1 Introduction
Given a (vertexweighted) graph and a set , the (Weighted) Subset Feedback Vertex Set problem asks for a vertex set of minimum (weighted) size that intersects all cycles containing a vertex of . It was introduced by Even et al. who obtained a constant factor approximation algorithm for its weighted version [13]. Interestingly Subset Feedback Vertex Set for also coincides with the NPcomplete Multiway Cut problem [16] in which the task is to disconnect a predescribed set of vertices [3, 18]. Cygan et al. [10] and Kawarabayashi and Kobayashi [25] independently showed that Subset Feedback Vertex Set is fixedparameter tractable (FPT) parameterized by the solution size, while Hols and Kratsch provided a randomized polynomial kernel for the problem [21]. There has been a considerable amount of work to obtain faster, still exponentialtime, algorithms even when restricted to particular graph classes [5, 16, 15, 19].
As a generalization of the classical Feedback Vertex Set for which , the problem remains NPhard on bipartite graphs [35] and planar graphs [17]. On the positive side, Weighted Subset Feedback Vertex Set can be solved in polynomial time on interval graphs, permutation graphs, and cobipartite graphs [31], the latter being a subclass of graphs of independent set size at most two. However a notable difference between the two problems regarding their complexity status is the class of split graphs: Feedback Vertex Set is known to be polynomialtime solvable on split graphs [7, 33], whereas Subset Feedback Vertex Set remains NPhard on split graphs [16]. This gives evidence to the fact that the Subset Feedback Vertex Set seems more difficult to attack than the classical setting of the problem. Thus it is interesting to explore and obtain further (in)tractability results for Subset Feedback Vertex Set.
Towards such a direction it is reasonable to consider structural parameters of graphs that may lend themselves to provide a unified approach. In terms of parameterized complexity Feedback Vertex Set is known to be FPT, when parameterized by treewidth [8] and cliquewidth [2] which implies that Feedback Vertex Set can be solved in polynomial time on graphs of bounded such parameters. Although Feedback Vertex Set is W[1]hard parameterized by the size of the independent set^{3}^{3}3In Section 4 we give a different and simpler reduction from the Multicolored Independent Set problem., it can be solved in polynomial time on graphs of bounded maximum induced matching (i.e., Feedback Vertex Set belongs in XP parameterized by the size of the maximum induced matching) [24]. Only very recently, Jaffke et al. proposed an algorithm that solves Weighted Feedback Vertex Set in time where is the maximum induced matching width of the given graph [23]. Despite their relevant name, graphs of bounded maximum induced matching are not related to graphs of bounded maximum induced matching width as indicated in [34].
The approach of [23] provides a powerful mechanism, as it unifies polynomialtime algorithms for Weighted Feedback Vertex Set on several graph classes such as interval graphs, permutation graphs, circulararc graphs, and Dilworth graphs among others. Such a mechanism raises the question of whether the algorithm given in [23] can be extended to the more general setting of Weighted Subset Feedback Vertex Set. However the proposed algorithm is based on the crucial fact that the forest of a solution has bounded number of internal nodes which is not necessarily true for the forest of Weighted Subset Feedback Vertex Set. Thus it seems difficult to control the size of the solution whenever . As this observation does not rule out any positive answer, here we develop the first step towards such an approach by considering graphs of bounded independent set number which form candidate relevant graphs. Notice that graphs of bounded independent set number are not related to graphs of bounded maximum induced matching width. However graphs of bounded independent set number form the first natural class of bounded structural parameter that are interesting to explore regarding the complexity of Subset Feedback Vertex Set. Although Weighted Feedback Vertex Set can be solved in time on graphs of maximum induced matching at most [24], Subset Feedback Vertex Set is already NPcomplete on graphs of maximum induced matching equal to one (i.e., split graphs) [16].
In this work we show that the complexity behaviour of the weighted version of the problem is completely different from the behaviour of the unweighted variant on graphs with bounded , where is the size of a maximum independent set in a graph .

