Graph theoretic and algorithmic aspect of the equitable coloring problem in block graphs
aa r X i v : . [ c s . D M ] S e p Graph theoretic and algorithmic aspect of the equitable coloringproblem in block graphs a, ∗ , Vahan Mkrtchyan b a Institute of Informatics, Faculty of Mathematics, Physics and Informatics,University of Gda´nsk, Gda´nsk, Poland b Gran Sasso Science Institute, L’Aquila, Italy
Abstract
An equitable coloring of a graph G = ( V, E ) is a (proper) vertex-coloring of G , such thatthe sizes of any two color classes differ by at most one. In this paper, we consider theequitable coloring problem in block graphs. Recall that the latter are graphs in which each2-connected component is a complete graph. The problem remains hard in the class of blockgraphs. In this paper, we present some graph theoretic results relating various parameters.Then we use them in order to trace some algorithmic implications, mainly dealing with thefixed-parameter tractability of the problem. Keywords: block-graph, equitable coloring, fixed-parameter tractability, W[1]-hardness
1. Introduction
In this paper, we consider finite, undirected graphs. They do not contain loops or paralleledges. Two vertices of a graph G are independent if there is no edge joining them. A set ofvertices is independent, if any two its vertices are independent. Let α ( G ) be the cardinalityof a largest independent set in a graph G . Similarly, two edges of a graph are independent,if they do not share a vertex. A matching is a subset of edges of a graph such that any twoedges in it are independent. Let ν ( G ) be the size of a largest matching of G . A matching isperfect if it covers all the vertices of the graph. A vertex cover is a subset of vertices whoseremoval results into a graph with no edge. The size of a smallest vertex cover is denoted by τ ( G ). In any graph G we have ν ( G ) ≤ τ ( G ) ≤ ν ( G ) . The work of the second author has been partially supported by the Italian MIUR PRIN 2017 ProjectALGADIMAR “Algorithms, Games, and Digital Markets.” ∗ I am corresponding author
Email addresses: [email protected] (Hanna Furma´nczyk), [email protected] (Vahan Mkrtchyan)
Preprint submitted to Sample Journal September 29, 2020 clique of a graph G is a maximal complete subgraph of G . For a graph G let ω ( G ) be thesize of a largest clique of G . A vertex v is a cut-vertex, if G − v contains more connectedcomponents than G . A block of a graph G is a maximal 2-connected subgraph of G .A graph G is a block graph, if every block of G is a clique. If G is a block graph, thena vertex is simplicial, if it is not a cut-vertex. Clearly, the neighbors of a simplicial vertexare in the same clique. A clique in a block graph is pendant if it contains one cut-vertex of G . Let P ( G ) be the number of pendant cliques of G . Let S ( G ) be the number of simplicialvertices of G . Clearly, for any block graph G , we have P ( G ) ≤ S ( G ).If G is a connected graph then let d ( u, v ) denote the length of a shortest path con-necting the vertices u and v . For a vertex u , its eccentricity ǫ G ( u ) is defined as ǫ G ( u ) =max v ∈ V { d ( u, v ) } . The radius of G is rad ( G ) = min v ∈ V { ǫ G ( v ) } and diam ( G ) = max v ∈ V { ǫ G ( v ) } .The center of a graph is the subset of vertices whose eccentricity is equal to the radius ofthe graph. For any graph G , we have rad ( G ) ≤ diam ( G ) ≤ · rad ( G ) . Following [11], we define a cluster in a graph G as a subgraph such that each of its compo-nents is a clique. The distance to the cluster, denoted by dc ( G ), is the smallest number ofvertices of G , whose removal results into a cluster. A set D is called a dc -set, if | D | = dc ( G )and G − D is a cluster. We refer to [13] for non-defined concepts on graphs.This paper deals with the one of variants of classical Vertex Coloring problem,namely
Equitable Coloring . If the set of vertices of a graph G can be partitionedinto k (possibly empty) classes V , V , . . . , V k such that each V i is an independent set andthe condition || V i | − | V j || ≤ i, j ), then G is said to be equitablyk-colorable . The smallest integer k for which G is equitably k -colorable is known as the equitable chromatic number of G and denoted by χ = ( G ) [17]. This model of graph coloringhas attracted attention of many graph theory specialists for almost 50 years. The conductedstudies are mainly focused on the proving known conjectures for particular graph classes,analysis of the problems complexity, designing exact algorithms for polynomial cases, andapproximate algorithms or heuristics for hard cases.We know that in general case, the Equitable Coloring problem is NP-complete, asa particular case of
Vertex Coloring . Note that the
Bin Packing Problem withConflicts (BPC) is closely related to
Equitable Coloring . The first one is definedas follows. We are given a set V of n items of weights w , w , . . . , w n , k identical bins ofcapacity c . Two items i and j are said to be conflicting if and only if they cannot be assignedto the same bin. The problem is to assign all items in the least possible number of binswhile ensuring that the total weight of all items assigned to a bin does not exceed c andthat no bin contains conflicting items. Note that the problem with c = n/k and weightsequal to one is equivalent to an equitable coloring of the corresponding conflict graph. Someexemplary heuristics for solving BPC can be found in [10, 18].An interesting overview of the results of studies over equitable coloring can be found in[9] and [16]. This issue is very important due to its many applications (creating timetables,task scheduling, transport problems, networks, etc.) (see for example [7, 8]). Very recently2here have been published a few papers investigating the parameterized complexity of theproblem of equitable coloring (see [2, 5, 6, 11, 12]). Recall that if Π is an algorithmic problem,then it is fixed-parameter tractable with respect to a parameter t , if it can be solved withan exact algorithm, whose running-time is g ( t ) · poly ( size ). Here g is some (computable)function of t , size is the length of the input and poly is a polynomial function. Usually, a(parameterized) problem is paraNP-hard, if it remains NP-hard even when the parameterunder consideration is constant. The reader can learn more about this topic from [3], thatcan be a good guide for algorithmic concepts that are not defined in this paper.Fellows et al. in [5] showed that the Equitable Coloring problem is W [1]-hard,parameterized by the treewidth plus the number of colors. Fiala et al. [6] considered anotherstructural parameters - vertex cover. They showed that the problem is FPT with respect toit. Gomes et al. [11] established new results for some other parameters: W [1]-hardness forpathwidth and feedback vertex set, and fixed parameter tractability for distance to cluster.Gomes et al. in their another paper [12] considered parameterized complexity of EquitableColoring problem for subclasses of perfect graphs. They showed W [1]-hardness for blockgraphs when parameterized by the number of colors, and for K , -free interval graphs whenparameterized by treewidth, number of colors and maximum degree.In this paper, we continue the line of the research by considering the problem of equi-table coloring of block graphs as a non-trivial subclass of chordal graphs, thus also perfectgraphs. For block graphs, it is shown in [12] that the problem is W[1]-hard with respectto the treewidth, diameter and the number of colors. This, in particular, means that un-der the standard assumption FPT =W[1] in parameterized complexity theory, the problemis not likely to be polynomial time solvable in block graphs. In this paper we investigateparameterized complexity of Equitable Coloring of block graphs with respect to manyother parameters thus completing the state of art in this area.The paper is organized as follows. In Section 2, we investigate the problem with respectto some parameters that are related to the independence of vertices and edges of graphs.In Section 3, we consider other parameters and related them in block graphs. Finally, weconclude the paper in Section 4, where we also present some open problems that we feeldeserve to be investigated.
