Better Approximation Algorithms for Minimum Weight Vertex-Connectivity Problems Vincenzo Aulettay
Mimmo Parentey
Abstract
Given a k vertex connected graph with weighted edges, we study the problem of nding a minimum weight spanning subgraph which is k vertex-connected, for k = 2; 3; 4. The problem is known to be NP-hard for any k 2, even when edges have no weight. In this paper we provide a 2 approximation algorithm for the cases k = 2; 3 and a 3 approximation algorithm for the case k = 4.
Keywords: Design and Analysis of Algorithms, Approximation Algorithms, Graph
Algorithms, Connectivity.
1 Introduction Let G = (V; E ) be an undirected graph and, for each edge e, let w(e) denote the non negative weight of e. The Minimum Weight k Vertex Connected problem, shortly kMWVC, consists of nding a minimum weight spanning subgraph of G that remains connected in presence of up to k ? 1 vertex failures. The computation of minimum weight graphs having a given connectivity degree is a fundamental issue in the design of communication networks. In fact, we can think of the weights of the edges as the costs of the realization of connections between any two nodes of a given network. In this case the connectivity degree can be seen as a measure of the tolerance to failures occurring in the nodes of the network. For more motivations see e.g. [8]. The k-MWVC problem is NP-hard, for k 2, even in presence of unweighted edges. Various approximation algorithms for its solution have been proposed in [10, 14, 4]. An approximation algorithm with factor , is a polynomial time algorithm that returns a solution whose cost is within a factor from the cost of an optimal solution. The best approximation algorithm for the k-MWVC problem is due to Ravi and P k Williamson [14] and it obtains a factor 2H (k), where H (k) = i=1 1=i. In [10] it is given A Preliminary version of this paper has been presented at STACS'97, see [1]. Dipartimento di Informatica ed Applicazioni, \R.M. Capocelli", Universita di Salerno, 84081 Baronissi (Italy). e-mail: fauletta,
[email protected]. y
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2 an approximation algorithm that obtains a factor not greater than 2 + 2(k ? 1)=n when the edge weights satisfy the triangle inequality. Better algorithms have been proposed for particular values of k. In [14] and [4] two algorithms are described that obtain a factor 3 for the case k = 2. In [10], Khuller and Raghavachari improved on this result obtaining a factor of 2 + 1=n, where n is the number of vertices in the graph. Recently, Penn and Shasha-Krupnik have given in [13] (announced to appear) a 2 approximation algorithm for the case k = 2 and a 3 approximation algorithm for the case k = 3. In this paper we study the k-MWVC problem for small values of k without any assumption on the weights of the edges of the graph. We give a 2 approximation algorithm for the cases k = 2; 3 and a 3 approximation algorithm for the case k = 4. Similar results have been obtained, independently from us, in [6]. The rest of the paper is organized as follows: in the next section we recall some basic de nitions on graphs; in section 3 we describe the algorithm Conn that solves the cases k = 2; 3; in section 4 we describe the algorithm 4-Conn, that uses Conn as a subroutine and solves the case k = 4.
