Approximation Algorithms for Connected Maximum Cut and Related Problems
Abstract
An instance of the Connected Maximum Cut problem consists of an undirected graph and the goal is to find a subset of vertices that maximizes the number of edges in the cut such that the induced graph is connected. We present the first nontrivial approximation algorithm for the Connected Maximum Cut problem in general graphs using novel techniques. We then extend our algorithm to edge weighted case and obtain a polylogarithmic approximation algorithm. Interestingly, in contrast to the classical MaxCut problem that can be solved in polynomial time on planar graphs, we show that the Connected Maximum Cut problem remains NPhard on unweighted, planar graphs. On the positive side, we obtain a polynomial time approximation scheme for the Connected Maximum Cut problem on planar graphs and more generally on bounded genus graphs.
1 Introduction
Submodular optimization problems have, in recent years, received a considerable amount of attention [1, 2, 3, 4, 7, 15, 27] in algorithmic research. In a general Submodular Maximization problem, we are given a nonnegative submodular^{1}^{1}1A function is called submodular if for all . function over the power set of a universe of elements, and the goal is to find a subset that maximizes so that satisfies certain prespecified constraints. In addition to their practical relevance, the study of submodular maximization problems has led to the development of several important theoretical techniques such as the continuous greedy method and multilinear extensions [4] and the double greedy [2] algorithm, among others.
In this study, we are interested in the problem of maximizing a submodular set function over vertices of a graph, such that the selected vertices induce a connected subgraph. Motivated by applications in coverage over wireless networks, Kuo et al. [25] consider the problem of maximizing a monotone, submodular function subject to connectivity and cardinality constraints of the form and provide an approximation algorithm. For a restricted class of monotone, submodular functions that includes the covering function^{2}^{2}2In this context, a covering function is defined as where is the closed neighborhood of the set of vertices , Khuller et al. [23] give a constant factor approximation to the problem of maximizing subject to connectivity and cardinality constraints.
In the light of these results, it is rather surprising that no nontrivial approximation algorithms are known for the case of general (nonmonotone) submodular functions. Formally, we are interested in the following problem, which we refer to as Connected Submodular Maximization (CSM): Given a simple, undirected graph and a nonnegative submodular set function , find a subset of vertices that maximizes such that is connected. We take the first but important step in this direction and study the problem in the case of one of the most important nonmonotone submodular functions, namely the Cut function. Formally, given an undirected graph , the goal is to find a subset , such that is connected and the number of edges that have exactly one end point in , referred to as the cut function , is maximized. We refer to this as the Connected Maximum Cut problem. Further, we also consider an edge weighted variant of this problem, called the Weighted Connected Maximum Cut problem, where function to be maximized is the total weight of edges in the cut .
We now outline an application to the image segmentation problem that seeks to identify “objects” in an image. Graph based approaches for image segmentation [16, 28] represent each pixel as a vertex and weighted edges represent the dissimilarity (or similarity depending on the application) between adjacent pixels. Given such a graph, a connected set of pixels with a large weighted cut naturally corresponds to an object in the image. Vicente et al. [33] show that even for interactive image segmentation, techniques that require connectivity also perform significantly better that cut based methods alone.
Related Work
MaxCut is a fundamental problem in combinatorial optimization that finds applications in diverse areas. A simple randomized algorithm that adds each vertex to independently with probability gives a approximate solution in expectation. In a breakthrough result, Goemans and Williamson [18] gave a approximation algorithm using semidefinite programming and randomized rounding. Further, Khot et al. [22] showed that this factor is optimal assuming the Unique Games Conjecture. Interestingly, the MaxCut problem can be optimally solved in polynomial time in planar graphs by a curious connection to the matching problem in the dual graph [20]. To the best of our knowledge, the Connected Maximum Cut problem has not been considered before our work. Haglin and Venkatesan [21] showed that a related problem, where we require both sides of the cut, namely and , to be connected, is NPhard in planar graphs.
We note that the well studied Maximum Leaf Spanning Tree (MLST) problem (e.g. see [31]) is a special case of the Connected Submodular Maximization problem. We also note that recent work on graph connectivity under vertex sampling leads to a simple constant approximation to the Connected Submodular Maximization for highly connected graphs, i.e., for graphs with vertex connectivity. Proofs of these claims are presented in the Appendix B and C respectively.