2. Equitable Coloring and independent sets of block graphs
In this section, we consider block graphs and the
Equitable Coloring problem fromthe perspective of independent sets of vertices and edges. In Subsection 2.1, we work mainlywith the parameter α min , which has tight connections with the size of the largest independentset of vertices of a block graph. In Subsection 2.2, our focus is on matchings of block graphs.In particular, we view the problem from the angle of the number of vertices that a maximummatching of a block graph does not cover.Before we start presenting our results, we list some minor observations and corollariesfrom some results in the literature. Lemma ([19]) . Let Π be an algorithmic problem, and let k and k be some parameters.Assume that there is a (computable) function g : N → N , such that for any instance I of Π , e have k ( I ) ≤ g ( k ( I )) . Then if Π is FPT with respect to k , then it is FPT with respectto k . Theorem ([12]) . EQUITABLE COLORING of block graphs of diameter at least fourparameterized by the number of colors and treewidth is W [1] -hard. Theorem ([11]) . EQUITABLE COLORING is FPT when parameterized by the distanceto cluster.
Theorem ([6]) . EQUITABLE COLORING is FPT when parameterized by vertex cover.
Theorem ([5]) . EQUITABLE COLORING is W [1] -hard, parameterized by treewidth. Theorem ([12]) . EQUITABLE COLORING is FPT when parameterized by the treewidthof the complement graph.
Directly, we have
Corollary . EQUITABLE COLORING of complements of block graphs with fixed cliquenumber is polynomialy solvable.2.1. Independent sets and the parameter α min Recall that for a graph
G α ( G ) denotes the size of a largest independent set in G . Let α ( G, v ) be the size of the largest independent set of G that contains the vertex v . Define: α min ( G ) = min v ∈ V ( G ) α ( G, v ) . In [4] we presented a conjecture, which implies that there are only two possible values forthe equitable chromatic number of block graphs.
Conjecture . ([4]) For any block graph G , we have: max (cid:26) ω ( G ) , (cid:24) | V ( G ) | + 1 α min ( G ) + 1 (cid:25)(cid:27) ≤ χ = ( G ) ≤ (cid:26) ω ( G ) , (cid:24) | V ( G ) | + 1 α min ( G ) + 1 (cid:25)(cid:27) . While the lower bound on χ = ( G ) is true, even for general graph, the upper bound is onlyproven for well-covered block graphs, some symmetric block graphs, and block graphs with α min ∈ { , } [4]. In this section we give full characterization of block graphs with α min = r , r ≥ Equitable Coloring of block graphs is FPT when parametrized by α min . We start with some preliminaries. Proposition . Let G be a block graph and let w be a simplicial vertex. Then α ( G, w ) = α ( G ) .Proof. Let I be an independent set of G of size α ( G ). If w ∈ I , then we are done. Thus,we can assume that w / ∈ I , hence there is a vertex u ∈ I that lies in the unique clique Q containing w . Consider the set I ′ obtained from I by replacing u with w . Observe that I ′ is an independent set of size α ( G ) and it contains w . The proof is complete.4 emma . Let G be a block graph, v be a cut vertex and let w be any simplicial vertex of G . Then α ( G, v ) ≤ α ( G, w ) .Proof. The statement follows directly from Proposition 8.The lemma implies
Corollary . For any block graph G containing a cut-vertex, there is a cut vertex v , suchthat α ( G, v ) = α min ( G ) .Proof. If α min ( G ) = α ( G ), there is nothing to prove. On the other hand, if α min ( G ) < α ( G ),then Lemma 9 implies that the minimum of α ( G, z )s is attained on cut-vertices. The proofis complete.
