The minimum G c cut problem

Transcription

1 The minimum G c cut problem Abstract In this paper we define and study the G c -cut problem. Given a complete undirected graph G = (V ; E) with V = n, edge weighted by w(v i, v j ) 0 and an undirected cluster graph, G c = (V c, E c ), with V c = k, a k-cut is a partition V 1,..., V k of V (G) such that V i for i = 1,..., k. The G c -cut problem is to compute a k-cut minimizing (i,j) E c w(v i, V j ) where w(v i, V j ) = p V i,q V j w(p, q). Denote G c as cluster graph and its vertices as clusters. We show that the G c cut problem is NP-hard and even not approximable in the general case even for k = 4. We study the partial G c cut problem and classify cases where the problem is polynomial or NP-hard. We show how to extend a given G c to G c such that if the problem is NP-hard on G c it stays NP-hard on G c, and give examples of small G c s for which the problem is NP-hard. We characterize a special families of G c s for which the problem is polynomial. We focus on the restriction of the G c -cut problem for which the edge weights of G satisfy the triangle inequality, and with additional requirement on the sizes of the clusters. We study the unweighted G c -cut problem. We show that this problem is polynomial and we can use its solution to approximate the G c -cut problem in ratio which depend on the weights of G when the weights are strictly positive. We also study the minimum k-cut problem on hypergraphs and we propose approximation algorithms where the ratio depends on the rank r(g), which is the size of the largest cardinality hyperedge of G. 1

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3 1 Introduction In this paper we define and study the G c -cut problem. Definition 1.1 [G c -cut] Let G = (V, E) be a complete undirected graph with V = n and edge weights w(v i, v j ) 0. Given is also an undirected cluster graph, G c = (V c, E c ), with V c = k. A k-cut is a partition V 1,..., V k of V (G), such that V i, i = 1,..., k. The G c -cut problem is to compute a k-cut minimizing (i,j) E c w(v i, V j ), where w(v i, V j ) = p V i w(p, q). The restriction of the G c -cut problem to unweighted graphs (i.e., w(e) = 1) q V j is called the unweighted G c -cut problem while the restriction to metric distance w (i.e., satisfying triangular inequality 1 ) is called the metric G c -cut problem. G c -cut is a generalization of k-cut which is special case where G c is a k-clique. If k is part of the input it is well known that the minimum k-cut problem is NP-hard (Garey and Johnson [?]), and so is the minimum G c -cut problem. When k is fixed we show that (in contrast to the k-cut problem) in general the problem is also NP-hard. We will show special cases for which the problem can be solved in polynomial time. The version of the minimum k-cut problem with the additional requirement that each cluster V i has a size of p i N, where k i=1 p i = V, is studied in [?] where it is shown that if the weights satisfy the triangle inequality and k is constant, it possible to obtain an approximation of at most three time the optimal value. We extend this result to the G c case, in two steps: 1. First, we demonstrate the general idea when G c is a k-vertices path graph. 2. Second, we apply the same arguments when G c is a general graph on k vertices. We also study the more general minimum k-cut problem where G is a hypergraph. The minimum k-cut problem is an extension of the minimum cut problem, and has applications in the area of VLSI system design, parallel computing systems, clustering, network reliability and finding cutting planes for the traveling salesman problems. The minimum k-cut problem has been well studied in the literature. Goldschmidt and Hochbaum 1 x, y, z V, w(x, y) w(x, y) + w(y, z). 3

4 [?] proved that the problem in ordinary graphs is NP-hard when k is part of the input and gave the first polynomial-time algorithm for fixed k. The running time of their algorithm has been improved by Kamidoi et al. [?] and Xiao [?]. Currently, the best results are the O(n 2(k 1) log n 3 )-time Monte Carlo algorithm due to Karger and Stein [?] and the O(n 2k )- time deterministic algorithm due to Thorup [?]. The minimum k-cut problem has also drawn much attention in the literature for small values of k. The minimum 2-cut problem is commonly known as the minimum cut problem. Another version, the minimum 2-way cut problem, is the minimum (s, t) cut problem, which asks us to find a minimum cut that separates two given vertices s and t. These two problems are fundamental problems in the subject of graph connectivity. For ordinary graphs, the minimum cut problem can be solved in O(mn + n 2 log n) time by Nagamochi and Ibaraki s algorithm [?] or Stoer and Wagner s algorithm [?], and the minimum (s, t) cut problem can be solved in O(mn log n2 m ) time by Goldberg and Tarjan s algorithm [?]. For hypergraphs, there are also some good results on these two problems. Klimmek and Wagner [?] and Mak and Wong [?] extended Stoer and Wagner s algorithm [?] to hypergraphs and gave an O(dn + n 2 log n)- time algorithm for the minimum cut problem in hypergraphs, where d is the sum of the cardinalities of all hyperedges. Lawler [11] showed that a minimum (s, t) cut in a hypergraph can be computed by using one maximum flow computation in an auxiliary digraph with n + 2m vertices and 2d + m edges. Then a minimum (s, t) cut in a hypergraph can be found in O(dm) time. For the minimum 3 cut problem in ordinary graphs, Kapoor [?] and Kamidoi et al. [?] showed that it can be solved by using O(n 3 ) maximum flow computations. Burlet and Goldschmidt [?] and Nagamochi and Ibaraki [?] improved the result to O(n 2 ). Furthermore, Nagamochi et al. [?],[?] proved that the minimum k-way cut problem can be solved in O(mn k ) time for k = 4, 5, 6. Unfortunately, we do not find many results in hypergraphs for k 3. Currently the frequently used algorithms in VLSI system design to partition a hypergraph are some heuristic algorithms without any theoretic guarantee, such as 4

5 algorithms based on the multilevel partitioning framework [?]. Effective algorithms for minimum 3-way cuts in hypergraphs have potential applications in VLSI system design and so on. A deep understanding of the structure of 3-way cuts will help us to investigate the structure of the general k-way cut problem. In this paper, we study the complexity and the approximability of the G c -cut problem according the structure of the graph G c. 1.1 Our results Below is a list of our main results. We show that the G c cut problem is NP-hard and even not in APX by polynomial reduction from the 2K2 problem. We show for the graphs G c with at most 4 vertices, the G c -cut problem is NP-hard if and only if G c = 2K 2. We study the partial G c cut problem and classify cases where the problem in P and when the problem become NP-hard. We show how to construct recursively small hard cases. We characterize a special families of G c s for which the problem is in P. We focus on the modification of the original G c problem for which the edge weights of G satisfy the triangle inequality, and with additional requirement on the sizes of the clusters. We show that the G c -cut on this modified version remains NP-hard. We show 3-approximation algorithm for the modified version for fixed k. We show that even without the additional constraint on the cluster sizes we have 3-approximation algorithm for the G c -cut when the edge weights of G satisfy the triangle inequality. 5