We conclude this section by noting that connected variants of many classical combinatorial problems have been extensively studied in the literature and have been found to be useful. The best example for this is the Connected Dominating Set problem. Following the seminal work of Guha and Khuller [19], the problem has found extensive applications (with more than a thousand citations) in the domain of wireless ad hoc networks as a virtual backbone (e.g. see [9, 12]). Few other examples of connected variants of classic optimization problems include Group Steiner Tree [17] (which can be seen as a generalization of a connected variant of Set Cover), Connected Domatic Partition [5, 6], Connected Facility Location [13, 32], and Connected Vertex Cover [8].
Contribution and Techniques
Our key results can be summarized as follows.
1. We obtain the first approximation algorithm for the Connected Maximum Cut (CMC) problem in general graphs. Often, for basic connectivity problems on graphs, one can obtain simple approximation algorithms using a probabilistic embedding into trees with stretch [14]. Similarly, using the cutbased decompositions given by Räcke [29], one can obtain approximation algorithms for cut problems (e.g. Minimum Bisection). Interestingly, since the CMC problem has the flavors of both cut and connectivity problems simultaneously, neither of these approaches are applicable. Our novel approach is to look for thick trees, which are basically subtrees with “high” degree sum on the leaves.
2. For the Weighted Connected Maximum Cut problem, we obtain an approximation algorithm. The basic idea is to group the edges into logarithmic number of weight classes and show that the problem on each weight class boils down to the special case where the weight of every edge is either or .
3. We obtain a polynomial time approximation scheme for the CMC problem in planar graphs and more generally in bounded genus graphs. This requires the application of a stronger form of the edge contraction theorem by Demaine, Hajiaghayi and Kawarabayashi [11] that may be of independent interest.
4. We show that the CMC problem remains NPhard even on unweighted, planar graphs. This is in stark contrast with the regular MaxCut problem that can be solved optimally in planar graphs in polynomial time. We obtain a polynomial time reduction from a special case of 3SAT called the Planar Monotone 3SAT (PM3SAT), to the CMC problem in planar graphs. This entails a delicate construction, exploiting the so called “rectilinear representation” of a PM3SAT instance, to maintain planarity of the resulting CMC instance.
2 Approximation Algorithms for General Graphs
In this section, we consider the Connected Maximum Cut problem in general graphs. In fact, we provide an approximation algorithm for the more general problem in which edges can have weight 0 or 1 and the objective is to maximize the number of edges of weight 1 in the cut. This generalization will be useful later in obtaining a polylogarithmic approximation algorithm for arbitrary weighted graphs.
We denote the cut of a subset of vertices in a graph , i.e., the set of edges in that are incident on exactly one vertex of by or when is clear from context, just . Further, for two disjoint subsets of vertices and in , we denote the set of edges that have one end point in each of and , by or simply . The formal problem definition follows 
Problem Definition. {0,1}Connected Maximum Cut (bCMC): Given a graph and a weight function , find a set that maximizes such that induces a connected subgraph.
We call an edge of weight , a 0edge and that of weight , a 1edge. Further, let denote the weight of the cut, i.e., the number of 1edges in the cut. We first start with a simple reduction rule that ensures that every vertex has at least one 1edge incident on it.
Given a graph , we can construct a graph in polynomial time, such that every has at least one 1edge incident on it and has a bCMC solution of weight at least if and only if has a bCMC solution of weight at least .
Claim 1
Proof
Let be a vertex in that has only 0edges incident on it and let denote the set of its neighbors. Consider the graph obtained from by deleting along with all its incident edges and adding 0edges between every pair of its neighbors such that . Let denote a feasible solution of weight in . If , then clearly is the required solution in . If , we set and we claim that is connected if is connected and . The latter part of the claim is true since all the edges that we delete and add are 0edges. To prove the former part, notice that if is not a cut vertex in then must be connected. On the other hand, even if is a cut vertex, the new edges added among all pairs of ’s neighbors ensure that is connected. Finally, to prove the other direction, suppose we have a feasible solution of weight in . Now, if is connected, then is a feasible solution in of weight . Otherwise, set . Since creates a path between all pairs of its neighbors, is connected if is connected and is thus a feasible solution of the same weight. The proof of the lemma follows from induction. ∎
From now on, we will assume, without loss of generality, that every vertex of has at least one 1edge incident on it. We now introduce some new definitions that would help us to present the main algorithmic ideas. We denote by the total weight of edges incident on a vertex in , i.e., . In other words, is total number of 1edges incident on . Further let be the total number of 1edges in the graph. The following notion of an thick tree is a crucial component of our algorithm.
Definition 1 (Thick Tree)
Let be a graph with vertices and 1edges. A subtree (not necessarily spanning), with leaf set , is said to be thick if .
