Close relatives of Feedback Vertex Set without single-exponential algorithms parameterized by treewidth
Benjamin Bergougnoux, Édouard Bonnet, Nick Brettell, O-joung Kwon
CClose relatives of Feedback Vertex Set withoutsingle-exponential algorithms parameterized by treewidth
Benjamin Bergougnoux
Department of Informatics, University of Bergen, Bergen, [email protected]
Édouard Bonnet
Univ Lyon, CNRS, ENS de Lyon, Université Claude Bernard Lyon 1, LIP UMR5668, [email protected]
Nick Brettell
School of Mathematics and Statistics, Victoria University of Wellington, New [email protected]
O-joung Kwon
Department of Mathematics, Incheon National University, Incheon, South KoreaDiscrete Mathematics Group, Institute for Basic Science (IBS), Daejeon, South [email protected]
Abstract
The Cut & Count technique and the rank-based approach have lead to single-exponential
FPT algorithms parameterizedby treewidth, that is, running in time 2 O ( tw ) n O (1) , for Feedback Vertex Set and connected versions of the classicalgraph problems (such as
Vertex Cover and
Dominating Set ). We show that
Subset Feedback Vertex Set , SubsetOdd Cycle Transversal , Restricted Edge-Subset Feedback Edge Set , Node Multiway Cut , and
MultiwayCut are unlikely to have such running times. More precisely, we match algorithms running in time 2 O ( tw log tw ) n O (1) withtight lower bounds under the Exponential-Time Hypothesis (ETH), ruling out 2 o ( tw log tw ) n O (1) , where n is the number ofvertices and tw is the treewidth of the input graph. Our algorithms extend to the weighted case, while our lower boundsalso hold for the larger parameter pathwidth and do not require weights. We also show that, in contrast to Odd CycleTransversal , there is no 2 o ( tw log tw ) n O (1) -time algorithm for Even Cycle Transversal under the ETH.
Theory of computation → Graph algorithms analysis; Theory of computation → Fixedparameter tractability
Keywords and phrases
Subset Feedback Vertex Set, Multiway Cut, Parameterized Algorithms, Treewidth, GraphModification, Vertex Deletion Problems
Acknowledgements
This work was initiated while the authors attended the “2019 IBS Summer research program onAlgorithms and Complexity in Discrete Structures”, hosted by the IBS discrete mathematics group. The third authorreceived support from the Leverhulme Trust (RPG-2016-258). The fourth author is supported by the National ResearchFoundation of Korea (NRF) grant funded by the Ministry of Education (No. NRF-2018R1D1A1B07050294) and supportedby the Institute for Basic Science (IBS-R029-C1).
Many NP -hard graph problems admit polynomial-time algorithms on graphs with bounded treewidth , a measureof how well a graph accommodates a decomposition into a tree-like structure. In fact, Courcelle’s Theorem [9]states that any problem definable in MSO logic can be solved in linear time on graphs of bounded treewidth.To obtain a more fine-grained perspective on the dependence on treewidth for certain problems, it is usefulto study the parameterized complexity with respect to treewidth. In particular, we can ask: what is the“smallest” function f for which we can obtain an algorithm that, given a graph with treewidth tw , has runningtime f ( tw ) n O (1) ? For Feedback Vertex Set , standard dynamic programming techniques can be used toobtain an algorithm running in 2 O ( tw log tw ) n O (1) time, and for a while many believed this to be essentiallybest possible. However, this changed in 2011 when Cygan et al. [12] developed the Cut&Count technique,by which they obtained a single-exponential tw n O (1) -time randomized algorithm. Following this, Bodlaenderet al. [3] showed there is a deterministic 2 O ( tw ) n O (1) -time algorithm, using a rank-based approach and theconcept of representative sets. The same year, Pilipczuk [26] exhibited a logic fragment whose model checkingadmits a single-exponential algorithm parameterized by the treewidth of the input graph, thereby providing ascaled-down but more fine-grained version of Courcelle’s theorem. Moreover, also in 2011, Lokshtanov et al. [22] a r X i v : . [ c s . D S ] J u l Close relatives of FVS without single-exponential algorithms in treewidth developed a framework yielding 2 Ω( tw log tw ) n O (1) -time lower bounds under the Exponential Time Hypothesis(ETH). Recall that the ETH asserts that there is a real number δ > -SAT cannot be solved intime 2 δn on n -variable formulas [18]. Lokshtanov et al.’s paper prompted several authors to investigate the exacttime-dependency on treewidth for a variety of graph modification problems.For a vertex-deletion problem , the task is to delete at most k vertices so that the resulting graph is in sometarget class. Feedback Vertex Set can be viewed as a vertex-deletion problem where the graphs in thetarget class consist of blocks with at most two vertices (a block is a maximal subgraph H such that H has nocut vertices). Bonnet et al. [6] considered the class of problems, generalizing Feedback Vertex Set , wherethe target graphs are those consisting of blocks each of which has a bounded number of vertices, and is in somefixed hereditary, polynomial-time recognizable class P . They showed that such a problem is solvable in time2 O ( tw ) n O (1) precisely when each graph in P is chordal (when P does not satisfy this condition, an algorithmwith running time 2 o ( tw log tw ) n O (1) would refute the ETH). Baste et al. [2] studied another generalization of Feedback Vertex Set : the vertex-deletion problem where the target graphs are those having no minorisomorphic to a fixed graph H . They showed a single-exponential parameterized algorithm in treewidth ispossible precisely when H is a minor of the banner (the cycle on four vertices with a degree-1 vertex attached toit), but H is not P (the path graph on five vertices), assuming the ETH holds.So-called slightly superexponential parameterized algorithms , running in time 2 O ( tw log tw ) n O (1) , are by nomeans a formality for problems that are FPT in treewidth. For instance, Pilipczuk [26] showed that deciding if agraph has a transversal of size at most k hitting all cycles of length exactly ‘ (or length at most ‘ ) for a fixed value ‘ cannot be solved in time 2 o ( tw ) n O (1) , unless the ETH fails. This lower bound matches a dynamic-programmingbased algorithm running in time 2 O ( tw ) n O (1) . Cygan et al. [10] investigated the more general problem of hittingall subgraphs H of a given graph G , for a fixed pattern graph H , again parameterized by treewidth. For various H , they found algorithms running in time 2 O ( tw u ( H ) ) n O (1) , and proved ETH lower bounds in 2 Ω( tw ‘ ( H ) ) n O (1) , forvalues 1 < ‘ ( H ) (cid:54) u ( H ) depending on H . Another recent example is provided by Sau and Uéverton [27] whoprove similar results for the analogous problem where “subgraphs” is replaced by “induced subgraphs”. Finally,for the vertex-deletion problem where the target class is a proper minor-closed class given by the non-empty listof forbidden minors, it is still open if the double-exponential dependence on treewidth is asymptotically bestpossible [1].Sometimes, only a seemingly slight generalization of Feedback Vertex Set can result in problems with nosingle-exponential algorithm parameterized by treewidth. Bonamy et al. [5] showed that
Directed FeedbackVertex Set can be solved in time 2 O ( tw log tw ) n O (1) but not faster under the ETH, where tw is the treewidthof the underlying undirected graph. In this paper, we consider another collection of problems that generalize Feedback Vertex Set , and that do not have single-exponential algorithms parameterized by treewidth. Anequivalent formulation of
FVS is to find a transversal of all cycles in a given graph. We consider problemswhere the goal is to find a transversal of some subset of the cycles of a given graph. If this subset of cycles isthose that intersect some fixed set of vertices S , we obtain the following problem: Subset Feedback Vertex Set ( Subset FVS ) Parameter: tw ( G ) Input:
A graph G , a subset of vertices S ⊆ V ( G ), and an integer k . Question:
Is there a set of at most k vertices hitting all the cycles containing a vertex in S ?If we further restrict this set of cycles to those that are odd, we obtain the next problem: Subset Odd Cycle Transversal ( Subset OCT ) Parameter: tw ( G ) Input:
A graph G , a subset of vertices S ⊆ V ( G ), and an integer k . Question:
Is there a set of at most k vertices hitting all the odd cycles containing a vertex in S ?Both of these problems are NP -complete. By setting S = V ( G ), one sees that the latter problem generalizes Odd Cycle Transversal , for which Fiorini et al. [16] presented a 2 O ( tw ) n O (1) -time algorithm.Alternatively, one can require a transversal of even cycles. We first consider the problem of finding atransversal of all even cycles since, to the best of our knowledge, the fine-grained complexity of this problemparameterized by treewidth has not previously been studied. Even Cycle Transversal ( ECT ) Parameter: tw ( G ) Input:
A graph G and an integer k . Question:
Is there a set of at most k vertices hitting all the even cycles of G ? . Bergougnoux, É. Bonnet, N. Brettell, and O. Kwon 3 We note that parameterizations by solution size have been studied for these three problems [13, 20, 24, 25, 28].We now move to edge variants of
FVS . Note that
Feedback Edge Set , where the goal is to find a setof edges of weight at most k that hits the cycles, can be solved in linear time, since it is equivalent to findinga maximum-weight spanning forest. Xiao and Nagamochi showed that the subset variants Vertex-SubsetFeedback Edge Set and
Edge-Subset Feedback Edge Set , where the deletion set only needs to hit cyclescontaining a vertex or an edge (respectively) of a given set S , can also be solved in linear time [29]. On theother hand, the latter problem becomes NP -complete when the deletion set cannot intersect S . This problem isknown as Restricted Edge-Subset Feedback Edge Set . Restricted Edge-Subset Feedback Edge Set ( RESFES ) Parameter: tw ( G ) Input:
A graph G , a subset of edges S ⊆ E ( G ), and an integer k . Question:
Is there a set of at most k edges of E ( G ) \ S whose removal yields a graph without any cyclecontaining at least one edge of S ?The final two NP -complete problems we consider are closely related to Subset Feedback Vertex Set and
Restricted Edge-Subset Feedback Edge Set , respectively (see the remark in Section 1.1). They arewell-established problems with an abundance of approximation and parameterized algorithms in the literature.
Node Multiway Cut
Parameter: tw ( G ) Input:
A graph G , a subset of vertices T ⊆ V ( G ), called terminals , and an integer k . Question:
Is there a set of at most k vertices of V ( G ) \ T hitting every path between a pair of terminals? Multiway Cut
Parameter: tw ( G ) Input:
A graph G , a subset of vertices T ⊆ V ( G ), called terminals , and an integer k . Question:
Is there a set of at most k edges hitting every path between a pair of terminals?The look-alike problem Multicut , where the task is to separate each pair of terminals in a given set ofpairs (rather than all the pairs in a given set), is NP -complete on trees [17]. Therefore a parameterization bytreewidth cannot help here. In the language of parameterized complexity, Multicut parameterized by treewidthis paraNP -complete.
With the exception of
Even Cycle Transversal , for which we provide only a lower bound, we show thatall the problems formally defined so far admit a slightly superexponential parameterized algorithm, and thatthis running time cannot be improved, unless the ETH fails. We leave as an open problem the existence of aslightly superexponential algorithm for (Subset) Even Cycle Transversal parameterized by treewidth. Wenote that Deng et al. [14] have already shown that
Multiway Cut can be solved in time 2 O ( tw log tw ) n O (1) . Ouralgorithms work for treewidth and weights, while our lower bounds hold for the larger parameter pathwidth anddo not require weights.On the algorithmic side we show the following: (cid:73) Theorem 1.
The following problems can be solved in time O ( tw log tw ) n O (1) on n -vertex graphs with treewidth tw : Subset Feedback Vertex Set , Subset Odd Cycle Transversal , Restricted Edge-Subset Feedback Edge Set , and
Node Multiway Cut . We provide algorithms having the claimed running time for the weighted versions of each of the four problemsin Theorem 1. In these weighted versions, the input graph is given with a weight function w on the vertices whenthe problem is to find a set of vertices, or on the edges when the problem is to find a set of edges. Furthermore,in the weighted versions, the problem asks for a solution of weight at most k .On the complexity side, the main conceptual contribution of the paper is to show that problems seeminglyquite close to Feedback Vertex Set do not admit a single-exponential algorithm parameterized by treewidth,under the ETH. (cid:73)
Theorem 2.
Unless the ETH fails, the following problems cannot be solved in time o ( pw log pw ) n O (1) on n -vertexgraphs with pathwidth pw : Close relatives of FVS without single-exponential algorithms in treewidth
Subset Feedback Vertex Set , Subset Odd Cycle Transversal , Even Cycle Transversal , Restricted Edge-Subset Feedback Edge Set , Node Multiway Cut , and
Multiway Cut . For the last two problems, our reductions build instances where the number of terminals | T | is Θ( pw ). Thuswe also rule out a running time of | T | o ( pw ) . All the reductions are from k × k -(Permutation) IndependentSet/Clique following a strategy suggested by Lokshtanov et al. [23] (see for instance, [2, 5–7, 15]). Theseproblems cannot be solved in time 2 o ( k log k ) , unless the ETH fails. k × k -Independent Set Parameter: k Input:
A graph H with vertex set V ( H ) = [ k ] for some integer k . Question:
An independent set of size k hitting each column exactly once. k × k -Permutation Independent Set Parameter: k Input:
A graph H with vertex set V ( H ) = [ k ] for some integer k . Question:
An independent set of size k hitting each column and each row exactly once.A row is a set of vertices of the form { ( i, , ( i, , . . . , ( i, k ) } ⊂ V ( H ) for some i ∈ [ k ], while a column is aset { (1 , j ) , (2 , j ) , . . . , ( k, j ) } ⊂ V ( H ) for some j ∈ [ k ]. The problem k × k -(Permutation) Clique is definedanalogously, where the solution is required to be a clique rather than an independent set. Roadmap for the lower bounds.