2 Preliminaries and de nitions Throughout all the paper we will refer to graphs having nonnegative weighted edges. Let G be a graph having n vertices and m edges. The weight of G, denoted by w(G), is the sum of the weights of all its edges. Given a set X of vertices of G, we denote by G=X the graph obtained from G by deleting all the vertices of X and the edges incident to vertices of X (when X = fxg we will abuse of the notation G=x). A subgraph H of G which contains all the vertices of G is said to be a spanning subgraph. Two (directed) paths from vertex u to vertex v are independent if they have only the vertices u and v in common. As in [10], we can generalize this notion and consider paths between u and a set of k vertices A: we say that there are k independent (directed) paths from u to A if there is a path from u to each vertex in A and all the k paths have only the vertex u in common. A (directed) graph is k-vertex connected, or simply k-connected if deleting up to k ? 1 vertices the remaining graph is (strongly) connected. Equivalently, a (directed) graph is k-connected if for each pair of vertices u and v, there exist k independent paths that connect u to v (and v to u). In the rest of the paper G denotes a graph with connectivity at least k. A set of vertices of G whose removal makes G not connected is said cut-set. A cut-set of cardinality k is called a k{cut-set. The Minimum Weight k Vertex-Connectivity Problem for a graph G, consists in nding a k-connected spanning subgraph of G of minimum weight. A similar problem is the computation of of minimal augmentation set, see [2]. Let G0 be a (k ? 1)-connected spanning subgraph of G and let F E . If the graph obtained by adding the edges of F to G0 is k-connected, then F is said to be an augmentation set for G0 . The minimum k-augmentation problem consists in nding an augmentation set for G0 of minimum weight. The directed version of G, is the digraph obtained by replacing every edge e by two anti-parallel arcs, each with the same weight of e. Symmetrically, given a digraph D with the same weight on each anti-parallel arcs, the undirected version of D is the undirected
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Input: A weighted k-connected graph G = (V; E ). Output: A k degree vertex root and a (d k2 e +1)-connected spanning subgraph H such that w(H ) 2w(H ). begin 1) Let D be the directed version of G. 2) for each vertex v of V do 3) for each set of k arcs F = fe1; e2; : : : ; ek g, outgoing from v in D do 4) Run the FT algorithm on the graph Dv (F ), the vertex v and the integer k;
end end
5) Let DFT be the minimum weight subgraph returned by the FT algorithms and let root be its distinguished vertex. 6) return the graph H , that is the undirected version of DFT and the vertex root.
end
Figure 1: The algorithm Conn
graph obtained by eliminating an arc for each pair of parallel arcs and by ignoring the orientation of all the arcs Let D be a digraph and let F be a set of arcs of D outgoing from a vertex v: we de ne the reduced graph of D with respect to F , denoted by Dv (F ), as the graph obtained from D by deleting all the arcs outgoing from v except those in F . The Frank and Tardos Algorithm. Following the strategy used in [10], our algorithms use the polynomial time algorithm FT, proposed in [3] by Andras Frank and E va Tardos, as a subroutine to compute an approximated solution of the k-MWVC problem. The FT algorithm takes in input a weighted digraph D, a distinguished vertex root and an integer k and returns a minimum weight spanning subgraph of D (if it exists) that contains k independent paths from root to each other vertex of the graph.
3 Two and Three Connectivity Let G be a k vertex connected graph and let H be a minimum weight k-connected spanning subgraph of G. In this section we describe the algorithm Conn that computes a spanning subgraph H of G that is at least (d k2 e +1)-connected and has weight w(H ) 2w(H ). Thus, Conn gives a 2 approximated solution to the k-MWVC problem, when k = 2; 3. Moreover, Conn will be used as a subroutine by the algorithm 4-Conn that solves the case k = 4 and that will be presented in the next section. The algorithm Conn is described in Figure 1. It takes in input a k-connected graph G and returns a distinguished vertex root and a subgraph H that has k independent paths between root and each other vertex. We notice that the running time of Conn depends on the time required to execute the O(nk+1 ) calls of the FT algorithm. Using the implementation of the FT algorithm given in [5], requiring time O(kn2m), it is easily seen that Conn has a polynomial running time, for any constant k. We will prove that
4 the graph returned by Conn is at least (d k2 e + 1)-connected and has weight not greater than 2w(H ). We say that a k-connected digraph D is minimally connected if the graph obtained from D by removing any arc is not k-connected. We observe that each minimum weight k-connected spanning subgraph of D is still a minimally connected digraph.
Lemma 1 If D is a k-connected digraph, then there exists a vertex v and a set F of k
arcs outgoing from v such that in the reduced graph Dv (F ) there are k independent paths from v to each other vertex.
Proof: Let Dmin be a minimally connected subgraph of D. From a classical result of graph theory due to Mader [12], (see also [9]) we have that there is at least a vertex of Dmin that has out-degree k. Let v be such a vertex and let F be the set of arcs outgoing from v. The graph Dvmin (F ) is a subgraph of Dv (F ) and it has k independent paths from v to each other vertex.
By the previous Lemma, Conn always returns a subgraph H of G along with a kdegree distinguished vertex root that is connected with k independent paths to each other vertex of H . To prove that this subgraph is at least (d k2 e + 1)-connected we use the following result from [10].
Lemma 2 ([10]) For each vertex v of a graph G, if v has degree k and there are k
independent paths from v to each other vertex of G, then the graph G=v is at least d k2 e-connected.
