Independent Feedback Vertex Set for \(P_5\)Free Graphs
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Abstract
The NPcomplete problem Feedback Vertex Set is that of deciding whether or not it is possible, for a given integer \(k\ge 0\), to delete at most k vertices from a given graph so that what remains is a forest. The variant in which the deleted vertices must form an independent set is called Independent Feedback Vertex Set and is also NPcomplete. In fact, even deciding if an independent feedback vertex set exists is NPcomplete and this problem is closely related to the 3Colouring problem, or equivalently, to the problem of deciding whether or not a graph has an independent odd cycle transversal, that is, an independent set of vertices whose deletion makes the graph bipartite. We initiate a systematic study of the complexity of Independent Feedback Vertex Set for Hfree graphs. We prove that it is NPcomplete if H contains a claw or cycle. Tamura, Ito and Zhou proved that it is polynomialtime solvable for \(P_4\)free graphs. We show that it remains polynomialtime solvable for \(P_5\)free graphs. We prove analogous results for the Independent Odd Cycle Transversal problem, which asks whether or not a graph has an independent odd cycle transversal of size at most k for a given integer \(k\ge 0\). Finally, in line with our underlying research aim, we compare the complexity of Independent Feedback Vertex Set for Hfree graphs with the complexity of 3Colouring, Independent Odd Cycle Transversal and other related problems.
Keywords
Independent feedback vertex set Nearbipartiteness Independent odd cycle transversal 3Colouring1 Introduction
Many computational problems in the theory and application of graphs can be formulated as modification problems: from a graph G, some other graph H with a desired property must be obtained using certain permitted operations. The number of graph operations used (or some other measure of cost) must be minimised. The computational complexity of a graph modification problem depends on the desired property, the operations allowed and the possible inputs; that is, we can prescribe the class of graphs to which G must belong. This leads to a rich variety of different problems, which makes graph modification a central area of research in algorithmic graph theory.
A set S of vertices in a graph G is a feedback vertex set of G if removing the vertices of S results in an acyclic graph, that is, the graph \(GS\) is a forest. The Feedback Vertex Set problem asks whether or not a graph has a feedback vertex set of size at most k for some integer \(k\ge 0\) and is a wellknown example of a graph modification problem: the desired property is that the obtained graph is acyclic and the permitted operation is vertex deletion. The directed variant of Feedback Vertex Set was one of the original problems proven to be NPcomplete by Karp [26]. The proof of this implies NPcompleteness of the undirected version even for graphs of maximum degree 4 (see [17]). We refer to the survey of Festa et al. [16] for further details of this classic problem.
We survey known results on Independent Feedback Vertex Set below.
1.1 Related Work
One way is to consider the problem from a parameterized point of view. Taking k as the parameter, Misra et al. [35] proved that Independent Feedback Vertex Set is fixedparameter tractable by giving a cubic kernel. This is in line with the fixedparameter tractability of the general Feedback Vertex Set problem (see [27] for the fastest known FPT algorithm). Later, Agrawal et al. [1] gave a faster FPT algorithm for Independent Feedback Vertex Set and also obtained an upper bound on the number of minimal independent feedback vertex sets of a graph.
Another way to obtain tractability results is to restrict the input to special graph classes in order to determine graph properties that make the problem polynomialtime solvable. We already mentioned some classes for which Independent Feedback Vertex Set is NPcomplete. In a companion paper [6], we show that the problem is polynomialtime solvable for graphs of diameter 2, and as stated above, the problem is NPcomplete on graphs of diameter 3. Tamura et al. [43] showed that Independent Feedback Vertex Set is polynomialtime solvable for chordal graphs, graphs of bounded treewidth and for cographs. The latter graphs are also known as \(P_4\)free graphs (\(P_r\) denotes the path on r vertices and a graph is Hfree if it has no induced subgraph isomorphic to H), and this strengthened a result of Brandstädt et al. [8], who proved that NearBipartiteness is polynomialtime solvable for \(P_4\)free graphs.
1.2 Our Contribution
The Independent Feedback Vertex Set problem is equivalent to asking for a (proper) 3colouring of a graph, such that one colour class has at most k vertices and the union of the other two induces a forest. We wish to compare the behaviour of Independent Feedback Vertex Set with that of the 3Colouring problem. It is well known that the latter problem is also NPcomplete [31] in general and polynomialtime solvable on many graph classes (see, for instance, the surveys [20, 40]). We also observe that 3Colouring is equivalent to asking whether or not a graph has an independent odd cycle transversal (of any size). However, so far very few graph classes are known for which Independent Feedback Vertex Set is tractable and our goal is to find more of them. For this purpose, we consider Hfree graphs and extend the result [43] for \(P_4\)free graphs in a systematic way.
The class of \(P_5\)free graphs is a wellstudied graph class. For instance, Hoàng et al. [24] proved that for every integer k, k Colouring is polynomialtime solvable for \(P_5\)free graphs, whereas Golovach and Heggernes [19] showed that Choosability is fixedparameter tractable for \(P_5\)free graphs when parameterized by the size of the lists of admissible colours. Lokshantov et al. [30] solved a longstanding open problem by giving a polynomialtime algorithm for Independent Set restricted to \(P_5\)free graphs (recently, their result was extended to \(P_6\)free graphs by Grzesik et al. [22]).
Our main result is that Independent Feedback Vertex Set is polynomialtime solvable for \(P_5\)free graphs. This is proved in Sects. 3 and 4: in Sect. 3 we give a polynomialtime algorithm for NearBipartiteness on \(P_5\)free graphs, and in Sect. 4 we show how to extend this algorithm to solve Independent Feedback Vertex Set in polynomial time for \(P_5\)free graphs.
In Sect. 5 we consider the related problem Independent Odd Cycle Transversal. We prove that our results for Independent Feedback Vertex Set also hold for Independent Odd Cycle Transversal.
In Sect. 6, we compare the complexities of Independent Feedback Vertex Set and Independent Odd Cycle Transversal for Hfree graphs with the complexity of 3Colouring and several other related problems, such as Feedback Vertex Set, Vertex Cover, Independent Vertex Cover and Dominating Induced Matching. We also survey some related open problems.
2 Hardness When H Contains a Cycle or Claw
Before stating the results in this section, we first introduce some necessary terminology. The line graph L(G) of a graph \(G=(V,E)\) has the edge set E of G as its vertex set, and two vertices \(e_1\) and \(e_2\) of L(G) are adjacent if and only if \(e_1\) and \(e_2\) share a common endvertex in G. The claw is the graph shown in Fig. 2. It is well known and easy to see that every line graph is clawfree. A graph is (sub)cubic if every vertex has (at most) degree 3.
Theorem 1
NearBipartiteness is NPcomplete for line graphs of planar subcubic bipartite graphs.
Proof
As the problem is readily seen to be in NP, it suffices to prove NPhardness. We reduce from the Hamilton Cycle Through Specified Edge problem. Given a graph G and an edge e of G, this problem asks whether G has a Hamilton cycle through e. Labarre [29] observed that Hamilton Cycle Through Specified Edge is NPcomplete for planar cubic bipartite graphs by noting that it follows easily from the analogous result of Akiyama et al. [2] for Hamilton Cycle. Let (G, e) be an instance of Hamilton Cycle Through Specified Edge. By the aforementioned result, we may assume that G is a planar cubic bipartite graph. Let \(u_1\) and \(u_2\) be the two endvertices of e. Delete the edge \(u_1u_2\) and add two new vertices \(v_1\) and \(v_2\), and two new edges \(e_1=u_1v_1\) and \(e_2=v_2u_2\). Let G be the resulting graph. We note that both \(v_1\) and \(v_2\) have degree 1 in \(G'\), whereas every other vertex of \(G'\) has degree 3. Hence \(G'\) is subcubic. Since G is planar and bipartite, it follows that \(G'\) is planar and bipartite. Moreover, we make the following observation.
