v and the open neighborhood of X is N G (X := ( v X N G(v \ X. 3 Exponentia Time Agorithms for Approximating Bandwidth We first estabish two simpe emmas that show that constructing a bucket arrangement can approximate the bandwidth of a graph. Lemma 1. Let G be a graph on n vertices, and et C be an (n, -capacity vector. If there exists a C-bucket arrangement for G then the bandwidth of G is at most 2 1. Proof. Given a C-bucket arrangement for G, create a inear arrangement respecting the bucket arrangement (if u appears in a smaer numbered bucket than v, then L(u <

4 L(v, where vertices in the same bucket are numbered in an arbitrary order. As the capacity of each bucket is at most and each edge spans at most two consecutive buckets, the maximum edge stretch in the constructed inear arrangement is at most 2 1. Lemma 2. Let G be a graph on n vertices, and et C be an (n, -capacity vector. If there exists no C-bucket arrangement for G then the bandwidth of G is at east + 1. Proof. Suppose there exists a inear arrangement L of G of bandwidth at most. Construct a bucket arrangement pacing the first C[1] vertices of L into the first bucket, the next C[2] vertices of L into the second bucket, and so on. In the resuting bucket arrangement, no edge spans more than two consecutive buckets. Therefore, a C-bucket arrangement exists for G, a contradiction. We wi use the previous fastest 2-approximation agorithm of Cygan et a. [5] as a subroutine. For competeness, we describe this simpe agorithm here. Proposition 1 ([5]. There is a poynomia space 2-approximation agorithm for the Bandwidth probem that takes O (3 n time on connected graphs with n vertices. Proof. Let G be a connected graph on n vertices. For increasing from 1 to n/2, the agorithm does the foowing. Let C be an (n, -capacity vector. The agorithm goes over a the k = n choices for assigning the first vertex to some bucket. The agorithm then chooses an unassigned vertex u which has at east one neighbor that has aready been assigned to some bucket. Assume that a neighbor of u is assigned to the bucket i. Now there are at most three choices of buckets (i 1, i, and i + 1 for assigning vertex u. Some of these choices may be invaid either because of the capacity constraints of the bucket or because of the previous assignments of (other neighbors of u. If the choice is vaid, the agorithm recurses by assigning u to that bucket. Let be the smaest integer for which the agorithm succeeds, in some branch, to pace a vertices of G into buckets in this way. Then, by Lemma 1, G has bandwidth at most 2 1 and by Lemma 2, G has bandwidth at east. Thus, the agorithm outputs 2 1, which is a 2-approximation for the bandwidth of G. As the agorithm branches into at most 3 cases for each of the n vertices (except the first one, and a other computations ony contribute poynomiay to the running time of the agorithm, this agorithm runs in worst-case O (3 n time using ony poynomia space. We now show another simpe agorithm based on a divide-and-conquer strategy that given an (n, -capacity vector C, decides whether a C-bucket arrangement exists for a connected graph G. Proposition 2. Let G be a connected graph on n vertices and C be an (n, -capacity vector with < n/2. ( There exists an agorithm that can decide if G has a C-bucket (n ( arrangement in O n/ n/4 time. Proof. Let k = n be the number of buckets in the C-bucket arrangement. Number the buckets from 1 to k from eft to right according to the bucket arrangement. Seect a bucket index i such that the sum of the capacities of the buckets numbered stricty

