Grundy Coloring & friends, Half-Graphs, Bicliques
Pierre Aboulker, Édouard Bonnet, Eun Jung Kim, Florian Sikora
GGrundy Coloring & friends, Half-Graphs, Bicliques
Pierre Aboulker
DI/ENS, PSL University, Paris, [email protected]
Édouard Bonnet
Univ Lyon, CNRS, ENS de Lyon, Université Claude Bernard Lyon 1, LIP UMR5668, [email protected]
Eun Jung Kim
Université Paris-Dauphine, PSL University, CNRS UMR7243, LAMSADE, Paris, [email protected]
Florian Sikora
Université Paris-Dauphine, PSL University, CNRS UMR7243, LAMSADE, Paris, Francefl[email protected]
Abstract
The first-fit coloring is a heuristic that assigns to each vertex, arriving in a specified order σ , thesmallest available color. The problem Grundy Coloring asks how many colors are needed for themost adversarial vertex ordering σ , i.e., the maximum number of colors that the first-fit coloringrequires over all possible vertex orderings. Since its inception by Grundy in 1939, Grundy Coloring has been examined for its structural and algorithmic aspects. A brute-force f ( k ) n k − -time algorithmfor Grundy Coloring on general graphs is not difficult to obtain, where k is the number of colorsrequired by the most adversarial vertex ordering. It was asked several times whether the dependencyon k in the exponent of n can be avoided or reduced, and its answer seemed elusive until now. Weprove that Grundy Coloring is W[1]-hard and the brute-force algorithm is essentially optimalunder the Exponential Time Hypothesis, thus settling this question by the negative.The key ingredient in our W[1]-hardness proof is to use so-called half-graphs as a buildingblock to transmit a color from one vertex to another. Leveraging the half-graphs, we also provethat b -Chromatic Core is W[1]-hard, whose parameterized complexity was posed as an openquestion by Panolan et al. [JCSS ’17]. A natural follow-up question is, how the parameterizedcomplexity changes in the absence of (large) half-graphs. We establish fixed-parameter tractabilityon K t,t -free graphs for b -Chromatic Core and Partial Grundy Coloring , making a step towardanswering this question. The key combinatorial lemma underlying the tractability result might beof independent interest.
Theory of computation → Graph algorithms analysis; Theory ofcomputation → Fixed parameter tractability
Keywords and phrases
Grundy coloring, parameterized complexity, ETH lower bounds, K t,t -freegraphs, half-graphs Acknowledgements
A part of this work was done while the authors attended the “2019 IBS Summerresearch program on Algorithms and Complexity in Discrete Structures”, hosted by the IBS discretemathematics group. The third and the fourth authors were partially supported by the ANR project“ESIGMA” (ANR-17-CE23-0010)
A coloring is said proper if no two adjacent vertices receive the same color. The chromaticnumber of a graph G denoted by χ ( G ) is the minimum number of colors required to properlycolor G . Let us now consider a natural heuristic to build a proper coloring of a graph G .Given an ordering σ of the vertices of G , consider each vertex of G in the order σ and assign a r X i v : . [ c s . CC ] J a n to the current vertex the smallest possible color (without creating any conflict), i.e., thesmallest color not already given to one of its already colored neighbors. The obtained coloringis obviously proper and it is called a first-fit or greedy coloring. The Grundy number, denotedby Γ( G ), is the largest number of colors used by the first-fit coloring on some ordering of thevertices of G . Thus Γ( G ) is an upper-bound to the output of a first-fit heuristic.The Grundy number has been introduced in 1939 [22], but was formally defined onlyforty years ago, independently by Christen and Selkow [10] and by Simmons [38]. GrundyColoring in directed graphs already appears as an NP-complete problem in the monographof Garey and Johnson [19]. The undirected version remains NP-hard on bipartite graphs [25]and their complements [41], chordal graphs [37] and line graphs [24]. When the input isa tree,
Grundy Coloring can be solved in linear time [27]. This result is generalizedto bounded-treewidth graphs with an algorithm running in time k O ( w ) O ( wk ) n = O ( n w )for graphs of treewidth w and Grundy number k [39], but this cannot be improved to O ∗ (2 o ( w log w ) ) under the ETH [4]. It is also possible to solve Grundy Coloring in time O ∗ (2 . n ) [4].In 2006, Zaker [42] observed that since a minimal witness (we will formally define awitness later) for Grundy number k has size at most 2 k − , the brute-force approach gives analgorithm running in time f ( k ) n k − , that is, an XP algorithm in the words of parameterizedcomplexity. Since then it has been open whether Grundy Coloring can be solved in FPTtime, i.e., f ( k ) n O (1) (where the exponent does not depend on k ). FPT algorithms wereobtained in chordal graphs, claw-free graphs, and graphs excluding a fixed minor [4], orwith respect to the dual parameter n − k [25]. The parameterized complexity of GrundyColoring in general graphs was raised as an open question in several papers in the pastdecade [37, 26, 20, 4].Closely related to Grundy coloring is the notion of partial Grundy coloring and b -coloring .We say that a proper coloring V ] · · · ] V k is a partial Grundy coloring of order k if thereexists v i ∈ V i for each i ∈ [ k ] such that v i has a neighbor in every V j with j < i . The problem Partial Grundy Coloring takes a graph G and a positive integer k , and asks if thereis a partial Grundy coloring of order k . Erdős et al. [16] showed that the partial Grundynumber coincides with the so-called upper ochromatic number. This echoes another result ofErdős et al. [15] that Grundy number and ochromatic number (introduced by Simmons [38])are the same.The b -chromatic core of order k of a graph G is a vertex-subset C of G with the followingproperty: C admits a partition into V ] · · · ] V k such that there is v i ∈ V i for each i ∈ [ k ]which contains a neighbor in every V j with j = i . The goal of the problem b -ChromaticCore is to determine whether an input graph G contains a b -chromatic core of order k . Thisnotion was studied in [14, 36] in relation to b -coloring , which is a proper coloring such that forevery color i , there is a vertex of color i which neighbors a vertex of every other color. Themaximum number k such that G admits a b -coloring with k colors is called the b -chromaticnumber of G . In [36], it was proven that deciding whether a graph G has b -chromatic numberat least k is W [1]-hard parameterized by k . The problem might be even harder since nopolytime algorithm is known when k is constant. The authors left it as an open questionwhether b -Chromatic Core is W[1]-hard or FPT. Our contribution: the half-graph is key.
We prove that
Grundy Coloring is W[1]-complete, thus settling the open question posed in [37, 26, 20, 4]. More quantitatively weshow that the double-exponential XP algorithm is essentially optimal. Indeed we provethat there is no computable function f such that Grundy Coloring is solvable in time f ( k ) n o (2 k − log k ) , unless the ETH fails. This further answers by the negative an alternative . Aboulker, É. Bonnet, E. J. Kim, F. Sikora 3 question posted in [37, 26], whether there is an algorithm in time n k O (1) .A key element in the hardness proof of Grundy Coloring is what we call a half-graph (definition in Section 2.1). The main obstacle encountered when one sets out to proveW[1]-hardness of
Grundy Coloring is the difficulty of propagating a chosen color from avertex to another while keeping the Grundy number low (i.e., bounded by a function of k ).Employing half-graphs turns out to be crucial to circumvent this obstacle, which we furtherexamine in Section 4. Leveraging half-graphs as color propagation apparatus, we also provethat b -Chromatic Core is W[1]-complete (albeit with a very different construction). Thissettles the question posed by [36]. Our contribution: delineating the boundary of tractability.
All three problems,
Grundy Coloring , Partial Grundy Coloring , and b -Chromatic Core are FPT onnowhere dense graphs, for the parameter k being the number of desired colors. The existenceof each induced witness can be expressed as a first-order formula on at most 2 k − variablesin the case of Grundy Coloring , and on at most k variables in the case of PartialGrundy Coloring and b -Chromatic Core . The problem is therefore expressible infirst-order logic as a disjunction of the existence of every induced witness while the numberof induced witnesses is bounded by 2 k − . And first-order formulas can be decided in FPTtime on nowhere dense graphs [21]. The next step is K t,t -free graphs, i.e., those graphswithout a biclique K t,t as a (not necessarily induced) subgraph, which is a dense graph classthat contains nowhere dense graphs and graphs of bounded degeneracy. In the realm ofparameterized complexity, K t,t -free graphs have been observed to admit FPT algorithms forotherwise W[1]-hard problems [40].We prove that Partial Grundy Coloring and b -Chromatic Core are fixed-parametertractable on K t,t -free graphs, even in the parameter k + t , now assuming that t is not a fixedconstant. To this end, a combinatorial lemma plays a crucial role by letting us rule out thecase when many vertices have large degree: if there are many vertices of large degree in a K t,t -free graph, one can find a collection of k vertex-disjoint and pairwise non-adjacent starson k -vertices, which is a witness for b -Chromatic Core and Partial Grundy Coloring .Now, we can safely confine the input instances to have bounded degrees, save a few vertices.We present an FPT algorithm that works under this setting.
Limits and further questions.
Noting that the W[1]-hardness constructions for
GrundyColoring and b -Chromatic Core heavily rely on (large) half-graphs as a unit for colorpropagation, one may wonder how the parameterized complexity shall be affected if largehalf-graphs are excluded. The aforementioned tractability of b -Chromatic Core is a steptoward this end: recall that excluding a K t,t implies excluding a large half-graph H t,t (butnot vice versa). We point out that our key combinatorial lemma for K t,t -free graphs collapseson H t,t -free graphs. Therefore, whether Grundy Coloring and b -Chromatic Core are fixed-parameter tractable on H t,t -free graphs remains open. For Grundy Coloring ,whether it admits an FPT algorithm on K t,t -free graphs is unsettled.The parameterized complexity of Partial Grundy Coloring on general graphs resistedour efforts so far. Again the main barrier toward a W[1]-hardness proof is to come up witha right unit for color propagation. Note that half-graphs cannot be used for
PartialGrundy Coloring as the partial Grundy number is already high on a large half-graph.This phenomenon suggests that the class of H t,t -free graphs is a promising zone for inspectionin order to establish the parameterized complexity of Partial Grundy Coloring ongeneral graphs.
Organization of the rest of the paper.
