1 Introduction
Given a graph and a graph property , the edgedeletion problem consists in determining the minimum number of edges required to be removed in order to obtain a graph satisfying [12]. Given an integer , the edgedeletion decision problem asks for a set with , such that the obtained graph by the removal of satisfies . Both versions have received widely attention on the study of their complexity, where we can cite [38, 37, 12, 33, 2, 23, 26, 29] and references therein for applications. When the obtained graph is required to be bipartite, the corresponding edge (vertex) deletion problem is called edge (vertex) bipartization [14, 1, 22] or edge (vertex) frustration [39]. Choi, Nakajima, and Rim [14] showed that the edge bipartization decision problem is complete even for cubic graphs.
Furmańczyk, Kubale, and Radziszowski [22] considered vertex bipartization of cubic graphs by the removal of an independent set. In this paper we study the analogous edge deletion decision problem, that is, the problem of determining whether a finite, simple, and undirected graph admits a removal of a set of edges that is a matching in in order to obtain a bipartite graph. Formally, for a set of edges of a graph , let be the graph with vertex set and edge set . For a matching , we say that is an odd decycling matching of if is bipartite. Let denote the set of all graphs admitting an odd decycling matching. We deal with the complexity of the following decision problem.
Odd Decycling Matching Input: A finite, simple, and undirected graph . Question: Does ?
A more restricted version of this problem is considered by Schaefer [35]. He deals with the problem of determining whether a given graph admits a coloring of the vertices so that each vertex has exactly one neighbor with same color as itself. We can see that the removal of the set of edges whose endvertices have same color, which is a perfect matching of , generates a bipartite graph. Schaefer proved that such a problem is complete even for planar cubic graphs.
With respect to the minimization version, the edgedeletion decision problem in order to obtain a bipartite graph is analogous to Simple Max Cut, which was proved to be complete by Garey, Johnson and Stockmeyer [23]. Yannakakis [37] proved its completeness even for cubic graphs.
Cowen et al. [20] studied the colorings, called defective, that are coloring of the vertices such that each color class has maximum degree . They proved that it is complete to determine whether a given graph is colorable even for graphs with maximum degree 4 and even for planar graphs with maximum degree 5. Angelini et al. [3] determined that Odd Decycling Matching can be solved in lineartime for partial trees, where it is known that tree graphs have treewidth at most , for any .
Odd Decycling Matching can also be seen as another problem. A graph is colorable if can be partitioned into , such that the induced subgraph has maximum degree at most , for all . This is a generalization of the classical proper coloring, when every , and the improper coloring, when every . It is clear to see that if and only if is colorable. Lovász [30] proved that if a graph satisfies then is colorable, where denotes the maximum degree of . This result shows that every subcubic graph is colorable and thus belongs to . Borodin, Kostochka, and Yancey [7] studied the colorable graphs with respect to the sparseness parameter . They proved that every graph with is colorable, where this bound is sharp. Moreover, they defined the parameter , such that . They showed that is colorable if . Finally, they also proved that every planar graph with girth (the size of the smallest cycle of ) at least is colorable. This is the best result concerning coloring of planar graphs.
In this work we summarize our results as follows. We prove that Odd Decycling Matching is complete even for colorable planar graphs with maximum degree , which improves the previous result by Cowen et al. [20]. As positive results, we show polynomial time algorithms for (claw, paw)free graphs, graphs that have only triangles as odd cycles, and graphs that have a small dominating set. We also show that graphs in can be expressed in monadic second order logic. Hence, using Courcelle’s metatheorems [15, 19, 16] we prove that Odd Decycling Matching is fixedparameter tractable when parameterized by cliquewidth, which improves the previous result by Angelini et al. [3]. We also show an exact algorithm, where is the vertex cover number of . Finally, for a generalization of Odd Decycling Matching, we show a kernel with at most vertices when such a more general problem is parameterized by neighborhood diversity number, .
1.1 Preliminaries
Let be a graph with vertices and edges. Given a subgraph of , we denote by the induced subgraph of by . Let be the neighborhood of in . Moreover, let be the closed neighborhood of in . The degree of a vertex , , is denoted by , and let be the maximum degree of .
Let and be a path and a cycle of length , respectively. Furthermore, we denote by and the complete graphs of order and the complete bipartite graphs with parts of order and , respectively.
A diamond is the graph obtained by removing one edge from the . Let be the wheel graph of order , that is, the graph containing a vertex , called central, and a cycle of order , such that is adjacent to all vertices of .
We say that a graph is a pool if it is formed by triangles edgedisjoint whose bases induce a . Formally, a pool is obtained from a cycle , such that the oddindexed vertices induce a cycle , called internal cycle of the pool, where , . The evenindexed vertex is the thborder of the pool, where and is taken modulo . Fig. 0(c) and Fig. 0(d) represent the pool and pool, respectively.
The claw () and the paw (a triangle plus an edge) graphs are the unique ones with degree sequences and , respectively.
Clearly, every graph admits a proper coloring. Hence every graph in is free. More precisely, every graph in is free, which is depicted in Fig 0(a). Hence some proper colorable graphs do not admit an odd decycling matching. Fig. 1 shows some others examples of forbidden subgraphs. Lemma 1 collects some properties of graphs in .
Lemma 1.
For a graph and an odd decycling matching of , the following assertions are true.