5 smaer than i and the one for the buckets numbered stricty arger than i are both at most n/2. The agorithm goes over a possibe ( n choices of fiing bucket i with vertices. Let X be a set of vertices assigned to the bucket i. Given a connected component of G \ X, note that a the vertices of this connected component must be paced either ony in buckets 1 to i 1 or buckets i + 1 to k. Note that each connected component of G \ X contains at east one vertex that is adjacent to a vertex in X (as G is connected. Therefore, for each connected component of G \ X, at east one vertex is paced into the bucket i 1 or i + 1. As the capacity of each bucket is at most, G \ X has at most 2 connected components, otherwise there is no C-bucket arrangement where X is assigned to the bucket i. Thus, there are at most 2 2 choices for assigning connected components of G\X to the buckets 1 to i 1 and i+1 to k. Some of these assignments might be invaid as they might vioate the capacity constraints of the buckets. We discard these invaid assignments. For each choice of X and each vaid assignment of the connected components of G\ X to the eft or right of bucket i, we have now obtained two independent subprobems: one subprobem for the buckets {1,..., i 1} and one subprobem for the buckets {i + 1,..., k}. These subprobems have sizes at most n/2. Consider the subprobem for the buckets {1,..., i 1} (the other one is symmetric and et Y be the set of vertices associated to these buckets. Let Z Y be the set of vertices in Y that have at east one neighbor in X. Now, add edges to the subgraph G[Y ] such that Z becomes a cique. This does not change the probem, as a the vertices in Z must be assigned to the bucket i 1, and G[Y ] becomes connected. This subprobem can be soved recursivey, ignoring those soutions where vertices in Z are not a assigned to the bucket i 1. The agorithm performs the above recursion unti it reaches subprobems of size at most n/4, which corresponds to two eves in the corresponding search tree. On instances of size at most n/4, the agorithm invokes the agorithm of Proposition 1, which takes worst-case O (3 n/4 time. Let T (n be the running time needed for the above procedure to check whether a graph with n vertices has a bucket arrangement for an (n, -capacity vector. Then, T (n ( n 2 2 ( n/2 (( n n/4 n O(1 = O ( n/ n/4. This competes the proof of the proposition. Combining Proposition 2 with Lemmas 1 and 2, we have the foowing coroary for 2-approximating the bandwidth of a graph. Coroary 1. There is an agorithm that, for a connected graph G on n vertices and an integer( n can decide whether the bandwidth of G is at east + 1 or at most 2 1 (n ( in O n/ n/4 time. Proof. If n/2, the bandwidth of G is at most 2 1. Otherwise, use Proposition 2 with G and some (n, -capacity vector C to decide if there exists a C-bucket arrangement for G. If so, then the bandwidth of G is at most 2 1 by Lemma 1. If not, then the bandwidth of G is at east + 1 by Lemma 2.

6 The running time of the agorithm of Coroary 1 is interesting for sma vaues of. Namey, if n/26, the running time is O( n. In the remainder of this section, we improve Proposition 2. We now concentrate on the cases where k = n/ 26. Let C be an (n, -capacity vector. A partia C-bucket arrangement of an induced subgraph G of G is a pacement of vertices of G into buckets such that: (a each vertex in G is assigned to a bucket or to a union of two consecutive buckets, (b the endpoints of each edge in G are either in the same bucket or in two consecutive buckets, and (c at most C[i] vertices are paced in each bucket i. Let B be a partia C-bucket arrangement of an induced subgraph G. We say that a bucket i is fu in B if the number of vertices that have been assigned to it equas its capacity (= C[i]. We say that two consecutive buckets i and i + 1 are jointy fu in B if a vertex subset Y of cardinaity equa to the sum of the capacities of i and i+1 have been assigned to these buckets (i.e., each vertex v Y is restricted to beong to the union of buckets i or i + 1, but which among these