Lemma . Let G be a block graph obtained from a block graph H by adding a clique K toa vertex u of H . Then for any vertex v = u of H , we have α ( G, v ) ≤ α ( H, v ) .Proof. If I is an independent set of G of size α ( G, v ) containing v , then clearly I can containat most one vertex of K . Thus, we consider the set I minus this vertex, then it is anindependent set of size α ( G, v ) − H . Thus, α ( G, v ) − ≤ α ( H, v ), or equivalently, α ( G, v ) ≤ α ( H, v ). The proof is complete.Our next result is dealing with the characterization of connected block-graphs with α min = r . Our theorem implicitly relies on the fact that an induced subgraph of a blockgraph is a block graph. Observe that the same property does not hold for usual subgraphs.Also, observe that if G is a block graph and we want to check whether a particular ver-tex z belongs to all independent sets in G of size α min ( G ) it suffices to check whether α min ( G ) = α min ( G − z ). Since G − z is a block graph and we can find the size of the largestindependent set in a block graph in polynomial time, we have that this property we cancheck in polynomial time. Theorem . Assume that r ≥ is an integer. Let G be a connected block graph and let v be a cut-vertex with α ( G, v ) = α min ( G ) (Corollary 10). Then α min ( G ) = r if and only ifthere is a sequence G ⊂ G ⊂ ... ⊂ G r = G of induced subgraphs of G , such that ( a ) G = G [ N [ v ]] ( G is the subgraph of G induced on the closed neighborhood of v ) , ( b ) α min ( G i +1 ) = α min ( G i ) + 1 , for i = 1 , ..., r − , ( c ) α ( G i , v ) = α min ( G i ) for i = 1 , ..., r and the induced subgraph G i +1 can be obtained from G i in the following way (1) add a clique to any cut-vertex of G i that is different from v and is adjacent to a pendantclique, (2) Let Q be a clique in G i that contains at most one vertex z that lies in a non-pendantclique ( except Q ) . . If there are at least three simplicial vertices in Q , add a clique to one of them. (2 . If there are exactly two simplicial vertices in Q , then let G ′ i be the graph obtainedfrom G i by adding two cliques to these two vertices, one clique per simplicialvertex of Q . If z belongs to all independent sets of size α ( G ′ i , v ) = α min ( G ′ ) ,then (2 . a ) let G i +1 be the graph obtained from G i by adding one clique to one ofsimplicial vertices of Q , otherwise (2 . b ) let G i +1 be G ′ i . (2 . If there is exactly one simplicial vertex in Q , which implies that z belongs to allindependent sets of size α ( G i , v ) = α min ( G ) that contain v , then add a clique tothe unique simplicial vertex x . (3) If at least one of cliques added in the above steps ( steps (1) , (2 . - (2 . is K , with V ( K ) = { w , w } , then let w be the simplicial vertex. If for each independent set I ,with v ∈ I and | I | = α min ( G ) , we have w / ∈ I , then we may add a clique of any sizeto w . (4) The resulting graph is G i +1 .Proof. We start with the sufficiency. Let us assume that we have the desired sequence ofinduced subgraphs. Let us show that α min ( G ) = r . Observe that by definition of G , thissubgraph is a clique-star, hence α min ( G ) = α min ( G , v ) = 1. By ( b ), we have α min ( G ) = 2, α min ( G ) = 3,.., α min ( G r ) = r . Since G r = G , we have α min ( G ) = r .Thus, in order to complete the proof of the theorem, it suffices to show that if α min ( G ) = r , then we can find the corresponding sequence of induced subgraphs G i . Let us prove thisstatement by induction on α min ( G ). If α min ( G ) = 1, then let G = G . Observe that thesequence of subgraphs comprised of just G is the required sequence for this case. Thus, byinduction, we can assume that our statement is true for all block graphs with α min ≤ r − G with α min ( G ) = r ≥
2, and we can assume that v isa cut-vertex attaining α min ( G ).First, let us assume that G contains a cut-vertex x = v , such that x is adjacent to atleast two pendant cliques. Let J be one of these pendant cliques. Consider the subgraph H = G − ( V ( J ) − x ) of G . Observe that H is an induced subgraph of G . Moreover, G can beobtained from H by applying the operation on (1). Let us show that α min ( H ) = α min ( G ) − I of G of size α ( G, v ) and containing the vertex v ,cannot contain the vertex x , as otherwise, we could have replaced x with one simplicialvertex from each pendant clique, incident to x and get a larger independent set containing v . This implies that any independent set containing the vertex v , must contain a vertexfrom V ( J ) − x , hence we have that α min ( H ) ≤ α ( H, v ) ≤ α ( G, v ) − α min ( G ) − . The final equality follows from the choice of v . On the other hand, by Lemma 11, α min ( H )cannot decrease by two or more. Thus, we have α ( H, v ) = α min ( H ) = α min ( G ) − r −
1. By induction, we have that there is the corresponding sequence of induced subgraphs6 , .., G r − = H . Now, if we define G r = G , then clearly G , .., G r meets our constraintsfrom the statement of the theorem.The above implies that without loss of generality, we can assume that each cut vertex x = v of G is adjacent to at most one pendant clique. Let Q be a non-pendant clique in G that contains at most one vertex z that lies in a non-pendant clique (except Q ). If there isno such vertex z , (then G looks like clique plus some pendant cliques) we can define z to beany cut vertex of Q . Clearly, Q must contain a cut vertex as α min ( G ) ≥ | Q | ≥
3. Now, if Q contains a simplicial vertex y , thenlet x be a cut vertex of Q different from z . Since Q is non-pendant, x exists. By ourassumption, x is adjacent to exactly one pendant clique J (Figure 1). zyx QJ Figure 1: The clique Q with simplicial vertex y and a cut vertex x that is adjacent to the pendant clique J . Consider the graph H = G − ( V ( J ) − x ). Observe that any independent set of size α ( G, v ) containing v contains a vertex from Q (as otherwise we could have added y to it).Thus, by replacing this vertex with y , we can obtain an independent set that contains y .This implies that α ( H, v ) = α min ( H ) = α min ( G ) − r −
1. By induction, we have thatthere is the corresponding sequence of induced subgraphs G , .., G r − = H . Now, if define G r = G , then clearly G , .., G r meets our constraints from the statement of the theorem.Moreover, observe that G can be obtained from H either from step (2 .
1) or step (2 . a ) sincethe clique Q contains at least two simplicial vertices in H .Thus, we can assume that | Q | ≥ Q different from z is adjacent toexactly one pendant clique. Since | Q | ≥ Q contains two cut vertices x and x , differentthan z , that are adjacent to pendant cliques J and J , respectively. If z belongs to allindependent sets of size α min ( G ) containing v , then consider the graph H = G − ( V ( J ) − x ).Observe that H is an induced subgraph of G . Moreover, α ( H, v ) = α min ( H ) = α min ( G ) − r −
1. By induction, we have that there is the corresponding sequence of induced subgraphs G , .., G r − = H . Now, if we define G r = G , then clearly G , .., G r meets our constraints fromthe statement of the theorem. Moreover, observe that G can be obtained from H from step(2 . a ) since the clique Q contains at least two simplicial vertices in H . On the other hand,assume that z does not belong to at least one independent set I of size α min ( G ) = α ( G, v )containing v . Let us show that without loss of generality, we can assume that x ∈ I .Assume that I does not contain x . We consider two cases. First, if I does not containa vertex of Q , then I must contain a simplicial vertex of J . Replace this vertex with x .Observe that the resulting set of vertices is independent, it contains the vertex v and it is7f size α ( G, v ). However, it contains x . On the other hand, if I contains a vertex w of Q ,then by our assumption, w is incident to a pendant clique J w . Observe that I must containa simplicial vertex of J . Now, consider a set of vertices obtained from I by removing thevertex w and the simplicial vertex of J , and adding a simplicial vertex of J w and the vertex x . Observe that the resulting set of vertices is independent, it contains the vertex v and itis of size α ( G, v ). However, it contains x .Consider the graph H = G − ( V ( J ) − x ) − ( V ( J ) − x ). Observe that H is an inducedsubgraph of G . Moreover, α ( H, v ) = α min ( H ) = α min ( G ) − r −
1. By induction, wehave that there is the corresponding sequence of induced subgraphs G , .., G r − = H . Now,if we define G r = G , then clearly G , .., G r meets our constraints from the statement of thetheorem. Moreover, observe that G can be obtained from H from step (2 . b ) since the clique Q contains at least two simplicial vertices in H .Thus, it remains to consider the case when Q = K . Let x be the other vertex of Q different from z . Let J be the pendant clique adjacent to x . If z belongs to all independentsets of size α min ( G ) containing v , then consider the graph H = G − ( V ( J ) − x ). Observethat H is an induced subgraph of G . Moreover, α ( H, v ) = α min ( H ) = α min ( G ) − r −
1. By induction, we have that there is the corresponding sequence of induced subgraphs G , .., G r − = H . Now, if we define G r = G , then clearly G , .., G r meets our constraintsfrom the statement of the theorem. Moreover, observe that G can be obtained from H fromstep (2 .