In Section 2, we lay out the terminology and preliminary results. Section 3 exhibits the main issue in establishing the parameterizedhardness of
Grundy Coloring . In Section 4, we show how to overcome this issue and provethat
Grundy Coloring is W[1]-hard. In Section 5, being even more restricted in our gadgetchoice, we show that b -Chromatic Core is W[1]-hard. The tractability of b -ChromaticCore and Partial Grundy Coloring in K t,t -free graphs is finally presented in Section 6. For any integer i, j , we denote by [ i, j ] the set of integers that are at least i and at most j ,and [ i ] is a short-hand for [1 , i ]. We use the standard graph notations [12]: for a graph G , V ( G ) denotes the set of vertices of G , E ( G ) denotes the set of edges. A vertex u is a neighborof v if uv is an edge of G . The open neighborhood of a vertex v is the set of all neighbors of v and N [ v ] denotes the closed neighborhood of v defined as N ( v ) ∪ { v } . The open (closed,respectively) neighborhood of a vertex-set S is S v ∈ S N ( v ) \ S ( S v ∈ S N ( v ) ∪ S , respectively).For a vertex-set Y ⊆ V ( G ), we denote N ( v ) ∩ Y ( N [ v ] ∩ Y , respectively) simply as N Y ( v )( N Y [ v ], respectively) and the same applies the open and closed neighborhood of a vertex-set S . For two disjoint vertex-sets X and Y , we say that X is (anti-)complete with Y if everyvertex of X is (non-)adjacent with every vertex of Y . We call anti-matching the complement of an induced matching. The anti-matching of height t is the complement of t edges. It will soon be apparent that all the coloring numbers consideredin this paper are lowerbounded by t , in presence of an anti-matching of height t . Thereforein the subsequent FPT reductions, we will not have the luxury to have anti-matchings ofunbounded size. This will constitute an issue since they are useful to propagate choices.Imagine we have two sets A and B of size unbounded by the parameter, and we want torelate a choice in A to the same choice in B . Let us put an antimatching between A and B . Trivially independent sets of size 2 will correspond to consistent choices. So it all boilsdown to expressing our problem in terms of finding large enough independent sets. Now thisoption is not available, another way to propagate choices is to use half-graphs .We call half-graph a graph whose vertices can be partitioned into ( A, B ) such that thereis no induced 2 K in the graph induced by the edges with one endpoint in A and the otherendpoint in B , and G [ A ] and G [ B ] are both edgeless. These graphs are sometimes called bipartite chain graphs . Equivalently we say that ( A, B ) induces, or by a slight abuse ofnotation, is a half-graph if A and B can be totally ordered, say a , . . . , a | A | and b , . . . , b | B | such that N B ( a ) ⊇ N B ( a ) ⊇ . . . ⊇ N B ( a t ) and if a i b j is an edge then for every j ∈ [ j +1 , t ], a i b j is also an edge. The orderings a , . . . , a | A | and b , . . . , b | B | are called orders of the half-graph. Note that the orders of A and B “orient the half-graph from A to B ”. What we meanis that the same orderings do not testify that ( B, A ) is a half-graph. For that, we wouldneed to take the reverse orders. This technicality will have its importance in limiting thenotion of “path” (or “cycle”) of graphs, when we will impose that the orientation is the samethroughout the path (or cycle).
The half-graph of height t is a bipartite graph with partition ( A = { a , . . . , a t } , B = { b , . . . , b t } ) such that there is an edge between a i and b j if and only if i < j . We denote thisgraph by H t,t . The level of a vertex v ∈ A ( v ∈ B ) in the half-graph of height t is its indexin the ordering of A (or B ). Note that this is not uniquely defined for a half-graph in general,but it is for the (canonical) half-graph of height t . Any half-graph can be obtained fromthe half-graph of height t (for some t ) by duplicating some vertices. The name half-graph . Aboulker, É. Bonnet, E. J. Kim, F. Sikora 5 actually comes from Erdős and Hajnal (see for instance [17]). More precisely what Erdősdefines as a half-graph corresponds in this paper to the (canonical) half-graph of height t .We now define “path” and “cycle” of half-graphs. A length- ‘ path of half-graphs is agraph H whose vertex-set can be partitioned into ( H , H , . . . , H ‘ +1 ) such that the threefollowing conditions hold:(i) there is no edge between H i and H j when | i − j | (cid:62) i ∈ [ ‘ ], H [ H i ∪ H i +1 ] is a half-graph with bipartition ( H i , H i +1 ), and(iii) for every i ∈ [2 , ‘ ], the ordering of H i in the half-graph induced by ( H i − , H i ) is thesame as in the half-graph ( H i , H i +1 ).A length- ( ‘ + 1) cycle of half-graphs is the same but there are edges between H and H ‘ +1 ,namely a half-graph respecting the ordering of H and H ‘ +1 in the half-graphs inducedby ( H , H ) and ( H ‘ , H ‘ +1 ). We denote by H ‘ × t the length- ‘ path of half-graphs whereeach half-graph is isomorphic to H t,t , and H ∗ ‘ × t the length- ‘ cycle of half-graphs where eachhalf-graph is isomorphic to H t,t . The graph H ∗ ‘ × t is also a convenient way of propagating a“one-among- t ” consistent choice, since the only maximum independent sets of H ∗ ‘ × t take all ‘ + 1 vertices at the same level. A proper coloring of G with k color classes V ] · · · ] V k is a Grundy coloring of order k iffor each i ∈ [2 , k ], every vertex v i ∈ V i has a neighbor in every V j with j < i . The Grundynumber Γ( G ) is defined as the largest k such that G admits a Grundy coloring of order k .We say that an induced subgraph H of G is a witness achieving (color) k if H has a Grundycoloring of order at least k ; in this case, we simply say that H is a k -witness (also called atom by Zaker [42] or critical [23]). We say that a k -witness is minimal if there is no properinduced subgraph of it whose Grundy number is at least k . A graph G has Grundy numberat least k if and only if it contains a minimal k -witness as an induced subgraph [42]. Thisgives us an equivalent formulation of Grundy Coloring , which we frequently employ as aworking definition.
Grundy Coloring
Parameter: k Input:
An integer k >
0, a graph G . Question:
Is there a vertex-subset S ⊆ V ( G ) such that G [ S ] is a Grundy minimal k -witness?Notice that in a minimal k -witness, exactly one vertex receives color k . It is not difficultto see that a minimal k -witness is connected. A colored witness is a witness together with acoloring of its vertices. The coloring can equivalently be given by the vertex ordering, by thecoloring function, or by the partition into color classes. When k is not specified, a witnessfor G is the witness which allows a Grundy coloring of order Γ( G ).Let V ] · · · ] V k be a Grundy coloring of order k . We say that a vertex u colored c supports v colored c if u and v are adjacent and c < c . A vertex v colored in c is said to be supported if the colors of the vertices supporting v span all colors from 1 to c − k -witness uses 2 k − vertices [42]. These witnessesare implemented by a family of rooted trees called binomial trees (see for instance [4]). Theset of binomial trees ( T k ) k (cid:62) is defined recursively as follows: T consists of a single vertex, declared as the root of T . T k consists of two binomial trees T k − such that the root of the first one is a child of theroot of the other. The root of the latter is declared as the root of T k . Figure 1
The binomial tree T , where the labels denote the color of each vertex in a first-fitcoloring achieving the highest possible color. We outline some basic properties of k -witnesses and binomial trees T k . The first observa-tion is straightforward. (cid:73) Observation 1.
Any subset of k color classes of a k -witness, with k < k , induces a k -witness. The following is shown in a more general form in Lemma 7 of [4]. (cid:73)
Lemma 2.
Let i ∈ [2 , k − , X ⊆ V ( T k ) be a subset of roots of T i whose parent is a rootof T i +1 , and T k be a tree obtained from T k by removing the subtree T i − of every vertex in X . We assume that T k is an induced subgraph of a graph G and that N ( V ( G ) \ V ( T k )) = X .Then the three following conditions are equivalent in G :(i) There is a Grundy coloring that colors k the root of T k .(ii) There is a Grundy coloring that colors i every vertex of X without coloring theirparent in T k first.(iii) There is a Grundy coloring that colors i − at least one neighbor of each vertex of X without coloring any vertex of T k first. Proof. (iii) implies (ii), and (ii) implies (i) are a direct consequence of the optimum Grundycoloring of a binomial tree, as depicted in Figure 1. We show that (i) implies (ii). Thisis equivalent to showing that the only way for a Grundy coloring of T k to color its root k ,even when there is joker that enables us to give any color to a vertex of X , is to respectthe coloring of Figure 1. This holds since coloring a vertex of X with a color greater than i prevents from coloring its parent w with color i + 1. Indeed in that case w cannot find aneighbor colored i (which is not its own parent). Coloring a vertex of X with a color smallerthan i , simply will not work, since the Grundy coloring of T k that gives color k to its root isunique. Finally (ii) implies (iii), since for every vertex of X , its only neighbor that can obtaincolor i − T k . For a complete proof, see Lemma 7 of [4]. (cid:74)(cid:73) Lemma 3. If u and v are false twins in G , i.e., N G ( u ) = N G ( v ) , then Γ( G ) = Γ( G − { v } ) . Proof.
Let V ] · · · ] V k be an arbitrary Grundy coloring. We first observe that u and v must be in the same color class. Suppose the contrary, and assume that the color c of u ishigher than the color c of v . Since u is supported by a vertex of color c and N ( u ) = N ( v ), v has a neighbor of the same color. This contradicts that the Grundy coloring is a propercoloring. Finally, observe that u and v support and are supported by the same set of vertices,which proves the lemma. (cid:74)(cid:73) Lemma 4.
Let H be an induced subgraph of G such that all the vertices of N ( V ( H )) havedegree at most s . Then no vertex of V ( H ) can get a color higher than Γ( H ) + s in a Grundycoloring of G . Proof.
