A Polynomial Kernel for 3-Leaf Power Deletion, Master’S Thesis, KAIST, South Korea, 2020

arXiv:1911.04249v3 [cs.DS] 9 Jul 2020 ihmr,Rge n hlks[1 nrdcdan introduced [31] Thilikos and Ragde, Nishimura, Introduction 1 sas upre yteNtoa eerhFudto fKor NRF-2018R1D1A1B07050294). of (No. Foundation Education Research of National Ministry the the by supported also is oudrtn h tutr fpyoeei re ncomput in trees phylogenetic of structure the understand to ugoAhn Jungho 3 1 1 A K n Oaespotdb h nttt o ai Scienc Basic for Institute the by supported are SO and OK, JA, eateto optrSine oa olwy nvriyof University Holloway, Royal Science, Computer of Department iceeMteaisGop nttt o ai cec (IBS), Science Basic for Institute Group, Mathematics Discrete mi addresses: Email oyoilkre o -efpwrdeletion power 3-leaf for kernel polynomial A T and rbe.Mr pcfial,w rsn oyoiltm algorithm polynomial-time a instance present input we an for specifically, More problem. in -efpwro oetree some of power 3-leaf hte hr saset a is there whether uhthat such 4 eateto ahmtc,IcenNtoa University, National Incheon Mathematics, of Department if T o o-eaieinteger non-negative a For w V sa most at is 2 p of eateto ahmtclSine,KAIST, Sciences, Mathematical of Department G 2,1 q G k dadEiben Eduard , seult h e flae of leaves of set the to equal is 1 r daeti n nyi h itnebetween distance the if only and if adjacent are [email protected] ď odn ga,Uie Kingdom United Egham, London, k ℓ [email protected] ie graph a Given . and G ajo,SuhKorea South Daejeon, Korea South Daejeon, nho,SuhKorea South Incheon, p S ,k G, 1 a tmost at has Ď T uy1,2020 10, July q V epoieaplnma enlfrthis for kernel polynomial a provide We . ootu neuvln instance equivalent an output to 3 p Abstract G -on Kwon O-joung , ℓ graph a , q , [email protected] fsz tmost at size of G 1 3 , O La oe Deletion Power -Leaf p k G 14 , q [email protected] T san is vertices. n itntvertices distinct and , 4,1 k ℓ la oe fatree a of power -leaf uhthat such n agi Oum Sang-il and , a(R)gatfne by funded grant (NRF) ea ℓ la oe fatree a of power -leaf ISR2-1.OK (IBS-R029-C1). e toa biology. ational G v p z and G S 1 asks k , sa is w 1 v q , 1,2 For a non-negative integer ℓ, a graph G is an ℓ-leaf power of a tree T if V G is equal to the set of leaves of T , and distinct vertices v and w of G p q are adjacent if and only if the distance between v and w in T is at most ℓ, where the distance between vertices x and y in a graph H is the length of a shortest path in H from x to y. We say that G is an ℓ-leaf power if G is an ℓ-leaf power of some tree. Note that an ℓ-leaf power could have more than one component. For instance, an ℓ-leaf power of a path of length at least ℓ 1 has two components. We remark that a graph is a 2-leaf power if and ` only if it is a disjoint union of cliques, and is a 3-leaf power if and only if it is a (bull, dart, gem)-free chordal graph [14], where a bull, a dart, and a gem are depicted in Figure 1. There are linear-time algorithms to recognize 4- and 5-leaf powers [7, 9] and a polynomial-time algorithm to recognize 6-leaf powers [16]. For each ℓ 7, there is a linear-time algorithm to recognize ě ℓ-leaf powers for graphs of bounded degeneracy [18]. We are interested in the following vertex deletion problem, which gener- alizes the corresponding recognition problem. ℓ-Leaf Power Deletion Input : A graph G and a non-negative integer k Parameter : k Question : Is there a set S V G with |S| k such that G S is a Ď p q ď z ℓ-leaf power? Vertex deletion problems include some of the best studied NP-hard problems in theoretical computer science, including Vertex Cover and Feed- back Vertex Set. In general, the problem asks whether it is possible to delete at most k vertices from an input graph so that the resulting graph belongs to a specified graph class. Lewis and Yannakakis [28] showed that every vertex deletion problem to a non-trivial1 and hereditary2 graph class is NP-hard. Since the class of ℓ-leaf powers is non-trivial and hereditary for every non-negative integer ℓ, it follows that ℓ-Leaf Power Deletion is NP-hard. Vertex deletion problems have been investigated on various graph classes through the parameterized complexity paradigm [13, 15], which measures the performance of algorithms not only with respect to the input size but also with respect to an additional numerical parameter. The notion of vertex deletion allows a highly natural choice of the parameter, specifically the

1 A class of graphs C is non-trivial if both C and the complement of C contain inﬁnitely many non-isomorphic graphs. 2 A class of graphs is hereditary if it is closed under taking induced subgraphs.

2 Kernel ℓ Running time Remark (The number of vertices) 0 O 1.2738k kn [10] 2k Ω log k [27, 29] Equivalent to Vertex Cover p ` q ´ p q 1 O 1.2738k kn [10] 2k Ω log k [27, 29] Reduced to Vertex Cover p ` q ´ p q 2 O 2kk m?n log n [23] O k5{3 [20] Equivalent to Cluster Deletion p ¨ q p q 3 O 37k n7 n m [3] O k14 [Theorem 1.1] - p ¨ p ` qq p q Table 1: The current best known running time of a ﬁxed-parameter algorithm and the current best known upper bound for the number of vertices in a kernel for ℓ-Leaf Power Deletion when ℓ is small. We denote by n the number of vertices and by m the number of edges of an input graph.

size of the deletion set k. A decidable parameterized problem Π is fixed- parameter tractable if it can be solved by an algorithm with running time f k nOp1q where n is input size and f : N N is a computable function. It p q¨ Ñ is well known that Π is fixed-parameter tractable if and only if it admits a kernel [15]. A kernel is basically a polynomial-time preprocessing algorithm that transforms the given instance of the problem into an equivalent instance whose size is bounded above by some function f k of the parameter. The p q function f k is usually referred to as the size of the kernel. A polynomial p q kernel is then a kernel with size bounded above by some polynomial in k. For a decidable fixed-parameter tractable problem, one of the most natural follow-up questions in parameterized complexity is whether the problem admits a polynomial kernel. The existence of polynomial kernels for vertex deletion problems has been widely investigated; see [21]. We are going to survey known results of ℓ-Leaf Power Deletion for small values of ℓ; see Table 1. When ℓ 0, ℓ-Leaf Power Deletion is “ identical to Vertex Cover. Currently, the best known fixed-parameter algorithm for Vertex Cover runs in time O 1.2738k k|V G | , by Chen, p ` p q q Kanj, and Xia [10], and 2k Ω log k is the best known upper bound for ´ p q the number of vertices in kernels for Vertex Cover, independently by Lampis [27] and Lokshtanov, Narayanaswamy, Raman, Ramanujan, and Saurabh [29]. When ℓ 1, since a graph is a 1-leaf power if and only if it either is “ isomorphic to K2, or has no edges, one can easily reduce ℓ-Leaf Power Deletion to Vertex Cover. Thus, 1-Leaf Power Deletion can be solved in time O 1.2738k kn and admits a kernel with 2k Ω log k p ` q ´ p q vertices. When ℓ 2, ℓ-Leaf Power Deletion was studied under the name of “

3 Cluster Deletion. Hüffner, Komusiewicz, Moser, and Niedermeier [23] showed that Cluster Deletion is fixed-parameter tractable by present- ing an algorithm with running time O 2kk |E G | |V G | log|V G | , and p ¨ p q p q p q q Fomin, Le, Lokshtanov, Saurabh, Thomassé, anda Zehavi [20] presented a kernel with O k5{3 vertices for Cluster Deletion. p q Now, we investigate when ℓ 3. Dom, Guo, Hüffner, and Nieder- “ meier [14] already showed that 3-Leaf Power Deletion is fixed-parameter tractable. The algorithm in [17] can be modified to a single-exponential fixed-parameter algorithm for 3-Leaf Power Deletion, that is an algorithm with running time αk nOp1q for input size n and some constant α 1; ¨ ą see [3]. Here is our main theorem. Theorem 1.1. 3-Leaf Power Deletion admits a kernel with O k14 p q vertices. As another motivation, our result is motivated by vertex deletion problems for chordal graphs and distance-hereditary graphs, which are super- classes of 3-leaf powers. For vertex deletion problems of chordal graphs and distance-hereditary graphs, fixed-parameter algorithms and polynomial kernels have been recently obtained [30, 8, 17, 24, 1, 26]. Roughly speaking, our first step is to find a “good” approximate solution, called a good modulator of an input graph G, that is a set S V G of Ď p q size O k2 such that G S v is a 3-leaf power for every vertex v in S. p q zp z t uq This technique of computing a good modulator has been used in several kernelization algorithms [24, 25, 26, 2]. To bound the number of components of G S, we introduce two concepts; a complete split of a graph G, which is a z special type of a clique cut-set of G, and a blocking pair for a set X V G , Ď p q which determine whether X,V G X is a complete split of G. A key p p qz q property, Lemma 4.4, of a blocking pair is that two components of G S z blocked by the same pair in S always contain an obstruction. Through a marking process with pairs in S, we show that if there are many components of G S blocked by some pairs in S, then we can safely remove all edges z inside some of the components. Afterward, we bound the number of isolated vertices of G S through another marking process, and then design a series z of reduction rules to bound the size of the remaining components of G S, z which utilize a tree-like structure of 3-leaf powers, introduced by Brandstädt and Le [6]. We organize this paper as follows. In Section 2, we summarize some terminologies in graph theory and parameterized complexity, and introduce 3-leaf powers. In Section 3, we introduce a good modulator of a graph, and then present an algorithm that either confirms that an input instance G, k p q

4 is a no-instance, or constructs a small good modulator of G. In Sections 4 and 5, we design a series of reduction rules that allows us to bound the number of vertices outside of a good modulator of a graph. In Section 6, we combine the above steps to prove our main result. In Section 7, we conclude this paper with some open problems.