To prove Theorem 2, we start by designing a gadget specification for genericvertex-deletion problems. We show that any such problem, allowing for gadgets respecting the specification,has the lower bound given in Theorem 2. This is achieved by a meta-reduction from k × k -PermutationIndependent Set . We give gadgets for Subset FVS , Subset OCT , and
ECT that comply with thespecification. We thus obtain the first three items of the theorem in a unified way, with simple and reusablegadgets. This mini-framework may in principle be useful for other vertex-deletion problems.In order to show a stronger lower bound for
Node Multiway Cut , with the number of terminals in Θ( k ),we depart from the previous specification slightly, although we still use some shared notation and arguments tobound the pathwidth, where convenient. This reduction is from k × k -Independent Set .Finally, the reduction to Multiway Cut is more intricate. For this problem it is surprisingly challenging todiscourage the undesirable solutions “cutting close” to every terminal but one, where the deletion set yieldsa very large connected component for one terminal, and small components for the rest of the terminals. Inparticular, the trick used for the
Node Multiway Cut lower bound cannot be replicated. We overcome thisissue by designing a somewhat counter-intuitive edge gadget which encourages the retention of as many pairs ofendpoints linked to two (distinct) terminals as possible. This uses the simple fact that, in a ∆-regular graph, aclique of size k minimizes the number of edges covered by k vertices: ∆ k − (cid:0) k (cid:1) vs ∆ k for an independent setof size k . We then reduce from k × k -Permutation Clique . We discuss why getting the same lower boundfor a regular variant of k × k -Permutation Clique is technical, and bypass that difficulty by encoding a degree-equalizer gadget directly in the Multiway Cut instance. As a side note, we nevertheless prove that asemi-regular variant of k × k -Clique also has the slightly superexponential lower bound. This proof uses aconstructive version of the Hajnal-Szemerédi theorem on equitable colorings. A remark on parameter-preserving reductions between the problems.
There is an easy reduction from
NodeMultiway Cut to Weighted Subset Feedback Vertex Set ( WSFVS , for short). It consists of addinga vertex v of “infinite” weight adjacent to all the terminals of the Multiway Cut instance, which also allget “infinite” weight. The set S of the WSFVS instance is { v } . The same process yields a reduction from Multiway Cut to Restricted Edge-Subset Feedback Edge Set , where now the set S of the Restricted Observe that we switch the columns and the rows compared to the original definition of k × k -Clique [23]. While this is ofcourse equivalent, it will make the representation of some gadgets slightly more conducive to the page layout. . Bergougnoux, É. Bonnet, N. Brettell, and O. Kwon 5 Edge-Subset Feedback Edge Set instance contains all the edges incident to v (recall that these edges arethus undeletable).From the latter reduction, we can immediately derive the lower bound for Restricted Edge-SubsetFeedback Edge Set from the lower bound for
Multiway Cut (see Theorem 13). However, the hardnessresult for
Node Multiway Cut (see Theorem 11) does not imply anything for
Subset Feedback VertexSet . Indeed, to encode the “infinite” weight that makes v and its neighbors undeletable, one would have toduplicate these vertices many times. This would result in a large biclique, or at the very least a large bicliqueminor, and would thereby make the pathwidth or treewidth large. Therefore Theorem 10 is necessary and cannotbe obtained by a simple modification of Theorem 11. Finally, we observe that the straightforward reductionfrom Multiway Cut to Node Multiway Cut requires vertex weights, or blows up the treewidth. So againTheorem 11 cannot be derived from Theorem 12.
Roadmap for the algorithms.
To prove Theorem 1, we first present a 2 O ( tw log tw ) n -time algorithm for theweighted variant of Subset OCT . With a few modifications, this algorithm can solve the weighted variant of
Subset FVS . We obtain algorithms for the other problems in Theorem 1 by reducing these problems to theweighted variant of
Subset FVS .Let us explain our approach for
Subset OCT on a graph G with S ⊆ V ( G ). We solve Subset OCT indirectly by finding a set X ⊆ V ( G ) of maximum weight that induces a graph with no odd cycles traversing S (we call such a graph S -bipartite). We prove that a graph has no odd cycle traversing S if and only if for eachblock C , either C is bipartite or C has no vertex in S . From this characterization, we prove that it is enough tostore 2 O ( tw log tw ) partial solutions at each bag B of a tree decomposition.Let B be a bag of the tree decomposition of G and G B be the graph induced by the vertices in B andits descendant bags in the tree decomposition. A partial solution of G B is a set X ⊆ V ( G B ) that induces an S -bipartite graph. We design an equivalence relation ≡ B on the partial solutions of G B such that for every X ≡ B Y and W ⊆ V ( G ) \ V ( G B ), G [ X ∪ W ] is S -bipartite if and only if G [ Y ∪ W ] is S -bipartite. Consequently,it is enough to keep a partial solution of maximum weight for each equivalence class of ≡ B . Intuitively, theequivalence relation ≡ B is based on the information: (1) how the blocks of G [ X ] intersecting B are connected,(2) whether important blocks (that have the possibility to create an S -traversing odd cycle later) contain a vertexof S , and (3) the parity of the paths between the vertices in B . Since ≡ B has 2 O ( tw log tw ) equivalence classes,we deduce from this equivalence relation a 2 O ( tw log tw ) n -time algorithm with standard dynamic programmingoperations. The polynomial factor n appears because we can test X ≡ B Y in time O ( n ).For the weighted variant of Subset FVS , we can use the same equivalence relation without (3). We reducethe weighted variant of
Node Multiway Cut to Subset FVS as explained in the previous subsection: byadding a vertex v of infinite weight adjacent to the set of terminals, setting S = { v } , and also giving infiniteweights to the terminals. Furthermore, we reduce the weighted variant of Restricted Edge-Subset FeedbackEdge Set to the weighted variant of
Subset FVS by subdividing each edge, setting S as the set of subdividedvertices corresponding to the given subset of edges, and giving infinite weights to the original vertices and thevertices in S . These two reductions show that both problems admit 2 O ( tw log tw ) n -time algorithms. The rest of the paper is organized as follows. In Section 2 we give the required graph-theoretic definitions andnotation. In Section 3 we prove all the ETH lower bounds of Theorem 2. More precisely, in Section 3.1 weintroduce a gadget specification for a generic vertex-deletion problem, and we show the slightly superexponentiallower bound for any problem complying with the gadget specification. In Section 3.2 we design gadgets for
Subset FVS , Subset OCT , ECT , and thus obtain the first three items of Theorem 2. In Sections 3.3 and 3.4we present specific reductions for
Node Multiway Cut and
Multiway Cut , respectively. In Section 4 weprove that the weighted variants of
Subset OCT , Subset FVS , Restricted Edge-Subset Feedback EdgeSet , and
Node Multiway Cut admit 2 O ( tw log tw ) n -time algorithms. Close relatives of FVS without single-exponential algorithms in treewidth
We assume all graphs have no loops or parallel edges. Let G be a graph. We denote the vertex set and the edgeset of G by V ( G ) and E ( G ), respectively. For a vertex v in G , we use G − v to denote the deletion of v from G , that is, the graph obtained by removing v and its incident edges. For X ⊆ V ( G ), we denote by G − X thegraph obtained by removing all vertices in X and their incident edges. For X ⊆ V ( G ), we denote by G [ X ] thesubgraph induced by the vertex set X . A subgraph H of G is an induced subgraph of G if H = G [ X ] for somevertex subset X of G . For two graphs G and G , G ∪ G is the graph with the vertex set V ( G ) ∪ V ( G ) andthe edge set E ( G ) ∪ E ( G ), and G ∩ G is the graph with the vertex set V ( G ) ∩ V ( G ) and the edge set E ( G ) ∩ E ( G ). A set X ⊆ V ( G ) is a clique if G has an edge between every pair of vertices in X . A graph withvertex set X ∪ Y that has an edge between every vertex x ∈ X and y ∈ Y is called a biclique , and is denoted K | X | , | Y | .For a vertex v in G , we denote by N G ( v ) the set of neighbors of v in G , and N G [ v ] := N G ( v ) ∪ { v } .For X ⊆ V ( G ), we let N G ( X ) := ( S v ∈ X N G ( v )) \ X , and say N G ( X ) is the (open) neighborhood of X . For u, v ∈ V ( G ), we say that u and v are twins if N ( u ) = N ( v ). If N [ u ] = N [ v ], then we also say that u and v are true twins; whereas when u and v are non-adjacent twins, we say that u and v are false twins.A vertex v of G is a cut vertex if the deletion of v from G increases the number of connected components.We say G is if it is connected and has no cut vertices. Note that every connected graph on at mosttwo vertices is 2-connected. A block of G is a maximal 2-connected subgraph of G .Let G be a graph. A walk in G is a sequence of vertices where every consecutive pair of vertices is an edgeof G . The first and last vertices in a walk are called end-vertices . A walk is closed if its two end-vertices arethe same. Given two walks W = ( v , . . . , v t ) and W = ( v t , v t +1 , . . . , v k ) whose internal vertices are pairwisedistinct, we denote by W · W the walk ( v , . . . , v t , v t +1 , . . . , v k ). We say that a walk is odd (resp. even) if thenumber of edges used by the the walk is odd (resp. even). Given S ⊆ V ( G ), we say that a walk is S -traversing if it contains at least one vertex in S . For a graph H and subgraph B of H , we say that a walk W in H isa B -walk if the endpoints of W are in B and the internal vertices of W are not in B . A path of a graph is awalk where each vertex is used at most once. A cycle of a graph is a closed walk where each vertex, except theend-vertices, is used at most once. A tree decomposition of a graph G is a pair ( T, B ) consisting of a tree T and a family B = { B t } t ∈ V ( T ) of sets B t ⊆ V ( G ), called bags , satisfying the following three conditions: V ( G ) = S t ∈ V ( T ) B t , for every edge uv of G , there exists a node t of T such that u, v ∈ B t , and for t , t , t ∈ V ( T ), B t ∩ B t ⊆ B t whenever t is on the path from t to t in T .The width of a tree decomposition ( T, B ) is max {| B t | − t ∈ V ( T ) } . The treewidth of G is the minimum widthover all tree decompositions of G . A path decomposition is a tree decomposition ( P, B ) where P is a path. The pathwidth of G is the minimum width over all path decompositions of G . We denote a path decomposition ( P, B )as ( B v , . . . , B v t ), where P is a path v v · · · v t .To design a dynamic programming algorithm, we use a convenient form of a tree decomposition known asa nice tree decomposition. A tree T is said to be rooted if it has a specified node called the root . Let T be arooted tree with root node r . A node t of T is called a leaf node if it has degree one and it is not the root. Fortwo nodes t and t of T , t is a descendant of t if the unique path from t to r contains t . If a node t is adescendant of a node t and t t ∈ E ( T ), then t is called a child of t .A tree decomposition ( T, B = { B t } t ∈ V ( T ) ) is a nice tree decomposition with root node r ∈ V ( T ) if T is arooted tree with root node r , and every node t of T is one of the following: a leaf node : t is a leaf of T and B t = ∅ ; an introduce node : t has exactly one child t and B t = B t ∪ { v } for some v ∈ V ( G ) \ B t ; a forget node : t has exactly one child t and B t = B t \ { v } for some v ∈ B t ; or a join node : t has exactly two children t and t , and B t = B t = B t . . Bergougnoux, É. Bonnet, N. Brettell, and O. Kwon 7 ( G, X )( H, X ) (
G, X ) ⊕ ( H, X ) Figure 1
An example of the sum (
G, X ) ⊕ ( H, X ). (cid:73) Theorem 3 (Bodlaender et al. [4]) . Given an n -vertex graph G and a positive integer k , one can either outputa tree decomposition of G with width at most k + 4 , or correctly answer that the treewidth of G is larger than k ,in time O ( k ) n . (cid:73) Lemma 4 (folklore; see Lemma 7.4 in [11]) . Given a tree decomposition of an n -vertex graph G of width w , one can construct a nice tree decomposition ( T, B ) of width w with | V ( T ) | = O ( wn ) in time O ( k · max( | V ( T ) | , | V ( G ) | )) . For a graph G and X ⊆ V ( G ), the pair ( G, X ) is called a boundaried graph . Two boundaried graphs (
G, X )and (
H, X ) are said to be compatible if V ( G − X ) ∩ V ( H − X ) = ∅ and G [ X ] = H [ X ]. For two compatibleboundaried graphs ( G, X ) and (
H, X ), the sum of two graphs is the graph obtained from the disjoint union of G and H by identifying each vertex of X in G with the same vertex in H and removing an edge from multipleedges that appear in X . We denote the resulting graph by ( G, X ) ⊕ ( H, X ). See Figure 1 for an example.