Suppose that a vertex v of H gets a color at least Γ( H ) + s + 1 in a Grundy coloringof G . Consider a minimal colored witness H of G achieving Γ( H ) + s + 1 in which v is the . Aboulker, É. Bonnet, E. J. Kim, F. Sikora 7 unique vertex getting the color Γ( H ) + s + 1. The vertices of N ( V ( H )), having degree atmost s , can get color at most s + 1. Since we are dealing with a minimal colored witnessachieving a color strictly larger than s + 1, they can get color at most s . Now we remove thefirst s color classes from our colored witness. By Observation 1 we obtain a Grundy coloringachieving color Γ( H ) + 1, and by the previous remark, it no longer intersects N ( V ( H )). Theconnected component containing v of that new witness achieves color Γ( H ) + 1 inside H , acontradiction. (cid:74) An immediate corollary of Lemma 4 is that a vertex with only few neighbors of high-degreecannot receive a high color. (cid:73)
Corollary 5.
In any greedy coloring, a vertex with at most t neighbors that have degree atmost s cannot receive a color higher than s + t + 1 . A proper coloring of G with k color classes V ] · · · ] V k is a partial Grundy coloring of order k if, for each i ∈ [2 , k ], there exists v i ∈ V i such that v i has at least one neighbor in every V j with j < i . A proper coloring of G with k color classes V ] · · · ] V k is a b -coloring oforder k if, for each i ∈ [ k ], there exists v i ∈ V i such that v i has at least one neighbor inevery V j with j = i . The b -chromatic number of G is the maximum k such that G allowsa b -coloring of order k . A vertex-set S of G is called a b -chromatic core of order k if G [ S ]admits a b -coloring of order k . It is easy to see that admitting a partial Grundy coloring oforder k is monotone under taking an induced subgraph. (cid:73) Observation 6.
A graph G admits a partial Grundy coloring of order at least k if andonly if there exists a vertex-set S ⊆ V ( G ) such that G [ S ] admits a partial Grundy coloringof order k . Proof.
The forward direction is trivial. Suppose that G [ S ] has a partial Grundy coloring V ] · · · ] V k of order k . For each uncolored vertex v ∈ V \ S , let c ∈ [ k ] be the highest colorsuch that for every color i ∈ [ c ], v has a neighbor colored in i . We assign color c + 1 to v . Itis straightforward to see that the coloring of S ∪ { v } is a partial Grundy coloring of order atleast k . (cid:74) Following from this observation, we can formally define
Partial Grundy Coloring as:
Partial Grundy Coloring
Parameter: k Input:
An integer k >
0, a graph G . Question:
Is there a vertex-subset S ⊆ V ( G ) such that G [ S ] admits a partial Grundycoloring of order k ?On the other hand, b -coloring is not monotone under taking induced subgraphs. That is, G might contain a b -chromatic core of order k but G does not allow a b -coloring of order atleast k . This leads us to the following monotone problem, which is distinct from decidingwhether the b -chromatic number of G is at least k . b -Chromatic Core Parameter: k Input:
An integer k >
0, a graph G . Question:
Is there a vertex-subset S ⊆ V ( G ) such that G [ S ] admits a b -coloring oforder k ? It is not difficult to see that there is a χ ( G )-coloring of G which is a b -coloring as well, where χ ( G ) isthe chromatic number of G . For both
Partial Grundy Coloring and b -Chromatic Core , the subgraph of G induced by S is referred to as a k -witness if S ⊆ V ( G ) is a solution to the instance ( G, k ). A k -witness H is called a minimal k -witness if H − v is not a k -witness for every v ∈ V ( H ).Let V ] · · · ] V k be a proper coloring of G . In the context of partial Grundy coloring( b -coloring, respectively), we say that a vertex v colored c is supported by u if uv ∈ E ( G )and u is colored c < c ( c = c , respectively). In the partial Grundy coloring ( b -coloring,respectively), a vertex v colored c is supported if the colors of the supporting vertices of v span all colors from 1 to c − k ] \ c , respectively). Such a vertex v is alsocalled a center . A color c is said realized if a vertex v colored c is supported. That vertex v is then realizing color c . Notice the crucial difference with Grundy colorings that these c − k -witness of Partial Grundy Coloring or b -Chromatic Core , eachcolor class contains a supported vertex, which we call a center . As each center requests atmost k − k -witnesses of Partial Grundy Coloring or b -Chromatic Core has size bounded by k [14]. We note that a leaf of a center may verywell be a center itself. A caricatural example of that is a k -clique, where all the vertices arecenters and leaves. Therefore, both Partial Grundy Coloring and b -Chromatic Core can be seen as finding at most k vertices and a proper k -coloring of them to realize all k colors. We denote by Γ ( G ), respectively Γ b ( G ), the maximum integer k such that G admitsa k -witness for Partial Grundy Coloring , respectively b -Chromatic Core . The
Exponential Time Hypothesis (ETH) is a conjecture by Impagliazzo et al. [28] that assertsthat there is no 2 o ( n ) -time algorithm for on n -variable instances. This conjectureimplies, for instance, that FPT = W[1]. Lokshtanov et al. [29] survey conditional lowerbounds under the ETH.We present some classic W[1]-hard problems which will be used as starting points ofour reductions. In the k -Multicolored Independent Set problem, one is given a graph G , whose vertex-set is partitioned into k sets V , . . . , V k , and is asked if there exists anindependent set I ⊆ V ( G ) such that | I ∩ V i | = 1 for every i ∈ [ k ]. The k -MulticoloredIndependent Set problem is a classic W[1]-hard problem when parameterized by k [13], andunless the ETH fails, cannot be solved in time f ( k ) | V ( G ) | o ( k ) for any computable function f [8]. A related problem is k -Multicolored Subgraph Isomorphism whose definition goesas follows: k -Multicolored Subgraph Isomorphism Parameter: k Input:
An integer k >
0, a graph G whose vertex-set is partitioned into k sets V , . . . , V k ,and a graph H with V ( H ) = [ k ]. Question:
Is there φ : i ∈ [ k ] v i ∈ V i such that for all ij ∈ E ( H ), φ ( i ) φ ( j ) ∈ E ( G )?Even when H is a 3-regular graph, k -Multicolored Subgraph Isomorphism cannotbe solved in time f ( k ) | V ( G ) | o ( k/ log k ) = f ( | E ( H ) | ) | V ( G ) | o ( | E ( H ) | / log | E ( H ) | ) , unless the ETHfails (see Theorem 5.5 in [33]). One can remove unnecessary edges in G (in those E ( V i , V j ) with ij / ∈ E ( H )) such that φ ([ k ]) and H are actually isomorphic in a solution. k -MulticoloredSubgraph Isomorphism allows stronger ETH lower bounds than a reduction from k -Multicolored Independent Set when the size of the new parameter depends on thenumber of edges of the sought pattern.The k -by- k Grid Tiling , or simply
Grid Tiling , introduced by Marx [30, 31], isa W[1]-hard problem (see for instance [11]) which is a convenient starting point to show . Aboulker, É. Bonnet, E. J. Kim, F. Sikora 9 parameterized hardness for geometric problems [30, 18]. It may turn out useful in othercontexts, such as H -free graphs [3], and as we will see in this paper for b -Chromatic Core . k -by- k Grid Tiling (or
Grid Tiling ) Parameter: k Input:
Two integers k, n >
0, and a set { P i,j } i,j ∈ [ k ] of k subsets of pairs in [ n ] × [ n ]. Question:
Can one select exactly one pair ( x i,j , y i,j ) in each P i,j such that for every i, j ∈ [ k ], x i,j = x i,j +1 and y i,j = y i +1 ,j ?Observe that the usual (equivalent) definition requires instead x i,j = x i +1 ,j and y i,j = y i,j +1 . We prefer the other formulation since it is closer to the geometric interpretation ofpacking squares in the plane. It is not difficult to see that deciding if a fixed vertex can get color k in a greedy coloring isW[1]-hard. Let us call this problem Rooted Grundy Coloring . (cid:73) Observation 7.
Rooted Grundy Coloring is W[1]-hard.
Proof.
We design an FPT reduction from k -Multicolored Independent Set to RootedGrundy Coloring . Let H be an instance of k -MIS with partition V , . . . , V k . We build anequivalent instance G of Rooted Grundy Coloring in the following way. We copy H in G and we add a clique C of size k + 1. We call v a fixed vertex of C and we add a pendantneighbor v to v . We number the vertices of C \ { v } , v , . . . , v k , and we make v i adjacent toall the vertices of V i for each i ∈ [ k ]. A greedy coloring can color v by k + 2 if and only ifthere is a k -multicolored independent set in H . (cid:74) Of course this reduction does not imply anything for
Grundy Coloring . Indeed thevertices of V ( H ) could get much higher colors than v . This is precisely the issue with showingthe parameterized hardness of Grundy Coloring .1 21 21 21 21 2 (a)
Biclique (b)
Anti-matching (c)
Half-graph short cycle ∗∗∗∗ ∗∗∗ ∗∗ ∗ ∗∗∗∗ ∗∗∗ ∗∗ ∗ (d)
Half-graph long path
Figure 2
The barriers in a propagation gadget for
Grundy Coloring . The biclique has smallGrundy number but do not propagate, nor it imposes a unique choice. The three other propagationshave arbitrary large Grundy number, rendering them useless for a parameterized reduction.
A reduction starting from any W[1]-hard problem has to “erase” the potentially largeGrundy number of the initial structure. This can be done by isolating it with low-degreevertices. However the degree ∆ of the graph should be large, and a large chunk of the instance should have degree unbounded in k since Grundy Coloring is FPT parameterizedby ∆ + k [37, 4]. Besides, as it is the case with W[1]-hardness reductions where inducedsubgraphs of the initial instance have to be tamed, we crucially need to propagate consistentlyone choice among a number of alternatives unbounded in the parameter.A natural idea for encoding one choice among t ≫ k is to have a set S of t vertices, one ofwhich, the selected vertex, receiving a specific color, say, 1. Then a mechanism should ensurethat one cannot color 1 two or more vertices of S . Note that we cannot force that propertyby cliquifying S , as this would elevate the Grundy number to at least t . Furthermore, byRamsey’s theorem, there will be independent sets of size 2 Ω( log tk ) in S . Thus we might as wellassume that S is an independent set, and look for another way of preventing two verticesfrom getting color 1, than by adding edges inside S .We are now facing the following task: Given a bipartite graph, or a “path” or “cycle” ofbipartite graphs whose partite sets are copies of S , ensure that exactly one vertex can receivecolor 1 in each partite set, and that this corresponds to a single vertex in S . A bicliquecertainly has low Grundy number (see Figure 2a) but does not propagate nor it actuallyforces a unique choice. Anything more elaborate seems to have large Grundy number, beit the complement of an induced matching, or anti-matching , (see Figure 2b), a “cycle” ofhalf-graphs (see Figure 2c), or even a long “path” of half-graphs (see Figure 2d). We remindthe reader that, as detailed in Section 2, half-graphs and anti-matchings are (the) two waysof propagating a consistent independent set. It might be guessed from the previous section that the solution will come from a constant-length “path” of half-graphs. It is easy to see that half-graphs (that can be seen as length-onepath of half-graphs) have Grundy number at most 3. Due to the 2 K -freeness of thehalf-graph, there cannot be both color 1 and color 2 vertices present on both sides of thebipartition, say ( A, B ). If A is the side missing a 1 or a 2 among its colors, then B in turncannot have a 3 (nor a 4). The absence of vertices colored 3 in B prevents vertices colored 4in A . Overall, no vertex with color 4 can exist.It takes more time to realize that a length-two path of half-graphs have constant Grundynumber. We keep this proof to convey a certain intuition behind the boundedness of theGrundy number of such graphs. However we will then generalize this statement to anyconstant-length path. (cid:73) Lemma 8.