2 Preliminaries

In this paper, all graphs are ﬁnite and simple. For a vertex v and a set X of vertices of a graph G, let NG v be the set of neighbors of v in V G , NG X p q p q p q be the set of vertices not in X that are adjacent to some vertices in X, and NG X : NG X X. We may omit the subscripts of these notations if it is r s “ p qY clear from the context. For disjoint sets X and Y of vertices of G, we say that X is complete to Y if each vertex in X is adjacent to all vertices in Y , and X is anti-complete to Y if each vertex in X is non-adjacent to all vertices in Y . By G X we denote the graph obtained from G by removing all vertices in X z and all edges incident with some vertices in X, and G X : G V G X . r s “ zp p qz q We may write G v instead of G v for each vertex v of G. For a set T of z z t u edges of G, let G T be a graph obtained from G by removing all edges in T . z A graph G is trivial if |V G | 1, and non-trivial, otherwise. A clique p q ď is a set of pairwise adjacent vertices. A graph is complete if every pair of distinct vertices is adjacent, and incomplete, otherwise. An independent set is a set of pairwise non-adjacent vertices. Distinct vertices v and w of G are twins in G if NG v w NG w v . Twins v and w in G are true if v p qz t u “ p qz t u and w are adjacent, and false if v and w are non-adjacent. A twin-set in G is a set of pairwise twins in G. A twin-set is true if it is a clique, and false if it is an independent set. A vertex v of a graph G is a cut-vertex if G v has more components than z G. A set X of vertices of G is a vertex cut if G X has more components z than G. A clique cut-set is a vertex cut which is a clique. A vertex set S is a feedback vertex set of G if G S is a forest. z A vertex of a graph is isolated if it has no neighbors. A node of a tree is a leaf if it has exactly one neighbor, and is branching if it has at least three neighbors. For graphs G1,...,Gm, a graph G is G1,...,Gm -free if G has p q no induced subgraph isomorphic to one of G1,...,Gm. We say that a reduction rule is safe if each input instance is equivalent to the resulting instance obtained from the input instance by applying the rule.

5 Figure 1: A bull, a dart, a gem, a house, and a domino.

2.1 Parameterized problems and kernels For a fixed finite set Σ of alphabets, an instance is an element in Σ˚ N. ˆ For an instance I, k , k is called a parameter. A parameterized problem is a p q set Π Σ˚ N. A parameterized problem Π is fixed-parameter tractable if Ď ˆ there is an algorithm, called a fixed-parameter algorithm for Π, that correctly confirms whether an input instance I, k is contained in Π in time f k nOp1q p q p q where n is the size of I and f is some computable function from N to N. A fixed-parameter algorithm for a parameterized problem is single-exponential if it takes αk nOp1q time for some constant α 1. ¨ ą For a decidable parameterized problem Π, an instance I, k is a yes- p q instance for Π if I, k Π, and a no-instance for Π, otherwise. Instances p q P I, k and I1, k1 are equivalent with respect to Π if I, k is a yes-instance p q p q p q for Π if and only if I1, k1 is a yes-instance for Π. A kernel for Π is a p q polynomial-time preprocessing algorithm that given an instance I, k , out- p q puts an instance I1, k1 equivalent to I, k with respect to Π such that p q p q |I1| k1 g k for some computable function g : N N. Such a function ` ď p q Ñ g k is the size of the kernel. A polynomial kernel for Π is a kernel for Π p q with the size as a polynomial in k. We may omit the term “for Π” and “with respect to Π” of all these definitions if it is clear from the context. There is a relationship between the fixed-parameter tractability and the existence of a kernel for decidable parameterized problems.

Theorem 2.1 (See Downey and Fellows [15] and Flum and Grohe [19]). A parameterized problem Π is ﬁxed-parameter tractable if and only if Π is decidable and admits a kernel.

2.2 Characterizations of 3-leaf powers Brandst¨adt and Le [6] presented a linear-time algorithm to recognize 3-leaf powers.

Theorem 2.2 (Brandstädt and Le [6, Theorem 15]). Given a graph G, one can either confirm that G is not a 3-leaf power, or find a tree of which G is a 3-leaf power in linear time.

6 The first three graphs in Figure 1 are called a bull, a dart, and a gem, respectively. A hole is an induced cycle of length at least 4. A graph is chordal if it has no holes. Dom, Guo, Hüffner, and Niedermeier [14] presented the following characterization of 3-leaf powers.

Theorem 2.3 (Dom, Guo, H¨uﬀner, and Niedermeier [14, Theorem 1]). A graph G is a 3-leaf power if and only if G is (bull, dart, gem)-free and chordal.

We say that a graph H is an obstruction if H either is a hole, or is isomorphic to one of the bull, the dart, and the gem. An obstruction H is small if |V H | 5. We have the following seven observations about p q ď obstructions.

(O1) No obstructions have true twins.

(O2) No small obstructions have an independent set of size at least 4.

(O3) No obstructions have K4 or K2,3 as a subgraph. (O4) No obstruction H has a cut-vertex v such that H v has exactly two z components H1 and H2 with |V H1 | |V H2 |. p q “ p q (O5) False twins in an obstruction H have degree 2 in H.

(O6) If a vertex v of an obstruction H has exactly one neighbor w in V H , p q then w has degree at least 3 in H.

(O7) A graph H is an obstruction having three distinct vertices of degree 2 in H if and only if H is a hole.

Brandst¨adt and Le [6] showed that a graph G is a 3-leaf power if and only if G is obtained from some forest F by substituting each node u of F with a non-empty clique Bu of arbitrary size. We rephrase this characterization by using the following deﬁnition. A tree-clique decomposition of a graph G is a pair F, Bu : u V F p t P p quq of a forest F and a family Bu : u V F of non-empty subsets of V G t P p qu p q satisfying the following two conditions.

(1) Bu : u V F is a partition of V G . t P p qu p q (2) Distinct vertices x and y of G are adjacent if and only if F has either a node u such that x,y Bu, or an edge vw such that x Bv and t u Ď P y Bw. P

7 We call Bu a bag of u for each node u of F . We say that B is a bag of G if B is a bag of some node of F . Note that each bag is a clique by (2). Theorem 2.4 (Brandst¨adt and Le [6, Theorem 14]). A graph G is a 3-leaf power if and only if G has a tree-clique decomposition. One can construct a tree-clique decomposition of a 3-leaf power in polynomial time. Moreover, if G is a connected incomplete 3-leaf power, then G has a unique tree-clique decomposition. We remark that every connected incomplete 3-leaf power has at least three bags. Brandst¨adt and Le [6] showed that for a connected incomplete 3-leaf power G, distinct vertices v and w of G are in the same bag of G if and only if v and w are true twins in G. Thus, for such a graph G, B is a bag of G if and only if B is a maximal true twin-set in G.

2.3 Characterizations of distance-hereditary graphs A graph G is distance-hereditary if for every connected induced subgraph H of G and vertices v and w of H, the distance between v and w in H is equal to the distance between v and w in G. Figure 1 shows two graphs called a house and a domino. Bandelt and Mulder [5] presented the following characterization of distance-hereditary graphs. Theorem 2.5 (Bandelt and Mulder [5, Theorem 2]). A graph G is distance- hereditary if and only if G is (house, domino, gem)-free and has no holes of length at least 5. Since both the house and the domino have a hole of length 4, every 3-leaf power is distance-hereditary by Theorems 2.3 and 2.5. The following lemma presents a necessary condition to be distance- hereditary. A proof of the following lemma is readily derived from the deﬁnition of a distance-hereditary graph. Lemma 2.6. Let P be an induced path of length at least 3 in a graph G. If G has a vertex v adjacent to both ends of P , then G V P v is not r p q Y t us distance-hereditary.

3 Good modulators

A set S of vertices of a graph G is a modulator of G if G S is a 3-leaf z power. A modulator S of a graph G is good if G S v is a 3-leaf power zp z t uq for each vertex v in S. Note that if G has a modulator S, then for every

8 induced subgraph G1 of G, S V G1 is a modulator of G1. This means that X p q if G, k is a yes-instance, then so is G1, k . We remark that if G has an p q p q obstruction H and a good modulator S, then H has at least two vertices in S. To find a small good modulator, we first find a modulator by combining a maximal packing of small obstructions with an outcome of an approximation algorithm for the following problem. Weighted Feedback Vertex Set Input : A graph G, a map w : V G Q 0, , and a non-negative p q Ñ X r 8q rational number k Parameter : k Question : Is there a set S V G with w v k such that G S Ď p q vPS p q ď z is a forest? ř Bafna, Berman, and Fujito presented a 2-factor approximation algorithm for Weighted Feedback Vertex Set.

Theorem 3.1 (Bafna, Berman, and Fujito [4]). Given a graph G, a map w : V G Q 0, , and a positive rational number k, one can either p q Ñ X r 8q confirm that G has no feedback vertex set S V G with w v k, or Ď p q vPS p qď find a feedback vertex set S V G with w v 2křin time bounded Ď p q vPS p q ď above by a polynomial in the input size. ř Lemma 3.2. Given an instance G, k with k 0, one can either confirm p q ą that G has no modulator of size at most k, or find a modulator of G having at most 7k vertices in time bounded above by a polynomial in |V G | k. p q ` To prove Lemma 3.2, we will use the following two lemmas.

Lemma 3.3 (Dom, Guo, H¨uﬀner, and Niedermeier [14, Lemma 2]). Given a graph G, let H be an induced subgraph of G obtained by taking exactly one vertex from each maximal true twin-set in G. Then H is triangle-free if and only if G has no induced subgraph isomorphic to one of the bull, the dart, and the gem.

Lemma 3.4. If a graph G has a modulator S and a true twin-set X such that X S is non-empty, then S X is a modulator of G. z z Proof. We may assume that S X is non-empty. Suppose that G S X X zp z q has an obstruction H. Since G S is a 3-leaf power, H has at least one z vertex in S X. Then H has exactly one vertex v in S X by (O1). Let X X w be a vertex in X S. Then G V H v w is isomorphic to H, a z rp p qz t uqYt us contradiction, because it is an induced subgraph of G S. z

9 Now, we prove Lemma 3.2.

Proof of Lemma 3.2. We can ﬁnd a maximal packing H1,...,Hm of vertex- disjoint small obstructions in G in time O |V G |6 . If m k 1, then we p p q q ě ` conﬁrm that G has no modulators of size at most k. Thus, we may assume that m k. Let X : V Hi . Note that |X| 5k and G X has ď “ iPt1,...,mu p q ď z no small obstructions. Therefore,Ť a set S of vertices is a modulator of G X z if and only if G X S is chordal. zp Y q Let H be an induced subgraph of G X obtained by taking exactly one z vertex from each maximal true twin-set in G X. We remark that if we say z that for vertices u and v of G X, u v if and only if u v or u and v are z „ “ true twins in G X, then it is easily deduced that is an equivalence relation z „ of V G X. Therefore, the set of maximal true twin-sets in G X forms a p qz z partition of V G X. For each vertex v of G X, let Bv be the maximal true p qz z twin-set in G X containing v, and w : V H Z be a mapping that maps z p qÑ v V H into |Bv|. P p q Claim 1. . G X, k is a yes-instance for 3-Leaf Power Deletion if p z q and only if H,w,k is a yes-instance for Weighted Feedback Vertex p q Set.

