Computation of and Related Contraction Problems: Tight Lower Bounds Fedor V. Fomin University of Bergen, Bergen, Norway, [email protected] Daniel Lokshtanov University of California, Santa Barbara, USA, [email protected] Ivan Mihajlin University of California, San Diego, California, USA [email protected] Saket Saurabh Department of Informatics, University of Bergen, Norway, and The Institute of Mathematical Sciences, HBNI and IRL 2000 ReLaX, Chennai, India, [email protected] Meirav Zehavi Ben-Gurion University of the Negev, Beer-Sheva, Israel, [email protected]

Abstract We prove that the Hadwiger number of an n-vertex graph G (the maximum size of a minor in G) cannot be computed in time no(n), unless the Exponential Time Hypothesis (ETH) fails. This resolves a well-known open question in the of exact exponential algorithms. The technique developed for resolving the Hadwiger number problem has a wider applicability. We use it to rule out the existence of no(n)-time algorithms (up to ETH) for a large class of computational problems concerning edge contractions in graphs.

2012 ACM Subject Classification Design and analysis of algorithms → Exact algorithms, Design and analysis of algorithms → Graph algorithm analysis

Keywords and phrases Hadwiger Number, Exponential-Time Hypothesis, Exact Algorithms, Edge Contraction Problems

Digital Object Identifier 10.4230/LIPIcs.CVIT.2016.23

Funding Fedor V. Fomin: Research Council of Norway via the project MULTIVAL.

Daniel Lokshtanov: European Research Council (ERC) under the European Union’s Horizon 2020

research and innovation programme (grant no. 715744), and United States - Israel Binational Science

Foundation grant no. 2018302.

Saket Saurabh: European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme (grant no. 819416), and Swarnajayanti Fellowship grant DST/SJF/MSA- 01/2017-18. Meirav Zehavi: Israel Science Foundation grant no. 1176/18, and United States – Israel Binational

arXiv:2004.11621v1 [cs.DS] 24 Apr 2020 Science Foundation grant no. 2018302.

© Fedor V. Fomin, Daniel Lokshtanov, Ivan Mihajlin, Saket Saurabh and Meirav Zehavi; licensed under Creative Commons License CC-BY 42nd Conference on Very Important Topics (CVIT 2016). Editors: John Q. Open and Joan R. Access; Article No. 23; pp. 23:1–23:25 Leibniz International Proceedings in Informatics Schloss Dagstuhl – Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany 23:2 Tight Lower Bounds on the Computation of Hadwiger Number

1 Introduction

The Hadwiger number h(G) of a graph G is the largest number h for which the Kh is a minor of G. Equivalently, h(G) is the maximum size of the largest complete graph that can be obtained from G by contracting edges. It is named after , who conjectured in 1943 that the Hadwiger number of G is always at least as large as its chromatic number. According to Bollobás, Catlin, and Erdős, this conjecture remains “one of the deepest unsolved problems in ” [4]. The Hadwiger number of an n-vertex graph G can be easily computed in time nO(n) by brute-forcing through all possible partitions of the vertex set of G into connected sets, contracting each set into one vertex and checking whether the resulting graph is a complete graph. The question whether the Hadwiger number of a graph can be computed in single- exponential 2O(n) time was previously asked in [1, 6, 14]. Our main result provides a negative answer to this open question.

I Theorem 1. Unless the Exponential Time Hypothesis (ETH) is false, there does not exist an algorithm computing the Hadwiger number of an n-vertex graph in time no(n).

The interest in the complexity of the Hadwiger number is naturally explained by the recent developments in the area of exact exponential algorithms, that is, algorithms solving intractable problems significantly faster than the trivial exhaustive search, though still in exponential time [8]. Within the last decade, significant progress on upper and lower bounds of exponential algorithms has been achieved. Drastic improvements over brute-force algorithms were obtained for a number of fundamental problems like [3] and Hamiltonicity [2]. On the other hand, by making use of the ETH, lower bounds could be obtained for 2-CSP [16] or for Subgraph Isomorphism and Graph Homomorphism [6]. (deciding whether a graph G contains a graph H as a minor) is a fundamental problem in graph theory and graph algorithms. Graph Minor could be seen as special case of a general graph embedding problem where one wants to embed a graph H into graph G. In what follows we will use n to denote the number of vertices in G and h to denote the number of vertices in H. By the theorem of Robertson and Seymour [15], there exists a computable function f and an algorithm that, for given graphs G and H, checks in time f(h) · n3 whether H is a minor of G. Thus the problem is fixed-parameter tractable (FPT) being parameterized by H. On the other hand, Cygan et al. [6] proved that unless the ETH fails, this problem cannot be solved in time no(n) even in the case when |V (G)| = |V (H)|. Other interesting embedding problems that are strongly related to Graph Minor include the following problems. Subgraph Isomorphism: Given two graphs G and H, decide whether G contains a subgraph isomorphic to H. This problem cannot be solved in time no(n) when |V (G)| = |V (H)|, unless the ETH fails [6]. In the special case called Clique, when H is a clique, a brute-force algorithm checking for every vertex subset of G whether it is a clique of size h solves the problem in time nO(h). The same algorithm also runs in single-exponential time O(2nn2). It is also known that Clique is W[1]-hard parameterized by h and cannot be solved in time f(h) · no(h) for any function f unless the ETH fails [7, 5]. Graph Homomorphism: Given two graphs G and H, decide whether there exists a homomorphism from G to H. (A homomorphism G → H from an undirected graph G to an undirected graph H is a mapping from the vertex set of G to that of H such that the image of every edge of G is an edge of H.) This problem is trivially solvable in time hO(n), and an algorithm of running time ho(n) for this problem would yield the failure of F.V. Fomin, D. Lokshtanov, I. Mihajlin, S. Saurabh and M. Zehavi 23:3

the ETH [6]. However, for the special case of H being a clique, Graph Homomorphism is equivalent to h-Coloring (deciding whether the chromatic number of G is at most h), and thus is solvable in single-exponential time 2n · nO(1) [3, 13]. When the graph G is a complete graph, the problem is equivalent to finding a clique of size n in H, and then is solvable in time 2h · hO(1). Topological Graph Minor: Given two graphs G and H, decide whether G contains H as a topological minor. (We say that a graph H is a subdivision of a graph G if G can be obtained from H by contracting only edges incident with at least one vertex of degree two. A graph H is called a topological minor of a graph G if a subdivision of H is isomorphic to a subgraph of G.) This problem is, perhaps, the closest “relative” of Graph Minor. Grohe et al. [10] gave an algorithm of running time f(h) · n3 for this problem for some computable function f. Similar to Graph Minor and Subgraph Isomorphism, this problem cannot be solved in time no(n) when |V (G)| = |V (H)|, unless the ETH fails [6]. However for the special case of the problem with H being a complete graph, Lingas and Wahlen [14] gave a single-exponential algorithm solving the problem in time 2O(n).

Thus all the above graph embedding “relatives” of Graph Minor are solvable in single-exponential time when graph H is a clique. However, from the perspective of exact exponential algorithms, Theorem 1 implies that finding the largest clique minor is the most difficult problem out of them all. This is why we find the lower bound provided by Theorem 1 surprising. Moreover, from the perspective of , finding a clique minor of size h, which is FPT, is actually easier than finding a clique (as a subgraph) of size h, which is W[1]-hard, as well as from finding an h-coloring of a graph, which is para-NP-hard. Theorem 1 also answers another question of Cygan et al. [6], who asked whether deciding if a graph H can be obtained from a graph G only by edge contractions, could be resolved in single-exponential time. By Theorem 1, the existence of such an algorithm is highly unlikely even when the graph H is a complete graph. Moreover, the technique developed to prove Theorem 1, appears to be extremely useful to rule out the existence of no(n)-time algorithms for various contraction problems. We formalize our results with the following F-Contraction problem. Let F be a graph class. Given a graph G and t ∈ N, the task is to decide whether there exists a subset F ⊆ E(G) of size at most t such that G/F ∈ F (where G/F is the graph obtained from G by contracting the edges in F ). We prove that in each of the cases of F-Contraction where F is the family of chordal graphs, interval graphs, proper interval graphs, threshold graphs, trivially perfect graphs, split graphs, complete split graphs and perfect graphs, unless the ETH fails, F-Contraction is not solvable in time no(n). For lack of space, these results are relegated to Appendix C.

Technical Details. A summary of the reductions presented in this paper is given in Fig. 1. To prove our lower bounds, we first revisit the proof of Cygan et al. [6] for the ETH-hardness of a problem called List Subgraph Isomorphism. Informally, in this problem we are given two graphs G and H on the same number of vertices, as well as a list of vertices in H for each vertex in G, and we need to find a copy of G in H so that each vertex u in G is mapped to a vertex v in H that belongs to its list (i.e. v belongs to the list of u). We prove that the instances produced by the reduction (after some modification) of [6] have a very useful property that we crucially exploit later. Specifically, we construct a proper coloring of G as well as a proper coloring of H, and show that every vertex v in H that belongs to the list of some vertex u is, in fact, of the same color as u. Having proved the above, we turn to prove the ETH-hardness of a special case of Clique

C V I T 2 0 1 6 23:4 Tight Lower Bounds on the Computation of Hadwiger Number

Properly Colored List Subgraph Homomorphism

Properly Colored List Subgraph Isomorphism

Cross Matching

Noisy Structured Clique Contraction special case of Clique Contraction

Hadwiger Number Split Contraction, Perfect Contraction F-Contraction Complete Split Contraction special case of Chordal Contraction, Interval Contraction, Proper Interval Contraction, Threshold Contraction, Trivially Perfect Contraction

Figure 1 A summary of the problems considered in this paper, and the reductions between them.

Contraction where the input graph is highly structured. To this end, we introduce an intermediate problem called Cross Matching. Informally, in this problem we are given a graph L with a partition (A, B) of its vertex set, and need to find a perfect matching between A and B whose contraction gives a clique. To see the connection between this problem and List Subgraph Isomorphism, think of the subgraph of L induced by one side of the partition—say, A—as a representation of the complement of G, and the subgraph of L induced by the other side of the partition as a representation of H. Then, the edges that go across A and B in a perfect matching can be thought of as a mapping of the vertices of G to the vertices of H. The crossing edges of L are easily defined such that necessarily a vertex of G can only be matched to a vertex in its list. In particular, we would like to enforce that every “non-edge” of the complement of G (which corresponds to an edge of G) would have to be mapped to an edge of H in order to obtain a clique. However, the troublesome part is that non-edges of the complement of G may also be “filled” (to eventually get a clique) using crossing edges rather than only edges of H. To argue that this critical issue does not arise, we crucially rely on the proper colorings of G and H. Now, for the connection between Cross Matching and Clique Contraction, note that a solution to an instance of Cross Matching is clearly a solution to the instance of Clique Contraction defined by the same graph, but the other direction is not true. By adding certain vertices and edges to the graph of an instance of Cross Matching, we enforce all solutions to be perfect matchings between A and B. In particular, we construct the instances of Clique Contraction in a highly structured manner that allows us to derive not only the ETH-hardness of Clique Contraction itself, but to build upon them and further derive ETH-hardness for a wide variety of other contraction problems. In particular, we show that the addition of “noise” (that is, extra vertices and edges) to any structured instance of Clique Contraction has very limited effect. Roughly speaking, we show that the edges in the “noise” and the edges going across the “noise” and core of the graph (that is, the original vertices corresponding to the structured instance of Clique Contraction) are not “helpful” when trying to create a clique on the core (i.e. it is not helpful to try to use these edges in order to fill non-edges between vertices in the core). Depending on the contraction problem at hand, the noise is slightly different, but the proof technique stays the same—first showing that the core must yield a clique, and then using the argument above F.V. Fomin, D. Lokshtanov, I. Mihajlin, S. Saurabh and M. Zehavi 23:5

(in fact, in all cases but that of perfect graphs, we are able to invoke the argument as a black box) to show that the noise is, in a sense, irrelevant. Preliminaries. As we only use standard notations, we relegate them to Appendix A.

2 Lower Bound: Prop-Colored List Subgraph Isomorphism

In this section we build upon the work of Cygan et al. [6] and show a lower bound for a problem called Properly Colored List Subgraph Isomorphism (Prop-Col LSI). Intuitively, Prop-Col LSI is a variant of Spanning Subgraph Isomorphism where given two graphs G and H, we ask whether G is isomorphic to some spanning subgraph of H. The input to the variant consists also of proper colorings of G and H and an additional labeling of vertices in G by subsets of vertices in H of the same color, so that each vertex in G can be mapped only to vertices in H contained in its list. Formally, it is defined as follows.

Properly Colored List Subgraph Isomorphism (Prop-Col LSI) Input: Graphs G and H with proper colorings cG : V (G) → {1, . . . , k} and cH : V (H) → {1, . . . , k} for some k ∈ N, respectively, and a function ` : V (G) → 2V (H) such that for every u ∈ V (G) and v ∈ `(u), cG(u) = cH (v). Question: Does there exist a bijective function ϕ : V (G) → V (H) such that (i) for every {u, v} ∈ E(G), {ϕ(u), ϕ(v)} ∈ E(H), and (ii) for every u ∈ V (G), ϕ(u) ∈ `(u)? Notice that as the function ϕ above is bijective rather than only injective, we seek a spanning subgraph. Our objective is to prove the following statement.

I Lemma 2. Unless the ETH is false, there does not exist an algorithm that solves Prop-Col LSI in time no(n) where n = |V (G)|.

In [6], the authors considered the two problems defined below. Intuitively, the second is defined as Prop-Col LSI when no proper colorings of H and G are given (and hence the labeling of vertices in G is not restricted accordingly); the first is defined as the second when we seek a homomorphism rather than an isomorphism (i.e., the sought function ϕ may not be injective) and also |V (G)| may not be equal to |V (H)| (thus ϕ may neither be onto).

List Subgraph Homomorphism (LSH) Input: Graphs G and H, and a function ` : V (G) → 2V (H) . Question: Does there exist a function ϕ : V (G) → V (H) such that (i) for every {u, v} ∈ E(G), {ϕ(u), ϕ(v)} ∈ E(H), and (ii) for every u ∈ V (G), ϕ(u) ∈ `(u)?

List Subgraph Isomorphism (LSI) Input: Graphs G and H where |V (G)| = |V (H)|, and a function ` : V (G) → 2V (H). Question: Does there exist a bijective function ϕ : V (G) → V (H) such that (i) for every {u, v} ∈ E(G), {ϕ(u), ϕ(v)} ∈ E(H), and (ii) for every u ∈ V (G), ϕ(u) ∈ `(u)? The proof of hardness of LSI consists of two parts:

Showing ETH-hardness of LSH. Giving a fine-grained reduction from LSH to LSI.

We cannot use the hardness of LSI as a black box because Prop-Col LSI is a special case of LSI. Nevertheless, we will prove that the instances generated by the reduction (with a minor crucial modification) of Cygan et al. [6] have the additional properties required to make them instances of our special case.

C V I T 2 0 1 6 23:6 Tight Lower Bounds on the Computation of Hadwiger Number

Lower Bound: Properly Colored Subgraph Homomorphism. Adapting the scheme of Cygan et al. [6] to our purpose, we will first show that finding a homomorphism remains hard if it has to preserve a given proper coloring:

Properly Colored List Subgraph Homomorphism (Prop-Col LSH) Input: Graphs G and H with proper colorings cG : V (G) → {1, . . . , k} and cH : V (H) → {1, . . . , k} for some k ∈ N, respectively, and a function ` : V (G) → 2V (H) such that for every u ∈ V (G) and v ∈ `(u), cG(u) = cH (v). Question: Does there exist a function ϕ : V (G) → V (H) such that (i) for every {u, v} ∈ E(G), {ϕ(u), ϕ(v)} ∈ E(H), and (ii) for every u ∈ V (G), ϕ(u) ∈ `(u)? In [6], the authors gave a reduction from the 3-Coloring problem on n-vertex graphs of degree 4 (which is known not to be solvable in time 2o(n) unless the ETH fails), which generates equivalent instances (G0,H0, `) of LSH where both |V (G0)| and |V (H0)| are bounded by n o(n) O( log n ). This proves that LSH is not solvable in time n where n = max{|V (G)|, |V (H)|} unless the ETH fails. For their reduction, Cygan et al. [6] considered the notion of a grouping (also known as quotient graph) Ge of a graph G is a graph with vertex set V (Ge) = {B1,B2,...,Bt} where (B1,B2,...,Bt) is a partition of V (G) for some t ∈ N and for any distinct i, j ∈ {1, . . . , t}, the vertices Bi and Bj are adjacent in Ge if and only if there exist u ∈ Bi and v ∈ Bj that are adjacent in G. Specifically, they computed a grouping with a coloring having specific properties as stated in the following lemma (see also Fig. 5 in Appendix B.).