The Grundy number of a length-two path of half-graphs is at most 5.
Proof.
We denote by A , B , and C the tripartition of the vertex-set of H , a path of half-graphsof length two. A first observation is that for a color i (cid:62) A ∪ C ,there should be all the colors from 1 to i − B . In particular for theGrundy number to exceed 5, one needs to have all the colors up to 4 present in B . Withoutloss of generality, we assume that one vertex v ∈ B colored 2 is supported by a vertex u ∈ A colored 1.We claim that there cannot be a vertex colored 2 in A . Indeed this vertex would needto be supported by a vertex colored 1 in B , thereby creating with uv an induced 2 K inthe half-graph H [ A ∪ B ]. This absence implies that every vertex x ∈ B colored 3 has itssupporting vertex colored 2, say y , in C . In turn it implies that there is no vertex colored 3in C , since this vertex and its supporting 2 in B would form with xy an induced 2 K in thehalf-graph H [ B ∪ C ]. Now a vertex colored 4 in B has to be supported by a 3 in A , and by . Aboulker, É. Bonnet, E. J. Kim, F. Sikora 11 a 2 in C . Similarly, these two facts prevent the existence of 4 in A , and in C . In that case,there cannot be a 5 nor a 6 in B . The absence of a 5 in B makes it impossible to have a 6 in A ∪ C . Overall there cannot be a 6 appearing in H . (cid:74) We could stop here and make a W[1]-hardness using path of half-graphs of length one ortwo. For the sake of curiosity, but also in order to present a more transparent constructionusing length-four path of half-graphs, we show that, in all generality, the Grundy number ofa length- ‘ path of half-graphs is a “constant” depending only on ‘ . (cid:73) Lemma 9.
The Grundy number of a length- ‘ path of half-graphs is at most ‘ . Proof.
Achieving a (more) reasonable upper bound –the Grundy number of such graphs ismost likely polynomial or even linear in ‘ – proves to be not so easy. We choose here to givea short proof of an admittingly bad upper bound.We show this bound by induction on ‘ . Note that the statement trivially holds for ‘ = 0, and that we previously verified it for ‘ = 1. Assume that the Grundy number of anylength-( ‘ −
1) path of half-graphs is at most 4 ‘ − , for any ‘ (cid:62) G be a length- ‘ path of half-graphs, with partition V ( G ) = V ] V ] · · · ] V ‘ where G [ V i ∪ V i +1 ] is a half-graph for each i ∈ [ ‘ − G − V and G − V ‘ are bothlength-( ‘ −
1) path of half-graphs. Let H be a colored witness of G achieving color Γ( G ). Wedistinguish some cases based on the number of colors of H appearing in V or in V ‘ . In eachcase, we conclude with Observation 1. No more than 4 ‘ − colors of H can be missing in V (resp. in V ‘ ). Otherwise by Observation 1, the corresponding color classes form a k -witness G − V (resp. in G − V ‘ ) with some k > ‘ − , contradicting the induction hypothesis.So we may assume that at least Γ( G ) − ‘ − colors appear in V (resp. in V ‘ ). Thusat least (2Γ( G ) − · ‘ − ) − Γ( G ) = Γ( G ) − · ‘ − colors appears in both V and V ‘ . IfΓ( G ) > ‘ , then Γ( G ) − · ‘ − > ‘ − . We further claim that the corresponding colorclasses would form a witness in G − V , a contradiction. If not, it must be because a vertex x ∈ V colored i was adjacent to a vertex y ∈ V colored j < i , and is not adjacent to anyvertex colored j in G − V . But we know that V contains a vertex y colored i , which inturn must be adjacent to a vertex x ∈ V colored j , forming an induced 2 K in G [ V ∪ V ],a contradiction. Therefore, Γ( G ) (cid:54) ‘ .We observe that our proof works for a more general notion of “path of half-graphs” whereone does not impose the orders of the successive half-graphs to have the same orientation(see the second paragraph of Section 2.1). (cid:74) Combining the ideas in Lemmas 8 and 9, we obtain the following corollary, with a betterbound than the straightforward application of Lemma 9. (cid:73)
Corollary 10.
The Grundy number of a length-four path of half-graphs is at most 53.
Proof.
53 being 3 × (3 × (cid:74) We are now ready to present the hardness construction. We reduce from k -MulticoloredSubgraph Isomorphism whose definition can be found in the preliminaries. (cid:73) Theorem 11.
Grundy Coloring is W[1]-complete and, unless the ETH fails, cannotbe solved in time f ( q ) n o (2 q − log q ) (nor in time f ( q ) n o ( q ) ) for any computable function f , on n -vertex graphs with Grundy number q . Proof.
The membership to W[1] is given by the framework of Cesati [7], since there isalways a witness of size 2 q − . We show the W[1]-hardness of Grundy Coloring by reducing from k -Multicolored Subgraph Isomorphism with 3-regular pattern graphs.Let ( G = ( V , . . . , V k , E ) , H = ([ k ] , F )) be an instance of that problem. We further assumethat k is a positive even integer and there is no edge between V i and V j in G whenever ij / ∈ E ( H ). The goal is now to find v ∈ V , . . . , v k ∈ V k such that H is isomorphic to G [ { v , . . . , v k } ]. Even with these restrictions k -Multicolored Subgraph Isomorphism cannot be solved in time f ( k ) | V ( G ) | o ( k/ log k ) = f ( | E ( H ) | ) | V ( G ) | o ( | E ( H ) | / log | E ( H ) | ) , unlessthe ETH fails (see [32, 34], and Theorem 5.5 in [33]).We build an equivalent Grundy Coloring -instance ( G , q ) with q = d log k e + 55 asfollows. For each color class V i , we fix an arbitrary total ordering (cid:54) i on the vertices of V i ,and we write u < i u if u = u and u (cid:54) i u . Let i ∈ [ k ] and let i (1) , i (2) , i (3) ∈ [ k ] be thethree neighbors of i in H . Each V i is encoded by a length-4 path of half-graphs denoted by H i (see Figure 3). We now detail the construction of H i .We set V ( H i ) := L i ∪ V i,i (1) ∪ V i,i (2) ∪ V i,i (3) ∪ R i . The vertices of L i (resp. R i ) are inone-to-one correspondence with the vertices of V i . We denote by l ( u ) (resp. r ( u )) the vertexof L i (resp. R i ) corresponding to u ∈ V i . For each p ∈ [3], the vertices of V i,i ( p ) are inone-to-one correspondence with the edges of E ( V i , V i ( p ) ). We denote by z ( u, v ) the vertex of V i,i ( p ) corresponding to the edge uv ∈ E ( V i , V i ( p ) ) with u ∈ V i and v ∈ V i ( p ) .We set E ( H i ) := E ( L i , V i,i (1) ) ∪ E ( V i,i (1) , V i,i (2) ) ∪ E ( V i,i (2) , V i,i (3) ) ∪ E ( V i,i (3) , R i ): l ( u ) z ( u , v ) ∈ E ( L i , V i,i (1) ) if and only if u < i u for p ∈ [2], z ( u, v ) z ( u , v ) ∈ E ( V i,i ( p ) , V i,i ( p +1) ) if and only if u < i u z ( u, v ) r ( u ) ∈ E ( V i,i (3) , R i ) if and only if u < i u .For each pair of vertices u, u ∈ V i such that u < i u , we add an edge between l ( u ) and z ( u , v ) ∈ V i,i (1) , respectively z ( u, v ) ∈ V i,i (1) and z ( u , v ) ∈ V i,i (2) , respectively z ( u, v ) ∈ V i,i (2) and z ( u , v ) ∈ V i,i (3) , respectively z ( u, v ) ∈ V i,i (3) and r ( u ) (see Figure 3).For each ij ∈ E ( H ), we create | E ( V i , V j ) | copies of the binomial tree T . So these treesare in one-to-one correspondence with the edges of G between V i and V j , and we denoteby T ( uv ) the tree corresponding to uv ∈ E ( V i , V j ). We denote by β ( uv ) and γ ( uv ) thetwo children getting color 2 of the only two vertices colored 3, in the Grundy coloring of T ( uv ) which gives color 5 to its root. We remove the pendant neighbor of β ( uv ) and of γ ( uv ) (the two vertices getting color 1 and supporting β ( uv ) and γ ( uv )). This results in afourteen-vertex tree. We denote this set of trees by T i,j , and the | E ( V i , V j ) | roots of the T by R i,j . For each ij ∈ E ( H ) and for every pair z ( u, v ) ∈ V i,j , z ( v, u ) ∈ V j,i , we make z ( u, v )and β ( uv ) adjacent, and we make z ( v, u ) and γ ( uv ) adjacent.For every i ∈ [ k ], we create | V i | copies of the binomial tree T . These trees are inone-to-one correspondence with V i . Similarly as above, we denote by β ( u ) and γ ( u ) the twovertices getting color 2, whose parents are colored 3, in T ( u ) and we remove their pendantneighbor (colored 1). For every pair l ( u ) ∈ L i and r ( u ) ∈ R i , we link l ( u ) and β ( u ), and welink r ( u ) and γ ( u ). We denote this set of trees by T i , and the | V i | roots of the T by R i .We finally create one copy of the binomial tree T q . We observe that there are | E ( H ) | sets R i,j and | V ( H ) | sets R i . The binomial tree T q has at least | V ( H ) | + | E ( H ) | = 2 . k verticesgetting color 7 in the greedy coloring giving color q to the root. Indeed the number of verticescolored 7 is 2 q − , and it holds that q − (cid:62) log k + log 2 .