3) since the clique Q contains exactly one simplicial vertex in H .Finally, if z does not belong to at least one independent set of size α min ( G ) containing v , then consider the graph H = G − V ( J ). Observe that H is an induced subgraph of G .Moreover, α ( H, v ) = α min ( H ) = α min ( G ) − r −
1. By induction, we have that there is thecorresponding sequence of induced subgraphs G , .., G r − = H . Now, if we define G r = G ,then clearly G , .., G r meets our constraints from the statement of the theorem. Moreover,observe that G can be obtained from H from step (3). The proof is complete. v Figure 2: An example of a block graph with α min = 3 that requires to attach two cliques in one step. It is natural to wonder whether there is a block graph that in its construction fromabove theorem requires to attach two cliques in one step. The example from Figure 2 hasthis property. It is easy to see that in this example α min ( G ) = 3 and v is the only vertexattaining the minimum. 8ow, we pass to the investigation of parameterized complexity of our problem withrespect to α min . We claim that the parameter dc ( G ) can be bounded in terms of α min ( G )for any block graph G . Theorem . For any block graph G , dc ( G ) ≤ α min ( G ) .Proof. Our proof is by induction on | V | . The theorem is obvious when | V | ≤
2. Now, let G be any block graph of order at least 3. Clearly, we can assume that G is connected as if G has components G , ..., G t , then dc ( G ) = dc ( G ) + ... + dc ( G t ) ≤ α min ( G ) + ... + α min ( G t ) ≤ α min ( G ) . If G is a star of cliques, then clearly dc ( G ) = α min ( G ) = 1 , thus the statement is trivial for this case. Hence we can assume that G is not a star of cliques.Let v be a vertex with α ( G, v ) = α min ( G ). First let us show that any other ( = v ) cut-vertexof G is adjacent to at most one pendant clique. Assume, that a cut-vertex w = v is adjacentto two pendant cliques. Let J be one of them. Consider the block graph H = G − ( V ( J ) − w ).Observe that H is a block graph of order smaller than G . Hence we have dc ( H ) ≤ α min ( H ).Observe that if D is a set of vertices of H such that H − D is comprised of cliques, then byadding w to it, we will have such a set in G . Thus, dc ( G ) ≤ dc ( H ) + 1 ≤ α min ( H ) + 1= α min ( G ) . The equality α min ( H ) + 1 = α min ( G ) can be proved in the same way that we did in (1)of Theorem 12. Thus, without loss of generality we can assume that any cut-vertex of G different from v is adjacent to at most one pendant clique. Let Q be a non-pendant cliqueof G such that it contains at most one vertex z that is not adjacent to a non-pendant cliqueother than Q . All other vertices of Q are either simplicial or they are adjacent to exactlyone pendant clique. Since G is not a star of cliques, we have that this vertex z exists. Weconsider two cases.Case 1: Q contains at least one simplicial vertex y . Observe that in this case | Q | ≥ x be a cut-vertex of Q that is adjacent to a pendant clique J . Since Q is not pendant,the vertex x exists. Consider the graph H = G − ( V ( J ) − x ). Observe that H is a smallerblock graph than G . By induction, we have dc ( H ) ≤ α min ( H ) . Moreover, as in the characterization theorem (proof of Theorem 12, the paragraph afterFigure 1), we have α min ( H ) = α min ( G ) − . D H is a smallest set such that H − D H is a disjoint union of cliques, thenadding x to it, we will get such a set in G . Thus, dc ( G ) ≤ dc ( H ) ≤ α min ( H ) = α min ( G ) . Case 2: Q contains no simplicial vertices. Observe that this case includes the one when Q = K . In this case, all vertices of Q except z are adjacent to exactly one pendant clique.Consider the graph H obtained from G by removing the pendant cliques adjacent to verticesof Q , including their cut vertices. Observe that only the vertex z remains in H . Note that H is a block graph containing v . Hence by induction, dc ( H ) ≤ α min ( H ) . Now observe that if D H is a smallest set such that H − D H is a disjoint union of cliques,then adding all cut vertices of Q except z to it, we will get such a set in G . Thus, dc ( G ) ≤ dc ( H ) + ( | Q | − ≤ α min ( H ) + ( | Q | − ≤ α min ( G )since α min ( H ) ≤ α ( H, v ) ≤ α ( G, v ) − ( | Q | −
1) = α min ( G ) − ( | Q | − . The proof is complete.Theorem 3, combined with Lemma 1, implies the following result.
Proposition . EQUITABLE COLORING of block graphs is FPT when parameterizedby α min .2.2. Matchings in block graphs In this section, we prove that EQUITABLE COLORING is W [1]-hard in block graphswith a perfect matching. Recall that in any graph G , we have ν ( G ) ≤ τ ( G ) ≤ · ν ( G ) . Thus the parameterization with respect to τ is equivalent to that of with respect ν . Theorem4 and Lemma 1 imply Corollary . EQUITABLE COLORING is FPT when parameterized by the matchingnumber.