6 We study the minimum cardinality G c -cut problem. We show that the problem in P and we can use its solution to approximate the minimum G c cut in ratio which depend on the weights of G. We also study the minimum k-cut on hypergraph as a function of the rank r(g), which is the size of the largest cardinality hyperedge of G. For r(g) 3 we give an approximation algorithms in two cases: 1. For fixed k we give an approximation algorithm which depends on r(g). 2. For k as part of the input we give an approximation algorithm with approximation factor twice the approximation factor for the case fixed k. 1.2 Preliminaries Most of the existing works on the G c -cut problem are given when G c = K k with additional constraints on the sizes of the clusters. Let G = (V, E) be a complete undirected graph with edge weights w(p, q) 0, (p, q) E, and values d = d 1,..., d m, d i N such that m i=1 d i n. The minimum restricted d-cut problem is to partition V into disjoint subsets V 1,..., V m such that V i d i, i {1,..., m}, minimizing the total weight of the edges whose two endpoints belong to different subsets. We observe here that for fixed m the problem is polynomially solvable, and this will be a natural extension of the minimum m-cut problem proven to be in P by Goldschmidt and Hochbaum [?]. We repeat some of the definitions and results stated in [?] for the minimum k-cut problem and extend them to the minimum restricted d-cut problem. Definition 1.2 For a set E E we use w(e ) to denote the total weight of edges in E. Let OPT= V 1,..., V m be a minimum restricted d-cut in G = (V, E). Let C i be the edges in G separating V i from V \ V i, i = 1,..., m. The weight of the cut satisfies, w(op T ) = 1 m 2 i=1 w(c i). We label the components so that w(c 1 ) w(c m ). 6

7 We choose an optimal solution satisfying the maximality assumption: There is no other optimal solution V 1,..., Vm with w( V 1 ) w( V i ), i = 2,..., m and such that V 1 V 1. From now on we assume that V 1 max(m 2, d 1 ). The case when V 1 < max(m 2, d 1 ) is handled like in [?]. Definition 1.3 The core, S, is a set of k 2 vertices in V 1. The terminal sets, are subsets T i V i, i = 2,..., m such that T i = d i. We also denote T = m i=2 T i. Note that T = k d 1. A minimum (S, T ) cut, is a partition (V S, V T ) of V with minimum weight set of edges connecting V S and V T, such that S V S and T V T. V S and V T are called the source and sink set, respectively. A maximal minimum (S, T )-cut is a minimum (S, T )-cut such that V S is maximal (it is not contained in the source set of any other minimum (S, T )-cut). Theorem 1.4 For a given core S, T V, the maximal minimum (S, T )-cut is unique and it can be identified in polynomial time. Proof: see [?]. The proof for Theorem?? is given in [?], where the only differences are in the basic assumptions. In [?] : 1. V 1 m 2, 2. V i 1, i = 2,..., m, which will be needed for preserving at least one vertex for each of the m 1 clusters. We assume, 1. V 1 max(m 2, d 1 ) 2. V i d i, i = 2,..., m, which will be needed for preserving at least d i vertices for each of the m 1 clusters. 7

8 Theorem Suppose m 4 and V 1 max(m 2, d 1 ), then there exist S V 1, S = m 2, and T = m i=2 T i, T i V i, T i = d i, such that C 1 is the maximalminimum (S, T )-cut. 2. Suppose m = 3 and V 1 max(2, d 1 ), then there exist S V 1, S = 2, and T = m i=2 T i, T i V i, T i = d i, such that C 1 is the maximal-minimum (S, T )-cut. Proof: See [?]. 2 Basic Definitions and properties Throughout this document, we use the following notation for a weighted graph (G, w): For E E(G), w(e ) = e E w(e). For v V (G) and U V (G), w(v, U) = u U w(v, u). Let us start by giving some definitions: Definition 2.1 The partial G c -cut problem is the G c -cut problem where given I = (G, w), S V (G c ) and S vertices {v 1,..., v S } V, we want to find an optimal G c -cut V 1,..., V S on I such that v i V i for i S. The d-size restricted G c -cut problem is the G c -cut problem where an integer vector (d 1,..., d k ) ( V (G c ) = k) is given. goal is to find an optimal G c -cut (V 1,..., V k ) on I where n k i=1 d i and such that V i d i. Finally, the d-restricted G c -cut problem is the d-size restricted G c -cut problem when n = k i=1 d i. In other words, V i = d i for every i = 1,..., k. The Note that if G c is a complete r-partite K d1,...,d r (See definition??) for some r (in particular, a complete graph), then the partial G c -cut problem with S = 1 is equivalent to the G c -cut problem, and then is polynomial if V (G c ) is fixed. If G c is a complete graph, then the d-size restricted G c -cut problem is equivalent to solve the minimum restricted d-cut problem (see Section?? for a definition). In general case, the partial G c -cut problem is at least as hard as the G c -cut problem. 8

9 Lemma 2.2 Assume that G c has a fixed number k of vertices. If the partial G c -cut problem is polynomial for some S, then the G c -cut problem is polynomial. Proof: If the partial G c -cut problem is polynomial for some S, then we could solve the G c -cut problem in O(n k ) times by guessing v i V i for i S. Remark 2.3 Question: What is the complexity of the partial G c -cut problem in general graphs? In the following theorem we deal with the case S > 1 and G c = K k. Theorem 2.4 The complexity of the partial K k -cut when k is fixed is polynomial when S = 2 and NP-hard when S > 2. Proof: We divide the proof to the following sub cases. 1. S = k. (a) S = 2. The problem is equivalent to the min-2-cut so the problem is in P. (b) S >= 3. The problem is equivalent to the minimum multiway cut with S >= 3 so the problem is NP-hard. 2. where k > S. Denote l = k S > 0 (a) S > 2. Let G = (V, E), S V, S > 2 be an instance of the minimum multiway cut problem. Constructing an instance for the partial K k problem in the following manner: S = S. G = (V, E ). V = V {x 1,..., x l } E = E E 1 E 2. Where E 1 = {(x i, v), and w(x i, v) = 0 i 1,..., l, v V }. E 2 = {(x i, x j ) i 1,..., l, v V }. and w(x i, x j ) = 0 i, j 1,..., l, i j}. 9