Claim 1
The graph G has a Hamilton cycle through e if and only if \(G'\) has a Hamilton path from \(v_1\) to \(v_2\).
Let n be the number of vertices in G. Then the number of vertices in \(G'\) is \(n+2\), meaning that a Hamilton path in \(G'\) has \(n+1\) edges. Moreover, as G is cubic, G has \(\frac{3}{2}n\) edges, so \(G'\) has \(\frac{3}{2}n+1\) edges, implying that \(L(G')\) has \(\frac{3}{2}n+1\) vertices. Furthermore, \(e_1\) and \(e_2\) have degree 2 in \(L(G')\) and all other vertices in \(L(G')\) have degree 4, so \(L(G')\) has maximum degree at most 4.
Claim 2
The graph \(G'\) has a Hamilton path from \(v_1\) to \(v_2\) if and only if \(L(G')\) is nearbipartite.
To prove Claim 2, first suppose that \(G'\) has a Hamilton path P from \(v_1\) to \(v_2\). Then, as every vertex in \(G'\) apart from \(v_1\) and \(v_2\) has degree 3, it follows that \(G'E(P)\) consists of two isolated vertices (\(v_1\) and \(v_2\)) and a set S of isolated edges. Thus S is an independent set in \(L(G')\), and so \(L(G')\) is nearbipartite.
Now suppose that \(L(G')\) is nearbipartite. Then \(L(G')\) contains a set S of vertices, such that \(F=L(G')S\) is a forest. As \(L(G')\) is a line graph, \(L(G')\) is clawfree. This means that F is the disjoint union of one or more paths. Suppose that F contains more than one path. Then, as \(e_1\) and \(e_2\) are the only two vertices in \(L(G')\) that are of degree 2 and all other vertices of \(L(G')\) have degree 4, at least one path of F has an endvertex of degree 4 in \(L(G')\). Let f be this vertex. As f is the endvertex of a path in F, we find that f has three neighbours \(f_1\), \(f_2\), \(f_3\) in S. As S is an independent set, \(\{f,f_1,f_2,f_3\}\) induces a claw in \(L(G')\). This contradiction tells us that F consists of exactly one path P, and by the same reasoning, \(e_1\) and \(e_2\) must be the endvertices of P.
By combining Claims 1 and 2 we have completed our hardness reduction and the theorem is proved.\(\square \)
Theorem 1 has the following immediate consequence (take \(k=n\)).
Corollary 1
Independent Feedback Vertex Set is NPcomplete for line graphs of planar subcubic bipartite graphs.
We will now prove that Independent Feedback Vertex Set is NPcomplete for graphs with no small cycles even if their maximum degree is small. The length of a cycle C is the number of edges of C. The girth g(G) of a graph G is the length of a shortest cycle of G; if G has no cycles then \(g(G)=\infty \). The subdivision of an edge \(e=uv\) in a graph deletes e and adds a new vertex w and edges uw and wv. We first need the following observation, which is well known. For completeness we give a short proof.
Lemma 1
(see e.g. [35]) Let uv be an edge in a graph G. Let \(G'\) be the graph obtained from G after subdividing uv. Then G has a feedback vertex set of size at most k if and only if \(G'\) does.
Proof
Let w denote the new vertex obtained from subdividing uv. Any feedback vertex set S of G is a feedback vertex set of \(G'\). Suppose \(S'\) is a feedback vertex set of \(G'\). If \(w\notin S'\), then \(S'\) is a feedback vertex set of G. Suppose \(w\in S'\). If at least one of u and v are in \(S'\) as well, then \(S'\setminus \{w\}\) is a feedback vertex set of G. If neither u nor v belong to \(S'\), then \((S'\setminus \{w\})\cup \{u\})\) is a feedback vertex set of G with the same size as \(S'\).\(\square \)
Lemma 1 implies that Feedback Vertex Set is NPcomplete for graphs of girth at least g for every constant \(g\ge 3\). We also use this lemma to prove our next result.
Proposition 1
For every constant \(g\ge 3\), Independent Feedback Vertex Set is NPcomplete for graphs of maximum degree at most 4 and girth at least g.
Proof
For a graph G, let \(G_s\) be the graph obtained from G after subdividing every edge of G; we say that \(G_s\) is a subdivided copy of G. Let \(\mathcal{G}_s\) be the graph class obtained from a graph class \(\mathcal{G}\) after replacing each \(G\in \mathcal{G}\) by its subdivided copy \(G_s\). It follows from Lemma 1 that if Feedback Vertex Set is NPcomplete for some graph class \(\mathcal{G}\), then it is also NPcomplete for \(\mathcal{G}_s\). By starting from the fact that Feedback Vertex Set is NPcomplete for line graphs of planar cubic bipartite graphs [37] and applying this observation a sufficient number of times, we find that for any constant \(g\ge 3\), Feedback Vertex Set is NPcomplete for graphs of maximum degree at most 4 and girth at least g. Moreover, any nonindependent feedback vertex set S of a subdivided copy \(G_s\) of a graph G contains two adjacent vertices, one of which has degree 2 in \(G_s\). Hence, we can remove such a degree 2 vertex from S to obtain a smaller feedback vertex set of \(G_s\). Thus all minimum feedback vertex sets of \(G_s\) are independent. This observation, which can also be found in [35], together with NPhardness for Feedback Vertex Set for graphs of maximum degree at most 4 and with arbitrarily large girth proves the proposition.\(\square \)
Recall that every line graph is clawfree. We also observe that for a graph H with a cycle C, the class of graphs of girth at least \(C+1\) is a subclass of the class of Hfree graphs. Hence, we can combine Corollary 1 and Proposition 1 to obtain the following result.
Corollary 2
Let H be a graph that contains a claw or a cycle. Then Independent Feedback Vertex Set is NPcomplete for Hfree graphs of maximum degree at most 4.
3 NearBipartiteness of \(P_5\)Free Graphs
In this section, we show that NearBipartiteness is polynomialtime solvable for \(P_5\)free graphs, that is, we give a polynomialtime algorithm for testing whether or not a \(P_5\)free graph has an independent feedback vertex set. To obtain a minimum feedback vertex set we need to first run this algorithm and then do the additional work described in Sect. 4.
A graph G is nearbipartite if and only if (G, L), with \(L(v)=\{1,2,3\}\) for all \(v \in V(G)\), is a yesinstance of List SemiAcyclic 3Colouring. To show that nearbipartite \(P_5\)free graphs can be recognised in polynomial time, we will prove the stronger fact that List SemiAcyclic 3Colouring is polynomialtime solvable for \(P_5\)free graphs.
A set of vertices in a graph G is dominating if every vertex of G is either in the set or has at least one neighbour in it. We will use a lemma of Bacsó and Tuza.
Lemma 2
[3] Every connected \(P_5\)free graph admits a dominating set that induces either a clique or a \(P_3\).
Lemma 2 implies that every connected 3colourable \(P_5\)free graph has a dominating set of size at most 3 (since it has no clique on more than three vertices). This was used by Randerath et al. [41] to show that 3Colouring is polynomialtime solvable on \(P_5\)free graphs. Their algorithm tries all possible 3colourings of a dominating set of size at most 3. It then adjusts the lists of the other vertices (which were originally set to \(\{1,2,3\}\)) to lists of size at most 2. As shown by Edwards [14], 2List Colouring can be translated to an instance of 2Satisfiability, which is well known and readily seen to be solvable in linear time. Hence this approach results in a polynomial (even constant) number of instances of the 2Satisfiability problem.
Our goal is also to apply Lemma 2 to a connected \(P_5\)free graph G and to reduce an instance (G, L) of List SemiAcyclic 3Colouring to a polynomial number of instances of 2Satisfiability. However, in our case this is less straightforward than in the case of 3Colouring restricted to \(P_5\)free graphs: the restriction of List SemiAcyclic 3Colouring to lists of size 2 turns out to be NPcomplete for general graphs even if every list consists of either colours 1 and 3 or only colour 2.
Theorem 2
List SemiAcyclic 3Colouring is NPcomplete even if \(L(v)\in \{\{1,3\},\{2\}\}\) for every vertex v in the input graph.
Proof
The problem is clearly in NP so we need only show that it is NPhard. We do this by reduction from Satisfiability.

