FPT and kernelization algorithms for the k-in-a-tree problem
Guilherme C. M. Gomes, Vinicius F. dos Santos, Murilo V. G. da Silva, Jayme L. Szwarcfiter
FFPT and kernelization algorithms for thek-in-a-tree problem
Guilherme C. M. Gomes , Vinicius F. dos Santos , Murilo V. G.da Silva , and Jayme L. Szwarcfiter Departamento de Ciência da Computação, Universidade Federal de Minas Gerais –Belo Horizonte, Brazil Departamento de Informática, Universidade Federal do Paraná – Curitiba, Brazil Universidade Federal do Rio de Janeiro – Rio de Janeiro, Brazil Universidade do Estado do Rio de Janeiro – Rio de Janeiro, Brazil
Abstract
The three-in-a-tree problem asks for an induced tree of the inputgraph containing three mandatory vertices. In 2006, Chudnovsky andSeymour [Combinatorica, 2010] presented the first polynomial time algo-rithm for this problem, which has become a critical subroutine in manyalgorithms for detecting induced subgraphs, such as beetles, pyramids,thetas, and even and odd-holes. In 2007, Derhy and Picouleau [DiscreteApplied Mathematics, 2009] considered the natural generalization to k mandatory vertices, proving that, when k is part of the input, the prob-lem is NP - complete , and ask what is the complexity of four-in-a-tree .Motivated by this question and the relevance of the original problem, westudy the parameterized complexity of k - in-a-tree . We begin by show-ing that the problem is W [1]- hard when jointly parameterized by the sizeof the solution and minimum clique cover and, under the ExponentialTime Hypothesis, does not admit an n o ( k ) time algorithm. Afterwards,we use Courcelle’s Theorem to prove fixed-parameter tractability undercliquewidth, which prompts our investigation into which parameteriza-tions admit single exponential algorithms; we show that such algorithmsexist for the unrelated parameterizations treewidth, distance to cluster,and distance to co-cluster. In terms of kernelization, we present a linearkernel under feedback edge set, and show that no polynomial kernel existsunder vertex cover nor distance to clique unless NP ⊆ coNP / poly . Alongwith other remarks and previous work, our tractability and kernelizationresults cover many of the most commonly employed parameters in thegraph parameter hierarchy. Given a graph G = ( V, E ) and a subset K ⊆ V ( G ) of size three – here calledthe set of terminal vertices – the three-in-a-tree problem consists of findingan induced tree of G that connects K . Despite the novelty of this problem, ithas become an important tool in many detection algorithms, where it usuallyaccounts for a significant part of the work performed during their executions.1 a r X i v : . [ c s . D S ] J u l t was first studied by Chudnovsky and Seymour [11] in the context of thetaand pyramid detection, the latter of which is a crucial part of perfect graphrecognition algorithms [12] and the former was an open question of interest [10].Across more than twenty pages, Chudnovsky and Seymour characterized allpairs ( G, K ) that do not admit a solution, which resulted in a O (cid:0) mn (cid:1) timealgorithm for the problem on n -vertex, m -edge graphs. Since then, three-in-a-tree has shown itself as a powerful tool, becoming a crucial subroutine for thefastest known even-hole [9], beetle [9], and odd-hole [14] detection algorithms;to the best of our knowledge, these algorithms often rely on reductions to mul-tiple instances of three-in-a-tree , e.g. the theta detection algorithm hasto solve O (cid:0) mn (cid:1) three-in-a-tree instances to produce its output [29]. De-spite its versatility, three-in-a-tree is not a silver bullet, and some authorsdiscuss quite extensively why they think three-in-a-tree cannot be used insome cases [13, 34]. Nevertheless, Lai et al. [29] very recently made a significantbreakthrough and managed to reduce the complexity of Chudnovsky and Sey-mour’s algorithm for three-in-a-tree to O (cid:0) m log n (cid:1) , effectively speeding upmany major detection algorithms, among other improvements to the number of three-in-a-tree instances required to solve some other detection problems.As pondered by Lai et al. [29], the usage of three-in-a-tree as a go-tosolution for detection problems may, at times, seem quite unnatural. In theaforementioned cases, one could try to tackle the problem by looking for con-stant sized minors or disjoint paths between terminal pairs and then resort toKawarabayashi et al.’s [28] quadratic algorithm to finalize the detection proce-dure. The problem is that neither the minors nor the disjoint paths are guar-anteed to be induced; to make the situation truly dire, this constraint makeseven the most basic problems NP - hard . For instance, Bienstock [1, 2] provedthat two-in-a-hole and three-in-a-path are NP - complete . As such, it isquite surprising that three-in-a-tree can be solved in polynomial time andbe of widespread importance. It is worth to note that the induced subgraphconstraint is also troublesome from the parameterized point of view. MaximumMatching , for instance, can be solved in polynomial time [22], but if we imposethat the matching must be induced subgraph, the problem becomes W [1]- hard when parameterized by the minimum number of edges in the matching [31].Naturally, we may wonder how far we may push for polynomial time algo-rithms when considering larger numbers of terminal vertices, i.e we are interestedin the complexity of k -in-a-tree for k ≥ . The first authors to examine thisproblem were Derhy and Picouleau, who proved in [18] that k -in-a-tree is NP - complete when the number of terminals is part of the input even on planarbipartite cubic graphs of girth four, but solvable in polynomial time if the girthof the graph is larger than the number terminals. A few years later, Derhy et al.[19] showed that four-in-a-tree is solvable in triangle-free graphs, while Liuand Trotignon [30] proved that so is k -in-a-tree on graphs of girth at least k ≥ ; their combined results imply that k -in-a-tree on graphs of girth atleast k is solvable in polynomial time. In terms of the k -in-a-path problem,Derhy and Picouleau [18] argued that their hardness reduction also applies tothis problem and showed that three-in-a-path is NP - complete even on graphsof maximum degree three. Fiala et al. [23] proved that k -in-a-path , k -InducedDisjoint Paths , and k -in-a-cycle can be solved in polynomial time on claw-free graphs for every fixed k , but all of them are NP - complete when k is part ofthe input even on line graphs; in fact, they proved that the previous problems2re in XP when parameterized by the number of terminals on claw-free graphs.Another related problem to k -in-a-tree is the well known Steiner Tree ,where we want to find a subtree of the input with cost at most w connecting allterminals. Being one of Karp’s 21 NP - hard problems [27], Steiner Tree hasreceived a lot of attention over the decades. Relevant to our discussion, however,is its parameterized complexity. When parameterized by the number of termi-nals, it admits a single exponential time algorithm [21]; the same was provento be true when treewidth [32] is the parameter [5]. On the other hand, whenparameterized by cliquewidth [16], it is paraNP - hard since it is NP - hard even oncliques: we may reduce from Steiner Tree itself and add, for each non-edgeof the input, an edge of cost w + 1 . As we see below, our first two results are incomplete contrast with the parameterized complexity of Steiner Tree . Our results.