10 Solving the partial G c cut where G c = K k for G, S optimally we must assign each x i, i 1,..., l to a single cluster of K k so that the vertices of V must arrange in an optimal minimum multiway cut on the rest of the S clusters of K k, and the result follow. (b) S = 2, S = {s, t}. We use the same construction as in [?] in a small modification that we enumerate over all the cores S such that s S, and terminals T such that t t instead of enumerating all the cores S and terminals T as done at [?]. The rest of the proof is exactly like in [?]. Definition 2.5 Let G = (V, E) be a simple graph on n 2 vertices and let S V (G). The contracted graph of G from S, denoted G(S), is the simple graph G(S) = (V, E ) where V = V \ S {v S } and (u, v) E if u, v / S {v S } and (u, v) E or if u = v S, v / S {v S } and s S with (s, v) E. The neighborhood of S, denoted N G (S) is the set of vertices adjacent to some vertices in S, i.e., N G (S) = {v V : u S, (u, v) E}. In particular, N 2 G (v) = N G(N G (v)). The degree of vertex v, denoted deg G (v), is the size of the neighborhood of v, i.e., deg G (v) = N G (v). A leaf is a vertex v such that deg G (v) = 1. Vertex u is a nested neighbor of vertex v if (u, v) / E and N G (u) N G (v). They are twins if (u, v) / E and N G (u) = N G (v). Remark 2.6 Assume that vertices 1 and k are twins of G c where V (G c ) = k. The G c -cut problem on I = (G, w) is equivalent to the d-size restricted G c-cut problem on I = (G, w) where G c = G c {k} and d 1 = 2, d i = 1, for i = 2,..., k 1. Proof: Trivial, because vertices 1 and k are twins of G c. 10

11 Theorem 2.7 Assume that vertex 1 is a nested neighbor of vertex 2 in G c, where V (G c ) = k. In any feasible G c -cut (V 1,..., V k ) of I = (G, w), one can assume that V 2 = 1. Proof: Let (V 1,..., V k ) be a G c -cut (V 1,..., V k ) of I. Assume that V 2 > 1 and let x V 2. Consider the G c -cut (V 1,..., V k ) where V i = V i if i 1, 2, V 1 = V 1 (V 2 \ {x}), V 2 = {x}. The value of (V 1,..., V k ) is not larger than the value of (V 1,..., V k ) because w is non-negative, (1, 2) / E c and N Gc (1) N Gc (2). Using Theorem??, we deduce the following result for the leafs: Corollary 2.8 Assume that vertex 1 is a leaf of G c where V (G c ) = k. In any feasible G c -cut (V 1,..., V k ) of I = (G, w), one can assume that V i = 1 for all i N 2 G c (1) \ {1}. Proof: If 1 is a leaf of G c, then for every i N 2 G c (1) \ {1}, vertex 1 is a nested neighbor of vertex i in G c. 3 Complexity results of the minimum G c -cut problem In this section we show that the complexity of G c -cut problem depends on the structure of G c. Definition 3.1 Consider a graph G = (V, E) such that V = k. 1. G is a matching graph, denoted wk 2, if k = 2w and deg(v) = 1 v V. 2. G is a k-path graph, denoted P k, if its a path with k vertices. 3. G is a k-cycle graph, denoted C k, if its a cycle with k vertices. 4. G = K d1,...,d r = (V, E) is a complete r-partite graph if there exists a partition L 1,..., L r of V with d i = L i, and r i=1 d i = k such that for every i j {1,..., r}, u, v L i (u, v) / E and u L i, v L j (u, v) E. Moreover, we require that r is the minimum integer such that the property holds. The sets L i are the color classes of G. 11

12 Consider V 1, V 2 V, V 1 V 2 =. Denote the edges with one end at V 1 and the other end at V 2 as E(V 1, V 2 ). A biclique is a complete bipartite graph. Definition 3.2 Biclique Vertex-Partition: Instance: A graph G and positive integer k. Question: Does G have a biclique vertex partition of size at most k consisting of mutually vertex-disjoint bicliques? (where the bicliques are (not necessarily vertex-induced) subgraphs of G). Remark 3.3 For every fixed k 3, biclique vertex-partition is NP-complete, and remains NP-complete for bipartite graphs, see [?]. The case k = 2 has been open since a long time, but very recently, biclique vertex-partition with k = 2 has been proved NP-complete, [?]. Because the case k = 1 is polynomial, the case k = 2 is equivalent to biclique vertexpartition of size exactly 2. Theorem 3.4 The 2K 2 -cut problem is NP-hard. Proof: We propose a polynomial reduction from biclique vertex-partition. Let G = (V, E) with V = n and k = 2 be an instance of biclique vertex-partition. Consider the complete graph (G, w) defined as follows: w(e) = 0 if e E and w(e) = 1 otherwise. We claim that there exists a 2K 2 -cut of G with value 0 iff G admits a biclique vertexpartition of size exactly 2. Let G i = (A i, B i ; E i ) with i = 1, 2 be a biclique vertex-partition of G. Clearly, V 2i 1 = A i, V 2i = B i for i = 1, 2 is a G c -cut of G with value 0. Conversely, let (V i ) i 4 be a a 2K 2 -cut of G with value 0. Thus, for every i 2, G i = (V 2i 1, V 2i ; E i ) is a biclique of G and then, (G 1, G 2 ) is a biclique vertex-partition of G of size 2. Remark 3.5 Note that the problem of finding a minimum K 2 -cut is polynomial. Corollary 3.6 The 2K 2 -cut problem is not approximable. 12

13 Proof: In proof of Theorem??, we have shown that for the 2K 2 -cut problem, it is NPcomplete to distinguish between opt 0 and opt > 0, where opt is the value of an optimal G c -cut. So, the result follows. Definition 3.7 In the following items we will define two type of extension for a given graph G. 1. For a given graph G and a path graph P = 1,..., i on i vertices, denote G + P an open P -extension of G if 1 V (P ) connect to V (G) by at least one edge and there is no edge between G and T = {2,..., i} V (P ). Denote T as the terminals set of G + P. 2. For a given graph G and a path graph P = 1,..., i on i vertices, denote G + P an closed P -extension of G if G + P connected and contained no leafs, and no edges between V (G) and T = {2,..., i 1} V (P ). Denote T as the terminals set of G + P. For instance, Figure?? gives an open P 3 -extension of 2K 2 with P 3 = (u 1, u 2, u 3 ) and terminal T = {u 2, u 3 }. Actually, the only induced subgraphs of G isomorphic to P 3 satisfying the condition of?? are G 1 = (v 1, v 2, u 1 ), G 2 = (v 3, v 4, u 1 ), and G 2 = (u 1, u 2, u 3 ). Finally, for every i = 1, 2, 3, we have: 2K 2 = G G i. The tree G will be called the 3- star of length 2 and denoted by S 2 3 (more generally, the p-star of length 2 is given by Sp 2 = {(r, v1 i ), (vi 1, vi 2 ) : i = 1,..., p}). 13

14 a b a b g h i e f e f Figure 1: Example of open P 3 -extension where G = 2K 2, H = P 3 = (u 1, u 2, u 3 ), G = S 2 3 and T = {u 2, u 3 }. a b a b j i e f e f g h Figure 2: Example of closed P 4 -extension where G = 2K 2, H = P 4 = (u 1, u 2, u 3, u 4 ) and T = {u 2, u 3 }. Figures?? and?? give a closed P 4 -extension of 2K 2 or 3K 2 with P 4 = (u 1, u 2, u 3, u 4 ) and terminal T = {u 2, u 3 }. 14