For each clause C of \(\phi \) create a (2C)cycle and assign lists \(\{1,3\}\) and \(\{2\}\) alternately to vertices around the cycle. Let the literals of the clause be represented by distinct vertices with lists \(\{1,3\}\).

For each variable x, choose a clause containing the positive literal x and let \(v_x\) be the vertex representing x in the corresponding cycle. For every other occurrence (if there are any) of the positive literal x, let the corresponding vertex be adjacent to a new middle vertex that is also joined by an edge to \(v_x\). Assign the list \(\{1,3\}\) to the middle vertex. For every occurrence of the negative literal \(\overline{x}\), add an edge so that the corresponding vertex is adjacent to \(v_x\).
First suppose that G has a semiacyclic 3colouring that respects L, and let us show that a satisfying assignment for \(\phi \) can be found. For each variable x, if \(v_x\) is coloured 1, then let x be true; if it is coloured 3, let x be false. Note that every other vertex corresponding to the positive literal x must be coloured the same as x, and every vertex corresponding to an instance of \(\overline{x}\) is coloured differently, so each literal is coloured 1 if and only if it is true. Thus every clause contains a true literal, otherwise in the corresponding cycle every vertex would be coloured 2 or 3 and the colouring would not be semiacyclic.
Now suppose that \(\phi \) has a satisfying assignment. If a literal is true in this assignment, colour the corresponding vertex 1, otherwise colour it 3. Colour each middle vertex with the colour not used on its neighbours (which must be coloured alike). Clearly this is a 3colouring, and each cycle corresponding to a clause contains a vertex coloured 1 as it contains a true literal. No other cycle in the graph is coloured with only 2 and 3, as the only edges that do not belong to the cycles representing the clauses each join a vertex coloured 1 to a vertex coloured 3. Thus the colouring is semiacyclic.\(\square \)
By Theorem 2, to prove that List SemiAcyclic 3Colouring is polynomialtime solvable on \(P_5\)free graphs, we need to refine our analysis and exploit \(P_5\)freeness beyond the use of Lemma 2. We adapt the approach used by Hoàng et al. [24] to show that k Colouring is polynomialtime solvable on \(P_5\)free graphs for all \(k\ge 3\) (extending the analogous result of Randerath et al. [41] for 3Colouring). Let us first outline the proof of [24].
The approach of Hoàng et al. [24] to solve k Colouring for \(P_5\)free graphs for any integer k uses Lemma 2 as a starting point, just as the approach of Randerath et al. [41] does for the \(k=3\) case. Lemma 2 implies that every kcolourable \(P_5\)free graph G has a dominating set D of size at most k (as the clique number of a kcolourable graph is at most k). Let the vertex set of D be \(\{v_1,\ldots ,v_{D}\}\). Then decompose the set of vertices not in D into D “layers” so that the vertices in a layer i are adjacent to \(v_i\) (and possibly to \(v_j\) for \(j>i\)) but not to any \(v_h\) with \(h<i\). Using the \(P_5\)freeness of G to analyse the adjacencies between different layers, it is possible to branch in such a way that a polynomial number of instances of \((k1)\) Colouring are obtained. Hence, by repetition, a polynomial number of instances of 2Colouring are reached, which can each be solved in polynomial time due to the result of [14].
The algorithm of [24] works by considering the more general List k Colouring problem, where each vertex v is assigned a list \(L(v) \subseteq \{1,\ldots ,k\}\) of permitted colours and the question is whether there is a colouring in which each vertex is assigned a colour from its list. The algorithm immediately removes any vertices whose lists have size 1 at any point (and then adjusts the lists of admissible colours of all neighbours of such vertices). We will follow the approach of [24]. In our case \(k=3\), but we cannot remove any vertices whose lists contain a singleton colour if this colour is 2 or 3. To overcome this extra complication we carefully analyse the 4vertex cycles in the graph after observing that these cycles are the only obstacles that may prevent a 3colouring of a \(P_5\)free graph from being semiacyclic.
For a subset \(S\subseteq V(G)\) of the vertex set of a graph G, we let G[S] denote the subgraph of G induced by S.
Theorem 3
List SemiAcyclic 3Colouring is solvable on \(P_5\)free graphs in \(O(n^{16})\) time.
Proof
Consider an input (G, L) for the problem such that G is \(P_5\)free. Since the problem can be solved componentwise, we may assume that G is connected. If G contains a \(K_4\), then it is not 3colourable and the input is a noinstance. As we can test whether or not G contains a \(K_4\) in \(O(n^4)\) time, we now assume that G is \(K_4\)free. We may also assume that G contains at least three vertices, otherwise the problem can be trivially solved.
 Rule 1.

If \(u,v \in V(G)\) are adjacent and \(L(u)=1\), set \(L(v) := L(v) \setminus L(u)\).
 Rule 2.

If \(L(v)=\emptyset \) for some \(v \in V(G)\), return no.
 Rule 3.

If \(G_{\lnot i}\) is not bipartite for some \(i \in \{1,2,3\}\), return no.
 Rule 4.

If \(G_{\lnot 1}\) contains an induced \(C_4\), return no.
 Rule 5.