5. We map 2 . k distinct verticescolored 6 in T q , that are children of vertices colored 7, in a one-to-one correspondence with V ( H ) ∪ E ( H ). Let f ( i ) be the vertex mapped to i ∈ V ( H ) and f ( ij ) be the vertex mappedto ij ∈ E ( H ). We further remove the subtree T of each of these 2 . k vertices colored 6. Forevery i ∈ V ( H ), we link f ( i ) to all the vertices in R i . Similarly for every ij ∈ E ( H ), welink f ( ij ) to all the vertices in R i,j . This finishes the construction of the graph G . Solving Grundy Coloring in time f ( q ) n o (2 q − log q ) = f ( d log k e + 55) n o ( k/ log k ) would give the same . Aboulker, É. Bonnet, E. J. Kim, F. Sikora 13 l ( u ) r ( u ) l ( u ) r ( u ) l ( u ) r ( u ) l ( u ) r ( u ) l ( u ) r ( u ) L i R i z ( u , v ) z ( u , v ) z ( u , v ) z ( u , v ) z ( u , v ) z ( u , v ) z ( u , v ) z ( u , v ) z ( u , v ) z ( u , w ) z ( u , w ) z ( u , w ) z ( u , w ) z ( u , w ) z ( u , w ) z ( u , w ) z ( u , w ) z ( u , w ) z ( u , x ) z ( u , x ) z ( u , x ) z ( u , x ) z ( u , x ) z ( u , x ) z ( u , x ) z ( u , x ) z ( u , x ) V i,i (1) V i,i (2) V i,i (3) l ( u ) r ( u ) z ( u , v ) z ( u , w ) z ( u , x ) Figure 3
The encoding H i of one V i ordered u < u < u < u < u . In bold, a possibleindependent set intersecting the five sets and containing a consistent pair l ( u ) , r ( u ). running time for k -Multicolored Subgraph Isomorphism , which is ruled out under theETH. We now prove that the reduction is correct. A solution to k -Multicolored Subgraph Isomorphism implies Γ ( G ) (cid:62) q . Let v ∈ V , v ∈ V , . . . , v k ∈ V k be a fixed solution to the k -Multicolored SubgraphIsomorphism -instance (the colored isomorphism being i ∈ [ k ] v i ). We say that each edge v i v j is in the solution (for i = j ∈ [ k ]). We color 1 all the vertices of G corresponding to edgesin the solution, that is, all the vertices z ( v i , v j ), as well as all vertices of G correspondingto vertices in the solution, that is l ( v i ) and r ( v i ). This is possible since the five vertices l ( v i ) , z ( v i , v i (1) ), z ( v i , v i (2) ), z ( v i , v i (3) ) , r ( v i ) form an independent set since ¬ ( v i < i v i ).We can now color 2 the vertices β ( v i ) and γ ( v i ). Therefore the root of T ( v i ) can receivecolor 5. Moreover, for every ij ∈ E ( H ) we can color 2 the vertices β ( v i v j ) and γ ( v i v j ).Therefore the root of T ( v i v j ) can receive color 5. Since one vertex in each R i , and onevertex in each R i,j get color 5, the vertices f ( i ) and f ( ij ) can all get color 6. Finally theroot of T q can receive color q . Γ ( G ) (cid:62) q implies a solution to k -Multicolored Subgraph Isomorphism. We first show that only the two vertices of T q with degree q − q . Besidesthese two vertices, the only vertices of T q with sufficiently large degree to get color q arethe vertices f ( i ) and f ( ij ). But these vertices have at most one neighbor of degree morethan 5. So according to Corollary 5, they cannot receive a color higher than 7 < q . Now weuse Corollary 10 to bound the color reachable outside of T q . For every i ∈ [ k ], the induced subgraph G [ H i ] is a length-four path of half-graphs. Thus by Lemma 3 and Corollary 10,Γ( G [ H i ]) (cid:54)
53. All the vertices in the open neighborhood of V i,i (1) ∪ V i,i (2) ∪ V i,i (3) havedegree at most 2. So by Lemma 4 vertices outside T q cannot receive a color beyond 55 < q .We now established that if Γ( G ) (cid:62) q (actually Γ( G ) = q ), then either one of the twopossible roots of T q shall receive color q . By Lemma 2, this implies that all the vertices f ( i )and f ( ij ) receive color 6, and that in each R i and each R i,j there is at least one vertexreceiving color 5. For every i ∈ [ k ], let T ( u i ) be one T of T i whose root gets color 5. Wewill now show that { u , . . . , u i , . . . , u k } is a solution to the k -Multicolored SubgraphIsomorphism -instance. Again by Lemma 2, this is only possible if β ( u i ) and γ ( u i ) both getcolor 2, and their unique neighbor outside T ( u i ) gets color 1. It means that l ( u i ) and r ( u i )both get color 1.Since every R i,j contains at least one vertex colored 5, Lemma 2 implies that every V i,i ( p ) (for each p ∈ [3]) gets at least one vertex colored 1. Let z ( u, v ) ∈ V i,i (1) , z ( u , v ) ∈ V i,i (2) , and z ( u , v ) ∈ V i,i (3) three vertices getting color 1. As { l ( u i ) , z ( u, v ) , z ( u , v ) , z ( u , v ) , r ( u i ) } should be an independent set, we have u i (cid:62) i u (cid:62) i u (cid:62) i u (cid:62) i u i . This implies that u i = u = u = u . In turn that implies that no vertex z ( u ∗ , v ) ∈ V i,i (1) ∪ V i,i (2) ∪ V i,i (3) with u ∗ = u i can get color 1. Indeed l ( u i ) prevents a 1 “above” z ( u i , v ) ∈ V i,i (1) and z ( u i , v ) ∈ V i,i (2) prevents a 1 “below” z ( u i , v ). The same goes for the color classes V i,i (2) and V i,i (3) . Thus the only trees T ( uv ) ∈ T i,j that can get color 5 at their root are the onessuch that { u, v } ⊂ { u , . . . , u k } . As all the 1 . k sets T i,j have such a tree, it implies that { u , . . . , u k } is a solution to the k -Multicolored Subgraph Isomorphism -instance. (cid:74) b -Chromatic Core A length-two path of half-graphs have arbitrary large b -chromatic core. Indeed the coloringof Figure 2c yields a b -chromatic core achieving the height of the half-graphs number ofcolors (even without the edges of the “cycle”). Nevertheless a simple half-graph only admits b -chromatic cores of bounded size. We show how to still build a W[1]-hardness constructionin this furtherly constrained situation. (cid:73) Theorem 12. b -Chromatic Core is W[1]-complete. Proof.
The inclusion in W[1] is immediate by the characterization of Cesati [7], and thefacts that minimal witnesses have size at most k , and that given the subgraph induced bya minimal witness one can check if it is solution. To show W[1]-hardness, we reduce from k -by- k Grid Tiling . We recall that, in this problem, given k sets of pairs over [ n ], say,( P i,j ⊆ [ n ] × [ n ]) i,j ∈ [ k ] × [ k ] , "displayed in a k -by- k grid", one has to find one pair ( x i,j , y i,j )in each P i,j such that x i,j = x i,j +1 and y i,j = y i +1 ,j , for every i, j ∈ [ k − Construction.