Lemma 3 Let G be a k-connected graph and let H and root be the subgraph and the
vertex returned by Conn on input G. If H=root is -connected, with < k, then H is ( + 1)-connected.
Proof: The proof is by contradiction. Suppose that H is not ( + 1)-connected and let X = fx1; x2; : : :; xg be a cut-set of H . We notice that if root 2 X , then the other
vertices of X should form a cut-set of H=root . But this is a contradiction since H=root is -connected by hypothesis. On the other hand, if root 2= X then X is not a cut-set of H . In fact, root is connected to each other vertex of H through k independent paths and X can cut at most of these paths. Thus, all the vertices of H=X are connected to root and H=X is connected.
From Lemma 2 and 3, the next corollary easily follows. Corollary 1 The subgraph H returned by Conn on input the k-connected graph G is at least (d k2 e + 1)-connected. We can state now the main result of the section. Theorem 1 The algorithm Conn computes a spanning subgraph H of a weighted kconnected graph G, with k > 1, such that H is (d k2 e +1)-connected and w(H ) 2w(H ), where H is the minimum weight k-connected spanning subgraph of G.
5 Proof: By Corollary 1, H is at least (d k2 e + 1)-connected and we have only to show that w(H ) 2w(H ). Let D be the directed version of G, Dmin be a minimumweight minimally k-connected spanning subgraph of D, and let DFT be the digraph computed at step 5 of Conn. We observe that DFT has minimum weight among the spanning subgraphs of D that have a vertex with out-degree k which is connected with k independent paths to all the other vertices. Since Dmin has a vertex of out-degree k, which is connected to all the other vertices with k independent paths, we have that w(DFT ) w(Dmin ). Moreover, we have that w(Dmin ) 2w(H ), since the directed version of H is minimally connected, and w(H ) w(DFT ), since H is the undirected version of DFT . Therefore the lemma holds. Corollary 2 The algorithm Conn solves the k-MWVC problem, for k = 2; 3, with an approximation factor 2.
4 Four Connectivity In this section we describe the algorithm 4-Conn that computes a minimum weight 4-connected spanning subgraph of G with an approximation factor 3. This algorithm uses Conn to nd a distinguished vertex root and a subgraph H of G that contains 4 independent paths between root and each other vertex of H . By Corollary 1, H is at least 3 connected. Thus, if H is 4 connected the algorithm halts; instead, if H is 3 connected, then 4-Conn computes an augmentation set for H . In the rest of the section we suppose that H is 3 connected. The augmentation algorithm is based on some properties of the graph H , described in section 4.1. De ne a k{rooted graph K as a graph that satis es the following two conditions: it contains a set RT (K ) of k distinct vertices RT (K ) = fv1; ; vk g, called root vertices, such that for each vertex u we have that: if u 2= RT (K ) there are k independent paths from u to RT (K ); if u = vi, for each i, there are k ? 1 independent paths from the root vertex vi to the other vertices of RT (K ). It can be easily seen that the graph J = H=root is a 4{rooted graph where the 4 adjacent vertices of root are the root vertices. By Lemmas 2 and 3, J is 2{connected and each augmentation set for J is also an augmentation set of H . Thus, we have reduced our problem to the computation of an augmentation set of the 4-rooted graph J . 4.1
Properties of
4{rooted graphs
Let us rst give some de nitions. A k-connected component C of G is maximal if adding any vertex of G=C it results not k-connected. In this section we call cut-set a 2{cut-set and we call block a set of vertices that is either a maximal 3 connected component or it is a connected component with less than 5 vertices such that at least one of these vertices does not belong to any cut-set. We say that two cut-sets X and Y are adjacent if and only if each vertex of Y either belongs to X or it is adjacent to a vertex of X . A block B , instead, is adjacent to the cut-set X if and only if X B . In the sequel we refer to a component of J meaning either a block or a cut-set of J .