We show that Weighted Subset Feedback Vertex Set can be solved in polynomial time on graphs with .
Such graphs consist of the complements of trianglefree graphs; recall that for trianglefree graphs Feedback Vertex Set remains NPhard [35]. Even on such graphs Weighted Subset Feedback Vertex Set requires a structural characterization of the solution with respect to the vertices that are close to .

We further provide a dichotomy result showing that Weighted Subset Feedback Vertex Set remains NPcomplete on graphs with .
Thus we enlarge our knowledge on the complexity difference of the two problems with respect to a structural graph parameter.

In order to complement our results we show that Subset Feedback Vertex Set can be solved in time , where .
Thus we provide a complexity difference between the weighted and the unweighted versions of the problem with respect to a natural structural parameter. Our main findings concerning Subset Feedback Vertex Set are summarized in Table 1.
Moreover, we demonstrate how our ideas can be extended to other terminal set problems on graphs of bounded independent set size. In these type of problems we are given a graph , a terminal set , and a nonnegative integer and the goal is to find a set with which intersects all “structures” (such as cycles or paths) passing through the vertices in [6]. The (unweighted) Node Multiway Cut problem is concerned with finding a set of size at most such that any path between two different terminals intersects . Node Multiway Cut is known to be in FPT parameterized by the solution size [4, 29] and even above guaranteed value [9]. For further results on variants of Node Multiway Cut we refer to [3, 18, 27]. We completely characterize the complexity of Node Multiway Cut with respect to the size of the maximum independent set.

In particular, we show that for Node Multiway Cut can be solved in polynomial time, whereas for it remains NPcomplete by adopting the reduction for Weighted Subset Feedback Vertex Set with .
We further consider a relaxed variation of Node Multiway Cut in which we are allowed to remove terminal vertices, called Node Multiway Cut with Deletable Terminals (also known as Unrestricted Node Multiway Cut).

We show that the (unweighted) Node Multiway Cut with Deletable Terminals problem can be solved in polynomial time on graphs of bounded independent set number, using an idea similar to the polynomialtime algorithm for the Subset Feedback Vertex Set problem.

We also consider its nodeweighted variation and provide a dichotomy complexity result showing that Weighted Node Multiway Cut with Deletable Terminals can be solved in on graphs with , whereas it becomes NPcomplete on graphs with .
It should be noted that the polynomialtime algorithm for the weighted variation is obtained by invoking our algorithm for Weighted Subset Feedback Vertex Set on graphs with .
Bounded Structural Parameter  

Max. Independent Set ()  Max. Induced Matching ()  
Weighted FVS  [24]  
Weighted SFVS  Theorem 1  
NPcomplete  Theorem 2  
Unweighted SFVS  Theorem 3  NPcomplete [16] 
2 Preliminaries
We refer to [1, 11, 20] for our standard graph terminology. For , and . A weighted graph is a graph, where each vertex is assigned a weight that is a positive integer number. We denote by the weight of each vertex . For a vertex set , the weight of , denoted by , is .
Given a graph , the independent set number, denoted by , is the size of the maximum independent set in . In terms of forbidden subgraph characterization, note that if and only if does not contain as an induced subgraph. We say that a graph has bounded independent set size if there exists a positive integer such that . The clique cover number of , denoted by , is the smallest number of cliques needed to partition into such that is a clique. A vertex cover is a set of vertices such that every edge of is incident to at least one vertex of the set. A matching is a set of edges having no common endpoint. An induced matching, denoted by , is a matching of edges such that is isomorphic to . The maximum induced matching number, denoted by , is the largest number of edges in any induced matching of . It is not difficult to see that for any graph , holds.
Here we consider the following problem.
As remarked, we distinguish between the weighted and the unweighted version of the problem. In the unweighted version of the problem note that all weights are equal and positive. The classical Feedback Vertex Set (FVS) problem is a special case of Subset Feedback Vertex Set with . A vertex of is simply called vertex. An induced cycle of is called cycle if an vertex is contained in the cycle. We define an forest to be the subgraph of induced by the vertex set for which no cycle in is an cycle. It is not difficult to see that the problem of computing a minimum weighted subset feedback vertex set is equivalent to the problem of computing a maximum weighted forest.
Let us give a couple of observations on the nature of Subset Feedback Vertex Set on graphs of bounded independent set size. Let be a graph and let be a positive integer such that every independent set of has at most vertices. Firstly note that the boundedsize independent set is a hereditary property, meaning that for every induced subgraph of , we have . Moreover for any clique of , any forest of contains at most two vertices of .
Observation 1.
Let be a graph with and let .