Our reductions for
Subset FVS , Subset OCT , and
ECT , in Section 3.2, will have the same skeleton. In orderto avoid repeating the same arguments, we show in Section 3.1 the lower bound of Theorem 2 for a meta-problem.We prove the lower bound for
Node Multiway Cut in Section 3.3, and the lower bounds for
Multiway Cut and
Restricted Edge-Subset Feedback Edge Set in Section 3.4.
The scope of application of Theorem 2 is any hereditary vertex-deletion problem Π; that is, if G − X satisfiesa problem instance P (Π), then G − X also satifies P (Π) for every X ⊇ X . The main part of the input isa graph G and a non-negative integer k . In addition, we allow any sort of labelings of G , be it subsets ofvertices S , S , . . . ⊆ V ( G ), of edges E , E , . . . , ⊆ E ( G ), pairs of vertices P , P , . . . ⊆ (cid:0) V ( G )2 (cid:1) , etc. The goal isto find a subset X ⊆ V ( G ) of k vertices such that a property P (Π), dependent on Π, is satisfied on G − X withits induced labeling. A subset of vertices A ⊆ V ( G ) is a Π -obstruction if G [ A ] does not satisfy P (Π). A set X ⊆ V ( G ) is Π -legal if G − X satisfies P (Π) (in particular, solutions are Π-legal sets of size k ). As P (Π) isassumed hereditary, a Π-legal set intersects every Π -obstruction . Finally a Π -legal s -deletion within Y is a set X ⊆ Y of size at most s such that G [ Y \ X ] satisfies P (Π). Common base
The meta-result of Theorem 5 concerns hereditary vertex-deletion problems admitting four types of gadgets.These gadgets, which will eventually depend on Π, are attached to a common problem-independent base. Wefirst describe the common base. H • is a set of 2 k vertices, for some implicit positive integer k . We denote thesevertices by v • ( i, j, z ) for each i ∈ [ k ], j ∈ [ k ], and z ∈ [2]. We imagine the vertices of H • being displayed in a k -by- k grid with v • ( i, j,
1) and v • ( i, j,
2) side by side in the i -th row and j -th column. Close relatives of FVS without single-exponential algorithms in treewidth
The base consists of copies of H • that we denote by H , H , . . . and typically index by p . The verticesof H p are denoted by v p ( i, j, z ). The vertices v p ( i, j,
1) and v p ( i, j,
2) are said to be homologous . We set C p,j := S i ∈ [ k ] ,z ∈ [2] { v p ( i, j, z ) } and refer to it as the j -th column of H p . Similarly R p,i := S j ∈ [ k ] ,z ∈ [2] { v p ( i, j, z ) } is called the i -th row of H p . We can attach to the base a list of gadgets as detailed now. The vertices added tothe base are called additional or new . Column selector gadget A k -column selector gadget has the following specification. Its vertex set is a single column C p,j plus O ( k )additional vertices C sel ( p, j ). The only restriction on the edge set of the gadget is that homologous vertices shouldremain non-adjacent. Other than that, any edge can be added within C p,j . However the open neighborhood of C sel ( p, j ) has to be contained in C p,j .A problem Π admits a column selector gadget if, for every positive integer k , one can build in time k O (1) a k -column selector such that the only Π-legal (2 k − C p,j ∪ C sel ( p, j ) are one of the k sets: C p,j \ { v p (1 , j, , v p (1 , j, } , C p,j \ { v p (2 , j, , v p (2 , j, } , . . . , C p,j \ { v p ( k, j, , v p ( k, j, } . Row selector gadget
In order to keep small balanced separators, our k -row selector gadget is quite different from the k -column selector.Its vertex set is a single row R p,i plus O (1) additional vertices R sel ( p, i ). Furthermore no edge can be addedwithin R p,i . Again the open neighborhood of R sel ( p, i ) has to be contained in R p,i .A problem Π admits a row selector gadget if, for every positive integer k , one can build in time k O (1) a k -row selector such that, for every j = j ∈ [ k ], R sel ( p, i ) ∪ { v p ( i, j, , v p ( i, j, , v p ( i, j , , v p ( i, j , } is aΠ-obstruction. Edge gadget
The vertex set of an edge gadget is of the form { v p ( i, j, , v p ( i, j, , v p ( i , j , , v p ( i , j , } ∪ E p ( i, j, i , j ) where i = i ∈ [ k ], j = j ∈ [ k ], and E p ( i, j, i , j ) is a set of O ( k ) vertices . There is no restriction on the edge set. Asusual the open neighborhood of E p ( i, j, i , j ) has to be contained in { v p ( i, j, , v p ( i, j, , v p ( i , j , , v p ( i , j , } .A problem Π admits an edge gadget if one can build in time k O (1) an edge gadget such that E p ( i, j, i , j ) ∪{ v p ( i, j, , v p ( i, j, , v p ( i , j , , v p ( i , j , } is a Π-obstruction. Propagation gadget
The vertex set of a propagation gadget is of the form H p ∪ H p +1 ∪ P p where P p is a set of k O (1) vertices. There isa subset P p ⊆ P p of size O ( k ) such that each vertex of P p \ P p has at most one neighbor in H p ∪ H p +1 and therest of its neighborhood in P p . This fairly technical condition aims to give some extra flexibility while keepingsufficiently small separators between H p and H p +1 . In particular, if P p is itself of size O ( k ), then the condition istrivially met with P p = P p . The propagation gadget has no edge with both endpoints in H p ∪ H p +1 . Everythingelse is permitted, but the open neighborhood of P p has to be contained in H p ∪ H p +1 .A problem Π admits a propagation gadget if one can build in time k O (1) a propagation gadget such that forevery i, j = j ∈ [ k ], P p ∪ { v p ( i, j, , v p ( i, j, , v p +1 ( i, j , , v p +1 ( i, j , } is a Π-obstruction. Intended-solution property
A hereditary vertex-deletion problem Π and a description of the four above gadgets for Π have the intended-solution property if the following holds. On any graph G built by adding to the base H ∪ . . . ∪ H p ∪ . . . H m at mostone edge gadget in each H p , one propagation gadget between consecutive pairs H p and H p +1 , and some columnand row selector gadgets, every deletion set S p ∈ [ m ] ,i ∈ [ k ] ,j ∈ [ k ] \{ j i } ,z ∈ [2] { v p ( i, j, z ) } (with { j , j , . . . , j k } = [ k ])intersecting every edge gadget is Π-legal.We can now state the lower bound for the generic hereditary vertex-deletion problems. O (1) vertices will actually suffice for all the gadgets of Section 3.2. . Bergougnoux, É. Bonnet, N. Brettell, and O. Kwon 9 (cid:73) Theorem 5.
Unless the ETH fails, every vertex-deletion problem Π admitting a column selector, a rowselector, an edge, and a propagation gadget, satisfying the intended-solution property, cannot be solved in time o ( pw log pw ) n O (1) on n -vertex graphs with pathwidth pw . Proof.
From any instance H of k × k -Permutation Independent Set , we build an equivalent Π-instance( G, k = k O (1) ) of size k O (1) with pathwidth in O ( k ). Since under the ETH there is no algorithm solving k × k -Permutation Independent Set in time 2 o ( k log k ) k O (1) , we derive the claimed lower bound. Construction.
We number the edges in E ( H ) as e , . . . , e m . We start with a base consisting of m copies of H • ,labelled H p for p ∈ [ m ] (see description of the common base). The vertices v p ( i, j,
1) and v p ( i, j,
2) encode thevertex ( i, j ) ∈ V ( H ); recall that we call such a pair homologous . We attach to each column C p,j , for p ∈ [ m ] and j ∈ [ k ], a column selector gadget (for Π), with additional vertices C sel ( p, j ). For each pair p ∈ [ m ] , i ∈ [ k ], weadd a row selector gadget to R p,i , with additional vertices R sel ( p, i ).For each edge e p = ( i p , j p )( i p , j p ) ∈ E ( H ) ( p ∈ [ m ]), we attach an edge gadget, with additional vertices E p ( i p , j p , i p , j p ), to { v p ( i p , j p , , v p ( i p , j p , , v p ( i p , j p , , v p ( i p , j p , } . For each p ∈ [ m − H p and H p +1 , with additional vertices P p . This finishes the construction of G . We set k := 2( k − km . Correctness.
We first assume that there is a solution I to k × k -Permutation Independent Set . Thatis, I is an independent set of H with exactly one vertex per column and per row. Say the vertices of I are(1 , j ) , (2 , j ) , . . . ( k, j k ) with { j , j , . . . , j k } = [ k ]. Then X := [ p ∈ [ m ] H p \ ∪ i ∈ [ k ] { v p ( i, j i , , v p ( i, j i , } is a solution to Π. Indeed it is Π-legal since it intersects every edge gadget (if not, the edge gadget wouldbe between two vertices of I , a contradiction) and Π satisfies the intended-solution property, by assumption.Furthermore | X | = 2 mk ( k −
1) = k .We now assume that the Π-instance ( G, k ) admits a solution (of size k ), say X . The graph G has km disjoint Π-obstructions C p,j ∪ C sel ( p, j ). For each of these sets, at least s := 2( k −
1) vertices must be deleted,by the specification of the column sector gadget. Since globally only k = kms vertices can be deleted, X intersects each C p,j ∪ C sel ( p, j ) at a set C p,j \ { v p ( i j,p , j, , v p ( i j,p , j, } for some i j,p ∈ [ k ]. Moreover, the k row selector gadgets attached to each H p enforce that { i ,p , i ,p , . . . , i k,p } = [ k ], and the propagation gadget P p enforces that i j,p = i j,p +1 for every j ∈ [ k ]. This implies that i j, = i j, = . . . = i j,m for every j ∈ [ k ],and we simply denote this common value by i j . We claim that { ( i , , ( i , , . . . , ( i k , k ) } is a solution tothe k × k -Permutation Independent Set instance. We have already argued that { i , i , . . . , i k } = [ k ].Finally there cannot be an edge e p = ( i j , j )( i j , j ) ∈ E ( H ) since then the Π-obstruction E p ( i j , j, i j , j ) ∪{ v p ( i j , j, , v p ( i j , j, , v p ( i j , j , , v p ( i j , j , } would be disjoint from X . Pathwidth in O ( k ) . Let P p be the O ( k ) vertices of P p with strictly more than one neighbor in H p ∪ H p +1 . Forevery p ∈ [ m − Y p := P p ∪ E p ( i p , j p , i p , j p ) ∪ C p,j p ∪ C sel ( p, j p ) ∪ C p,j p ∪ C sel ( p, j p ) ∪ S i ∈ [ k ] R sel ( p, i ), andwe observe that | Y p | = O ( k ) (this is where it is important that each R sel ( p, i ) has constant size). For each p ∈ [ m ]and j ∈ [ k − Z p,j be C p,j ∗ ∪ C sel ( p, j ∗ ) where j ∗ is the j -th index, by increasing value, in [ k ] \ { j p , j p } .Again we notice that | Z p,j | = O ( k ).Here is a path-decomposition of G of width O ( k ) in case every P p \ P p is empty: Y , Y ∪ Z , , Y ∪ Z , , . . . , Y ∪ Z ,k − , Y ∪ Y , Y ∪ Y ∪ Z , , Y ∪ Y ∪ Z , , . . . , Y ∪ Y ∪ Z ,k − , Y ∪ Y , . . . , Y p − ∪ Y p − , Y p − ∪ Y p − ∪ Z p − , , Y p − ∪ Y p − ∪ Z p − , , . . . , Y p − ∪ Y p − ∪ Z p − ,k − , Y p − , Y p − ∪ Z p, , Y p − ∪ Z p, , . . . , Y p − ∪ Z p,k − .Indeed the maximum bag size is O ( k ) and each edge of G appears in at least one bag. Two crucial propertiesused in this path-decomposition are that (1) the removal of P p ∪ P p +1 , so in particular of Y p ∪ Y p +1 , disconnects H p +1 from the rest of G , and (2) there is no edge between Z p,j and Z p,j for j = j ∈ [ k −
2] and p ∈ [ m ].In the general case, a path-decomposition of width O ( k ) for G is obtained from the previous decompositionby observing the following rule. Each time a vertex of H p appears in a bag for the first time, we introduce andimmediately remove each of its neighbors in P p \ P p one after the other. (cid:74) We now build specific gadgets for
Subset Feedback Vertex Set , Subset Odd Cycle Transversal , and
Even Cycle Transversal . For these problems, we always use S to denote the prescribed subset of verticesthrough which no cycle, no odd cycle, or no even cycle should go, respectively. We begin with the column selector gadget G ( C ) used for Subset FVS and
Subset OCT , followed by thegadget G ( C ) used for ECT . The column selector gadget G ( C ) attached to a column C p,j is defined as follows.It comprises 3 k additional vertices. These 3 k vertices are all added to S , and they form an independent set.Each of the first k vertices, d p,j (1 , , . . . , d p,j ( k, ∈ S , are adjacent to all vertices in S i ∈ [ k ] { v p ( i, j, } , sothese vertices induce a biclique. The next k vertices, d p,j (1 , , . . . , d p,j ( k, ∈ S , also twins, are adjacent toall vertices in S i ∈ [ k ] { v p ( i, j, } . We add d p,j (1) , . . . , d p,j ( i ) , . . . , d p,j ( k ) and, for each i ∈ [ k ], we link d p,j ( i ) toall the vertices in { v p ( i, j, } ∪ S i ∈ [ k ] \{ i } { v p ( i , j, } . Finally we make every distinct pair v p ( i, j, z ) , v p ( i , j, z )adjacent, except if i = i . See Figure 2 for an illustration. v p (1 , j, d p,j (1 , v p (1 , j, d p,j (1 , v p (2 , j, d p,j (2 , v p (2 , j, d p,j (2 , v p (3 , j, d p,j (3 , v p (3 , j, d p,j (3 , d p,j (1) d p,j (2) d p,j (3) Figure 2
The column selector gadget G ( C ). Doubly-circled vertices are in S . Blue edges linking boxes denote bicliquesbetween the two surrounded vertex sets. The gadget G ( C ) is obtained by subdividing each red edge once, and adding afalse twin to d p,j ( k,
1) (or equivalently, any d p,j ( i, d p,j ( k, We obtain the column selector gadget G ( C ) from G ( C ) by adding, for each z ∈ [2], a vertex d p,j ( k + 1 , z )adjacent to all vertices in S i ∈ [ k ] { v p ( i, j, z ) } , and by subdividing each edge d p,j ( i ) v p ( i, j,
1) once. (cid:73)
Lemma 6. G ( C ) is a column selector gadget for Subset Feedback Vertex Set and
Subset Odd CycleTransversal , and G ( C ) is a column selector gadget for Even Cycle Transversal . Proof.