two buckets v beongs is not fixed. We say that a bucket is empty in B if no vertices have been assigned to it. Proposition 3. Let G be a graph on n vertices and C be a capacity vector of size n and ength k, where k is an integer constant. Let B = B(G be a partia C-bucket arrangement of some induced subgraph G of G such that in B some buckets are fu, some pairs of consecutive buckets are jointy fu, and a other buckets are empty. If in B no 3 consecutive buckets are empty, then it can be decided if B can be extended to a C-bucket arrangement in poynomia time. Proof Outine. Let G = (V, E and G = (V, E. Let r be the number of connected components of G \ V (the graph induced on V \ V, and et V represent the set of vertices in the th connected component of G \ V. If the bucket i is fu in B, et X i denote the set of vertices assigned to it. If the buckets i and i + 1 are jointy fu in B, et X i,i+1 denote the set of vertices assigned to the union of buckets i and i+1. We use dynamic programming to start from a partia bucket arrangement satisfying the above conditions to construct a C-bucket arrangement. During its execution, the agorithm assigns vertices to the buckets which are empty in B. We ony present an outine of the dynamic programming agorithm here. The dynamic programming agorithm constructs a tabe T [... ], which has the foowing indices. An index p, representing the subprobem on the first p connected components of G \ V. For every empty bucket i in B such that both the buckets i 1 and i + 1 are fu, it has an index s i, representing the number of vertices assigned to the bucket i. For every two consecutive empty buckets i and i + 1 in B, it has indices t i,i+1, x i, and x i+1. The index t i,i+1 represents the tota number of vertices assigned to the buckets i and i + 1. The index x i represents the number of vertices assigned to the buckets i and i + 1 that have at east one neighbor in the bucket i 1. The index x i+1 represents the number of vertices assigned to the buckets i and i + 1 that have at east one neighbor in the bucket i + 2. For every two consecutive buckets i, i + 1 which are jointy fu in B, it has indices f i and f i+1 representing the number of vertices assigned to these buckets that have at east one neighbor in the bucket i 1 (f i or in the bucket i + 2 (f i+1.

7 Tabe T [...] is initiaized to fase everywhere, except for the entry corresponding to a-zero indices, which is initiaized to true. The rest of the tabe is buit by increasing vaues of p as described beow. Here, we ony write those indices that differ in the ooked-up tabe entries and the computed tabe entry (i.e., indices in the tabe that pay no roe in a given recursion are omitted. We aso ignore the expicit checking of the invaid indices in the foowing description. The agorithm ooks at the vertices which are neighbors (in G of the vertices in V p and have aready been assigned. If the vertices in V p have at east one neighbor in each of the fu buckets i 1 and i + 1, have no neighbors in any other buckets, and bucket i is empty in B, then T [p, s i,...] = T [p 1, s i V p,...]. If the vertices in V p have at east one neighbor in the fu buckets i 1 and i + 2, have no neighbors in any other buckets, and the buckets i and i + 1 are both empty in B, then T [p, t i,i+1, x i, x i+1,...] = fase if N G (X i 1 N G (X i+2, T [p 1, t i,i+1 V p, x i V p N G (X i 1, x i+1 V p N G (X i+2,...] otherwise. If the vertices in V p have at east one neighbor in the jointy fu buckets i 2 and i 1, and at east one neighbor in the jointy fu buckets i + 1 and i + 2, but have no neighbors in any other buckets, and bucket i is empty in B, then T [p, s i, f i 1, f i+1,... ] = T [p 1, s i V p, f i 1 N G (V p X i 2,i 1, f i+1 N G (V p X i+1,i+2,... ]. The recursion for the other possibiities where V p has neighbors in two distinct buckets can now easiy be deduced. We now consider the cases where V p has ony neighbors in one bucket. Again, we ony describe some key-cases, from which a other cases can easiy be deduced. If the vertices in V p have ony neighbors in the fu bucket i 1, and the buckets i 2 and i are both empty in B, but the buckets i 3 and i + 1 are either fu or non-existing, then T [p, s i 2, s i,...] = T [p 1, s i 2 V p, s i,...] T [p 1, s i 2, s i V p,...]