For any set of vertices, there is a cycle in .

Any forest of has at most vertices from .
Proof.
Let be a set of vertices. Assume that is a forest. As an induced subgraph of , any independent set of has size at most . Since is acyclic, there is a proper 2coloring of the vertices of such that . By the fact that , we conclude that , leading to a contradiction that . Thus contains a cycle.
For the second statement, let be an forest of . By the first statement, if has at last vertices then there is a cycle in that passes through a vertex of , which implies an cycle in . Thus . ∎
We note that Observation 1 directly implies that any vertices of induce an cycle, which allows us to construct by brute force all possible subsets of belonging to any forest in time .
3 Weighted SFVS on Graphs of Bounded Independent Set
Here we consider the Weighted Subset Feedback Vertex Set and we show a dichotomy result with respect to the size of the maximum independent set. We first provide a polynomialtime algorithm on graphs of independent set size at most three and then we show that Weighted Subset Feedback Vertex Set is NPcomplete on graphs of independent set size at most four.
Let be an instance of Weighted Subset Feedback Vertex Set for which is a graph of independent set size at most . In the forthcoming arguments, instead of directly computing a solution for Weighted Subset Feedback Vertex Set, we consider the equivalent problem of computing an forest of having weight at least .
Let be an forest of . We partition the forest into two induced subgraphs and as follows:

is the subgraph of induced by the vertices of ; the vertices of are at distance at most one from and are denoted by .

is the graph and contains vertices that are at distance at least two from .
Such a partition is called distance partition of , denoted by . The set of edges of having one endpoint in and the other in are called the cut with respect to and . Notice that a vertex of that is adjacent to a vertex of belongs to .
Let be a partition of the vertices of such that each induces a connected component in . Because is an induced subgraph of , it is clear that . Let be a tuple of subsets of , i.e., each holds. We say that the cut satisfies the tuple if for any vertex , we have . The notion of an distance partition of with the corresponding cut is illustrated in Figure 1.
We now utilize the distance partition of in order to construct an algorithm that solves Weighted Subset Feedback Vertex Set on graphs of independent set size at most and subsequently show that this algorithm is efficient for . Our general approach relies on the following facts:

By Observation 1 (2) we try all subsets of with at most vertices and keep those sets that induce an forest. This step is responsible for constructing the graph . We will show that the number of the produced such subsets is bounded by .

For each of the potential subsets constructed in the previous step, and for each , we determine all possible tuples in with that are only satisfied by cuts of distance partitions of induced subgraphs of with that are forests. We show why considering only these tuples is sufficient in Lemma 1.

Up to that point we can show that all steps can be executed in polynomial time regardless of . However for the next and final step we can only achieve polynomial running time if we restrict ourselves to due to the number of connected components of . Then for each tuple selected in the previous step we find connected components of maximum weight such that the cut of satisfies the tuple.
We begin by showing that the distance partition of provides a useful tool towards computing a maximum forest. Given a set of vertices and subsets of , we construct the graph that is obtained from by adding vertices such that every vertex is adjacent to all the vertices of . In what follows, we always assume that is a graph having independent set size at most .
Lemma 1.
Let be an forest of with distance partition such that . Then there is a tuple with such that