I Lemma 3 (Lemma 3.2 in [6]). For any constant d ≥ 1, there exist positive integers λ = λ(d), n0 = n0(d) and a polynomial time algorithm that for a given graph G on n ≥ n0 vertices of p n maximum degree d and a positive integer r ≤ 2λ , finds a grouping Ge of G and a coloring ec : V (Ge) → [λr] with the following properties: 1. |V (Ge)| ≤ |V (G)|/r; 2 1 2. The coloring ec is a proper coloring of Ge ; 3. Each vertex of Ge is an independent set in G; 4. For any edge {Bi,Bj} ∈ E(Ge), there exists exactly one pair (u, v) ∈ Bi × Bj such that {u, v} ∈ E(G).

Now, we describe the reduction of [6]. Here, without loss of generality, it is assumed that G has no isolated vertices, else they can be removed. An explanation of the intuition behind this somewhat technical definition is given below it.

I Definition 4. For any instance G of 3-Coloring where G has degree d and a positive p integer r = o( |V (G)|), the instance reduce(G) = (G,e H,e `) of LSH is defined as follows. The graph Ge. Let Ge and ec : V (Ge) → {1, 2,...,L} be the grouping and coloring given by Lemma 3 where L = λ(d)r. Additionally, for each B ∈ V (Ge), define φB : {1, 2,...,L} → B ∪ {0} as follows: for any i ∈ {1, 2,...,L}, if there exists (u, v, B0) such that u ∈ B 0 0 2 and v ∈ B , {u, v} ∈ E(G) and ec(B ) = i, then φB(i) = u, and otherwise φB(i) = 0. The graph He. Let V (He) = {(R, l): R ∈ {0, 1, 2, 3}L, l ∈ L},3 and E(He) = {{(R, l), (R0, l0)} : R[l0] 6= R0[l]}.

1 The square G2 of a graph G is the graph on vertex set V (G) and edge set {{u, v} : {u, v} ∈ E(G) or there exists w ∈ V (G) with {u, w}, {v, w} ∈ E(G)}. 2 The uniqueness of u (if it exists), and thus the validity of φB , follows from Properties 2 and 4 in Lemma 3. 3 That is, R is a vector with L entries where each entry is 0, 1, 2 or 3. F.V. Fomin, D. Lokshtanov, I. Mihajlin, S. Saurabh and M. Zehavi 23:7

The labeling `. For any B ∈ V (Ge), let `(B) contain all vertices (R, l) ∈ V (He) such that ec(B) = l, and there exists f : B → {1, 2, 3} such that for all i ∈ {1, 2,...,L}, either φB(i) = R[i] = 0 or both φB(i) 6= 0 and f(φB(i)) = R[i].

Intuitively, for every vertex B ∈ V (Ge), the function φB can be interpreted as follows. It is the assignment, for every possible color i ∈ {1,...,L}, of the unique vertex u within the vertex set identified with B itself that is adjacent to some vertex in the vertex subset identified with some vertex B0 ∈ V (Ge) colored i, if such a vertex u exists (else the assignment is of 0). In a sense, B thus stores the information on the identity of each vertex within it that is adjacent (in G) to some vertex outside of it, where each such internal vertex is uniquely accessed by specifying the color of the vertex in Ge whose identified vertex set contains the neighbor. With respect to the graph He and labeling `, we interpret each vertex (R, l) ∈ V (He) as a “placeholder” (i.e. potential assignment of the sought function ϕ) for any vertex B ∈ V (Ge) that “complies with the pattern encoded by the pair (R, l)” as follows. First and straightforwardly, B must be colored l. Here, we remind that the colors of vertices in Ge belong to {1,...,L}, while vertices in G are colored 1, 2 or 3 only. Then, the second requirement is that we can recolor (by f) the vertices in B so that the color of each vertex in B that is adjacent (in G) to some vertex outside B is as encoded by the vector R—that is, for each color i ∈ {1,...,L}, if the vertex φB(i) is defined (i.e., φB(i) 6= 0), then its color (which is 1, 2 or 3) must be equal to the i-th entry of R. (Further intuition is given in Fig. 5 in Appendix B.) Now, we state the correctness of the reduction.

I Lemma 5 (Lemma 3.3 in [6]). For any instance G of 3-Coloring where G is an n-vertex p graph of degree d, and a positive integer r = o( |V (G)|), the instance reduce(G) = (G,e H,e `) is computable in time polynomial in the sizes of G, Ge and He, and has the following properties. G is a Yes-instance of 3-Coloring if and only if (G,e H,e `) is a Yes-instance of LSH. |V (Ge)| ≤ n/r, and |V (He)| ≤ γ(d)r where γ is some computable function of d.

We next prove that we can add colorings to the instance reduce(G) = (G,e H,e `) of LSH in order to cast it as an instance of Prop-Col LSH while making a minor mandatory modification to the graph He.

I Lemma 6. Given an instance reduce(G) = (G,e H,e `) of LSH, an equivalent instance 0 0 (G,e He , c , c 0 , `) of Prop-Col LSH, where He is a subgraph of He, is computable in Ge He polynomial time.

Proof. Define c = c where c is the coloring of Ge in Definition 4. Additionally, let He 0 be the Ge e e subgraph of He induced by the vertex set {(R, l) ∈ V (He): there exists B ∈ V (Ge) such that (R, l) ∈ `(B)}. Then, define c : V (H0) → {1, 2,...,L} as follows: for any (R, l) ∈ V (H0), H0 e e e 0 define c 0 ((R, l)) = l. Notice that, by the definition of V (He ), every set assigned by ` is He subset of V (He 0). 0 First, we assert that (G,e He , c , c 0 , `) is an instance of Prop-Col LSH. To this end, Ge He we need to verify that the three following properties hold. 1. c is a proper coloring of G. G e e 0 2. c 0 is a proper coloring of He . He 3. For every B ∈ V (Ge) and (R, l) ∈ `(B), it holds that c (B) = c 0 ((R, l)). Ge He By the definition of c , it is a proper coloring of Ge2, which is a supergraph of Ge. Thus, Ge c is a proper coloring of Ge. Ge

C V I T 2 0 1 6 23:8 Tight Lower Bounds on the Computation of Hadwiger Number

Now, we argue that c is a proper coloring of H0. To this end, consider some edge H0 e 0 0 0 e 0 0 {(R, l), (R , l )} ∈ E(He ). We need to show that c 0 ((R, l)) 6= c 0 ((R , l )). By the definition He He of c , we have that c ((R, l)) = l and c ((R0, l0)) = l0, and therefore it suffices to show that He0 He0 He0 l 6= l0. By the definition of E(He) (which is a superset of E(He 0)), we have that R[l0] 6= R0[l]. Thus, necessarily at least one among R[l0] and R0[l] is not 0, and so we suppose w.l.o.g. that R[l0] is not 0. Furthermore, since (R, l) ∈ V (He 0), we have that there exists B ∈ E(Ge) such that (R, l) ∈ `(B). Thus, ec(B) = l. There exists f : B → {1, 2, 3} such that for all i ∈ {1, 2,...,L}, either φB(i) = R[i] = 0 or both φB(i) 6= 0 and f(φB(i)) = R[i]. 0 0 From the second property, and because R[l ] 6= 0, we necessarily have that both φB(l ) 6= 0 0 0 0 and f(φB(l )) = R[l ]. In particular, by the definition of φB, having φB(l ) 6= 0 means 0 0 0 0 that there exists (u, v, B ) such that u ∈ B, v ∈ B , {u, v} ∈ E(G) and ec(B ) = l . By the definition of Ge as a grouping of G, having u ∈ B, v ∈ B0 and {u, v} ∈ E(G) implies that 0 0 {B,B } ∈ E(Ge). Because ec is a proper coloring of Ge, this means that ec(B) 6= ec(B ). Since c(B) = l and c(B0) = l0, we derive that l 6= l0. Hence, c is indeed a proper coloring of H0. e e H0 e 0 e To conclude that (G,e He , c , c 0 , `) is indeed an instance of Prop-Col LSH, it remains Ge He to assert that for every B ∈ V (Ge) and (R, l) ∈ `(B), it holds that c (B) = c 0 ((R, l)). To Ge He this end, consider some B ∈ V (Ge) and (R, l) ∈ `(B). By the definition of ` (recall Definition 4), (R, l) ∈ `(B) implies that c(B) = l. As c = c, we have that c (B) = l. Moreover, the e Ge e Ge definition of c directly implies that c ((R, l)) = l. Thus, c (B) = c ((R, l)). H0 H0 G H0 e e e e 0 Finally, we argue that (G,e H,e `) is a Yes-instance of LSH if and only if (G,e He , c , c 0 , `) Ge He is a Yes-instance of Prop-Col LSH. In one direction, because He 0 is a subgraph of He, it is 0 immediate that if (G,e He , c , c 0 , `) is a Yes-instance of Prop-Col LSH, then so is (G,e H,e `). Ge He For the other direction, suppose that (G,e H,e `) is a Yes-instance of LSH. Thus, there exists a function ϕ : V (Ge) → V (He) such that (i) for every {B,B0} ∈ E(Ge), {ϕ(B), ϕ(B0)} ∈ E(He), and (ii) for every B ∈ V (Ge), ϕ(B) ∈ `(B). In particular, directly by the definition of V (He 0), the second condition implies that for every B ∈ V (Ge), it holds that ϕ(B) ∈ V (He 0). Thus, because He 0 is an induced subgraph of He, it holds that for every {B,B0} ∈ E(Ge), 0 0 0 {ϕ(B), ϕ(B )} ∈ E(He ). Therefore, ϕ witnesses that (G,e He , c , c 0 , `) is a Yes-instance of Ge He Prop-Col LSH. J

We are now ready to assert the hardness of Prop-Col LSH. The proof, based on Lemmas 3, 5 and 6, can be found in Appendix B.

I Lemma 7. Unless the ETH is false, there does not exist an algorithm that solves Prop-Col LSH in time no(n) where n = max(|V (G)|, |V (H)|).

From Graph Homomorphism to Subgraph Isomorphism. In this part, we observe that the reduction of [6] from LSH to LSI can be essentially used as is to serve as a reduction from Prop-Col LSH to Prop-Col LSI. For the sake of completeness, we give the full details (and the conclusion of the proof of Lemma 2) in Appendix B.

3 Lower Bound for the Cross Matching Problem

In this section, towards the proof of a lower bound for Clique Contraction, we prove a lower bound for an intermediate problem called Cross Matching that somewhat resembles Clique Contraction, and which is defined as follows. Lemma 3.1. Unless the ETH is false, there does not exist an algorithm that solves Prop-Col- LSI in time no(n) where n = V (G) . | | TODO