We concern ourselves with the parameterized complexity of k -in-a-tree . We begin by presenting some algorithmic results for k -in-a-tree inSection 3, showing that the latter is W [1]- hard when simultaneously parameter-ized by the number of vertices in the solution and size of a minimum clique coverand, moreover, does not admit an n o ( k ) time algorithm unless the ExponentialTime Hypothesis [26] (ETH) fails. This partially answers a (generalization) ofDerhy and Picouleau’s question about the complexity of k -in-a-tree , in thesense that there is very little hope of obtaining an algorithm that runs in poly-nomial time only on the size of the input graph. On the positive side, we provetractability under cliquewidth using Courcelle’s Theorem [15], which promptsus, in Section 4, to turn our attention to which parameters allow us to devisesingle exponential time algorithms for k -in-a-tree . Using Bodlaender et al.’sdynamic programming optimization machinery [5], we show that such algo-rithms exist under treewidth, distance to cluster, and distance to co-cluster. InSection 5, we present a kernel with q vertices and q edges when we param-eterize k -in-a-tree by the size q of a minimum feedback edge set. In Section 6we prove that the problem does not admit a polynomial kernel when parameter-ized by bandwidth, nor when simultaneously parameterized by the size of thesolution, diameter, and distance to any graph class of your choosing. In particu-lar, the latter shows that k -in-a-tree does not admit a polynomial kernel whenparameterized by vertex cover nor when parameterized by distance to clique.All our negative kernelization results are obtained assuming NP (cid:42) coNP / poly .In terms of tractability and kernelization, our results encompass most of thecommonly employed parameters of Sorge and Weller’s graph parameter hier-archy [33]; we present a summary of our results in Figure 1. To see why thedistance to solution parameter sits between vertex cover and feedback vertexset, we refer to the end of Section 3. We refer the reader to [17] for basic background on parameterized complex-ity, and recall here only some basic definitions. A parameterized problem is alanguage L ⊆ Σ ∗ × N . For an instance I = ( x, q ) ∈ Σ ∗ × N , q is called the parameter . A parameterized problem is fixed-parameter tractable ( FPT ) if thereexists an algorithm A , a computable function f , and a constant c such thatgiven an instance ( x, q ) , A correctly decides whether I ∈ L in time bounded by f ( q ) · | I | c ; in this case, A is called an FPT algorithm . A fundamental concept3igure 1: Hasse diagram of graph parameters and associated results for k -in-a-tree . Parameters surrounded by shaded ellipses have both single exponentialtime algorithms and polynomial kernels. Solid boxes represent parameters underwhich the problem is FPT but does not admit polynomial kernels; if the box isshaded, we have a single exponential time algorithm for that parameterization.A single dashed box corresponds to a W [1]- hard parameterization, while doubledashed boxes surround parameters under which the problem is paraNP - hard .Aside from the paraNP - hardness for genus, maximum degree, and distance tobipartite, all results are original contributions proposed in this work.in parameterized complexity is that of kernelization ; see [24] for a recent bookon the topic. A kernelization algorithm, or just kernel , for a parameterizedproblem Π takes an instance ( x, q ) of the problem and, in time polynomial in | x | + q , outputs an instance ( x (cid:48) , q (cid:48) ) such that | x (cid:48) | , q (cid:48) (cid:54) g ( q ) for some function g ,and ( x, q ) ∈ Π if and only if ( x (cid:48) , q (cid:48) ) ∈ Π . Function g is called the size of thekernel and may be viewed as a measure of the “compressibility” of a problemusing polynomial-time pre-processing rules. A kernel is called polynomial (resp. quadratic, linear ) if g ( q ) is a polynomial (resp. quadratic, linear) function in q . A breakthrough result of Bodlaender et al. [3] gave the first framework forproving that some parameterized problems do not admit polynomial kernels, byestablishing so-called composition algorithms . Together with a result of Fort-now and Santhanam [25], this allows to exclude polynomial kernels under theassumption that NP (cid:42) coNP / poly , otherwise implying a collapse of the polyno-mial hierarchy to its third level [35].All graphs in this work are finite and simple. We use standard graph theorynotation and nomenclature for our parameters; for any undefined terminologyin graph theory we refer to [6]. We denote the degree of vertex v on graph G by deg G ( v ) , and the set of natural numbers { , , . . . , t } by [ t ] . A graph is a cluster graph if each of its connected components is a clique, while a co-clustergraph is the complement of a cluster graph. The distance to cluster ( co-cluster )of a graph G , is the size of the smallest set U ⊆ V ( G ) such that G \ U is acluster (co-cluster) graph. As defined in [8], a set U ⊆ V ( G ) is an F - modulator of G if G \ U belongs to the graph class F . When the context is clear, we omitthe qualifier F . For cluster and co-cluster graphs, one can decide if G admits a4odulator of size q in time FPT on q [7]. While it has been known for some time that k -in-a-tree is NP - complete evenon planar bipartite cubic graphs, it is not known to be even in XP when pa-rameterized by the natural parameter, the number of terminals. We take a firststep with a negative result about this parameterization, ruling out the existenceof an FPT algorithm unless
FPT = W [1] ; in fact, we show for stronger parame-terization: the maximum size of the induced tree that should contain the set of k terminal vertices K and the size of a minimum clique cover. Theorem 1. k -in-a-tree is W [1] - hard when simultaneously parameterized bythe number of vertices of the induced tree and size of a minimum clique cover.Moreover, unless ETH fails, there is no n o ( k ) time algorithm for k -in-a-tree. Proof.
Our reduction is from
Multicolored Independent Set parameter-ized by the number of color classes (cid:96) . Formally, let H be the input to our sourceproblem such that V ( H ) is partitioned into (cid:96) color classes { V , . . . , V (cid:96) } and each V i induces a clique on H . Our instance of k -in-a-tree ( G, K ) is such that K = { v , . . . , v (cid:96) } , V ( G ) = K ∪ V ( H ) , every edge of H is also in G , each v i ∈ K has N G ( v i ) = V i , and N ( v ) = V ( H ) , so k = (cid:96) + 1 .If I is a solution to Multicolored Independent Set , I ∪ K is a solutionto ( G, K ) : there are no cycles since I is also an independent set of G , v connectsall vertices of I , and each other terminal is connected to exactly one vertex of I . For the converse, if T is a solution to ( G, K ) , we claim that I = T \ K is anindependent set of size (cid:96) . To see that this is the case, note that: (i) T ∩ V ( H ) must be independent, otherwise they would form a triangle with v ; and (ii) | T ∩ V i | = 1 because each V i is a clique and the only way to connect v i to v isby picking at least one vertex of V i . Since | K | = (cid:96) + 1 we have that the solutionsize is at most k + 1 and, since Q i = V i ∪ { v i } is a clique, we can cover G withthe k + 1 cliques (cid:83) i ∈ [ k ] { Q i } . The second statement follows directly fromthe fact that Multicolored Independent Set has no n o ( k ) time algorithmunder ETH [17] and that our reduction exhibits a linear relation between theparameters of the source and destination problems.Since the natural parameters offer little to no hope of fixed-parameter tractabil-ity, to obtain parameterized algorithms we turn our attention to the broad classof structural parameters. Our first positive result is a direct application oftextbook MSO formulae. Theorem 2. k -in-a-tree parameterized by cliquewidth is in FPT .Proof.
Let ( G, K ) be the input to k -in-a-tree , R be the binary relation rep-resented by the formula ϕ R ( u, v, Y ) = u ∈ Y ∧ v ∈ Y ∧ ( e ( u, v ) ∨ e ( v, u )) , and T C [ R ; x, y ] be the reflexive and transitive closure of R . As such, conn ( Y ) = ∀ x, y ( x ∈ Y ∧ y ∈ Y ⇒ T C [ R ; x, y ]) is true if and only if G [ Y ] is con-nected [15]. Similarly, formula cycle ( X ) = ∃ x, y, z ∈ X ( x (cid:54) = y ∧ y (cid:54) = z ∧ x (cid:54) = z ∧ e ( x, z ) ∧ e ( y, z ) ∧∃ Y ⊂ X ( z / ∈ Y ∧ x ∈ Y ∧ y ∈ Y ∧ conn ( Y (cid:48) )) is true if and onlyif G [ X ] has a cycle [15]. Putting the previous two formulae together, formula indtree ( K ) = ∃ S ( K ⊆ S ∧ conn ( S ) ∧ ¬ cycle ( S )) is true if and only if there is5ome superset of K that is connected and acyclic. By Courcelle’s Theorem [15],if G has cliquewidth at most q , then one can determine in f ( q ) n O (1) time if G satisfies indtree ( K ) for some set K of terminals.Towards showing that the minimum number of vertices we must delete toobtain a solution sits between feedback vertex set and vertex cover in Figure 1,let S ⊂ V ( G ) be such that G \ S is a solution. First, note that S is a feedbackvertex set of G ; for the other inequality, take a vertex cover C of G and note thatplacing two vertices of G \ C with the same neighborhood in C either generatesa cycle in the solution or only one of them suffices – even if we have manyterminals, we need to keep only two of them – so | S | ≤ | C | + 2 | C | +1 . In termsof paraNP - hardness results, we can easily show that k -in-a-tree is paraNP - hard when parameterized by bisection width : to reduce from the problem to itself,we pick any terminal of the input and append to it a path with as many verticesas the original graph to obtain a graph with bisection width one. Similarly, whenparameterizing by the size of a minimum dominating set and again reducingfrom k -in-a-tree to itself, we add a new terminal adjacent to any vertex of K and a universal vertex, which can never be part of the solution since it forms atriangle with the new terminal and its neighbor. All results in this section rely on the rank based approach of Bodlaender et al. [5],which requires the additional definitions we give below. Let U be a finite set, Π( U ) denote the set of all partitions of U , and (cid:118) be the coarsening relationdefined on Π( U ) , i.e given two partitions p, q ∈ Π( U ) , p (cid:118) q if and only if eachblock of q is contained in some block of p . It is known that Π( U ) together with (cid:118) form a lattice, upon which we can define the usual join operator (cid:116) and meet operator (cid:117) [5]. The join operation p (cid:116) q outputs the unique partition z wheretwo elements are in the same block of z if and only if they are in the same blockof p or q . The result of the meet operation p (cid:117) q is the unique partition suchthat each block is formed by the non-empty intersection between a block of p and a block of q . Given a subset X ⊆ U and p ∈ Π( U ) , p ↓ X ∈ Π( X ) is thepartition obtained by removing all elements of U \ X from p , while, for Y ⊇ U , p ↑ Y ∈ Π( Y ) is the partition obtained by adding to p a singleton block for eachelement in Y \ U . For X ⊆ U , we shorthand by U [ X ] the partition where oneblock is precisely { X } and all other are the singletons of U \ X ; if X = { a, b } ,we use U [ ab ] .A set of weighted partitions of a ground set U is defined as A ⊆ Π( U ) × N .To speed up dynamic programming algorithms for connectivity problems, theidea is to only store a subset A (cid:48) ⊆ A that preserves the existence of at leastone optimal solution. Formally, for each possible extension q ∈ Π( U ) of thecurrent partitions of A to a valid solution, the optimum of A relative to q isdenoted by opt ( q, A ) = min { w | ( p, w ) ∈ A , p (cid:116) q = { U }} . A (cid:48) represents A if opt ( q, A (cid:48) ) = opt ( q, A ) for all q ∈ Π( U ) . The key result of Bodlaender et al. [5]is given by Theorem 3. The width of a bipartition ( A, B ) of V ( G ) is the number of edges between the parts. Thebisection width is the minimum width of all bipartitions of V ( G ) such that | A | ≤ | B | ≤ | A | +1 . heorem 3 (3.7 of [5]) . There exists an algorithm that, given A and U , com-putes A (cid:48) in time |A| ( ω − | U | | U | O (1) and |A (cid:48) | ≤ | U |− , where ω is the matrixmultiplication constant. A function f : 2 Π( U ) × N × Z (cid:55)→ Π( U ) × N is said to preserve representation if f ( A (cid:48) , z ) = f ( A , z ) for every A , A (cid:48) ∈ Π( U ) × N and z ∈ Z ; thus, if one candescribe a dynamic programming algorithm that uses only transition functionsthat preserve representation, it is possible to obtain A (cid:48) . In the following lemma,let rmc ( A ) = { ( p, w ) ∈ A | (cid:64) ( p, w (cid:48) ) ∈ A , w (cid:48) < w } . Lemma 4 (Proposition 3.3 and Lemma 3.6 of [5]) . Let U be a finite set and A ⊆ Π( U ) × N . The following functions preserve representation and can becomputed in |A| · |B| · | U | O (1) time. Union.