15 g h i j i j a b a b c d c d e f e f g h i j d f g h Figure 3: Another example of closed P 4 -extension where G = 3K 2, H = P 4 = (u 1, u 2, u 3, u 4 ) and T = {u 2, u 3 }. On the top, the Ψ-graph (left and top) and the κ- graph (right and top). On the bottom, the κ-graph minus G 2 = (v 2, v 1, u 1, u 2 ). We get 3K 2 κ-graph G 2. 15

16 In Figure??, for the Ψ-graph, the only induced P 4 satisfying the condition of?? are G 0 = (u 1, u 2, u 3, u 4 ) and G 1 = (u 1, v 1, v 2, u 4 ) and we get Ψ-graph G i = 3K 2 for i = 0, 1 while for the κ-graph, the only induced P 4 satisfying the hypothesis are G 0 = (u 1, u 2, u 3, u 4 ), G 1 = (u 1, v 1, v 2, u 4 ), G 2 = (v 2, v 1, u 1, u 2 ) and G 3 = (u 3, u 2, u 1, v 1 ). Moreover, we have κ-graph G i = 3K 2 for i = 0, 1 and 3K 2 κ-graph G i for i = 2, 3. Now, we present some polynomial reductions preserving approximation from the G c - cut problem to itself depending on the structure of graph G c. Theorem 3.8 There exists a polynomial reduction preserving approximation from the G c -cut problem to the G c-cut problem in the following cases: 1. Assume that G c has k vertices and the smallest connected component of G c has s 2 vertices. G c = G c + H assume that H is a connected graph of at least 2 vertices and at most s vertices, is disconnected from G c and if V (H) = s, H G i (which means E(H) E(G i )) for every connected component of G c with exactly s vertices. 2. G c = G c + P 0 is an open P i -extension of G c with i 3, P i = (k + 1,..., k + i) and terminal T = {k + 2,..., k + i}. 3. G c = G c + P 0 is an closed P i -extension of G c with i 4, P i = (k + 1,..., k + i) and terminal T = {k + 2,..., k + i 1}. Proof: 1. Let G c and H be two graphs satisfying the condition at?? and consider G c = G c +H where V (H) = {k+1,..., k+p}, 2 p s. Let (K n, w) be an instance of the G c -cut problem and consider the instance (K n+p, w ), V (K n+p ) \ V (K n ) = {u 1,..., u p } of the G c-cut problem defined as follows: if u, v V (K n ), then w (u, v) = w(u, v). If u V (K n ) and v / V (K n ), w (u, v) = + 2 Finally, w (u i, u j ) = 0 if (i, j) E(H) and w (u i, u j ) = otherwise. 2 In the rest of the paper, we set + in order to simplify, but the right value will be for instance (n + 1)w max where w max = max e E(G) w(e). 16

17 Clearly, any G c -cut of (K n, w) can be converted into a G c-cut of (K n+p, w ) with same value by setting V k+i = {u i }. Conversely, consider any G c-cut (V 1,..., V k+p ) of (K n+p, w ). From the previous part, we can assume that this G c-cut has a finite value. Assume u 1 V i1. We get V i1 V (K n ) = because each connected component of G c has a size at least two and V i2 {u 1,..., u p } for every (i 1, i 2 ) E(G c). Hence, we deduce V j {u 1,..., u p } if V j {u 1,..., u p } =. Now, we must get V ij = {u j } for j = 1,..., p because each connected component G c has a size at least 2 and at most p. Hence, the subgraph G induced by {i 1,..., i p } is a connected component of G c. If p < s or G is isomorphic to H, then clearly, we must get G = H and the restriction of this G c-cut to (K n, w ) is a G c -cut of (K n, w) with same value. Now, assume p = s and G H. Since, by assumption E(H) E(K n ), we get E(K n ) \ E(H) and then the value of the G c-cut restricted to H has an infinite value, leading to contradiction. Hence, G = H and the result follows. 2. We first prove the case i = 3. Let (K n, w) be an instance of the G c -cut problem where G c = (V c, E c ) is a graph with V c = k 1 vertices, and let P 0 = (k + 1, k + 2, k + 3). Now, let G c = (V c, E c) = G c + P 0 be any P 3 -extension of G c with terminal T = {k+2, k+3} (which means that the edges between the P 0 and G c are only connected to endpoint k+1). Consider the following instance (K n+3, w ) of the G c-cut problem: V (K n+3 ) \ V (K n ) = {u 1, u 2, u 3 } and w (u, v) = w(u, v) for u, v V, w (u 1, v) = 0, w (u 2, v) = w (u 3, v) = + for v V, and w (u 1, u 2 ) = w (u 3, u 2 ) = 0, and w (u 1, u 3 ) = +. Any G c -cut of (K n, w) can be converted into a G c-cut of (K n+3, w ) with same value by setting V k+i = {u i } for i = 1, 2, 3. Conversely, assume that (V 1,..., V k+3 ) is a G c- cut of (K n+3, w ) with finite value. Assume that u 2 V i2 and (i 3, i 2 ) E c (because G c is a connected graph with at least 4 vertices). We get V i3 V (K n ) =, V i2 {u 1, u 3 } = and V i3 {u 1, u 3 } because by construction w (u 2, v) = w (u 3, v) = + for v V and w (u 1, u 3 ) = +. Hence, we deduce V i2 = {u 2 } since V i2 V (K n ) =. 17

18 If V i3 = {u 1, u 3 }, then vertex i 1 must be a leaf of G c and vertex i 2 has a neighbor i 1 i 3 in G c (because G c is connected with at least 4 vertices). But V i1 V (K n ) and w (u 3, v) = + for v V (K n ), contradiction. Now, since w (u 3, v) = w (u 3, u 1 ) = + for v V (K n ) and G c is connected with at least 4 vertices we get V i3 = {u 3 } and vertex i 3 is a leaf of G c. Because i 3 is a leaf of G c, then vertex i 2 must get exactly one neighbor i 1 i 3 and V i1 = {u 1 }. So, P = (i 3, i 2, i 1 ) is an induced P 3 of G c with terminal {i 2, i 3 }. Since, G c is an P 3 -extension of G c, then the value of the G c-cut is minimum if V k+ i = {u i } for i = 1, 2, 3 (because G c P 0 = G c. Actually, if we flip the sets corresponding to P by the sets corresponding to P 0, the value of the G c-cut does not increase). Hence, the restriction of this G c-cut to (K n, w ) is a G c -cut of (K n, w) with same value. For the general case, let P 0 = (k + 1,..., k + i) with i 3. We replace (K n+3, w ) by (K n+i, w ) where: V (K n+i ) \ V (K n ) = {u 1,..., u i } and w (u, v) = w(u, v). For u, v V (K n ), w (u 1, v) = 0, w (u j, v) = + for v V and j = 2,..., i. Finally, w (u j, u j+1 ) = 0, for j = 1,..., i 1 and w (u j, u j ) = + otherwise. The rest of the proof is completely similar to the previous one. 3. We first prove the case i = 4. Let (K n, w) be an instance of the G c -cut problem where G c = (V c, E c ) has k 1 vertices and let P 0 = (k + 1,..., k + 4). Now, let G c = (V c, E c) = G c + P 0 be any P 4 -extension of G c with terminal T = {k + 2, k + 3} such that G c is without a leaf. Consider the following instance (K n+4, w ) of the G c-cut problem: V (K n+4 ) \ V (K n ) = {u 1,..., u 4 }, and w (u, v) = w(u, v) for u, v V (K n ). Moreover, w (u j, v) = 0 for j = 1, 4, and w (u j, v) = + for j = 2, 3. Finally, w (u j, u j+1 ) = 0, for j = 1,..., 3, and w (u j, u j ) = + otherwise. 18