Proof of Claim 1. Firstly, we show that if S is a minimal modulator of G X z containing at most k vertices, then S V H is a feedback vertex set of X p q H with w v k. Since S is a modulator of G X, G X S vPSXV pHq p q ď z zp Y q is chordal.ř Then since H S is an induced subgraph of G X S , H S z zp Y q z is chordal. By Lemma 3.3, since G X has no small obstructions, H is z triangle-free, and so is H S. Therefore, H S is acyclic, that is, S V H is z z X p q a feedback vertex set of H. By Lemma 3.4, Bv S for every vertex v S. Ď P Thus, |S| w v k. “ vPSXV pHq p qď Conversely,ř we show that if T is a feedback vertex set of H with w v vPT p qď k, then Bv is a modulator of G X having at most k vertices.ř Let vPT z BT : ŤvPT Bv. Note that |BT | vPT w v k. Suppose that BT is not “ “ p q ď 1 a modulatorŤ of G X. Then G X řBT has a hole H , and therefore z zp Y q

¨ Bv˛ V H X p q vPVďpH1q ˝ ‚ is a hole of H T , contradicting that T is a feedback vertex set of H. z We apply Theorem 3.1 for H, w, and k. If the algorithm in Theorem 3.1 confirms that H has no feedback vertex set T with w v k, then vPT p q ď ř 10 we confirm that G has no modulator of size at most k. Thus, we may assume that the algorithm in Theorem 3.1 outputs a set T V H with Ď p q w v 2k. Note that |BT | w v 2k. Then X BT is a vPT p q ď “ vPT p q ď Y řmodulator of G with |X BT | |X| ř |BT | 5k 2k 7k. Y “ ` ď ` ď With a modulator of size O k at hand, we are ready to find a small p q good modulator. We note that, in principle, a small good modulator might not exist, but if that is the case, we are able to identify a vertex that has to be in every modulator of size at most k. Then we can remove it from the input graph, and decrease the parameter k by 1.

Reduction Rule 1 (R1). Given an instance G, k with k 0, if G has k 1 p q ą ` obstructions H1,...,Hk 1 and a vertex v of G such that V Hi V Hj ` p qX p q“ v for every distinct i and j in 1,...,k 1 , then replace G, k with t u t ` u p q G v, k 1 . p z ´ q Proof of Safeness. It suﬃces to show that if G has a modulator S of size at most k, then S contains v. Suppose not. Then S contains at least one vertex of Hi v for each i 1,...,k 1 . Therefore, |S| k 1, a z P t ` u ě ` contradiction.

To ﬁnd the obstructions H1,...,Hk`1, we make use of the following lemma, which we slightly rephrase for our application.

Lemma 3.5 (Jansen and Pilipczuk [24, Lemma 1.3]). Given a graph G, a non-negative integer k, and a vertex v, if G v is chordal, then one can z ﬁnd either holes H1,...,Hk 1 in G such that V Hi V Hj v for ` p qX p q “ t u every distinct i and j in 1,...,k 1 , or a set S V G v of size at t ` u Ď p qz t u most 12k such that G S is chordal in time bounded above by a polynomial z in |V G | k. p q ` Now, we present an algorithm that outputs an equivalent instance with a small good modulator of the graph in the resulting instance.

Lemma 3.6. Given an instance G, k with k 0, one can find an equiv- p q ą alent instance G1, k1 and a good modulator of G1 having size at most p q 84k2 7k such that |V G1 | |V G | and k1 k in time bounded above ` p q ď p q ď by a polynomial in |V G | k. p q ` Proof. We first try to find a modulator S of G having size at most 7k by using Lemma 3.2. If it fails, then G, k is a no-instance, and therefore we take 1 1 p q 1 K2,2, 0 as G , k and V K2,2 as a good modulator of G . Otherwise, for p q p q p q v v each vertex v in S, let Gv : G S v , and F ,...,F be a maximal “ zp z t uq 1 mpvq

11 v v packing of small obstructions in Gv such that V Fi V Fj v for p qX1 p q “ t uv every distinct i and j in 1,...,m v . Finally, let G : Gv V F t p qu v “ zpp p 1 qY V F v v . If m v k 1 for some vertex v S, then we apply ¨¨Ÿ p mpvqqqz t uq p qě ` P our algorithm recursively for G v, k 1 . This is safe from the safeness of p z ´ q (R1). Therefore, we may assume that m v k for every vertex v S. p qď P By Lemma 3.5 for G1 , k m v , and v, we can either v ´ p q (1) find k m v 1 holes Hv,...,Hv in G1 such that V Hv ´ p q` 1 k´mpvq`1 v p i qX V Hv v for every distinct i and j in 1,...,k m v 1 , or p j q “ t u t ´ p q` u (2) find a set S1 V G1 v of size at most 12 k m v such that v Ď p vqz t u p ´ p qq G1 S1 is chordal. vz v If (1) holds, then we apply our algorithm recursively for G v, k 1 . p z ´ q Therefore, we may assume that (2) holds for every vertex v in S. Then let v v 1 Sv : V F V F S v . Note that |Sv| 4m v 12 k “ p p 1 qY¨¨Ÿ p mpvqqY vqz t u ď p q` p ´ m v 12k and Gv Sv is a 3-leaf power. p qq ď z 1 1 We take G, k as G , k and X : S Sv as a good modulator p q p q “ Y vPS of G. Clearly, |X| |S| 12k|S| 84k2 7kŤ. It remains to argue that X ď ` ď ` is a good modulator of G. Suppose that H is an obstruction in G. Since S is a modulator of G, H has a vertex v S. If |V H S| 1, then H is P p qX “ an induced subgraph of Gv, and therefore H has at least one vertex in Sv. Since Sv and S are disjoint, H has at least two vertices in X. Therefore, X is a good modulator of G.

4 Bounding the number of components outside of a good modulator

Let S be a good modulator of a graph G. We bound the number of components of G S. In Subsection 4.1, we introduce a complete split of a graph, z and present two lemmas observing obstructions with a complete split of a graph. Then we deﬁne a blocking pair for a set of vertices, and present a characterization of a complete split of a graph and a lemma observing obstructions with a common blocking pair for two sets of vertices. All lemmas introduced in this subsection will be used in the next subsection to bound the number of non-trivial components of G S. In Subsection 4.2, we par- z tition S into S` and S´, and bound the number of components of G S z having neighbors of S´. Afterward, we design a reduction rule to bound the number of non-trivial components of G S having no neighbors of S´. z In Subsection 4.3, we bound the number of isolated vertices of G S. z

12 4.1 Complete splits and blocking pairs Cunningham [12] introduced a split of a graph. A split of a graph G is a partition A, B of V G such that |A| 2, |B| 2, and N A is complete p q p q ě ě p q to N B . We say that a split A, B of G is complete if N A N B is a p q p q p qY p q clique. If a graph has a complete split, then obstructions must satisfy some conditions which we prove in the following two lemmas.

Lemma 4.1. Let A, B be a complete split of a graph G. If G has a hole p q H, then V H A or V H B . p qX “H p qX “H Proof. Suppose not. Since N A N B is a clique, H has at most two p qY p q vertices in N A N B , because otherwise H has K3 as a subgraph. Since p qY p q both V H A and V H B are non-empty and H is connected, H has p qX p qX vertices x1 in N A and x2 in N B . Therefore, H has exactly two vertices p q p q x1 and x2 in N A N B , a contradiction, because H x1x2 is disconnected. p qY p q z

Lemma 4.2. Let A, B be a complete split of a graph G. If G has an p q obstruction H having exactly two vertices in A, then H is isomorphic to the bull.

Proof. We denote by a1 and a2 the vertices in V H A. Suppose that both p qX a1 and a2 have neighbors in B. Since N A N B is a clique, a1 and a2 p qY p q are adjacent, and have the same set of neighbors in B. Then a1 and a2 are true twins in H, contradicting (O1). Therefore, either a1 or a2, say a1, has no neighbors in B. Since H is connected, a1 is adjacent to a2. Thus, a1 has degree 1 in H. By (O3), a2 has at most three neighbors in H. By (O6), a2 has at least three neighbors in H. Therefore, a2 has degree 3 in H, and the bull is the only possible obstruction for H.

Now, we deﬁne a blocking pair for a set X V G . A blocking pair for Ď p q X is an unordered pair v, w of distinct vertices in N X such that if v and t u p q w are adjacent and N v X N w X, then N v X is not a clique. p qX “ p qX p qX We say that X is blocked by v, w if v, w is a blocking pair for X. We t u t u remark that if N X has a blocking pair v, w for some subset of X, then p q t u X is blocked by v, w as well. This deﬁnition is motivated by the following t u lemma.

Lemma 4.3. Let A, B be a partition of the vertex set of a graph G such p q that |A| 2 and |B| 2. Then A, B is a complete split of G if and only ě ě p q if N B is a clique and B has no blocking pairs for A. p q

13 Proof. It is clear that if A, B is a complete split of G, then N B is a p q p q clique, and B has no blocking pairs for A. Conversely, suppose that N B is a clique, and B has no blocking pairs p q for A. We may assume that |N A | 2, because otherwise N A N B is p q ě p qY p q a clique and A, B is a complete split of G. Since B has no blocking pairs p q for A, N A is a clique, because if N A has a non-edge vw, then v, w is a p q p q t u blocking pair for A. Moreover, N v A N w A for each pair of distinct p qX “ p qX vertices v and w in N A , because otherwise v, w is a blocking pair for A. p q t u This means that N A is complete to N B . Therefore, N A N B is a p q p q p qY p q clique and A, B is a complete split of G. p q The following lemma shows that if there is a blocking pair v, w for a t u set X V G such that G X has two distinct components whose vertex Ď p q r s sets are blocked by v, w , then G is not a 3-leaf power. t u Lemma 4.4. Let A, B be a partition of the vertex set of a graph G such p q that |A| 2 and |B| 2. If G A has distinct components C1 and C2 ě ě r s such that both V C1 and V C2 are blocked by v, w of vertices in B, then p q p q t u G V C1 V C2 v, w is not a 3-leaf power. r p qY p q Y t us Proof. If NG v V C1 NG w V C1 , then we may assume that C1 p qX p q ‰ p qX p q has a vertex u1 adjacent to v and non-adjacent to w, because otherwise we may swap v and w. Let u2 be a neighbor of v in V C2 , and P be an induced p q path in G V C1 V C2 w from u1 to u2. Note that the length of P r p qY p q Y t us is at least 3, because P must intersect w that is non-adjacent to u1. Since v is adjacent to both ends of P , G V P v is not distance-hereditary by r p q Y t us Lemma 2.6. Therefore, G V C1 V C2 v, w is not a 3-leaf power. r p qY p q Y t us Thus, we may assume that NG v V Ci NG w V Ci for i 1, 2. p qX p q“ p qX p q “ If v and w are non-adjacent, then for a neighbor u1 of v in V C1 and a p q neighbor u2 of v in V C2 , G v, w, u1, u2 is a hole. Otherwise, since v, w p q rt us t u is a blocking pair for V C1 , NG v V C1 has a non-edge u1u2. Let P p q p qX p q be an induced path in C1 from u1 to u2. Since v is adjacent to both ends of P , we may assume that the length of P is exactly 2 by Lemma 2.6. Let u3 be a common neighbor of u1 and u2 in V P , and u4 be a neighbor p q of v in V C2 . Then G v, u1, u2, u3, u4 is isomorphic to the dart if u3 is p q rt us adjacent to v, and has a hole of length 4 if u3 is non-adjacent to v. Therefore, G V C1 V C2 v, w is not a 3-leaf power. r p qY p q Y t us 4.2 The number of non-trivial components Let S` be the set of vertices v in S such that for each component C of ´ ` G S, NG v V C is a true twin-set in C, and S : S S . The following z p qX p q “ z

14 proposition shows that G S has at most |S´| components having neighbors z of S´.