For
B ∈ Π( U ) × N , A (cid:93) B = rmc ( A ∪ B ) . Insert.
For X ∩ U = ∅ , ins ( X, A ) = { ( p ↑ X ∪ U , w ) | ( p, w ) ∈ A} . Shift.
For any integer w (cid:48) , shift ( w (cid:48) , A ) = { ( p, w + w (cid:48) ) | ( p, w ) ∈ A} . Glue.
Let ˆ U = U ∪ X , then glue ( X, A ) = rmc (cid:16)(cid:110) ( ˆ U [ X ] (cid:116) p ↑ ˆ U , w ) | ( p, w ) ∈ A (cid:111)(cid:17) . Project. proj ( X, A ) = rmc (cid:16) { ( p ↓ X , w ) | ( p, w ) ∈ A , ∀ u ∈ X : ∃ v ∈ X : p (cid:118) U [ uv ] } (cid:17) ,if X ⊆ U . Join. If ˆ U = U ∪ U (cid:48) , A ⊆ Π( U ) × N and B ∈ Π( U (cid:48) ) × N , then join ( A , B ) = rmc (cid:16) { ( p ↑ ˆ U (cid:116) q ↑ ˆ U , w + w (cid:48) ) | ( p, w ) ∈ A , ( q, w (cid:48) ) ∈ B} (cid:17) . Even though our problem is unweighted, we found it convenient to solvea weighted version and of k -in-a-tree . We state this slightly more generalproblem below. Light Connecting Induced Subgraph
Instance : A graph G , a set of k terminals K ⊆ V ( G ) , and two integers (cid:96), f . Question : Is there a connected induced subgraph of G on (cid:96) + k vertices and atmost f edges that contains K ?Note that an instance ( G, K ) of k -in-a-tree is positive if and only if thereis some integer (cid:96) where the Light Connecting Induced Subgraph instance ( G, K, (cid:96), (cid:96) + k − is positive. Our goal is to use the number of edges in thesolution to Light Connecting Induced Subgraph as the cost of a partialsolution in a dynamic programming algorithm. This shall be particularly usefulfor join nodes during our treewidth algorithm, as we may resort to the optimalityof the solution to guarantee that the resulting induced subgraph of a (cid:116) operationis acyclic. A tree decomposition of a graph G is a pair T = ( T, B = { B j | j ∈ V ( T ) } ) , where T is a tree and B ⊆ V ( G ) is a family where: (cid:83) B j ∈B B j = V ( G ) ; for every edge uv ∈ E ( G ) there is some B j such that { u, v } ⊆ B j ; for every i, j, q ∈ V ( T ) , if q is in the path between i and j in T , then B i ∩ B j ⊆ B q . Each B j ∈ B is calleda bag of the tree decomposition. G has treewidth has most t if it admits a tree7ecomposition such that no bag has more than t vertices. For further propertiesof treewidth, we refer to [32]. After rooting T , G x denotes the subgraph of G induced by the vertices contained in any bag that belongs to the subtree of T rooted at bag x . An algorithmically useful property of tree decompositions isthe existence of a nice tree decomposition that does not increase the treewidthof G . Definition 5 (Nice tree decomposition) . A tree decomposition T of G is saidto be nice if its tree is rooted at, say, the empty bag r ( T ) and each of its bagsis from one of the following four types:1. Leaf node : a leaf x of T with B x = ∅ .2. Introduce vertex node : an inner bag x of T with one child y such that B x \ B y = { u } .3. Forget node : an inner bag x of T with one child y such that B y \ B x = { u } .4. Join node : an inner bag x of T with two children y, z such that B x = B y = B z . Theorem 6.
There is an algorithm for
Light Connecting Induced Sub-graph that, given a nice tree decomposition of width t of the n -vertex inputgraph G rooted at the forget node for some terminal r ∈ K , runs in time O ( t ) n O (1) .Proof. Let ( G, K, (cid:96), f ) be an instance of Light Connecting Induced Sub-graph . For each bag x , we compute the table g x ( S, (cid:96) (cid:48) ) ⊆ Π( S ) × N , where S ⊆ B x contains the vertices of B x that must be present in a solution and (cid:96) (cid:48) isthe number of vertices we allow in the induced subgraphs of G x ; each weightedpartition ( p, w ) ∈ g x ( S, (cid:96) (cid:48) ) corresponds to a choice of an induced subgraph of G x with (cid:96) (cid:48) vertices, w + | E ( G [ S ]) | edges, and connected components given bythe blocks of p . If | S | > (cid:96) (cid:48) , we define g x ( S, (cid:96) (cid:48) ) = ∅ . After every operation, weapply the algorithm of Theorem 3. Leaf node.
Since B x = ∅ , the only possible connecting induced subgraph isprecisely the empty graph, so we define: g x ( ∅ , (cid:96) (cid:48) ) = (cid:40) { ( ∅ , } , if (cid:96) (cid:48) = 0 ; ∅ , otherwise. Introduce node.
Let y be the child bag of x and B x \ B y = { v } . We compute g x ( S, (cid:96) (cid:48) ) as follows, where A y ( S, (cid:96) (cid:48) , v ) = ins ( { v } , g y ( S \ { v } , (cid:96) (cid:48) − : g x ( S, (cid:96) (cid:48) ) = glue ( N [ v ] ∩ S, A y ( S, (cid:96) (cid:48) , v )) , if v ∈ S ; g y ( S, (cid:96) (cid:48) ) , if v / ∈ S ∪ K . ∅ , otherwise.On the third case above, if v ∈ K but v / ∈ S , we are not including a terminalinto the induced subgraph, thus we cannot accept any partition with support S as valid. If v / ∈ K ∪ S , then no changes are necessary, since the only vertexof G x not in G y is not considered for the solution. Finally, for the first case,since v ∈ S , we must extend each partition of g y ( S \ { v } , (cid:96) (cid:48) − to the groundset S (which we achieve by the insert operation); however, since we are looking8or an induced subgraph, we must use every edge between v and the neighborsof v in S , merging the connected components containing them. Unlike in someconnectivity problems such as Steiner Tree , we only count the edges within S while processing forget nodes; this shall simplify the join operation considerably,as we do not need to worry about repeatedly counting edges inside the currentbag. Forget node.