4 Lower Bound for the Cross Matching Problem

In this section, towards the proof of a lower bound for Clique Contraction,weprovealower bound for an intermediate problem called Cross Matching that somewhat resembles Clique Contraction, and which is defined as follows. Lemma 3.1. Unless the ETH is false, there does not exist an algorithm that solves Prop-Col- Cross Matching LSI in time no(n) where n = V (G) . | | Input: A graph G withLemmaLemma a partition 3.1. 3.1.UnlessUnless ( theA, the ETH B ETH) of is is false,V false,(G there)where there does does notA not exist= exist anB an algorithm. algorithm that that solves solvesProp-Col-Prop-Col- o(n)TODO LSILSIinin time timenno(nwhere) wherenn==VV(G(G) .) . | | | | Question: Does there exist a perfect matching| | |M| in G such that every edge in M has one endpoint in ALemmaand the 3.1.TODO otherUnlessTODO thein ETHB, and is false,G/M thereis does a not clique? exist an algorithm that solves Prop-Col- LSI in time no(n) where n = V (G) . 4| Lower| Bound for the Cross Matching Problem Lemma 3.1. Unless the ETH is false, there does not exist an algorithm that solves Prop-Col- Our objectiveTODO is to prove the following statement. LSI in time no(n) where44n Lower= LowerV (G) . Bound Bound for for the theCrossCross Matching MatchingProblemProblem | In| this section, towards the proof of a lower bound for Clique Contraction,weprovealower Lemma 4.1.TODOUnless theIn this ETH section,bound is towards false, for an the there intermediate proof does of a lower not problem bound exist for calledanClique algorithmCross Contraction Matching that solves,weprovealowerthatCross somewhat resembles Clique 4 LoweroIn(n this) Bound section, towards for the theCross proof of Matching a lower boundProblem for Clique Contraction,weprovealower Matching in time n boundboundwhere for for an anContractionn intermediate intermediate= A . problem problem, and called which calledCross isCross defined Matching Matching as follows.thatthat somewhat somewhat resembles resemblesCliqueClique In this section, towards the proof| | of a lower bound for Clique Contraction,weprovealower 4 Lower BoundContractionContraction for theCross,Cross and, and which which Matching Matching is is defined definedProblem as as follows. follows. Proof. Targetingbound a contradiction, for an intermediate suppose problem called thatCross there Matching existsthat an somewhat algorithm, resembles denotedClique by Matchin- In this section, towardsCross the proof Matching of a lower bound for Clique Contraction,weprovealower ContractionCross, and Matching whichInput: is definedA as graph follows.o(n)G with a partition (A, B) of V (G)where A = B . gAlg, thatbound solves for anCross intermediate MatchingInput: problemA graph calledinCrossG timewith Matchingn a partition.that We ( somewhatA, will B) showof V resembles(G that)whereClique thisA implies= B . the existence| | | | of Cross MatchingInput: A graphQuestion:G with aDoes partition there (A, exist B) of aV perfect(G)where matchingA| =| B|M.| in G such that every edge in M has one Contraction, and whichQuestion: is definedDoes as follows. there exist a perfect matching M in G such|o(|n) that| | every edge in M has one an algorithm, denotedInput: A byQuestion: graphLSIAlgG with,Does thatendpoint a partition there solves exist in (A,AProp-Col B aand) perfect of V the(G)where matching other LSI inAMinB=,in timeB andG. suchG/Mn thatis, thereby a every clique? edge contradicting in M has one Cross Matching endpoint in A and the other in B, and G/M| is| a clique?| | Question:endpointDoes there in A existand a the perfect other matching in B, andM inG/MG suchis a that clique? every edge in M has one Lemma 3.1Input:andA hence graph G completingwith a partition ( theA,Our B) proof. ofobjectiveV (G)where is toA = proveB . the following statement. endpoint in AOurand objective the other isin toB, prove and G/M the followingis| a| clique?| | statement. Question: Does thereOur exist objective a perfect matching is to proveM in theG such following that every statement. edge in M has one We define theOur execution objective is of toLSIAlg prove theas following follows. statement. Given an instance (G, H, cG,cH , `) of Prop- Col LSI, endpointLSIAlg inconstructsA and theLemma other an in 4.1.BLemma instance, andUnlessG/M 4.1.is the ( aL, clique? ETH A,Unless B is) false, of theCross ETH there does is Matching false, not exist there an doesas algorithm follows not exist that (see an solvesFig. algorithmCross?): that solves Cross Our objectiveLemma isLemma 4.1. toMatching proveUnless the 4.1. following theMatchinginUnless ETH time statement. isn theo( false,n)in ETHwhere time there isn false,ndoes=o(nA) not. therewhere exist doesn an= algorithmnotA exist. that an algorithm solves Cross that solves Cross MatchingMatchingin time noin(n) timewherenno(n=) whereA . n = A| .| | | V (LLemma)=V 4.1.(G)UnlessVProof.( theH ETH). Targeting is false, there a contradiction, does| | not exist suppose| an| algorithm that that there solves existsCross an algorithm, denoted by Matchin- Matching in time no(n) where n =Proof.A . Targeting a contradiction, suppose that there exists an algorithm, denoted by Matchin- • Proof. [TargetingProof.gAlg,Targeting that a contradiction, solves| | aCross contradiction, suppose Matching that suppose therein time exists thatno an( theren). algorithm, We exists will show an denoted algorithm, that by thisMatchin- implies denoted the by existenceMatchin- of gAlg, that solves Crosso(n) Matching in time no(n). We will show that this implies the existence of Proof. TargetinggAlg, that a contradiction, solves Cross suppose Matching that therein time existsn an. We algorithm, willo show(n denoted) that this by Matchin- implies the existenceo(n) of E(L)=E(G) gAlgEan(H, algorithm, that) solvesu, denotedCross v : u byMatchingLSIAlgV (G,) that,vin time solvesL(nuProp-Col) .. We willo(n LSI show) in that time thisn implies, thereby the contradicting existence of an algorithm, denoted by LSIAlg,o that(n) solves Prop-Col LSI in time n , thereby contradicting o(n) • gAlg, that solves[CrossanLemma algorithm, Matching[3.1{{anand denotedinalgorithm, time hence} n by completing.2LSIAlg We denoted will, showthat the that solvesby2 proof.LSIAlg thisProp-Col implies}, that the solvesexistence LSI inProp-Col time of no(n), thereby LSI in contradicting time n , thereby contradicting an algorithm,Lemma denotedF.V.3.1 Fomin,and by LSIAlg hence D. Lokshtanov,, that completing solves I.Prop-Col Mihajlin, the proof. S. Saurabh LSI in time and M.no(n Zehavi), thereby contradicting 23:9 LemmaWe3.1 defineandLemma the hence execution3.1 completingand of henceLSIAlg the completingproof.as follows. the Given proof. an instance (G, H, cG,cH , `) of Prop- A =LemmaV (G3.1) andandWe hence defineB = completing theV ( executionH). the proof. of LSIAlg as follows. Given an instance (G, H, cG,cH , `) of Prop- • ColWe LSI define, LSIAlg theWe executionconstructs define of the anLSIAlg executioninstanceas follows. (L, of A,LSIAlg B Given) of Crossas an follows. instance Matching Given (G, H,as c an follows,c instance, `) (see of Prop-Fig. (G,?): H, cG,cH , `) of Prop- We defineCol the LSI execution, LSIAlg constructs of LSIAlg as an follows. instance Given (L, an A, instanceB) of Cross (G, H, Matching cG,cH , `) of asProp- follows (see Fig. G?): H Col LSI, LSIAlg constructsCol LSI an, LSIAlg instanceColconstructs ( LSIL, A,, BLSIAlg) of anCross instanceconstructs Matching (L, A, anas B follows instance) of Cross (see Fig. (L, Matching A,?): B) of Crossas follows Matching (see Fig. ?):as follows (see Fig. ?): Then, LSIAlg calls MatchingAlgV (L)=V (GV)(L)=Vwith(HV).(G ()L,V A,(H B). ) as input, and returns the answer of this call. • • [ [ Denote nV (=L)=VV ((GG) ) V.(H First,). note that by construction, A = B = n. Thus, because • E(L[)=E V(GE(L)()=L)=E(VHE()G(G)V) (u,VLE v()=H(H:).u)V (VG()Gu,) v,vV :(HuL().u)V .(G),v L(u) . |• | • • o(n[) •[ {{[[ } 2[ {{ [ } 2 2 } |2 | }| | MatchingAlg Eruns(L)= inE(G time) E(Hn ) u,,sodoes v : u V (GLSIAlg),v L(u.) . • A =[V (G)E andA([L{{=)=BV=E(}GV(G)(H and)2E).(ELB()=H=)2VE(H(Gu,).}) v E: u(H)V (G),vu, v L:(uu) . V (G),v L(u) . • • [ [ {{ } 2 2 } For the correctnessA = V (G) and B of=•the V (H). algorithm,• first suppose[ that[ {{ (G,} H, c2G,cH , `)isa2 Yes} -instance of • Then, LSIAlg calls MatchingAlg with (L, A, B) as input, and returns the answer of this call. Prop-Col LSI. This meansThen, A =LSIAlg thatV (G)calls there andAMatchingAlgB== existsVV(G(H)). and awith bijectiveB (=L,V A,( BH function)). as input, and' : returnsV (G) the answerV (H) of such this call. that Then, LSIAlgDenotecalls FigureMatchingAlgn• 2= TheV construction(Gwith) . (L, First, of A, an B instance) note as input, thatof Cross and by Matching construction,returnsin the the answer proofA of Lemma of= thisB 8. call.= n. Thus, because Denote| |n =• V (G) . First, note that by| construction,| | | A = !B = n. Thus, because (i) for everyDenoteu,MatchingAlg vn = V E(G)(runsG. ), First, in time' note(un) thato,(n'),sodoes( byv) construction,LSIAlgE(H. ),A and= B (ii)= n.for Thus, every becauseu V (G), '(u) L(u). | Then,| LSIAlg calls MatchingAlg| | o(n)with| (L,| A,| B|) as input, and returns| | the| answer| of this call. MatchingAlg runs in timeMatchingAlgno(n),sodoesThen,runsLSIAlg inLSIAlg time. ncalls,sodoesMatchingAlgLSIAlg.with (L, A, B) as input, and returns the answer of this call. { For} 2 the correctness { of the algorithm,} 2 first suppose that (G, H, c ,c , `)isaYes2 -instance of 2 Having ' at hand, weCross will MatchingDenote shown that= V ( (GL,) . A, First, B)isa noteYes that-instance, by construction,G whichH A will= B imply= n. that Thus, the because call For the correctness ofFor the algorithm,the correctness first suppose of the that algorithm, (G, H, cG,c firstH , `)isa supposeYes-instance that (G, of H, cG,cH , `)isaYes-instance of Prop-ColInput: LSIA graph. ThisG with means aDenote partition| that there|(A,n Bo)( exists=nof) V (GV) a(whereG bijective) .|A| = First,|B function|. note' : thatV (G) | by|V construction,(H|) such| that A = B = n. Thus, because Prop-Col LSI. ThisMatchingAlgProp-Col means thatruns there LSI. existsin This time a means bijectiven ,sodoes that| function there| LSIAlg' exists: V (G.) a bijectiveV (H) such function that! ' : V (G) V|(H|) such| that| to MatchingAlg(i)withfor everyQuestion: (L, A,u,Does vB) there asE(G exist input), a' perfect(u) returns, ' matching(v) MEYes(inHG),,such and thato(ii)(n hence every) for! edge everyLSIAlg in Muhas V (returnsG), '(u) YesL(u.). 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By the definition of E(L), suchthis that (i) meansfor every thatsu{u, v}cesu, ∈ E v to(G)/, show{Eϕ(u(G), ϕ) that(v and2)} ∈ Eu hence(Hand), andvu,(ii)are vfor adjacent everyE(uG∈).V (G inBy), L Conditionor u0 and(i)v0above,are adjacent we derive in L (or both). To this ϕ(u) ∈ L(u). Having ϕend,at hand,{ suppose we} 2 will show that thatu(L,and A, B) vis a{areYes-instance, not} 2 adjacent which will in L, else we are done. By the definition of E(L), imply that the call to MatchingAlg with (L, A, B) as input returns Yes2 , and hence LSIAlg returns Yes. this means that u, v / E(G) and hence u, v E(G). By Condition (i) above, we derive Based on ϕ, we define a subset M ⊆ E(L) as{ follows:} M2= {{u, ϕ(u)} : u ∈ A}. Notice{ } 2 that the containment of M in E(L) follows from the definition of E(L) and Condition (ii) 2 above. Moreover, by the definition of A, B and because ϕ is bijective, it further follows that M is a perfect matching in L such that every edge in M has one endpoint in A and the other in B. Thus, to conclude that (L, A, B) is a Yes-instance, it remains to argue that L/M is a clique. To this end, we consider two arbitrary vertices x and y of L/M, and prove that they are adjacent in L/M. Necessarily x is a vertex that replaced two vertices u ∈ A and u0 ∈ B such that {u, u0} ∈ M, and y is a vertex that replaced two vertices v ∈ A \{u} and v0 ∈ B \{u0} such that {v, v0} ∈ M. By the definition of contraction, to show that x and y are adjacent in L/M, it suffices to show that u and v are adjacent in L or u0 and v0 are

C V I T 2 0 1 6 23:10 Tight Lower Bounds on the Computation of Hadwiger Number

adjacent in L (or both). To this end, suppose that u and v are not adjacent in L, else we are done. By the definition of E(L), this means that {u, v} ∈/ E(G) and hence {u, v} ∈ E(G). By Condition (i) above, we derive that {ϕ(u), ϕ(v)} ∈ E(H). By the definition of M, we know that u0 = ϕ(u) and v0 = ϕ(v), therefore {u0, v0} ∈ E(H). In turn, by the definition of E(L), we get that {u0, v0} ∈ E(L). Thus, the proof of the forward direction is complete. Now, suppose that LSIAlg returns Yes, which means that the call to MatchingAlg with (L, A, B) returns Yes. Thus, (L, A, B) is a Yes-instance, which means that there exists a perfect matching M in G such that every edge in M has one endpoint in A and the other in B, and G/M is a clique. We define a function ϕ : A → B as follows. For every u ∈ V (G), let ϕ(u) = v where v is the unique vertex in B such that {u, v} ∈ M; the existence and uniqueness of v follows from the supposition that M is a perfect matching such that every edge in M has one endpoint in A and the other in B. Furthermore, by the definition of A, B and the edges in E(L) with one endpoint in A and the other in B, it directly follows that ϕ is a bijective mapping between V (G) and V (H) such that for every u ∈ V (G), it holds that ϕ(u) ∈ L(u). Thus, it remains to argue that for every edge {u, v} ∈ E(G), it holds that {ϕ(u), ϕ(v)} ∈ E(H). To this end, consider some arbitrary edge {u, v} ∈ E(G), and denote u0 = ϕ(u) and v0 = ϕ(v). Because L/M is a clique and M is a matching that, by the definition of ϕ, necessarily contains both {u, u0} and {v, v0}, we derive that at least one of the following four conditions must be satisfied: (i) {u, v} ∈ E(L); (ii) {u0, v0} ∈ E(L); (iii) {u, v0} ∈ E(L); (iv) {v, u0} ∈ E(L). Because {u, v} ∈ E(G), we have that {u, v} ∈/ E(G) and therefore {u, v} ∈/ E(L). Thus, we are left with Conditions (ii), (iii) and (iv). Now, we will crucially rely on the proper colorings of G and H to rule out the satisfaction of Conditions (iii) and (iv).

0 0 B Claim 9. For any two edges {x, x }, {y, y } ∈ E(L) such that {x, y} ∈ E(G) and x0, y0 ∈ V (H), it holds that neither {x, y0} nor {y, x0} belongs to E(L).

Proof of Claim 9. Because cG is a proper coloring of G and {x, y} ∈ E(G), it holds that 0 0 0 0 cG(x) 6= xG(y). Because {x, x }, {y, y } ∈ E(L), x, y ∈ V (G) and x , y ∈ V (H), and by the 0 0 0 definition of E(L), it holds that x ∈ L(x) and y ∈ L(y), and therefore cG(x) = cH (x ) and 0 0 0 0 cG(y) = cH (y ). Thus, cG(x) 6= cH (y ) and cG(y) 6= cH (x ), implying that y ∈/ L(x) and x0 ∈/ L(y). In turn, by the definition of E(L), this means that neither {x, y0} nor {y, x0} belongs to E(L). This completes the proof of the claim. 

We now return to the proof of the lemma. By Claim 9, we are only left with Condition (ii), that is, {u0, v0} ∈ E(L). However, by the definition of E(L), this means that {u0, v0} ∈ E(H). As argued earlier, this completes the proof of the reverse direction. J

4 Lower Bounds: Clique Contraction and Hadwiger Number

In this section, we prove a lower bound for Clique Contraction and consequently for Hadwiger Number, defined as follows.

Clique Contraction Input: A graph G and t ∈ N. Question: Is there a subset F ⊆ E(G) of size at most t such that G/F is a clique?

Hadwiger Number Input: A graph G and h ∈ N. Question: Is the Hadwiger number of G at least as large as h? Clique Contraction Input: A graph G and t N. 2 Question: Does there exist a subset F E(G) of size at most t such that G/F is a clique? Clique ContractionClique Contraction ✓ Input: A graph GCliqueInput:and t A ContractionN graph. G and t . Clique2 Hadwiger Contraction NumberN Question: DoesClique thereInput: existContractionA graph a subsetCliqueG andF ContractiontE2(GN.) of size at most t such that G/F is a clique? Question:Input:Input:DoesA graph thereAG graph existand2 t aG subsetNand. hF NE.(G) of size at most t such that G/F is a clique? Input:Question:A graphDoesInput:G thereand t✓ existA graphN. a2 subsetG andF 2t✓E(GN). of size at most t such that G/F is a clique? Question:Question:Does there2Is the exist Hadwiger a subset✓F 2 numberE(G) of of sizeG atat most leastt such as thatlargeG/F as ish? a clique? Hadwiger NumberQuestion:HadwigerDoes NumberQuestion: there exist aDoes subset thereF existE✓(G a) subset of sizeF at mostE(Gt)such of size that at mostG/F ist such a clique? that G/F is a clique? Hadwiger NumberOurClique objective Contraction is to prove✓ the following statement,✓ where the analogous statement for Had- Input: A graph GInput:and hHadwigerAN graph. NumberG and h N. Input: A2 graph Input:G andCliquehA graph2 Contraction. G and t N. Question: IsHadwiger theQuestion: HadwigerInput: Numberwiger numberIsAHadwiger the graph Number Hadwiger of G andat leastNumberNhwill numberN as follow. large of2G as asat ah? corollary. least as large as h? Question:Input:2 DoesA2 graph thereG existand t a subsetN. F E(G) of size at most t such that G/F is a clique? Question:Question:Is theIs Hadwiger the Hadwiger number number of G ofatG2at least least as as large large as ashh?? Our objectiveInput: is toOur proveA objective graph the followingInput:G isand toQuestion: provehA statement, graphN the. Does followingG whereand thereh exist the statement,N analogousa. subset✓ whereF statementE( theG) of analogous size for atHad- most statementt such that forG/FHad-is a clique? Our objectiveTheorem is to prove 5.1.2 theUnless following the ETH statement,2 is false, where there✓ the does analogous not exist statement an algorithm for Had- that solves Clique wiger NumberQuestion:willwiger follow NumberOurIs as thea objective corollary.Question: HadwigerwillHadwiger follow is to prove numberIs as Number the a the corollary. Hadwiger following ofo(nG) at statement, least number as largewhere of G astheath leastanalogous? as large statement as h for? Had- wiger NumberContractionwill follow asin a time corollary.n where n = V (G) . wiger NumberInput:HadwigerwillA follow graph as NumberG aand corollary.h N. | | Our objective is toOur prove objective the following is to prove2 statement, the following where statement, the analogous where statement the analogous for Had- statement for Had- Theorem 5.1. 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ThisthatB inFC isisGevery a[ aDA contradiction,u perfect, isXB vertex at]. mostmatching ThisC inA isD thusG a inB[A contradiction,GX we[A1+]/FB getBC,]. that andCD / indeed2+ inD thus particularXX we=4 every]|/F getn| , that vertexand1+ of [Xu indeed in, inin| should[ particularAG[[A everyB| [B be vertex 5C ofn uD,1+ in shouldXA].X BecauseB bein 5un 1+ X in [ [ [ [that [Fandis thereby[ a perfect[ replaced[ matchingG by[A[ a singleB [| inC vertex.[G[[AD| [ WeBX[]. refer].|[ This[ to the[ is| a special contradiction,[| | case ofNoisy thus| Structured| we get[| that[|  indeed[ [ every[| vertex[|[ | | in A B | | is incident toG at[A leastis incidentB oneC edgethat toD inG at[FXFAis least.] notis/F FromB a incidentone. perfect However,where[C this, edge[ND to because= matching in[ any∅XF becauseas[ edge.]/F From.F in However, inF no this,,G itsn vertex[Aand degree becauseB becauseA in]. in AGBF[Ais=2 no adjacentBnn vertex,wederiveandC. NoteDA to in thatX aAB] vertex/Fis=2 adjacentis atn in,wederive mostD itsand to degree a no vertex in[ in D and no Clique Contractionis incident to at least[Structured one| | edge 4 in Clique[F .| From Contraction[ | this,|| [ because[ [ |F[[ n| and A B =2n,wederive that F is a perfectthat[ matchingStructuredF[ is[ a in perfectG[[ CliqueAG[[A matchingB].B Contraction[ C[ inDG[[XAis]. also ThisB]. a special is a contradiction, case of Clique thus Contraction we get that| . |  indeed every| [ vertex| in A B vertex in B is adjacentvertex[that[ to in aBF[ vertexis[ a adjacent perfect[ in D matching, to the a degree vertex4 in G[A inof anyDB]., the vertex degree in A of anyB, andvertex in inparticularA B, and of[ in particular of Solutions tois instances incident of Noisy to at Structuredleast one[ edge Clique in F . Contraction From[ this, becauseexhibit4 theF following[ n and A B =2[n,wederive u, is at most A uB, is at1+ mostC AD /B2+ X1+=4C n D1+/2+XXin=4G[nA |1+B| CX inD|G[X[A].| B BecauseC Du X]. Because u property,| which[ that| will beF is crucial| a perfect[|4 in[ the| matching proof| | of| Theorem| in G[[4A 10| B as]. well| | as| | results4 in[ Section[| C.|[ [ [ [ [ [ is not incident tois any not edge incident in F , to its any degree edge in inGF[A,[ itsB degreeC inDG[AX]/FB isC at mostD X its]/F degreeis at in most its degree in I Lemma 11. Let F be a solution to an instance (G, A, B, C,[ D, N,[ n) of[Noisy[ Structured[ [ [ [ G[A B C DG[AX].B ThisC is aD contradiction,X]. This is thus a contradiction, we get4 that thus indeed we every get that vertex indeed in A everyB vertex in A B [ Clique[ [ Contraction[ [ .[ Then,[F is[ a matching of size n in G such that each edge in F has [ [ is incidentone to endpoint at leastis in incidentA oneand the edge to other at in in leastFB.. From one edge this, in becauseF . FromF this,n becauseand A FB =2nnand,wederiveA B =2n,wederive | |  | [| || | [ | that F isProof. a perfectWethat first matching argueF is that a perfect in everyG[ vertexA matchingB in].A ∪ B inisG incident[A B to]. at least one edge in F . Targeting a contradiction, suppose that[ there exists a vertex u[∈ A ∪ B that is not incident 4 4