the cut of satisfies and

every induced subgraph of with distance partition that satisfies is an forest.
Proof.
Let be a partition of the vertices of such that every induces a connected component in . We define a tuple in which every . Clearly since every vertex is at distance at least two from . Thus, by construction, the cut of satisfies the tuple .
For the next claim we first show that with respect to and the tuple is an forest. Assume for contradiction that there is an cycle in . Since does not contain any cycle, contains a vertex and at least two vertices from . By the fact that , there is a vertex in of that is adjacent to and there is a vertex in of that is adjacent to . Together with a path between and in the connected component , we construct a path in with endvertices and that is completely contained in . This means that if we replace every vertex of then we obtain an cycle in , leading to a contradiction. Thus is an forest.
Let be an induced subgraph of with distance partition that satisfies . Observe that as they are induced subgraphs of the same vertex set of . Thus does not contain any cycle, because is an forest. Since the cut of satisfies , there is a partition in such that is a connected component of and . We show that is indeed an forest. For contradiction, assume an cycle in . There are no cycles in which implies that for some . For every such set we replace the part by a vertex . Denote by the resulting graph. Notice that is a subgraph of because . This, however, implies an cycle in which gives the desired contradiction. Therefore is an forest. ∎
Next we show how to bound the vertices of .
Lemma 2.
Let be an forest of with distance partition such that .

If then .

If then .
Proof.
Let be such an forest of with . By Observation 1 (2), we know that . To ease the presentation, we let . We consider separately the two cases of the claim.
Case 1. Let . Assume for contradiction that . We show that contains a matching with at least edges. Observe that . Applying Observation 1 (1) shows that there is a cycle in . Since is an forest, this is not an cycle, so all vertices contained in are vertices of . Iteratively picking the two endpoints of an edge from as long as , constructs edges of having no common endpoints. Thus contains a matching with at least edges.
Let be the connected components of . Notice that because is an induced subgraph of a graph with maximum independent set size at most . By construction, every vertex of is adjacent to at least one vertex of . If the endpoints of an edge of in are adjacent to vertices of the same component then there is an cycle in since every vertex of belongs to . Thus the endpoints of every edge of are adjacent to different connected components of . Now obtain a bipartite graph by contracting every component into a single vertex and every edge of into a single vertex and keep only the adjacencies between the components and the edges of . Let be the bipartition of the resulting bipartite graph such that contains the components of and contains the edges of . Since and every vertex of is adjacent to at least two vertices of , there is a cycle in the bipartite graph. Then it is not difficult to see that the cycle of the contracted vertices corresponds to an cycle in . Therefore there is an cycle in an forest, leading to a contradiction.
Case 2. Let . Assume for contradiction that . This means that contains at least one vertex. We pick a nonempty subset of as follows. If then consists of any two vertices of . If then consists of an arbitrary vertex of . In both cases, notice that by the fact . Then Observation 1 (1) implies that there is a cycle in . Since has at most two vertices, we conclude that the induced cycle of has at least one vertex from , hence it is an cycle in . Therefore we reach a contradiction which implies that . ∎
Lemma 2 shows that we can compute all possible candidates for in polynomial time as follows.

We first try by brute force all subsets of having at most vertices, according to Observation 1 (2).

Then, for each such subset we incorporate the neighbors of for which either or according to Lemma 2.

Given the described sets and , we check if induces an forest and, if so, we include them into a list containing all candidates for .
The correctness follows from Observation 1 and Lemma 2. Regarding the running time notice that we create at most subsets for each of and . Thus in time we can compute a list that contains all possible subsets of the vertices corresponding to . Notice that such vertices are enough to build the part .
Let be a set of . We now focus on the graph that contains the vertices that are at distance at least two from . Notice that all possible vertices in an forest that are in distance at most one from are described in . Let be the number of connected components of . It is clear that . In fact, if contains at least one vertex then , since the vertices of are at distance at least two from . Moreover, observe that if then is a trivial solution. From now on, we assume that so that .
By brute force, we find all tuples such that the following hold:

and

the graph with respect to and is an forest.
Notice that by Lemma 1 it is sufficient to consider only such tuples. Since , , and , the number of tuples is , so that we can obtain the desired set of tuples that satisfy both conditions in polynomial time.
In what follows, we consider the case for . By the previous arguments we are given a set and tuples of the form or which are subsets of . Our task is to compute a subset of the vertices of such that the vertices of induce a maximum forest and the cut satisfies or , respectively. We distinguish the two cases.
Lemma 3.
Let and let be a subset of such that both and with respect to and are forests. There exists a polynomialtime algorithm that computes a maximum forest with an distance partition having a cut satisfying .
Proof.
Since is a fixed forest of , we need to determine the vertices of that are included in . By the desired cut of we are restricted to the vertices of that have neighbors only to . Those vertices can be described as follows:
Notice that contains vertices that are at distance at least two from the vertices of . Since the cut satisfies a single subset , we have at most one connected component of in . In order to choose the correct connected component of , we try to include each of them in and select the one having the maximum total weight. Notice that adding any component of into cannot create any cycle because with respect to and is an forest. Thus by Lemma 1 we correctly compute a maximum forest with the desired properties. Clearly the set can be constructed in polynomial time. Since the number of connected components is at most two, all steps can be executed in polynomial time. ∎
Next we consider the tuple .
Lemma 4.
Let and let be subsets of such that both and with respect to and are forests. There exists a polynomialtime algorithm that computes a maximum forest with an distance partition having a cut satisfying .
Proof.
Similar to the proof of Lemma 3, we first construct the sets that contain vertices of and satisfy the cut obtained from :
and  
As the desired cut of satisfies , there are two connected components of which are subsets of these two sets respectively. Let and be the connected components of such that and . Now observe that there should be two nonadjacent vertices and that belong to and , respectively. We iterate over all possible pairs of nonadjacent vertices and in time. Assuming a given choice for and , observe the following:

Since and are vertices of different connected components of , the components themselves are further restricted to be subsets of and , respectively. That is, and .

Since has at least one vertex of , are nonadjacent, and by the fact , we have that and induce cliques in . Thus and , respectively.
Then by the second statement it is not difficult to see that and are disjoint. Let and . Now in order to find the maximum induced forest under the stated conditions and our assumption that and belong to the two connected components of it suffices to find the maximum subset of such that there are no edges between the vertices of and the vertices of . This boils down to compute a minimum weighted vertex cover on the bipartite graph obtained from and removing the edges inside and . By maximum flow standard techniques, we compute a minimum weighted vertex cover on in polynomial time [30]. Therefore contains the connected components and , as required. ∎
Now we are equipped with our necessary tools in order to obtain our main result, namely a polynomialtime algorithm that solves Weighted Subset Feedback Vertex Set on graphs of independent set of size at most .
Theorem 1.
Weighted Subset Feedback Vertex Set on graphs of independent set of size at most can be solved on time .
Proof.
Let us briefly explain such an algorithm for computing a maximum forest of a graph having independent set size at most three. Let . Initially we set . Then for every set with such that is an forest, we try by brute force all subsets and with such that with respect to and or is an forest. For each of such subsets we find a maximum forest with an distance partition having a cut satisfying or , respectively, by applying the algorithms described in Lemma 3 and Lemma 4. At each step, we maintain the maximum weighted forest by comparing with . Finally we provide the vertices as the set with the minimum total weight that are removed from .
By Lemma 2, it is sufficient to consider the described subsets . Since every induced subgraph of contains at most two connected components, Lemma 1 implies that all possible subsets or with the described properties are enough to consider. Thus the correctness follows from Lemmata 2–4. Regarding the running time, notice that whether a graph contains an cycle can be tested in linear time. Thus we can construct all described and valid subsets in time. Therefore the total running time of the algorithm takes time , since each of the algorithms given in Lemma 3 and Lemma 4, respectively, requires polynomial time. ∎
Let us now show that extending Theorem 1 to graphs of larger independent sets is not possible. More precisely with the following result we show that Weighted Subset Feedback Vertex Set is paraNPcomplete parameterized by .
Theorem 2.
Weighted Subset Feedback Vertex Set is NPcomplete on graphs of independent set of size at most .
Proof.
We will provide a polynomial reduction from the Vertex Cover (VC) problem on tripartite graphs which is NPcomplete [17]. Let be a tripartite graph where is the partition of . We construct a weighted graph from in polynomial time as follows.