Let y be the child bag of x and v be the forgotten vertex. Thetransition is directly computed by: g x ( S, (cid:96) (cid:48) ) = g y ( S, (cid:96) (cid:48) ) (cid:93) shift ( | N ( v ) ∩ S | , proj ( { v } , g y ( S ∪ { v } , (cid:96) (cid:48) ))) If v is not used in a partial solution, g y ( S, (cid:96) (cid:48) ) already correctly contains allthe partial solutions where v is not used; on the other hand, if v was used in somesolution, we must eliminate v from the partitions where it appears; however, weonly keep the partitions that do not lose a block, otherwise we would have aconnected component (represented by v ) that shall never be connected to theremainder of the subgraph and, thus, cannot be extended to a valid solution. Join node. If y, z are the children of bag x , we compose its table by theequation: g x ( S, (cid:96) (cid:48) ) = (cid:93) (cid:96) + (cid:96) = (cid:96) (cid:48) + | S || S | ≤ (cid:96) ,(cid:96) join ( g y ( S, (cid:96) ) , g z ( S, (cid:96) )) Where the union operator runs over all integer values satisfying the sys-tem (cid:96) + (cid:96) = (cid:96) (cid:48) + | S | , | S | ≤ (cid:96) , (cid:96) ; since we do not know how many verticeswere used on the partial solutions of each subtree, we must try every combi-nation to obtain all the partial solutions for the subtree rooted at x . Since weforce the vertices in S to be present in the solutions to the subtree rooted atbag x , combining two partial solutions, represented by ( p, w ) ∈ g y ( S, (cid:96) ) and ( q, w (cid:48) ) ∈ g y ( S, (cid:96) ) , corresponds to uniting the set of edges of the respective par-tial solutions G ( p ) , G ( q ) , which results in a merger of connected components.Since the edges of S have not been counted towards the weights w, w (cid:48) , G ( p (cid:116) q ) has exactly w + w (cid:48) + | E ( G [ S ]) | edges. This is precisely the definition of the join operation.In order to obtain the answer to the problem, we look at the child x of theforget node for terminal r , that is, the child of the root of the tree, and checkif g x ( { r } , (cid:96) + | K | ) (cid:54) = ∅ . In the affirmative, note that there is only one entry ( , w ) ∈ g ( { r } , (cid:96) + | K | ) and that the graph that connects all the terminalsusing (cid:96) + | K | vertices has exactly w edges, since E ( G [ { r } ])0 ∅ .For an introduce bag x with child y , the time taken to compute all entries of g x is of the order of (cid:96) (cid:80) | B x | i =0 (cid:0) | B x | i (cid:1) ωi t O (1) ≤ n (1 + 2 ωi ) t t O (1) ; the term ωi comesfrom the time needed to execute the algorithm of Theorem 3 upon a initial setof size i . For join nodes, the intermediate (cid:116) operation may yield a set of size i ,so we have that the tables can be computed in time (cid:96) (cid:80) | B x | i =0 (cid:0) | B x | i (cid:1) ( ω +1) i t O (1) ≤ (1 + 2 ( ω +1) ) t n O (1) . Corollary 7.
There is an algorithm for k -in-a-tree that, given a nice treedecomposition of width t of the n -vertex input graph G rooted at the forget nodefor some terminal r ∈ K , runs in time O ( t ) n O (1) . .2 Distance to cluster We now show that parameterizing by the distance to cluster q also yields an FPT algorithm. Throughout this section, G is the input graph, U is the clustermodulator, and C = { C , . . . , C r } are the maximal cliques of G \ U . We alsouse the framework developed by Bodlaender et al. [5] to optimize our dynamicprogramming algorithm. Theorem 8.
There is an algorithm for k -in-a-tree that runs in time O ( q ) n O (1) on graphs with distance to cluster at most q graphs.Proof. Suppose we are given the instance ( G, K ) and the q -vertex cluster mod-ulator U ⊆ V ( G ) . We begin by guessing a subset K ∩ U ⊆ S ⊆ U of verticesthat will be present in a solution for the problem. Now, given S , we execute thefollowing pre-processing step: for each clique C i ∈ C , we discard all but one ver-tex of each maximal set of true twins; this way, we limit the size of C ∗ i = C i \ K to q .An entry of our dynamic programming table f S ( i, c, (cid:96) ) ⊆ Π( S ) × N is a set ofpartial solutions of G i = G [ K ∪ S (cid:83) ij =1 C j ] , each of which uses exactly c verticesof C ∗ i and induces a subgraph of G i on (cid:96) vertices. Note that we cannot usemore than two vertices of each clique, so we only consider c ∈ { , , } . In each ( p, w ) ∈ f S ( i, c, (cid:96) ) , each block of p corresponds to the vertices of S that lie inthe same connected component of G i , and w is the number of edges used in therespective induced subgraph. Our transition is given by the following equation,where W ( S, X ) = | N ( X ) ∩ ( S ∪ K ) | + | E ( G [ X ]) | ; if either c + | K ∩ C i | > , c > (cid:96) , or there is some vertex in K ∩ C i that has no neighbor in S and c = 0 ,we define f S ( i, c, (cid:96) ) = ∅ . f S ( i, c, (cid:96) ) = (cid:93) j =0 (cid:93) X ∈ ( C ∗ ic ) glue W ( S,X ) ( N S ( X ) , f S ( i − , j, (cid:96) − c )) The above defines f S for all ( i, c, (cid:96) ) ∈ [ r ] × { , , } × [ n ] ; we extend f S to include the base case f S (0 , , | S ∪ K | ) = { ( p ( S, K ) , E ( G [ S ∪ K ]) } , where p ( S, K ) ∈ Π( S ) is the partition obtained by gluing together the connected com-ponents of G [ S ] which have a common neighbor in K ; for all other entries, f S ( r + 1 , c, (cid:96) ) = ∅ . Our goal now is to show that there is a solution to ourproblem using the vertices of S if and only if ( { S } , w ) ∈ f S ( r, c, (cid:96) ) , for somepair (cid:96) ≥ | S ∪ K | , c ≥ | K ∩ C | , such that w = (cid:96) − . To do so, we first provethat f ( i, c, (cid:96) ) contains all partitions of S that represent all possible induced sub-graphs of G i on (cid:96) vertices that use c vertices of C ∗ i , and that use as few edgesas possible.By induction, suppose that this holds for every entry f S ( i − , a, b ) . If c + | K ∩ C i | > or c > (cid:96) , either we want to use more than two vertices of C i ,which certainly implies that there is a copy of K in the solution, or we wantto use more than (cid:96) vertices of C ∗ i , which is equally impossible, so f S ( i, c, (cid:96) ) = ∅ . If c = 0 , we have that a weighted partition ( p, w ) is valid in G i if andonly if it is valid in G i − , since we use no vertices in V ( G i ) \ V ( G i − ; thus, ( p, w ) ∈ f S ( i, , (cid:96) ) if and only if ( p, w ) ∈ f S ( i − , j, (cid:96) ) for some j ∈ { , , } .Otherwise, suppose we want to add a subset of vertices X ⊆ C ∗ i to a partialsolution H , which is represented by ( p, w ) ∈ f S ( i − , a, (cid:96) − | X | ) . In this case,since N ( X ) \ C i ⊆ U and U ∩ V ( H ) ⊆ S , we have that X can only reduce the10umber of connected components of H if N ( X ) intersects two distinct blocks of p . Thus, the connected components of G [ V ( H ) ∪ X ] are represented, precisely,by glue ( N S ( X ) , p ) and the only new edges used are those between X and S ∪ K ,and the ones internal to X ; this is precisely the shift accounted by W ( S, X ) . Theminimality of w for an entry ( p, w ) ∈ f ( i, c, (cid:96) ) is guaranteed by the rmc operationimbued in both glue and (cid:93) . Note that if there is some entry ( p, w ) ∈ f S ( r, c, (cid:96) ) for some c such that p = { S } and w = (cid:96) − , this means that there is aninduced subgraph of graph that has S contained in a connected component,uses (cid:96) vertices and (cid:96) − edges, and thus, must be an induced tree of G . Thatit connects all vertices of K follows from the fact that f S (0 , , | S ∪ K | ) contains p ( S, K ) and, in every clique C i that contains a vertex of K with no neighbor in S , we force that at least one vertex of C i must be picked in a solution.In terms of complexity, after every glue or (cid:93) operation, we apply the algo-rithm of Theorem 3. For each tuple ( S, i, c, (cid:96), j ) , we do so up to | C ∗ i | ≤ q times per tuple, which implies in a time requirement time of the order of q · q − ( ω − q q O (1) ≤ ( ω +3) q q O (1) . Since we have q n O (1) tuples, our al-gorithm runs in ( ω +4) q n O (1) time. We can use the result on distance to cluster to solve the problem on graphs withdistance to co-cluster at most q without much effort, as we see in the followingproposition. Theorem 9.