If G contains an induced \(C_4\), and exactly one vertex v of this cycle has a list containing the colour 1, set \(L(v)=\{1\}\).
We must show that these rules are safe. That is, that when they modify the instance they do not affect whether or not it is a yesinstance or a noinstance, and when they return the answer no, this is correct and no semiacyclic colouring that respects the lists can exist. This is trivial for Rules 1 and 2. We may apply Rule 3 since in any 3colouring of G every pair of colour classes must induce a bipartite graph. We may apply Rules 4 and 5 since in every solution, every induced \(C_4\) must contain at least one vertex coloured with colour 1. In fact, if there is a 3colouring of G with an induced cycle made of vertices coloured only 2 and 3, then this induced cycle must be an even cycle. Since G is \(P_5\)free, such an induced cycle must in fact be isomorphic to \(C_4\). Hence the problem, when restricted to \(P_5\)free graphs, is equivalent to testing whether G has a 3colouring respecting the lists such that every induced \(C_4\) contains at least one vertex coloured with colour 1.
By Lemma 2, G has a dominating set S that either is a clique or induces a \(P_3\). If S is a clique, then it has at most three vertices, as G is \(K_4\)free, so we can find such a set in \(O(n^4)\) time. Thus, adding vertices arbitrarily if necessary, we may assume \(S=\{a_1,a_2,a_3\}\). We consider all possible combinations of colours that can be assigned to the vertices in S, that is, we branch into at most \(3^3\) cases, in which \(a_1\), \(a_2\) and \(a_3\) have each received a colour, or equivalently, have had their list of permissible colours reduced to size exactly 1. In each case we proceed as follows.
Our strategy is to reduce the instance (G, L) to a polynomial number of instances \((G,L')\), in which there are no edges between any two distinct sets \(V_i'\) and \(V_j'\) (defined with respect to \(L'\)). We will do this by branching on possible partial colourings in such a way that afterwards there are no edges between \(V_i''\) and \(V_j'''\), no edges between \(V_i''\) and \(V_j''\) and no edges between \(V_i'''\) and \(V_j'''\) for every pair \(i,j\in \{1,2,3\}\) with \(i\ne j\). As the branching procedure is similar for each of these possible combinations, we pick an arbitrary pair, namely \(V_1''\) and \(V_2''\). As we shall see, we do not remove any edges between \(V_1''\) and \(V_2''\). Instead, we decrease the lists of some of their vertices to size 1, so that these vertices will leave \(V_1'\cup V_2'\) by definition of \(V_1'\) and \(V_2'\) (and therefore leave \(V_1''\) and \(V_2''\) by definition of \(V_1''\) and \(V_2''\)).
We choose an arbitrary colour \(c' \in \{1,2,3\} \setminus \{c_1,c_2\}\). Note that if \(c_1 \ne c_2\) then this choice is unique and otherwise there are two choices (as we will show, it suffices to branch on only one choice). Also note that every vertex in \(V_1''\) and \(V_2''\) has colour \(c'\) in its list.
We now branch over \(k+1\) possibilities, namely the possibilities that vertex \(u_i\) is the first vertex coloured with colour \(c'\) (so vertices \(u_1,\ldots ,u_{i1}\), if they exist, do not get colour \(c'\)) and the remaining possibility that no vertex of \(V_1''\) is coloured with colour \(c'\). To be more precise, for branch \(i=1\) we set \(L(u_1)=\{c'\}\), for each branch \(2 \le i \le k\) we remove colour \(c'\) from each of \(L(u_1),\ldots ,L(u_{i1})\) and set \(L(u_i)=\{c'\}\) and for branch \(i=k+1\) we remove colour \(c'\) from each of \(L(u_1),\ldots ,L(u_{k})\). If \(i=k+1\), all vertices of \(V_1''\) will have a unique colour in their list and thus leave \(V_1'\) and thus \(V_1''\) by definition of \(V_1'\). Hence, \(V_1''\) becomes empty and thus, as required, we no longer have edges between \(V_1''\) and \(V_2''\). Otherwise, if \(i \le k\), then all of \(u_1,\ldots ,u_i\) will have a list containing exactly one colour, so they will leave \(V_1'\) and therefore \(V_1''\). By Rule 1 all neighbours of \(u_i\) in \(V_2''\) will have \(c'\) removed from their lists, so they will leave \(V_2'\) and therefore \(V_2''\). By the ordering of neighbourhoods of vertices in \(V_1''\), this means that no vertex remaining in \(V_1''\) has a neighbour remaining in \(V_2''\), so if \(i\le k\), then it is also the case that we no longer have edges between \(V_1''\) and \(V_2''\).
Note that removing all the edges between distinct sets \(V_i'\) and \(V_j'\) in the above way involves branching into \(O(n^{12})\) cases. We consider each case separately, and for each case we proceed as below.
By the above branching we may assume that there are no edges between any two distinct sets \(V_i'\) and \(V_j'\). We say that an induced \(C_4\) is tricky if there exists a (proper) colouring of it (not necessarily extendable to all of G) using only the colours 2 and 3 such that every vertex receives a colour from its list. We say that a vertex in an induced \(C_4\) is good for this induced \(C_4\) if its list contains the colour 1. By definition of tricky, every good vertex for a tricky induced \(C_4\) must belong to \(V_1' \cup V_2' \cup V_3'\). By Rules 4 and 5, every tricky induced \(C_4\) must contain at least two good vertices. If an induced \(C_4\) contains two good vertices that are adjacent, then they must belong to the same set \(V_i'\) (since there are no edges between any two distinct sets \(V_i'\) and \(V_j'\)), so they must have the same list. This means that in every colouring of this induced \(C_4\) that respects the lists, one of the good vertices in this induced \(C_4\) will be coloured with colour 1, contradicting the definition of tricky. We conclude that every tricky induced \(C_4\) must contain exactly two good vertices, which must be nonadjacent.

\(2 \in L(v_1)\), \(3 \in L(v_2)\), \(2 \in L(v_3)\) and \(3 \in L(v_4)\) or

\(3 \in L(v_1)\), \(2 \in L(v_2)\), \(3 \in L(v_3)\) and \(2 \in L(v_4)\).