C V I T 2 0 1 6 Proof. We first argue that every vertex in A B is incident to at least one edge in F . Targeting [ a contradiction, suppose that there exists a vertex u A B that is not incident to any edge 2 [ in F . Because A B C D =6n, F n and G[A B C D X]/F is a clique | [ [ [ | | |  [ [ [ [ (where the last two properties follow from the supposition that F is a solution), it holds that Proof. We first argue that every vertex in A B is incident to at least one edge in F . Targeting G[A B C D X]/F is a clique[ on at least 5n + X vertices. Hence, the degree of a contradiction,[ [ suppose[ that[ there exists a vertex u A B that| | is not incident to any edge every vertex in G[A B C D X]/F , and2 in particular[ of u, should be 5n 1+ X in in F . Because A B C[ D[ =6[ n,[ F n and G[A B C D X]/F is a clique| | G[A B |C [D [X]/F[. However,| because| |  no vertex in[A is[ adjacent[ [ to a vertex in D and no (where the[ last[ two[ properties[ follow from the supposition that F is a solution), it holds that G[A vertexB C in BDis adjacentX]/F is to a a cliquevertex inonD at, the least degree 5n + ofX anyvertices. vertex in A Hence,B, and the in degree particular of of [ [ [ [ | | [ every vertexu, is at in mostG[AA BB C1+DC XD]//F2+, andX =4 in particularn 1+ X ofinuG,[ shouldA B beC 5nD 1+X]. BecauseX in u | [[ [| [ | [[ | | | | | [ [ [ [ | | G[A isB notC incidentD X to]/F any. However, edge in F because, its degree no vertex in G[A in AB isC adjacentD X to]/F a vertexis at most in D itsand degree no in [ [ [ [ [ [ [ [ vertexG in[AB isB adjacentC D toX a]. vertex This in is aD contradiction,, the degree of thus any we vertex get thatin A indeedB, and every in particular vertex in A of B [ uProof., is atWe most first[ argueA[ thatB[ every1+[ vertexC in AD B/2+is incidentX =4 to atn least1+ one edgeX ininFG. Targeting[A B C D X]. Because u[ is incident to at least one edge[ in F . From this, because F n and A B =2n,wederive a contradiction, suppose| [ that| there| exists[ a vertex| u| A| B that is not| incident| to any[ edge[ [ [ is not incident to any edge in F , its degree2 [ in G[A B C D| |X]/F is at| most[ | its degree in in F . Becausethat F Ais aB perfectC D matching=6n, F inn Gand[AG[AB].B C[ D[ X]/F[ is[ a clique G[A B C| [D [ X[]. This| is| a| contradiction, [ [ thus[ [ we[ get that indeed every vertex in A B (where[ the last[It remains two[ properties[ to argue follow from that the every supposition edge in thatFFhasis a one solution), endpoint it holds in thatA and the other in B. Targeting[ isG[A incidentB C toD atX least]/F is one a clique edge on in atF least. From 5n + X this,vertices. because Hence,F the degreen and of A B =2n,wederive [ a[ contradiction,[ [ suppose that this is false.| | Because F| is|  a perfect| matching[ | in G[A B], this thatevery vertexF is a in perfectG[A B matchingC D X] in/F ,G and[A in particularB]. of u, should be 5n 1+ X in [ [ [ [ | | [ G[A BmeansC D thatX]/F there. However, exist because two novertices vertex[ ina,A ais0 adjacentA such to a that vertex ina,D a0and noF . By the definition of Noisy It[ remains[ [ [ to argue that every edge in F has one endpoint in A and the other in B. Targeting vertex in B is adjacent to a vertex in D, the degree of any vertex2 in A B, and in{ particular} 2 of a contradiction,Structured suppose Clique that Contraction this is false.,neither Becausea[Fnorisa a0 is perfect adjacent matching to any vertex in G[A in DB.], Moreover, this u, is at most A B 1+ C D /2+ X =4n 1+ X in G[A B C D X]. Because u note| that[ | D |V[(G|[A |B| C D | X| ]/F ).[ In[ particular,[ [ the vertex of G[A B [C D X]/F meansis not incident that to there any edge exist in F two, its degree vertices in G[Aa, aB0 CA DsuchX]/F thatis ata, most a0 its degreeF . By in the definition of Noisy ✓ [ [ [ [ [2[ [ [ { } 2 [ [ [ [ StructuredG[A ByieldedC D by CliqueX]. the This contraction is Contraction a contradiction, of a, thus a,neither0 weis get not thata adjacentnor indeeda0 everyis to adjacentany vertex vertex in A to anyB of D vertexin G[A in DB. Moreover,C D X]/F . [ [ [ [ { } [ [ [ [ [ noteis incident that toD at leastV one(G[ edgeA inBF .C FromD this,X because]/F ).F In particular,n and A B the=2n vertex,wederive of G[A B C D X]/F However, this is a contradiction because| | G[A |B[ C| D X]/F is a clique. that F is a perfect✓ matching in[ G[A[ B[]. [ [ [ [ [ [ [ [ [ yielded by the contraction of[ a, a0 is not adjacent to any vertex of D in G[A B C D X]/F . Clique ContractionIt remains to argue that every edge in{ F has} one endpoint in A and the other in B. Targeting [ [ [ [ Clique ContractionClique ContractionCliqueHowever, ContractionWe this are is a now contradiction ready to prove because a lowerG[ boundA B forCStructuredD X]/F Cliqueis a clique. Contraction. Because CliqueInput: ContractionA grapha contradiction,CliqueG and Contractiont suppose. that this is false. Because F is a perfect[ matching[ [ in [G[A B], this Input: AInput: graph AG graphandInput:t GANand graph. t GNand. Nt . [ means2 thatthis2 there problem2 exist twoN is vertices a speciala, a0 caseA such of thatCliquea, a0 ContructionF . 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However,[ isC a[ contradiction,D to[ anyX because thus[ edge]/F we. in However,F getno thus, its vertex that degree we indeedget because in in thatAG[A everyis indeed no adjacentB vertex vertexC everyD to in in vertexX aA] vertex/FisB adjacentis in at inA mostDB itsand to degree a no vertex in[ in D and no [ [ [ Proof.[ [Targeting[ a contradiction,is incident to suppose at least[ that one there edge exists4 in anF . algorithm, From this, denoted[ because[ by[CliCon-F[ n[and A [B =2n,wederive is incidentis toincident at[ least[ to one at[ least edge[ G one in[[A F edgeB.[ FromC in[ FD this,.[ FromX because]. This this, is because aF contradiction,n andF A thusn andB we=2A getn that,wederive|B | =2 indeedn,wederive every| [ vertex| in A B vertex inAlgB, thatis solvesadjacentvertexStructuredthat[ to in aBF[ vertexis Clique[ a adjacent perfect[ in ContractionD matching, to the a degree vertex| 4 inin| G time[A inofn|o anyDB(n|].,) where| the vertex[ degreen is| inthe| A number[ of anyB| , andvertex in inparticularA B, and of[ in particular of that F isthat a perfectF is a matching perfect matching inisG incident[A inB to].G at[A leastB]. one edge in F . From[ this, because F [ n and A B =2[n,wederive u, is atof most verticesA inu theB, is input at1+ graph.most[ C WeAD will[/B2+ showX1+ that=4 thisC n impliesD1+/2+ theX existenceXin=4G of[nA an algorithm,|1+B| CX inD|G[X[A].| B BecauseC Du X]. Because u denoted| by MatchingAlg[ that| F,is that| a perfect solves[| [| matching| | | | in G[[Ain| timeB]. n| o(|n|) where| 4 n is[ the size[| of|[ [ [ [ [ [ is not incident tois any not edge incident in F , to itsCross any degree edge Matching in inGF[A,[ itsB degreeC inDG[AX]/FB isC at mostD X its]/F degreeis at in most its degree in the set A in the input, thereby contradicting4 Lemma4 8 and[ hence[ completing[ [ the proof.[ [ [ [ G[A B C DG[AX].B ThisC is aD contradiction,X]. This is thus a contradiction, we get4 that thus indeed we every get that vertex indeed in A everyB vertex in A B [ [We define[ the[ execution[ [ of MatchingAlg[ [ as follows. Given an instance (G, A, B) of Cross [ [ is incidentMatching to at, MatchingAlg leastis incident oneconstructs edge to at in least anF instance. From one(H, edge this, A, B, in because C,F D,. n) Fromof StructuredF this,n becauseand CliqueA FB =2nnand,wederiveA B =2n,wederive as follows (see Fig. 4): | |  | [| || | [ | that F Contractionis a perfectthat matchingF is a perfect in G[A matchingB]. in G[A B]. Let n = |A|, and K be a clique on 4[n new vertices. Let (C,D[ ) be a partition of V (K) such that |C| = |D|. V (H) = V (G) ∪ V (K). 4 4 F.V. Fomin, D. Lokshtanov, I. Mihajlin, S. Saurabh and M. Zehavi 23:13

E(H) = E(G) ∪ E(K) ∪ {{a, c} : a ∈ A, c ∈ C} ∪ {{b, d} : b ∈ B, d ∈ D}. Then, MatchingAlg calls CliConAlg with (H, A, B, C, D, n) as input, and returns the answer. First, note that by construction, |V (H)| = 6n. Thus, because CliConAlg runs in time |V (H)|o(|V (H)|) ≤ no(n), it follows that MatchingAlg runs in time no(n). For the correctness of the algorithm, first suppose that (G, A, B) is a Yes-instance of Cross Matching. This means that there exists a perfect matching M in G such that every edge in M has one endpoint in A and the other in B, and G/M is a clique. By the definition of E(H), M ⊆ E(H). We will show that H/M is a clique. As |M| = n, this will mean that (H, A, B, C, D, n) is a Yes-instance of Structured Clique Contraction, which will mean, in turn, that the call to CliConAlg with (H, A, B, C, D, n) as input returns Yes, and hence MatchingAlg returns Yes. Note that V (H/M) = V (K) ∪ V (G/M). To show that H/M is a clique, we consider two arbitrary vertices u, v ∈ V (H/M), and show that they are adjacent in H/M. If u, v ∈ V (K), then because K is a clique, it is clear that {u, v} ∈ E(H/M). Moreover, if u, v ∈ G/M, then because G/M is a clique, it is clear that {u, v} ∈ E(H/M). Thus, one of the vertices u and v belongs to V (G/M) and the other belongs to V (K). We suppose w.l.o.g. that u∈ / V (K). Because M is a perfect matching in G such that every edge in M has one endpoint in A and the other in B, it follows that u resulted from the contraction of the edge between some a ∈ A and some b ∈ B. If v ∈ C, then {a, v} ∈ E(H), and otherwise v ∈ D and so {b, v} ∈ E(H). Thus, by the definition of contraction, we conclude that {u, v} ∈ E(H/M). This completes the proof of the forward direction. Now, suppose that MatchingAlg returns Yes, which means that the call to CliConAlg with (H, A, B, C, D, n) returns Yes. Thus, (H, A, B, C, D, n) is a Yes-instance, which means that there exists a subset F ⊆ E(H) of size at most n such that H/F is a clique. We will show that F is a perfect matching in G such that every edge in F has one endpoint in A and the other in B. Because H/F is a clique, this will imply that G/F is a clique and thus that (G, A, B) is a Yes-instance of Cross Matching. To achieve this, notice that by Lemma 11, F is a matching of size n in H such that each edge in F has one endpoint in A and the other in B. Because G = H[A ∪ B], we have that F is a perfect matching in G. Thus, the proof of the reverse direction is complete. J

I Corollary 13. Unless the ETH is false, there does not exist an algorithm that solves Hadwiger Number in time no(n) where n = |V (G)|.