19 Any G c -cut of (K n, w) can be converted into a G c-cut of (K n+4, w ) with same value by setting V k+i = {u i } for i = 1,..., 4. Conversely, assume that (V 1,..., V k+4 ) is a G c-cut of (K n+4, w ) with finite value. Assume that u 3 V i3 and (i 3, i 4 ), (i 3, i 2 ) E c (because G c has minimum degree 2). By construction, we get V i3 V = because otherwise V ij V = for j = 1, 3 and V ij {u 4, u 2 } = for every j = 2, 4 (thus, this G c-cut will get an infinite value because either u 1 V ij or u 2 V ij for some j = 2, 4). Hence, we deduce V ij {u 4, u 2 } for j = 2, 4 and then we can assume V ij = {u j } for j = 2, 4. Moreover, i 3 must have a degree 2 in G c and (i 2, i 4 ) / E(G c). Now, because i 2 has a degree has at least 2 in G c, there is an edge (i 1, i 2 ) E(G c) with i 1 / {i 3, i 4 }. Thus, V i1 = {u 1 }, and on the one hand i 2 must have a degree 2 in G c, and on the other hand (i 1, i 4 ) / E(G c). Hence P = (i 1,..., i 4 ) is an induced P 4 of G c with terminal {i 2, i 3 }. Finally, since G c is a P 4 -extension of G c with terminal {k + 2, k + 3}, we can assume that V k+i = {u i } for i = 1,..., 4. In conclusion, the restriction of this G c-cut to (K n, w ) is a G c -cut of (K n, w) with same value. For the general case, let P 0 = (k + 1,..., k + i) with i 4. We replace the instance (K n+4, w ) by (K n+i, w ) where: V (K n+i ) \ V (K n ) = {u 1,..., u i }. w (u, v) = w(u, v) for u, v V (K n ). w (u j, v) = 0 for j = 1, i. w (u j, v) = + for v V (K n ) and j = 2,..., i 1. Finally, w (u j, u j+1 ) = 0, for j = 1,..., i 1 and w (u j, u j ) = + otherwise. The rest of the proof is completely similar to the previous one. We saw at all the above constructions that the new i added vertices placed at the new i added clusters in an optimal solution and the original vertices must placed in an optimal 19

20 way at the original clusters. Since the construction can perform in polynomial time the result are follow. Definition 3.9 A spider graph is a tree with at most one vertex of degree more than two, called the center of spider (if no vertex of degree more than two, then any vertex can be the center). A leg of a spider is a path from the center to a vertex of degree one. Corollary 3.10 The G c -cut problem is NP-hard and not approximable in the following cases: (i) G c A collection of disjoint paths each with either 2 vertices or at least 5 vertices. (ii) G c is a spider graph with at least 3 legs and at list 2 vertices in any leg. (iii) G c is a tree. (iiii) G c consisting of two triangles connected by a path with at least one internal vertex. Proof: (i). Starting from G c = 2K 2 we can apply part 1 of Theorem?? to add new disjoint P 2 path s and we can lengthen each by applying part 1 of Theorem?? and the result follows. (ii). We use (i) to get collection of disjoint paths each with either 2 vertices or at least 5 vertices and then by applying part 2 of?? we can connect all the obtained collection of disjoint paths with another path graph with length 3 and to obtain the required spider graph. (iii). Since spider is a private case of a tree we deduce the claimed result. (iiii).applying part 3 of?? on 2K 2 the result follows. Remark 3.11 The G c -cut problem is NP-hard and not approximable in the following cases: (i).when G c = Ψ-graph or G c = κ-graph. (ii).when G c is a pk 2 graph. 20

21 (iii).when G c is a p-star of length 2. Proof: (i). The Ψ-graph and the κ-graph are P 4 -extensions of 3K 2 and P 0 = (7, 8, 9, 10) with terminal T = {8, 9} and are without leaf. Hence, using part 3 of Theorem?? and Theorem??, the result follows. (ii) Starting from 2K 2 and applying p 2 part 1 of Theorem?? and the pk 2 graph obtained. (iii) Starting from (p 1)K 2 graph proved to be NP-hard at (ii) and then applying part 2 of Theorem?? join one side of each of the K 2 graphs with one side of new added P 3 graph, the result follows. In part 1 of Theorem??, we have proved that the complexity of the G c -cut problem does not depend on the connectivity of G c as long as, the size of each connected component is a at least 2. In Section??, we will see that the G c -cut problem is polynomial time solvable if G c has a fixed number of vertices and at least one isolated vertex. So now, we will assume that G c is connected. In Corollary??, all of the different connected graphs G c such that the G c -cut problem is NP-hard have a maximum degree at least 3. Here, we prove the prove that this result remains true for the partial G c -cut problem on a connected graphs G c of maximum degree 2. Theorem 3.12 The partial P k -cut(defined at??) problem is NP-hard and not approximable in the following cases: (i) k = 5 or k 8, even when only one vertex is specified (i.e., S = 1). (ii) k = 6, even when only two vertices are specified (i.e., S = 2). Proof: We give reduction preserving approximation from the 2K 2 -cut problem proved NP-hard and not approximable in Theorem?? and Corollary??. Let I = (K n, w) be an instance of the 2K 2 -cut problem. 21