One can try to strengthen this result. Since in any graph τ ( G ) − ν ( G ) ≤ ν ( G ) ,
10e can ask about the parameterization with respect to τ ( G ) − ν ( G ). Equitable coloring isNP-hard for bipartite graphs [1]. In these graphs, the difference τ ( G ) − ν ( G ) is zero, thusthe problem is paraNP-hard with respect to τ ( G ) − ν ( G ), and hence less likely to be FPTwith respect to it.Observe that in any graph G , ν ( G ) ≤ | V | − ν ( G ) . From Corollary 15 and Lemma 1, we have that equitable coloring is FPT with respect to | V | − ν ( G ). Thus one can try to do the next step trying to show that it is FPT with respectto | V | − ν ( G ). We consider the restriction of the problem to block graphs with a perfectmatching, i.e. with | V | − ν ( G ) = 0.Below we observe that equitable coloring is NP-hard for graphs containing a perfectmatching. In order to demonstrate this, we will need: Theorem ([20]) . Let G be a connected, claw-free graph on even number of vertices.Then G has a perfect matching. Observation . Every line graph is claw-free.
The classical result by Holyer [14] states that the problem of testing a given bridgelesscubic graph for 3-edge-colorability is an NP-complete problem. One can always assume thatthe bridgeless cubic graph in this problem contains even number of edges. For otherwise, justreplace one vertex with a triangle. The resulting graph is a bridgeless cubic graph on evennumber of edges and it is 3-edge-colorable if and only if the original graph is 3-edge-colorable.
Observation . Equitable coloring is NP-hard for 4-regular graphs containing a perfectmatching.Proof.
We start with the 3-edge-coloring problem for the connected bridgeless cubic graphswith even number of edges. For such a graph G , consider its line graph L ( G ). Observethat G is 3-edge-colorable if and only if L ( G ) is 3-vertex-colorable. Moreover, since in any3-edge-coloring of G , the color classes must form a perfect matching, we have that the colorclasses in G have equal size. Thus, the color classes in any 3-vertex-coloring of L ( G ) musthave equal size, too. Thus, G is 3-edge-colorable if and only if χ = ( L ( G )) = 3.Now, observe that L ( G ) is connected, since G is connected. Moreover, it is 4-regular,since G is cubic. Finally, | V ( L ( G )) | is even since G has even number of edges. Thus, L ( G )has a perfect matching via Theorem 16. The proof is complete. Corollary . Equitable coloring is paraNP-hard with respect to | V | − ν ( G ) in the classof 4-regular graphs. Now, we are going to show that the problem remains hard even in block graphs with aperfect matching. In [12], some results are obtained about the parameterized complexity ofthe equitable coloring problem in block graphs. We will use them in order to obtain somemore results. In this paper, the
Bin Packing problem is defined as follows: given a set11f natural numbers A = { a , ..., a n } , two natural numbers k and B , the goal is to checkwhether A can be partitioned into k parts such that the sum of numbers in each part isexactly B . In [15], it is shown that Bin Packing remains W [1]-hard with respect to k evenwhen the numbers are represented in unary. Below we prove the following: Observation . Bin Packing remains W [1] -hard when the parity of k is fixed.Proof. We reduce
Bin Packing to the
Bin Packing with fixed parity of k . Let I bean instance of Bin Packing . If we are happy with the parity of k in I , then we outputthe same instance. Assume that we are unhappy with the parity of k . Then consider theinstance I ′ = ( A ′ , k ′ , B ′ ) defined as follows: A ′ = A ∪ { , B − } , k ′ = k + 1 , B ′ = B. Observe that I ′ can be constructed from I in polynomial time. Moreover, in I ′ k ′ has adifferent parity than k . Let us show that I is a yes instance if and only if I ′ is a yes instance.If I is a yes instance, then we can add { , B − } as a new bin and we will have a ( k + 1)partition of A ′ such that the sum in each partition is B . Now, assume that I ′ is a yesinstance. Observe that 1 and B − k sets in the partition form a k -partition in A . Thus, I is ayes-instance. The proof is complete. Corollary . Equitable coloring remains W [1] -hard in block graphs with odd valuesof k .Proof. In [12], the authors reduce the instance I = ( A, k, B ) of (unary)
Bin Packing tothe equitable ( k + 1)-colorability of a block graph G I . By Observation 20, we can apply thesame reduction only to instances of (unary) Bin Packing when k is even. Thus, we willhave that ( k + 1) is odd for the resulting instances of equitable coloring in block graphs.The proof is complete. Observation . Equitable Coloring remains W [1] -hard in block graphs with a perfectmatching.Proof. Let us start with any instance of
Equitable Coloring in block graphs where k isodd. We will construct graph G ′ being an instance for the same problem, but G ′ will havea perfect matching such that G has equitable k -coloring if and only if G ′ does.If G has a perfect matching, then G ′ := G and we are done. So, let G does not havea perfect matching. Let M ( G ) be a matching of the largest size in G and V M ( G ) = { v ∈ V ( G ) : v e, for any e ∈ M ( G ) } , i.e. V M ( G ) is a subset of vertices of G such that they arenot end vertices of any edge belonging to M ( G ), they are not covered by M . Recall that k is odd. We construct G ′ from G by adding to every vertex of G not covered by M ( G ) anedge with a pendant clique K k (cf. Figure 3).Since in every k -coloring of the added gadgets every color is used exactly the samenumber of times, then we immediately get the equivalence of equitable k -coloring of G and G ′ . The proof is complete. 12 G Figure 3: The construction of G ′ from G : we add an edge with a clique of size k to every vertex of G notcovered by the maxmimum matching. In this example, k = 3. Corollary . If FPT = W [1] , then Equitable Coloring in block graphs is not FPTwith respect to | V | − ν ( G ) .Proof. If Equitable Coloring in block graphs is FPT with respect to | V | − ν ( G ), thenit is polynomial time solvable for block graphs with a perfect matching. Hence it is FPT forblock graphs containing a perfect matching. By the previous observation it is W[1]-hard forblock graphs with a perfect matching. Hence FPT=W[1].