We turn the three independent sets , and into cliques by adding all necessary edges and we give all vertices unary weight.

We add a vertex that is adjacent to all of the vertices of and we assign weight to . In a completely symmetric way, we add vertices and with respect to the sets and , respectively.

We add a vertex that is adjacent to all three vertices and we assign weight to .
This completes the construction of . Observe that all vertices of have weight equal to one. It is not difficult to verify that the constructed graph is a graph having an independent set at most , since is a vertexdisjoint union of three cliques.
Next we claim that has a vertex cover of size at most if and only if with has a subset feedback vertex set of weight at most . Assume a vertex cover of . By definition, covers all edges of , so that is an independent set. This means that is a vertexdisjoint union of cliques. Since is nonadjacent to any vertex of and is an independent set, every cycle of contains a vertex of and with at least two vertices from and , respectively. Thus is a connected forest. Therefore is a subset feedback vertex set of of size at most .
For the opposite direction, assume a subset feedback vertex set of . If is not a subset of , then its sum of weights is greater or equal to . Then is not a minimum subset feedback vertex set of , since minus a single vertex is trivially a subset feedback vertex set of of total weight . Thus is indeed a subset of . Assume that is not a vertex cover of . By definition, there is an edge of that remains uncovered. Without loss of generality, assume that this edge has its endpoints on the vertices and . Then is an induced cycle of , which contradicts the fact that is a subset feedback vertex set of . Therefore is a vertex cover of . ∎
We stress that Theorem 2 further implies that the NPcompleteness result carries along to graphs of clique cover number at most four, since the constructed graph given in the proof can be partitioned into four disjoint cliques.
4 SFVS on Graphs of Bounded Independent Set
Here we show that despite the complexity dichotomy result for the Weighted Subset Feedback Vertex Set, whenever the weights of the vertices are equal Subset Feedback Vertex Set can be solved in polynomial time on graphs of bounded independent set number.
Theorem 3.
Subset Feedback Vertex Set on graphs of independent set of size at most can be solved in time .
Proof.
Let be a graph with and let . Denote by a minimum subset feedback vertex set of . Let be a maximum forest of . By Observation 1 (2), the vertices of that belong to are at most . Thus for every optimum solution , the set has at most vertices.
Now we claim that it is enough to consider subsets of for which . To see this, observe that if has order more than , then has more vertices than , leading to a contradiction to the optimality of . Hence, has at most vertices. In order to find an optimal solution, it suffices to consider all such candidates for and for . To check whether an induced subgraph of consists an forest takes time. Since the number of such sets is at most and the number of the considered sets is at most , the total running time is bounded by . Therefore in time we compute a minimum subset feedback vertex set showing the claimed result. ∎
Regarding the dependence of the exponent in the running time of the algorithm given in Theorem 3, note that we can hardly avoid this fact, since Feedback Vertex Set is W[1]hard parameterized by the independent set number as explicitly given in [24]. At the same time such an observation follows from the W[1]hardness result from the construction given in [22] with respect to the maximum induced matching width. In the following result, we provide a different and simpler reduction from the Multicolored Independent Set problem [14, 32] which shows an interesting connection with graphs of bounded independent set size.
Theorem 4.
Feedback Vertex Set is W[1]hard when parameterized by the clique cover number.
Proof.
The reduction comes from the Multicolored Independent Set problem: given a graph and a partition of , decide whether contains an independent set of size using exactly one vertex from each . It is known that Multicolored Independent Set is W[1]hard parameterized by [14, 32]. Let be an instance of Multicolored Independent Set. From we construct a graph as follows.