The gadgets G ( C ) and G ( C ) add 3 k and 4 k + 2, respectively, new vertices, thus O ( k ). Their edge setrespects the specification of the column selector.We first show that the only Π-legal (2 k − G ( C ) are the sets C p,j \ { v p ( i, j, , v p ( i, j, } (for i ∈ [ k ]), for Π ∈ { Subset FVS , Subset OCT } . For every p ∈ [ m ], j ∈ [ k ], and z ∈ [2], the biclique K k,k between S i ∈ [ k ] { v p ( i, j, z ) } and S i ∈ [ k ] { d p,j ( i, z ) } ⊆ S forces the removal of all but at most one vertex of S i ∈ [ k ] { v p ( i, j, z ) } , or all the vertices in S i ∈ [ k ] { d p,j ( i, z ) } . Indeed, recall that the former set is a clique, while thelatter set is an independent set and is contained in the prescribed set S . Hence keeping at least one vertex in S i ∈ [ k ] { d p,j ( i, z ) } and at least two in S i ∈ [ k ] { v p ( i, j, z ) } results in an odd cycle (a triangle) going through at leastone vertex of S . Thus the only Π-legal (2 k − G ( C ) have to remove exactly k − S i ∈ [ k ] { v p ( i, j, } and exactly k − S i ∈ [ k ] { v p ( i, j, } . Let Y denote such a deletion set, and observethat Y ∩ S = ∅ . We further claim that if v p ( i, j,
1) is not in Y , then v p ( i, j,
2) is also not in Y . Assume, forthe sake of contradiction, that v p ( i, j,
1) and v p ( i , j,
2) are two (adjacent) vertices, not in Y , with i = i . Then d p,j ( i ) ∈ S forms a surviving triangle with v p ( i, j,
1) and v p ( i , j, Y = C p,j \ { v p ( i, j, , v p ( i, j, } forsome i ∈ [ k ]. . Bergougnoux, É. Bonnet, N. Brettell, and O. Kwon 11 This finishes the proof that G ( C ) is a column selector gadget for Subset FVS and
Subset OCT . Wenow adapt the arguments for G ( C ) and Π = ECT . Now the biclique K k,k +1 between S i ∈ [ k ] { v p ( i, j, z ) } and S i ∈ [ k +1] { d p,j ( i, z ) } ⊆ S forces the removal of all but at most one vertex of S i ∈ [ k ] { v p ( i, j, z ) } , or all but at mostone vertex of S i ∈ [ k +1] { d p,j ( i, z ) } , otherwise there would be a surviving even cycle C . Since only k − S i ∈ [ k ] { v p ( i, j, z ) } ∪ S i ∈ [ k +1] { d p,j ( i, z ) } ⊆ S (with z ∈ [2]), the onlyΠ-legal (2 k − G ( C ) remove all but one vertex in S i ∈ [ k ] { v p ( i, j, } and in S i ∈ [ k ] { v p ( i, j, } .The end of the proof is similar to the previous paragraph since the triangle d p,j ( i ) v p ( i, j, v p ( i , j,
2) is now a C (recall that we subdivided the edge d p,j ( i ) v p ( i, j,
1) once). (cid:74)
The row selector G ( R ), attached to R p,i , consists of two additional vertices r ( p, i ) , r ( p, i ) ∈ S made ad-jacent to every vertex in S j ∈ [ k ] { v p ( i, j, } . The row selector G ( R ) consists of three additional vertices r ( p, i ) , r ( p, i ) , r ( p, i ), each adjacent to all vertices in S j ∈ [ k ] { v p ( i, j, } . We put only r ( p, i ) in S , and we addan edge between r ( p, i ) and r ( p, i ). (cid:73) Lemma 7. G ( R ) is a row selector gadget for Subset Feedback Vertex Set and
Even Cycle Trans-versal , and G ( R ) is a row selector gadget for Subset Odd Cycle Transversal . Proof.
The gadgets G ( R ) and G ( R ) add 2 and 3 new vertices, respectively, thus O (1). Their edge set respectsthe specification of the row selector.The set { r ( p, i ) , r ( p, i ) , v p ( i, j, , v p ( i, j , } is a Π-obstruction, for every pair j = j ∈ [ k ], for everyproblem Π ∈ { Subset FVS , ECT } . Indeed it induces an even cycle (a C ) and, in the case of Subset FVS ,we note that this cycle goes through two vertices of S . The set { r ( p, i ) , r ( p, i ) , r ( p, i ) , v p ( i, j, , v p ( i, j , } is a Π-obstruction, for every pair j = j ∈ [ k ], for Π = Subset OCT . Indeed it contains an odd cycle r ( p, i ) v p ( i, j, r ( p, i ) v p ( i, j , r ( p, i ) going through r ( p, i ) ∈ S . (cid:74) Crucially for the intended-solution property, the odd cycle r ( p, i ) v p ( i, j, r ( p, i ) does not contain any vertexof S . Let G ( E ) be the following edge gadget, that we present for e p = ( i, j )( i , j ). We add an edge between v p ( i, j, v p ( i , j , s p adjacent to both v p ( i, j,
1) and v p ( i , j , s p to the set S ⊆ V ( G ).The edge gadget G ( E ) is obtained from G ( E ) by subdividing the edge s p v p ( i , j ,
1) once. (cid:73)
Lemma 8. G ( E ) is an edge gadget for Subset Feedback Vertex Set and
Subset Odd Cycle Trans-versal , and G ( E ) is an edge gadget for Even Cycle Transversal . Proof.
Both gadgets introduce a constant number of additional vertices (1 and 2, respectively, so O ( k )), andtheir edge set respects the specification. The gadget G ( E ) is an odd cycle (a triangle) with a vertex in S , hencean obstruction for Subset Feedback Vertex Set and
Subset Odd Cycle Transversal . The gadget G ( E ) is an even cycle (a C ), hence an obstruction for Even Cycle Transversal . (cid:74) We present G ( P ), a propagation gadget inserted between H p and H p +1 . We first add an independent set of2 k vertices. Among them, the k vertices r p, , . . . , r p,k represent the row indices in H p and H p +1 , while the k other vertices c p, , . . . , c p,k represent the column indices. We link r p,i to all the vertices in S j ∈ [ k ] { v p ( i, j, } ∪ S j ∈ [ k ] { v p +1 ( i, j, } . Similarly, we link c p,j to all the vertices in S i ∈ [ k ] { v p ( i, j, } ∪ S i ∈ [ k ] { v p +1 ( i, j, } . Finally,we add a vertex c p ∈ S adjacent to all the vertices c p, , . . . , c p,k .The gadget G ( P ) is defined similarly, except that we subdivide the edge r p,i v p ( i, j,
2) once, for each i, j ∈ [ k ].Finally the gadget G ( P ) adds to G ( P ), a vertex c p,j , for each j ∈ [ k ]. The vertex c p,j is linked to c p,j and to c p . (cid:73) Lemma 9. G ( P ) is a column selector gadget for Subset Feedback Vertex Set , G ( P ) is a columnselector gadget for Subset Odd Cycle Transversal , and G ( P ) is a column selector gadget for Even CycleTransversal . Proof.
Let P p := { r p, , . . . , r p,k , c p, , . . . , c p,k , c p } . The gadget G ( P ) adds to the base the set P p of size 2 k + 1,thus O ( k ). Hence it trivially satisfies the technical condition of the propagation gadget. The gadget G ( P ) addsa further k vertices, stemming from the subdivision of the edges r p,i v p ( i, j, H p ∪ H p +1 and the rest of their neighbors in P p , so satisfy the specification. For the same reason, G ( P ) also satisfies the specification. We denote by P p the set of 2 k + 1 + k vertices consisting of P p plus thesubdivision vertices, and P p the set of 3 k + 1 + k vertices added in G ( P ). The edge sets of G ( P ) , G ( P ) , G ( P )respect the specification of the propagation selector.For every i, j = j ∈ [ k ], P p ∪ { v p ( i, j, , v p +1 ( i, j , } is a Π-obstruction for Π = Subset FVS . Indeed r p,i v p ( i, j, c p,j c j c p,j v p +1 ( i, j ,
1) is a cycle (a C ) going through c j ∈ S . Similarly P p ∪ { v p ( i, j, , v p +1 ( i, j , } is a Π-obstruction for Π = Subset OCT , the same cycle being now of odd length (a C ), due to thesubdivision of r p,i v p ( i, j, P p ∪ { v p ( i, j, , v p +1 ( i, j , } is a Π-obstruction for Π = ECT since r p,i w p ( i, j, v p ( i, j, c p,j c p,j c j c p,j v p +1 ( i, j ,
1) is an even cycle (a C ), where w p ( i, j,
2) is the subdivided vertexstemming from the edge r p,i v p ( i, j, (cid:74) (cid:73) Theorem 10.
Unless the ETH fails, the following problems cannot be solved in time o ( pw log pw ) n O (1) on n -vertex graphs with pathwidth pw : Subset Feedback Vertex Set , Subset Odd Cycle Transversal , and
Even Cycle Transversal . Proof.
We need to check that these problems satisfy the preconditions of Theorem 5. Sections 3.2.1 to 3.2.4 andLemmas 6 to 9 show how to build the four types of gadgets. Which problem uses which version of the gadget issummarized in Table 1. See Figure 3 for a schematic representation of the construction for
Subset FVS . column selector row selector edge gadget propagation gadget Subset Feedback Vertex Set G ( C ) G ( R ) G ( E ) G ( P ) Subset Odd Cycle Transversal G ( C ) G ( R ) G ( E ) G ( P ) Even Cycle Transversal G ( C ) G ( R ) G ( E ) G ( P ) Table 1
The different gadgets used for the different problems.
Finally we have to check that the problems have the intended-solution property. We shall prove that everyset X := S p ∈ [ m ] ,i ∈ [ k ] ,z ∈ [2] { v p ( i, j i , z ) } , with { j , . . . , j k } = [ k ] and intersecting all the edge gadgets is Π-legal inany graph G obtained by attaching to the base the four types of gadgets with respect to their specification ofSection 3.1. The set X is a solution to Π ∈ { Subset FVS , Subset OCT , ECT } , if and only if no 2-connectedcomponent (i.e., a block of size at least 3) of G − X is a Π-obstruction. Indeed no cycle can go through acut-vertex.We first note that there is no 2-connected component within G ( C ) , G ( C ) , G ( R ) , G ( E ) , G ( E ) restricted to G − X . For the latter two gadgets, this is because, by assumption, X intersects every edge gadget. In a gadget G ( R ) restricted to G − X , there is one 2-connected component, namely a triangle; but none of its verticesbelongs to S .We now observe that every vertex c p is a cut-vertex in G ( P ), G ( P ), and G ( P ) restricted to G − X . Sothe remaining 2-connected components of G − X are induced cycles C of the form r p,i v p ( i, j, c p,j v p +1 ( i, j, G ( P ) is used, or induced C when G ( P ) is used, or triangle and induced cycle C when G ( P ) is used.In the first two cases, none of the vertices of the cycles belongs to S . In the third case, no cycle is even. Thisestablishes that Subset FVS , Subset OCT , and
ECT with their respective combination of gadgets have theintended-solution property. (cid:74) . Bergougnoux, É. Bonnet, N. Brettell, and O. Kwon 13 s s s rowcolumn c rowcolumn c C sel C sel C sel C sel C sel C sel C sel C sel C sel Figure 3
Example of the overall picture for
Subset Feedback Vertex Set . The first three edges (in green) in thereduction from k × k -Permutation Independent Set , with k = 3, to Subset FVS . The doubly-circled vertices arevertices in S . The column selector gadget C sel , of size O ( k ), forces that only one pair of homologous vertices is retained ineach column (see Figure 2). We did not represent the row selector gadget. For
Node Multiway Cut we will also start from the base S p ∈ [ m ] H p but we will deviate from the gadgetspecification of Section 3.1. We will “communalize” the selector, edge, and propagation gadgets. That way, weare able to show the claimed lower bound even when the number of terminals is linearly tied to the pathwidth.This is unlike our constructions for Subset FVS and
Subset OCT in Theorem 10 where the size of theprescribed subsets S is significantly larger than the pathwidth. (cid:73) Theorem 11.