The following lemma shows that this notion of an thick tree is intimately connected with the bCMC problem. \thmt@toks\thmt@toks For any , given a polynomial time algorithm that computes an thick tree of a graph , we can obtain an approximation algorithm for the bCMC problem on .
Lemma 1
Proof
Given a graph and weight function , we use Algorithm to compute an thick tree , with leaf set . Let denote the number of 1edges in , the subgraph induced by in the graph . We now partition into two disjoint sets and such that the number of 1edges in . This can be done by applying the standard randomized algorithm for MaxCut (e.g. see [26]) on after deleting all the 0edges. Now, consider the two connected subgraphs and . We first claim that every 1edge in belongs to either or . Indeed, any 1edge in , belongs to one of the four possible sets, namely , , and . In the first two cases, belongs to while in the last two cases, belongs , hence the claim. Further, every 1edge in belongs to both and . Hence, we have 
(1)  
(2) 
Hence, the better of the two solutions or is guaranteed to have a cut of weight at least , where is the total number of 1edges in . To complete the proof we note that for any optimal solution , . ∎
Thus, if we have an algorithm to compute thick trees, Lemma 2 provides an approximation algorithm for the bCMC problem. Unfortunately, there exist graphs that do not contain thick trees for any nontrivial value of . For example, let be a path graph with vertices and 1edges. It is easy to see that for any subtree , the sum of degrees of the leaves is at most 4. In spite of this setback, we show that the notion of thick trees is still useful in obtaining a good approximation algorithm for the bCMC problem. In particular, Lemma 2 and Theorem 2 show that path graph is the only bad case, i.e., if the graph does not have a long induced path, then one can find an thick tree. Lemma 2 shows that we can assume without loss of generality that the bCMC instance does not have such a long induced path.
Shrinking Thin Paths.
A natural idea to handle the above “bad” case is to get rid of such long paths that contain only vertices of degree two by contracting the edges. We refer to a path that only contains vertices of degree two as a d2 path. Further, we define the length of a d2 path as the number of vertices (of degree two) that it contains. The following lemma shows that we can assume without loss of generality that the graph contains no “long” d2 paths.
Given a graph , we can construct, in polynomial time, a graph with no d2 paths of length such that has a bCMC solution of cut weight () at least if and only if has a bCMC solution of cut weight at least . Further, given the solution of , we can recover in polynomial time.
Lemma 2
Proof
We may assume that is connected, because otherwise we can handle each component separately. We further assume that is not a simple cycle, otherwise it is trivial to solve such an instance. If does not have a d2 path of length , then trivially we have . Otherwise, let be a path in such that and have degree two and . Note that such a path must exist as is not a simple cycle. We now perform the following operation on to obtain a new graph : Delete these elements . Add a new vertex and edges and . Since and , we are guaranteed that and hence we do not introduce any multiedges. The weights on the new edges are determined as follows  Let denote the number of 1edges in . If , we set . If , then we set and . Otherwise, we set . We claim that has a bCMC solution of cut weight at least if and only if has a solution of cut weight at least .
Let us first assume that there is a set in that is a solution to the bCMC problem with cut weight . We now show that there exists a in that is a solution to the bCMC problem with cut weight at least . The proof in this direction is done for three possible cases, based on the cardinality of . We note that is , since must be connected.
Case 1. . Note that since is connected, we must have either (i) or (ii) . In the former case, we set and the claim follows by the definition of and . In the latter case, we set . Since and are vertices of degree two, is connected. Further, every edge also belongs to . The claim follows once we observe that both and are in .
Case 2. . In this case, we must have either or but not both. Let us first assume . We set . It is clear that if is connected, so is . Due to the removal of and , we have for some edge . On the other hand, due to the addition of , we have and the claim follows since for any . Now assume that . In this case, we set . Since , we again have and the proof follows as above.
Case 3. . In this case, one of the following holds, either (i) or (ii) . If the latter is true, the proof is trivial by setting . In the former case, we set . The addition of maintains connectivity between and and hence since is connected, so is . Further, we have since no edge in in incident on or .
In order to prove the other direction, we assume that is a solution to the bCMC problem on with a cut weight of . We now construct a set that is a solution to bCMC on of weight at least . The proof proceeds in three cases similarly.
Case 1. Both and . One of the following holds  (i) or (ii) . In the former case, let be the subset of having the largest weight cut. By construction, we have that weight of the cut is at least the sum of weights of and . For the latter, let to be the best among and and the proof follows as above.