Let ( P i,j ⊆ [ n ] × [ n ]) i,j ∈ [ k ] × [ k ] be the instance of Grid Tiling . For each ( i, j ), we have theset of pairs P i,j with | P i,j | = t . For each ( i, j ), we add a biclique K t,q − ( i, j ) := K t,q − , where q := 14 k . The part of K t,q − ( i, j ) with size t is denoted by A i,j and the other part by B i,j (see Figure 4). We denote by A i,j the t vertices to the left of K t,q − ( i, j ) on Figure 4, and by B i,j , the q − A i,j are in one-to-one correspondencewith the pairs of P i,j . We denote by a i,j ( x, y ) ∈ A i,j the vertex corresponding to ( x, y ) ∈ P i,j .We make the construction “cyclic”, or rather “toroidal”. So in what follows, every occurrenceof i + 1 or j + 1 should be interpreted as 1 in case i = k or j = k . . Aboulker, É. Bonnet, E. J. Kim, F. Sikora 15 For every vertically (resp. horizontally) consecutive pairs ( i, j ) and ( i + 1 , j ) (resp. ( i, j )and ( i, j + 1)) we add a half-graph H ( i → i + 1 , j ) (resp. H ( i, j → j + 1)) with bipartition H ( i → i + 1 , j ) ∪ H ( i → i + 1 , j ) (resp. H ( i, j → j + 1) ∪ H ( i, j → j + 1)). Both sets H ( i → i + 1 , j ) and H ( i, j → j + 1) are in one-to-one correspondence with the verticesof A i,j , while the set H ( i → i + 1 , j ) is in one-to-one correspondence with the verticesof A i +1 ,j , and H ( i, j → j + 1), with the vertices of A i,j +1 . We denote by h i → i +1 ,j ( x, y )(resp. h i → i +1 ,j ( x , y )) the vertex corresponding to a i,j ( x, y ) (resp. a i +1 ,j ( x , y )). Similarlywe denote by h i,j → j +1 ( x, y ) (resp. h i,j → j +1 ( x , y )) the vertex corresponding to a i,j ( x, y )(resp. a i,j +1 ( x , y )). Every vertex in a half-graph H ( i → i + 1 , j ) or H ( i, j → j + 1) ismade adjacent to its corresponding vertex in A i,j ∪ A i +1 ,j ∪ A i,j +1 . Thus a i,j ( x, y ) is linkedto h i → i +1 ,j ( x, y ), h i − → i,j ( x, y ), h i,j → j +1 ( x, y ), and h i,j − → j ( x, y ). Note that underlinednumbers are used to distinguish names, and to give information on its neighborhood. Wecall vertical half-graph an H ( i → i + 1 , j ), and horizontal half-graph an H ( i, j → j + 1). Wenow precise the order of the half-graphs. In vertical half-graphs, we put an edge between h i → i +1 ,j ( x, y ) and h i → i +1 ,j ( x , y ) whenever y < y . In horizontal half-graphs, we put anedge between h i,j → j +1 ( x, y ) and h i,j → j +1 ( x , y ) whenever x < x . ai,j (1 , hi,j − → j (1 , hi,j − → j (1 , hi,j → j +1(1 , hi,j → j +1(1 , ai,j (1 , hi,j − → j (1 , hi,j − → j (1 , hi,j → j +1(1 , hi,j → j +1(1 , ai,j (2 , hi,j − → j (2 , hi,j − → j (2 , hi,j → j +1(2 , hi,j → j +1(2 , ai,j (2 , hi,j − → j (2 , hi,j − → j (2 , hi,j → j +1(2 , hi,j → j +1(2 , ai,j (3 , hi,j − → j (3 , hi,j − → j (3 , hi,j → j +1(3 , hi,j → j +1(3 , ai,j (3 , hi,j − → j (3 , hi,j − → j (3 , hi,j → j +1(3 , hi,j → j +1(3 , ai,j (3 , hi,j − → j (3 , hi,j − → j (3 , hi,j → j +1(3 , hi,j → j +1(3 , ai,j (4 , hi,j − → j (4 , hi,j − → j (4 , hi,j → j +1(4 , hi,j → j +1(4 , ai,j (4 , hi,j − → j (4 , hi,j − → j (4 , hi,j → j +1(4 , hi,j → j +1(4 , ai,j (5 , hi,j − → j (5 , hi,j − → j (5 , hi,j → j +1(5 , hi,j → j +1(5 , A i,j H i,j − → j H i,j → j +1 B i,j hi,j − → j (1 , hi,j − → j (1 , hi,j − → j (1 , hi,j − → j (1 , hi,j − → j (1 , hi,j − → j (1 , hi,j − → j (2 , hi,j − → j (2 , hi,j − → j (3 , hi,j − → j (3 , hi,j − → j (3 , hi,j − → j (3 , hi,j − → j (3 , hi,j − → j (3 , hi,j − → j (4 , hi,j − → j (4 , hi,j − → j (5 , hi,j − → j (5 , hi,j − → j (5 , hi,j − → j (5 , hi,j → j +1(1 , hi,j → j +1(1 , hi,j → j +1(1 , hi,j → j +1(1 , hi,j → j +1(2 , hi,j → j +1(2 , hi,j → j +1(2 , hi,j → j +1(2 , hi,j → j +1(3 , hi,j → j +1(3 , hi,j → j +1(4 , hi,j → j +1(4 , hi,j → j +1(4 , hi,j → j +1(4 , hi,j → j +1(5 , hi,j → j +1(5 , hi,j → j +1(5 , hi,j → j +1(5 , hi,j → j +1(5 , hi,j → j +1(5 , H i,j − → j H i,j → j +1 Figure 4
The biclique K t,q − ( i, j ) encoding the pairs P i,j , and its connection to the two neigh-boring horizontal half-graphs, with n = 5, t = 10, and q = 14. We then add a global clique C of size q − k . We attach k private neighbors to eachvertex of C . Among the q − k vertices of C , we arbitrarily distinguish 33 vertices: a set D = { d , . . . , d } of size 18, and three sets C , C − , C + each of size 5. We fully link d z toevery B i,j if z takes one of the following values:3( j mod 3 −
1) + i mod 3 , succ(3( j mod 3 −
1) + i mod 3) , i mod 3 −
1) + j mod 3 + 9 , succ(3( i mod 3 −
1) + j mod 3 + 9) , where the modulos are always taken in { , , } , and succ( x ) := x + 1 if x is not dividable by3 and succ( x ) := x − B i,j is linked with d z fortwo successive (indicated by the operator succ( x )) integers z in the range of [1 , ,
6] or [7 ,
9] depending on the coordinate j modulo 3. Likewise, each B i,j is linked with d z for twosuccessive integers z in the range of [10 , ,
15] or [17 ,
18] depending on the coordinate i modulo 3. , , , , , , , , , , , , , , , , , ,
45 6 4564 4 , , , , , , , , , , , ,
78 9 7897 7 , , , , , , , , , , , ,
12 3 1 2311 , , , , , , , , , , , ,
45 6 4 5644 , , , , , ,
45 6 45647 , , , , , ,
78 9 7 8977 , , , , , ,
78 9 7897
Figure 5
The rounded rectangles represent the B i,j , and the numbers therein, the z ∈ [18] suchthat d z is fully linked to it. These are the “colors” that a center in A i,j will have to fight for. Thecircles represent the half-graphs, and the number therein, the only z such that d z is not linked toit. Red integers are offset by 9 (1 = 10, 2 = 11, and so on). The edges represent the non-emptyinteraction between the A i,j and the half-graphs. The structure is glued like a torus. We observe that B i,j and B i +1 ,j are linked to exactly one common d z ( z ∈ [18]), and wefully link the half-graph H ( i → i + 1 , j ) to D \ { z } . Similarly we fully link the half-graph H ( i, j → j + 1) to D \ { z } where z is the unique integer of [18] such that d z is fully linked toboth B i,j and B i,j +1 . We fully link each A i,j to C , each H ( i → i + 1 , j ) and H ( i, j → j + 1)to C − , and each H ( i → i + 1 , j ) and H ( i, j → j + 1) to C + . This is just to prevent that oneuses a vertex of a B i,j or of a half-graph as a center. As we will see, intended solutions haveall their centers in C ∪ S i,j ∈ [ k ] A i,j .This ends our polytime construction. We denote by G the obtained graph. We ask for a b -chromatic core achieving color q . This polytime construction is of polynomial size in the k -by- k Grid Tiling -instance, and q is a computable function of k . We thus only need toshow that the reduction is correct. A solution to k -by- k Grid Tiling implies Γ b ( G ) (cid:62) q . We color the clique C with all the colors in [ q − k ], and the k private neighbors of each vertexof C with the corresponding remaining colors in [ q − k + 1 , q ]. The coloring of D uses all thecolors in [18] and respects Figure 5. The colors in [ q − k ] are already realized . The remaining k colors in [ q − k + 1 , q ] will be realized by the k centers corresponding to the solutionof the Grid Tiling -instance. Let { p i,j = ( x i,j , y i,j ) ∈ P i,j } i,j ∈ [ k ] be a grid-tiling solution.We arbitrarily color the a i,j ( x i,j , y i,j ) with all the colors of [ q − k + 1 , q ]. The intended . Aboulker, É. Bonnet, E. J. Kim, F. Sikora 17 q -witness shall take the vertices a i,j ( x i,j , y i,j ) as the centers of the top k colors. Note thateach a i,j ( x i,j , y i,j ) has precisely q neighbors: four vertices corresponding to a i,j ( x i,j , y i,j ) ineach of the four half-graphs, q − B i,j and five vertices in C . Therefore, in orderfor a i,j ( x i,j , y i,j ) to become a center, we color the set B i,j with the q − q ]which are the only available colors after excluding: the five colors of C that are already seenby A i,j , and the four colors of [18] already seen by B i,j .The four colors left to be seen by a i,j ( x i,j , y i,j ) shall be provided by the four half-graphs. Let z ∈ [18] be the common missing color of a i,j ( x i,j , y i,j ) and a i +1 ,j ( x i +1 ,j , y i +1 ,j )(resp. a i,j +1 ( x i,j , y i,j +1 )): this is the color adjacent to both B i,j and B i +1 ,j (resp. B i,j and B i,j +1 ). We color by z the neighbor of a i,j ( x i,j , y i,j ) in H ( i → i + 1 , j ) and theneighbor of a i +1 ,j ( x i +1 ,j , y i +1 ,j ) in H ( i → i + 1 , j ) (resp. a i,j ( x i,j , y i,j ) in H ( i, j → j + 1)and the neighbor of a i,j +1 ( x i,j +1 , y i,j +1 ) in H ( i, j → j + 1)). This is a proper coloringsince the k pairs form a solution for Grid Tiling , and by construction of the half-graphs. Indeed in horizontal (resp. vertical) half-graphs H i,j → j +1 (resp. H i → i +1 ,j ) the twoneighbors are h i,j → j +1 ( x i,j , y i,j ) and h i,j → j +1 ( x i,j +1 , y i,j +1 ) (resp. h i → i +1 ,j ( x i,j , y i,j ) and h i → i +1 ,j ( x i +1 ,j , y i +1 ,j )) which are non-adjacent since x i,j = x i,j +1 (resp. y i,j = y i +1 ,j ). Γ b ( G ) (cid:62) q implies a solution to k -by- k Grid Tiling.
We set C d := D ∪ C ∪ C − ∪ C + ⊂ C . Let us first show that a vertex of C \ C d has to be acenter. We thus upperbound the maximum number of centers outside C \ C d . We need thefollowing lemma which is true in any graph, not only in the constructed graph G . Informallyit says that vertices with few private neighbors cannot provide too many centers. (cid:73) Lemma 13.
Let p be a non-negative integer, A be a subset of vertices, and Y be theircommon neighborhood, that is T v ∈ A N ( v ) . If for every v ∈ A , | N ( v ) \ Y | (cid:54) p , then themaximum number of centers in A with distinct colors is at most p + 1 . Proof.
For the sake of contradiction, assume that there are p + 2 centers v , v , . . . , v p +2 withdistinct colors in A , say, without loss of generality, 1 , , . . . , p + 2. Recall that in the contextof b -Chromatic Core (unlike Grundy colorings) all the colors play the same role. Wereach a contradiction by considering any v i , say v . Vertex v colored 1, has to be adjacentto some vertices colored 2 , , . . . , p + 2. None of these vertices can be in Y since otherwise,the coloring would not be proper. But v has at most p other neighbors, to accommodatesupporting vertices with p + 1 colors. (cid:74) Observe that the vertices in an A i,j have only four neighbors (one per neighboringhalf-graph) that are not in the common neighborhood of A i,j . Hence Lemma 13 impliesthat each A i,j has at most 5 centers (with distinct colors). The vertices in a B i,j have noprivate neighbor, so by the same lemma, they can contain at most one center. However if thecorresponding A i,j has a center, then no vertex of B i,j can be a center. Indeed for a vertexin A i,j to be a supported center, one needs to color at least two vertices of B i,j with distinctcolors, say c and c . This prevents any vertex of B i,j to be supported, since they would misseither c or c . So the maximum number of centers in A i,j ∪ B i,j is still 5. Thus the totalnumber of centers from all the A i,j ∪ B i,j is at most 5 k .We now observe that Lemma 13 also works with a half-graph instead of a biclique between A and Y . (cid:73) Lemma 14.