. If the vertices in V p have ony neighbors in the fu bucket i 1, and the buckets i 3, i 2, i, and i + 1 are a empty in B, then T [p, t i 3,i 2, x i 2, t i,i+1, x i,...] = T [p 1, t i 3,i 2 V p, x i 2 V p N G (X i 1, t i,i+1, x i,...] T [p 1, t i 3,i 2, x i 2, t i,i+1 V p, x i V p N G (X i 1,...]. If the vertices in V p have ony neighbors in the jointy fu buckets i and i + 1, and the buckets i 1 and i + 2 are both empty in B, but the buckets i 2 and i + 3 are either

8 fu in B or non-existing, then T [p, s i 1, s i+2, f i, f i+1,...] = T [p 1, s i 1 V p, s i+2, f i N G (V p X i,i+1, f i+1,...] T [p 1, s i 1, s i+2 V p, f i, f i+1 N G (V p X i,i+1,...]. The fina answer (true or fase produced by the agorithm is a disjunction over a tabe entries whose indices are as foows: p = r, s i = C[i] for every index s i, t i,i+1 = C[i] + C[i + 1] for every index t i,i+1, x i C[i] for every index x i, and f i C[i] for every index f i. Remark 1. The dynamic programming agorithm in Proposition 3 can easiy be modified to construct a C-bucket arrangement (from any partia bucket arrangement B satisfying the stated conditions, if one exists. If the number of buckets is a constant, the foowing proposition wi be crucia in speeding up the procedure for assigning connected components to the right or the eft of a bucket fied with a vertex set X. Denote by sc(g the set of a connected components of G with at most n vertices and by c(g the set of a connected components of G with more than n vertices. Let V (sc(g and V (c(g denote the set of a vertices which are in the connected components beonging to sc(g and c(g, respectivey. We now make use of the fact that if there are many sma components in G \ X, severa of the assignments of the vertices in V (sc(g \ X to the buckets are equivaent. Let C be a capacity vector of size n (i.e., i C[i] = n and et B be a partia C- bucket arrangement of an induced subgraph G of G. Let C be the capacity vector obtained from C by decreasing the capacity C[i] of each bucket i by the number of vertices assigned to the bucket i in B. We say that B produces the capacity vector C. Proposition 4. Let G = (V, E be a graph on n vertices. Let C be a capacity vector of size n and ength k, where k is an integer constant. Let j be a bucket and X V be a subset of C[j] vertices. Consider a capacity vectors which are produced by the partia C-bucket arrangements of G[V (sc(g \ X X] where the vertices in X are aways assigned to the bucket j. Then, there exists an agorithm which runs in O (3 n time and takes poynomia space, and enumerates a (distinct capacity vectors produced by these partia C-bucket arrangements. Proof. Let V be the vertex set of the th connected component in sc(g \ X. Let L p denote the ist of a capacity vectors produced by the partia C-bucket arrangements of G[ 1 p V X] where the vertices in X are aways assigned to the bucket j. Note that since k is a constant, the number of distinct vectors in L p is poynomia (at most n k. Then, L 1 can be obtained by executing the agorithm of Proposition 1 on the graph G[V 1 ] with a capacity vector C which is the same as C except that C [i] = 0. In genera, L p can be obtained from L p 1 by executing the agorithm of Proposition 1 on the graph G[V p ] for every capacity vector in L p 1. As the size of each connected component in sc(g \ X is at most n, the resuting running time is O (3 n.