If has a diamond as a subgraph, then contains no edge incident to only one vertex of degree three of .

cannot contain two disjoint , for every .

cannot contain a as a subgraph, for all .

cannot contain a pool as a subgraph, for all odd .
Proof.
(i) Let be a graph that contains a diamond as a subgraph, such that and . We can see that equals to exactly one of the following sets: , , . For each of such sets, both and are matched by . Hence cannot contain any edge incident to only or .
(ii) Let such that contains two disjoint , and . It follows that and are diamonds that share a vertex of degree at least three. By (i) the statement holds.
(iii) Suppose for a contradiction that contains a subgraph isomorphic to a wheel graph , . Let , such that is adjacent to all vertices of the cycle . If , then contains two disjoint in its neighborhood, and thus it follows by (ii) that . In this case, it can be easily verified that and are forbidden subgraphs.
(iv) Suppose, for a contradiction, that contains a subgraph isomorphic to a pool, for some odd . Let be its internal cycle and let be the vertices of the border of , such that , for all modulo . Clearly must contain some edge of and one edge of every triangle . W.l.o.g., consider . This implies that contains no edge in . Therefore, and must be in , which forbids two more edges from the triangles and . Continuing this process, it follows that , which is at the same distance of and in , must contain two incident edges in , a contradiction. ∎
1.2 A linear Time Algorithm for Subcubic Graphs
Bondy and Locke [6] presented the following lemma, which was also obtained by Erdős [21] by induction on .
Lemma 2.
(Bondy and Locke [6]) Let be a graph and let be a largest bipartite subgraph of . Then , for every .
Lemma 2 shows that every subcubic graph admits an odd decycling matching, since every vertex has at most one incident edge not in a largest bipartite subgraph of . This result was also obtained by Lovász [30] with respect to improper coloring of graphs with maximum degree at most .
Consider a bipartition of into sets and . For every vertex , we say that is of type if and , where is the part (either or ) which contains . We present a linear algorithm to find an odd decycling matching of subcubic graphs, Algorithm 1.
Theorem 3.
Algorithm 1 returns in lineartime an odd decycling matching for subcubic graphs.
Proof.
Let be a maximal independent set of . Let . In this case, every vertex of is of type and there is no vertex in of type , . Therefore, if there exists a vertex of type with , then it must be in and be of type . In order to prove the correctness of Algorithm 1, it is sufficient to show that the operations on lines 7–8 and 10–11 do not generate vertices of type with .
Let . If is of type , then is moved from to by lines 7–8. In this case, it follows that both and are vertices of type after the line 8. If is not of type , then the lines 10–11 modify the types of and .

If is of type , then and are modified to type and , respectively;

If is of type , then and are modified to type and , respectively;

If is of type , then and are modified to type and , respectively;