Assume that p is a non-negative integer and ( A, Y ) induces a half-graph. Iffor every v ∈ A , | N ( v ) \ Y | (cid:54) p , then the maximum number of centers in A with distinctcolors is at most p + 1 . Proof.
Let p + 2 centers in A : v , v , . . . , v p +2 with distinct colors, say, 1 , , . . . , p + 2, andthat v p +2 has the highest level in the half-graph among { v , v , . . . , v p +2 } . We recall that itmeans that N ( v p +2 ) ∩ Y ⊆ N ( v i ) ∩ Y for every i ∈ [ p + 2]. Then the colors 1 , , . . . , p + 1,supporting v p +2 cannot come from Y . A contradiction since v p +2 has only at most p otherneighbors. (cid:74) Each set H i,j → j +1 , H i,j → j +1 , H i → i +1 ,j , H i → i +1 ,j has, by Lemma 14, at most 2 supportedcenters with distinct colors. Indeed, outside the other side of their half-graphs and theshared neighborhood C − or C + , they have only one private neighbor. So the total numberof supported centers in all the half-graphs is at most 8 k .Obviously the private neighbors of the vertices of C cannot be centers, since they havedegree 1. Overall the number of supported centers with distinct colors is 5 k + 8 k + [ C d | =13 k + 33 < k = q . This implies that a vertex of C \ C d is indeed a center. Such a vertexhas degree q −
1, so all its vertices should be colored (with distinct colors). In particular,the whole clique C should be colored. Without loss of generality, let us assume that theclique is colored with [ q − k ], and that D receives the colors from 1 to 18, consistently withFigure 5. By coloring appropriately the private neighbors of the vertices C , all the colors in[ q − k ] are realized by C . Since Γ b ( G ) (cid:62) q , actually now Γ b ( G ) = q , we know that there are k additional centers in V ( G ) \ C .We now prove that the remaining centers are in the A i,j . A vertex in a B i,j cannot beadjacent to a color used for C , since C and A i,j are fully adjacent and N ( B i,j ) ⊆ D ∪ A i,j .Hence such a vertex cannot be a supported center. Similarly vertices in the half-graphscannot be centers. Indeed, on one side of the half-graph, they would not find all five colorsused for C + which are excluded in the other side of the half-graph. And vice versa, onthe other side, they would miss some colors of C − which are unavailable for the first side.Therefore all the k remaining centers are in the A i,j .Let v ∈ A i,j be a supported center colored c ∈ [ q − k + 1 , q ]. We claim that there cannotbe another center with a distinct color in the same A i,j . The only way for v to be a supportedcenter is that B i,j is colored with the q − C \ ( C ∪ D i,j ) where D i,j ⊂ D is the set of four vertices fully linked to B i,j . (Then the four neighbors of v in thehalf-graphs have to receive all the colors of D i,j .) In particular, the only colors that anothervertex v = v ∈ A i,j could get are already realized colors. This shows that each A i,j containsexactly one supported center.We denote by a i,j ( x i,j , y i,j ) the unique center in A i,j . We already observed that thesupporting vertices with colors in D i,j have to come from the half-graphs. Looking at Figure 5,for a fixed color of D i,j only one of the four neighboring half-graphs can provide such acolor. This is due to how we linked the d z with B i,j and the half-graphs. Furthermore thecommon missing color between two consecutive A i,j has to come from the shared half-graph.Again since there is a solution to the b -Chromatic Core -instance, the neighbors of thecenters in the half-graphs form an independent set. This means that x i,j (cid:54) x i,j +1 and y i,j (cid:54) y i +1 ,j for every i, j ∈ [ k ]. Due to the cyclicity of the construction, it implies that x i,j = x i,j +1 and y i,j = y i +1 ,j for every i, j ∈ [ k ]. That in turn implies a solution for the Grid Tiling -instance. (cid:74) b -Chromatic Core on K t,t -free graphs In the following subsection, we prove that both b -Chromatic Core and Partial GrundyColoring can be solved in FPT time when all but a bounded number of vertices havebounded degree. This is a preparatory step to show the tractability in K t,t -free graphs. . Aboulker, É. Bonnet, E. J. Kim, F. Sikora 19 The technique of random separation [6, 5], inspired by the color coding technique [2], comeshandy when one wants to separate a latent vertex-subset of small size from the rest of thegraph. Consider two disjoint vertex-sets A and B . By sampling a vertex-subset S , withprobability 1 / S , the probability that S contains A andavoids B is at least 2 −| A |−| B | . Especially if we choose B as the open neighborhood of A ,this means that with a reasonably large probability, a random sample S will precisely carveout the connected components of G [ A ] as long as both A and B have bounded size. Noticethat some of the connected component of G [ S ], upon a successful sample S , might not be aconnected component of G [ A ] and some extra works would be needed to extract the desiredset A . Nonetheless, the knowledge that the components components of G [ A ] are capturedas connected components of G [ S ] can be substantially useful. A derandomized version ofrandom separation can be obtained with the so-called splitters of Naor et al. [35] (see alsoChitnis et al. [9]). For two disjoint sets A and B of a universe U , we say that S ⊆ U is an( A, B ) -separating set if A ⊆ S and B ∩ S = ∅ . (cid:73) Lemma 15 (Chitnis et al. [9]) . Let a and b be non-negative integers. For an n -elementuniverse U , there exists a family F of O (min( a,b ) log ( a + b )) log n subsets of U such that for anydisjoint subsets A, B ⊆ U with | A | (cid:54) a and | B | (cid:54) b , there exists an ( A, B ) -separating set S in F . Furthermore, such a family F can be constructed in time O (min( a,b ) log ( a + b )) n log n . (cid:73) Theorem 16.
Let G be a graph in which at most s vertices have degree larger than d .Then whether G has a k -witness for b -Chromatic Core ( Partial Grundy Coloring ,respectively) can be decided in FPT time parameterized by k + d + s . Proof.
Let X be the set of s vertices of degree larger than d . In order to explain the algorithmand prove its correctness, it is convenient to assume that G does contain a k -witness H for b -Chromatic Core (or Partial Grundy Coloring ) as an induced subgraph. We define I := V ( H ) ∩ X, A := V ( H ) \ X, and B := N ( A ) \ X. We can guess I by considering at most 2 s subsets of X . To find A , we use Lemma 15.From the fact that every vertex of V \ X has degree at most d and that H is a minimal k -witness, we have | A | (cid:54) k and | B | (cid:54) dk . Hence, by Lemma 15 with universe V ( G ) \ X , wecan compute in time 2 O ( k log ( k + dk )) n log n a family F with 2 O ( k log ( k + dk )) log n subsetsof V ( G ) \ X , that contains an ( A, B )-separating set.We guess this (
A, B )-separating set by iterating over all elements of F . Let S be a correctguess, i.e., S is an ( A, B )-separating set. So A ⊆ S and S ∩ B = ∅ . Observe that everyconnected component of G [ A ] appears in G [ S ] as a connected component.Let C S be the set of connected components of G [ S ] of size at most k . Since | A | (cid:54) k ,larger connected component of G [ S ] are clearly disjoint from A . Moreover, by definition of B and since S is disjoint from B , each connected component of G [ A ] is an element of C S .Since each element of C S has at most k vertices, the number of equivalence classes of C S under graph isomorphism is bounded by a function of k . In fact, the number of equivalenceclasses under a stronger form of isomorphism is bounded by a function of k . We define alabeling function ‘ : S → I as ‘ ( v ) := N ( v ) ∩ I . Let ∼ S be a relation on C S such that, forevery C, C ∈ C S : C ∼ S C if and only if there is a graph isomorphism φ : C → C with ‘ ( v ) = ‘ ( φ ( v )) for every v ∈ C . Let [ ∼ S ] be the partition of C S into equivalence classesunder ∼ S . As members of C S have cardinality at most k and there are 2 | I | (cid:54) k labels, C S has at most 2 k equivalence classes under ∼ S . And thus we can compute [ ∼ S ] in time2 k n . The definition of ∼ S clearly implies that two equivalent sets C and C under ∼ S are exchangeable as a connected component of H − X . That is, for any induced subgraph D of G with V ( D ) ∩ X = I , if C is a connected component of D − I , then G [( V ( D ) \ C ) ∪ C ] isisomorphic to D .We will now guess, by doing an exhaustive search, how many connected components H − I takes from each part of the partition [ ∼ S ]. There are 2 k k possible such guesses andfrom the fact that the number of connected components in H − I is at most k . Choose anelement (i.e. a connected vertex-set) from each part of [ ∼ S ] as many times as the currentguess suggests (if this is impossible, then discard the current guess) and let W be the unionof the chosen connected vertex-sets. We can now verify by brute-force that G [ W ∪ I ] isa k -witness for b -Chromatic Core or Partial Grundy Coloring , depending on theproblem at hand.To complete the proof of correctness, note that if we find a k -witness for some choice of I , S ∈ F and W , the input graph G clearly admits a k -witness. One can easily observe thatthe running time is FPT in k + d + s . (cid:74) K t,t -free graphs In this subsection, we present an FPT algorithm on graphs which do not contain K t,t as asubgraph. A key element of this algorithm is a combinatorial result (Proposition 20), whichstates that if there are many vertices of large degree, then one can always find a k -witness.Therefore, we may assume that the input graph has a small number of vertices of largedegree. Using the FPT algorithm on almost bounded-degree graphs given in Section 6.1, thisimplies an FPT algorithm on K t,t -free graphs. Most of this subsection is thus devoted toprove our combinatorial tool.We begin with a few technical lemmas. The first lemma is a simple application of aRamsey-type argument. (cid:73) Lemma 17.