Proposition 4.5. Let S be a good modulator of a graph G, v be a vertex in S, and C be a component of G S. If NG v V C contains distinct z p qX p q vertices w1 and w2 that are not true twins in C, then no components of G S z diﬀerent from C have neighbors of v.

Proof. Suppose that there is a component of G S diﬀerent from C having z a neighbor w of v. If w1 and w2 are adjacent, then C has a vertex w3 adjacent to exactly one of w1 and w2, because w1 and w2 are not true twins in C. Then G v, w, w1, w2, w3 is isomorphic to the dart if w3 is adjacent rt us to v, and the bull if w3 is non-adjacent to v, a contradiction, because it has exactly one vertex v in S, which is a good modulator of G. Therefore, w1 and w2 are non-adjacent. Let P be an induced path in C from w1 to w2. Since v is adjacent to both ends of P , and S is a good modulator of G, the length of P is exactly 2 by Lemma 2.6. Let w3 be a common neighbor of w1 and w2 in V P . Then G v, w, w1, w2, w3 is p q rt us isomorphic to the dart if w3 is adjacent to v, and has a hole of length 4 if w3 is non-adjacent to v, a contradiction, because it has exactly one vertex v in S.

We present a reduction rule to bound the number of non-trivial components of G S having no neighbors of S´. For that, we will use the following z deﬁnition. Let X be a set of vertices of a graph Q. For a non-negative integer ℓ, a set M E Q is an X,ℓ -matching of Q if each vertex in X is incident Ď p q p q with at most ℓ edges in M, and each vertex in V Q X is incident with at p qz most one edge in M.

Reduction Rule 2 (R2). Given an instance G, k with k 0 and a non- p q ą empty good modulator S of G, let S` be the set of vertices u in S such that for each component C of G S, NG u V C is a true twin-set in C, z p qX p q X be the set of 2-element subsets of S`, and Y be the set of non-trivial components of G S having no neighbors of S S`. Let Q be a bipartite graph z z on X 1, 2, 3 ,Y such that the following three statements are true. p ˆ t u q (1) Elements v, w , 1 X 1 and C Y are adjacent in Q if and only pt u q P ˆ t u P if V C is blocked by v, w . p q t u (2) Elements v, w , 2 X 2 and C Y are adjacent in Q if and only pt u q P ˆ t u P if C has a vertex adjacent to both v and w.

15 (3) Elements v, w , 3 X 3 and C Y are adjacent in Q if and only pt u q P ˆ t u P if C has an edge xy such that x is adjacent to both v and w, and y is non-adjacent to both v and w.

If Q has a maximal X 1, 2, 3 , k 2 -matching M avoiding some element p ˆt u ` q U in Y , then replace G, k with G E U , k . p q p z p q q Proof of Safeness. Let G1 : G E U . Firstly, we show that if G, k is a “ z p q p q yes-instance, then so is G1, k . Suppose that G has a modulator S1 of size p q at most k, and G1 S1 has an obstruction H. Since G S1 is a 3-leaf power, z 1 z 1 H has vertices b1 and b2 such that b1b2 E U S . Thus, |V U S | 2. P p z q p qz ě Claim 2. V U S1,V G V U S1 is a split of G1 S1. p p qz p qzp p qY qq z Proof of Claim 2. We first show that |V G V U S1 | 2. If H is a hole p qzp p qY q ě of length 4, then H has at most two vertices of U S1, because V U S1 is an z p qz independent set of G1 S1, and no holes of length 4 have an independent set z of size at least 3. Therefore, H has at least two vertices of G V U S1 . zp p qY q Thus, we may assume that |V H | 5. By (O2), if H is small, then H has p q ě at most three vertices of U S1, and therefore H has at least two vertices z of G V U S1 . If H is a hole of length at least 6, then H has at most zp p qY q t|V H | 2u vertices of U S1, and therefore H has at least r|V H | 2s 2 p q { z p q { ě vertices of G V U S1 . zp p qY q Therefore, |V G V U S1 | 2. Now, suppose that V U S1,V G V U p qzp p qY q ě p p qz p qzp p qY S1 is not a split of G1 S1. Then G V U S1 has vertices v and w such qq z zp p qY 1 q 1 that both v and w have neighbors in V U S , and NG v V U S 1 p qz p q X p p 1qz q ‰ NG w V U S . Thus, v, w is a blocking pair for V U S , so for p q X p p qz q t u p qz V U . Then U is adjacent to v, w , 1 in Q. Since M is maximal, Y has p q pt u q distinct elements C1,...,Ck`2 different from U such that V Ci is blocked 1 p q by v, w for each i 1,...,k 2 . Since |S | k, two of them, say C1 and t u P t 1 ` u ď C2, have no vertices in S . Then G V C1 V C2 v, w is not a 3-leaf r p qY p q Y t us power by Lemma 4.4, a contradiction, because it is an induced subgraph of G S1. z Since V U S1 is an independent set of G1 S1, and H is connected, both p qz z1 b1 and b2 have neighbors in V G V U S . Then by Claim 2, b1 and 1 1 p qzp p qY q b2 are false twins in G S . By (O5), both b1 and b2 have degree 2 in H. z Let z1 and z2 be the neighbors of b1 in V H S. Then U is adjacent to p qX 1 1 z1, z2 , 2 in Q. Since M is maximal, Y has distinct elements C1,...,Ck`2 pt u q 1 different from U such that Ci has a vertex adjacent to both z1 and z2 for each i 1,...,k 2 . Since |S1| k, two of them, say C1 and C1 , have no P t ` u ď 1 2

16 vertices in S1. Note that S1 has no vertices of H, because H is an induced subgraph of G1 S1. z 1 1 If z1 and z2 are non-adjacent, then G V C V C z1, z2 has a r p 1qY p 2q Y t us hole of length 4, a contradiction, because it is an induced subgraph of G S1. z Therefore, z1 and z2 are adjacent. Since G b1, z1, z2 is isomorphic to K3, rt us H is not a hole, and therefore |V H | 5. Let a be a vertex of H different p q “ 1 from b1, b2, z1, and z2. We may assume that a is not in V C1 , because 1 1 p1 q otherwise we may swap C1 and C2. Let c be a vertex of C1 adjacent to both z1 and z2. Note that G b1, b2, z1, z2 is isomorphic to K4 b1b2. Since rt us z the dart and a hole of length 4 are the only obstructions having false twins, H is isomorphic to the dart. Thus, NH a z1 or NH a z2 . Then p q “ t u p q “ t u G a, b1, c, z1, z2 is isomorphic to the gem if c is adjacent to a, and the dart rt us if c is non-adjacent to a, a contradiction, because it is an induced subgraph of G S1. Therefore, if G, k is a yes-instance, then so is G1, k . z p q p q Secondly, we show that if G1, k is a yes-instance, then so is G, k . p q p q Suppose that G1 has a modulator S1 of size at most k, and G S1 has an z obstruction H. Since G1 S1 is a 3-leaf power, H has an edge of U S1. Thus, z z |V U S1| 2. Since S is a good modulator of G, H has at least two vertices p qz ě in S S1. Then |V G V U S1 | 2, since S S1 V G V U S1 . z p qzp p qY q ě z Ď p qzp p qY q Claim 3. V U S1,V G V U S1 is a complete split of G S1. p p qz p qzp p qY qq z Proof of Claim 3. Suppose not. We first show that V G V U S1 has p qzp p qY q a blocking pair for V U S1. Since U is a component of G S, and has p qz z no neighbors of S S`, it suffices to show that S` S1 has a blocking pair z z for V U S1. We may assume that for all vertices v and w in S` S1 having p qz z neighbors in V U S1, v and w are adjacent, and have the same set of neigh- p qz bors in V U S1, because otherwise v, w is a blocking pair for V U S1. p qz t u p qz For each vertex v in S` S1 having neighbors in V U S1, the set of neigh- z p qz bors of v in V U S1 is a true twin-set in U S1, that is, a clique. Therefore, ` 1 p qz 1 z1 ` 1 NG S S V U S is a clique of U S . Thus, by Lemma 4.3, S S p z q X p p qz q z z has a blocking pair v, w for V U S1, so for V U . t u p qz p q Since V U is blocked by v, w , U is adjacent to v, w , 1 in Q. Since p q t u pt u q M is maximal, Y has distinct elements C1,...,Ck`2 different from U such 1 that V Ci is blocked by v, w for each i 1,...,k 2 . Since |S | k, p q t u P t 1 ` u ď two of them, say C1 and C2, have no vertices in S . Then G V C1 V C2 r p qY p qY v, w is not a 3-leaf power by Lemma 4.4, a contradiction, because it is an t us induced subgraph of G1 S1. z Since both U S1 and G V U S1 have vertices of H, H is not a z zp p q Y q hole by Lemma 4.1 and Claim 3, and therefore |V H | 5. Let t1,...,tp p q “

17 1 be the vertices of H in V U S , and s1,...,sq be the vertices of H in p qz V G V U S1 . Note that both p and q are at least 2. Since |V H | 5, p qzp p qY q p q “ p,q 3, 2 or p,q 2, 3 . p q “ p q p q “ p q If p,q 3, 2 , then we may assume that NH s1 s2 and NH s2 p q “ p q p q “ t u p q“` s1,t1,t2 by Lemma 4.2 and Claim 3. Since U has no neighbors of S S , t u` 1 z s2 is in S . Thus, t1 and t2 are true twins in U S , contradicting (O1). z Therefore, p,q 2, 3 . By Lemma 4.2 and Claim 3, we may assume p q “ p q that NH t1 t2 and NH t2 t1,s1,s2 . Note that s1 and s2 are in 1 p q “ t u p q “ t u S S . Then U is adjacent to s1,s2 , 3 in Q. Since M is maximal, Y has z 2 2 pt u q 2 distinct elements C1 ,...,Ck`2 diﬀerent from U such that Ci has an edge xiyi such that xi is adjacent to both s1 and s2, and yi is non-adjacent to 1 both s1 and s2 for each i 1,...,k 2 . Since |S | k, two of them, 2 2 P t 1` u ď say C1 and C2 , have no vertices in S . We may assume that s3 is not in V C2 , because otherwise we may swap C2 and C2. We remark that the p 1 q 1 2 bull is the only possible graph to which H is isomorphic. Thus, s1 and s2 are adjacent, and s3 is adjacent to exactly one of s1 and s2 in H. Then G x1,y1,s1,s2,s3 is isomorphic to the gem if both x1 and y1 are adjacent rt us to s3, the bull if both x1 and y1 are non-adjacent to s3, and the dart if x1 is adjacent to s3 and y1 is non-adjacent to s3, and has a hole of length 4 if x1 is non-adjacent to s3 and y1 is adjacent to s3, a contradiction, because it is an induced subgraph of G1 S1. Therefore, if G1, k is a yes-instance, then z p q so is G, k . p q Proposition 4.6. Given an instance G, k with k 0 and a non-empty p q ą good modulator S of G, if (R2) is not applicable to G, k , then G S has at p q z most 2 k 2 |S|2 non-trivial components. p ` q Proof. Let S` be the set of vertices u in S such that for each component ´ ` C of G S, NG u V C is a true twin-set in C, and S : S S . By z p qX p q “ z Proposition 4.5, each vertex in S´ is adjacent to at most one component of G S. Therefore, G S has at most |S´| non-trivial components having z z neighbors of S´. Let Q and M be deﬁned as in (R2). Since (R2) is not applicable to G, k , each non-trivial component of G S having no neighbors of S´ is p q z incident with exactly one edge in M. Since each edge in M is incident with some element in X 1, 2, 3 , and each element in X 1, 2, 3 is incident ˆ t u ˆ t u ` with at most k 2 edges, |M| k 2 |X 1, 2, 3 | k 2 3 |S | ` ď p ` q ¨ ˆ t u ď p ` q ¨ 2 ď 3 k 2 |S|2 2. Then |S´| |M| |S| 3 k 2 |S|2 2 k 2 |`S|2 2˘ p ` q { ` ď ` p ` q { ď p ` q { ` 3 k 2 |S|2 2 2 k 2 |S|2, and therefore G S has at most 2 k 2 |S|2 p ` q { “ p ` q z p ` q non-trivial components.