There is an algorithm for k -in-a-tree that runs in time O ( q ) n O (1) where q is the distance to co-cluster.Proof. Suppose we are given a co-cluster modulator U of the input graph G andlet I = { I , . . . , I r } be the family of independent sets of G − U . Since G − U is acomplete multipartite graph, if we pick vertices of three distinct elements of I ,we will form a K in the induced subgraph. Moreover, for each pair I i , I j ∈ I ,at most one of them may have more than one vertex in any solution, the inducedsubgraph would contain a C . This implies that K can intersect V ( G − U ) in atmost three vertices and at most two independent sets. If this intersection hassize three, for each K ∩ U ⊆ S ⊆ U , we can easily verify in polynomial time if G [ S ∪ K ] is a tree. Otherwise, for each pair I i , I j ∈ I such that K ⊆ U ∪ I i ∪ I j ,we guess which one of them will have more than one vertex in the solution, say I i , and which vertex v ∈ I i will be in the solution, with the restriction that K ⊆ U ∪ I j ∪ { v } . Now, the graph G (cid:48) = G [ U ∪ I j ∪ { v } ] has a cluster modulator U ∪ { v } , and we can apply the algorithm of Theorem 8 on it to decide if there isan induced tree of G (cid:48) connecting K . It follows from the observations that thereis a valid induced subtree of G if and only if for some choice I i , I j and v ∈ I i G (cid:48) has one such induced subgraph. In this section, we prove that k -in-a-tree admits a linear kernel when parame-terized by the size q of a minimum feedback edge set. Throughout this section,we denote our input graph by G , the set of terminals by K , and the tree ob-tained by removing the edges of a minimum size feedback edge set F by T ( F ) .11ote that, if G is connected and F is of minimum size, G \ F is a tree; wemay safely assume the first, otherwise we either have that ( G, K ) is a negativeinstance if K is spread across multiple connected components of G , or theremust be some edge of F that merges two connected components of G \ F anddoes not create a cycle, contradicting the minimality of F . The kernelizationalgorithm we describe works in two steps: it first finds a feedback edge set F that minimizes the number of edges incident to vertices of degree two in T ( F ) ,then compresses long induced paths of G . We denote the set of leaves of a tree H by leaves ( H ) . Reduction Rule 1. If G has a vertex v of degree one, remove v and, if v ∈ K ,add the unique neighbor of v in G to K .Proof of safeness of Rule 1. Safeness follows from the fact that a degree onevertex is in the solution if and only if its unique neighbor also is.
Observation 10.
After exhaustively applying Rule 1, for every minimum feed-back edge set F of G , T ( F ) has at most q leaves. Moreover, T ( F ) has at mostas many vertices of degree at least three as leaves. We begin with any minimum feedback edge set F of G . We partition T ( F ) \ leaves ( T ( F )) into ( D , D ∗ ) according to the degree of the vertices of G in T ( F ) : v ∈ D if and only if deg T ( F ) ( v ) = 2 . For u, v, f ∈ V ( G ) , we say that u F -links v to f if v = u or if T ( F ) \ { u } has no v − f path. We say that vertices u, f are an F -pair if the set of internal vertices of the unique u − f path P F ( u, f ) of T ( F ) is entirely contained in D ; we denote the set of internal vertices by P ∗ F ( u, f ) . Reduction Rule 2.
Let u, f, w , w ∈ V ( G ) be such that u, f form an F -pair, w (cid:54) = w (cid:54) = u (cid:54) = w , w is the unique neighbor of f that F -links it to u and w F -links w to u . If f w ∈ F , remove edge f w from F and add edge f w to F .Proof of safeness of Rule 2. Let F (cid:48) = F \ { f w } and note that w F -links f to w ; as such, edge f w is in the unique cycle of G \ F (cid:48) , so F (cid:48)(cid:48) = F (cid:48) ∪ { f w } isa feedback edge set of G of size q . Furthermore, w is the only vertex that hasfewer neighbors in T ( F (cid:48)(cid:48) ) than in T ( F ) ; since w had two neighbors in T ( F ) ,and F (cid:48)(cid:48) is a minimum feedback edge set of G , w is a leaf of T ( F (cid:48)(cid:48) ) , so it holdsthat leaves ( T ( F )) ⊂ leaves ( T ( F (cid:48)(cid:48) )) .Reduction Rule 2 guarantees that there are no edges in F between verticesof the paths between F -pairs, otherwise we could increase the number of leavesof our tree. Reduction Rule 3.
Let f, u, v ∈ V ( G ) be such that v / ∈ leaves ( T ( F )) ∪ P F ( u, f ) , u, f form an F -pair, and | P F ( u, f ) | ≥ . If there are adjacent vertices w , w ∈ P ∗ F ( u, f ) with vw ∈ F and w F -linking v and w , remove edge vw from F and add edge w w to F .Proof of safeness of Rule 3. Let F (cid:48) = F \ { w w } . Since G \ F (cid:48) has one moreedge than T ( F ) and w F -links v and w (see Figure 2), the unique cycle of G \ F (cid:48) contains edge w w , so F (cid:48)(cid:48) = F (cid:48) ∪ { vw } is a feedback edge set of G ofsize q . Since neither v nor w are leaves of T ( F ) and w ∈ D , deg T ( F (cid:48)(cid:48) ) ( w ) = 1 ,so it holds that | leaves ( T ( F )) | < | leaves ( T ( F (cid:48)(cid:48) )) | .12 w fvuD D ∗ Figure 2: Example for Reduction Rule 3, where the thick edge vw is removedfrom F and the dotted edge w w added to F .Note that, in Reduction Rule 3, the only properties that we exploit are that v, w , w / ∈ leaves ( T ( F )) and that there is at least one pair of vertices between v and f in D . So even if v = u , u ∈ D , or f ∈ leaves ( T ( F )) , we canapply Rule 3. Essentially, if Rule 3 is not applicable, for each edge e ∈ F thatcontains a vertex w of D as an endpoint, either e has a leaf of T ( F ) as itsother endpoint, or w is adjacent to two vertices not in D . Reduction Rule 4.
Let f, u, v, z, w ∈ V ( G ) be such that v ∈ leaves (( T ( F )) \{ f } , w ∈ D is the unique neighbor of v in T ( F ) , u, f and z, v are F -pairs,and u F -links f to z . If there is some w ∈ P ∗ F ( u, f ) with vw ∈ F , remove vw from F and add vw .Proof of safeness of Rule 4. Let F (cid:48) = F \{ vw } . Since T ( F ) is a tree and w F -links w and v , edge vw is contained in the unique cycle of G \ F (cid:48) ; consequently, F (cid:48)(cid:48) = F (cid:48) ∪ { vw } is a feedback edge set of G of size q , but it holds that thedegrees of v and w in T ( F (cid:48)(cid:48) ) are equal to one. Since w / ∈ leaves ( T ( F )) , wehave that leaves ( T ( F )) ⊂ leaves ( T ( F (cid:48)(cid:48) )) .Our analysis for Rule 4 works even if u = z or z = w : what is truly crucialis that w ∈ D and that v (cid:54) = f . We present an example of the general case inFigure 3. w w fvuz D D ∗ leaves ( T ( F )) Figure 3: Example for Reduction Rule 4, where the thick edge vw is removedfrom F and the dotted edge vw is added to F . Reduction Rule 5.
Let f, u, v, z, w , w ∈ V ( G ) be such that v ∈ leaves (( T ( F )) , w ∈ N T ( F ) ( u ) ∩ P F ( u, f ) , w w , vz ∈ E ( T ( F )) , u, f is an F -pair, z ∈ D ∗ , and u F -links f to v . If v is adjacent to some w ∈ P F ( u, f ) \ { w } that F -links w to f , remove vw from F and add w w to F .Proof of safeness of Rule 5. Let F (cid:48) = F \ { vw } . Since w F -links w to f , wehave that w F -links w to v , so edge w w belongs to the unique cycle of G \ F (cid:48) and, consequently, F (cid:48)(cid:48) = F (cid:48) ∪ { w w } is a feedback edge set of G of size q . Since deg T ( F ) ( w ) = deg T ( F ) ( w ) = 2 , deg T ( F (cid:48)(cid:48) ) ( w ) = deg T ( F (cid:48)(cid:48) ) ( w ) = 1 , however,13 is adjacent to both z and w in T ( F (cid:48)(cid:48) ) , so it holds that leaves ( T ( F (cid:48)(cid:48) )) = leaves ( T ( F )) ∪ { w , w } \ { v } . w w w fvzu D D ∗ leaves ( T ( F )) Figure 4: Example for Reduction Rule 5, where the dotted edge w w is addedto F and the thick edge vw is removed from F .Our next lemma guarantees that the exhaustive application of rules 2 through 5finds a set of paths in T ( F ) that have many vertices of degree two in G ; essen-tially, at this point, we are done minimizing the number of incident edges tovertices of D . Lemma 11.