Proof. Targeting a contradiction, suppose that there exists an algorithm, denoted by Hadwi- gerAlg, that solves Hadwiger Number in time no(n) where n is the number of vertices in the input graph. We will show that this implies the existence of an algorithm, denoted by CliConAlg, that solves Clique Contraction in time no(n) where n is the number of vertices in the input graph, thereby contradicting Theorem 10 and hence completing the proof. We define the execution of CliConAlg as follows. Given an instance (G, t) of Clique Contraction, if G is not connected, then CliConAlg returns No, and otherwise it returns Yes if and only if HadwigerAlg returns Yes when called with (G, |V (G)| − t) as input. Because the call to HadwigerAlg with input (G, |V (G)| − t) runs in time no(n) where n = |V (G)|, we have that CliConAlg runs in time no(n) as well. For the correctness of the algorithm, first observe that if G is not connected, then no sequence of edge contractions can yield a clique, and hence it is correct to return No. Thus, now assume that G is connected. First, suppose that (G, t) is a Yes-instance of Clique Contraction. This means that there exists a sequence of at most t edge contractions that transforms G into a clique. In particular, this clique must have at least |V (G)| − t vertices,

C V I T 2 0 1 6 23:14 Tight Lower Bounds on the Computation of Hadwiger Number

and therefore the Hadwiger number of G is at least as large as |V (G)| − t. By the correctness of HadwigerAlg, its call with (G, |V (G)| − t) returns Yes, and therefore CliConAlg returns Yes. Now, suppose that CliConAlg returns Yes, which means that the call to HadwigerAlg with (G, |V (G)| − t) returns Yes. By the correctness of HadwigerAlg, the clique Kh for h = |V (G)| − t is a minor of G. This means that there is a sequence of vertex deletions, edge deletions and edge contractions that transforms G into Kh. In particular, this sequence can contain at most t vertex deletions and edge contractions in total. Furthermore, by replacing each vertex deletion for a vertex v by an edge contraction for some edge e incident to v (which exists because G is connected) and dropping all edge deletions, we obtain another sequence that transforms G into Kh. Because this sequence contains only edge contractions, and at most t of them, we conclude that (G, t) is a Yes-instance of Clique Contraction. J

References 1 Akanksha Agrawal, Daniel Lokshtanov, Saket Saurabh, and Meirav Zehavi. Split contraction: The untold story. In 34th Symposium on Theoretical Aspects of Computer Science (STACS 2017). Schloss Dagstuhl-Leibniz-Zentrum fuer Informatik, 2017. 2 Andreas Björklund. Determinant sums for undirected hamiltonicity. SIAM J. Comput., 43(1):280–299, 2014. 3 Andreas Björklund, Thore Husfeldt, and Mikko Koivisto. Set partitioning via inclusion– exclusion. SIAM J. Computing, 39(2):546–563, 2009. 4 B. Bollobás, P. A. Catlin, and P. Erdős. Hadwiger’s conjecture is true for almost every graph. European J. Combin., 1(3):195–199, 1980. URL: https://doi.org/10.1016/S0195-6698(80) 80001-1, doi:10.1016/S0195-6698(80)80001-1. 5 Jianer Chen, Benny Chor, Michael R. Fellows, Xiuzhen Huang, David Juedes, Iyad A. Kanj, and Ge Xia. Tight lower bounds for certain parameterized NP-hard problems. Information and Computation, 201(2):216 – 231, 2005. URL: http://www.sciencedirect.com/science/ article/pii/S0890540105000763, doi:http://dx.doi.org/10.1016/j.ic.2005.05.001. 6 Marek Cygan, Fedor V. Fomin, Alexander Golovnev, Alexander S. Kulikov, Ivan Mihajlin, Jakub Pachocki, and Arkadiusz Socala. Tight lower bounds on graph embedding problems. J. ACM, 64(3):18:1–18:22, 2017. URL: https://doi.org/10.1145/3051094, doi:10.1145/ 3051094. 7 Rodney G. Downey and Michael R. Fellows. Parameterized complexity. Springer-Verlag, New York, 1999. 8 Fedor V. Fomin and Dieter Kratsch. Exact Exponential Algorithms. Springer, 2010. An EATCS Series: Texts in Theoretical Computer Science. 9 Martin Charles Golumbic. Algorithmic Graph Theory and Perfect Graphs. North-Holland Publishing Co., Amsterdam, The Netherlands, The Netherlands, 2004. 10 Martin Grohe, Ken-ichi Kawarabayashi, Dániel Marx, and Paul Wollan. Finding topological subgraphs is fixed-parameter tractable. In Proceedings of the 43rd ACM Symposium on Theory of Computing, STOC 2011, San Jose, CA, USA, 6-8 June 2011, pages 479–488, 2011. 11 Russell Impagliazzo and Ramamohan Paturi. Complexity of k-sat. In Proceedings of the 14th Annual IEEE Conference on Computational Complexity, Atlanta, Georgia, USA, May 4-6, 1999, pages 237–240, 1999. 12 Russell Impagliazzo, Ramamohan Paturi, and Francis Zane. Which problems have strongly exponential complexity? J. Comput. Syst. Sci., 63(4):512–530, 2001. 13 Eugene L. Lawler. A note on the complexity of the chromatic number problem. Information Processing Letters, 5(3):66–67, 1976. 14 Andrzej Lingas and Martin Wahlen. An exact algorithm for subgraph homeomorphism. J. Discrete Algorithms, 7(4):464–468, 2009. URL: http://dx.doi.org/10.1016/j.jda.2008.10. 003, doi:10.1016/j.jda.2008.10.003. F.V. Fomin, D. Lokshtanov, I. Mihajlin, S. Saurabh and M. Zehavi 23:15

15 Neil Robertson and Paul D. Seymour. Graph minors. XIII. The disjoint paths problem. J. Combinatorial Theory Ser. B, 63(1):65–110, 1995. 16 Patrick Traxler. The time complexity of constraint satisfaction. In Parameterized and Exact Computation, pages 190–201. Springer, 2008.

A Preliminaries

For a vector R with L entries and i ∈ {1,...,L}, let R[i] be the value of the i-th entry of R. Unless specified otherwise, bases of logarithms are assumed to be 2. Given a graph G, let V (G) and E(G) denote its vertex set and edge set, respectively. Given a subset U ⊆ V (G), let G[U] denote the subgraph of G induced by U, that is, V (G[U]) = U and E(G[U]) = {{u, v} ∈ E(G): u, v ∈ U}. Given a subset F ⊆ E(G), let V (F ) denote the set of vertices that are incident in G to at least one edge in F , and let G[F ] = G[V (F )]. We say that G contains a graph H as an induced subgraph if there exists U ⊆ V (G) such that G[U] and H are identical up to relabelling vertices (more precisely, isomorphic). The set of neighbors of a vertex u ∈ V (G) is denoted by NG(u), that is, NG(u) = {v ∈ V (G): {u, v} ∈ E(G)}. When G is clear from context, we drop it from subscripts of notations. A matching M in G is subset of E(G) such that no two edges in M share an endpoint. In case every vertex in V (G) is an endpoint of an edge in M, that is, |M| = |V (G)|/2, it is said that M is perfect. A function c : V (G) → N is a proper coloring of G if for every edge {u, v} ∈ E(G), c(u) 6= c(v). The complement of G, denoted by G, is the graph with vertex set V (G) and edge set {{u, v} ∈/ E(G): u, v ∈ V (G), u 6= v}. Given an edge e = {u, v} ∈ E(G), the contraction of e in G is the operation that replaces u and v by a new vertex that is adjacent to all vertices previously adjacent to u or v (or both), where the resulting graph is denoted by G/e. In other words, V (G/e) = (V (G) \{u, v}) ∪ {x} for some new vertex x, and E(G/e) = {{s, t} ∈ E(G): s, t∈ / {u, v}} ∪ {{s, x} : s ∈ N(u) ∪ N(v)}. More generally, given a subset F ⊆ E(G), the contraction of F in G is the operation that replaces each connected component C of G[F ] by a new vertex xC that is adjacent to all vertices previously adjacent to at least one vertex in C, where the resulting graph is denoted by G/F . A graph H is said to be a minor of a graph G if H can be obtained from G by a series of vertex deletions, edge deletions and edge contractions. For any h ∈ N, the clique on h vertices is denoted by Kh, and the cycle on h vertices is denoted by Ch. The Hadwiger number of a graph G is the largest h ∈ N such that Kh is a minor of G. To obtain (essentially) tight conditional lower bounds for the running times of algorithms, we rely on the Exponential-Time Hypothesis (ETH) [11, 12]. To formalize its statement, we remind that given a formula ϕ in conjuctive normal form (CNF) with n variables and m clauses, the task of CNF-SAT is to decide whether there is a truth assignment to the variables that satisfies ϕ. In the p-CNF-SAT problem, each clause is restricted to have at most p literals. Then, ETH asserts that 3-CNF-SAT cannot be solved in time 2o(n).

B Details Omitted from Section 2

We first present the proof of Lemma 7.

Proof of Lemma 7. Targeting a contradiction, suppose that there exists an algorithm, denoted by LSHAlg, that solves Prop-Col LSH in time no(n) where n = max(|V (G)|, |V (H)|) for input graphs G and H. We will show that this implies the existence of an algorithm, denoted by ColAlg, that solves 3-Coloring on graphs of maximum degree 4 in time 2o(n)

C V I T 2 0 1 6 23:16 Tight Lower Bounds on the Computation of Hadwiger Number

B B’ z u x y v w

Figure 5 The reduction in Definition 4. The vertices of G are depicted by black shapes, where each distinct shape represents a different color (say, square is 1, rectangle is 2 and oval is 3), and the vertices of Ge are depicted by circles enclosing the vertex sets identifies with them, where the color of a vertex is the color of its circle (say, black is 1, green is 2, yellow is 3, red is 4, blue is 5 and grey is 6). Edges (of both graphs) are depicted by black lines. (The graph He is not shown). Then, the function φB is defined as follows: φB (1) = z, φB (2) = φB (5) = w, φB (3) = x, φB (4) = 0, and φB (6) = y. Moreover, the function φB0 is defined as follows: φB0 (1) = φB0 (2) = φB0 (4) = 0 u, φB0 (3) = v, and φB0 (5) = φB0 (6) = 0. With respect to B and B , the labeling ` is defined as follows: `(B) = {(R, 4) : R[1] 6= 0,R[2] = R[5] 6= 0,R[3] 6= 0,R[4] = 0,R[6] 6= 0}, and `(B0) = {(R, 5) : R[1] = R[2] = R[4] 6= 0,R[3] 6= 0,R[5] = R[6] = 0}.

where n is the number of vertices of the input graph, which contradicts the ETH and hence completes the proof. The execution of ColAlg is as follows. Given an instance G of 3-Coloring on graphs of maximum degree 4, ColAlg constructs the instance reduce(G) = (G,e H,e `) of LSH in Defini- tion 4 with r = dlogγ(4)(n/ log n)e where n = |V (G)|. By Lemma 5, reduce(G) = (G,e H,e `) is computable in time polynomial in the sizes of G, Ge and He, and has the following properties: G is a Yes-instance of 3-Coloring if and only if (G,e H,e `) is a Yes-instance of LSH. |V (Ge)| ≤ n/r = O(n/ log n), and |V (He)| ≤ γ(4)r = O(n/ log n). Then, ColAlg calls the polynomial-time algorithm in Lemma 6 with (G,e H,e `) to construct an equivalent instance (G, H0, c , c , `) of , where H0 is a subgraph of H. e e G H0 Prop-Col LSH e e e e 0 Lastly, ColAlg calls LSHAlg with (G,e He , c , c 0 , `) as input, and returns its answer. Ge He Since the instance G of 3-Coloring was argued above to be equivalent to the instance 0 (G,e He , c , c 0 , `) of Prop-Col LSH, the correctness of ColAlg directly follows. For the Ge He running time, denote M = max(|V (Ge)|, |V (He)|), and notice that M ≤ O(n/ log n). Thus, because LSHAlg runs in time M o(M) ≤ (n/ log n)o(n/ log n) ≤ 2o(n), it follows that ColAlg runs o(n) in time 2 . This completes the proof. J F.V. Fomin, D. Lokshtanov, I. Mihajlin, S. Saurabh and M. Zehavi 23:17

In the rest of this appendix, we provide the details omitted from Section 2 regarding the transition Prop-Col LSH to Prop-Col LSI. We begin by adapting the Turing reduction of [6] from LSH to LSI.

O(n) I Lemma 14. There is an 2 -time algorithm that, given an instance (G, H, cG, cH , `) of Prop-Col LSH, returns 2O(n) instances of Prop-Col LSI having input graphs on at most n vertices for n := max(|V (G)|, |V (H)|), such that (G, H, cG, cH , `) is a Yes-instance of Prop-Col LSH if and only if at least one of the returned instances is a Yes-instance of Prop-Col LSI.

Proof. Given an instance (G, H, cG, cH , `) of Prop-Col LSH, the algorithm works as follows. |V (H)| Without loss of generality, suppose that V (H) = {1, 2,..., |V (H)|}. Let P = {P ∈ N0 : P|V (H)| i=1 P [i] = |V (G)|}. That is, P contains every vector with |V (H)| entries that are non-negative integers whose sum is |V (G)|. Then, for each P ∈ P, the algorithm returns one instance (G, HP , cG, cHP , `P ) of Prop-Col LSI that is constructed as follows. The graph HP is constructed from H by replacing each vertex v ∈ V (H) with P [v] copies of it, denoted v1, v2 . . . vP [v]. (Note that P [v] can be equal to 0). Then, we connect two vertices vi to uj in HP if and only if v is connected to u in H. That is, V (HP ) = {vi : v ∈ V (H), i ∈ {1, 2,...,P [v]}} and E(HP ) = {{ui, vj} : {u, v} ∈ E(H), ui, uj ∈ V (HP )}.

For every vertex ui ∈ V (HP ), let cHP (ui) = cH (u). For every vertex u ∈ V (G), let `P (u) = {vi ∈ V (HP ): v ∈ `(u)}. This completes the description of the algorithm.

First, we consider some P ∈ P and assert that (G, HP , cG, cHP , `P ) is indeed an instance P|V (H)| of Prop-Col LSI. By the construction of V (HP ) and since i=1 P [i] = |V (G)|, we have that |V (G)| = |V (HP )|. Clearly, as (G, H, cG, cH , `) is an instance of Prop-Col LSH, we have that cG is a proper coloring of G. Now, consider an edge {ui, vj} ∈ E(HP ). Then, {u, v} ∈ E(H), and since cH is a proper coloring of H (as (G, H, cG, cH , `) is an instance of Prop-Col LSH), this means that cH (u) 6= cH (v). By definition, cHP (ui) = cH (u) and cHP (vi) = cH (v), and therefore cHP (ui) 6= cHP (vj). Thus, cHP is a proper coloring of HP . Lastly, consider some vertices u ∈ V (G) and vi ∈ `P (u). By the definition of `P , we have that v ∈ `(P ). Therefore, as (G, H, cG, cH , `) is an instance of Prop-Col LSH, cG(u) = cH (v).

Thus, because cHP (vi) = cH (v), we have that cG(u) = cHP (vi). Now, we consider the number of instances returned by the algorithm along with its |V (G)|+|V (H)|−1 n running time. Towards this, first note that |P| = |V (H)|−1 ≤ 4 . As the number of returned instances equals |P|, it is upper bounded by 2O(n) as required. Because each instance is computed in polynomial time, we also get that the running time of the algorithm is bounded by 2O(n). Finally, we consider the correctness of the algorithm. In one direction, suppose that at least one of the returned instances is a Yes-instance of Prop-Col LSI. Then, there exists P ∈ P such that (G, HP , cG, cHP , `P ) is a Yes-instance of Prop-Col LSI. Thus, there exists a bijective function ϕP : V (G) → V (HP ) such that (i) for every {u, v} ∈ E(G), {ϕP (u), ϕP (v)} ∈ E(HP ), and (ii) for every u ∈ V (G), ϕP (u) ∈ `P (u). We define a function ϕ : V (G) → V (H) as follows: for every u ∈ V (G), let ϕ(u) = v where v ∈ V (H) is the vertex for which there exists i ∈ {1, 2,...,P [v]} such that ϕP (u) = vi. We now verify that ϕ is a solution to the instance (G, H, cG, cH , `) of Prop-Col LSH. Firstly, by item (i) above, for every {u, v} ∈ E(G), we have that {xi, yi} ∈ E(HP ) where xi = ϕP (u) and yi = ϕP (v); by the definition of HP , this means that {x, y} ∈ E(H), and as x = ϕ(u) and y = ϕ(v) (by the definition of ϕ), we get that {ϕ(u), ϕ(v)} ∈ E(H). Secondly, by item (ii) above, for every u ∈ V (G), vi ∈ `P (u) where vi = ϕP (u); by the definition of `P , we have that v ∈ `P (u),

C V I T 2 0 1 6 23:18 Tight Lower Bounds on the Computation of Hadwiger Number

and by the definition of ϕ, we have that v = ϕ(u), therefore ϕ(u) ∈ `(u). Thus, we conclude that (G, H, cG, cH , `) is a Yes-instance of Prop-Col LSH. In the other direction, suppose that (G, H, cG, cH , `) is a Yes-instance of Prop-Col LSH. Then, there exists a function ϕ : V (G) → V (H) such that (i) for every {u, v} ∈ E(G), {ϕ(u), ϕ(v)} ∈ E(H), and (ii) for every u ∈ V (G), ϕ(u) ∈ `(u). Let P be the vector with −1 P|V (H)| |V (H)| entries where for each i ∈ {1, 2,..., |V (H)|}, P [i] = |ϕ (i)|. Then, i=1 P [i] = P|V (H)| −1 i=1 |ϕ (i)| = |V (G)|, and therefore P ∈ P. Choose some arbitrary order < on V (G). Now, we define a function ϕP : V (G) → V (HP ) as follows: for every u ∈ V (G), let ϕP (u) = vi where v = ϕ(u) and i = |{w ∈ V (G): w ≤ u, v = ϕ(w)}|. It should be clear that ϕP is a bijection. Moreover, analogously to the previous direction, we assert that (i) for every {u, v} ∈ E(G), {ϕP (u), ϕP (v)} ∈ E(HP ), and (ii) for every u ∈ V (G), ϕP (u) ∈ `P (u). Thus,

(G, HP , cG, cHP , `P ) is a Yes-instance of Prop-Col LSI, which means that at least one of the returned instances is a Yes-instance of Prop-Col LSI. J We are ready to complete the proof of Lemma 2.