If is of type , then and are modified to type and , respectively.
We can see that each neighbor of in the same part of loses exactly one neighbor (that is ) in . Moreover, receives at most one new neighbor (that is ) in . The same occurs for every neighbor of in . Therefore, in any case it is not obtained vertices of type with , which implies that the Algorithm 1 finishes. ∎
2 NPCompleteness for Odd Decycling Matching
In this section we prove that Odd Decycling Matching is complete even for planar graphs of maximum degree at most . We organize the proof in three parts. In the first one we show some polynomial time reductions from NotAllEqual 3SAT (NAE3SAT) [35] and Positive Planar 1In3SAT [32]. In the second part we prove that Odd Decycling Matching is complete for graphs with maximum degree at most . This proof is a more intuitive and easier to understand the gadgets and construction of the next part. The third part presents a proof that Odd Decycling Matching is complete even for planar graphs with maximum degree at most . Finally, the proof finishes as a corollary from the previous results by just slightly modifying the used gadgets.
2.1 Preliminaries
Let be a Boolean formula in CNF with set of variables and set of clauses . The associated graph of , , is the bipartite graph such that there exists a vertex for every variable and clause of , where is a bipartition of into independent sets. Furthermore, there exists an edge if and only if contains either or . We say that is planar if its associated graph is planar. In order to obtain a polynomial reduction, we consider the following decision problems, which are complete.
NotAllEqual 3SAT (NAE3SAT) [35] Input: A Boolean formula in CNF, . Question: Is there a truth assignment to the variables of , in which each clause has one literal assigned true and one literal assigned false?
Positive Planar 1In3SAT [32] Input: A planar Boolean formula in CNF, , with no negated literals. Question: Is there a truth assignment to the variables of , in which each clause has exactly one literal assigned true?
In order to prove the completeness of Odd Decycling Matching, we first present a polynomial time reduction from NAE3SAT and Positive Planar 1In3SAT to the following decision problems, respectively:
NAE3SAT Input: A Boolean formula in CNF, , where each clause has either or literals, each variable occurs at most times, and each literal occurs at most twice. Question: Is there a truth assignment to the variables of in which each clause has at least one literal assigned true and at least one literal assigned false?
Planar 1In3SAT Input: A planar Boolean formula in CNF, , where each clause has either or literals and each variable occurs at most times. Moreover, each positive literal occurs at most twice, while every negative literal occurs at most once in . Question: Is there a truth assignment to the variables of in which each clause has exactly one true literal?
Theorem 4.

NAE3SAT is complete.

Planar 1In3SAT is complete.
Proof.
Since verifying whether a graph is planar can be done in lineartime [27], as well as whether a formula in CNF has a truth assignment, both problems are in .
Let be a Boolean formula in CNF such that denotes
the set of variables and is the set of clauses of .
We construct a formula from as follows.
For a vertex , let be the degree of in .
For such a variable with , we create new clauses
of size , and new variables as follows:
In addition, we replace the () occurrence of the variable by an occurrence of a variable , where a literal (resp. ) is replaced by a literal .
Let be the set of all vertices with . For such a vertex , let and .
Note that, the associated graph can be obtained from by replacing the corresponding vertex of by a cycle of length induced by the corresponding vertices of the new clauses in and the new variables in . In addition, for each and an edge is added in , such that every corresponding vertex has exactly one neighbor . Fig. 2 shows an example of the transformation for a Boolean formula.
As we can see, every variable occurs at most times in the clauses of , since every variable with is replaced by new variables that are in exactly clauses of . By the construction, each literal occurs at most twice. Moreover, if has no negative literals, then only the new variables have a negated literal and each one occurs exactly once in .
Now, it remains to show that if is planar then we can construct as a planar formula. Consider a planar embedding of , we construct replacing each corresponding vertex by a cycle of length , as described above. After that, in order to preserve the planarity, we can follow the planar embedding to add a matching between vertices corresponding to variables in such a cycle and vertices corresponding to clauses and that . Such a matching indicates in which clause of a given new variable will replace in . Thus, without loss of generality, if is planar then we can assume that is planar as well.
Let be an instance of NAE3SAT (resp. Positive Planar 1In3SAT) such that denotes its set of variables and its set of clauses. Let be the formula obtained from by the above construction. As we can observe, for any truth assignment of , all (for a given variable of ) have the same value. Therefore, any clause of containing exactly two literals has true and false values. At this point, it is easy to see that has a notallequal (resp. 1in3) truth assignment if and only if has a notallequal (resp. 1in3) truth assignment. ∎
Now we show the completeness of Odd Decycling Matching. Let us call the graph depicted in Fig 2(a) by head. Vertex is the neck of the head. Given a graph , the next lemma shows that such a structure is very useful to ensure that some edges cannot be in any odd decycling matching of . The next simple lemma is used in the correctness of our reductions.
Lemma 5.
Let be a graph that contains an induced subgraph isomorphic to a head graph, whose neck is . Then all edges not in incident to cannot be in any odd decycling matching of . Moreover admits only one odd decycling matching.
Proof.
Let be an odd decycling matching of . Suppose for a contradiction that there exists an edge incident to , such that contains an endvertex not in . In this case, we get that and does not belong to , which implies that . By the triangle , it follows that must be in . Hence the cycle remains in , a contradiction.
Now suppose that . In this case, the edge cannot be in , otherwise the cycle survives in . In the same way, the edge , otherwise the cycle is not destroyed by . Therefore we get that must be in , which implies that . Hence the cycle belongs to . Since the triangle has no edge in , it is not destroyed by , a contradiction.
Finally, we get that must be in , which implies that as well. Therefore, it follows that must be in . Hence also must be in , which turns the graph bipartite. Since all choices of the edges of are necessary, we get that there is only one possible odd decycling matching of , which is perfect. Fig. 2(b) shows such a matching. This concludes the proof. ∎
2.2 Completeness for Graphs with Maximum Degree at Most 4
With Lemma 5 we can establish the completeness of Odd Decycling Matching. Remember that graphs in are all colorable. The next results show that the completeness is also obtained even for colorable bounded degree graphs. First we present a more intuitive proof by a reduction from NAE3SAT, next we present a more complex proof that also preserves the planarity. The circles with an in the figures represent an induced subgraph isomorphic to the head graph, whose neck is the vertex touching the circle. By simplicity, this pattern will be used in the remaining figures whenever possible.
Theorem 6.
Odd Decycling Matching is complete even for colorable graphs with maximum degree at most .
Proof.
We prove that Odd Decycling Matching is complete by a reduction from NAE3SAT Let be an instance of NAE3SAT, with and be the sets of variables and clauses of , respectively. We construct a graph as follows:

For each variable , we construct a variable gadget . Such a gadget consists on a diamond with a head, whose neck is the vertex of degree two in . The vertices of degree three in , and , represent the literal , while the last one, , of degree two represents the negative literal . Fig. 4 shows the variable gadget .

For each clause , we associate a clause gadget . If contains three literals, then is a triangle with vertices , , and . Moreover, each vertex is adjacent to a linking vertex , , which is a neck of a head . Such clause gadget is showed in Fig. 4(b). In a similar way, if has size two, then is as depicted in Fig. 4(a), where and are the vertices that connect to the gadgets of the variables contained in .

We link a clause gadget to a variable gadget , such that , as follows. If contains the positive literal , then add one edge between a linking vertex to either or , otherwise we add the edge , for some .
Since the Head graph is colorable, clearly the above construction generates also a colorable graph. Next we prove that has a truth assignment if and only if the graph obtained form the above construction has an odd decycling matching. If has a truth assignment , then each clause contains at least one true literal and at least one false literal. For such a clause, we associate true to if and only if its corresponding literal is true in . In the same way, for every variable gadget , we associate true to and if and only if the positive literal is true in . Therefore, we can construct a bipartition of into sets and , that represent the literal assigned true and false, respectively, as follows.

For each clause gadget of literals, remove the edge if , ;

For each clause gadget of literals, remove either the edge or ;

For every variable gadget , remove the edge ;

For each induced head , remove edges as in Fig. 2(b).
It is not hard to see that the obtained graph is bipartite, since each linking vertex is in the opposite set of and of , such that . Moreover, and are in opposite sets, for every clause of length . Since the removed edges are clearly a matching in , it follows that .
Now we consider that . By Lemma 5, it follows that must be in any odd decycling matching of , for every variable gadget . Analogously, either or and exactly one edge must be included in any odd decycling matching of , for every clause gadget of and literals, respectively. Therefore, for an odd decycling matching of , we can associate to the parts of the bipartition of as true and false. Thus, it follows that:

and are in the same part, while is in the opposite one, for every variable gadget ;

and are in different parts, for every clause gadget of length ;