Let a, b ∈ V ( G ) be an F -pair such that a, b / ∈ D , | P ∗ F ( a, b ) | ≥ ,and let w be one of its inner vertices at distance at least three from both a and b . If none of the rules between Rule 2 and Rule 5 are applicable, then deg G ( w ) = deg T ( F ) ( w ) .Proof. Suppose that this is not the case, and let v ∈ N G ( w ) \ N T ( F ) ( w ) . • If v ∈ P F ( a, b ) , suppose w.l.o.g. that w F -links v to a ; moreover, let w be the unique neighbor of v that F -links it to w . In this case, Rule 2 isapplicable: a, v are an F -pair with the required properties, w has thesame role here as in the definition of the rule, and we may set w as w . • If v / ∈ leaves ( T ( F )) , we may assume, w.l.o.g., that w F -links v to b . Wecan apply Rule 3: v is not adjacent to w in T ( F ) , so w has one neighbor w ∈ D that F -links it to v . • If v ∈ leaves ( T ( F )) and its unique neighbor is w ∈ D \ P F ( a, b ) , weagain may assume w.l.o.g. that w F -links a and v . In this case, Rule 4is applicable: there is some z / ∈ leaves ( T ( F )) (possibly z ∈ { a, w } ) thatforms an F -pair with v , where P F ( z, v ) \ { z, v } may be empty if z = w . • If v ∈ leaves ( T ( F )) and z ∈ D ∗ is its unique neighbor in T ( F ) , then, sincethere are at least two other vertices between w and each of the endpointsof P F ( a, b ) , Rule 5 is applicable; to see that this is the case, set w to w inthe definition of the rule and w , w as appropriate to depending on whichendpoint of P F ( a, b ) F -links w to v .Thus, we conclude that v cannot exist and that the statement holds.At this point, paths between F -pairs are mostly the same as in G : only theto vertices closest to each endpoint may be adjacent to some leaves of T ( F ) ,while all others have degree two in G . We say that u, f are a strict F -pair if forevery w ∈ P ∗ F ( u, f ) , deg G ( w ) = 2 . 14 eduction Rule 6. Let u, f ∈ V ( G ) be a strict F -pair. If there are adjacentvertices w , w ∈ P ∗ F ( u, f ) such that either w , w ∈ K or w , w / ∈ K , add anew vertex w ∗ to G that is adjacent to N G ( w ) ∪ N G ( w ) \ { w , w } and removeboth w , w from G . If w , w ∈ K , set w ∗ as a terminal vertex.Proof of safeness of Rule 6. Correctness follows directly from the hypothesesthat w ∈ K if and only if w ∈ K and that both are degree two vertices. So,in a minimal solution H to ( G, K ) , either both vertices are in H or neither is in H . For the converse, any minimal solution H (cid:48) to the reduced instance ( G (cid:48) , K (cid:48) ) either has w ∗ , in which H is obtained by replacing w ∗ with both w and w , or w ∗ / ∈ V ( H ) , in which case H (cid:48) itself is a solution to ( G, K ) . Reduction Rule 7.
Let u, f ∈ V ( G ) be a strict F -pair such that P ∗ F ( u, f ) ≥ .If Rule 6 is not applicable, replace P ∗ F ( u, f ) with three vertices a, t, b so that a is adjacent to u , b to f , and t to both a and b . Furthermore, t is a terminal ofthe new graph if and only if K ∩ P ∗ F ( u, f ) (cid:54) = ∅ .Proof of safeness of Rule 7. Let G (cid:48) and K (cid:48) be, respectively, the graph and setof terminals obtained after the application of the rule. Suppose H is a minimalsolution to the k -in-a-tree instance ( G, K ) , i.e every vertex of H is containedin a path between two terminals. Note that, if P ∗ F ( u, f ) ∩ V ( H ) = ∅ , H is also asolution to the instance ( G (cid:48) , K (cid:48) ) ; as such, for the remainder of this paragraph,we may assume w.l.o.g. that P ∗ F ( u, f ) ∩ V ( H ) (cid:54) = ∅ and that u ∈ V ( H ) . If P F ( u, f ) \ { u } (cid:42) V ( H ) , H (cid:48) = H ∪ { a, t } \ P ∗ F ( u, f ) is a solution to ( G (cid:48) , K (cid:48) ) :since at least one vertex of P F ( u, f ) is not in V ( H ) and every w ∈ P ∗ F ( u, f ) has degree two in G , the subpaths of P F ( u, f ) in H are used solely for thecollection of terminal vertices of P F ( u, f ) ; consequently, H (cid:48) is an induced treeof G (cid:48) that contains all elements of K (cid:48) . On the other hand, if P F ( u, f ) ⊆ V ( H ) , H (cid:48) = H ∪ { a, t, b } \ P ∗ F ( u, f ) is a solution to ( G (cid:48) , K (cid:48) ) ; to see that this is the case,note that H \ P ∗ F ( u, f ) is a forest with exactly two trees where u and f are indifferent connected components since P F ( u, f ) is the unique path between themin H , and K (cid:48) ⊆ V ( H (cid:48) ) since P ∗ F ⊆ V ( H ) and K \ P ∗ F ( u, f ) ⊆ V ( H ) \ P ∗ F ( u, f ) .For the converse, let H (cid:48) be a minimal solution to ( G (cid:48) , K (cid:48) ) . If { a, t, b } ⊆ V ( H (cid:48) ) , H = H (cid:48) ∪{ P ∗ F ( u, f ) }\{ a, t, b } is a solution of ( G, K ) , as we are replacingone path consisting solely of degree two vertices with another that satisfiesthe same property. If a ∈ V ( H (cid:48) ) but b / ∈ V ( H (cid:48) ) , then t ∈ K (cid:48) (recall that H (cid:48) is minimal) and u ∈ V ( H ) , implying that there is at least one terminalvertex in P ∗ F ( u, f ) . We branch our analysis in the following subcases, where v ∈ P ∗ F ( u, f ) ∩ N G ( f ) : • If v / ∈ K , then H = H (cid:48) ∪ P ∗ F ( u, f ) \ { v } \ { a, t } is a solution to ( G, K ) : allterminals of P ∗ F ( u, f ) are contained in P ∗ F ( u, v ) and no cycle is generatedsince all vertices of the path P ∗ F ( u, v ) have degree two. • If v ∈ K but f / ∈ V ( H (cid:48) ) , H = H (cid:48) ∪ P ∗ F ( u, f ) \ { a, t } is a solution to ( G, K ) : we cannot create any new cycle since f / ∈ V ( H ) and u, f form astrict F -pair, moreover all terminals of P ∗ F ( u, f ) are contained in H . • If v ∈ K and f ∈ V ( H (cid:48) ) , there is at least one non-terminal vertex w ∈ P ∗ F ( u, v ) since Rule 6 is not applicable to P F ( u, f ) . As such, we set H = H (cid:48) ∪ P ∗ F ( u, w ) ∪ P ∗ F ( w, f ) \ { a, t } and obtain a solution to ( G, K ) .15inally, if { a, t, b } ∩ V ( H (cid:48) ) = ∅ , it follows immediately from the assumption that H (cid:48) is a solution to ( G (cid:48) , K (cid:48) ) that H (cid:48) is also a solution to ( G, K ) .We are now ready to state our kernelization theorem. Theorem 12.