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Figure 2: An example of a decomposition of a 4{rooted graph J in its components. In this subsection we prove (in Lemma 4 and 5) that each component of J that does not contain root vertices partitions J into two connected components, each one containing two root vertices. As a consequence, we have that the root vertices of J do not belong to any cut-set. We also prove that all these components partition the set RT (J ) in the same two subsets, denoted by RT (J )l and RT (J )r, (when there is no ambiguity we simply write RT l and RT r ) where RT l is the subset containing v1. Thus, we can de ne the left and the right side of a component X : the left side Xl is the connected component of J=X that contains RT l; the right side Xr is the connected component containing RT r . If X is a block that contains vertices of RT l, then we show that Xl is empty and Xr is equal to J=X . Similarly, if RT r X then Xl is equal to J=X and Xr is empty. In Figure 2 an example of a decomposition of a 4{rooted graph into its components, is given.
Lemma 4 Each cut-set partitions J into two connected components, each one contain-
ing two root vertices of RT (J ). Moreover, all the cut-sets of J partition the set RT (J ) in the same two subsets.
Proof: Let X = fx1; x2g be a cut-set of J and let C be a connected component of J=X .
We observe that C contains at least two root vertices, since X can cut at most two of the paths connecting any vertex y of C to RT (J ). On the other hand, C cannot contain more than two root vertices, otherwise it would be the unique connected component of J=X , contradicting the hypothesis that X is a cut-set. Thus, we can state that J=X
7 consists of two connected components: Xl, containing the root vertices RT Xl = fv1; v2g, and Xr , containing the root vertices RT Xr = fv3; v4g. Now, suppose by contradiction that there exists a cut-set Y = fy1; y2g, distinct from X , that partitions RT (J ) in RT Yl = fv1; v3g and RT Yr = fv2; v4g. Without loss of generality assume that x1 2= Y and y1 2 Xr . We observe that, as RT Xl 2 Xl, all the paths between y1 and RT Xl go through vertices of X . There exists a path from y1 to v2 2 Yr going through x1 and not containing any vertex of Yl. Thus x1 2 Yr . Similarly there exists a path from y1 to v1 2 Yl going through x1 and not containing any vertex of Yr . Thus x1 2 Yl as well. But, this is a contradiction, as Yl and Yr are disjoint.
Lemma 5 Each block B of J that does not contain root vertices partitions J into two
connected components containing RT l and RT r , respectively. Moreover, there exists a unique cut-set X adjacent to B that cuts all the paths between vertices of B=X and the vertices of RT l (resp. RT r ).
Proof: Let B be a block of J and let X be a cut-set adjacent to B and such that X
cuts all the paths between RT l and B=X . Such a cut-set exists and it is unique, since B is a block that does not contain root vertices. Similarly, there exists a unique cut-set Y , adjacent to B , that cuts all the paths between B=Y and RT r . Thus Xl and Yr are disjoint and J=B = Xl [ Yr . Now the lemma follows as, by Lemma 4, Xl is a connected component containing RT l and Yr is a connected component containing RT r .
We notice that if B contains root vertices, then it contains either RT r or RT l or both. In this case the graph J=B either is empty, if RT B , or it is connected, if it contains only one of RT l and RT r . 4.2
The block-cut graph
The block-cut graph ?(J ) is a weighted digraph associated to the 4{rooted graph J . Roughly speaking, the block cut graph ?(J ) has a vertex for each block of J and a vertex for each vertex of J belonging to a cut-set. More formally, for each vertex x 2 J there exists a vertex (x) 2 ?(J ) called the representative of x. Moreover, for each pair of distinct vertices x; y 2 J , if x; y do not belong to cut-sets and they are in the same block, then (x) = (y), otherwise (x) 6= (y). We say that a vertex x in J corresponds to a vertex in ?(J ) if = (x). We denote by and the block vertices representing the vertices of RT l and RT r , respectively. For each edge (x; y) in J there is an arc in ?(J ) between (x) and (y) if and only if
(x) 6= (y) and fx; yg is not a cut-set. Let X be a component containing x: if y 2 Xl then the arc is directed from (y) to (x); if, instead, y 2 Xr then it is directed from
(x) to (y). Notice that if x belongs to X and y is in the right side of X , then y is in the right side of all the other components containing x. There is also a second rule for adding arcs to ?(J ): for each cut-set Z , if there exist two cut vertices x and y adjacent to both the vertices of Z and such that x 2 Zl , y 2 Zr and fx; yg is not a cut-set, then there exists an arc in ?(J ) from (x) to (y). All the arcs of the block-cut graph have weight 0. In Figure 3 an example of a block-cut graph is given. To construct the block
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Figure 3: The block-cut graph associated to the graph J . cut graph ?(J ) we have to enumerate all the components of J . This list is obtained by using the following simple strategy: compute the block containing RT l, then iteratively, compute for each block B the unique cut-set X adjacent to B separating B=X from RT r and for each cut-set X compute all the components adjacent to X in X [ Xr . It can be easily seen that this construction takes polynomial time.