\(L(v_1)=\{1,2\}\), \(L(v_2)=\{3\}\), \(L(v_3)=\{1,2\}\) and \(L(v_4)=\{3\}\) or

\(L(v_1)=\{1,3\}\), \(L(v_2)=\{2\}\), \(L(v_3)=\{1,3\}\) and \(L(v_4)=\{2\}\).
We say that an induced \(C_4\) is strongly tricky if its vertices have lists of this form (see also Fig. 6). Note that, by the above arguments, we may assume that all tricky induced \(C_4\)s in the instances we consider are in fact strongly tricky.
Recall that every tricky induced \(C_4\) is strongly tricky, and is therefore entirely contained in either \(G[L_2 \cup L_{1,3}]\) or \(G[L_3 \cup L_{1,2}]\). By Rule 3, \(G_{\lnot 1}\) and therefore \(G[L_{2,3}]\) is bipartite. Hence we can colour the vertices of \(L_{2,3}\) with colours from their lists such that no vertex in \(L_{2,3}\) is adjacent to a vertex of the same colour in G and no induced \(C_4\)s are coloured with colours alternating between 2 and 3 (indeed, recall that induced \(C_4\)s cannot exist in \(G[L_{2,3}]\) by Rule 4). It therefore remains to check whether the vertices of \(G[L_2 \cup L_{1,3}]\) (and \(G[L_3 \cup L_{1,2}]\)) can be coloured with colours from their lists so that no pair of adjacent vertices in \(L_{1,3}\) (resp. \(L_{1,2}\)) receive the same colour and every strongly tricky \(C_4\) has at least one vertex coloured 1. By symmetry, it is sufficient to show how to solve the \(G[L_2 \cup L_{1,3}]\) case. Hence we have reduced the original instance (G, L) to a polynomial number of instances of a new problem, which we define below after first defining the instances.
Definition 1
A graph \(G=(V,E)\) is troublesome if every vertex v in G has list either \(L(v)=\{2\}\) or \(L(v)=\{1,3\}\), such that \(L_2\) is an independent set and \(L_{1,3}\) induces a bipartite graph.
In particular, for each of our created instances the set \(L_2\) is independent due to Rule 1 and \(L_{1,3}\) induces a bipartite graph by Rule 3. Note that by definition of troublesome, all tricky induced \(C_4\)s in a troublesome graph are strongly tricky.
Definition 2
Let G be a troublesome graph. A 3colouring of the graph G is troublefree if each vertex receives a colour from its list, no two adjacent vertices of G are coloured alike and at least one vertex of every strongly tricky induced \(C_4\) of G receives colour 1.
We can encode an instance of TroubleFree Colouring as an instance of 2Satisfiability as follows. For each vertex \(u \in L_{1,3}\), we create two variables \(u_{1}\) and \(u_{3}\). If we assign \(u_1\) or \(u_3\) to be true, this means that u will be assigned colour 1 or 3, respectively. Hence, we need to add the clauses \((u_1\vee u_3)\) and \((\overline{u_1}\vee \overline{u_3})\). To ensure that two adjacent vertices are not coloured alike, for each pair of adjacent vertices \(u,v \in L_{1,3}\) we add the clauses \((\overline{u_1} \vee \overline{v_1})\) and \((\overline{u_3} \vee \overline{v_3})\). For each strongly tricky \(C_4\) with good vertices u and v, we add the clause \((u_1 \vee v_1)\) to ensure that at least one of them will be assigned colour 1. Let \(\mathcal{I}\) be the resulting instance of 2Satisfiability. We claim that G has a troublefree colouring if and only if \(\mathcal{I}\) is satisfiable.
First suppose that G has a troublefree colouring. Then setting \(x_c\) to be true if the vertex x receives colour c gives a satisfying assignment for \(\mathcal{I}\). Now suppose that \(\mathcal{I}\) has a satisfying assignment. Then we colour the vertex \(v \in L_{1,3}\) with colour c if \(v_c\) is set to true in this assignment. For \(v \in L_2\) we colour v with colour 2. No two adjacent vertices of \(L_{1,3}\) are assigned the same colour, because \(\mathcal{I}\) is satisfied. No vertex of \(L_2\) is assigned the same colour as one of its neighbours, since \(L_2\) is an independent set and every vertex of \(L_{1,3}\) is assigned colour 1 or 3. Therefore the obtained colouring is a 3colouring of G. Since \(\mathcal{I}\) is satisfied, every strongly tricky \(C_4\) contains at least one vertex coloured 1. Hence G has a troublefree colouring.
So, by branching, we have reduced the original instance (G, L) of List SemiAcyclic 3Colouring to a polynomial number of instances of 2Satisfiability. If we find that one of the instances of the latter problem is a yesinstance, then we obtain a corresponding yesinstance of TroubleFree Colouring. We therefore solve TroubleFree Colouring on \(G[L_2 \cup L_{1,3}]\) and (after swapping colours 2 and 3) on \(G[L_3 \cup L_{1,2}]\). If one of these two instances of TroubleFree Colouring is a noinstance, then we return no for this branch and try the next one. If both of these are yesinstances, then we return yes and obtain a semiacyclic 3colouring by combining the colourings on \(G[L_1 \cup L_{2,3}]\), \(G[L_2 \cup L_{1,3}]\) and (after swapping colours 2 and 3 back) \(G[L_3 \cup L_{1,2}]\). If every branch returns no then the original graph has no semiacyclic 3colouring. This completes the proof of the correctness of the algorithm and it remains to analyse its runtime.
Let n be the number of vertices in G. Recall that we can check if G is \(K_4\)free in \(O(n^4)\) time and if it is then we can find a dominating set of size at most 3 in \(O(n^4)\) time. Rule 1 can be applied in \(O(n^2)\) time. Rule 2 can be applied in O(n) time. Rule 3 can be applied in \(O(n^2)\) time. Rules 4 and 5 can be applied in \(O(n^4)\) time. We first branch up to \(3^3\) times and then subbranch \(O(n^{12})\) times and in each case we apply the rules. We find all (strongly) tricky \(C_4\)s and the two corresponding good vertices in \(O(n^4)\) time. It is readily seen that every created instance of 2Satisfiability is solvable in \(O(n^2)\) time (see also Edwards [14]). This leads to a total runtime of \(O(n^4)+O(1)\times O(n^{12}) \times (O(n^4)+O(n^4) + O(n^2)) =O(n^{16})\).\(\square \)
As mentioned, Theorem 3 has the following consequence.
Corollary 3
NearBipartiteness can be solved in \(O(n^{16})\) time for \(P_5\)free graphs.
Proof
Let G be a graph. Set \(L(v)=\{1,2,3\}\) for all \(v \in V(G)\). Then G is nearbipartite if and only if (G, L) is a yesinstance of List Semiacyclic 3Colouring. In particular, the vertices coloured 1 by a semiacyclic colouring of G form an independent feedback vertex set of G. The corollary follows by Theorem 3.\(\square \)
4 Independent Feedback Vertex Sets of \(P_5\)Free Graphs
In this section we prove that Independent Feedback Vertex Set is polynomialtime solvable for \(P_5\)free graphs by extending the algorithm from Sect. 3: the first part of our proof uses the proof of Theorem 3, as we will explain in the proof of Lemma 3. As such, we heavily use Definitions 1 and 2. Let \(G=(V,E)\) be a troublesome \(P_5\)free graph. For a troublefree colouring c of G, let \(t_c(G)=\{u\in V\; \; c(u)=1\}\) denote the number of vertices of G coloured 1 by c. Let t(G) be the minimum value \(t_c(G)\) over all troublefree colourings c of G, and set \(t_c(G)=\infty \) if no such colouring exists.
Lemma 3
Let G be a nearbipartite \(P_5\)free graph. In \(O(n^{16})\) time it is possible to reduce the problem of finding the smallest independent feedback vertex set of G to finding the value \(t(G')\) of \(O(n^{12})\) instances of TroubleFree Colouring, all on induced subgraphs of G.
Proof
Let G be a nearbipartite \(P_5\)free graph, that is, we assume that G has an independent feedback vertex set. We may assume that G is connected, otherwise, we solve the problem componentwise. We set \(L(v)=\{1,2,3\}\) for all \(v \in V(G)\) and run the algorithm of Theorem 3. As can be seen from the proof of Theorem 3, this algorithm branches up to \(3^3\) times and then subbranches \(O(n^{12})\) times. Each branch gives us, after some preprocessing in \(O(n^4)\) time, either a no answer, in which case we discard this branch, or two vertexdisjoint instances of TroubleFree Colouring (one on \(G[L_2 \cup L_{1,3}]\) and one (after swapping colours 2 and 3) on \(G[L_3 \cup L_{1,2}]\)) and we will denote these two instances by \(G'\) and \(G''\), respectively. Such instances consist of a troublesome graph \(G'\) or \(G''\), which is an induced subgraph of G and whose vertices have lists of admissible colours determined by the branching.
As explained in the proof of Theorem 3, in any branch that we did not discard, \(G[L_1 \cup L_{2,3}]\) will have a semiacyclic 3colouring that respects the lists and \(L_1\) will be the set of vertices that are coloured 1 in any such colouring. Therefore, given troublefree colourings c and \(c'\) of \(G'\) and \(G''\), respectively, we can obtain an independent feedback vertex set \(S(c,c',G',G'')\) by taking the union of the set \(L_1\) of \(G[L_1 \cup L_{2,3}]\) and the sets of vertices in \(G'\) and \(G''\) that c or \(c'\) colour with colour 1.
Now let \(c^*\) and \(c^{**}\) be such that \(t(G')=t_{c^*}(G')\) and \(t(G'')=t_{c^{**}}(G'')\). If we know \(t(G')\) and \(t(G'')\), we can compute the size \(s(G',G'')=S(c^*,c^{**},G',G'')\) in O(1) time (and, if we know \(c^*\) and \(c^{**}\), we can compute the corresponding independent feedback vertex set \(S(c^*,c^{**},G',G'')\) in O(n) time). Let \(\hat{s}\) be the minimum \(s(G',G'')\) over all branches of our procedure. As our procedure had \(O(n^{12})\) branches, given the values of \(t(G')\) and \(t(G'')\) for every branch, we can compute \(\hat{s}\) in \(O(n^{12})\) time. As we branched in every possible way, \(\hat{s}\) is the size of a minimum independent feedback vertex set of G.\(\square \)
We still need a polynomialtime algorithm that computes t(G) for a given troublesome \(P_5\)free graph. We present such an algorithm in the following lemma (in the proof of this lemma we again use Definitions 1 and 2).
Lemma 4
Let G be a troublesome \(P_5\)free graph on n vertices. Determining t(G) can be done in \(O(n^3)\) time.
Proof
Let \(G=(V,E)\) be a troublesome \(P_5\)free graph. Note that in G, an induced \(C_4\) on vertices \(v_1\), \(v_2\), \(v_3\), \(v_4\), in that order, is strongly tricky if \(v_1,v_3 \in L_{1,3}\) and \(v_2,v_4 \in L_2\).
We construct an auxiliary graph H as follows. We let \(V(H)=L_{1,3}\). Every edge of \(G[L_{1,3}]\) belongs to H. We say that such edges are red. For nonadjacent vertices \(v_1,v_3 \in L_{1,3}\), if there is a strongly tricky induced \(C_4\) on vertices \(v_1\), \(v_2\), \(v_3\), \(v_4\) with \(v_2,v_4 \in L_2\), we add the edge \(v_1v_3\) to H. We say that such edges are blue. Note that H is a supergraph of \(G[L_{1,3}]\) and that there exists at most one edge, which is either blue or red, between any two vertices of H.
 (i)
no red edge is monochromatic, that is, the two endvertices of every red edge must be coloured, respectively, \( 1 \& 3\) or \( 3 \& 1\);
 (ii)
the two endvertices of every blue edge must be coloured, respectively, \( 1 \& 3\), \( 3 \& 1\) or \( 1 \& 1\) (the only forbidden combination is \( 3 \& 3\), as in this case we obtain a strongly tricky induced \(C_4\) in G with colours 2 and 3).
 Rule 1.