9 3.1 Exponentia Time 2-Approximation Agorithm for Bandwidth Let G = (V, E be the input graph. Our agorithm tests a bucket sizes from 1 to n/2 unti it finds an (n, -capacity vector C such that G has a C-bucket arrangement. For a given, et k = n denote the number of buckets. Our agorithm uses various strategies depending on the vaue of k. The case of k = 1 is trivia. If = n/2, we have at most two buckets and any partition of the vertex set of G into sets of sizes and n is a vaid C-bucket arrangement. If k 27, Coroary 1 gives a running time of O( n. For a other vaues of k, we wi obtain running times in O( n. Let I k be the set of a integers ying between n/(k 1 and n/k. The basic idea (as iustrated in Proposition 2 is quite simpe. The agorithm tries a possibe ways of assigning vertices to the midde bucket. Once the vertex set X assigned to the midde bucket is fixed and the agorithm has decided for each connected component of G \ X if the connected component is to be assigned to the buckets to the eft or to the right of the midde bucket, the probem breaks into two independent subprobems on buckets which are to the eft and to right of the midde bucket. To get the caimed running time, we buid upon this idea to design individuaized techniques for different ks (between 3 and 26. For each case, if G has at east one C-bucket arrangement for an (n, -capacity vector C, then one such arrangement is constructed. We know that if G has no C-bucket arrangement for an (n, -capacity vector C then the bandwidth of G is at east + 1 (Lemma 2, and if it has one then its bandwidth is at most 2 1 (Lemma 1. If k = 8, 10, or 12, the agorithm uses a eft-packed (n, -capacity vector C, and otherwise, the agorithm uses a baanced (n, -capacity vector C. k = 3. The agorithm goes over a subsets X V of cardinaity X = C[3] (n /2 with I 3. X is assigned to the bucket 3. If the remaining vertices can be assigned to the buckets 1 and 2 in a way such that a vertices which are neighbors of the vertices in X (in G are assigned to the bucket 2, then G has a C-bucket arrangement where C has ength 3. The worst-case running time for this case is max I3 O ( ( n X. k = 4 or k = 5. The agorithm goes over a subsets X V with X = and I k. X is assigned to the bucket 3. Then, we can concude using the dynamic programming agorithm outined in Proposition 3 (see aso the remark foowing it. The worst-case running time for these cases are max Ik O ( ( n. k = 6. If k = 6, the agorithm goes through a subsets X V with X = 2 and I 6. X is assigned to the union of buckets 3 and 4 (i.e., some non-specified vertices from X are assigned to the bucket 3, and the remaining vertices of X are assigned to the bucket 4. Then, we can again concude by the agorithm outined in Proposition 3. The worst-case running time for this case is max I6 O (( n 2. k = 7. The agorithm goes through a subsets X V with X = and I 7. X is assigned to the bucket 4. For each such X, the agorithm uses Proposition 4 to enumerate a possibe capacity vectors produced by the partia C-bucket arrangements of G[V (sc(g \ X X] (with X assigned to the bucket 4. This step can be done in O (3 n time. There are ony poynomiay many such (distinct capacity vectors. For

10 each of these capacity vector C, the agorithm goes through a choices of assigning each connected component in c(g \ X to the buckets 1 to 3 or to the buckets 5 to 7. Thus, we obtain two independent subprobems on the buckets 1 to 3 and on the buckets 5 to 7. As the number of number of components in c(g \ X is at most n (as each connected component has at east n vertices, going through a possibe ways of assigning each connected component in c(g \ X to the buckets numbered smaer or arger than 4 takes O (2 n time. Some of these assignments may turn out to be invaid. For each vaid assignment, et V 1 denote the vertex set assigned to the buckets 1 to 3. Then, the vertices of V 1 are assigned to the buckets 1 to 3 as described in the case with 3 buckets with the capacity vector (C [1], C [2], C [3] and with the additiona restriction that a vertices in V 1 which are neighbors of the vertices in X need to be assigned to the bucket 3. The number of vertices in V 1 is at most (n /2 (as C is baanced. Now the size of bucket 1 is C [1] (n 5/2. Let n 1 = (n /2 and 1 = (n 5/2. If V 1 has at east one vaid bucket arrangement into 3 buckets (with vertices in V 1 neighboring the vertices in X assigned