Lemma 6 Let be a vertex of ?(J ) and let x be a vertex corresponding to and X a
component containing x. For each vertex in ?(J ) there is a directed path from to if and only, for each vertex y 2 J corresponding to , it holds that y 2 Xr and fx; yg is not a cut-set.
Proof: By the properties of the 4-rooted graph and by the de nition of the block cut
graph, it can be easily derived that for each vertex of ?(J ) there is a path from to . The proof is by induction on the length of the path from to . We observe that the right side of the block represented by is empty. Moreover, since has no outgoing arcs, there is no vertex reachable from . Thus, when = the lemma holds. Suppose now that the lemma holds for all the vertices of ?(J ) whose distance from is less than h and let be a vertex such that the path from to has length h. By the de nition of block-cut graph, there are no arcs from representatives of vertices of X [ Xr to representatives of vertices of X [ Xl. Thus, if y 2 X [ Xl , then there is no path in ?(J ) from to . Therefore, we have only to prove that when y 2 Xr , there is a path from to if and only if fx; yg is not a cut-set. Let be a vertex adjacent to and lying on a path from to . By inductive hypothesis the lemma holds for . Denote by z a vertex corresponding to and denote by Z a component containing z. By de nition of block-cut graph, fx; zg is not a cut-set. We will consider only the case in which X and Z are cut-sets with a vertex in common. All the other cases are similar. Let X = fx; wg and Z = fz; wg. We observe that X [ Xr = fxg [ Z [ Zr and x is adjacent to both the vertices of Z . In fact, x has at least two adjacent vertices in X [ Xr , but no adjacent vertices in Zr , since Z is a cut-set. If part: If y = z the result trivially holds. If fy; zg is not a cut-set, by inductive hypothesis there is a path from to that goes through . If Y = fy; zg is a cut-set, then y is adjacent to both the vertices of Z (y has at least two adjacent vertices in Y [ Yl = fyg [ Z [ Zl but no adjacent vertices in Zl). Thus, from the second rule to de ne arcs of ?(J ), we have that there exists an arc from to .
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Only if part: By the de nition of block-cut graph if and are adjacent then fx; yg
is not a cut-set. Thus, we suppose that the path from to has length greater than one and assume that is the rst vertex of the path. As observed previously, z is the unique vertex of Xr adjacent to x. Thus, we have that either = or = (v), for some v that is adjacent to the vertices of Z and such that fx; vg is not a cut-set (by the second rule of block-cut graph). Suppose hence that Y = fx; yg is a cut-set. If = then there is a path from to and, by inductive hypothesis, fy; zg is not a cut-set. We have that w 2 Yl and it has no adjacent vertices in Yr = fzg [ Zr ? fyg. But, this is a contradiction, since w has at least two adjacent vertices in Z [ Zr . Thus, Y cannot be a cut-set. On the other hand, if = (v) we have that v 2 Yl and z 2 Yr and Y is not a cut-set. Since v is adjacent to zY is not a cut-set.
By the previous Lemma, we have that for each vertex of ?(J ) there exists a path from to . Let ? be a weighted digraph obtained from ?(J ) by adding some arcs and consider the following colouring of the arcs of ?: the arcs belonging to ?(J ) are coloured white, the added arcs are black. We say that a b-w path is a path that does not contain two consecutive black arcs. We prove that the problem of computing an augmentation set for J can be reduced to the problem of constructing a graph from ?(J ) that has a b-w path from to . We de ne the following relation between edges added to J and the black arcs of ?. For each edge e = (x; y) added to J , if there exists at least a component X containing x and such that y 2 Xr then the representative of e in ? is the black arc ( (x); (y)). For each black arc (; ) of ?, the corresponding edge in the augmentation set of J is the edge (x; y) that has minimum weight among the edges that are represented by (; ). The weight of the arc (; ) is equal to the weight of its corresponding edge. Notice that if for each component X containing x it holds that y 2 X , then the edge (x; y) has no representative in ?. However, it can be easily seen that this edge does not belong to any minimum weight augmentation set of J . In fact this edge does not cross any cut set of J and, thus, can be removed from an augmentation set, still having an augmentation set.