If there is a blue edge in H between two vertices \(u,v\in X_i\) or two vertices \(u,v\in Y_i\), then assign colour 1 to u and v.
 Rule 2.

If there is a blue edge e in H between a vertex \(u\in X_i\) and a vertex \(v\in Y_i\), then delete e from H.
 Rule 3.

If there are blue edges uv and \(u'v'\) such that either \(u,u'\in X_i\) or \(u,u'\in Y_i\) (where \(u=u'\) is possible), while \(v\in X_j\) and \(v'\in Y_j\) for some \(j\ne i\), then assign colour 1 to u and \(u'\).
 Rule 4.

If an uncoloured vertex u is adjacent to a vertex with colour 3 via a blue edge, then assign colour 1 to u.
 Rule 5.

If an uncoloured vertex u is adjacent to a coloured vertex v via a red edge, then assign colour 1 to u if v has colour 3 and assign colour 3 to u otherwise.
 Rule 6.

If there is a red edge with endvertices both coloured 1 or both coloured 3, or a blue edge with endvertices both coloured 3, then return no.
 Rule 7.

Remove all vertices that have received colour 1 or colour 3, keeping track of the number of vertices coloured 1.
Since each \(R_i\) is connected and bipartite, in every feasible colouring of H, for all i either all vertices in the set \(X_i\) must be coloured 1 and all vertices in the set \(Y_i\) must be coloured 3, or vice versa. Therefore we may safely apply Rules 1 and 2. Suppose that there exist vertices u and \(u'\) (we allow the case where \(u=u'\)) in some \(X_i\) or in some \(Y_i\), such that u is incident with a blue edge uv and \(u'\) is incident with a blue edge \(u'v'\) for two vertices v and \(v'\) that belong to different partition classes of the same red component \(R_j\) for some \(j\ne i\). Then, as either v or \(v'\) must get colour 3 in every feasible colouring of H and \(u,u'\) must be coloured alike, we find that u and \(u'\) must receive colour 1. Hence Rule 3 is also safe to apply. Rules 4–6 are also safe; this follows immediately from the definition of a feasible colouring. If a vertex v is assigned colour 3, then by Rule 4 all its neighbours along blue edges get colour 1, so Property (ii) of a feasible colouring is satisfied for all blue edges with endvertex v. If a vertex v is assigned a colour, then by Rule 5 all its neighbours along red edges get a different colour, so Property (i) of a feasible colouring is satisfied. We conclude that Rule 7 is safe.
By Rules 1 and 2, if two vertices are in the same red component \(R_i\), we may assume that they are not connected by a blue edge. Hence, we may assume from now on that red components contain no blue edges in H. By Rule 3, we may also assume that if \(i \ne j\) then either no vertex in \(X_j\) is joined by a blue edge to a vertex of \(X_i\) or no vertex in \(Y_j\) is joined by a blue edge to a vertex of \(X_i\) (and similarly with \(X_i\) replaced by \(Y_i\)).
From H we construct another auxiliary graph \(H^*\) as follows. First, we replace each red component \(R_i\) on more than two vertices by an edge \(x_iy_i\), which we say is a red edge. Hence, the set of red components of H is reduced to a set of red components in \(H^*\) in such a way that each red component of \(H^*\) is either an edge or a single vertex. Next, for \(i\ne j\) we add an edge, which we say is a blue edge, between two vertices \(x_i\) and \(x_j\) if and only if there is a blue edge between a vertex in \(X_i\) and a vertex in \(X_j\). Similarly, for \(i\ne j\) we add a blue edge, between two vertices \(y_i\) and \(x_j\) (resp. \(y_j\)) if and only if there is a blue edge between a vertex in \(Y_i\) and a vertex in \(X_j\) (resp. \(Y_j\)).
Recall that, by Rules 1 and 2, no two vertices in the same component \(R_i\) are connected by a blue edge in H. So every feasible colouring of H corresponds to a feasible colouring of \(H^*\) and vice versa. To keep track of the number of vertices coloured 1, we introduce a weight function \(w: V(H^*)\rightarrow {\mathbb Z}_+\) by setting \(w(x_i)=X_i\) and \(w(y_i)=Y_i\). Our new goal is to find a feasible colouring c of \(H^*\) that minimises the sum of the weights of the vertices coloured 1, which we denote by w(c).
Since in the graph H, for \(i \ne j\) either no vertex in \(X_j\) is joined by a blue edge to a vertex of \(X_i\) or no vertex in \(Y_j\) is joined by a blue edge to a vertex of \(X_i\) (and similarly with \(X_i\) replaced by \(Y_i\)), we find that \(H^*\) contains no triangle consisting of one red edge and two blue edges. As red edges induce a disjoint union of isolated edges, this means that the only triangles in \(H^*\) consist of only blue edges. Let \(B_1,\ldots , B_q\) be the components of the graph obtained from \(H^*\) after removing all red edges. We say that these are blue components (this includes the case where they are singletons).
We will now show that all blue components of \(H^*\) are complete.
Claim 1
Each \(B_i\) is a complete graph.
We prove Claim 1 as follows. For contradiction, suppose there is a blue component \(B_i\) that is not a complete graph. Then \(B_i\) contains three vertices u, v, w such that uv and vw are blue edges and uw is not a blue edge. As uv and vw are blue edges, v is not in the same red component of \(H^*\) as u or w. As no triangle in \(H^*\) can have two blue edges and one red edge, u and w are not adjacent in \(H^*\), meaning that u, v, w in fact belong to three different red components in H. Let \(u'v'\) and \(v''w'\) be blue edges of H corresponding to the edges uv and vw, respectively. As uw is not a blue edge in \(H^*\), we find that \(u'w'\) is not a blue edge in \(H'\). We distinguish between two cases and show that neither of them is possible.
Case 1
\(v'=v''\).
As \(u'v'\) is a blue edge in H, we find that in G, the vertices \(u'\) and \(v'\) must have at least two common neighbours in \(L_2\). For the same reason, in G, the vertices \(v'\) and \(w'\) must have at least two common neighbours in \(L_2\). Since \(u'w'\) is not a blue edge in H, we find that in G, the vertices \(u'\) and \(w'\) have at most one common neighbour in \(L_2\). Therefore G contains two vertices \(p,q \in L_2\) such that p is adjacent to \(u'\) and \(v'\) but nonadjacent to \(w'\) and q is adjacent to \(v'\) and \(w'\) but nonadjacent to \(u'\). As \(L_2\) is an independent set in G, p is nonadjacent to q. Now \(G[\{u', p, v', q, w'\}]\) is a \(P_5\), which is a contradiction.
Case 2
\(v'\ne v''\).
Let \(R_i\) be the red component of H containing \(v'\) and \(v''\). Then, due to the way the red edges of \(H^*\) are constructed, either \(v'\) and \(v''\) both belong to \(X_i\), or they both belong to \(Y_i\). As \(R_i\) is bipartite, connected and \(P_5\)free, \(R_i\) must contain a vertex s that is adjacent to both \(v'\) and \(v''\). Just as in Case 1, in G the vertices \(u'\) and \(v'\) have at least two common neighbours \(p,p' \in L_2\), and \(v'\) and \(w'\) also have at least two common neighbours \(q,q' \in L_2\). As \(L_2\) is independent in G, it follows that \(\{p,p',q,q'\}\) is also an independent set (which may have size smaller than 4).
 Rule 8.