Unless the ETH fails,
Node Multiway Cut cannot be solved in time o ( p log p ) n O (1) on n -vertexgraphs where p = pw + | T | is the sum of the pathwidth of the input graph and the number of terminals. Proof.
We now reduce from k × k -Independent Set . Again let H be an m -edge k × k -Independent Set instance. We build an equivalent Node Multiway Cut instance (
G, T, k := 2( k − km ), with | T | = k + 2, byadding only 2 k + 2 new vertices to the base S p ∈ [ m ] H p . We link every non-homologous pair of vertices within eachcolumn C p,j (for p ∈ [ m ] , j ∈ [ k ]). We add two terminals t, t ∈ T . For every edge e p = ( i p , j p )( i p , j p ) ∈ E ( H ),we make v p ( i p , j p ,
2) and v p ( i p , j p ,
2) adjacent. We also link t to v p ( i p , j p , t to v p ( i p , j p , k terminals r , . . . , r k ∈ T . We link every vertex on an i -th row ( R p,i ) to r i , except if the vertex isalready adjacent to t or t . This exception concerns the vertices v p ( i p , j p ,
2) and v p ( i p , j p , k (non-terminal) vertices c , . . . , c k . For each p ∈ [ m ] , j ∈ [ k ] , i ∈ [ k ], we add an edge between v p ( i, j,
1) and c i .This finishes the construction of G . The set of terminals is T := { t, t , r , . . . , r k } . We ask for a deletion set of size k := 2( k − km . The pathwidth of G is O ( k ), since it is obtained by adding 2 k + 2 vertices ( { t, t , r , . . . , r k } )to a graph satisfying the gadget specification of Section 3.1 (with “empty” row selector and propagation gadgets).We now show the correctness of this reduction. Assume that the graph H admits an independent set I := { ( i , , ( i , , . . . , ( i k , k ) } . We claim that X := S p ∈ [ m ] ,j ∈ [ k ] ,z ∈ [2] H m \ { v p ( i j , j, z ) } is a solution to the Node Multiway Cut instance. We first observe that the connected component of G − X containing t (andsimilarly t ) does not contain any other terminal. Indeed, since I is an independent set, at most one of v p ( i p , j p , v p ( i p , j p ,
2) is preserved in G − X when ( i p , j p )( i p , j p ) is an edge of H . Hence each vertex v p ( i p , j p ,
2) or v p ( i p , j p ,
2) that exists in G − X has degree 1: it is adjacent only to t or t . So the connected component in G − X of t (resp. t ) is a star centered at t (resp. t ) whose leaves are all in S p ∈ [ m ] H p , hence are non-terminals. We nowobserve that there is no path between r i and r i (with i = i ) in G − X . Such a path would have to go through avertex c j . Indeed, no edge within a column C p,j is preserved in G − X (nor the edges v p ( i p , j p , v p ( i p , j p , c j is adjacent to a single row in G − X ,since we kept only one pair v p ( i, j, , v p ( i, j,
2) per column C p,j , and we made the same choice in every H p .Let us now assume that X is a solution to the Node Multiway Cut instance (
G, T, k ). A first observationis that no edge within a column C p,j can be present in G − X , otherwise there is a 3-edge path between a pair ofterminals in { r , . . . , r k , t, t } , since every edge within C p,j is between non-homologous vertices, and every vertex in C p,j is adjacent to a terminal. This implies that for each p ∈ [ m ] and j ∈ [ k ], we have { v p ( i p,j , j, , v p ( i p,j , j, } ⊆ C p,j \ X for some i p,j ∈ [ k ]. In fact, since at least 2( k − k vertices of H p must be removed for each p ∈ [ m ],and the solution X has size at most 2( k − km , we have C p,j \ X = { v p ( i p,j , j, , v p ( i p,j , j, } . In particular, X ⊆ S p ∈ [ m ] H p , so c i / ∈ X for each i ∈ [ k ]. We now show that the i p,j ’s coincide for each p ∈ [ m ]. Assumefor the sake of contradiction that v p ( i, j,
1) and v p ( i , j,
1) are both present in G − X with p = p and i = i .Then r i v p ( i, j, c j v p ( i , j, r i is a path in G − X , a contradiction. Therefore i ,j = . . . = i m,j . Let i j denotethis common value. We claim that { ( i , , . . . , ( i k , k ) } is an independent set in H . Suppose there is an edge( i j , j )( i j , j ) ∈ E ( H ) for distinct j, j ∈ [ k ]. Then there is a path tv p ( i j , j, v p ( i j , j , t in G for some p ∈ [ m ],between the terminals t and t , a contradiction. (cid:74) To obtain the lower bound for
Multiway Cut , we reduce from k × k -Permutation Clique .However, we note that reducing from Semi-Regular k × k -Permutation Clique , where all the verticesof a column have the same degree towards another column, and there is no edge with both endpoints in thesame row, would make the construction cleaner. So the first reflex is to try and show the same 2 o ( k log k ) lowerbound for this variant. Ensuring the semi-regularity condition can be done rather smoothly; it requires revisitingthe grouping technique from, say, 3 -Coloring , and using known results on equitable colorings. An interestedreader can find a complete proof in the appendix. Nonetheless, getting rid of the “horizontal” edges (with bothendpoints in the same row) in order to obtain an instance k × k -Permutation Clique , while preserving thesemi-regularity, is unnecessarily complex. In particular, the reduction from k × k -Clique to k × k -PermutationClique presented in the seminal paper [23] does not preserve semi-regularity. To prove the next theorem, we willinstead directly reduce from k × k -Permutation Clique and “regularize” the degree by some ad hoc gadgetry. (cid:73) Theorem 12.
Unless the ETH fails,
Multiway Cut cannot be solved in time o ( p log p ) n O (1) on n -vertexgraphs where p = pw + | T | is the sum of the pathwidth of the input graph and the number of terminals. Proof.
We reduce from an instance H of k × k -Permutation Clique , so we may assume that there is no edgeof H with both endpoints in the same row. Let µ be the number of edges of H , and let ∆ be the maximumdegree of vertices of H . We associate each v ∈ V ( H ) to the non-negative integer δ ( v ) := ∆ − d H ( v ), where d H ( v )is the degree of v . It is useful to consider the graph H obtained from H by attaching δ ( v ) pendant leaves toeach v ∈ V ( H ), where each vertex in V ( H ) has degree ∆ in H . We set m := k ∆, and observe that m (cid:62) µ corresponds to the number of edges in H .We build an equivalent Multiway Cut instance (
G, T, k ), with | T | = k + 1, by adding a polynomial numberof vertices to the base S p ∈ [ µ + k ] H p . We do not need the vertices v p ( i, j, v p ( i, j,
1) intosimply v p ( i, j ). Now R p,i is the set { v p ( i, , . . . , v p ( i, k ) } and C p,j is { v p (1 , j ) , . . . , v p ( k, j ) } .We encode weighted edges in the following way. When we say that we add an edge of weight w ∈ N betweentwo vertices u, v , we mean that we add w “parallel” 2-edge paths between u and v . None of the introducedvertices are terminals, so the instance behaves equivalently as with the weighted edge. Thus, for the sake ofsimplicity, we will treat these parallel paths as a weighted edge. If we require an edge to be “undeletable”,we give it weight k + 1, just above the total budget. All the weights of the construction are encoded with . Bergougnoux, É. Bonnet, N. Brettell, and O. Kwon 15 polynomially many vertices and unit-weight edges. Therefore the lower bound does apply to the unweightedversion of Multiway Cut .We set h := 12 m − k ∆ − (cid:0) k (cid:1) and k := ( h + 1)( k − k ( µ + k ) + h . We add k terminals r , . . . , r k ∈ T ,and we link every vertex in the i -th row (that is, in a set R p,i ) to r i by an edge of weight h + 1. We add k non-terminals c , . . . , c k and for each p ∈ [ µ + k ] , j ∈ [ k ] , i ∈ [ k ], we add an edge of weight k + 1 (an undeletableedge) between v p ( i, j ) and c i . For every e p = ( i p , j p )( i p , j p ) ∈ E ( H ) with p ∈ [ µ ], we add the following edgegadget between v p ( i p , j p ) and v p ( i p , j p ). We first build a 5-vertex path where the edge weights are, from oneendpoint to the other, 5 , , ,
5. We denote by z p the central vertex, and we link the second vertex, x p , and thefourth vertex, y p , by an edge of weight 3. We link the first vertex (one endpoint) of the gadget to v p ( i p , j p )and to r i p by edges of weight 3, and the last vertex (the other endpoint) to v p ( i p , j p ) and to r i p by edgesof weight 3. Finally we link z p to an additional terminal t (common to every p ∈ [ µ ]) by an edge of weight k + 1. See Figure 4 for an illustration of the edge gadget and how it is attached to the terminals. So far we z p y p x p v p ( i p , j p ) v p ( i p , j p )3 335 53 3 r i p r i p th + 1 3 h + 13 k + 1 Figure 4
The edge gadget for
Multiway Cut . have added edge gadgets to the first µ copies H , . . . , H µ . We now describe what we (potentially) add to thelast k copies H µ +1 , . . . , H µ + k . We put an arbitrary total order over V ( H ), say, the natural (cid:54) where ( i, j ) isinterpreted as n ( i, j ) = i + ( j − k . We attach to v µ + n ( i,j ) ( i, j ), r i , and t the following simple gadget, called a degree-equalizer , which can be seen as a degenerate case of an edge gadget with multiplicity δ (( i, j )) (henceforthwe simply write δ ( i, j ) for the sake of legibility). We add a vertex w µ + n ( i,j ) ( i, j ), and link it to t by an edge ofweight 11 δ ( i, j ), and to v µ + n ( i,j ) ( i, j ) and r i by edges of weight 6 δ ( i, j ) each. This finishes the construction of( G, T := { r , . . . , r k , t } , k := ( h + 1)( k − k ( µ + k ) + h ).The pathwidth of G is O ( k ) following the arguments for the Node Multiway Cut construction.We now show the correctness of the reduction. Assume that there is a clique C := { ( i , , . . . , ( i k , k ) } in H ,with { i , . . . , i k } = [ k ]. We build the following edge deletion-set X for the Multiway Cut instance. We startby including in X all the edges of weight h + 1 between r i and v p ( i, j ) ( p ∈ [ µ + k ]) such that ( i, j ) / ∈ C . Thisrepresents k ( k − µ + k ) weighted edges, and ( h + 1) k ( k − µ + k ) unit-weight edges.We distinguish three cases for the edge gadget of every e p = ( i p , j p )( i p , j p ) ( p ∈ [ µ ]). If { ( i p , j p ) , ( i p , j p ) }∩ C = ∅ (i.e., e p has no endpoint in C ), we add to X the four weight-3 edges incident to v p ( i p , j p ), r i p , v p ( i p , j p ), and r i p ;a total of 12 edges. If |{ ( i p , j p ) , ( i p , j p ) } ∩ C | = 1, say, without loss of generality, that ( i p , j p ) ∈ C , then we addthe weight-5 edge incident to x p and the two weight-3 edges incident to v p ( i p , j p ) and r i p . This consists of 11edges in total. In the symmetric case ( i p , j p ) ∈ C , we would remove the weight-5 edge incident to y p and the twoweight-3 edges incident to v p ( i p , j p ) and r i p . Finally if |{ ( i p , j p ) , ( i p , j p ) } ∩ C | = 2, we add the 9 edges of theweighted triangle x p y p z p to X .For every degree-equalizer gadget attached to H µ + n ( i,j ) , we add to X the weight-11 δ ( i, j ) edge incident to t if ( i, j ) ∈ C , and the two weight-6 δ ( i, j ) edges incident to w µ + n ( i,j ) ( i, j ) if ( i, j ) / ∈ C . Note that these numbers ofedges correspond to what we would remove in δ ( i, j ) copies of an edge gadget where the other endpoint is not in C . This finishes the construction of X .There are (cid:0) k (cid:1) edges of H with both endpoints in C , there are k ∆ − (cid:0) k (cid:1) edges with exactly one endpointin C , and m − k ∆ + (cid:0) k (cid:1) edges with no endpoint in C . So there are 9 (cid:0) k (cid:1) + 11( k ∆ − (cid:0) k (cid:1) ) + 12( m − k ∆ + (cid:0) k (cid:1) ) = 12 m − k ∆ − (cid:0) k (cid:1) = h edges added to X from edge and degree-equalizer gadgets. Thus X has size( h + 1) k ( k − µ + k ) + h = k as imposed. Let G be the graph ( V ( G ) , E ( G ) \ X ). We show that every connected component of G contains at most one terminal. Observe that in G − { t } , each vertex z p is in aconnected component contained in the edge gadget of e p (and, in particular, not containing a terminal). Since t isonly adjacent (by weighted edges) to the vertices z p and w µ + n ( i,j ) ( i, j ), it follows that the connected componentin G containing t has no other terminals. Note furthermore that the removal of the edges in X disconnectsevery pair v p ( i p , j p ) , v p ( i p , j p ) in the edge gadget of e p = ( i p , j p )( i p , j p ) for p ∈ [ µ ]. Thus the vertices reachablefrom r i in G are { c j } ∪ S p ∈ [ m ] C p,j , such that j is unique integer of [ k ] with i j = i , as well as some non-terminalvertices in some edge and degree-equalizer gadgets. In particular there is no path between r i and r i , with i = i ,in G . Thus X is a solution.Let us now assume that the Multiway Cut instance (