When parameterized by the size q of a feedback edge set, k -in-a-tree admits a kernel with q vertices and q edges that can be computed in O (cid:0) q + qn (cid:1) time.Proof. Let G be our n -vertex input graph and K a set of terminals. We beginby applying Rule 1 until no degree one vertex remains in G . Then, we take anyfeedback edge set F of G – we can obtain one in O ( n ) time by listing the set ofback edges of a depth-first search tree of G – and construct P F in O ( n ) time.Let P F be the set of paths between all F -pairs such that, for each P F ( u, f ) ∈P F it holds that u, f / ∈ D . By Observation 10, we have at most q leaves in T ( F ) and q vertices in D ∗ , so there are at most q paths in P F .Each iteration of the first part of the algorithm is described below. If thereis some path P F ( u, f ) ∈ P F with f ∈ leaves ( T ( F )) and P ∗ F ( u, f ) (cid:54) = ∅ , check ifthere is an edge in F between f and one of the vertices of P ∗ F ( u, f ) ∪ { u } ; ifthis is the case, apply Rule 2 in O ( n ) time and move on to the next iteration;by Observation 10, |P F | ≤ q , so we can inspect each path and perform thisstep in O ( q + n ) time. Otherwise, let vw ∈ F be such that w / ∈ leaves ( T ( F )) ,and w ∈ P F ( u, f ) ∈ P F . If v / ∈ leaves ( T ( F )) and w is adjacent to some w ∈ P ∗ F ( u, f ) that F -links to v , apply Rule 3; we can check if these conditionsare satisfied in O ( n ) time, in particular, F -linking is a matter of testing if w and v are in the same connected component of T ( F ) \ { w } . If, however, v ∈ leaves ( T ( F )) \ { f } , v forms an F -pair with z ∈ V ( G ) , w ∈ D ∩ N T ( F ) ( v ) , and u F -links f to z , Rule 4 is applicable in O ( n ) time. Finally, if v ∈ leaves ( T ( F )) \{ f } is adjacent to some z ∈ D ∗ which F -links u and v , w ∈ P F ( u, f ) ∩ N T ( F ) ( u ) is adjacent to w which in turn is F -linked to f by w , Rule 5 is applicable is O ( n ) time.If none of the conditions stated in the previous paragraph is satisfied, westop the algorithm: the number of leaves of T ( F ) cannot be increased by singleedge swaps. As to the number of iterations, rules 2 through 5 guarantee that,when applicable, the number of leaves increases by exactly one. Since each oneof their applications can be performed in O ( q + n ) time and we have at most max F | leaves ( T ( F )) | ≤ q iterations, the above algorithm can be executed in O (cid:0) q + qn (cid:1) time.Let P α be the set of paths of P F whose endpoints are strict F -pairs. Foreach path P F ( u, f ) in P F \ P α , let u (cid:48) , f (cid:48) ∈ P F ( u, f ) be the strict F -pair thatmaximizes | P F ( u (cid:48) , f (cid:48) ) | . Note that | P F ( u, f ) \ P F ( u (cid:48) , f (cid:48) ) | ≤ if there are leavesadjacent to each of the vertices at distance two from the endpoints; we refer toFigure 5 for an example. Finally, define P β as P F ( u (cid:48) , f (cid:48) ) for each P F ( u, f ) ∈P F \ P α .Now, for each path in P α ∪ P β , we can apply Rule 6; since at each step weremove one vertex from G , all applications of the rule amount to O ( n ) -time.Afterwards, we apply Rule 7 to compress the paths as much as possible; again,this entire process is feasibly done in O ( n ) steps. As such, each path in P α ∪ P β has size at most three; consequently each path in P F has size at most seven. Atfirst glance, this would yield a kernel of size q + 7 · q = 32 q ; however, we canobserve that, for each edge in F incident to a vertex in some path of P β , we are16 f (cid:48) u (cid:48) D ∗ fz D ∗ z u leaves ( T ( F )) v v Figure 5: Example of a path P F ( u, f ) and its longest subpath between a strict F -pair u (cid:48) , f (cid:48) ; dotted edges belong to F .essentially reducing the number of leaves of T ( F ) and large degree vertices in oneunit each : the bound of q leaves is only met with equality if every edge of F isincident to two leaves of T ( F ) . Therefore, if β = |P β | , the kernel’s size is givenby | leaves ( T ( F )) | + | D ∗ | +3 |P α | +7 |P β | ≤ (2 q − β )+(2 q − β )+3(4 q − β )+7 β =16 q − β , which is maximized when β = 0 . Regarding the number of edges, thecontracted graph has q vertices and a feedback edge set of size q , so it has atmost q edges. Finally, since the first part of the algorithm runs in O (cid:0) q + qn (cid:1) time and the latter in O ( n ) time, we have a total complexity of O (cid:0) q + qn (cid:1) time. In this section, we apply the cross-composition framework of Bodlaender et al. [4]to show that, unless NP ⊆ coNP / poly , k -in-a-tree does not admit a polynomialkernel under bandwidth, nor when parameterized by the distance to any graphclass with at least one member with t vertices for each integer t , which wecollectively call non-trivial classes. We say that an NP - hard problem R OR-cross-composes into a parameterized problem L if, given t instances { y , . . . , y t } of R , we can construct, in time polynomial in (cid:80) i ∈ [ t ] | y i | , an instance ( x, k ) of L that satisfies k ≤ p (max i ∈ [ t ] | y i | + log t ) and admits a solution if and only if atleast one instance y i of R admits a solution; we say that R AND-cross-composes into L if the first two conditions hold but all ( x, k ) has a solution if and only ifall t instances of R admit a solution. Theorem 13.
When parameterized by bandwidth, k -in-a-tree does not admita polynomial kernel unless NP ⊆ coNP / poly .Proof. We are going to show that k -in-a-tree AND-cross-composes into itself.Let H = { ( H , K ) , . . . , ( H t , K t ) } be a set of instances of k -in-a-tree whereeach graph has n vertices, (cid:96) ≥ of which are terminals. The input ( G, K ) to k -in-a-tree parameterized by bandwidth is constructed as follows: G is initiallythe disjoint union of the t input graphs and K = (cid:83) i ∈ [ t ] K i ; now, for each i ∈ [ t ] ,take two distinct terminals v ( i ) , v ( i ) and add edge v ( i ) v ( i + 1) for every i ∈ [ t − . Essentially, we are organizing the H i ’s in a path.Suppose now that every H i has a solution T i and note that T = (cid:83) i ∈ [ t ] V ( T i ) is a solution to ( G, K ) : T is a tree and every terminal in K has a path to17nother. For the converse, take a solution T to ( G, K ) and let T i = T ∩ V ( H i ) .To see that T i is in fact a solution to ( H i , K i ) , it suffices to observe that therecan be no path between two vertices of H i that contains vertices that do notbelong to H i . As to the bandwidth, we claim that it is at most n − : we canset each vertex of H i in the interval [ n ( i − , ni − arbitrarily as long as v ( i ) is place at n ( i − and v ( i ) at ni − , obtaining the mapping f . Consequently,every edge ab ∈ E ( H i ) satisfies | f ( a ) − f ( b ) | ≤ n and each edge v ( i ) v ( i + 1) satisfies f ( v ( i + 1)) − f ( v ( i )) = n ( i + 1 − − ( ni −
1) = 1 . In this section, we show that
Hamiltonian Path on cubic graphs OR-cross-composes into k -in-a-tree parameterized by vertex cover and number of ter-minals. Our construction, however, can be trivially adapted to different param-eterizations, such as distance to clique. In both cases, we heavily rely on theoriginal gadget by Derhy and Picouleau [18], but make some modifications tosuit our needs. Let H be an instance of Hamiltonian Path on cubic graphs.The
Derhy-Picouleau graph of H , which we denote by DP ( H ) , is constructed asfollows: for each v i ∈ V ( H ) , add to DP ( H ) one copy T i of the gadget depictedin Figure 6 and, for each edge v i v j ∈ E ( H ) , connect one of the black vertices of T i to one of the black vertices of T j so that the degree of each black vertex of DP ( H ) is three. We say that T i and T j are adjacent if there is an edge betweena black vertex α i of T i and a black vertex β j of T j , where { α, β } ⊂ { a, b, c } .The set of mandatory vertices of DP ( H ) is the set of gray vertices. a i b i c i s i Figure 6: Vertex gadget T i for vertex v i . Vertex s i is the only terminal of thisgadget; white vertices are part of an independent set of maximum size.Before presenting the composition itself, we need to make some slight mod-ifications to DP ( H ) , to obtain what we dubbed the representative graph of H .Ultimately, our goal is to overlay the multiple instances of Hamiltonian Path and, by applying an instance selector gadget, force the graph representing theselected instance to emerge from the confounding structure.
Our key modification to DP ( H ) is to replace the edge between black verticeswith edge gadgets. Suppose that v i v j ∈ E ( H ) , i < j , and that α i β j ∈ DP ( H ) .We replace the latter edge with the four vertex gadget e ( i, j, α, β ) as in Figure 7.Note that e ( i, j, α, β ) and e ( i, j, β, α ) are different gadgets whenever α (cid:54) = β . By18oing this for every edge of H , we obtain the representative graph of H , denotedby Rep ( H ) . Intuitively, if c i , b j are in the solution of the k -in-a-tree instancegiven by DP ( H ) , then g cbij is not in the solution of the instance whose input is Rep ( H ) . If either c i or b j are not in the solution, g cbij acts as a garbage collectorand is used to connect s j and s cbij . a i b i c i s i a j b j c j s j s cbij g cbij p cbij q cbij Figure 7: Edge gadget e ( i, j, c, b ) for edge v i v j ∈ E ( H ) ( i < j ). Lemma 14.