If there exist (red) edges \(u_1v_1\) and \(u_2v_2\) for \(u_1,u_2\in B_i\) and \(v_1,v_2\in B_j\) \((i \ne j)\), then assign colour 1 to every vertex in \((B_i\cup B_j)\setminus \{u_1,u_2,v_1,v_2\}\).
Since Rules 4 and 5 can be safely applied on H, they can be safely applied on \(H^*\). It follows that Rules 6 and 7 can also be safely applied on \(H^*\). We may also safely apply Rule 8: the red edges \(u_1v_1\) and \(u_2v_2\) force \(u_i\) and \(v_i\) to have different colours for \(i\in \{1,2\}\), whereas the blue components forbid \(u_1,u_2\) both being coloured 3 and \(v_1,v_2\) both being coloured 3. Hence, exactly one of \(u_1,u_2\) and exactly one of \(v_1,v_2\) must be coloured 3. Because at most one vertex in any blue component may be coloured 3, this implies that all vertices in \((B_i\cup B_j)\setminus \{u_1,u_2,v_1,v_2\}\) must be coloured 1.
As every vertex is incident with at most one red edge in \(H^*\), we obtain a resulting graph that is an induced subgraph of \(H^*\) with the following property: if there exist (red) edges \(u_1v_1\) and \(u_2v_2\) for \(u_1,u_2\in B_i\) and \(v_1,v_2\in B_j\), then \(\{u_1,u_2,v_1,v_2\}\) induces a connected component of \(H^*\). We can colour such a 4vertex component in exactly two ways and we remember the colouring with minimum weight (either \(w(u_1)+w(v_2)\) or \(w(u_2)+w(v_1)\) depending on whether \(u_1\) gets colour 1 or 3, respectively). Hence, from now on we may assume that the resulting graph, which we again denote by \(H^*\), does not have such components. That is, there is at most one red edge between any two blue components of \(H^*\). As we can colour \(H^*\) componentwise, we may assume without loss of generality that \(H^*\) is connected.
For each \(B_i\) we define the subset \(B_i'\) to consist of those vertices of \(B_i\) not incident with a red edge, and we let \(B_i''=B_i\setminus B_i'\). We note the following. If we colour every vertex of some \(B_i''\) with colour 1, then every neighbour of every vertex of \(B_i''\) in any other blue component \(B_j\) must be coloured 3 by Rule 5 (recall that vertices in different blue components are connected to each other only via red edges). As soon as one vertex u in some blue component \(B_j\) has colour 3, all other vertices in \(B_ju\) must get colour 1 by Rule 4. In this way we can use Rules 4 and 5 exhaustively to propagate the colouring to other vertices of \(H^*\) where we have no choice over what colour to use.
Recall that no vertex of \(H^*\) is incident with more than one red edge. This is a crucial fact: it implies that propagation to other blue components of \(H^*\) happens only via red edges vw between two blue components, one endvertex of which, say v, is first coloured 1, which implies that the other endvertex w of such an edge must get colour 3; this in turn implies that all other vertices in the blue component containing w must get colour 1 and so on. Hence, as \(H^*\) was assumed to be connected, colouring every vertex of a set \(B_i''\) with colour 1 propagates to all vertices of \(H^*\) except for the vertices of \(B_i'\). Note that we may still colour (at most) one vertex of \(B_i'\) with colour 3.
Due to the above, we now do as follows for each \(i\in \{1,\ldots ,q\}\) in turn: We colour every vertex of \(B_i''\) with colour 1 and propagate to all vertices of \(H^*\) except for the vertices of \(B_i'\). If we obtain a monochromatic red edge or a blue edge whose endvertices are coloured 3, we discard this option (by Rule 6). Otherwise, we assign colour 3 to a vertex \(u\in B_i'\) with maximum weight w(u) over all vertices in \(B_i'\) (if \(B_i'\ne \emptyset \)). We store the resulting colouring \(c_i\) that corresponds to this option.
 (i)
Colour all vertices of every \(B_i'\) with colour 1 (doing this does not cause any propagation).
 (ii)
If some \(B_i''\) consists of a single vertex, then colour this vertex with colour 3, and afterwards propagate by using Rule 5 exhaustively.
 (iii)
Remove coloured vertices using Rule 7.
By our procedure, every vertex of every blue component \(B_i\) is incident with a red edge, so the total number of outgoing red edges for each \(B_i\) is equal to \(B_i\ge 2\), and all outgoing red edges go to \(B_i\) different blue components. Hence the graph \(H'\) obtained from \(H^*\) by contracting each blue component to a single vertex has minimum degree at least 2. As \(H'\) has minimum degree at least 2, we find that \(H'\) contains an edge that is not a bridge (a bridge in a connected graph is an edge whose removal disconnects the graph). Let uv be the corresponding red edge in \(H^*\), say u belongs to \(B_i\) and v belongs to \(B_j\).
We have two options to colour u and v, namely by 1, 3 or 3, 1. We try them both. Suppose we first give colour 1 to u. Then we propagate in the same way as before. Because uv is not a bridge in \(H'\), eventually we propagate back to \(B_i\) by giving colour 3 to an uncoloured vertex of \(B_i\). When that happens we have “identified” the colour3 vertex of \(B_i\) and then need to colour all other vertices of \(B_i\) with colour 1. This means that we can in fact propagate to all blue components of \(H^*\), just as before. If at some point we obtain a monochromatic red edge or a blue edge with endvertices coloured 3, then we discard this option (by Rule 6). Next, we give colour 1 to v and proceed similarly.
At the end we have at most \(q+2\) different feasible colourings of \(H^*\). We pick the one with minimum weight and translate the colouring to a feasible colouring of H. Finally, we translate the feasible colouring of H to a troublefree colouring of the original graph G.
It remains to analyse the runtime. Let n be the number of vertices in G. Given two nonadjacent vertices in \(L_{1,3}\), we can test whether they have have two common neighbours in \(L_2\) in O(n) time. Therefore we can construct H in \(O(n^3)\) time.
Applying Rules 1 and 2 takes \(O(n^2)\) time. Applying Rule 3 takes \(O(n^3)\) time. Rules 1–3 only need to be applied exhaustively once, just after H is first constructed. Rules 4 and 5 can be applied exhaustively in \(O(n^3)\) time. Rule 6 can be applied in \(O(n^2)\) time. Rule 7 can be applied in O(n) time.
Constructing \(H^*\) takes \(O(n^2)\) time. By Claim 1, in \(H^*\) every blue component is a clique, so Rule 4 can be applied exhaustively on \(H^*\) in \(O(n^2)\) time. By construction, every red component of \(H^*\) contains at most one edge, so applying Rules 5 and 8 on \(H^*\) can be done in \(O(n^2)\) time. Therefore, Rules 4–8 can be applied to \(H^*\) in \(O(n^2)\) time. It follows that each option of colouring the vertices of some \(B_i''\) with colour 1 and then doing the propagation and colouring the vertices of \(B_i'\) takes \(O(n^2)\) time. Since there are \(q\le n\) blue components, the total time for this is \(O(n^3)\). Then afterwards we consider the situation where each blue component of \(H^*\) has exactly one vertex coloured 3.
We construct \(H'\) in \(O(n^2)\) time and also identify a nonbridge of \(H'\) in \(O(n^2)\) time. Colouring the corresponding red edges in both ways and doing the propagation takes \(O(n^2)\) time again. Then, if there is at least one possibility for which we did not return a noanswer, then we have obtained O(n) different feasible colourings of \(H^*\). Finding the colouring with minimum weight and translating this colouring into a feasible colouring of H and then into a troublefree colouring of the original graph G also takes \(O(n^2)\) time.\(\square \)
We are now ready to state and prove the main result of our paper.
Theorem 4
The size of a minimum independent feedback vertex set of a \(P_5\)free graph on n vertices can be computed in \(O(n^{16})\) time.
Proof
Let G be a \(P_5\)free graph on n vertices. As we can check in \(O(n^{16})\) time whether or not G is nearbipartite, we may assume without loss of generality that G is nearbipartite. Then, by Lemma 3, in \(O(n^{16})\) time we can reduce the problem finding the value \(t(G')\) of \(O(n^{12})\) instances of TroubleFree Colouring, all on induced subgraphs of G (which have at most n vertices). By Lemma 4 we can compute \(t(G')\) in \(O(n^3)\) time for each such instance. This gives a total runtime of \(O(n^{16})\). The result follows.\(\square \)
Remark 1
From our proof, we can find in polynomial time not just the size of a minimum independent feedback vertex set, but also the set itself. The corresponding algorithm can also be adapted to find in polynomial time a maximum independent feedback vertex of a \(P_5\)free graph, or an independent feedback vertex set of arbitrary fixed size (if one exists).
5 Independent Odd Cycle Transversal
Recall that an (independent) set \(S\subseteq V\) of a graph G is an (independent) odd cycle transversal if \(GS\) is bipartite. We also recall that a graph G has an independent odd cycle transversal if and only if G is 3colourable. This means that if 3Colouring is NPcomplete for a graph class \(\mathcal{G}\), then so is Independent Odd Cycle Transversal. Hence, as 3Colouring is NPcomplete for graphs of girth at least g for any constant \(g\ge 3\) [15] (see also [28, 32]) and for line graphs [25], we find the following result.
Proposition 2