to the bucket 3, then the above step wi construct one in worst-case O ( ( n 1 1 time. The agorithm uses a simiar approach for V 2 = V \ (V 1 X with the buckets 5 to 7. Since, the agorithm tries out every subset X for bucket 4, the worst-case running time for this case is max O I 7 (( n ( 3 n + 2 n ( n1 1 = max O I 7 (( n 2 O( n k = 8. The agorithm uses a eft-packed (n, -capacity vector C for this case. The agorithm goes through a subsets X V with X = and I 8. X is assigned to the bucket 4. The remaining anaysis is simiar to the case with 7 buckets. Buckets 1 to 3 have a joint capacity of 3 (as C is eft-packed and the buckets 5 to 8 have a joint capacity of n 4. The worst-case running time for this case is max O I 8 (( n 2 O( n max {( 3, ( n 4 }. The terms in the max expression come from the cases with 3 and 4 buckets. k = 9 or k = 11. The agorithm goes through a subsets X V with X = and I k. X is assigned to the bucket k/2. As in the previous two cases, Proposition 4 is invoked for G[V (sc(g \ X X] (with X assigned to the bucket k/2. For each capacity vector generated by Proposition 4, the agorithm ooks at every possibe way of assigning each connected component in c(g\x to the buckets 1 to k/2 1 or to the buckets k/2 + 1 to k. Each assignment gives rise to two independent subprobems one on vertices V 1 assigned to the buckets 1 to (k 1/2, and one on vertices V 2 assigned to the buckets (k + 3/2 to k (with vertices in V 1 and V 2 neighboring the vertices in X assigned to the buckets (k 1/2 and (k + 3/2, respectivey. The agorithm soves these subprobems recursivey as in the cases with 4 or 5 buckets. Let n 1 = (n /2. Then, the worst-case running times are max Ik O ( ( n 2 O( n (n 1. k = 10 or k = 12. The agorithm uses a eft-packed (n, -capacity vector C for these cases. The agorithm goes through a subsets X V with X = and I k. X is ( n1 1.

11 assigned to the bucket k/2. The remaining anaysis is simiar to the previous cases. For k = 10, the worst-case running time is max I10 O ( ( n 2 O( n (n/2. For k = 12, the worst-case running time is max I12 O ( ( n 2 O( n max{ ( ( 5, n 6 }. 13 k 26. The agorithm enumerates a subsets X V with X = and I k. X is assigned to the bucket k/2. As in the previous cases, Proposition 4 is invoked for G[V (sc(g \ X X]. For each capacity vector generated by Proposition 4, the agorithm ooks at every possibe way of assigning each connected component in c(g\ X to the buckets 1 to k/2 1 or to the buckets k/2 + 1 to k. Each assignment gives rise to two independent subprobems. For each of these two subprobems, the agorithm proceeds recursivey unti reaching subprobems with at most 2 consecutive empty buckets, which can be soved by Proposition 3 in poynomia time. If k 23, this recursion has depth 3, giving a running time of (( ( ( n max O 2 O( n n/2 2 O( n n/4 2 O( n. I k If 24 k 26, the recursion has depth 4, giving a running time of (( ( ( ( n max O 2 O( n n/2 2 O( n n/4 2 O( n n/8 2 O( n I k k 27. By Proposition 2 the running time of the agorithm is bounded in this case by (( ( n n/2 max O n/4. I k Main Resut. Putting together a the above arguments and using numerica vaues (see [13] for the compete detais we get our main resut (Theorem 1. The running time is dominated by the cases where k = 7 and k = 8. The agorithm outputs 2 1, where is the smaest integer such that G has a bucket arrangement with an (n, - capacity vector. The agorithm requires ony poynomia space. If G is disconnected, the agorithm finds for each connected component G i = (V i, E i the smaest i such that G i has a bucket arrangement corresponding to a ( V i, i -capacity vector and outputs 2 m 1, where m = max i { i }. Theorem 1 (Main Theorem. There is a poynomia space 2-approximation agorithm for the Bandwidth probem that takes O( n time on graphs with n vertices. 4 Concusion For finding exact soutions, it is known that many probems (by subexponentia time preserving reductions do not admit subexponentia time agorithms under the Exponentia Time Hypothesis [15] (a stronger hypothesis than P NP. The Exponentia Time Hypothesis supposes that there is a constant c such that 3-SAT cannot be soved in time O(2 cn, where n is the number of variabes of the input formua. We conjecture that the Bandwidth probem has no subexponentia time 2-approximation agorithm, uness the Exponentia Time Hypothesis fais. 2.

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