22 For (i) and k = 5, consider the instance I = (K n+1, w ) where V (K n+1 )\V (K n ) = {x}, w (u, v) = w(u, v) if u, v x, and w (u, x) = 0 for u V (K n ). Let S = {3} with x V 3. Assume that G c = P 5 = (1, 2, 3, 4, 5). Let (V 1, V 2, V 3, V 4 ) be any 2K 2 -cut of I. (V 1, V 2, V 3, V 4, V 5 ) with V 1 = V 1, V 2 = V 2, V 4 = V 3, V 5 = V 4 and V 3 = {x} is a P 5-cut of I with same value. Conversely, let (V 1, V 2, V 3, V 4, V 5 ) be any P 5-cut of I such that x V 3. Using Corollary?? with leaf 1 and N 2 P 5 (1)\{1} = {3}, we know that we can assume that V 3 = {x}. Hence, (V 1, V 2, V 3, V 4 ) where V 1 = V 1, V 2 = V 2, V 3 = V 4, V 4 = V 4 is a 2K 2-cut of I with the same value. For k 8, using the P i -extension of P 5 for i 3 given in part 2 of Theorem?? and the result given above for the partial P 5 -cut problem, the result follows. For (ii) and k = 6, consider the instance I = (K n+2, w ) where V (K n+2 ) \ V (K n ) = {x, y}, w (u, v) = w(u, v) if u, v x, u, v y and w (u, x) = w (u, y) = 0 for u V, Let S = {3, 4} with x V 3 and y V 4. Assume that G c = P 6 = (1, 2, 3, 4, 5, 6). Since, vertices 1 and 6 are leafs of P 6 and NP 2 6 ({1, 6}) \ {1, 6} = {3, 4}, the same proof as previously gives the expected result. In Section??, we will see that the P k -cut problem is polynomial if k 4 (note that these results also holds the partial P k -cut problem). Remark 3.13 The problems of finding a minimum partial P 7 -cut or minimum P k -cut for k 5 are open. The complexity of the C k -cut problem or the partial C k -cut problem where k 5 are open problems. Nevertheless, these restrictions are NP-hard when the number of vertices is unbounded because the C n -cut problem on (G, w) (resp., P n -cut) is clearly equivalent to solve the traveling salesman problem on (G, w), (resp., Hamiltonian path problem). 22

23 1 2 3 Figure 4: An H 3 = K 2 + K 1 graph To conclude this section, we give a list of G c s for which the G c -cut problem are still open because we will see in Section?? that the C k -cut problem for k 4 and the P k -cut problem for k 4 are polynomial time solvable. 1. C k, for some k 5, 2. P k, for k 5. 4 Polynomially solvable cases 4.1 When G c contains twins Here, we mainly show that if G c is a complete r-partite graph K d1,...,d r (see Definition??) where k = r i=1 d i is fixed, then the G c -cut problem can be solved in polynomial time, using an extension of the algorithm of [?]. Some simple complete r-partite graphs are the following: the stable graph K n (ie., E = ) is complete 1-partite, and the complete graph is complete n-partite. We also look at the case where G c is a restricted split graphs. Let us begin by some properties of complete r-partite. As we will see, these graphs are recognizable within polynomial-time. (although deciding if a graph is r-partite is a NP-complete for any fixed r 3). Definition 4.1 H 3 = K 2 + K 1, that is the graph G = (V, E) with V = 3 and E = 1 (See Figure??). Lemma 4.2 G = (V, E) is complete r-partite if and only if G is H 3 -free. 23

24 Proof: Suppose that G is complete r-partite. It is clear from the definition that u, v, w V we have three cases, none of which defines an H 3 graph: 1. u, v, w reside at different color classes, so the graph induced by them is a 3-clique. 2. u, v reside at the same color class, and w belongs to a different color class, so the graph induced by them contains the edges (u, w), (v, w). 3. u, v, w reside at same color class so the graph induced by them contains no edge. The opposite direction is done by induction on n = V (G). Assume that any graph G with less than n vertices which does not contain H 3 as an induced subgraph is complete r-partite for some r > 0, and consider a graph G = (V, E) with n vertices which does not contain H 3 as an induced subgraph. Let v V. G = G v is also H 3 -free, and then by inductive hypothesis is a complete r-partite graph. We study two cases. 1. deg G (v) = n 1. We add v in a new color class L r+1. Obviously, G is complete (r + 1)-partite. 2. deg G (v) < n 1. So, there is u L i such that (u, v) / E. We add v in the color class L i. Let us prove that G is complete r-partite. First, L i {v} is a stable set because L i is a stable set and G is H 3 -free. Second, j i, u L j, (u, v) E. Otherwise, u L j with (u, v) / E. Let w L i. The graph induced by {u, v, w} is isomorphic to H 3, a contradiction. Using Lemma??, it is clear that we can check in O( V (G) 3 ) whether a graph is complete r-partite. Actually, a careful analysis of Lemma?? gives a O( E(G) ) time algorithm to construct such graphs. Theorem 4.3 The K d1,...,d r -cut problem can be solved in polynomial time if k = r i=1 d i is fixed and is NP-hard, if there exists d i and d j unbounded. 24

25 Proof: Assume k = r i=1 d i fixed and G c = K d1,...,d r. Since G c is a complete r-partite graph, any two vertices u, v L i of the same layer are twins. Hence, the K d1,...,d r -cut problem is equivalent to solve the d -size restricted K r -cut problem on I = (G, w) where d i = d i, for i = 1,..., r which is itself equivalent to computing a minimum r-cut (V 1,..., V r ) of G with the constrains V i d i. It follows from Theorem?? that this problem can computed in polynomial time. Now, we first prove that the case of complete bipartite graph is NP-complete if k = d 1 + d 2 is unbounded. The bisection graph problem consists of finding a cut of minimum size of a graph such that the two cut sets have the same size. Assume n even and consider d 1 = d 2 = n/2. The K d1,d 2 -cut problem on (G, w) is equivalent to solve the bisection graph problem on G = (V, E) (by setting w(e) = 1 if e E and w(e) = 0 if e / E) which is known to be NP-hard [?]. Now, we reduce the K d1,d 2 case to the K d1,d 2,d 3 case. An inductive proof on r allows us to conclude the proof. Given an instance I = (K n, w) of the K d1,d 2 -cut problem, consider the following instance I = (K n+d3, w ) of the K d1,d 2,d 3 -cut problem: V (K n+d3 ) \ V (K n ) = {u 1,..., u d3 } and if u, v V (K n ), w (u, v) = w(u, v). If u V (K n ) and v / V (K n ), w (u, v) = 0. Finally, w (u i, u j ) =. Consider a solution of the K d1,d 2,d 3 -cut problem with color classes L 1, L 2, L 3. By construction, {u 1,..., u d3 } belongs to the same color class, say L i because G c is complete 3-partite and w (u i, u j ) =. We study two case: L i d 3. Wlog., we can assume that d 3 sets of L i are such that V i = {u i } with i = 1,..., d 3 because G c is 3-partite. Let L j = d 3. We flip the sets of L i different to V i in L j (as new sets). We obtain a new solution of the K d1,d 2,d 3 -cut problem. L i = d 3. Wlog., we can assume that {u i } V i for i = 1,..., d 3, because G c is 3-partite. We move the vertices of L i \ {u 1,..., u d3 } to the color class L j with j i. Again, we obtain a new solution of the K d1,d 2,d 3 -cut problem with value. 25