graphs of girth at least g for any constant \(g\ge 3\);

for line graphs.
As shown by Chiarelli et al. [13], Odd Cycle Transversal is also NPcomplete for graphs of girth at least g for any constant \(g\ge 3\) and for line graphs. Hence, both problems are NPcomplete for Hfree graphs if H contains a cycle or a claw.
Our algorithm for Independent Feedback Vertex Set restricted to \(P_5\)free graphs can be adapted in the following way to solve Independent Odd Cycle Transversal for \(P_5\)free graphs. We follow the proof of Theorem 3 but remove Rules 4 and 5 used in that proof. For each branch, we still obtain the situation displayed in Fig. 7, and it remains to colour the vertices in \(L_{1,3}\) and \(L_{1,2}\) greedily and componentwise, such that the number of vertices with colour 1 is minimized.
Theorem 5
The size of a minimum independent odd cycle transversal of a \(P_5\)free graph on n vertices can be computed in \(O(n^{16})\) time.
6 Conclusions
Our main result is that Independent Feedback Vertex Set is polynomialtime solvable for \(P_5\)free graphs. As explained in Sect. 5, our algorithm can be readily adapted to also solve Independent Odd Cycle Transversal for \(P_5\)free graphs in polynomial time. We also proved that Independent Feedback Vertex Set is NPcomplete for Hfree graphs if H contains a cycle or a claw. As discussed, the same hardness results were known for Feedback Vertex Set and 3Colouring, and the hardness results for 3Colouring immediately transfer across to Independent Odd Cycle Transversal.
The complexity of 3Colouring and eight related problems for \(P_r\)free graphs, where \(r\ge 1\) is a fixed integer (the columns \(r\le 4\) and \(r\ge 8\) represent multiple cases)
\(r\le 4\)  \(r=5\)  \(r=6\)  \(r=7\)  \(r=8\)  \(r\ge 9\)  

3Colouring  P  P [41]  P [39]  P [7]  ?  ? 
Vertex Cover  P  P [30]  P [22]  ?  ?  ? 
Independent Vertex Cover  P  P  P  P  P  P 
Feedback Vertex Set  P [9]  ?  ?  ?  ?  ? 
Independent Feedback Vertex Set  P [43]  P  ?  ?  ?  ? 
NearBipartiteness  P [8]  P  ?  ?  ?  ? 
Odd Cycle Transversal  P [9]  ?  ?  ?  ?  ? 
Independent Odd Cycle Transversal  P  P  ?  ?  ?  ? 
Dominating Induced Matching  P  P  P  P [10]  P [11]  ? 
Apart from Independent Vertex Cover, the complexities of the other problems that we discussed are not settled for Hfree graphs when H is a linear forest (disjoint union of one or more paths), or even when H is a path. Randerath and Schiermeyer [39] proved that 3Colouring is polynomialtime solvable for \(P_r\)free graphs for \(r=6\), and more recently, Bonomo et al. [7] proved this for \(r=7\). The complexity of 3Colouring for \(P_r\)free graphs is not known for \(r\ge 8\) (we refer to [20] for further details on k Colouring for \(P_r\)free graphs).
The problems Feedback Vertex Set and Odd Cycle Transversal are polynomialtime solvable for the class of permutation graphs [9], which contains the class of \(P_4\)free graphs [9], but their complexity is not known for \(P_r\)free graphs when \(r\ge 5\). This is in contrast to Independent Feedback Vertex Set and Independent Odd Cycle Transversal due to our result on \(P_5\)free graphs. For these two problems we do not know their complexity for \(r\ge 6\). As mentioned, Lokshantov et al. [30] proved that Vertex Cover (or equivalently, Independent Set) is polynomialtime solvable for \(P_5\)free graphs, and this result was recently extended to \(P_6\)free graphs by Grzesik et al. [22]. The computational complexity of Vertex Cover for \(P_r\)free graphs is not known for \(r\ge 7\).
We refer to Table 1 for a summary of the above problems. In this table we also added the Dominating Induced Matching problem, which is also known as the Efficient Edge Domination problem. This problem is that of deciding whether or not a graph G has an independent set S such that \(GS\) is an induced matching, that is, the disjoint union of a set of isolated edges. Cardoso et al. [12] proved that \(GS\) is in fact a maximum induced matching. We note that every graph G whose vertex set allows a partition into an independent set and an induced matching is 3colourable. Grinstead et al. [21] proved that Dominating Induced Matching is NPcomplete. Later, the problem was shown to be NPcomplete or polynomialtime solvable for various graph classes. In particular, Brandstädt and Mosca [10] proved that Dominating Induced Matching for \(P_r\)free graphs is polynomialtime solvable if \(r=7\). Later they extended their result to \(r=8\) [11]. The complexity status of Dominating Induced Matching is unknown for \(r\ge 9\). Hertz et al. [23] conjectured that the problem is polynomialtime solvable for Hfree graphs whenever H is a forest, each connected component is a subdivided claw, a path or an isolated vertex.
Completing Table 1 is a highly nontrivial task. In particular, we note that no NPhardness results are known for any of the problems in Table 1 when restricted to \(P_r\)free graphs. As such, it would be interesting to know whether the problem of determining whether or not a \(P_r\)free graph has an independent feedback vertex set (or equivalently, whether or not a \(P_r\)free graph is nearbipartite) is polynomially equivalent to the 3Colouring problem restricted to \(P_{f(r)}\)free graphs for some function f.
To solve Independent Feedback Vertex on \(P_r\)free graphs for \(r\in \{6,7,8\}\), one could try to exploit the techniques used to solve 3Colouring for \(P_r\)free graphs, just as we did for the \(r=5\) case in this paper. However, this seems difficult due to additional complications and a different approach may be required.
Finally, we point out that the connected problem variants Connected Feedback Vertex Set, Connected Odd Cycle Transversal, and Connected Vertex Cover, which each require the desired set S of size at most k to induce a connected graph, are also known to be NPcomplete for line graphs and graphs of arbitrarily large girth. This was shown by Chiarelli et al. [13] for Connected Feedback Vertex Set and Connected Odd Cycle Transversal, whereas Munaro [36] proved that Connected Vertex Cover is NPhard for line graphs (of planar cubic bipartite graphs) and for graphs of arbitrarily large girth. Moreover, for these three problems the complexity has not yet been settled for Hfree graphs when H is a linear forest (see [13] for some partial results in this direction).
Notes
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