We make every set clique by adding all necessary edges.

For each we add two vertices that are adjacent to every vertex of .

We add a vertex that is adjacent to all the vertices of .
This completes the construction of . Observe that . Let and . Then forms an independent set in of size . Notice also that the vertices of induce a clique, so that has a clique partition of size . Thus the clique cover number of is at most which implies that has a clique cover number that is linearly dependent on . We claim that has a multicolored independent set if and only if has a feedback vertex set of size at most .
Let be vertices from each of that form an independent set in . We describe an induced forest of that contains vertices starting from the vertices of . For each vertex of we add in both vertices and . Notice that contains disjoint trees of the form . Since is nonadjacent to every vertex of , we can safely include in . Thus is an induced forest with vertices, so that constitutes a feedback vertex set of size .
For the opposite direction, let be a feedback vertex set of size . Then is an induced forest of that has at least vertices. We claim that from each there is at most one vertex in and . Assume for contradiction that at least two vertices from are contained in . Since is a clique in , there are exactly two vertices from in . Then neither nor is included in because they are both adjacent to and . Then is an induced forest of , as are nonadjacent to any vertex of . This, however, shows that there is an induced forest with at least vertices, leading to a contradiction. Thus which implies that . Then observe that is an independent set in and no vertex from induces a cycle with the vertices from , so that . If then . Hence and . Since contains at least vertices, each of contains exactly one vertex in . Moreover the vertices of are pairwise nonadjacent because is a vertex of and is adjacent to every vertex of . Therefore the vertices of each of form an independent set in . ∎
5 Extending to other Terminal Set Problems
Let us now consider further terminal set problems that are related to Subset Feedback Vertex Set. In these type of problems we are given a graph , a terminal set , and a nonnegative integer and the goal is to find a set with which intersects all “structures” (such as cycles or paths) passing through the vertices in [6]. In this setting Subset Feedback Vertex Set is a particular terminal set problem when the objective structure is a cycle. We show that the ideas that we developed for Subset Feedback Vertex Set on graphs of bounded independent set size, can be extended to further terminal set problems when the objective structure is a path instead of a cycle.
The (unweighted) Node Multiway Cut problem is formulated as follows.
Notice that in this problem we are not allowed to remove any terminal. For graphs having independent set size at most we completely characterize the complexity of Node Multiway Cut. In particular, for we can adopt the reduction given in Theorem 2.
Theorem 5.
Let be a graph of independent set of size at most . If then Node Multiway Cut can be solved on time . Otherwise, Node Multiway Cut is NPcomplete on graphs of independent set of size at most .
Proof.
Let be an instance of Node Multiway Cut. If contains an edge then we conclude that is a noinstance, since we are not allowed to remove any vertex from . In what follows we assume that is an independent set. If there are at most two terminals, so that , and we can solve the problem by standard maximum flow techniques [30].
For , we give a reduction from the NPcomplete Vertex Cover problem on tripartite graphs, similar to the one given in Theorem 2. Let be a tripartite graph where is the partition of . We construct a graph from by making the three independent sets , and into cliques and adding three new vertices , that are adjacent to every vertex of , , and , respectively. It is clear that has independent set size . We let and claim that has a vertex cover of size at most if and only if has a set of size at most which intersects every path between the vertices of . Removing a vertex cover from results in a vertexdisjoint union of three cliques in in which each of the vertices belongs to a separate clique. Thus is a solution for Node Multiway Cut on . For the opposite direction, observe that cannot contain any of the three vertices . Assume that is not a vertex cover of . Then there is an edge that is not covered by where and belong to different partitions of . Let and be the terminal vertices of which are adjacent to and , respectively, in . Then it is clear that there is a path between the terminals and in , leading to a contradiction. Therefore, is a vertex cover of of size at most . ∎
Due to the difficulty of Node Multiway Cut even for the unweighted version and with a small size of independent set, we consider a relaxed variation in which we are allowed to remove terminal vertices.