Case 2. Either or but not both. Let be the edge of maximum weight in . The edge splits the path into two connected components one containing , call it and the other containing , call it . Now to construct , we delete from (if it contains it) and add the component if or the component if . Again connectivity is clearly preserved. We now argue that the cut weight is also preserved. Indeed, this is true since we have that and the rest of the cut edges in remain as they are in .
Case 3. None of belong to . In this case, if , then trivially works. Otherwise, we set . It is easy to observe that both connectivity and all the cut edges are preserved in this case.
Now, to construct , we repeatedly apply the above contraction as long as possible. This will clearly take polynomial time as in each iteration, we reduce the number of degree2 vertices by 1. Hence we have the claim. ∎
Spanning Tree with Many Leaves.
Assuming that the graph has no long d2 paths, the following lemma shows that we can find a spanning tree that has leaves. Note that Claim 2 now guarantees that there are 1edges incident on the leaves of .
Given a graph with no d2 paths of length , we can obtain, in polynomial time, a spanning tree with at least leaves.
Lemma 3
Proof
Let be any spanning tree of . We note that although does not have d2 paths of length 3, such a guarantee does not hold for paths in . Suppose that there is a d2 path of length 7 in . Let the vertices of this path be numbered and consider the vertices . Since does not have any d2 path of length 3, there is a vertex such that . We now add an edge in to the tree . The cycle that is created as a result must contain either the edge or the edge . We delete this edge to obtain a new spanning tree . It is easy to observe that the number of vertices of degree two in is strictly less than that in . This is because, although the new edge can cause to have degree two in , we are guaranteed that the vertex will have degree three and vertices and (or and ) will have degree one. Hence, as long as there are d2 paths of length 7 in , the number of vertices of degree two can be strictly decreased. Thus this process must terminate in at most steps and the final tree obtained does not have any d2 paths of length .
We now show that the tree contains leaves by a simple charging argument. Let the tree be rooted at an arbitrary vertex. We assign each vertex of a token and redistribute them in the following way : Every vertex of degree two in gives its token to its first non degree two descendant, breaking ties arbitrarily. Since there is no d2 path of length , each non degree two vertex collects at most 7 tokens. Hence, the number of vertices not having degree two in is at least . Further, since the average degree of all vertices in a tree is at most 2, a simple averaging argument shows that must contain at least vertices of degree one, i.e., leaves. ∎
Obtaining an Approximation
We now have all the ingredients required to obtain the approximation algorithm. We observe that if the graph is sparse, i.e. (for a suitable constant ), then the tree obtained by using Lemma 2 is an thick tree and thus we obtain the required approximate solution in this case. On the other hand, if the graph is sparse, then we use Lemma 2 to obtain a spanning tree, delete the leaves of this tree, and then repeat this procedure until we have no more vertices left. Since, we delete a constant fraction of vertices in each iteration, the total number of iterations is . We then choose the “best” tree out of the trees so obtained and show that it must be an thick tree, with . Finally, using Lemma 2, we obtain an approximate solution as desired. We refer to Algorithm 1 for the detailed algorithm.
Algorithm 1 gives an approximate solution for the bCMC problem.
Theorem 1
Proof
Let us assume that (for some constant ). Now, Lemma 2 and Claim 2 together imply that . Further, since we have , is an thick tree for some . Hence, we obtain an approximate solution using Lemma 2.
On the other hand, if , we show that at least one of the trees obtained by the repeated applications of the Lemma 2 is an thick tree of for . We first observe that the While loop in Step 1 runs for at most iterations. This is because we delete leaves in each iteration and hence after iterations, we get . We now count the number of 1edges “lost” in each iteration. We recall that is the total number of 1edges incident on in a graph . In an iteration , the number of 1edges lost at Step 1 is at most . In addition, we may lose a total of at most edges due to the contraction of degree two vertices in Step 1. Suppose for the sake of contradiction that where is a suitable constant. Then the total number of 1edges lost in iterations is at most
The equality follows for a suitable constant as . The final inequality holds for a suitable choice of the constants and . But this is a contradiction since we have .
Since we choose to be the best iteration, we have for some constant . Hence the tree is an thick tree of for and the theorem follows by Lemma 2. ∎
General Weighted Graphs
We now consider the Weighted Connected Maximum Cut (WCMC) problem. Formally, we are given a graph and a weight function . The goal is to find a subset of vertices that induces a connected subgraph and maximizes the quantity . We obtain a approximation algorithm for this problem. Our basic strategy is to group edges having nearly the same weight into a class and thus create classes. We then solve the bCMC problem for each class independently and return the best solution.