All the vertices are not in the same part, for every clause gadget of length ;
Hence, every clause has at least one true and one false literal, which implies that is satisfiable. ∎
Since NAE3SAT is polynomial time solvable for planar graphs [31], the previous construction cannot be planar. Moreover planar graphs are classical colorable graphs. Hence it is interesting to know what happens in such a class. The next Subsection deals with this problem.
2.3 Completeness for Planar Graphs
Now we will show that Odd Decycling Matching remains complete even for colorable planar graphs with maximum degree . We prove the completeness by a reduction from Planar 1In3SAT. In order to prove this result, next we give a useful lemma.
Lemma 7.
Let be a border of an odd pool graph , such that and are its neighbors in . It follows that every odd decycling matching of must contain exactly one edge of the internal cycle, which is different from . Moreover, there is only one odd decycling matching for such an edge.
Proof.
Let be the internal cycle of and let be the thborder of , such that , . Since has odd length, it follows that every odd decycling matching of contains at least one edge of .
Suppose for a contradiction that has an odd decycling matching containing . In this case, we get that the edges in cannot be in . Therefore must contain the edges and . In the same way, we can see that the edges are not in . Hence, it can be seen that all edges indent to are forbidden to be in , which implies that the triangles and have no edge in , a contradiction by the choice of .
Let be an edge of contained in an odd decycling matching of . In a same fashion, the edges in cannot be in . Following this pattern, we can see that every edge must be in , for every . Furthermore, it follows that , for every . Since contains one edge of every triangle of , it follows that is unique, for every edge . Finally, such an odd decycling matching contains only one edge of . ∎
Theorem 8.
Odd Decycling Matching is complete even for colorable planar graphs with maximum degree at most .
Proof.
Let be an instance of Planar 1In3SAT, with and be the sets of variables and clauses of , respectively. We construct a planar graph of maximum degree as follows:

For each clause , we construct a gadget as depicted in Fig. 6. Such gadgets are just a pool and a pool less a border for clauses of size and , respectively. Moreover, for the alternate edges of the internal cycle we subdivide them twice and append a head graph to each such a new vertex. Finally, we add two vertices and , such that and , for . For such new vertices, we append a head graph to each one.

For each variable , we construct a gadget as depicted in Fig. 7. Such a gadget is a pool less a border, where we subdivide the edges , , , and twice, where every such a new vertex has a pendant head. We rename each border vertex () as and as , for . Moreover we add a new vertex adjacent to , which has a pendant head graph.

The connection between clause and variable gadgets are as in Fig. 6 and Fig. 7, where each pair of arrow head edges in a variable gadget corresponds to a pair of such edges in a clause gadget , such that . More precisely, if , then we add the edges and , for some and for some . On the other hand, if , then we add the edges and , for some .

If a variable occurs only twice in , then just consider those connections corresponding to the literals of in the clauses of , such that and represent .
Let be the graph obtained from by the above construction. We can see that has maximum degree , where the only vertices with degree are those , for each variable gadget . Furthermore, it is clear that is colorable.
It remains to show that if is planar (that is, if is planar), then is planar. Consider a planar embedding of . We replace each variable vertex of by a variable gadget , as well as every clause vertex of by a clause gadget . The clause gadgets correspond to clauses of length two or three, which depends on the degree of in . Since the clause and variable gadgets are planar, we just need to show that the connections among them keep the planarity. Given an edge , we connect and by duplicating such an edge as parallel edges and , for some and for some or and , for some , as previously discussed.
In order to prove that is satisfiable if and only if , we discuss some considerations related to odd decycling matchings of the clause and variable gadgets. By Lemma 7, we know that an odd pool graph less a border admits one odd decycling matching for each edge of the internal cycle, except that whose both endvertices are adjacent to the removed border. Furthermore, by Lemma 5 it follows that each external edge incident to the neck of an induced head cannot be in any odd decycling matching. In this way, Fig. 8 shows the possible odd decycling matchings , given by the stressed edges for the clause gadget of clauses of length three. The black and white vertex assignment represents the bipartition of . Notice that exactly one pair of vertices and () is such that they have the same color, while the other such pairs have opposite colors. More precisely, we can see that has the same color for each pair with opposite color vertices as well as , for each odd decycling matching of . In this way, we can associate one literal , , and to each pair of vertices and , . A similar analysis can be done for clause gadgets of clauses of length two.
In the same fashion as the clause gadgets, each variable gadget admits two possible odd decycling matchings as depicted in Fig. 9. We can see that the pair and has a different assignment for the other two pairs and , . Moreover, the last two pairs have the same assignment, as it can be seen in Fig. 8(a)