18 4.3 The number of isolated vertices We present a reduction rule to bound the number of isolated vertices of G S. z Reduction Rule 3 (R3). Given an instance G, k with k 0 and a non- p q ą empty good modulator S of G, let A be the set of ordered pairs A1, A2 of p q disjoint subsets of S such that 2 |A1| |A2| 4, and X be the set of ď ` ď isolated vertices of G S. For each A1, A2 A, let XA1,A2 be a maximal set z p q P of vertices v in X such that NG v A1 A2 A1 and |XA1,A2 | k 3. p q X p Y q“ ď ` If X has a vertex u XA1,A2 , then replace G, k with G u, k . R pA1,A2qPA p q p z q Ť Proof of Safeness. We need to show that if G u, k is a yes-instance, then p z q so is G, k . Suppose that G u has a modulator S1 of size at most k, and p q z G S1 has an obstruction H. Then u V H , because G S1 u is a 3-leaf z P p q zp Yt uq power. If H is a hole, then u has exactly two neighbors v1 and v2 in S V H X p q such that v1 is non-adjacent to v2. By the construction of Xtv1,v2u,H, Xtv1,v2u,H contains distinct vertices u1, . . . , uk`3 different from u. Note that H has at most one of u1, . . . , uk`3, because v1 and v2 have at most two common neighbors in V H including u. Then since |S1| k, two of them, p q1 ď say u1 and u2, are not in S V H . Thus, G v1, v2, u1, u2 is a hole, a Y p q rt us contradiction, because it is an induced subgraph of G S1 u . zp Y t uq Suppose that H is isomorphic to one of the bull, the dart, and the gem. Then 2 |S V H | 4, because H has exactly five vertices including ď X p q ď u, and S is a good modulator of G. Let B1 : S V H NG u , “ p X p qq X p q and B2 : S V H NG u . By the construction of XB1,B2 , XB1,B2 “ p X p qqz p q contains distinct vertices u1, . . . , uk 3 different from u. Since |V H | 5 ` p q “ and 2 |S V H | 4, H has at most three vertices in X including u. ď X p q ď 1 Thus, H has at most two of u1, . . . , uk`3. Then since |S | k, one of them, 1 ď say u1, is not in S V H . Thus, G V H u u1 is isomorphic to Y p q rp p qz t uq Y t us H, a contradiction, because it is an induced subgraph of G S1 u . zp Y t uq Proposition 4.7. Given an instance G, k with k 0 and a non-empty p q ą good modulator S of G, if (R3) is not applicable to G, k , then G S has at p q z most 2 k 3 |S|4 3 isolated vertices. p ` q {

Proof. Let A, X, and XA1,A2 be deﬁned as in (R3). Since S is a good modulator, we may assume that |S| 2. Let s : |S|. For each m-element ě “ m subset T of S with 2 m 4, A contains exactly 2 elements A1, A2 ď ď p q

19 such that T A1 A2. Therefore, |A| is at most “ Y s s s 2 4 24 23 22 s 1 4 s 1 3 2s s 1 ¨ ˆ4˙ ` ¨ ˆ3˙ ` ¨ ˆ2˙ ď 3p ´ q ` 3p ´ q ` p ´ q 2 s 1 2 s 1 2 2 s 1 2s s 1 “ 3p ´ q pp ´ q ` p ´ qq ` p ´ q 2 s 1 2s2 2s s 1 ď 3p ´ q ` p ´ q 2s s 1 s2 s 3 3 “ p ´ qp ´ ` q{ 2s s 1 s2 s 3 2s2 s2 1 3 2s4 3. ď p ´ qp ` q{ “ p ´ q{ ď {

For each element A1, A2 in A, |XA1,A2 | k 3. Thus, | XA1,A2 | p q ď ` pA1,A2qPA ď 2 k 3 |S|4 3. Since (R3) is not applicable to G, k , everyŤ isolated vertex p ` q { p q 4 of G S is in XA1,A2 . Therefore, G S has at most 2 k 3 |S| 3 z pA1,A2qPA z p ` q { isolated vertices.Ť

5 Bounding the size of components outside of a good modulator

Let S be a good modulator of a graph G. In this section, we bound the size of each component of G S. Subsection 5.1 is about complete components of z G S, and Subsection 5.2 is about incomplete components of G S. z z 5.1 The size of each complete component We present a reduction rule to bound the size of each complete component of G S, which proceed by a similar marking process with (R3). z Reduction Rule 4 (R4). Given an instance G, k with k 0 and a non- p q ą empty good modulator S of G, let A be the set of ordered pairs A1, A2 of p q disjoint subsets of S such that 2 |A1| |A2| 4, and C be a complete ď ` ď component of G S. For each A1, A2 A, let XA1,A2 be a maximal set of z p q P vertices v of C such that NG v A1 A2 A1 and |XA1,A2 | k 3. If p q X p Y q“ ď ` C has a vertex u XA1,A2 , then replace G, k with G u, k . R pA1,A2qPA p q p z q Proof of Safeness. WeŤ need to show that if G u, k is a yes-instance, then p z q so is G, k . Suppose that G u has a modulator S1 of size at most k, and p q z G S1 has an obstruction H. Then u V H , because G S1 u is a 3-leaf z P p q zp Yt uq power. Suppose that H is a small obstruction. Since H has at most ﬁve vertices including u, and S is a good modulator of G, 2 |S V H | 4. Let B1 : ď X p q ď “

20 S V H NG u , and B2 : S V H NG u . By the construction of p X p qq X p q “ p X p qqz p q XB1,B2 , XB1,B2 contains distinct vertices u1, . . . , uk`3 different from u. Since |V H | 5 and 2 |S V H | 4, H has at most three vertices of C p q ď ď X p q ď 1 including u. Thus, H has at most two of u1, . . . , uk`3. Then since |S | k, 1 ď one of them, say u1, is not in S V H . Thus, G V H u u1 Y p q rp p qz t uqYt us is isomorphic to H, a contradiction, because it is an induced subgraph of G S1 u . zp Y t uq Therefore, H is a hole of length at least 6. Note that H has at most two vertices of C, because C is complete. Suppose that H has exactly one vertex u of C. In this case, u is adjacent to distinct vertices v1 and v2 in V H S. Then H u is an induced path of length at least 4 from v1 to p qX z v2. By the construction of Xtv1,v2u,H, Xtv1,v2u,H contains distinct vertices 1 u1, . . . , uk`3 different from u. Since |S | k, one of them, say u1, is not in 1 ď S . Then G V H u u1 is not distance-hereditary by Lemma 2.6, rp p qz t uq Y t us a contradiction, because it is an induced subgraph of G S1 u . zp Y t uq Therefore, H has exactly two vertices u and u1 of C. In this case, u is 1 adjacent to a vertex v1 in V H S, and u is adjacent to another vertex v2 in 1 p qX V H S. Note that u is non-adjacent to v1. Then H u is an induced path p qX 1 z of length at least 4 from v1 to u . By the construction of Xtv1u,tv2u, Xtv1u,tv2u 1 contains distinct vertices u1, . . . , uk`3 different from u. Since |S | k, one 1 ď of them, say u1, is not in S . Then G V H u u1 is not distance- rp p qz t uq Y t us hereditary by Lemma 2.6, a contradiction, because it is an induced subgraph of G S1 u . Therefore, if G u, k is a yes-instance, then so is G, k . zp Yt uq p z q p q Proposition 5.1. Given an instance G, k with k 0 and a non-empty p q ą good modulator S of G, if (R4) is not applicable to G, k , then every com- p q plete component of G S has at most 2 k 3 |S|4 3 vertices. z p ` q {

Proof. Let A, C, and XA1,A2 be deﬁned as in (R4). Since (R4) is not applicable to G, k , every vertex of C is in XA1,A2 . Since S is a p q pA1,A2qPA good modulator, we may assume that |S| Ť2. For each m-element subset ě m T of S with 2 m 4, A contains exactly 2 elements A1, A2 such that ď ď 4 |S| 3 |S| 2 p |S| q 4 T A1 A2. Therefore, |A| 2 2 2 2|S| 3, as “ Y ď ¨ 4 ` ¨ 3 ` ¨ 2 ď { in the proof of Proposition 4.7. For` each˘ element` ˘A1, A2 ` in˘A, |XA1,A2 | p 4 q ď k 3. Therefore, | XA1,A2 | 2 k 3 |S| 3, and C has at most ` pA1,A2qPA ď p ` q { 2 k 3 |S|4 3 vertices.Ť p ` q { 5.2 The size of each incomplete component In this subsection, we present four reduction rules to bound the size of each incomplete component of G S. Firstly, we present a reduction rule to bound z

21 the size of a true twin-set in G.