26 g h a b e f Figure 5: A K 4 K 2. i Now, let S be the vertices which have been flipped. For each of the two cases, the new solution loses w(s, C j ) and wins w(s, {u 1,..., u d3 }) = 0. Hence, the new solution has a better cost and then we can assume the restriction of K n is a K d1,d 2 -cut with same value. For instance, K 4 K 2 the graph depicted in Figure?? is a complete 3-partite K 2,1,1 and then, the K 4 K 2 -cut problem is solvable in polynomial time. Remark 4.4 If only d 1 depends on the instance and r is fixed, then the K d1,...,d r -cut problem can is still an open problem. 4.2 When G c contains nested neighbors or leafs Definition 4.5 G = (S, K; E) is a split graph if V (G) = S K where K S =, S is a stable set and K is a maximum clique of G. It is called restricted split graph if deg G (v) < V (G) 1 for every v K. For instance, P 4 = (v 1, v 2, v 3, v 4 ) is a restricted split graph with K = {v 2, v 3 } and S = {v 1, v 4 }. Theorem 4.6 If G c = (S c, K c ; E c ) is restricted split graph where K c is upper bounded by a constant, then the G c -cut problem can be solved in polynomial time. 26

27 Proof: Let G c = (V c, E c ) be a restricted split graph on k 3 vertices and let OP T = (V 1,..., V k ) be an optimal G c-cut of I = (G, w). Then, for every i K, there exists i S such that i is nested neighbor of vertex i in G c because G c is a restricted split graph. Using Theorem??, we know that V i = 1 for all i K c. Hence, we can guess the K c vertices of V i = {v i } for i K c. After, consider the following complete bipartite graph BP = (S c, V \ {vi : i K c}; E(BP )), edge weighted by d where d(i, v) = j N Gc (i) w(v, v j )), and find a b-matching M saturating S c of minimum weight d (the algorithm is the same as finding a b-matching of maximum weight d where d (e) = d max d(e), d max = max e E(BP ) d(e)) with b (i) = 1 and b + (i) = V for i S c and b (v) = b + (v) = 1 for v V \ {v i : i K c}. Recall that a b-matching of a graph G = (V, E) is a subset M such that if G = (V, M), then v V, b (v) deg G (v) b + (v). A b-matching of maximum weight can be done in polynomial-time O( V (G) 3 ), see [?] section 21 page 337. Since any G c -cut corresponds to a b-matching M saturating S c with value d(m) (when v i for i K c have been guessed), the previous algorithm finds an optimal solution in time O(n Kc +3 ). In particular, the P 4 -cut problem on (G, w) can be solved in O(n 5 ) time. However, for the P 4 -cut problem on (G, w), we can improve the complexity to O(n 3 ). Instead of applying a b-matching algorithm, we apply the following greedy algorithm: each vertex v of V \ {v 2, v 3 } (here K c = {2, 3} and S c = {1, 4}) is assigned to the V i with i = 1, 4 minimizing its contribution (i.e., i = arg min s Sc j N Gc (s) w(v, v j )). Careful attention must be taken to avoid to get V1 = or V 4 =. For instance, if V 1 =, then we find v = arg min{d(1, v) d(4, v) : v V \{v2, v 3 }} and we add v to V1. The time complexity of this algorithm is O(n 3 ). Also note that in the same spirit of the proof of Theorem?? the result holds if S N 2 G c (S) = V (G c ) where S is the leafs of G c. Lemma 4.7 If G c has a fixed number of vertices and S N 2 G c (S) = V (G c ), where S is leafs of G c, then the G c -cut problem can be solved in polynomial time. 27

28 4.3 When G c contains isolated vertices Definition 4.8 G is an H 0 graph if it contains at least one isolated vertex v, i.e., deg G (v) = 0. Let S(G) = {v V : deg G (v) = 0}. Lemma 4.9 Let G c be an H 0 graph. The G c -cut problem can be solved in polynomial time iff V (G c ) \ S(G c ) is fixed. Proof: Consider the unbounded case of V (G c ) \ S(G c ). G c = C n + K 1 is an H 0 graph and the G c -cut problem is clearly equivalent to solve the minimum traveling salesman problem. Now, let G c be an H 0 graph such that V (G c ) \ S(G c ) = k 1 is fixed and consider an instance I = (G, w) of the G c -cut problem. Denote the clusters corresponding to vertices of S(G c ) as V k,..., V p and the k 1 clusters corresponding to V (G c ) \ S(G c ) by V 1,..., V k 1. Enumerate the ordered subsets H V, H = k 1. Insert the vertices from H into the clusters V 1,..., V k 1 so that each cluster contains exactly one vertex according to the order, and insert arbitrarily the vertices of V \ H into the remaining clusters V k,..., V p. Let S H be this solution and S ( ( be the minimum one. Since k 1 is fixed the complexity is O n ) k 1 (k 1)!(k 1) 2) = O(n k 1 ). The obtained solution is optimal since any assignment of the vertices of V into the clusters V (G c ) \ S(G c ) which results with one of these clusters having more than one vertex, can be improved by moving all these vertices except one from this cluster into any one cluster among V k,..., V p. This new solution is an improved solution to the original, since the subset of edges in the new solution is strictly contained in the original solution. Definition 4.10 A complete r H 0 graph is a graph G = (V, E) where V (G) = r i=1 L i such that (i) for every i {1,..., r} the graph induced by L i is an H 0 graph, and (ii) for every i, j {1,..., r} with i j, (u, v) E for every u L i, v L j t. In particular a complete 1 H 0 graph is a H 0 graph and K 3 ± K 3 2 (see Figure??) is 3 ± means that K 3 and K 2 share a common vertex. 28

29 Figure 6: A K 3 +K 2. a complete 2 H 0 graph (where L 1 = {1} and L 2 = {2, 3, 4}). The following result extends Lemma?? to complete 2 H 0 graphs. Theorem 4.11 Suppose that G c is a complete 2 H 0 cluster graph and V c = k c is constant. Then the G c -cut problem can be solved in polynomial time. Proof: Denote V c = V c1 V c2 such that the two induced graphs G c (V c1 ), G c (V c2 ) are H 0 - graphs, V c1 = k 1, and V c2 = k 2, where k 1 + k 2 = k. Let I = (G, w) be an instance and assume that optimal solution assigns the vertices V 1 V to the clusters of G c(v c1 ) and V 2 V (with V 1 V 2 = V ) to the clusters of G c(v c2 ). The optimal solution on V 1 must assign k 1 1 vertices to k 1 1 clusters and the rest of its vertices to the cluster represented by the isolated vertices. Similarly with V2. So again we iterate over all ordered subsets K 1, K 2 V such that K 1 = k 1, K 2 = k 2, and compute a min-cut on G which separates K 1 from K 2. Denote the cost of the assignment of K 1 vertices to the clusters V c1 according to the order as W 1. Denote the cost of the assignment of K 2 vertices to the clusters V c2 according to the order as W 2. Denote the cost of min-cut on G which separates K 1 from K 2 as W 3, and let W = W 1 + W 2 + W 3. Compute K 1, K 2 minimizing W. The clustering derived from them is the optimal solution and has a polynomial runtime complexity in V. 29