G, T, k ) has a solution X , and let G be ( V ( G ) , E ( G ) \ X ). A first observation is that there is a path in G between any pair of vertices in the j -th column, say v p ( i, j )and v p ( i , j ), since there are undeletable edges between c j and each vertex v p ( i, j ). Thus there is a componentof G containing S p ∈ [ µ + k ] C p,j , for each j ∈ [ k ], and this component contains at most one terminal. With abudget of ( h + 1) k ( k − µ + k ) + h , one can remove at most k ( k − µ + k ) edges of weight h + 1. Since notwo edges r i v p ( i, j ) and r i v p ( i , j ) can remain in G , for distinct i, i ∈ [ k ], j ∈ [ k ], and p ∈ [ µ + k ], at least k ( k −
1) edges of weight h + 1 incident to a vertex in H p must be in X , for each p ∈ [ µ + k ], for a total ofat least k ( k − µ + k ) edges of weight h + 1. Now the only possibility is that, for each j ∈ [ k ], there existsan i j ∈ [ k ] such that X contains all the edges of weight h + 1 from S p ∈ [ µ + k ] C p,j to { r , . . . , r k } except thoseincident to r i j . We set C := { ( i , , . . . , ( i k , k ) } , and we will now show that C is a clique in H . In particular { i , . . . , i k } = [ k ] since there is no edge of H with endpoints in the same row.First we consider an edge e p = ( i p , j p )( i p , j p ) ∈ E ( H ) such that { ( i p , j p ) , ( i p , j p ) } ∩ C = ∅ . Note that, in thiscase, v p ( i p , j p ) (resp. v p ( i p , j p )) is, in G , in the connected component of r i jp = r i p (resp. r i j p = r i p ). We thenneed to separate the seven pairs: ( r i p , v p ( i p , j p )), ( t, v p ( i p , j p )), ( r i p , t ), ( r i p , v p ( i p , j p )), ( t, v p ( i p , j p )), ( r i p , t ), and( r i p , r i p ). This requires 12 edge deletions.We now consider an edge e p = ( i p , j p )( i p , j p ) ∈ E ( H ) such that |{ ( i p , j p ) , ( i p , j p ) } ∩ C | = 1. We assumethat ( i p , j p ) ∈ C (the other case is symmetric). In this case, v p ( i p , j p ) is, in G , in the connected componentof r i j p = r i p . We then need to separate the six pairs: ( t, v p ( i p , j p )), ( r i p , t ), ( r i p , v p ( i p , j p )), ( t, v p ( i p , j p )), ( r i p , t ),and ( r i p , r i p ). This requires 11 edge deletions: the weight-5 edge incident to x p and the two weight-3 edgesincident to v p ( i p , j p ) and to r i p .Finally let us assume that e p = ( i p , j p )( i p , j p ) ∈ E ( H ) is such that |{ ( i p , j p ) , ( i p , j p ) } ∩ C | = 1. Here weneed to separate the five pairs: ( t, v p ( i p , j p )), ( r i p , t ), ( t, v p ( i p , j p )), ( r i p , t ), and ( r i p , r i p ). This requires 9 edgedeletions: the three weight-3 edges in the triangle x p y p z p .We now turn to the degree-equalizer gadgets. If ( i, j ) / ∈ C , then we need to separate the three pairs( r i , v µ + n ( i,j ) ( i, j )), ( t, v µ + n ( i,j ) ( i, j )), and ( r i , t ). This requires 12 δ ( i, j ) edge deletions (the weighted edges r i w µ + n ( i,j ) ( i, j ) and v µ + n ( i,j ) ( i, j ) w µ + n ( i,j ) ( i, j )). If on the contrary ( i, j ) ∈ C , we only need to separate the twopairs ( t, v µ + n ( i,j ) ( i, j )) and ( r i , t ). This requires 11 δ ( i, j ) deletions (the weighted edge tw µ + n ( i,j ) ( i, j )).We denote by s the number of edges in H [ C ]. Since the edge and degree-equalizer gadgets are pairwise edge-disjoint, what we have shown implies that X contains at least 9 s + 11( k ∆ − s ) + 12( m − k ∆ + s ) = 12 m − k ∆ − s edges in the edge gadgets. As X is of size at most k , we have that s has to be equal to (cid:0) k (cid:1) . This implies that C is a clique. (cid:74) By the simple reduction from
Multiway Cut to Restricted Edge-Subset Feedback Edge Set , givenin the introduction, we obtain the following as a corollary. (cid:73)
Theorem 13.
Unless the ETH fails,
Restricted Edge-Subset Feedback Edge Set cannot be solved intime o ( p log p ) n O (1) on n -vertex graphs where p = pw + | S | is the sum of the pathwidth of the input graph and thenumber of undeletable (terminal) edges. It is not difficult to adapt the construction of Theorem 12 for the directed variant of
Multiway Cut . (cid:73) Theorem 14.
Unless the ETH fails,
Directed Multiway Cut cannot be solved in time o ( pw log pw ) n O (1) on n -vertex directed graphs whose underlying undirected graph has pathwidth pw . . Bergougnoux, É. Bonnet, N. Brettell, and O. Kwon 17 In this section, we present 2 O ( tw log tw ) n -time algorithms for the weighted variants of the considered problemswith the exception of ECT .We first present in Theorem 17 a 2 O ( tw log tw ) n -time algorithm for Subset OCT . Then, we show that withsimple modifications this algorithm can solve
Subset FVS . We deduce the algorithms for the other problems byreducing these problems to the weighted variant of
Subset FVS .Let us focus on the
Subset OCT problem. For a graph G and a vertex set S of G , we say that G is S -bipartite if it has no odd cycle containing a vertex of S . Solving Subset OCT is equivalent to find an S -bipartite induced subgraph of maximum size. The following characterization of S -bipartite graphs will beuseful. (cid:73) Lemma 15.
A graph G is S -bipartite if and only if for every block B of G , either B has no vertex of S , or itis bipartite. Proof. ( ⇒ ) Assume toward a contradiction that G is S -bipartite and that a block B of G contains a vertex s ∈ S and B is not bipartite. Because B is not bipartite, there exists an odd cycle C in B . Since G is S -bipartiteby assumption, C does not contain s .Since B is 2-connected and has at least 3 vertices, there exist two paths P sc and P c s between s and twodistinct vertices c, c of C such that the internal vertices of P sc and P c s and the vertices of C are pairwisedistinct. Let P cc and b P cc be the two paths between c and c in C . The concatenations C = P sc · P cc · P c s and C = P sc · b P cc · P c s are two S -traversing cycles. Since C is an odd cycle, the parity of P cc and b P cc are not thesame. Hence, one of the two cycles C and C is an odd S -traversing cycle. This yields a contradiction.( ⇐ ) Assume that G is not S -bipartite. Then, G contains an odd S -traversing cycle C . This cycle is containedin a block B of G . Thus G has a block that is not bipartite and that contains at least one vertex in S . (cid:74) One can easily modify the proof of the first direction of Lemma 15 to prove the following fact. (cid:66)
Fact 16.
If a graph G is 2-connected and not bipartite, then there exists an odd path and an even pathbetween every pair of vertices. (cid:73) Theorem 17. ( Weighted ) Subset Odd Cycle Transversal can be solved in time O ( tw log tw ) n on n -vertex graphs with treewidth tw . Proof.
In the following, we fix a graph G , S ⊆ V ( G ), and a weight function w : V ( G ) → R . Using Theorem 3and Lemma 4, we obtain a nice tree decomposition of G of width at most 5 tw + 4 in time O ( c tw · n ) for someconstant c . Let ( T, { B t } t ∈ V ( T ) ) be the resulting nice tree decomposition. For each node t of T , let G t be thesubgraph of G induced by the union of all bags B t where t is a descendant of t .Let t be a node of T . A partial solution of G t is a subset X ⊆ V ( G t ) such that G [ X ] is S -bipartite. Inthe following, we introduce a notion of auxiliary graph in order to design an equivalence relation ≡ t betweenpartial solutions such that X ≡ t Y if, for every W ⊆ V ( G t ), G [ X ∪ W ] is S -bipartite if and only if G [ Y ∪ W ] is S -bipartite.Let X ⊆ V ( G ) (not necessarily contained in G t ). We denote by Inc( X ) the block-cut tree of G [ X ], that isthe bipartite graph whose vertices are the blocks and the cut vertices of G [ X ] and where a block B is adjacentto a cut vertex v if v ∈ V ( B ). Observe that Inc( X ) is by definition a forest.We say that a vertex v of Inc( X ) is active (with respect to t ) if: v is a cut vertex of G [ X ] in B t , v is a block of G [ X ] that contains at least two vertices in B t , or v is a block of G [ X ] that contains exactly one vertex in B t that is not a cut vertex.Note that every vertex in B t is an active cut vertex or it is in an active block of G [ X ]. Intuitively, the auxiliarygraph associated with a partial solution X needs to encode how the active blocks of Inc( X ) are connectedtogether.We construct the auxiliary graphs Aux p ( X, t ) and Aux(
X, t ) from Inc( X ) by the following operations: We remove recursively the leaves and the isolated vertices that are inactive. Let Aux p ( X, t ) be the resultinggraph ( p for “prototype”). For every maximal path P of Aux p ( X, t ) between u and v and with inactive internal vertices of degree 2, weremove the internal vertices of P and we add an edge between u and v (shrinking degree 2 nodes that areinactive).Figure 5 illustrates the constructions of Aux p ( X, t ) and Aux(
X, t ). Observe that Operation 1 removes theinactive blocks of G [ X ] that contain one vertex in B t . Thus, every block in Aux p ( X, t ) that contains verticesin B t is active. By construction, Aux( X, t ) is a forest whose vertices are the active vertices of Inc( X ) and theinactive vertices that have degree at least 3 in Aux p ( X, t ). An important remark is that the algorithm usesthe graphs Aux(
X, t ) for X ⊆ V ( G t ) and in the proof we will use Aux( X, t ) and Aux p ( X, t ) for X ⊆ V ( G t ) or X ⊆ B t ∪ V ( G t ).By Step 2, any edge uv of Aux( X, t ) corresponds to an alternating sequence P of cut vertices and blocks A , A , . . . , A x that forms a path from u = A to v = A x in Inc( X ). We define the graph M uv as the union ofthe blocks in P . Note that one of A and A is a cut vertex and one of A x − and A x is a cut vertex. We saythat these cut vertices are the endpoints of M uv . Inc( X ) G [ X ]Aux( X, t ) Figure 5
Example of graphs Inc( X ) and Aux( X, t ) constructed from a graph G [ X ]. The vertices in B t are white filled.The red vertices and edges in Inc( X ) are those we remove to obtain Aux p ( X, t ). Let X and Y be two partial solutions of G t . We say that X ≡ t Y if X ∩ B t = Y ∩ B t , and there is anisomorphism ϕ from Aux( X, t ) to Aux(
Y, t ) such that the following conditions are satisfied: For every vertex v in Aux( X, t ), v is active if and only if ϕ ( v ) is active. For every vertex v in Aux( X, t ), v is a block if and only if ϕ ( v ) is a block. For every active cut vertex v in Aux( X, t ), we have ϕ ( v ) = v . For every active block B in Aux( X, t ): a. V ( B ) ∩ B t = V ( ϕ ( B )) ∩ B t , b. V ( B ) ∩ S = ∅ if and only if V ( ϕ ( B )) ∩ S = ∅ , and c. B is bipartite if and only if ϕ ( B ) is bipartite. For every edge uv in Aux( X, t ): a. M uv is bipartite if and only if M ϕ ( u ) ϕ ( v ) is bipartite, and b. V ( M uv ) ∩ S = ∅ if and only if V ( M ϕ ( u ) ϕ ( v ) ) ∩ S = ∅ . For every pair ( u, v ) of vertices in B t ∩ X and every path P X between u and v in G [ X ], there exists a path P Y in G [ Y ] between u and v with the same parity as P X . (cid:66) Claim 18.
For every node t of T , the equivalence relation ≡ t has 2 O ( tw log tw ) equivalence classes. Proof.