Reduction Rule 5 (R5). Given an instance G, k with k 0, if G has a p q ą true twin-set X such that |X| k 2, then replace G, k with G v, k for ě ` p q p z q some vertex v X. P Proof of Safeness. We need to show that if G v, k is a yes-instance, then p z q so is G, k . Suppose that G v has a modulator S1 of size at most k, and p q z G S1 has an obstruction H. Then v V H , because G S1 v is a 3-leaf z P p q zp Yt uq power. By (O1), V H X v . Since |S1| k, X contains a vertex w not p qX “ t u ď in S1 v . Then G V H v w is isomorphic to H, a contradiction, Yt u rp p qz t uqYt us because it is an induced subgraph of G S1 v . zp Y t uq Later, we will apply (R5) only for true twin-sets in G that are subsets of V G S, which one can ﬁnd in polynomial time by Theorem 2.4. p qz In the following reduction rules, we start with computing a tree-clique decomposition of G S. We present a reduction rule to remove some bags of z G S which are anti-complete to S. z Reduction Rule 6 (R6). Given an instance G, k with k 0 and a non- p q ą empty good modulator S of G, let B be a maximal true twin-set in G S. If z G S B has a component D having no neighbors of S and V D NG B zp Y q p qz p q is non-empty, then replace G, k with G V D NG B , k . p q p zp p qz p qq q 1 Proof of Safeness. Let G : G V D NG B . We need to show that if “ zp p qz p qq G1, k is a yes-instance, then so is G, k . Suppose that G1 has a modulator p q p q S1 of size at most k, and G S1 has an obstruction H. Since G1 S1 is a 3- z z leaf power, H has at least one vertex of D NG B . Since H is connected, z p q and D has no neighbors of S, H has at least one vertex in V D NG B . p qX p q Thus, H has at least two vertices of D. Since V H S , V H B p qX ‰ H p qX is a clique cut-set of H, and therefore H is not a hole. Thus, |V H | 5. p q “ Since S is a good modulator of G, |V H S| 2, |V H B| 1, and p qX “ p qX “ |V H V D | 2, contradicting (O4). p qX p q “ We present two reduction rules to reduce the number of bags of G S. z Reduction Rule 7 (R7). Given an instance G, k with k 0 and a non- p q ą empty good modulator S of G, let B be a maximal true twin-set in G S. If z G S B has distinct components D1,...,Dk 4 such that NG V D1 zp Y q ` p p qq “ NG V Dk 4 , and either V D1 V Dk 4 NG B , or ¨¨¨“ p p ` qq p qY¨¨¨Y p ` qĎ p q H‰ V Di NG B V Di for each i 1,...,k 4 , then replace G, k with p qX p q‰ p q P t ` u p q G V D1 , k . p z p q q

22 To show that (R7) is safe, we will use the following two lemmas. Lemma 5.2 will be useful because it implies that for a good modulator S of G, a subset B of V G S is a true twin-set in G S if and only if it is a true twin-set in p qz z G. Lemma 5.2. Let G be a 3-leaf power having a vertex v such that G v is z connected and incomplete. Then vertices t1 and t2 in V G v are true p qz t u twins in G if and only if t1 and t2 are true twins in G v. z Proof. It is clear that if t1 and t2 are true twins in G, then so are in G v. z Conversely, suppose that t1 and t2 are true twins in G v, and v is adjacent z to t1, and non-adjacent to t2. Note that |NG t2 | 2, because otherwise p q ě G v is isomorphic to K2. z If NG t2 is a clique, then G v has at least one vertex not in NG t2 , p q z p q because otherwise G v is complete. Thus, G has an edge xy such that x is z adjacent to both t1 and t2, and y is non-adjacent to both t1 and t2, because G v is connected. Then G v,x,y,t1,t2 is isomorphic to the gem if both z rt us x and y are adjacent to v, the bull if both x and y are non-adjacent to v, and the dart if x is adjacent to v and y is non-adjacent to v, and has a hole of length 4 if x is non-adjacent to v and y is adjacent to v, a contradiction, because it is an induced subgraph of G. Therefore, t2 has distinct neighbors x and y such that x is non-adjacent to y. Then G v,x,y,t1,t2 has a hole of length 4 if both x and y are adjacent rt us to v, and is isomorphic to the gem if exactly one of x and y is adjacent to v, and the dart if both x and y are non-adjacent to v, a contradiction, because it is an induced subgraph of G.

Lemma 5.3. Let A, B be a complete split of a graph G, and S be a non- p q empty good modulator of G. If G has an obstruction H, and S B N A , Ď z p q then H has at most one vertex in A. Proof. Suppose not. Since S is a good modulator of G, H has at least two vertices in S. Thus, H has vertices in both A and B. Since A, B is a p q complete split of G, H is not a hole by Lemma 4.1, and therefore |V H | 5. p q “ Then |V H N A | 5 |V H A| |V H S| 5 2 2, contradicting p qX p q ď ´ p qX ´ p qX ď ´ ´ (O4).

Proof of Safeness for (R7). We need to show that if G V D1 , k is a yes- p z p q q 1 instance, then so is G, k . Suppose that G V D1 has a modulator S of size p1 q z p q 1 at most k, and G S has an obstruction H. Since G V D1 S is a 3-leaf z zp p qY q power, H has at least one vertex of D1. Since S is a good modulator of G, G S v is a 3-leaf power for each vertex v in S. Thus, if v has a neighbor zp z t uq

23 in a true twin-set X in G S, then v is complete to X by Lemma 5.2. This z t u means that every true twin-set in G S is a true twin-set in G as well. z We claim that (a) for each i 1,...,k 4 , V Di NG B is a true P t ` u p qX p q twin-set in G S. Suppose that V Di NG B contains two vertices x z p q X p q and y such that x is non-adjacent to y. Let P be an induced path in Di from x to y. By Lemma 2.6, the length of P is exactly 2 . Let z be a common neighbor of x and y in V P . Then z NG B , because p q P p q otherwise V P with a vertex in B induces a hole of length 4. Then for a p q 1 vertex v in B, and v in V Dj NG B for some j 1,...,k 4 i , p qX p q P t ` u z t u G v, v1,x,y,z is isomorphic to the dart, contradicting the assumption rt us that S is a modulator of G. Therefore, V Di NG B is a clique. Now, p qX p q suppose that G S has a vertex w adjacent to a vertex t1 V Di NG B z P p qX p q and non-adjacent to a vertex t2 V Di NG B . Note that w is a vertex P p qX p q 1 of Di NG B . Then for a vertex v in B and a vertex v of V Dj NG B z p q 1 p qX p q for some j 1,...,k 4 i , G v, v ,w,t1,t2 is isomorphic to the bull, P t ` u z t u rt us a contradiction, and this proves (a). Suppose that V D1 V Dk 4 NG B . By (O1), for each p qY¨¨¨Y p ` q Ď p q i 1,...,k 4 , Di has at most one vertex of H. By (O2), if H is small, P t ` u then at most three of D1,...,Dk`4 have vertices of H. If H is a hole of length at least 6, then at most two of D1,...,Dk`4 have vertices of H, because otherwise H has a vertex of degree at least 3 in H. Since |S1| k, 1 ď one of D2,...,Dk 4, say Dj, has no vertices in S V H . Let s be a vertex ` Y p q of H in D1 and t be a vertex in Dj. Since NG V D1 NG V Dj , s p p qq “ p p qq and t have the same set of neighbors in V H . Then G V H s t p q rp p qz t uq Y t us is isomorphic to H, a contradiction, because it is an induced subgraph of 1 G V D1 S . zp p qY q Therefore, V Di NG B V Di for each i 1,...,k 4 . We H‰ p qX p q‰ p q P t ` u claim that (b) Di NG B has no neighbors of S. Suppose that Di NG B z p q z p q has a neighbor pi of some vertex v in S. Let j 1,...,k 4 i . Since P t ` u z t u NG V Di NG V Dj , Dj has a neighbor pj of v. Since some vertex p p qq “ p p qq in B has neighbors in both Di and Dj, G S has a path P from pi to pj. z Note that the length of P is at least 3, because pi is not in NG B . Since v p q is adjacent to both ends of P , G V P v is not distance-hereditary by r p q Y t us Lemma 2.6, a contradiction, because it is an induced subgraph of G S v , zp z t uq and this proves (b). For each i 1,...,k 4 , since V Di NG B is a true twin-set in G, P t ` u p qX p q H has at most one vertex in V Di NG B by (O1). Let Di,1,...,D p qX p q i,mpiq be the components of Di NG B for each i 1,...,k 4 . We claim that z p q P t ` u (c) for each j 1,...,m i , if |V Di,j | 2, then V Di,j ,V G V Di,j P t p qu p q ě p p q p qz p qq is a complete split of G. Since V Di NG B is a true twin-set in G, and p qX p q

24 Di NG B has no neighbors of S, it suﬃces to show that NG NG B z p q p p qq X V Di,j is a clique. Suppose that NG NG B V Di,j contains vertices p q p p qq X p q x and y which are non-adjacent. Let P be an induced path in Di,j from x to y. By Lemma 2.6, the length of P is exactly 2. Let z be a common neighbor of x and y in V P . Then z NG NG B , because otherwise P p q P p p qq with a vertex v in NG B V Di induces a hole of length 4. Then for a p qX p q vertex v1 in B, G v, v1,x,y,z is isomorphic to the dart, a contradiction, rt us and this proves (c). Therefore, each component of Di NG B has at most one vertex of H z p q by Lemma 5.3. Each V Di NG B has at most one vertex of H, be- p qX p q cause V Di NG B is a true twin-set. Therefore, at most one compo- p qX p q nent of Di NG B has a vertex of H, because H cannot have false twins of z p q degree at most 1 by (O5). By (O2), if H is small, then at most three of D1,...,Dk`4 have vertices of H. If H is a hole of length at least 6, then at most two of D1,...,Dk`4 have vertices of H, because otherwise H has a vertex of degree at least 3 in H. Since |S1| k, one of 1 ď D2,...,Dk 4, say Di, has no vertices in S V H . Note that H has a ` Y p q vertex s1 in V D1 NG B , because D1 NG B has no neighbors of S, p qX p q z p q H is connected, and has vertices in both S and V D1 . Let t1t2 be an p q edge of Di such that t1 V Di NG B and t2 V Di NG B . Since P p qX p q P p qz p q NG V D1 NG V Di , and both V D1 NG B and V Di NG B p p qq “ p p qq p qX p q p qX p q are true twin-sets, s1 and t1 have the same set of neighbors in V H V D1 . p qz p q If H has a vertex s2 in V D1 NG B , then V D1 V H s1,s2 , be- p qz p q p qX p q “ t u cause both V D1 NG B and V D1 NG B have at most one vertex of p qX p q p qz p q H. Then G V H s1,s2 t1,t2 is isomorphic to H, a contradiction, rp p qz t uqYt us 1 because it is an induced subgraph of G V D1 S . Therefore, H has no zp p qY q vertices in V D1 NG B . Then G V H s1 t1 is isomorphic to H, p qz p q rp p qz t uq Y t us 1 a contradiction, because it is an induced subgraph of G V D1 S . zp p qY q Reduction Rule 8 (R8). Given an instance G, k with k 0 and a non- p q ą empty good modulator S of G, let B1,...,Bm be pairwise disjoint maximal true twin-sets in G S for m 6 such that NG Bi Bi 1 Bi 1 for z ě p q “ ´ Y ` each i 2,...,m 1 . Let ℓ be an integer in 3,...,m 2 such that P t ´ u 1 t ´ u |Bℓ| |Bi| for each i 3,...,m 2 , and G be a graph obtained from ď P t ´ u G B3 Bm 2 Bℓ by making Bℓ complete to B2 Bm 1. Then zpp Y¨¨¨Y ´ qz q Y ´ replace G, k with G1, k . p q p q To show that (R8) is safe, we will use the following lemma.