Lemma 7 Let AUG be a set of edges added to J and let AUG be the set of representatives of AUG. Denote by ? the graph obtained by adding the arcs of AUG to ?(J ). Then, AUG is an augmentation set for J if and only if there is a b-w path in ? from to .
Proof: If part: Let ? contain a b-w path P from to . For each cut-set X of
J there is a black arc e of P that connects the representative of a vertex of Xr to the representative of a vertex of Xl . The edge of AUG corresponding to e crosses the cut-set X . Thus, the graph obtained by augmenting J with the edges of AUG is 3-connected. Only if part: Let AUG be an augmentation set of J . Let B be the connected component containing RT r and let X0 be the unique cut set adjacent to B (see Lemma 5). In AUG there is at least one edge that crosses X 0 and connects a vertex of B to a vertex x1 2 Xl0.
Input: A weighted 4-connected graph G. Output: A 4-connected spanning subgraph H 0 of G such that w(H 0 ) 3w(H ). begin 1) Run algorithm Conn on input G and let H and root be the graph and the distinguished vertex returned, respectively. 2) if H is 4 connected, then stop.
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3) Construct the block-cut graph ?(J ) associated to J = H=root . 4) For each edge e of G not contained in H add to ?(J ) the arc corresponding to e. Let ? be the graph obtained. 5) Compute the minimum weight b-w path P in ? that connects to . Let AUG be the set of black arcs contained in p and let AUG be the set of edges corresponding to the arcs of AUG. 6) return the graph H 0 obtained by augmenting H with the edges of AUG.
end
Figure 4: Algorithm 4-Conn.
The corresponding black arc in AUG connects , that is the representative of the vertices of B , to (x1). Let X 1 be the component containing x1 and such that no vertex of Xr1 forms a cut set together with x1 (notice that such a component must exist). Then, by Lemma 6 we have that in ? there is a b-w path from to all the representatives of vertices of Xr1. Without loss of generality suppose that X 1 is a cut-set (otherwise consider the unique cut-set adjacent to X1 that cuts the paths to RT r ). Consider an edge in AUG that connects a vertex y of Xr1 to a vertex x2 2 Xl1. The corresponding black arc connects the vertex (y), reachable from (x1) using only white arcs due to Lemma 6, to (x2). Thus, there is a b-w path from to all the representatives of Xr2, where X2 is the component containing x2 such that no vertex of Xr2 forms a cut set together with x2. Iterating the above argument one can easily prove that in ? there is a b-w path from to . By this Lemma we obtain that the minimum weight augmentation set for J can be found computing the minimum weight set of arcs that, added to ?(J ), produces a b-w path from to . The algorithm 4-Conn, given in Figure 4, constructs the graph ? by adding to ?(J ) all the black arcs corresponding to the edges of G not contained in H . Then, it computes the minimum weight b-w path of ? . The edges corresponding to the black arcs of this path form a minimum weight augmentation set for H . It can be easily seen that 4-Conn takes polynomial time to compute a 4 connected spanning subgraph of G. Next Theorem shows that the weight of this subgraph is not greater than three times the weight of the minimum weight 4 connected spanning subgraph of G. Theorem 2 The algorithm 4-Conn solves the 4-MWVC problem with an approximation factor 3.
Proof: Let H be the minimum weight 4 connected spanning subgraph of G. From
Theorem 1, we have that the weight of the subgraph H obtained in the step 1 of the
11 algorithm 4-Conn is at most 2w(H ). Consider now the weight of the edges of the augmentation set AUG computed at step 5. Since H is 4 connected it exists a subset B of edges of H that is an augmentation set for J . On the other hand, the set AUG is the minimum weight augmentation set for J . Thus, we have that w(H [ AUG) = w(H ) + w(AUG) 3w(H ).
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