Let t be a node and X be a partial solution of G t . Let k = | B t | . In the following, we will upper boundthe number of possibilities for the conditions in the definition of ≡ t . Notice that there are at most 2 k possibilitiesfor X ∩ B t .Now, observe that the number of active blocks of G [ X ] is at most k . Note that if an active block containsone vertex of B t , then it is not a cut vertex of G [ X ], and if an active block intersects at least two vertices of B t , . Bergougnoux, É. Bonnet, N. Brettell, and O. Kwon 19 then it contains either two cut vertices of G [ X ] contained in B t , or it contains at least one vertex of B t thatis not a cut vertex of G [ X ]. We consider Inc( X ) as a rooted forest (where each tree has a root), and give aninjection φ from the set of active blocks to B t as follows. For each active block B ,if it contains a vertex in B t that is not a cut vertex of G [ X ], then choose such a vertex v and set φ ( B ) = v ,andif all vertices of B t ∩ V ( B ) are cut vertices of G [ X ], then | B t ∩ V ( B ) | ≥ v ∈ B t ∩ V ( B ) that is a child of B in Inc( X ), and set φ ( B ) = v .Clearly, φ is an injection, and it shows that the number of active blocks of G [ X ] is at most k . We deduce thatAux( X, t ) contains at most 2 k active vertices as there are at most k active blocks and at most k active cutvertices.By construction, all the vertices of degree at most 2 in Aux( X, t ) are active vertices of Inc( X ). In particular,the leaves of Aux( X, t ) are active vertices. The leaves connected to vertices of degree at least 3 induce anindependent set in Aux(
X, t ). We can easily show from this fact that there are at most k leaves connected tovertices of degree 3 (at most the number of part in a partition of k elements). Since Aux( X, t ) is a forest, thenumber of vertices of degree at least 3 is at most k . We deduce that Aux( X, t ) has at most 2 k active verticesand k inactive vertices. Hence, there are at most (2 k + 1)( k + 1) possibilities for Condition 1 as it is at mostthe number of ways of choosing two numbers one in [0 , k ] and one in [0 , k ]. By Cayley’s formula [8], thenumber of forests on 3 k labeled vertices is (3 k + 1) k − . Thus, there are 2 O ( tw log tw ) non-isomorphic graphs in { Aux(
W, t ) | W is a partial solution of G t } .For Condition 2, there are 2 possibilities for each vertex in Aux( X, t ): either it is a block or a cut vertex.Thus, there are at most 2 k possibilities for this condition.We claim that there are 2 O ( k log k ) possibilities for Conditions 3 and 4.a. Let v , . . . , v d be the cut vertices of G [ X ] in B t and X , . . . , X ‘ be the intersections between B t and the vertex sets of the active blocks of G [ X ].Note that for every distinct X i and X j , | X i ∩ X j | ≤
1. Moreover, since they came from Inc( X ), there is nocyclic structure; that is, X i − v i − X i · · · − v i α − − X i α where X i = X i α and each v i j only belongs to X i j and X i j +1 . This means that the number of possibilities for v , . . . , v d and X , . . . , X ‘ is the same as the numberof ways of partitioning a set of k vertices into blocks and cut vertices, as isolated vertices are single blocks.We claim that the number of ways of partitioning a set of k vertices into blocks and cut vertices is 2 O ( k log k ) .Let T be a set of k vertices. First take a partition P of T . There are at most 2 k log k possibilities for P . Chooseamong the singletons of P the cut vertices. There are at most 2 k possibilities. The other parts of P indicate thevertex set of blocks after removing cut vertices. We add k new dummy parts to P representing the possibleblocks that may contain only cut vertices. Now, we have at most 2 k parts in P . Observe that any forest betweenthe parts of P that represent the cut vertices and those that represent the blocks induces one way of decomposingthe k vertices into blocks and cut vertices. A dummy part adjacent to the singletons containing the vertices w , w , . . . , w ‘ indicate that { w , w , . . . , w ‘ } forms a block. By Cayley’s formula [8], the number of forests on r labeled vertices is ( r + 1) r − . So, there are at most (2 k + 1) k − ways. Hence, the number of ways of partitioninga set of k vertices into blocks and cut vertices is at most 2 O ( k log k ) .For Conditions 4.b and 4.c, there are 3 possibilities for each active block of Aux( X, t ). Indeed, since G [ X ] is S -bipartite and by Lemma 15, if a block is not bipartite, then it cannot contain vertices in S . Thus, there are atmost 3 k possibilities for Condition 4.b and 4.c.For Condition 5, there are 6 possibilities for each edge uv of Aux( X, t ): 3 possibilities for the parity of pathsbetween the endpoints of M uv and two for the existence of a vertex in S in M uv . Since Aux( X, t ) is a forestwith at most 3 k vertices, we have at most 6 k − possibilities for Condition 5.It remains to upper bound the number of possibilities for Condition 6 on the parities of the paths betweenthe vertices in B t . Let u and v be two vertices in X ∩ B t . If there is no path between u and v , then we can seethis in Aux( X, t ) as it implies that there is no path between the active vertices associated with u and v .Assume that u and v are connected in G [ X ]. Let P uv be a path between u and v in G [ X ]. Let B u be theblock of G [ X ] that contains u and its neighbor in P uv . Similarly, let B v be the block that contains v and itsneighbor P uv . Observe that we can have B u = B v if and only if u and v are in the same block. By construction,there exists a unique path P between B u and B v in Aux( X, t ) (this path can have length 0 if B u = B v ). If thereexists a block B in P that is not bipartite, then by Fact 16, we deduce that there exist an odd path and an evenpath between u and v . If such a non-bipartite block B exists, then either B = B u = B v or there exists an edge uv used by P such that M uv contains B . If B = B u = B v , then B is an active block of Inc( X ) since it contains at least two vertices in B t . In this case, Condition 4.b stores the information that B is not bipartite. Otherwise,if there is an edge uv used by P such that M uv contains B , then M uv is not bipartite and Condition 5 storesthis information.Suppose now that every block in P is bipartite. Let H be the subgraph that is the union of the bipartiteblocks in G [ X ], and let ( X , X ) be a bipartition of H . By assumption the union of the blocks in P is a subgraphof H . Thus, every path between u and v has the same parity and this only depends on whether u and v belongto the same part of ( X ∩ B t , X ∩ B t ). We deduce that the number of possibilities for Condition 6 are at most2 k . For each condition on ≡ t , we proved that the number of possibilities are 2 O ( k ) or 2 O ( k log k ) . Since k = B t ≤ tw + 4, we conclude that ≡ t has at most 2 O ( tw log tw ) equivalence classes. (cid:74)(cid:66) Claim 19.
Let t be a node of T and X, Y be two partial solutions associated with t . If X ≡ t Y , then, forevery Z ⊆ V ( G t ), the graph G [ X ∪ Z ] is S -bipartite if and only if G [ Y ∪ Z ] is S -bipartite. Proof.
Assume that X ≡ t Y and let Z ⊆ V ( G t ) such that G [ Y ∪ Z ] is S -bipartite. Let ϕ be the isomorphismfrom Aux( X, t ) to Aux(
Y, t ) given by ≡ t . We show that G [ X ∪ Z ] is S -bipartite. This will prove the claimbecause ≡ t is an equivalence relation. To prove that G [ X ∪ Z ] is S -bipartite, by Claim 15, it is sufficient toprove that every block B of G [ X ∪ Z ] is S -bipartite.Let G X := G [ X ] , G Y := G [ Y ], and G Z := G [ Z ∪ ( B t ∩ X )]. We observe that G [ X ∪ Z ] := ( G X , ( B t ∩ X )) ⊕ ( G Z , ( B t ∩ X )) and G [ Y ∪ Z ] := ( G Y , ( B t ∩ X )) ⊕ ( G Z , ( B t ∩ X )).Let B be a block of G [ X ∪ Z ]. Observe that if B is a block of G [ X ], then it is S -bipartite because X is apartial solution. Moreover, if B is a block of G [ Z ], then it is S -bipartite because G [ Y ∪ Z ] is S -bipartite.In the following, we assume that B contains vertices from X and Z . Consequently, B has at least 2 verticesin B t . Observe that for every block B of G X or G Z , either | V ( B ) ∩ V ( B ) | ≤
1, or B is fully contained in B .Thus, all the blocks of G X or G Z contained in B are in Aux p ( X, t ) or Aux p ( Z ∪ ( X ∩ B t ) , t ).We will take a corresponding 2-connected subgraph in G [ Y ∪ Z ]. Let B X be the set that contains the blocksand the cut vertices of Aux( X, t ) contained in B . We take the subset Y B of Y that contains (1) ϕ ( v ) for everycut vertex v in B Y , (2) V ( ϕ ( B )) for every blocks B in B X and (3) V ( M ϕ ( u ) ϕ ( v ) ) for every edge uv of Aux( X, t )with u, v ∈ B X . Let F := G [ Y B ∪ ( V ( B ) ∩ Z )]. Since all the blocks in Aux( X, t ) with vertices in S are active,Condition 4.a of ≡ t guarantees that V ( B ) ∩ B t = V ( F ) ∩ B t .We claim that F is a 2-connected induced subgraph of G [ Y ∪ Z ] such that F contains a vertex of S if and only if B contains a vertex of S , F is bipartite if and only if B is bipartite.By Lemma 15, this will imply that B is S -bipartite, because F is a subgraph of the S -bipartite graph G [ Y ∪ Z ].Let F Y := F ∩ G Y and F Z := F ∩ G Z .(1) ( F is 2-connected.) It is not difficult to see that F is connected from the construction of Y B and because Y ≡ t X implies that (1) V ( B ) ∩ B t = V ( ϕ ( B )) ∩ B t for every active block of G [ X ] and (2) two blocks B , b B ofAux( X, t ) are connected if and only if ϕ ( B ) and ϕ ( b B ) are connected in Aux( Y, t ).Assume towards a contradiction that F has a cut vertex c . Let a , a be the neighbors of c that are containedin distinct components of F − c . For each i ∈ { , } , let U i be a block of G Y or G Z containing a i c . Note that U and U are fully contained in F , because every block of G Y or G Z containing two vertices of F is containedin F . Therefore, U i appears in Aux p ( Y, t ) if it is a block of G Y and it appears in Aux p ( V ( G Z ) , t ) otherwise.Now, we choose U and U in Aux( Y, t ) and Aux( V ( G Z ) , t ) related to U and U , respectively.(Case 1. U , U are in the same part of Aux p ( Y, t ) or Aux p ( V ( G Z ) , t ).)Let us assume that both U and U are in Aux p ( Y, t ). A similar argument holds for the other case. As c isthe intersection of U and U which are blocks of G Y , c is a cut vertex of G Y , and it appears in Aux p ( Y, t ).Following the path of Aux p ( Y, t ) with direction from c to U i , we choose the first vertex U i in Aux( Y, t ).(Case 2. U , U are not in the same part of Aux p ( Y, t ) or Aux p ( V ( G Z ) , t ).)Without loss of generality, we assume that U is in Aux p ( Y, t ) and U is in Aux p ( V ( G Z ) , t ). We explain howto choose U . The symmetric argument is applied to U . In this case, U and U share c , and therefore,either c is an active cut vertex in G Y , or U is an active block of G Y . In the former case, following the pathof Aux p ( Y, t ) with direction from c to U , we choose the first vertex U in Aux( Y, t ). In the latter case, weset U := U . . Bergougnoux, É. Bonnet, N. Brettell, and O. Kwon 21 For each i ∈ { , } , if U i is block in Aux( Y, t ), let V i = ϕ − ( U i ), otherwise, if U i is a block in Aux( V ( G Z ) , t ),let V i = U i . Because of the construction, V and V are blocks contained in B . We choose a vertex c X in B corresponding to c in F . If c ∈ B t ∪ Z , then we set c X = c . Otherwise, c ∈ Y \ B t and c must be a cut vertex in G Y and either (A) c is a vertex of Aux( Y, t ) with neighbors U and U or (B) U U is an edge of Aux( Y, t ) and c is a vertex of M U U . If (A) holds, then we set c X = ϕ − ( c ) and if (B) holds, then we take c X a cut vertex in M ϕ − ( U ) ϕ − ( U ) (every M uv admits at least one cut vertex by definition).Since B is 2-connected, there is a path p p · · · p m from V − c X to V − c X in B − c X . This provides asequence B , B , . . . , B m of blocks that appear in Aux( X, t ) or Aux( V ( G Z ) , t ) such that B = V and B m = V and for every i ≤ m −
1, either B i B i +1 is an edge in Aux( X, t ) or in Aux( V ( G Z ) , t ), or B i and B i +1 arecontained in distinct parts of G X and G Z and they share a vertex in B t .For every i ≤ m , let b B i be B i if B i is a block in Aux( V ( G Z ) , t ) or ϕ ( B i ) if B i is a block in Aux( X, t ). Let i ≤ m −