Proof of Lemma 2. Targeting a contradiction, suppose that there exists an algorithm, de- noted by LSIAlg, that solves Prop-Col LSI in time no(n) where where n = max(|V (G)|, |V (H)|) for input graphs G and H. We will show that this implies the existence of an algorithm, de- noted by LSHAlg, that solves Prop-Col LSH in time no(n) where n = max(|V (G)|, |V (H)|) for input graphs G and H, which contradicts Lemma 7 and hence completes the proof. The execution of LSHAlg is as follows. Given an instance (G, H, cH , cG, `) of Prop-Col LSH, LSHAlg calls the algorithm in Lemma 14 so that in time 2O(n) it obtains 2O(n) instances of Prop-Col LSI having input graphs on at most n vertices for n := max(|V (G)|, |V (H)|), such that (G, H, cG, cH , `) is a Yes-instance of Prop-Col LSH if and only if at least one of the returned instances is a Yes-instance of Prop-Col LSI. Then, it calls LSIAlg on each of the returned instances, and returns Yes if and only if at least one of these calls returns Yes. o(n) It should be clear that LSHAlg runs in time n and that it is correct. J

C Lower Bounds for Contraction to Graph Classes Problems

In this section, we prove lower bounds for several cases of the F-Contraction problem, defined as follows. Here, F is a (possibly infinite) family of graphs.

F-Contraction Input: A graph G and t ∈ N. Question: Does there exist a subset F ⊆ E(G) of size at most t such that G/F ∈ F? Notice that Clique Contraction is the case of F-Contraction where F is the family of cliques. In this section, we consider the cases of F-Contraction where F is the family of chordal graphs, interval graphs, proper interval graphs, threshold graphs, trivially perfect graphs, split graphs, complete split graphs and perfect graphs, also called Chordal Contraction, Interval Contraction, Proper Interval Contraction, Threshold Contraction, Trivially Perfect Contraction, Split Contraction, Complete Split Contraction and Perfect Contraction, respectively. Before we define these classes formally, it will be more enlightening to first define only the class of chordal graphs as well as somewhat artificial classes of graphs that will help us prove lower bounds for many of the classes above in a unified manner.

I Definition 15 (Chordal Graphs). A graph is chordal if it does not contain C` for all ` ≥ 4 as an induced subgraph. Proof. We first argue that every vertex in A B is incident to at least one edge in F . Targeting Proof. We first argue that every vertex[ in A B is incident to at least one edge in F . Targeting a contradiction, suppose that there exists a vertex[u A B that is not incident to any edge a contradiction, suppose that there exists a vertex2 u[ A B that is not incident to any edge in F . Because A B C D =6n, F n and G[A2 B[ C D X]/F is a clique in F . Because| [ A[ B[ C| D =6| | n, F n and[G[A[ B[ C[ D X]/F is a clique (where the last two properties| [ [ follow[ from| the supposition| |  that F [is a solution),[ [ [ it holds that (where the last two properties follow from the supposition that F is a solution), it holds that G[A B C D X]/F is a clique on at least 5n + X vertices. Hence, the degree of [G[A[ B[ C[ D X]/F is a clique on at least| 5n| + X vertices. Hence, the degree of every vertex[ in G[[A [ B [C D X]/F , and in particular of |u,| should be 5n 1+ X in every vertex in[G[A[ B[ C[ D X]/F , and in particular of u, should be 5n | 1+| X in G[A B C D X]/F . However,[ [ because[ [ no vertex in A is adjacent to a vertex in D and no | | [ G[[A [B C[ D X]/F . However, because no vertex in A is adjacent to a vertex in D and no vertexClique in B[is adjacent[ Contraction[ to[ a vertex in D, the degree of any vertex in A B, and in particular of vertex in B is adjacent to a vertex in D, the degree of any vertex[ in A B, and in particular of u, is atInput: most A AB graph1+ CG Dand/2+t X =4. n 1+ X in G[A B C D[ X]. Because u u,Clique is at| most[ Contraction|A B | 1+[ |C D| /|2+N X=4n| |1+ X in[ G[[A [B C[ D X]. Because u is not incident to any| edge[ | in F , its| [ degree2| in |G[|A B C |D | X]/F is[ at[ most[ its[ degree in Question:isInput: not incidentA toDoes graph any edge thereG inandF exist, itst degree a subset. in[ G[[AF B[ EC[(G)D ofX size]/F is at at most most itst such degree that in G/F is a clique? G[A B C D X]. This is a contradiction,N thus we get[ that[ [ indeed[ every vertex in A B [ G[A[ B[ C[ D X]. This is a contradiction,2 thus✓ we get that indeed every vertex[ in A B is incidentQuestion: to[ at[ least[ oneDoes[ edge there in F . From exist this, a subset because FF En (andG)A of sizeB =2 atn,wederive most t such[ that G/F is a clique? is incident to at least one edge in F . From this, because| |✓ F n| and[ |A B =2n,wederive that FHadwigeris a perfect matching Number in G[A B]. | |  | [ | that F is a perfect matching in[ G[A B]. It remains to argue that every edge in F [has one endpoint in A and the other in B. Targeting Input:HadwigerIt remainsA graph to argue NumberG thatand everyh edgeN in. F has one endpoint in A and the other in B. Targeting a contradiction, suppose that this is false.2 Because F is a perfect matching in G[A B], this Question:aInput: contradiction,AIs graph suppose the HadwigerG thatand thish is numberfalse.N. Because of GF atis a least perfect as matching large as in[Gh[?A B], this means that there exist two vertices a, a0 A such that a, a0 F . By the definition of Noisy[ means that there exist two vertices2 a,2 a0 A such{ that} 2a, a0 F . By the definition of Noisy StructuredQuestion: Clique ContractionIs the Hadwiger,neither numbera2nor a0 is adjacent of G{ at to} least any2 vertex as large in D. Moreover, as h? StructuredOur objective Clique is Contraction to prove the,neither followinga nor a statement,is adjacent to any where vertex the in D analogous. Moreover, statement for Had- note that D V (G[A B C D X]/F ). In particular, the0 vertex of G[A B C D X]/F noteOur that✓ D objectiveV[(G[A is[B to[C proveD X] the/F ). following In particular, statement, the vertex of[ whereG[A [B the[C analogousD X]/F statement for Had- yieldedwiger by the Numbercontraction✓ ofwill[a, a[ follow0 is[ not[as adjacent a corollary. to any vertex of D in G[A B C[ D[ X[]/F[. yielded by the contraction{ of} a, a is not adjacent to any vertex of D in[ G[[A [B [C D X]/F . However,wiger this is Number a contradictionwill because follow{ 0}G[A as aB corollary.C D X]/F is a clique. [ [ [ [ However, this is a contradiction because[ G[A[ B[ C[ D X]/F is a clique. Theorem 5.1. Unless the ETH is false,[ [ there[ [ does not exist an algorithm that solves Clique We are now ready to prove a lower bound for Structured Clique Contraction. Because TheoremWe are now 5.1. readyUnless to prove ao the lower(n) ETH bound is for false,Structured there Clique does not Contraction exist an. algorithm Because that solves Clique thisContraction problem is a special casein time of Cliquen Contructionwhere n =, thisV (G will) . directly yield the correctness Contractionthis problem is a specialin time case ofnCliqueo(n) where Contructionn =| V ,(G this|) . will directly yield the correctness of Theorem 5.1. | | ofTo Theorem make5.1. our approach adaptable to extract analogous statements for other contraction Lemma 5.2.ToUnless make the our ETH approachis false, there adaptabledoes not exist an to algorithm extract that analogous solves Structured statements for other contraction Lemma 5.2. Unless the ETH is false, there does not exist an algorithm that solves Structured Cliqueproblems, Contraction we willin time firstno define(n) where an new= V problem(G) . (which will arise in Section 6) along with a special problems,Clique Contraction we will firstin time defineno(n) where a new| n = problem| V (G) . (which will arise in Section 6) along with a special case of it that is also a special case of| Clique| Contraction, prove a crucial property of Proof.caseTargeting of it a contradiction, that is also suppose a special that there case exists of anClique algorithm, Contraction denoted by CliConAlg,, prove a crucial property of Proof. Targeting a contradiction, suppose that there exists an algorithm, denoted by CliConAlg, thatinstances solves Structured of this Clique problem, Contraction and thenin time usingno(n this). We property will show that prove this implies Theorem 5.1 and its corollary. instancesthat solves Structured of this problem, Clique Contraction and then usingin time thisno(n) property. We will show prove that this Theorem implies 5.1 and its corollary. the existenceThe definition of an algorithm, of the denoted new problem by MatchingAlg is as, that follows solves (seeCrossFig. Matching?). in time F.V. Fomin, D. Lokshtanov,Thethe existence definition of I. an Mihajlin, of algorithm, the newS. denoted problemSaurabh by MatchingAlg andis as M. follows, Zehavi that solves (see CrossFig. ?). Matching in 23:19 time no(n), thereby contradicting Lemma 4.1 and hence completing the proof. no(n), thereby contradicting Lemma 4.1 and hence completing the proof. We define the execution of MatchingAlg as follows. Given an instance (G, A, B) of Cross NoisyNoisyWe define Structured Structured the execution Clique of CliqueMatchingAlg Contraction Contractionas follows. Given an instance (G, A, B) of Cross Matching, MatchingAlg constructs an instance (H, A, B, C, D, n) of Structured Clique Input:Matching A, MatchingAlg graph G onconstructs at least an 6instancen vertices (H, A, for B, C, some D, n)n of Structured, and a Cliquepartition (A, B, C, D, E) ContractionInput:as followsA graph (see Fig.G on?): at least 6n vertices for some n NN, and a partition (A, B, C, D, E) Contraction as follows (see Fig. ?): 22 ofofVV(G( G)) such such that thatAA==BB==n,n,CC==DD=2=2n,novertexinn,novertexinAAisis adjacent adjacent to to a a vertex vertex in in Let n = A , and K be a| clique|| | on| | 4|n|new vertices.| || | | Let| | (|C, D) be a partition of V (K) • DD, and,Let and | non| no= vertexA vertex, and inK inBbeB ais cliqueis adjacent adjacent on 4n tonew to a vertices. a vertex vertex Let in inD (C,D. D.) be a partition of V (K) such• that C =| D| . such| that| | C| = D . Question:Question: Does| Does| | there| there exist exist a a subset subsetFF EE(G(G)) of of size size at at most mostnnsuchsuch that thatGG[A[A BB CC V (H)= V (G) V (K). a a ✓✓ [[ [[ [[ Figure 6 A two-cliques• DD XV (X]H graph/F])=/F[isV (seeis(G a) a clique, Definition clique,V (K). where 16).whereXX== uu EE:thereexistsavertex:thereexistsavertexv v AA BB CC DDsuchsuch • [ E(H[)=[ E(G) E(K) a, c : a A, c C { { 2b,2 d : b B,d D . 22 [[ [[ [[ • thatthatEu(uHand)=and[Ev(vGbelong)belong[E{{(K to) to} the thea,2 c same same: a 2 A, connected} connected c[ {{C } componentb,2 componentd : b 2B,d} ofD ofG.G[F[F] ]? ? • [ [ {{ } 2 2 } [ {{ } 2 2 } } } Our firstThen, class ofMatchingAlg graphsa iscalls definedCliConAlg as followswith (H, (see A, B,Fig. C, D,6). n) as input, and returns the answer of Then,aNoteNoteMatchingAlg that thatF FmightcallsmightCliConAlg contain containwith edges edges (H, outside outsideA, B, C,G D,[GA[ nA)B asB input,CC andDD returnsXX].]. Then, theThen, answer we we slightly slightly of abuse abuse notation notation so so this call. [ [ [ [ [ [ [ [ Definition 16 (Two-Cliquesthatthisthat call.G[GA[A BB Graphs)CC DD X. X]A/F]/Ftwo-cliquesrefersrefers to toG graph[GA[A BisB aC graphC DDGXsuchX]/(]/F(F thatE(E thereG([GA[A BB CC DD XX])).])). I First, note that[[ by[[ construction,[[ [[ V (H) =6n. Thus,[ [ [ because[ [ [ [CliConAlg[ \runs\ in time[ [ [ [ [ [ [ [ o( VFirst,(H) ) noteo(n) that by construction,| |V (H) =6n. Thus,o(n) because CliConAlg runs in time exist A, B ⊆ V ((HG)) such| We|o that( V refer(Hn)A) ∪, to itBo follows( then=) V special( thatG), MatchingAlgG[A case] |and ofG|runsNoisy[B] inare time cliques,Structuredn . ando(n) there Clique do not Contraction where E = | |V We(H) refer| | ton the, specialit follows thatcaseMatchingAlg of Noisyruns Structured in time n . Clique Contraction where E = For| the correctness|  of the algorithm, first suppose that (G, A, B)isaYes-instance of Cross ; exist vertices a ∈ Aas\ StructuredForB and the correctnessb ∈ B \ A Clique ofsuch the algorithm, that Contraction{a, b} first ∈ E suppose(G). The that. Notetwo-cliques (G, A, that B)isaStructured classYes-instanceis the of Cross Clique Contraction;is Matchingas Structured. This means that Clique there exists Contraction a perfect matching. NoteM in thatG suchStructured that every edge Clique Contraction is class of all two-cliquesMatching graphs.. This means that there exists a perfect matching M in G such that every edge in M alsohas one a special endpoint case in A ofandClique the other Contraction in B, and G/M is. a clique. By the definition alsoin M a specialhas one endpoint case of inCliqueA and the Contraction other in B, and .G/M is a clique. By the definition It shouldof beE clear(H), MSolutions that theE(H two-cliques). to We instances will show class thatof isNoisy aH/M subclassis Structured a ofclique. the class As M of Clique chordal= n, this graphs.Contraction will mean exhibit the following of SolutionsE(H✓), M toE( instancesH). We will of showNoisy that StructuredH/M is a clique.| Clique As| M = Contractionn, this will mean exhibit the following Now, we furtherthat define(H,property, A, B, a C, family D, which n✓)isa of classes willYes-instance be of crucialgraphs of Structured as in follows. the proof Clique of Theorem Contraction| | 5.1,whichwillas well as results in Section 6. property,that (H, A, which B, C, D, will n)isa beYes crucial-instance in of theStructured proof of TheoremClique Contraction5.1 as well,whichwill as results in Section 6. I Definition 17 (LemmaNon-Trivial 5.1. ChordalLet F be Class) a solution. We5 say to that an ainstance class of graphs(G, A,F B,is C,non- D, E, n) of Noisy Structured Lemma 5.1. Let F be a solution to5 an instance (G, A, B, C, D, E, n) of Noisy Structured trivial chordal ifClique it is a subclass Contraction of the class. Then, of chordalF is graphs, a matching and a superclass of size n ofin theG such that each edge in F has two-cliques class. Cliqueone endpoint Contraction in A and. the Then, otherF inisB a. matching of size n in G such that each edge in F has Clearly, theone class endpoint of cliques in is notA and a non-trivial the other chordal in B. class, and the class of chordal Proof. We first argue that every vertex in A B is incident to at least one edge in F . Targeting graphs is a non-trivial chordal class. The rest of this section is divided[ as follows. First, in Section C.1, we proveProof.a contradiction, a lowerWe first bound argue suppose for any that non-trivial everythat there vertex chordal exists in A class. aB vertex Then,is incidentu in SectionA toB at C.2,that least is one not edge incident in F . to Targeting any edge [ 2 [ we prove a lowera boundin contradiction,F . for Because some graph supposeA classesB that thatC there areD not exists=6 non-trivialn, aF vertex chordal.nuandAG[AB thatB isC notD incidentX]/F tois any a edge clique | [ [ [ | | |  2 [ [ [ [ [ in(whereF . Because the lastA twoB propertiesC D follow=6n from, F then suppositionand G[A thatB FCis aD solution),X]/F itis holds a clique that | [ [ [ | | |  [ [ [ [ C.1 Non-Trivial(whereG[A Chordal theB lastC Graph twoD properties ClassesX]/F is follow a clique from on the at supposition least 5n + thatX vertices.F is a solution), Hence, it the holds degree that of [ [ [ [ | | Gevery[A B vertexC inDG[AX]/FB isC a cliqueD X on]/F at, and least in 5 particularn + X vertices. of u, should Hence, be 5 then degree1+ X ofin The main objective of[ this subsection[ [ is[ to[ prove[ the[ following[ theorem. Afterwards,| we| will | | everyG[A vertexB C in GD[A X]B/F .C However,D X because]/F , and no in vertex particular in A is of adjacentu, should to abe vertex 5n in1+D Xandin no derive lower bounds for[ several[ known[ [ graph[ [ classes[ as corollaries.[ | | Gvertex[A B in CB isD adjacentX]/F . to However, a vertex because in D, the no degree vertex of in anyA is vertex adjacent in A toB a, vertex and in in particularD and no of Theorem 18. Let F[ be[ any non-trivial[ [ chordal graph class. Unless the ETH is false, there [ I vertexu, is at in mostB is adjacentA B to1+ a vertexC D in/2+D, theX =4 degreen of1+ anyX vertexin G[A in AB BC, andD inX particular]. Because ofu does not exist an algorithm that solves| [ F|-Contraction | [ | in time| |no(n) where n |= ||V (G)|. [ [[ [ [ u,is is not at most incidentA toB any1+ edgeC in FD,/ its2+ degreeX =4 innG[A1+BX Cin GD[A XB]/FCis atD mostX]. its Because degreeu in | [ | | [ | | | [ | [| [ [[ [ [ [ For the proofis ofG not[ thisA incident theorem,B C to theD any followingX edge]. This in well-knownF is, its a contradiction, degree property in G of[A chordal thusB we graphsC getD that willX indeed]/F is at every most vertex its degree in A inB come in handy. This property[ [ is[ a direct[ consequence of the alternative[ characterization[ [ [ [ Gis[A incidentB C to atD leastX]. one This edge is a in contradiction,F . From this, thus because we getF thatn indeedand A everyB vertex=2n,wederive in A B of the class of chordal[ graphs[ as[ the[ class of graphs that admit clique- decompositions,| |  | [ | [ isthat incidentF is to a perfect at least matching one edge in inGF[A. FromB]. this, because F n and A B =2n,wederive see [9]. [ | |  | [ | that F is a perfect matching in G[A B]. 4 I Proposition 19. Let G be a chordal graph, and let u and[ v be two non-adjacent vertices in G. Then, G[N(u) ∩ N(v)] is a clique. 4 We are now ready to prove Theorem 18.