Theorem 2.1
Algorithm 2 gives a approximation guarantee for the Weighted Connected Maximum Cut problem.
Proof
Let be an optimal solution for a given instance of the problem and let . Also, let . Since we have that , we can reset the weights of those edges with weight to 0 and assume that where denotes the weight of the minimum (non zero) weight edge. Let be the set of edges such that and finally let . We now claim that . This immediately gives us that .
We now prove the claim. Consider solving the bCMC instance with weight function . Clearly is a feasible solution to this instance and we have . The previous inequality holds as is guaranteed to be an approximate solution by Theorem 2. Now, we have . Hence, the claim. ∎
3 CMC in Planar and Bounded Genus Graphs
In this section, we consider the CMC problem in planar graphs and more generally, in graphs with genus bounded by a constant. We show that the CMC problem has a PTAS in bounded genus graphs.
PTAS for Bounded Genus Graphs.
We use the following (paraphrased) contraction decomposition theorem by Demaine, Hajiaghayi and Kawarabayashi [11].
([11]) For a boundedgenus graph and an integer , the edges of can be partitioned into color classes such that contracting all the edges in any color class leads to a graph with treewidth . Further, the color classes are obtained by a radial coloring and have the following property: If edge is in class , then every edge such that is in class or or .
Theorem 2
Given a graph of constant genus, we use Theorem 3 appropriately to obtain a graph with constant treewidth. In Appendix A, we show that one can solve the CMC problem optimally in polynomial time on graphs with constant treewidth.
If the CMC problem can be solved optimally on graphs of constant treewidth, then there exists a polynomial time approximation algorithm for the CMC problem on bounded genus graphs (and hence on planar graphs).
Theorem 3
Proof
Let be the graph of genus bounded by a constant and let denote the optimal CMC of and be its size. Using Theorem 3 with , we obtain a partition of the edges into color classes namely . We further group three consecutive color classes into groups where . Let denote the group that intersects the least with the optimal connected max cut of , i.e., ^{3}^{3}3We “guess” by trying out all the possibilities. As the groups partition the edges, we have . Let , so that . Let denote the graph of treewidth obtained by contracting all edges of color .
We first show that has a CMC of size at least . For a vertex , let denote the set of vertices of that have merged together to form due to the contraction. We define a subset as . Note that because we contract edges (and not delete them), remains connected. We claim that . Let be an edge in . Now implies that at least one edge such that has been contracted. By the property guaranteed by Theorem 3, we have that . Hence we have, .
Finally, given a connected max cut of size in , we can recover a connected max cut of size at least in by simply uncontracting all the contracted edges. Hence, by solving the CMC problem on optimally, we obtain a approximate solution in . ∎
NPhardness in planar graphs
We now describe a nontrivial polynomial time reduction of a 3SAT variant known as Planar Monotone 3SAT (PM3SAT) to the CMC problem on a planar graph, thereby proving that the latter is NPhard. The following reduction is interesting as the classical MaxCut problem can be solved optimally in polynomial time on planar graphs using duality. In fact, it was earlier claimed that even CMC can be solved similarly [21].
An instance of PM3SAT is a 3CNF boolean formula such that 

A clause contains either all positive literals or all negative literals.

The associated bipartite graph ^{4}^{4}4 has a vertex for each clause and each variable and an edge between a clause and the variables that it contains is planar.
Given such an instance, the PM3SAT problem is to decide whether the boolean formula is satisfiable or not. Berg and Khosravi [10] show that the PM3SAT problem is NPcomplete.
The Reduction. Given a PM3SAT formula , with a rectilinear representation, we obtain a polynomial time reduction to a Planar CMC instance, there by showing that the latter is NPhard. Let denote the variables of the PM3SAT instance and denote the clauses. We construct a planar graph as follows. For every variable , we construct the following gadget: We create two vertices and corresponding to the literals and . Additionally, we have “helper” vertices, such that each is adjacent to both and . Further, for every we add a set of new vertices that are adjacent only to . Now, in the rectilinear representation of the PM3SAT, we replace each variable rectangle by the above gadget. For two adjacent variable rectangles in the rectilinear representation, say and , we connect the helpers and . For every clause , has a corresponding vertex with edges to the three literals in the clause. Finally, for each vertex , we add a set of new vertices adjacent only to . It is easy to observe that the reduction maintains the planarity of the graph. Figure 0(b) illustrates the reduction by an example.