3. Equitable coloring and other structural parameters
In Subsection 2.1, we used the parameter distance to cluster to bound from above α min for block graphs. In consequence we showed that Equitable Coloring is FPT for blockgraphs when parameterized by α min . Now, we establish some new similar results. Theorem . Let G be a connected, block graph. Then rad ( G ) ≤ α min ( G ) .Proof. Our proof is by induction on | V | . Clearly, the theorem is true when | V | ≤
2. Now,let G be a connected block graph with | V | ≥ G is a star of cliques, then clearly rad ( G ) = α min ( G ) = 1 , thus the statement is trivial for this case. Hence we can assume that G is not a star of cliques.Let v be a cut-vertex with α ( G, v ) = α min ( G ) (Corollary 10). Since v is a cut-vertex, it isadjacent to at least two cliques. Let us show that all cliques around v are non-pendant.Assume that there is a pendant clique J around v and let y be a simplicial vertex of J .Consider the block graph H = G − y . Observe that since ǫ G ( y ) ≥ ǫ G ( v ) and for all vertices u ∈ V ( H ), ǫ H ( u ) = ǫ G ( u ), we have rad ( G ) = rad ( H ). Thus, rad ( G ) = rad ( H ) ≤ α min ( H ) ≤ α ( H, v ) = α ( G, v ) = α min ( G ) . Thus, we can assume that all cliques around v are non-pendant. Now, let us show thatany other ( = v ) cut-vertex of G is adjacent to at most one pendant clique. Assume, that acut-vertex w = v is adjacent to two pendant cliques. Let J be one of them, while K is theother one. Consider the block graph H = G − ( V ( J ) − w ). Observe that H is a block graphof order smaller than G containing v . Hence we have rad ( H ) ≤ α min ( H ). Now, it is easy13o see that rad ( G ) = rad ( H ). Indeed, it is enough to notice that ǫ G ( q K ) = ǫ G ( q J ) for anyvertices q K ∈ V ( K ) and q J ∈ V ( J ), so min q ∈ V ( G ) ǫ G ( q ) = min p ∈ V ( H ) ǫ H ( p ), of course underthe assumption that G is not a star of cliques. Hence rad ( G ) = rad ( H ) ≤ α min ( H ) ≤ α ( H, v ) ≤ α ( G, v ) = α min ( G ) . Thus, without loss of generality we can assume that any cut vertex w = v of G is adjacentto at most one pendant clique. Let Q be a non-pendant clique of G such that it containsat most one vertex z that is not adjacent to a non-pendant clique other than Q . All othervertices of Q are either simplicial or they are adjacent to exactly one pendant clique. Note,that there is at least one non-simplicial vertex x in Q , except z . Since G is not a star ofcliques, we have that this vertex z exists. Let us show that we can assume that Q containsno simplicial vertices. If y is a simplicial vertex in Q , then consider the graph H = G − y containing v . Now, it is easy to see that rad ( G ) = rad ( H ). Indeed, it follows from the factsthat ǫ G ( y ) ≥ ǫ G ( z ), ǫ G ( y ) ≥ ǫ G ( x ), ǫ H ( x ) = ǫ G ( x ), ǫ H ( z ) = ǫ G ( z ) and Q is a clique. Hence rad ( G ) = rad ( H ) ≤ α min ( H ) ≤ α ( H, v ) ≤ α ( G, v ) = α min ( G ) . Thus, we can assume that Q contains no simplicial vertices. Hence all vertices of Q except z are adjacent to exactly one pendant clique. Let us show that Q is K . Assume that | Q | ≥ x be any cut vertex in Q different from z . Let J be the pendant clique adjacent to x .Define H = G − ( V ( J ) − x ) containing v . Now, it is easy to see that rad ( G ) = rad ( H ).This just follows from the fact that Q is a clique, ǫ G ( q J ) ≥ ǫ G ( x ) for any vertex q J ∈ J , and ǫ H ( u ) = ǫ G ( u ) for any vertex u ∈ V ( H ). Hence rad ( G ) = rad ( H ) ≤ α min ( H ) ≤ α ( H, v ) ≤ α ( G, v ) = α min ( G ) . Thus, we can assume that Q = K . Let J be the unique clique adjacent to the other ( = z )vertex x of Q . Let us show that J = K . If | J | ≥
3, then let y be a simplicial vertex in Q .Consider the graph H = G − y containing v . Now, it is easy to see that rad ( G ) = rad ( H ).The reasoning is similar to the previous one. This just follows from the fact that J is aclique, ǫ G ( y ) ≥ ǫ G ( x ), and ǫ H ( u ) = ǫ G ( u ) for any vertex u ∈ V ( H ). Hence rad ( G ) = rad ( H ) ≤ α min ( H ) ≤ α ( H, v ) ≤ α ( G, v ) = α min ( G ) . Thus, we have that J = K . Observe that this conclusion holds for every clique Q chosen asabove. Now, consider the graph H containing v obtained from G by removing the verticesof all such pendant cliques J . We remove the two vertices of J for each choice of J . Observethat rad ( G ) ≤ rad ( H ) + 2. In order to see this, let us observe that, by the definition of rad ( H ), we have that for some vertex w ∈ V ( H ), ǫ H ( w ) = rad ( H ). Now, by construction, rad ( G ) ≤ ǫ G ( w ) ≤ ǫ H ( w ) + 2 = rad ( H ) + 2 . Also, note that α ( G, v ) ≥ α ( H, v ) + 2. This can be proved as follows. Since the cut-vertex14 is adjacent to at least two cliques, and all cliques around v are non-pendant, then thereare at least two choices of cliques Q and their corresponding pendant cliques J . Thus, α ( G, v ) ≥ α ( H, v ) + 2. Hence, rad ( G ) ≤ rad ( H ) + 2 ≤ α min ( H ) + 2 ≤ α ( H, v ) + 2 ≤ α ( G, v ) = α min ( G ) . The proof is complete.In the next theorem we bound the radius of a block graph by a function of its dc ( G ).We precede the theorem with some simple observations concerning block graphs. Observation . For any vertices u and v of G , the internal vertices of any shortest u − v -path are cut-vertices. Observation . Let u be a vertex in G and let P be a u − v -path such that P is of length ǫ G ( u ) . Then all internal vertices of P are cut-vertices and v is a simplicial vertex. Observation . Assume that a vertex y is adjacent to vertices x and z such that xz / ∈ E .Then for any dc -set D , we have D ∩ { x, y, z } 6 = ∅ . Observation . Let H and G be two graphs with H ⊆ G . Then dc ( H ) ≤ dc ( G ) . Note that the difference dc ( G ) − rad ( G ) can be arbitrarily large. In order to see this, let G be a graph obtained from a star of at least two cliques K , sharing vertex v , by addingone clique Q = K k +2 to one of K in the star. The common vertex of Q and the star ofcliques is named by x . Finally, we add exactly one pendant clique, of size at least 3, toeach simplicial vertex of Q . Note that rad ( G ) = 2 - the center is formed by vertex x , while dc ( G ) = k + 2. Any dc -set D of G of size dc ( G ) is formed by vertices v , x and k cut-verticesof Q , excluding x (cf. Fig. 3). xv Figure 4: An exemplary block graph G with dc ( G ) − rad ( G ) = k with k = 7. heorem . Let G be a connected, block graph. Then rad ( G ) ≤ · dc ( G ) + 1 .Proof. Assume that the statement is wrong. Let G be a counter-example minimizing | V ( G ) | .Clearly, | V | ≥ | V | ≤