Proof of Theorem 18. Targeting a contradiction, suppose that there exists an algorithm, denoted by NonTrivChordAlg, that solves F-Contraction in time no(n) where n is the number of vertices in the input graph. We will show that this implies the existence of an algorithm, denoted by CliConAlg, that solves Structured Clique Contraction in time no(n) where n is the number of vertices in the input graph, thereby contradicting Lemma 12 and hence completing the proof. We define the execution of CliConAlg as follows. Given an instance (G, A, B, C, D, n) of Structured Clique Contraction, CliConAlg constructs an instance (H, n) of F- Contraction as follows (see Fig. 7):

C V I T 2 0 1 6 Proof. We first argue that every vertex in A B is incident to at least one edge in F . Targeting [ a contradiction, suppose that there exists a vertex u A B that is not incident to any edge 2 [ in F . Because A B C D =6n, F n and G[A B C D X]/F is a clique | [ [ [ | | |  [ [ [ [ (where the last two properties follow from the supposition that F is a solution), it holds that G[A B C D X]/F is a clique on at least 5n + X vertices. Hence, the degree of [ [ [ [ | | every vertex in G[A B C D X]/F , and in particular of u, should be 5n 1+ X in [ [ [ [ | | G[A B C D X]/F . However, because no vertex in A is adjacent to a vertex in D and no [ [ [ [ vertex in B is adjacent to a vertex in D, the degree of any vertex in A B, and in particular of [ u, is at most A B 1+ C D /2+ X =4n 1+ X in G[A B C D X]. Because u | [ | | [ | | | | | [ [ [ [ is not incident to any edge in F , its degree in G[A B C D X]/F is at most its degree in [ [ [ [ G[A B C D X]. This is a contradiction, thus we get that indeed every vertex in A B [ [ [ [ [ is incident to at least one edge in F . From this, because F n and A B =2n,wederive | |  | [ | that F is a perfect matching in G[A B]. [ It remains to argue that every edge in F has one endpoint in A and the other in B. Targeting a contradiction, suppose that this is false. Because F is a perfect matching in G[A B], this [ means that there exist two vertices a, a A such that a, a F . By the definition of Noisy 0 2 { 0} 2 Structured Clique Contraction,neithera nor a0 is adjacent to any vertex in D. Moreover, Proof. We firstProof. argueWe that first every argue vertex that in everyA B vertexis incident in A toB atis least incident one to edge at least in F . one Targeting edge in F . Targeting [ [ a contradiction,notea contradiction, suppose that thatD there supposeV exists(G that[A a vertexthereB existsu CA a vertexBDthatuX isA not]/F incidentB).that In is toparticular, not any incident edge to any the edge vertex of G[A B C D X]/F 2 [ 2 [ in F . Becausein AF . BecauseB C ✓DA =6B n,CF D[ n=6and[n, GF[A Bn[andC G[DA BX]/FC is aD cliqueX]/F is a clique [ [ [ [ yielded| [ by[ the[| |[ contraction[ | [| | of | a,| [ a [is not[ [ adjacent[ [ [ to any vertex of D in G[A B C D X]/F . (where the last(where two properties the last two follow properties from the follow supposition from{ the that supposition0}F is a solution), that F is it a holds solution), that it holds that [ [ [ [ Clique ContractionClique ContractionG[A B C G[AD BX]/FC isD a cliqueX]/F onis at a least clique 5n on+ atX leastvertices. 5n + Hence,X vertices. the degree Hence, of the degree of [ [However,[ [[ [ this[ is[ a contradiction| | because| G| [A B C D X]/F is a clique. every vertex ineveryG[A vertexB inC G[DA XB]/FC, andD inX particular]/F , and of inu particular, should be of 5un, should[1+ X[ bein 5n [1+ [X in Input: A graph Input:G and t A graphN. G and t [ N[. [ [[ [ [ [ | | | | G[A B C GD[A XB2]/F .C However,D X]/F because. However, no vertex because in A nois adjacentvertex in toA ais vertex adjacent in D toand a vertex no in D and no 2 [ [ [ [[ [ [ [ Question: DoesQuestion: there exist avertexDoes subset in thereB isFvertex adjacentWe existE in are( toBG a ais) vertexnow subset adjacent of size in readyD toF, at a the vertex most degreetoE prove( inG ofDt), anysuch the of a vertex degree size lower that inat ofA any boundG/FmostB vertex, andist in forsuch a particularA clique?StructuredB, that and of inG/F particularis of a Clique clique? Contraction. Because ✓ [ [ u, is at most Au,✓ isB at most1+ CA DB/2+1+XC=4Dn /2+1+XX =4in nG[A1+BX Cin GD[A XB]. BecauseC D uX]. Because u this| [ problem| || [[ is|| a special| || [ | case| | | | of Clique [ | [| [ Contruction[[ [ [ [ , this will directly yield the correctness is not incidentis to not any incident edge in toF , any its degreeedge in inF G, its[A degreeB C in GD[A XB]/F Cis atD mostX]/F its degreeis at most in its degree in Hadwiger NumberHadwiger Number [ [ [ [[ [ [ [ Clique ContractionCliqueClique Contraction ContractionG[A B CofG TheoremD[A XB]. ThisC isD5.1 a contradiction,X. ]. This is a contradiction, thus we get that thus indeed we get every that vertex indeed in everyA B vertex in A B Clique Contraction[ [ [ [[ [ [ [ [ [ Input: A graph Input:G and hA graphis. incidentG toand atis incident leasth oneN to edge. at least in F . one From edge this, in F because. From this,F becausen and AF B n=2andn,wederiveA B =2n,wederive Input: AInput: graphCliqueInput:AG graphandA Contraction graphCliquetNG and.G andt Contractiont . N. | |  | | [| | | [ | Input:2 thatANF graphis a perfectthatGNandF matching22is at perfectN in .G matching[A B]. in G[A B]. Question: Is theInput:Question:Question: HadwigerAInput: graph numberDoesIs2 the thereG HadwigerLemmaA ofand2 graphG existatt leasta 5.2. numberG2 subset.and asUnless[ largetF of G theE as.at(Gh ETH least[)? of size is as false, large at most there as th?such does that notG/F existis an a algorithm clique? that solves Structured Question:Question:Does thereQuestion:Does exist thereIt a remainsDoes subset exist to there argueItF a remains subset that existEN every to(G a argueF subset) edge of that in sizeEF✓ everyF(NhasG at) one edgeE of most( endpointG sizein)F ofhast at sizesuch in oneA mostand endpoint at that themostt othersuch inG/FtAsuch inandB that.is the Targeting that a otherG/F clique?G/F in Bis. Targetingis a a clique? clique? a contradiction,a contradiction, suppose✓2 that this suppose is✓ false. that2 Because this✓ is false.F is a Because perfecto(nF matching)is a perfect in G matching[A B], this in G[A B], this Our objective isQuestion: toOur prove objectiveQuestion: theDoes following isClique there toClique proveDoes statement, Contraction exist the Contraction there a following subset existwhereF a statement, the subsetinE analogous( timeG)F ofn where sizeE statement(whereG at the) most of analogous sizen = fort[ such atVHad-( mostG statement that) .[t G/Fsuch for thatis aHad- clique?G/F is a clique? means that theremeans exist that two there vertices exista, two a0 verticesA such✓ a, that a0 a,A asuch0 ✓F that. Bya, the a0 definitionF . 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[ [ [ [ [ [ [ [ TheoremQuestion: 5.1.yieldedIs2 theUnless by Hadwiger theyielded contraction2 the ETHby number the of2 contractiona, is a0 false,is of not ofG adjacent therea,at a0 leastis to not does any adjacent as vertex not large✓ of to existD anyin asG vertexh[A an? B of algorithmDC inDG[AX]/FB . thatC D solvesXo]/F(n). Clique TheoremQuestion: 5.1.Question:UnlessIs the theHadwigerQuestion:Is ETH the Hadwiger is numberIs false,that theQuestion: Hadwiger there of solvesnumberG at does{DoesStructured leastnumber of} notG there asat exist of large{ leastexistG } anat aas Cliqueas least algorithm subseth large? asF large as Contraction thathE as?(Gh)[ solves? of[ size[ Clique at[ in[ most[ time[t such[n that. 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ThisD [ isX a]. contradiction, This[ is[ a contradiction, thus we get thus that we indeed get[2 that every indeed vertex\ every in A vertexB in A B However,[ [notStructured there[[ exist do[ not[ aexist vertex[ Clique a vertex[ in in ContractionV (K(K1) 1and) and a vertex. a Furthermore, vertexin V (K2) that in V are since(K adjacent2)F that inn areand adjacentwe have already in H[, shown and for[ every vertex is incidentH, andis for toNow, incident every at least vertex notice to one in at thatV ( edgeK least1) (∪H,V in one( A,KF2) B,,. edge its From C, neighborhood D, in this,F E,. n From)where because outside this, thisEF because set= isV containedn(K|and1|)F AV (nKB2and) is=2 anAn,wederive instanceB =2n of,wederiveNoisy inthatVH(K[A1)B VC(KD2),X its]/F neighborhoodis a clique, we have outside that| F| thisis a solution set| [| is| contained to[ this| instance.| [ in |A Therefore,B C D.Thus,c1 in A Structured∪ B ∪ C ∪ D.[ Thus, Clique[ c1 [and [c Contraction2 must be non-adjacent. Furthermore, in H/F . However, since thisF is a n and we have already shown that F is athatby perfect LemmaF is matching a5.1[ perfect, F is inmatching a matchingG[A B in]. ofG size[A n Bin].H such that each| edge|  in F has one endpoint[ in[A [ contradictionthatandH to[Ac the2 mustB argumentC beD that non-adjacentX(H/F]/F)[N[isH/F a( clique,u) ∩ inNH/F[H/F we(v)] haveis. a clique.However, that FromF is this, a this solution we is a to contradiction this instance. Therefore,to the argument that and the other in B. In particular, F E(G) and hence X = . Because G = H[A B C D], (H/F )[[NH/F[ [(u)[ NH/F (v)] is a4 clique. From4 this, we derive that H[A B C D X]/F is by Lemma 5.1, F is a\ matching of size✓ n in H such that each; edge in F has one[ endpoint[[ [ [ in[A [ andindeed the other a clique. in B. In particular, F E(G) and8 hence X = . Because G = H[A B C D], ✓ ; [ [ [ Now, notice that (H, A, B, C, D, E, n)whereE = V (K1) V (K2) is an instance of Noisy 8 [ Structured Clique Contraction. Furthermore, since F n and we have already shown | |  that H[A B C D X]/F is a clique, we have that F is a solution to this instance. Therefore, [ [ [ [ by Lemma 5.1, F is a matching of size n in H such that each edge in F has one endpoint in A and the other in B. In particular, F E(G) and hence X = . Because G = H[A B C D], ✓ ; [ [ [ 8 F.V. Fomin, D. Lokshtanov, I. Mihajlin, S. Saurabh and M. Zehavi 23:21 derive that H[A ∪ B ∪ C ∪ D ∪ X]/F is indeed a clique. Now, notice that (H, A, B, C, D, N, n) where N = V (K1)∪V (K2) is an instance of Noisy Structured Clique Contraction. Furthermore, since |F | ≤ n and we have already shown that H[A ∪ B ∪ C ∪ D ∪ X]/F is a clique, we have that F is a solution to this instance. Therefore, by Lemma 11, F is a matching of size n in H such that each edge in F has one endpoint in A and the other in B. In particular, F ⊆ E(G) and hence X = ∅. Because G = H[A ∪ B ∪ C ∪ D], we thus derive that G/F is a clique. Thus, we conclude that (G, A, B, C, D, n) is a Yes-instance of Structured Clique Contraction. This completes the proof of the reverse direction. J Now, we give definitions for several classes of graphs for which lower bounds will follow from Theorem 19. First, a graph is an interval graph if there exists a set of intervals on the real line such that the vertices of the graph are in bijection with these intervals, and there exists edge between two vertices if and only if their intervals intersect. A graph is a proper interval graph if, in the former definition, we also add the constraint that all intervals must have the same length. A graph is a threshold graph if it can be constructed from a one-vertex graph by repeated applications of the following two operations: addition of a single isolated vertex to the graph; addition of a single vertex that is connected to all other vertices. A graph is trivially perfect if in each of its induced subgraphs, the maximum size of an independent set equals the number of maximal cliques. It is well-known that every graph that is a (proper) interval graph, or a threshold graph, or a trivially perfect graph, is also a chordal graph (see [9]). Moreover, it is immediate to verify that the two-cliques class is a subclass of the classes of (proper) interval graphs, threshold graphs and trivially perfect graphs. Thus, these classes are non-trivial chordal graphs classes, and therefore Theorem 18 directly implies lower bounds for them as state below.

I Corollary 20. Unless the ETH is false, none of the following problems admits an algorithm that solves it in time no(n) where n = |V (G)|: Chordal Contraction, Interval Con- traction, Proper Interval Contraction, Threshold Contraction and Trivially Perfect Contraction.

C.2 Other Graph Classes In Section 4, we have already proved a lower bound for a class of graphs that is not non-trivial chordal, namely, the class of cliques. In this section, we show that our approach can yield lower bounds for other classes of graphs that are not non-trivially chordal. For illustrative purposes, we consider the classes of Split Graphs, Complete Split Graphs and Perfect Graphs. A graph G is a split graph if there exists a partition (I,K) of V (G) such that G[I] is edgeless and G[K] is a clique. In case {{i, k} : i ∈ I, k ∈ K}, we further say that G is a complete split graph. Notice that the two-cliques class is not a subclass of the class of split graphs, and hence the class of (complete) split graphs is not non-trivially chordal. For the class of (complete) split graphs, we prove the following statement.