We show the following theorem that proves the Planar Connected Maximum Cut problem is NPhard.
Let denote an instance of the planar CMC problem corresponding to an instance of PM3SAT obtained as per the reduction above. Then, the formula is satisfiable if and only if there is a solution to the CMC problem on with .
Theorem 4
Proof
For brevity, we denote as in the rest of the proof.
Forward direction. Assume that is satisfiable under an assignment . We now show that we can construct a set with the required properties. Let be the set of literals that are true in . We define , i.e., the set of vertices corresponding to the true literals, all the clauses and all the helper vertices. By construction, the set of all helper vertices and one literal of each variable induces a connected subgraph. Further, since in a satisfying assignment every clause has at least one true literal, the constructed set is connected. We now show that . Indeed, contains all the edges corresponding to the one degree vertices incident on clauses and all the helpers. This contributes a profit of . Also, since no vertex corresponding to a false literal is included in but all helpers are in , we get an additional profit of for each variable. Hence, we have the claim.
Reverse direction. Assume that is a subset of vertices in such that is connected and . We now show that is satisfiable. We may assume that is an optimal solution (since optimal solution will satisfy these properties, if a suboptimal solution does). We first observe that at least one of the (two) literals for each variable must be chosen into . Indeed, if this is not the case for some variable, for to be remain connected, none of the helper vertices corresponding to that variable can be chosen. This implies that the maximum possible value for (this is the number of remaining edges) (since ) , a contradiction. We now show that every helper vertex must be included in . Assume that this is not true and let be some helper vertex not added to . We note that none of the degree one vertices in can be in because must be connected. Now, consider the solution formed by adding to . Since at least one vertices or is in , if is connected, so is . Further, the total number of edges in the cut increases by . This is a contradiction to the fact that is an optimal solution. Hence, every helper vertex belongs to the solution . We now show that, no two literals of the same variable are chosen into . Assume the contrary and let , both be chosen into . We claim that removing one of these two literals will strictly improve the solution. Indeed, consider removing from . Clearly, we gain all the edges from to all the helper vertices corresponding to this variable. Thus we gain at least edges. We now bound the loss incurred. In the worst case, removing from might force the removal of all the clause vertices due to the connectivity restriction. But this would lead to a loss of at most . Hence, we arrive at a contradiction that is an optimal solution. Therefore, exactly one literal vertex corresponding to each variable is included in . Finally, we observe that all the clauses must be included in . Assume this is not true and that clause vertices are in . Now the total cut is , which is again a contradiction. Now, the optimal solution gives a natural assignment to the PM3SAT instance: a literal is set to TRUE if its corresponding vertex is included in . Since, every clause vertex belongs to , which in turn is connected, it must contain a TRUE literal and hence the assignment satisfies . ∎
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Appendix A Dynamic program for constant treewidth graphs
The notion of tree decomposition and treewidth was first introduced by Robertson and Seymour [30]. Given a graph , its tree decomposition is a tree representation , where each (called as a bag) is associated with a subset such that the following properties hold:

.

For every edge , there is a bag , such that .

For every , the subgraph of , induced by bags that contain , is connected.
The width of a decomposition is defined as the size of the largest bag minus one. Treewidth of a graph is the minimum width over all the possible tree decompositions. In this section, we show that the CMC problem can be solved optimally in polynomial time on graphs with constant treewidth .
Notation. We denote the tree decomposition of a graph by . For a given bag of the decomposition , let denote the set of vertices of contained in and denote the set of vertices in the subtree of rooted at . As shown by Kloks [24], we may assume that is nice tree decomposition, that has the following additional properties.

Any node of the tree has at most children.

A node with no children is called a leaf node and has .

A node with two children and is called a join node. For such a node, we have .

A node with exactly one child is either a forget node or an introduce node. If is a forget node then for some . One the other hand, if is an introduce node then for .
We now describe a dynamic program to obtain the optimal solution for the CMC problem. Let denote the optimal solution. We first prove the following simple claim that helps us define the dynamic program variable.
Claim
For any bag , the number of components induced by in is at most .
Proof
Consider the induced subgraph and let be one of its components. We observe that has at least one vertex in , i.e., . Assume this is not true and . Now consider an edge such that and . Such an edge is guaranteed to exist owing to the connectivity of . By our assumptions, . This implies there is some bag not in the subtree of rooted at , that contains both and . But this in turn implies , a contradiction to the assumption . Now, since each vertex in belongs to at most one component, there can be at most components in