Lemma 5.4. Let G be a graph with disjoint true twin-sets B1,...,Bm for m 5 such that N Bi Bi 1 Bi 1 for each i 2,...,m 1 . Then G ě p q“ ´ Y ` P t ´ u

25 is a 3-leaf power if and only if G B3 Bm 2 is a 3-leaf power and zp Y¨¨¨Y ´ q has no paths from a vertex in B2 to a vertex in Bm´1.

Proof. It is clear that if G is a 3-leaf power, then G B3 Bm 2 is zp Y¨¨Ÿ ´ q a 3-leaf power, and has no paths from a vertex in B2 to a vertex in Bm´1, because otherwise G has a hole. Conversely, suppose that G B3 Bm 2 is a 3-leaf power, and has zp Y¨¨Ÿ ´ q no paths from a vertex in B2 to a vertex in Bm´1, and G has an obstruction H. Since G B3 Bm 2 is a 3-leaf power, H has at least one vertex zp Y¨¨Ÿ ´ q in B3 Bm 2. For each i 1,...,m , since Bi is a true twin-set Y¨¨Ÿ ´ P t u in G, H has at most one vertex in Bi by (O1). Then every vertex of H in B2 Bm 1 has degree at most 2 in H. If H has a vertex v in Bj Y¨¨Ÿ ´ for some j 3,...,m 2 , then both Bj 1 and Bj 1 have vertices of H P t ´ u ´ ` by (O6). This means that Bi contains exactly one vertex of H for each i 2,...,m 1 . Then H has vertices in each B1 and Bm as well by (O6). P t ´ u Thus, V H B2 Bm 1 contains at least three vertices of degree 2 in p qXp Y¨¨Ÿ ´ q H. Then H is a hole by (O7), a contradiction, because H B3 Bm 2 zp Y¨¨Ÿ ´ q is a path in G B3 Bm 2 from a vertex in B2 to a vertex in Bm 1. zp Y¨¨Ÿ ´ q ´ Therefore, G is a 3-leaf power.

Proof of Safeness for (R8). Firstly, we show that if G, k is a yes-instance, p q then so is G1, k . Suppose that G has a minimal modulator S1 of size at most p1 q 1 1 k. Since S is minimal, S Bi or S Bi Bi for each i 1,...,m X “H 1 X “ P t 1 u by Lemma 3.4. We claim that (a) if S B1 Bm is empty, then S is 1 1 Xp Y¨¨Ÿ q 1 a modulator of G . Since G S is a 3-leaf power, G B3 Bm 2 S is z zp Y¨¨Ÿ ´ Y q a 3-leaf power, and has no paths from a vertex in B2 to a vertex in Bm´1 by 1 1 1 Lemma 5.4. Since G B3 Bm 2 S is isomorphic to G Bℓ S , zp Y¨¨Ÿ ´ Y q zp Y q G1 S1 is a 3-leaf power by Lemma 5.4, and this proves (a). z 1 We claim that (b) if S B1 B2 Bm 1 Bm is non-empty, then X p Y Y ´ Y q S1 V G1 is a modulator of G1. Suppose that G1 S1 V G1 has an X p q 1 zp X p qq obstruction H. Since G B3 Bm´2 S is a 3-leaf power, and is 1 zp1 Y¨¨Ÿ1 1 Y 1 q 1 isomorphic to G Bℓ S V G , G Bℓ S V G is a 3-leaf power. zp Yp X p qqq zp Yp X p qqq Therefore, H has at least one vertex in Bℓ. For each i 1, 2,ℓ,m 1,m , 1 P t ´ u since Bi is a true twin-set in G , H has at most one vertex in Bi by (O1). Then every vertex of H in B2 Bℓ Bm 1 has degree at most 2 in H. Thus, Y Y ´ for each i 1, 2,ℓ,m 1,m , Bi contains exactly one vertex of H by (O6). P t ´ u Then S1 V G1 contains at least one vertex of H, a contradiction, because X p q H is an induced subgraph of G1 S1 V G1 , and this proves (b). zp1 X p qq Thus, we may assume that S B1 B2 Bm´1 Bm is empty and 1 X p Y Y 1 Y q S B3 Bm 2 is non-empty. Let T : S B3 Bm 2 Bℓ. X p Y¨¨Ÿ ´ q “ p zp Y¨¨Ÿ ´ qq Y

26 1 Since G B3 Bm´2 S is a 3-leaf power, and is isomorphic to 1 zp1 Y¨¨Ÿ Y q 1 G T , G T is a 3-leaf power. Since S B3 Bm´2 is non-empty, z z 1 X p Y¨¨Ÿ q 1 |T | |T B3 Bm 2 | |Bℓ| |S B3 Bm 2 | |Bℓ| |S | k. “ zp Y¨¨Ÿ ´ q ` ď zp Y¨¨Ÿ ´ q ` ď ď Therefore, if G, k is a yes-instance, then so is G1, k . p q p q Secondly, we show that if G1, k is a yes-instance, then so is G, k . p q p q Suppose that G1 has a minimal modulator S1 of size at most k. Since S1 is 1 1 minimal, S Bi or S Bi Bi for each i 1, 2,ℓ,m 1,m by X “ H X “ P t ´ u Lemma 3.4. It suffices to show that S1 is a modulator of G. 1 1 1 Since G S is a 3-leaf power, if S B1 B2 Bm´1 Bm is empty, 1 z 1 X p Y Y Y q then G Bℓ S is a 3-leaf power, and has no paths from a vertex in B2 zp Y q 1 1 to a vertex in Bm´1 by Lemma 5.4. Since G Bℓ S is isomorphic to 1 1 zp Y q G B3 Bm´2 S , G S is a 3-leaf power by Lemma 5.4. zp Y¨¨Ÿ Y q z 1 Thus, we may assume that S B1 B2 Bm´1 Bm is non-empty, 1 X p Y1 Y 1 Y q and G S has an obstruction H. Since G Bℓ S is a 3-leaf power, and is z 1 zp Y q 1 isomorphic to G B3 Bm 2 S , G B3 Bm 2 S is a 3-leaf zp Y¨¨Ÿ ´ Y q zp Y¨¨Ÿ ´ Y q power. Therefore, H has at least one vertex in B3 Bm 2. For each Y¨¨Ÿ ´ i 1,...,m , since Bi is a true twin-set in G, H has at most one vertex in P t u Bi by (O1). Then every vertex of H in B2 Bm 1 has degree at most Y¨¨Ÿ ´ 2 in H. If H has a vertex v in Bj for some j 3,...,m 2 , then both P t ´ u Bj´1 and Bj`1 have vertices of H by (O6). This means that Bi contains exactly one vertex of H for each i 2,...,m 1 by (O6), and both B1 P t ´ u 1 and Bm have vertices of H as well by (O6). Thus, S contains at least one vertex of H, a contradiction. Therefore, if G1, k is a yes-instance, then so p q is G, k . p q By applying aforementioned reduction rules exhaustively to an input instance G, k with a good modulator S of G, we can bound the size of p q each incomplete component of G S. z Proposition 5.5. Given an instance G, k with k 0 and a non-empty p q ą good modulator S of G, if none of (R2), (R5), (R6), (R7), and (R8) is applicable to G, k , then each incomplete component of G S has at most p q z k 1 k 4 |S| |S| 2k 15 vertices. p ` qp ` q p ` ` q To prove Proposition 5.5, we will use the following lemma, which we slightly rephrase for our application.

Lemma 5.6 (Brandst¨adt and Le [6, Corollary 11]). Let G be a 3-leaf power. If G has a vertex v of degree at least 1 such that G v is connected, then G v z z has a true twin-set B such that NG v B or NG v NG B . p q“ r s“ r s

27 Proof of Proposition 5.5. Let C be an incomplete component of G S with a z tree-clique decomposition F, Bu : u V F . Since S is a good modulator p t P p quq of G, G V C v is a 3-leaf power for each vertex v in S. Thus, if S r p q Y t us has a vertex w having a neighbor in a bag B of C, then w is complete to t u B by Lemma 5.2. This means that every bag of C is a true twin-set in G. Since (R5) is not applicable to G, k , each bag of C contains at most k 1 p q ` vertices. Therefore, in the remaining of this proof, we are going to bound the number of bags of C. Let X be the set of leaves of F whose bags are anti-complete to S.

Claim 4. If a node u of F X has degree at most 1 in F X, then Bu N S z z X p q‰ . H Proof of Claim 4. If NF u X, then Bu contains a neighbor of S, be- p q Ď cause otherwise C has no neighbors of S and (R2) is applicable to G, k . p q If NF u X is non-empty, then NF u X contains exactly one node u1, be- p qz p qz cause u has degree at most 1 in F X. If Bu contains no neighbors of S, then z (R6) is applicable to G, k by taking Bu1 as B. Therefore, Bu contains a p q neighbor of S.

Claim 5. The maximum degree of F is at most |S| 2k 7. ` ` Proof of Claim 5. Suppose that F has a node u of degree at least |S| 2k 8 ` ` in F . For each vertex w in S, if at least two components of C Bu have neigh- z bors of w, then all components of C Bu have neighbors of w by Lemma 5.6. z Thus, for each vertex w in S, we can choose a component of C Bu, say D, z such that either all other components of C Bu have neighbors of w, or no z other components of C Bu have neighbors of w. Since C Bu has at least z z |S| 2k 8 components, C Bu has distinct components D1,...,D2k 7 dif- ` ` z ` ferent from D such that for each vertex w in S, either all or none of them have neighbors of w. Thus, NG V D1 NG V D2k 7 . By the p p qq “ ¨¨¨ “ p p ` qq pigeonhole principle, V Di NG Bu or V Di NG Bu V Di p q Ď p q H ‰ p qX p q ‰ p q is satisﬁed by at least k 4 values of i, contradicting the assumption that ` (R7) is not applicable to G, k . p q For each vertex v in S, let Xv be the set of nodes of F X whose bags z contain neighbors of v, S1 be the set of vertices v in S such that Xv contains some leaf of F X, and S2 : S S1. Note that by Lemma 5.6, for each vertex z “ z v in S, if Xv is non-empty, then F X has a node, say p, such that Xv p or 1 z “ t u Xv NF zX p . Let F be a tree obtained from F X by contracting all edges “ r s 1 z in F Xv for each vertex v in S. By Claim 4, F has at most |S1| leaves, and r s therefore it has at most max |S1| 2, 0 branching nodes. Let Y be the set p ´ q

28 1 of nodes of F which come from Xv for some vertex v S, and Z be the set 1 P of branching nodes of F . Then |Y Z| |Y | |Z| |S| max |S1| 2, 0 Y ď ` ď ` p ´ qď 2|S|. Since (R8) is not applicable to G, k , each component of F 1 Y Z p q zp Y q has at most three nodes. Therefore, |V F 1 Y Z | 6|S|. Then by p zp Y qq ď Claim 5, |V F X | is at most p z q |Y | |S| 2k 8 |Z| |V F 1 Y Z | |S| |S| 2k 8 |S| 6|S| p ` ` q` ` p zp Y qq ď p ` ` q` ` |S| |S| 2k 15 . “ p ` ` q Since (R7) is not applicable to G, k , each node of F X is adjacent to p q z at most k 3 nodes in X. Thus, |V F | k 4 |S| |S| 2k 15 . ` p q ď p ` q p ` ` q By (R5), each bag of C has at most k 1 nodes. Therefore, |V C | ` p q ď k 1 k 4 |S| |S| 2k 15 . p ` qp ` q p ` ` q 6 A proof of the main theorem

In this section, we prove Theorem 1.1.