There is an induced tree connecting the terminal vertices of DP ( H ) if and only if there is an induced tree connecting the terminal vertices of Rep ( H ) .Proof. Let S be a solution to DP ( H ) ; we construct the solution S (cid:48) to Rep ( H ) asfollows. S (cid:48) contains every vertex in S . For every pair of adjacent black vertices α i , β j add s αβij to S (cid:48) ; if both α i and β j are in S , add { p αβij , q αβij } to S (cid:48) ; otherwiseadd g αβij to S (cid:48) ; this concludes the definition of S (cid:48) . To see that S (cid:48) induces atree of Rep ( H ) , note that each path (cid:104) s i , white vertex, α i , β j , white vertex, s j (cid:105) effectively had edge α i β j replaced by an induced P ; furthermore, g αβij is in S (cid:48) if and only if { α i , β j } (cid:42) S , so no cycle can be formed in the edge gadget;since S induces a tree of DP ( H ) , we conclude that S (cid:48) induces a tree of Rep ( H ) .Finally, S (cid:48) contains all terminal (gray) vertices of Rep ( H ) : all such vertices alsoin DP ( H ) were already connected, while the new ones are either included in theinduced P ’s with endpoints α i , β j , or are connected by g αβij to s j (assuming i < j ).For the converse, suppose S (cid:48) ⊂ V ( Rep ( H )) induces a tree of Rep ( H ) . Notethat we can assume that S (cid:48) is minimal; in particular, we may safely assumethat every black α i vertex in S (cid:48) is used to connect s i to some other s j or, at thevery least, to some terminal of an edge gadget. With this restriction in mind,we obtain our solution S to DP ( H ) by setting S := S (cid:48) ∩ V ( DP ( H )) ; note that,if S (cid:48) is not minimal, there could be a pair of vertices α i ∈ V ( T i ) , β j ∈ v ( T j ) with α i β j ∈ E ( DP ( H )) , which would imply that S was not acyclic. Towardsshowing that S induces a tree of DP ( H ) , let s i , s j be two terminals of DP ( H ) and P Rep ( i, j ) be the unique path between them in the subgraph of Rep ( H ) induced by S (cid:48) . We claim that there is no vertex g αβ(cid:96)r in P Rep ( i, j ) : g αβ(cid:96)r ∈ S (cid:48) implies that it is the unique neighbor of s αβ(cid:96)r in S (cid:48) , otherwise S (cid:48) would notinduce a tree of Rep ( H ) . Consequently, P Rep ( i, j ) ∩ V ( DP ( H )) induces a path of19 P ( H ) and, since (cid:83) i,j ∈ [ n ] P Rep ( i, j ) induce a tree of Rep ( H ) , we conclude that (cid:83) i,j ∈ [ n ] P Rep ( i, j ) ∩ V ( DP ( H )) induces a tree of DP ( H ) . For the remainder of this section, we assume that we are given t instances H = { H , . . . , H t } of Hamiltonian Path such that each H (cid:96) ∈ H is cubic andhas n vertices, and the input we construct to k -in-a-tree is the graph G . Whatwe are going to do now is overlay the representative graphs of H while avoidingadditional vertex gadgets and maintaining only a few additional copies of theedge gadgets. To this end, if α i β j ∈ E ( DP ( H (cid:96) )) , we say that edge v i v j of H (cid:96) is represented by the ordered pair ( α, β ) ; in a slight abuse of notation, we write Rep (cid:96) ( v i v j ) = ( α, β ) , where { α, β } ⊂ { a, b, c } .Formally, G initially has n copies { T , . . . , T n } of the vertex gadget givenin Figure 6. For each (cid:96) ∈ [ t ] , define E (cid:96) as the set of edge gadgets of Rep ( H (cid:96) ) ,i.e e ( i, j, α, β ) ∈ E (cid:96) if and only if v i v j ∈ E ( H ) and Rep (cid:96) ( v i v j ) = ( α, β ) . Weupdate G to include all vertex gadgets and all edge gadgets contained in some E (cid:96) . It is critical to note that (cid:83) (cid:96) ∈ [ t ] E (cid:96) has O (cid:0) n (cid:1) elements. For our instanceselector gadget, we have a copy of K ,t with bipartition ( X, Y ) , where eachof the t vertices of Y corresponds to one instance in H , both vertices of X are terminal vertices, and one of them is identified with the terminal vertex s of T . To conclude the construction of G , for each y (cid:96) ∈ Y and edge gadget e ( i, j, α, β ) / ∈ E (cid:96) , we add all edges between y (cid:96) and the four vertices of e ( i, j, α, β ) ,as we show in Figure 8. s cbij g cbij p cbij q cbij y (cid:96) s Figure 8: Interaction between the instance selector gadget and the edge gadget e ( i, j, c, b ) ( i < j ), if e ( i, j, c, b ) / ∈ E (cid:96) . Lemma 15.
The graph G has a vertex cover of size O (cid:0) n (cid:1) and at most O (cid:0) n (cid:1) terminals.Proof. Note that G has n vertices in vertex gadgets, at most n − n ) verticesin edge gadgets, and t + 1 vertices in the instance selector gadget (recall thatone vertex of X is identified with a terminal of T ). Since Y is an independentset, V ( G ) \ Y is a vertex cover with O (cid:0) n (cid:1) elements. For the last part of thestatement, it suffices to observe that the set of terminals of G is a subset of V ( G ) \ Y . Lemma 16.
There is no induced tree of G connecting all terminal vertices withzero or more than one vertex of Y . Moreover, if y (cid:96) ∈ Y is fixed, the graphobtained after removing X , Y , and all vertices that are in a triangle with y (cid:96) isprecisely Rep ( H (cid:96) ) . roof. If no vertex of Y is picked, then there is no path between the two ter-minals s , x ∈ X precisely because N ( x ) = Y . If two vertices y (cid:96) , y p ∈ Y arepicked, then { x, y (cid:96) , s , y p } is an induced C .Now, let e ( i, j, α, β ) be an edge gadget of G . For the second part of thestatement, recall that V ( e ( i, j, α, β )) ⊆ N ( y (cid:96) ) \ X if and only if e ( i, j, α, β ) / ∈E (cid:96) , i.e the vertices W (cid:96) of G that form a triangle with y (cid:96) are precisely thosethat belong to the edge gadgets e ( i, j, α, β ) not present in Rep ( H (cid:96) ) . As such,the induced subgraph of G that remains after the removal of W (cid:96) , X and Y iscomposed precisely of the vertex gadgets and E (cid:96) ; since no extra edges were addedwithin either group of gadgets or between them, we have that G \ ( W (cid:96) ∪ X ∪ Y ) = Rep ( H (cid:96) ) .Theorem 17 is a direct consequence of Lemmas 14, 15, and 16. Theorem 17. k -in-a-tree does not admit a polynomial kernel when param-eterized by the number of vertices of the induced tree, and size of a minimumvertex cover unless NP ⊆ coNP / poly . As for Corollary 18, we observe that there is nothing special about set Y ofour instance selector gadget being an independent set; the key feature is thatonly one of them may be used in a solution; as such, we may freely encode amember of whichever graph classes we are interested in G [ Y ] . Corollary 18.
For every non-trivial graph class G , k -in-a-tree does not admita polynomial kernel when parameterized by the number of vertices of the inducedtree and size of a minimum G -modulator unless NP ⊆ coNP / poly . In this work, we performed an extensive study of the parameterized complexityof k -in-a-tree and the existence of polynomial kernels for the problem, moti-vated by the relevance of three-in-a-tree in subgraph detection algorithmsand a question of Derhy and Picouleau [18] about the complexity of four-in-a-tree . We presented multiple positive and negative results on the problem,of which we highlight its W [1]- hardness under the natural parameter, the linearkernel under feedback edge set, and the nonexistence of a polynomial kernelunder vertex cover/distance to clique. The main question about the complexityof k -in-a-tree for fixed k , however, remains open; our hardness result showedthat there is no n o ( k ) time algorithm under ETH, but no XP algorithm is knownto exist. It is worthwhile to revisit some cases where three-in-a-tree hasnot been successful to identify possible applications for k -in-a-tree . There arealso no known running time lower bounds for the parameters we study, anddetermining whether or not we can obtain o ( q ) n O (1) time algorithms seems afeasible research direction; still in terms of algorithmic results, it would be quiteinteresting to see how we can avoid Courcelle’s Theorem to get an algorithmwhen parameterizing by cliquewidth. The natural investigation of k -in-a-tree on different graph classes may provide some insights on how to tackle particularcases, such as four-in-a-tree ; this study has already been started in [20] andin others – such as cographs – may even be trivial, but many other cases maybe quite challenging and much still remains to be done.21 eferences [1] Dan Bienstock. On the complexity of testing for odd holes and inducedodd paths. Discrete Mathematics , 90(1):85 – 92, 1991. ISSN 0012-365X.doi: https://doi.org/10.1016/0012-365X(91)90098-M. URL .[2] Dan Bienstock. Corrigendum: To: D. bienstock, “On the complexity oftesting for odd holes and induced odd paths” Discrete Mathematics 90(1991) 85–92.
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