2, then dc ( G ) = 0 and rad ( G ) ≤ J such that it contains a simplicial vertex y . Let x be a cut-vertex of J . Consider the block graph H = G − y . Observe that since ǫ G ( y ) ≥ ǫ G ( x ) and for all vertices u ∈ V ( H ), ǫ H ( u ) = ǫ G ( u ), we have rad ( G ) = rad ( H ).Thus, rad ( G ) = rad ( H ) ≤ · dc ( H ) ≤ · dc ( G ) . Thus, all non-pendant cliques contain only cut vertices. Next, let us show that each pendantclique contains exactly one simplicial vertex. In particular, this means that all pendantcliques in G are K s. On the opposite assumption, consider a pendant clique J with at leasttwo simplicial vertices. Consider the block graph H obtained from G by removing one ofthem. Observe that rad ( G ) = rad ( H ). Thus, rad ( G ) = rad ( H ) ≤ · dc ( H ) ≤ · dc ( G ) . Thus, all pendant cliques in G are K s. Now, let us show that there is no cut vertex w in G such that it is adjacent to two pendant cliques. Assume that w is adjacent to two pendantcliques J = { w, z } and J = { w, z } , where z and z are simplicial vertices. Consider thegraph H = G − z . Observe that rad ( G ) = rad ( H ). Thus, rad ( G ) = rad ( H ) ≤ · dc ( H ) ≤ · dc ( G ) . Thus, we can also assume that each cut vertex is adjacent to at most one pendant clique.Let Q be a non-pendant clique of G such that it contains at most one vertex z that is notadjacent to a non-pendant clique other than Q . All other vertices of Q are adjacent toexactly one pendant clique. Note, that there is at least one cut vertex x in Q , except z .Since Q contains no simplicial vertices, we have that all vertices of Q except z are adjacentto exactly one pendant clique. Let us show that Q is K . Assume that | Q | ≥
3. Let x beany cut vertex in Q different from z . Let J be the pendant clique adjacent to x . Define H = G − ( V ( J ) − x ). Now, it is easy to see that rad ( G ) = rad ( H ). This just follows fromthe fact that Q is a clique, ǫ G ( q J ) ≥ ǫ G ( x ) for any vertex q J ∈ J , and ǫ H ( u ) = ǫ G ( u ) for anyvertex u ∈ V ( H ). Hence rad ( G ) = rad ( H ) ≤ · dc ( H ) ≤ · dc ( G ) . Thus, we can assume that Q = K . Now let us assign levels to our cliques of G . We doit by the following algorithm: all pendant cliques are assigned level 1. Then we remove allpendant cliques except their cut vertices. In the resulting block graph, all pendant cliquesget level 2 in G . We repeat this process till all cliques of G get their level. It is easy to see16hat the cliques Q from the previous paragraph are level 2 cliques.If G contains no level 3 clique, then it is easy to see that rad ( G ) = 2 and dc ( G ) = 1.Thus, our inequality is true for this case. This means, that we can assume that G containsat least one level 3 clique R .Let us show that all such cliques R are K . Indeed, if R is a clique of size at least 3, weuse the following reasoning. Note, since R is a clique of level 3, by the definition, there isat most one cut vertex of R that is adjacent to cliques of level 3 or more. At least one ofits cut vertices x is adjacent to exactly one level 2 clique Q which in its turn is adjacent toa pendant clique J , by the definition of Q . We have that all vertices of R are cut vertices,so the rest of cut vertices are adjacent to a level 1 or 2 cliques. If one of these cut vertices w of R is adjacent to a pendant clique K = wz , then consider the graph H = G − z . Wehave ǫ G ( z ) ≥ ǫ G ( w ) and ǫ H ( u ) = ǫ G ( u ) for any vertex u ∈ V ( H ). Hence rad ( G ) = rad ( H ) ≤ · dc ( H ) ≤ · dc ( G ) . Next, if one of these cut vertices w is adjacent to a level 2 clique Q ′ = wz and z is adjacentto a pendant clique J ′ = zy , then consider the graph H = G − z − y . We have ǫ G ( y ) ≥ ǫ G ( w ), ǫ G ( z ) ≥ ǫ G ( w ) and ǫ H ( u ) = ǫ G ( u ) for any vertex u ∈ V ( H ). Hence rad ( G ) = rad ( H ) ≤ · dc ( H ) ≤ · dc ( G ) . Thus, in our counterexample graph G the clique R must be K . Moreover, the cut vertex x that belongs to both Q and R is of degree two.Now, let G contain at least two level 3 cliques R and R . Let cliques Q , J and cliques Q , J be the corresponding level 1 and level 2 cliques corresponding to cliques R and R .We have ( V ( J ) ∪ V ( Q )) ∩ ( V ( J ) ∪ V ( Q )) = ∅ . By Observation 27, any dc ( G ) set D intersects V ( J ) ∪ V ( Q ) and V ( J ) ∪ V ( Q ). Thus, ifwe define H = G − V ( J ) − V ( Q ) − V ( J ) − V ( Q ), then dc ( H ) ≤ dc ( G ) − . Hence rad ( G ) ≤ rad ( H ) + 3 ≤ · dc ( H ) ≤ · ( dc ( G ) −
2) = 1 + 32 · dc ( G ) . Thus, we are left with the case when there is exactly one level 3 clique R in G . It isnot hard to see that the graph G forms P and the two vertices of R form the center of G .Moreover, their eccentricity is 3 and rad ( G ) = 3. On the other hand, we can assume that dc ( G ) ≥
2. Hence rad ( G ) = 3 ≤ · dc ( G )17nd the proof is complete. Remark . The bound presented in the previous theorem is tight for infinitely many blockgraphs. Let P n be the path on n vertices. Observe that rad ( P n ) = j n k . Using Observation 27, it can be shown that dc ( P n ) = j n k . Thus, for n = 2(mod6) , we will have rad ( P n ) = · dc ( P n ) + 1 . In Theorem 24, we have shown that in any connected block graph G , rad ( G ) ≤ α min ( G ).Thus, one can try to strengthen the result about the parameterization of EQUITABLECOLORING with respect to α min ( G ), by showing that it is FPT with respect to rad ( G ).Unfortunately, it turns out that such a result is less likely to be true. In [12], it is shown thatequitable coloring is W[1]-hard with respect to diam ( G ) - the diameter of G (cf. Theorem2), for block graphs. Since in any graph G , not necessarily block graph, rad ( G ) ≤ diam ( G ) ≤ · rad ( G ) , from Lemma 1, we have that diam ( G ) and rad ( G ) are equivalent from the perspective ofFPT. Thus, [12] implies that equitable coloring is less likely to be FPT with respect to rad ( G ) even when the input is restricted to block graphs.