Algorithm 1 Kernelization for 3-Leaf Power Deletion 1: function Compress(G, k) 2: if k 0 then “ 3: if G is a 3-leaf power then return K1, 0 . p q 4: else return K2,2, 0 . p q 5: end if 6: else Find an instance G1, k1 equivalent to G, k , and a good mod- p q p q ulator S of G1 having size O k2 by Lemma 3.6. p q 7: if |S| k then return K1, 0 . ď p q 8: else if G1, k1 G, k then return Compress(G1, k1). p q ‰ p q 9: else if (Ri) for some i 2,..., 8 is applicable to G, k then P t u p q return Compress(G2, k2) where G2, k2 is the resulting instance ob- p q tained from G, k with S by applying (Ri). p q 10: else return G, k . p q 11: end if 12: end if 13: end function

Proof of Theorem 1.1. By Lemma 3.6, we can reduce an input instance to an equivalent instance with a good modulator having at most O k2 vertices in p q polynomial time. Each of (R2), . . . , (R8) can be applied in polynomial time by Theorem 2.4. Therefore, Algorithm 1 is a polynomial-time algorithm.

29 We are going to show that for the instance G, k returned in Line 10, p q G has at most O k2|S|6 vertices. Note that |S| k 1. By Proposi- p q ě ` tion 4.6, G S has at most 2 k 2 |S|2 non-trivial components. By Propo- z p ` q sition 5.1, each complete component of G S has at most 2 k 3 |S|4 3 z p ` q { vertices. By Proposition 5.5, each incomplete component of G S has at z most k 1 k 4 |S| |S| 2k 15 vertices. Therefore, each non-trivial p ` qp ` q p ` ` q component of G S has at most O k|S|4 vertices. Then the union of all non- z p q trivial components of G S has at most 2 k 2 |S|2 O k|S|4 O k2|S|6 z p ` q ¨ p q “ p q vertices. By Proposition 4.7, G S has at most 2 k 3 |S|4 3 isolated ver- z p ` q { tices. Thus, |V G | |S| 2 k 3 |S|4 3 O k2|S|6 O k2|S|6 . By p q ď ` p ` q { ` p q “ p q Lemma 3.6, |S| O k2 , and therefore |V G | O k14 . “ p q p q “ p q 7 Conclusions

In this paper, we show that 3-Leaf Power Deletion admits a kernel with O k14 vertices. It would be an interesting problem to significantly reduce p q the size of the kernel. Gurski and Wanke [22] stated that for every positive integer ℓ, ℓ-leaf powers have bounded clique-width. Rautenbach [32] presented a characterization of 4-leaf powers with no true twins as chordal graphs with ten forbidden induced subgraphs. This can be used to express, in monadic second-order logic, whether a graph is a 4-leaf power and whether there is a vertex set of size at most k whose deletion makes the graph a 4-leaf power. Therefore, by using the algorithm in [11], we deduce that 4-Leaf Power Deletion is fixed-parameter tractable when parameterized by k. It is natural to ask whether 4-Leaf Power Deletion admits a polynomial kernel. For ℓ 5, ě we do not know whether we can express ℓ-leaf powers in monadic second- order logic. If it is true for some ℓ, then not only ℓ-Leaf Power Deletion is fixed-parameter tractable, but also ℓ-Leaf Power Recognition can be solved in polynomial time, which is still open for ℓ 7. ě References

[1] Akanksha Agrawal, Daniel Lokshtanov, Pranabendu Misra, Saket Saurabh, and Meirav Zehavi, Feedback vertex set inspired kernel for chordal vertex deletion, ACM Trans. Algorithms 15 (2019), no. 1, Art. 11, 28. MR 3914442

30 [2] Akanksha Agrawal, Pranabendu Misra, Saket Saurabh, and Meirav Ze- havi, Interval vertex deletion admits a polynomial kernel, Proceedings of the Thirtieth Annual ACM-SIAM Symposium on Discrete Algorithms, SIAM, Philadelphia, PA, 2019, pp. 1711–1730. MR 3909572

[3] Jungho Ahn, A polynomial kernel for 3-leaf power deletion, Master’s thesis, KAIST, South Korea, 2020.

[4] Vineet Bafna, Piotr Berman, and Toshihiro Fujito, A 2-approximation algorithm for the undirected feedback vertex set problem, SIAM J. Dis- crete Math. 12 (1999), no. 3, 289–297. MR 1710236

[5] Hans-J¨urgen Bandelt and Henry Martyn Mulder, Distance-hereditary graphs, J. Combin. Theory Ser. B 41 (1986), no. 2, 182–208. MR 859310

[6] Andreas Brandst¨adt and Van Bang Le, Structure and linear time recognition of 3-leaf powers, Inform. Process. Lett. 98 (2006), no. 4, 133–138. MR 2211095

[7] Andreas Brandst¨adt, Van Bang Le, and R. Sritharan, Structure and linear-time recognition of 4-leaf powers, ACM Trans. Algorithms 5 (2009), no. 1, Art. 11, 22. MR 2479182

[8] Yixin Cao and D´aniel Marx, Chordal editing is ﬁxed-parameter tractable, Algorithmica 75 (2016), no. 1, 118–137. MR 3492058

[9] Maw-Shang Chang and Ming-Tat Ko, The 3-Steiner root problem, Graph-theoretic concepts in computer science, Lecture Notes in Com- put. Sci., vol. 4769, Springer, Berlin, 2007, pp. 109–120. MR 2428569

[10] Jianer Chen, Iyad A. Kanj, and Ge Xia, Improved upper bounds for vertex cover, Theoret. Comput. Sci. 411 (2010), no. 40-42, 3736–3756. MR 2724100

[11] Bruno Courcelle, Johann A. Makowsky, and Udi Rotics, Linear time solvable optimization problems on graphs of bounded clique-width, The- ory Comput. Syst. 33 (2000), no. 2, 125–150. MR 2001e:68146

[12] William H. Cunningham, Decomposition of directed graphs, SIAM J. Algebraic Discrete Methods 3 (1982), no. 2, 214–228. MR 655562

[13] Marek Cygan, Fedor V. Fomin,Lukasz Kowalik, Daniel Lokshtanov, D´aniel Marx, Marcin Pilipczuk, Micha l Pilipczuk, and Saket Saurabh, Parameterized algorithms, Springer, Cham, 2015. MR 3380745

31 [14] Michael Dom, Jiong Guo, Falk H¨uﬀner, and Rolf Niedermeier, Error compensation in leaf power problems, Algorithmica 44 (2006), no. 4, 363–381.

[15] Rodney G. Downey and Michael R. Fellows, Fundamentals of parameterized complexity, Texts in Computer Science, Springer, London, 2013. MR 3154461

[16] Guillaume Ducoﬀe, The 4-Steiner root problem, Graph-theoretic concepts in computer science, Lecture Notes in Comput. Sci., vol. 11789, Springer, Berlin, 2019, pp. 14–26.

[17] Eduard Eiben, Robert Ganian, and O-joung Kwon, A single-exponential ﬁxed-parameter algorithm for distance-hereditary vertex deletion, J. Comput. System Sci. 97 (2018), 121–146. MR 3832000

[18] David Eppstein and Elham Havvaei, Parameterized leaf power recognition via embedding into graph products, 13th International Symposium on Parameterized and Exact Computation, LIPIcs. Leibniz Int. Proc. Inform., vol. 115, Schloss Dagstuhl. Leibniz-Zent. Inform., Wadern, 2019, pp. Art. No. 16, 14. MR 3932984

[19] J. Flum and M. Grohe, Parameterized complexity theory, Texts in The- oretical Computer Science. An EATCS Series, Springer-Verlag, Berlin, 2006. MR 2238686

[20] Fedor V. Fomin, Tien-Nam Le, Daniel Lokshtanov, Saket Saurabh, St´ephan Thomass´e, and Meirav Zehavi, Subquadratic kernels for im- plicit 3-hitting set and 3-set packing problems, ACM Trans. Algorithms 15 (2019), no. 1, Art. 13, 44. MR 3914444

[21] Fedor V. Fomin, Daniel Lokshtanov, Saket Saurabh, and Meirav Zehavi, Kernelization, Cambridge University Press, Cambridge, 2019, Theory of parameterized preprocessing. MR 3887686

[22] Frank Gurski and Egon Wanke, The NLC-width and clique-width for powers of graphs of bounded tree-width, Discrete Appl. Math. 157 (2009), no. 4, 583–595. MR 2499471

[23] Falk H¨uﬀner, Christian Komusiewicz, Hannes Moser, and Rolf Nieder- meier, Fixed-parameter algorithms for cluster vertex deletion, Theory Comput. Syst. 47 (2010), no. 1, 196–217. MR 2643916

32 [24] Bart M. P. Jansen and Marcin Pilipczuk, Approximation and kernelization for chordal vertex deletion, SIAM J. Discrete Math. 32 (2018), no. 3, 2258–2301. MR 3852710

[25] Eun Jung Kim and O-joung Kwon, A polynomial kernel for block graph deletion, Algorithmica 79 (2017), no. 1, 251–270. MR 3673102

[26] , A polynomial kernel for distance-hereditary vertex deletion, Algorithms and data structures, Lecture Notes in Comput. Sci., vol. 10389, Springer, Cham, 2017, pp. 509–520. MR 3690913

[27] Michael Lampis, A kernel of order 2k c log k for vertex cover, Inform. ´ Process. Lett. 111 (2011), no. 23-24, 1089–1091. MR 2893938

[28] John M. Lewis and Mihalis Yannakakis, The node-deletion problem for hereditary properties is NP-complete, J. Comput. System Sci. 20 (1980), no. 2, 219–230, ACM-SIGACT Symposium on the Theory of Comput- ing (San Diego, Calif., 1978). MR 574592

[29] Daniel Lokshtanov, N. S. Narayanaswamy, Venkatesh Raman, M. S. Ramanujan, and Saket Saurabh, Faster parameterized algorithms using linear programming, ACM Trans. Algorithms 11 (2014), no. 2, Art. 15, 31. MR 3283570

[30] D´aniel Marx, Chordal deletion is ﬁxed-parameter tractable, Algorith- mica 57 (2010), no. 4, 747–768. MR 2629495

[31] Naomi Nishimura, Prabhakar Ragde, and Dimitrios M. Thilikos, On graph powers for leaf-labeled trees, J. Algorithms 42 (2002), no. 1, 69– 108. MR 1874637

[32] Dieter Rautenbach, Some remarks about leaf roots, Discrete Math. 306 (2006), no. 13, 1456–1461. MR 2237730