I Theorem 21. Unless the ETH is false, there does not exist an algorithm that solves Split Contraction (or Complete Split Contraction) in time no(n) where n = |V (G)|. Proof. Targeting a contradiction, suppose that there exists an algorithm, denoted by SplitAlg, that solves Split Contraction (or Complete Split Contraction) in time no(n) where n is the number of vertices in the input graph. We will show that this implies the existence

C V I T 2 0 1 6 we thus derive that G/F is a clique. Thus, we conclude that (G, A, B, C, D, n)isaYes-instance of Structured Clique Contraction. This completes the proof of the reverse direction.

Now, we giveProof. the commonWe first definitions argue that for every several vertex classes in ofA graphsB is incident for which to lower at least bounds one edge in F . Targeting will follow from Theorem 6.1. First, a graph is interval if there[ exists a set of intervals on the a contradiction, suppose that there exists a vertex u A B that is not incident to any edge real line such that the vertices of the graph are in bijection with these intervals,2 [ and there exists in F . Because A B C D =6n, F n and G[A B C D X]/F is a clique edge between two vertices if and only| [ if their[ intervals[ | intersect.| |  A graph if properly[ [ interval[ [ if, inwe the thus former derive definition,(where that G/F theis we last a clique.also two add Thus, properties the we constraint conclude follow that that from (G, all A,the intervals B, supposition C, D, n must)isaYes have that-instance theF is same a solution), it holds that of Structured Clique Contraction. This completes the proof of the reverse direction. length. A graph isG a[thresholdA B graphC Dif it canX]/F be constructedis a clique from on at a one-vertex least 5n + graphX byvertices. repeated Hence, the degree of [ [ [ [ | | applicationsNow, of we the giveevery following the vertex common two in definitionsoperations:G[A B for additionC severalD classes ofX a]/F single of, graphs and isolated in for particular which vertex lower to of bounds theu, graph; should be 5n 1+ X in will follow from Theorem 6.1. First, a[ graph[ is interval[ [ if there exists a set of intervals on the | | addition of a singleG[ vertexA B thatC isD connectedX]/F . to However, all other because vertices. no A vertex graph inis triviallyA is adjacent perfect toif a vertex in D and no in eachreal of line its inducedsuch that subgraphs, the[ vertices[ [ of the the[ maximum graph are insize bijection of an independent with these intervals, set equals and there the number exists of edge betweenvertex two vertices in B ifis and adjacent only if to their a vertex intervals in intersect.D, the degree A graph of if anyproperly vertex interval in A B, and in particular of maximal cliques. [ if, in the formeru, is definition, at most weA alsoB add1+ the constraintC D /2+ thatX all=4 intervalsn 1+ mustX havein G the[A sameB C D X]. Because u It is well-known that every graph| [ that| is a (proper)| [ | interval| | graph, or a threshold| | graph,[ or[ a [ [ length. A graphis not is a threshold incident graph to anyif it edge can be in constructedF , its degree from ain one-vertexG[A B graphC byD repeatedX]/F is at most its degree in triviallyapplications perfect graph, of the isfollowing also a chordal two operations: graph (see addition [2]). of Moreover, a single isolated it is immediate[ vertex[ to[ the to verify[ graph; that G[A B C D X]. This is a contradiction, thus we get that indeed every vertex in A B the two-cliquesaddition of class a single is[ a vertex subclass[ that[ of is the connected[ classes to of all (proper) other vertices. interval A graphs, graph is thresholdtrivially perfect graphsif and [ triviallyin each perfect of its graphs. inducedis incident Thus, subgraphs, to these at leastthe classes maximum one are edge non-trivial size in ofF an. independent From chordal this, graphs set because equals classes, theF numberandn thereforeand of A B =2n,wederive | |  | [ | Theoremmaximal6.1 directly cliques.that impliesF is a lower perfect bounds matching for them in G as[A stateB]. below. It is well-known that every graph that is a (proper) interval[ graph, or a threshold graph, or a It remains to argue that every edge in F has one endpoint in A and the other in B. Targeting Corollarytrivially 6.1. perfectUnless graph, the is ETHalso a chordalis false, graph none (see of [ the2]). following Moreover, itproblems is immediate admits to verify an algorithm that the two-cliquesa class contradiction, is a subclass supposeof the classes that of (proper)this is false. interval Because graphs, thresholdF is a perfect graphs and matching in G[A B], this that solves it in time no(n) where n = V (G) : Chordal Contraction, Interval Con- trivially perfect graphs. Thus, these classes are non-trivial chordal graphs classes, and therefore [ means that there exist| two vertices| a, a0 A such that a, a0 F . By the definition of Noisy tractionTheorem, Proper6.1 directly Interval implies lower Contraction bounds for them, Threshold as state2 below. Contraction{ and} 2Trivially Perfect ContractionStructured. Clique Contraction,neithera nor a0 is adjacent to any vertex in D. Moreover, Corollary 6.1.noteUnless that theD ETHV (G is[ false,A B noneC ofD the followingX]/F ). problemsIn particular, admits the an algorithm vertex of G[A B C D X]/F that solves it in time no(n) ✓where n =[ V ([G) :[Chordal[ Contraction, Interval Con- [ [ [ [ 6.2 Other Graphyielded Classes by the contraction| of | a, a0 is not adjacent to any vertex of D in G[A B C D X]/F . traction, Proper Interval Contraction{, Threshold} Contraction and Trivially [ [ [ [ However, this is a contradiction because G[A B C D X]/F is a clique. In SectionPerfect5, we Contraction have already. proved a lower bound for a class of[ graphs[ that[ is[ not non-trivial chordal, namely, the class of cliques. In this section, we show that our approach can yield lower 6.2 Other GraphWe are Classes now ready to prove a lower bound for Structured Clique Contraction. Because bounds for other classes of graphs that are not non-trivially chordal. For illustrative purposes, this problem is a special case of Clique Contruction, this will directly yield the correctness we considerIn Section the5 classes, we have of alreadySplit proved Graphs a lower, Complete bound for Split a class Graphs of graphsand thatPerfect is not non-trivial Graphs. Clique Contractionof Theorem 5.1. Clique ContractionClique ContractionACliquechordal, graph G Contraction namely,is a split the graph class ofifcliques. there exists In this a partitionsection, we (I,K show) that of V our(G) approach such that canG yield[I] is lower edgeless Input: AInput: graphCliqueInput:AG graphandboundsA Contraction graphCliquet G forNand.G otherandt Contraction classesNt . N. of graphs that are not non-trivially chordal. For illustrative purposes, and GInput:[K] is2A a graph clique.2G Inand2 caset N. i, k : i I,k K , we further say that G is a complete split Question:Question:DoesInput:Question: thereweDoesA exist considerInput: graphDoes there a theresubset theG existLemmaAand classes graph existF at subset aof 5.2.GE2N subsetSplit(.{{andGF)Unless of GraphstF} sizeE(NG theE2 at.)(,G of ETHmostComplete) size of2 sizet is atsuch} false, mostat Split most that theret such GraphsG/Ft such does thatis that not aandG/F clique?G/F existPerfectisis ana clique? a algorithm clique? Graphs. that solves Structured graphQuestion:. 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F supposeE(G) of size that at there most t existssuch that anG/F algorithm,is a clique? denoted by CliConAlg, Input: AInput: graphHadwigerAG graphandInput:theHadwiger classh G NumberANand of graph. (complete)h G NumberandN2. splith N graphs. is not2 non-trivially chordal. Question:TheoremIs2 6.2. the HadwigerUnlessthatQuestion:2 solves the number ETH2DoesStructured is there of false,G existat thereleast a Clique subset as does large✓F not Contraction asE exist(Gh?) of an size algorithm atin most timet such thatno(n that solves). WeG/FSplit willis a show clique? that this implies Question:Question:IsInput: the HadwigerQuestion:IsA theForInput: graph Hadwigerthe numberIs classGA theand graph of Hadwiger of number (complete)hG GatNand least.number of splitGh asat graphs, of large leastNG. at as aswe leasth large prove? as large asthe✓ h following as? h? statement. ContractionOur objectiveWe isHadwiger(or to aretheComplete prove nowexistence the Number2 ready following Split of an to2 algorithm, Contractionprove statement, Theorem denoted where)intime the6.1 by analogous.MatchingAlgno(n) where statement,n that= V for solves(G)Had-. 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Therefore, However, by Lemma this 11, is a contradiction to the argument that F is a matching of size n in H such that each edge in F has one endpoint in A and the other (H/F )[NH/F (u) NH/F (v)] is a4 clique. From4 this, we derive that H[A B C D X]/F is \ [ [ [ [ indeed a clique. Now, notice that (H, A, B, C, D, E, n)whereE = V (K ) V (K ) is an instance of Noisy 1 [ 2 Structured Clique Contraction. Furthermore, since F n and we have already shown | |  that H[A B C D X]/F is a clique, we have that F is a solution to this instance. Therefore, [ [ [ [ by Lemma 5.1, F is a matching of size n in H such that each edge in F has one endpoint in A and the other in B. In particular, F E(G) and hence X = . Because G = H[A B C D], ✓ ; [ [ [ 8 Then, CliConAlgThen, CliConAlgcalls SplitAlgcalls withSplitAlg (H,with n) as (H, input, n) as and input, returns and the returns answer the of answer this call. of this call. 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This is[ means a contradiction, that thus there we exists get thus a subset that weF indeed get⊆2E( thatG) everyof indeed vertex\ every in A vertexB in A B [ [not[[ exist[[ a vertex[ [ in V (K1) and a vertex[ in V (K2[) that are adjacent in H[, and for[ every vertex is incidentsizeis to at incident most at leastn such to one that at edgeG/F leastis in one a clique.F . edge From Now, in we this,F . will From because show that this,H/FF becauseis an perfectandF graph.A nBand=2An,wederiveB =2n,wederive Toin thisV end,(K consider1) V some(K induced2), its subgraph neighborhoodS of H/F . In outside case the| | maximum this 4 set| size| is| of contained[ a 4 | | [ in |A B C D.Thus,c1 that F isclique athat perfect in SFisis2, matching a then[ perfectS can contain inmatchingG[ atA mostB in four].G non-leaf[A B vertices:]. at most two vertices from [ [ [ and c2 must be non-adjacent[ in[H/F . However, this is a contradiction to the argument that (H/F )[NH/F (u) NH/F (v)] is a4 clique. From4 this, we derive that H[A B C D X]/F is \ [ [ [ [ indeed a clique. C V I T 2 0 1 6 Now, notice that (H, A, B, C, D, E, n)whereE = V (K ) V (K ) is an instance of Noisy 1 [ 2 Structured Clique Contraction. Furthermore, since F n and we have already shown | |  that H[A B C D X]/F is a clique, we have that F is a solution to this instance. Therefore, [ [ [ [ by Lemma 5.1, F is a matching of size n in H such that each edge in F has one endpoint in A and the other in B. In particular, F E(G) and hence X = . Because G = H[A B C D], ✓ ; [ [ [ 8 23:24 Tight Lower Bounds on the Computation of Hadwiger Number

K and at most two vertices from outside K ∪ I (because H[V (G)]/F is a clique); then, it is trivial to color S with number of colors equal to its maximum clique size—in fact, it is straightforward to verify that any graph on at most four vertices is perfect. Thus, in what follows, suppose that the maximum size of a clique in S is at least 3. Now, consider a clique Cb of maximum size in S, and observe that it must either consist only of vertices in K or of no vertex in K (in which case it can contain at most one vertex from I). In the first case, color each vertex in u0 ∈ V (Cb) by a distinct color, and note that all vertices in V (S) \ V (Cb) can be colored using the same set of colors so that a vertex and its tagged copy are assigned distinct colors. The second case is analogous. In either case, we obtain that the chromatic number of S equals its maximum clique size. Thus, (H, n) is a Yes-instance of Perfect Contraction, which means that the call to PerfectAlg with (H, n) returns Yes, and hence CliConAlg returns Yes. Now, suppose that CliConAlg returns Yes, which means that the call to PerfectAlg with (H, n) returns Yes. Thus, (H, n) is a Yes-instance of Perfect Contraction, which means that there exists a subset F ⊆ E(H) of size at most n such that H/F is a perfect graph. We first argue that there does not exist a vertex a ∈ A ∪ B such that neither a nor a0 is incident to at least one edge in F . Targeting a contradiction, suppose that there exists a ∈ A ∪ B such that neither a not a0 is incident to at least one edge in F . Assume that a ∈ A as the other case is symmetric. Because |F | ≤ n and |D| = 2n, there either exists a vertex d ∈ D such that neither d nor d0 is incident to at least one edge in F , or F is a perfect matching in either G[D] or G[D0], where in the latter case we let d denote some arbitrarily chosen vertex from D. Additionally, since I is an independent set of size n + 1, there exists a vertex i ∈ I that is not incident to any edge in F . Now, consider the cycle i − a − a0 − d0 − d − i (on five vertices) in H. This cycle is an induced cycle in H, because no vertex in A is adjacent to any vertex in D, and by the construction of H, i is not adjacent to a0 and d0, a is not adjacent to d0 and a0 is not adjacent to d. Furthermore, as i, a and a0 are not incident to any edge in F , and if any of d and d0 is incident to an edge in F , then F is a perfect matching in either G[D] or G[D0], we obtain that i − a − a0 − db− db0 − i is an induced cycle (on five vertices) in H/F , where db and db0 are the vertices yielded by the replacement of the connected components of H[F ] that contain d and d0, respectively, if such components exist (otherwise, db = d and db0 = d0). However, an induced cycle on five vertices has chromatic number 3 and maximum clique size 2, thus we derive a contradiction to the supposition that H/F is perfect. So far, we derived that there does not exist a vertex a ∈ A ∪ B such that neither a nor a0 is incident to at least one edge in F . As |F | ≤ n and |A| = |A0| = |B| = |B0| = n, this means that every edge in F has both endpoints in A ∪ A0 ∪ B ∪ B0 and that for each u ∈ A ∪ B, exactly one vertex among u and u0 is incident to an edge in F . Now, we will show that each vertex a ∈ A ∪ B is incident to at least one edge in F . Targeting a contradiction, suppose that there exists a vertex a ∈ A ∪ B that is not incident to any edge in F . Assume that a ∈ A, as the other case is symmetric. Denote i, d, d0, db and db0 as before, and again consider the induced cycle i − a − a0 − d0 − d − i in H. Unlike before, now a0 belongs to some connected component of H[F ], yet we know that this connected component consists only of a0 and some 0 0 vertex in B . Let ba be the vertex yielded by the replacement of this component. As no vertex 0 0 0 in B is adjacent to any vertex in I ∪D, we again have that i−a−ba −db −db−i is an induced cycle in H/F , which gives rise to a contradiction. Thus, as |F | ≤ n and |A| = |B| = n, we know that F is a perfect matching in G[A ∪ B]. Next, we will show that G/F is a clique. This will imply that (G, A, B, C, D, n) is a Yes-instance of Structured Clique Contraction and thereby complete the proof. Targeting a contradiction, suppose that G/F is not a clique, and therefore there exist two F.V. Fomin, D. Lokshtanov, I. Mihajlin, S. Saurabh and M. Zehavi 23:25 non-adjacent vertices u and v in G/F . As F is a matching in G[A ∪ B], we can let x and y be two vertices in A ∪ B that belonged to the connected components of H[F ] that yielded u and v, respectively. Notice that the only vertex in A ∪ B adjacent to x0 is x, and the analogous claim holds for y0 and y. As F is a matching in G[A ∪ B] that does not match x and y (since otherwise u and v would not be distinct vertices), we have that neither u is adjacent to y0 in H/F nor v is adjacent to x0 in H/F . From this, by the construction of H and since F is a matching in G[A ∪ B], we immediately derive that i − u − x0 − y0 − v − i is an induced cycle in H/F where i is some arbitrarily chosen vertex from I. However, as before, the existence of such a cycle contradicts the supposition that H/F is a perfect graph. Thus, the proof of the reverse direction is complete. J

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