30 k 1 k Figure 7: A complete 2 H0 Example. proposition 4.12 If G c has at most four vertices, then the G c -cut problem is polynomial iff G c 2K 2. Proof: The only cases to study are the graphs which contain a P 4 as partial subgraph, because the remaining cases are the H 0 -graphs, 2K 2, or the connected graphs on at most 3 vertices (and then isomorphic to K 1, K 2, K 3 or P 3 = K 1,2 ). Thus, the graphs which contain a P 4 are K 4, C 4 = K 2,2, K 3 ± K 2 (see Figure??), K 4 K 2 (see Figure??), but all are polynomial as proved previously. The following example demonstrate a G c for which the previous Theorems imply a polynomial algorithm. V c = {0,,..., k}, E c = {(0, 1),..., (0, k)} {(1, 2),..., (k 2, k 1)}. G c is a complete 2 H 0 graph where L 1 = {0}, L 2 = {1,..., k} depicted in Figure??. It is easy to see that complete r-partite graphs generalize the k-cut problem, because G c is a k-clique and we can look at each vertex as a different color class. It is also clear 30

31 that complete r H 0 graphs generalize complete r-partite graphs, because each color class is an H 0 -graph. We are still left with the open problem whether when G c is a complete r H 0 graph, the problem is in P or NP-hard for r > 2. 5 Approximation results 5.1 The d-restricted G c -cut problem The version of the k-cut problem with the additional requirement that cluster V i must have a size of d i N, where k i=1 d i = V, is studied in [?], and it is shown there that under the triangle inequality and fixed k it possible to obtain an approximation of at most three times the optimal value. We extend this result to the G c case, doing it in two steps: 1. First, we demonstrate the idea assuming that G c = C k, a cycle graph on k vertices. 2. Second, we apply the same arguments to general G c graphs on k vertices. We use an auxiliary problem, the min-adjacent-star problem, as explained in the next lines. For a given G c = (V c, E c ) and a given set of centers C = {c 1,..., c k } V with c i V i, we wish to arrange the vertices of V \ C into V c = k clusters. After the arrangement we will get V i = p i and we want to minimize k i=1 ( {j (i,j) E c} v V j \{c j } w(c i, v)). Thus, we want to arrange the clusters so that the arrangement yields the minimum sum of distances from the rest of the vertices to the given centers of the k stars according to the neighborhood relations between the stars. 5.2 Metric cyclic k-cut Let G = (V, E) be a complete undirected graph with V = n. The edges e E have nonnegative weights w(e) 0 that satisfy the triangle inequality (i.e., x, y, z V, w(x, y) w(x, y) + w(y, z)). k i=1 d i = n. Given is also a set of integers K = {d i } k i=1 such that 31

32 Definition 5.1 For any k 3, the metric cyclic k-cut problem computes, given an instance I = (G, w) satisfying the triangle inequality and k integers d i with k i=1 d i = n, k disjoint subsets of vertices V i V with size V i = d i for i k, minimizing the total weight of edges whose two ends are in the i and i + 1 sets for i = 1,..., k, where k Actually, the minimum cyclic k-cut problem is the metric d-restricted C k -cut problem as indicated in Definition??. The minimum cyclic 3-cut problem is NP-hard because it is the 3-cut problem with the additional requirement, proved NP-hard in [?]. Here, we strengthen this result by proving that it is the case even if the weights are either one or two. Theorem 5.2 For k = 3, 4, the metric cyclic k-cut problem is NP-hard, even if w(e) {1, 2}. Proof: We propose a polynomial reduction from the bisection graph problem to the metric cyclic k-cut problem. Recall that the bisection graph problem consists of finding a minimum cut of an unweighted graph such that the two cut sets have the same size. Let G = (V, E) be a graph on 2n vertices, instance of the bisection graph problem and consider several instances depending on the value of k and transform in polynomial time from the bisection instance defined above. 1. k = 3, actually the C 3 case. Let G(V, E), V = 2n, a weighted graph with edge weights {1, 2} be an instance of the bisection problem. We transform it to the C 3 instance with requirements (n, n, l), l > 0 in the following manner. Let G = (V, E ) where V = V {x 1,..., x l } and E = E {(x i, v), and w(x i, v) = 1 i (1,..., l), v V } {(x i, x j ) i, j (1,..., l), i j}. According to our construction U V, v V \U, i 1,..., l w(x i, U) w(v, U). Assume that the C 3 optimal solution with the requirements (n, n, l) is (V 1, V 2, V 3 ), V 1 = V 2 = n, V 3 = l. Assumed that some v V placed inside V 3, so it must be that for some i 1,..., l, x i placed into V 1 or V 2 (assume V 1 ). The contribution of v to the cut is w(v, V 1 {x i })+w(v, x 1 )+w(v, V 2 ), 32

33 and the contribution of x i to the cut is w(x i, V 2 ). In total it is w(v, V 1 {x i }) + w(v, x 1 ) + w(v, V 2 ) + w(x 1, V 2 ) w(x 1, V 1 {x i }) + w(v, x 1 ) + w(x 1, V 2 ) + w(v, V 2 ), which is the contribution of v, x i if we swap their placement. So we can assume that V 3 consists of x 1,..., x l and V 1, V 2 consist of 2n vertices of V. Hence,the optimal solution of C 3 must consist of the optimal bisection of V between V 1 and V 2. Since minimum bisection is NP-hard problem so is the restricted metric cyclic 3-cut problem. 2. k = 4. Consider the C 4 case with requirements (n, n, l, l) where l > 0. The structure of C 4 = (1, 2, 3, 4) is actually a biclique between {1, 3} and n {2, 4}. Adding the requirements of (n, n, l, l) we actually are in the bisection problem of (n + l, n + l) which is NP-hard and therefore so is the restricted metric cyclic 3-cut problem. Note that if k is unbounded, then the minimum cyclic k-cut problem is APX-hard because this problem contains the metric TSP problem. We demonstrate that for any fixed k it is possible to obtain in polynomial time an approximation of at most three times the optimal value. We start by defining a new problem which we solve optimally for a constant k, and then use its solution to approximate the metric cyclic k-cut problem. Definition 5.3 The min-adjacent-star problem finds vertices v 1,..., v k and a k-cut, such that v i V i, V i = d i, i = 1,..., k, and p i=1 ( ) d i w(v i, V i+1 ) + d i+1 w(v i+1, V i ) + d i d i+1 w(v i, v i+1 ) is minimized, where indices are (mod k). Theorem 5.4 Algorithm FindCyclicPartition (see Algorithm??) solves the min-adjacentstar problem. It can be executed in time O(n k+1 ). 33