Let t and N be two positive integers with N (cid:62) t , and let G be a graph on avertex-set A ] B not containing K t,t as a subgraph. If | A | (cid:62) N N + t and | B | (cid:62) N + t , thenthere exist two sets A ⊆ A and B ⊆ B , each of size at least N , such that there is no edgebetween A and B . Proof.
We may assume that B = { b , . . . , b N + t } by discarding some vertices of B if necessary.We are going to construct a sequence A ⊇ A ⊇ · · · ⊇ A N + t of subsets of A such that foreach i , the vertex b i is either complete or anti-complete with A i , and the last set A N + t contains at least N vertices. Take A as the set of neighbors of b if | N ( b ) ∩ A | (cid:62) | A | / b otherwise. Supposed that a sequence A ⊇ · · · ⊇ A i for i < N + t has been constructed. We define A i +1 as follows; if | N ( b i ) ∩ A i | (cid:62) | A i | /
2, then A i +1 = N ( b i ) ∩ A i , and A i +1 = A i \ N ( b i ) otherwise. As the step i + 1 halves A i in theworst case, the size of A ensures that A N + t (cid:62) N .Now, observe that any vertex b i of B which is complete with A i is also complete with A N + t . Since G is K t,t -free and | A N + t | (cid:62) N (cid:62) t , all but at most t − B areanti-complete with their associated set A i , and thus with A N + t . Therefore we can find atleast N vertices in B that are anti-complete with A N + t . We set B as these vertices and A := A N + t . (cid:74) The next lemma can be obtained as a corollary of the previous lemma. (cid:73)
Lemma 18.
For any integers k and t , there exists an integer M such that the followingholds: given a K t,t -free graph G and a partition A ] · · · ] A k of V ( G ) such that each A i . Aboulker, É. Bonnet, E. J. Kim, F. Sikora 21 contains at least M vertices, there exists either a clique on k vertices, or an independent setof size k which contains k vertices from each A i . Proof.
We assume that t is at least k ; if not, increase t appropriately. Clearly, G is K t,t -freefor the new value t . By (cid:0) k (cid:1) successive applications of Lemma 17, we can select for i = 1 , . . . , k (by choosing M large enough) a set A i ⊆ A i of 2 t vertices, such that there is no edge linking A j and A j for all j = j . Concretely, fix M = ( k t := 8 ·· t | {z } k times and apply Lemma 17 between A and A , which returns A ⊆ A and A i ⊆ A i of respectivesize at least ( k − t . Now let A ← A and A ← A , and execute the same procedurebetween A and A i for i = 3 up to i = k . Note that at the end of k − A is at least (8) t and A i has size at least ( k − t for each i ∈ [ k ] \ A is anti-complete with every A i for i ∈ [ i ] \
1. By induction on k , we areleft with A , . . . , A k which are pairwise anti-complete and each of which has size at least 8 t .Then one can apply Ramsey’s theorem on all A i ’s to get either a clique on k vertices in some A i , or an independent set of size 2 t (cid:62) k in every A i for i ∈ [ k ]. (cid:74) The following statement is proved in [1]. (cid:73)
Lemma 19 (Aboulker et al. [1]) . Let t be a positive integer and let (cid:15) ∈ (0 , . Then thereis an integer N ( t, (cid:15) ) that satisfies the following: if H = ( V, E ) is a hypergraph on at least N ( t, (cid:15) ) vertices, where all hyperedges have size at least (cid:15) | V | , and the intersection of any t hyperedges has size at most t − , then | E | < t/(cid:15) t . We are ready to prove the key combinatorial result on K t,t -free graphs. (cid:73) Proposition 20.
Let t, k be positive integers. Let G be a K t,t -free graph and let X ] Y bea partition of V ( G ) . There exist integers f ( t, k ) and g ( t, k ) such that the following holds: If | X | (cid:62) f ( t, k ) , and | N Y ( x ) | (cid:62) g ( t, k ) for every x ∈ X , then G contains kK ,k as an inducedsubgraph. In particular, G admits k -witnesses for b -Chromatic Core and thus for PartialGrundy Coloring . Proof.
We first observe that kK ,k (even kK ,k − ) is a k -witness for b -Chromatic Core :color the k centers with distinct colors, and assign colors from [ k ] \ { i } to the leaves ofthe center colored i . Let N ( t, /k ) and M be the integers defined in Lemmas 18 and 19respectively, and set M := max( M, N ( t, /k )) , f ( t, k ) := 2 t + k ( tk t + t ) , g ( t, k ) := 2 k ( tk t + t ) M .By Ramsey’s theorem, any graph on at least f ( t, k ) vertices admits either a clique of size2 t or an independent set of size k ( tk t + t ). Since G [ X ] (which has at least f ( t, k ) vertices) is K t,t -free, the former outcome is impossible, so it has an independent set of size k ( tk t + t ). Itshould be noted that the inductive proof of Ramsey’s theorem yields a greedy linear-timealgorithm which outputs a clique or an independent set of the required size. Hence weefficiently find an independent set of size k ( tk t + t ) in G [ X ]. Starting from j = 1, we nowprove the following claim inductively for all j (cid:54) k .( ? ) If | X | (cid:62) j ( tk t + t ) and | N Y ( x ) | (cid:62) j ( tk t + t ) M for every x ∈ X , then there are j vertices { b , . . . , b j } ⊆ X and a family of j disjoint vertex-sets A , . . . , A j ⊆ Y eachof size at least M , such that each A i are private neighbors of b i ; that is, the verticesof A i are adjacent with b i and not adjacent with any other vertices from { b , . . . , b j } . The claim ( ? ) trivially holds when j = 1. Suppose it holds for all integers smallerthan j , where 2 (cid:54) j (cid:54) k . We may assume that X has precisely j ( tk t + t ) vertices bydiscarding some vertices if its size exceeds the bound. For each ∅ 6 = I ⊆ X , we define N I = T v ∈ I N Y ( v ) ∩ T v ∈ X \ I Y \ N Y ( v ). Thus N I corresponds to all the vertices of Y whoseneighborhood in X is exactly I . Observe that the N I ’s partition Y and that N I correspondsto the set of vertices of Y that are complete with I and anti-complete with X − I .Choose a vertex x ∈ X that minimizes | N Y ( x ) | . As there are 2 j ( tk t + t ) possible subsets of X and | N Y ( x ) | (cid:62) j ( tk t + t ) M , there exists I ∗ ⊆ X such that x ∈ I ∗ and | N I ∗ | (cid:62) M .Let X x be the set of vertices in X adjacent with at least k -th fraction of N Y ( x ), that is, X x = { v ∈ X : | N Y ( v ) ∩ N Y ( x ) | (cid:62) | N Y ( x ) | k } . Set X = X − ( I ∗ ∪ X x ), Y = Y − N Y ( x ) and let G = G [ X ∪ Y ].We want to apply the induction hypothesis on G with respect to X and Y . Forthis, we need to make sure that it satisfies the conditions of ( ? ) for j −
1. To prove that | X | (cid:62) ( j − tk t + t ), we need to bound the size of I ∗ and X x . To bound the size of I ∗ ,notice that N I ∗ is complete with I ∗ . Since | N I ∗ | (cid:62) M (cid:62) t and G is K t,t -free, we concludethat | I ∗ | < t . To bound the size of X x , we apply Lemma 19 with (cid:15) = 1 /k to the hypergraphon the vertex-set N Y ( x ) and with hyperedge set { N Y ( v ) ∩ N Y ( x ) : v ∈ X } . Each hyperedgeof size at least | N Y ( x ) | k corresponds to a vertex in X x which gives us the bound | X x | (cid:54) tk t .Notice that Lemma 19 can be legitimately applied on this hypergraph as N Y ( x ) has at least M (cid:62) N ( t, /k ) vertices. Therefore, we have | X | (cid:62) | X | − t − tk t (cid:62) ( j − tk t + t ) . It remains to verify that each v ∈ X has at least 2 ( j − tk t + t ) M neighbors in Y . Indeed | N Y ( v ) | (cid:62) | N Y ( v ) | − | N Y ( v ) ∩ N Y ( x ) | (cid:62) | N Y ( v ) | − | N Y ( x ) | k (cid:62) | N Y ( v ) | − | N Y ( v ) | k (cid:62) k − k j ( tk t + t ) M (cid:62) ( j − tk t + t ) M . This proves that G meets the requirement to apply the induction hypothesis, and thuswe can find { b , . . . , b j } and sets A , . . . , A j in G as claimed in ( ? ). Observe now that N I ∗ isanticomplete to { b , . . . , b j } and recall that | N I ∗ | (cid:62) M . Hence, setting b = x and A = N I ∗ complete the proof of ( ? ). Now, applying Lemma 18 to the sets A ] · · · ] A k given by ( ? )gives us either a clique on k vertices or the announced set of stars. (cid:74) Combined with the main result of the previous subsection, this implies that b -ChromaticCore and Partial Grundy Coloring can be solved in FPT time on K t,t -free graphs.Observe that our algorithm is FPT in the combined parameter k + t , which is a strongerthan having an FPT algorithm in k when t is a fixed constant. (cid:73) Theorem 21.
There is a function h and an algorithm which, given a graph G = ( V, E ) notcontaining K t,t as a subgraph, decides whether G admits k b -Chromatic Core ( PartialGrundy Coloring , respectively) in time h ( k, t ) n O (1) . Proof.
Let X ⊆ B be the set of all vertices whose degree is at least g ( t, k ) + f ( t, k ), where g ( t, k ) and f ( t, k ) are the integers as in Proposition 20. If X contains at least f ( t, k ) vertices,then we there exists a k -witness in G by Proposition 20. If X contains less than f ( t, k )vertices, the algorithm of Theorem 16 can be applied to correctly decide whether G containsa k -witness. (cid:74) . Aboulker, É. Bonnet, E. J. Kim, F. Sikora 23 References Pierre Aboulker, Jørgen Bang-Jensen, Nicolas Bousquet, Pierre Charbit, Frédéric Havet,Frédéric Maffray, and José Zamora. χ -bounded families of oriented graphs. Journal of GraphTheory , 89(3):304–326, 2018. URL: https://doi.org/10.1002/jgt.22252 , doi:10.1002/jgt.22252 . Noga Alon, Raphael Yuster, and Uri Zwick. Color-coding.
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