4. Conclusion and future work
In this paper, we discussed the problem of
Equitable Coloring of block graphswith respect to many different parameters. Our research completes the approach given in[2, 6, 11, 12]. We presented some graph theoretic results that relate various parameters inblock graphs. We also discussed algorithmic implications of these results.Many parameters still remain open for the problem of
Equitable Coloring , even forblock graphs. Two of them are
M axLeaf and
M inLeaf - the maximum and minimumnumber of leaves in a spanning tree of G , respectively. Let us note that M axLeaf wasstated as an open problem already in [11]. These parameters are NP-hard to compute ingeneral graphs. Below we present two observations that imply that these two parameterscan be easily computed in the class of block graphs.
Proposition . Let G be a connected block graph. Then M inLeaf ( G ) coincides with thenumber of pendant cliques in G .Proof. Let P ( G ) be the number of pendant cliques of G . First observe that any spanningtree of G has at least one degree-one vertex in a pendant clique of G . Thus, M inLeaf ( G ) ≥ ( G ). Moreover, any simplicial vertex of a pendant clique can be made as a leaf in thespanning tree with smallest number of leaves.In order to show the converse inequality, let us proceed by induction. If G is a star ofcliques, then clearly M inLeaf ( G ) ≤ P ( G ) . Now, let us consider an arbitrary connected block graph G . Let J be a pendant clique in G .Consider the block graph H obtained from G by removing all the vertices of J except theunique cut-vertex x . Observe that H is a connected block graph of smaller order. Thus, M inLeaf ( H ) ≤ P ( H ) . Let T H be a spanning tree in H with smallest number of leaves. Attach a Hamiltonian pathof J to it to get T G . Observe that T G is a spanning tree of G . Moreover, x is no longer aleave even it were. We consider two cases.Case 1: x was a simplicial vertex of a pendant clique in H . In this case, we can assumethat x was a leaf in T H . Hence we will have M inLeaf ( G ) ≤ | Leaves ( T G ) | = | Leaves ( T H ) | = M inLeaf ( H ) ≤ P ( H ) ≤ P ( G ) , where Leaves ( T ) denotes the set of all leaves in tree T .Case 2: x was not a simplicial vertex of a pendant clique in H . This means that P ( G ) = P ( H ) + 1. Hence, M inLeaf ( G ) ≤ | Leaves ( T G ) | ≤ | Leaves ( T H ) | = 1 + M inLeaf ( H ) ≤ P ( H ) = P ( G ) . The proof is complete.
Proposition . Let G be a connected block graph. Then M axLeaf ( G ) coincides with thenumber of simplicial vertices in G .Proof. In order to prove the statement, let us proceed by induction on the number of vertices.If G is a star of cliques, then certainly the spanning tree with the largest number of leaveshas a form of a star. Then M axLeaf ( G ) = | V ( G ) | − S ( G ) . Now, let us consider any connected block graph not being a star of cliques. Let Q be a non-pendant clique such that it contains at most one vertex z that is not adjacent to pendantcliques other than Q . Since Q is not pendant, then there is at least one cut vertex x in Q ,except z . Let x , . . . , x t be cut vertices in Q , except z . Let H be the block graph obtainedfrom G by removing all pendant cliques that are adjacent to Q in x , . . . , x t , but withoutthese vertices. H is a connected block graph of smaller order. Thus, M axLeaf ( H ) = S ( H ) . T H be a spanning tree in H with the largest number of leaves. Now, vertices x , . . . , x t are leaves in T H , by the inductive assumption. Attach edges of form { x i , u } for all x i ∈{ x , . . . , x t } and u ∈ V ( G ) \ V ( H ) to T H to get T G . Observe that T G is a spanning tree of G . Moreover, all vertices that were added to H due to obtain G become leaves. In addition,vertices x , . . . , x t are no longer leaves, while each such a vertex u becomes simplicial in G .Since no cut vertex of G can be a leaf in a spanning tree of G , then we have M axLeaf ( G ) = | Leaves ( T G ) | = | Leaves ( T H ) | − t + | V ( H ) \ V ( G ) | == S ( H ) − t + | V ( H ) \ V ( G ) | = S ( G )and the proof is complete.Since M axLeaf ( G ) coincides with the number of simplicial vertices for block graph G ,then | V | − M axLeaf ( G ) is equal to the number of cut vertices. Note that removing all cutvertices from a block graph G leads to a union of cliques. Thus, dc ( G ) ≤ | V | − M axLeaf ( G ) ≤ | V | − M inLeaf ( G ) . Due to Theorem 3 and Lemma 1 we have the following results.
Proposition . Equitable Coloring in block graphs is FPT with respect to | V | − M axLeaf ( G ) . Proposition . Equitable Coloring in block graphs is FPT with respect to | V | − M inLeaf ( G ) . Even our problem is FPT with respect to complementary parameters | V | − M axLeaf and | V | − M inLeaf , its status with respect to
M axLeaf and
M inLeaf is open. We believethat these two parameters deserve a further investigation.
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