1. If B i and B i +1 is an edge of Aux( X, t ), then ϕ ( B i ) ϕ ( B i +1 ) = b B i b B i +1 is an edge in Aux( Y, t ). Now,suppose that B i and B i +1 are contained in distinct parts of G X and G Z and assume w.l.o.g. that U i is a blockin G Y . By Condition 4.a in the definition of ≡ t , we have V ( B i ) ∩ B t = V ( b B i ) ∩ B t . We deduce that b B i and B i +1 = b B i +1 share a vertex in B t . From the sequence b B , . . . , b B m , we conclude that there exists a path in F between U − c and U − c in F − c . This contradicts the assumption that c is a cut vertex.(2) ( F contains a vertex of S if and only if B contains a vertex of S .) Observe that for each block U ofAux( Y, t ), U contains a vertex of S if and only if ϕ ( U ) contains a vertex of S , and for every edge uv of Aux( Y, t ), M uv has a vertex of S if and only if M ϕ ( u ) ϕ ( v ) has a vertex of S . Since F ∩ G Z = B ∩ G Z , we obtain the result.(3) ( F is bipartite if and only if B is bipartite.) Suppose that B is bipartite. We take the bipartition ( L, R )of B . As a connected bipartite graph has a unique bipartition, this is unique up to changing L and R .As F ∩ G Z = B ∩ G Z , this gives a bipartition ( L , R ) of F ∩ G Z . Let u, v ∈ V ( F ) ∩ B t be vertices that arecontained in the same connected component of F ∩ G Y . Assume u, v are contained in the same part of L and R . Since u, v are also contained in the same connected component of B ∩ G X and they are in the same part of L and R , all the paths from u to v in B ∩ G X have even length. Note that the blocks containing edges of thepath from u to v are all contained in B , and those blocks appear in Aux p ( X, t ). As B is bipartite, each of theseblocks is bipartite.Since Aux( Y, t ) is isomorphic to Aux(
X, t ), there is a corresponding sequence of blocks whose last blockscontain u and v , respectively, and all these blocks are bipartite. By the last condition of the equivalence relation ≡ t , there is an even path from u to v in G Y . This shows that a bipartition of F ∩ G Y is compatible with thebipartition ( L , R ) of F ∩ G Z . Thus, F is bipartite. (cid:74) We are now ready to describe our algorithm. For each node t of T and I ⊆ B t , let P [ t, I ] be the set of allpartial solutions X of G t where X ∩ B t = I . A reduced set R [ t, I ] is a subset of P [ t, I ] satisfying thatfor every partial solution X ∈ P [ t, I ], there exists X ∈ R [ t, I ] where X ≡ t X and w ( X ) ≥ w ( X ), andno two partial solutions in R [ t, I ] are equivalent.We will recursively compute a reduced set R [ t, I ] for every node t of T and I ⊆ B t . Claim 18 guarantees that | S I ⊆ B t R [ t, I ] | = 2 O ( tw log tw ) .We describe how to compute a reduced set R [ t, I ] depending on the type of the node t , and prove thecorrectness and the running time of each procedure. We fix a node t and I ⊆ B t . For each leaf node t and I = ∅ ,we assign R [ t, I ] := ∅ . For A ⊆ V ( G t ) , we define reduce t ( A ) as the operation which removes the elements of A that does not induce S -bipartite graph and then returns a set that contains, for each equivalence class C of ≡ t over A , a partial solution of C of maximum weight. t is an introduce node with child t and B t \ B t = { v } : If v / ∈ I , then it is easy to see that R [ t , I ] is a reduced set of P [ t, I ] = P [ t , I ]. In this case, we take R [ t, I ] = R [ t , I ].Assume now that v ∈ I . We set R [ t, I ] = reduce t ( A ) with A the set that contains X ∪ { v } for every X ∈ R [ t , I \ { v } ].We claim that R [ t, I ] is a reduced set of P [ t, I ]. Let X ∈ P [ t, I ]. As v ∈ I , we have that X \ { v } ∈P [ t , I \ { v } ]. Since R [ t , I \ { v } ] is a reduced set of P [ t , I \ { v } ], there exists X ∈ P [ t , I \ { v } ] such that X ≡ t X \ { v } and w ( X ) ≥ w ( X \ { v } ). By the construction of R [ t, I ], we added b X to R [ t, I ] where b X ≡ t X ∪ { v } and w ( b X ) ≥ w ( X ∪ { v } ). It is not difficult to check that b X ≡ t X by considering G t and thesum ( G t , B t \ { v } ) ⊕ ( G [ B t ] , B t \ { v } ) and applying Claim 21. t is a forget node with child t and B t \ B t = { v } : We set R [ t, I ] = reduce t ( R [ t , I ] ∪ R [ t , I ∪ { v } ]). We easily deduce that R [ t, I ] is a reduced set of P [ t, I ]from the fact that, by definition, P [ t, I ] = P [ t , I ] ∪ P [ t , I ∪ { v } ]. t is a join node with two children t and t : We set R [ t, I ] = reduce t ( A ) where A is the set that contains X ∪ X for every X ∈ R [ t , I ] and X ∈ R [ t , I ].We claim that R [ t, I ] is a reduced set of P [ t, I ]. Let X ∈ P [ t, I ]. For each i ∈ { , } , let X i := X ∩ V ( G t i ).It is not difficult to see that X i ∈ P [ t i , I ], as it is a partial solution of G t i .Since R [ t i , I ] is a reduced set of P [ t i , I ], there exists X i ∈ R [ t i , I ] such that X i ≡ t X i and w ( X i ) ≥ w ( X i ).By construction, X ∪ X ∈ A , and there exists X ∈ R [ t, I ] such that X ≡ t X ∪ X and w ( X ) ≥ w ( X ∪ X ).It is not difficult to check that X ≡ t X by considering G t and the sum ( G t , B t ) ⊕ ( G t , B t ) and applyingClaim 21 twice. As w ( X ∪ X ) ≥ w ( X ∪ X ) = w ( X ), we conclude that R [ t, I ] is a reduced set.It remains to prove the correctness and the running time of our algorithm. We can assume w.l.o.g. that thebag B r associated with the root r of T is empty. By definition P [ t, ∅ ] contains an optimal solution. Since R [ t, ∅ ]is a reduced set, we conclude that R [ t, ∅ ] also contains an optimal solution.For the running time, we use the fact that we can compute reduce t ( A ) in time |A| O ( tw log tw ) n . This followsfrom the upper bound of Claim 18 on the number of equivalence classes of ≡ t and the fact that for every partialsolutions X, Y of G t , we can decide whether X ≡ t Y in time O ( n ).Let t be a node of T that is not a leaf. Observe that we set R [ t, I ] = reduce t ( A ) where A is some sets thatdepends on the type of the node t . Let k be maximum size of a reduced set R [ t , I ] for t ∈ V ( T ) and I ⊆ B t .By construction, if t is an introduce node, then the size of A is at most k . If t is a forget node, then the size of A is at most 2 k . And if t is a join node, then the size of A is at most k . From Claim 18, we have k ≤ O ( tw log tw ) .We deduce that, for every node t and I ⊆ B t , we can compute R [ t, I ] in time 2 O ( tw log tw ) · n . Since T has atmost O ( n · tw ) nodes and there are at most 2 O ( tw ) possibilities for I , the running time of our algorithm is2 O ( tw log tw ) n . (cid:74) The dependency n on the input size n in Theorem 17 was obtained because we keep the partial solutionsthemselves, and have to check their equivalences. We believe that with a careful argument by keeping onlyauxiliary graphs Aux( X, t ), we can reduce the dependence of the input size; however, for simplicity, we presentthe running time with n factor.Now, we solve the other problems. (cid:73) Theorem 20.
Subset Feedback Vertex Set , Restricted Edge-Subset Feedback Edge Set , NodeMultiway Cut , and their weighted variants can be solved in time O ( tw log tw ) n on n -vertex graphs with treewidth tw . Proof.
It is easy to check that a graph has no S -traversing cycle if and only if every block of size at least 3 hasno vertex of S . So, for Subset FVS , we can simply adapt the algorithm for
Subset OCT , to ignore checksfor bipartiteness. A partial solution at a node t is a subset Y of V ( G t ) such that G t [ Y ] is a graph having no S -traversing cycles.We use the same auxiliary graph Aux( Y, t ), and use the equivalence relation obtained by removing theconditions for bipartiteness in the equivalence relation for
Subset OCT . More specifically, we say that twopartial solutions are equivalent for
Subset FVS if X ∩ B t = Y ∩ B t , and there is an isomorphism ϕ fromAux( X, t ) to Aux(
Y, t ) such that the following conditions are satisfied:For every vertex v in Aux( X, t ), v is active if and only if ϕ ( v ) is active.For every vertex v in Aux( X, t ), v is a block if and only if ϕ ( v ) is a block.For every active cut vertex v in Aux( X, t ), we have ϕ ( v ) = v .For every active block B in Aux( X, t ): V ( B ) ∩ B t = V ( ϕ ( B )) ∩ B t , and V ( B ) ∩ S = ∅ if and only if V ( ϕ ( B )) ∩ S = ∅ . . Bergougnoux, É. Bonnet, N. Brettell, and O. Kwon 23 For every edge uv in Aux( X, t ): V ( M uv ) ∩ S = ∅ if and only if V ( M ϕ ( u ) ϕ ( v ) ) ∩ S = ∅ .It is straightforward to adapt an algorithm for Subset OCT to Subset FVS with this equivalence relation.We conclude that
Subset FVS admits a 2 O ( tw log tw ) n -time algorithm.For the weighted variant of Node Multiway Cut , we use the reduction to
Subset FVS described inthe introduction. Given an instance (
G, S, k ) of
Weighted Node Multiway Cut with weight function w : V ( G ) → R , we construct an equivalent instance ( G , S , k ) of Weighted Subset FVS with tw ( G ) ≤ tw ( G ) + 1as follows. We add a new vertex v to G adjacent to all the vertices in S . Let G be the resulting graph and S = { v } . Since we only add a single vertex, tw ( G ) ≤ tw ( G ) + 1. For every vertex v in G , we set the weightof v to be “infinite” if v ∈ S ∪ { v } and w ( v ) otherwise. Consequently, a solution cannot contain vertices in S ∪ { v } . One can easily prove that ( G, S, k ) is a yes-instance of
Weighted Node Multiway Cut if and onlyif ( G , S , k ) is a yes-instance of Weighted Subset FVS . Hence,
Node Multiway Cut and its weightedvariant admit 2 O ( tw log tw ) n -time algorithms.It remains to prove that the theorem holds for Weighted Restricted Edge-Subset Feedback EdgeSet ( WSFES for short). For doing so, we use a reduction to
Weighted Subset FVS . Let (
G, S, k ) be ainstance of
WSFES with weight function w : E ( G ) → R . We will construct a instance ( G , S , k ) of WeightedSubset FVS where tw ( G ) = tw ( G ).Let G be the graph obtained by subdividing the edges of G . Since subdivisions do not increase the treewidth,we have tw ( G ) = tw ( G ). For every edge e of G , we call v e the vertex in G created from the subdivision of e . Let S be the set { v e | e ∈ S } . We give “infinite” weights to the vertices of G that belong to V ( G ) ∪ S . Moreover,for every edge e ∈ E ( G ) \ S , we give to v e the same weight as e . Consequently, a solution cannot contain verticesin V ( G ) ∪ S . It is easy to prove that ( G, S, k ) is a yes-instance of
WSFES if and only ( G , S , k ) is a yes-instanceof Weighted Subset FVS . We conclude that
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Semi-Regular k × k -Clique , namely that it is not easier than k × k -Clique . We insist that we do not need this result in the paper, but we believe that it may be helpful in adifferent context. We recall that by semi-regular , we mean that every vertex of a given column has the samedegree towards another fixed column.The known parameterized reduction from k -Clique on general graphs to k -Clique on regular graphs willnot work here, since it would increase the parameter polynomially (which we cannot afford). We take a couple ofsteps back and show how to partition the vertex set V ( G ) of a hard instance of Bounded-Degree -Coloring in such a way that the seminal reduction from 3 -Coloring to k × k -Clique only produces semi-regular instances.Let us recall that the latter reduction builds one vertex per 3-coloring of a part of the partition, and links by anedge every pair of consistent partial colorings (i.e., the union of the colorings is proper in the graph induced bythe two parts). If the number of parts k is chosen so that 3 | V ( G ) | /k ≈ k , we obtain a “square” k -by- k instance of Clique . Now we want to ensure, in addition, that each 3-coloring of a part P has the same number of consistent3-colorings of another part P .It is a folklore consequence of the Sparsification Lemma [19] and known reductions that 3-coloring a bounded-degree n -vertex graph cannot be done in 2 o ( n ) . For instance Cygan et al. show the following. (cid:73) Theorem 21 (Lemma 1 in [10]) . Unless the ETH fails, on n -vertex graphs of maximum degree 4cannot be solved in o ( n ) . Ultimately this result is a reduction from 3 -SAT . The instances produced by that reduction serveas our starting point. (cid:73)
Theorem 22.
Unless the ETH fails,
Semi-Regular k × k -Clique cannot be solved in time o ( k log k ) . Proof.
Let G be a hard instance of with maximum degree 4, produced by the reduction ofTheorem 21. G , the square of G , (with an edge between two vertices at distance at most 2) has degree 16.By Hajnal-Szemerédi Theorem, G admits an equitable coloring using 17 colors. This equitable coloring canfurther be found in polynomial time, by Kierstead and Kostochka [21]. We refine this 17 classes arbitrarily into k classes of equal size, such that d k log k e = | V ( G ) | . Let us call P the obtained partition of V ( G ). We performthe reduction of 3 -Coloring to k × k -Clique with this equipartition P . The resulting instance is semi-regular.By design, for every pair of parts P = P ∈ P , G [ P ∪ P ] consists of some isolated edges between P and P , andisolated vertices. Indeed an edge within P , or within P , or a degree-2 vertex would all contradict the coloring of G (with 17 colors). Thus any 3-coloring of P is consistent with the same number of 3-colorings of P . Thisshared number is 3 times the number of isolated vertices of G [ P ∪ P ] in P times twice the number of edges in G [ P ∪ P ].].