Conflict-Free Coloring on Open Neighborhoods
aa r X i v : . [ c s . D M ] O c t Conflict-Free Coloring on Open Neighborhoods
Sriram Bhyravarapu and Subrahmanyam Kalyanasundaram
Department of Computer Science and Engineering, IIT Hyderabad { cs16resch11001, subruk } @iith.ac.in Abstract.
In an undirected graph, a conflict-free coloring (with respectto open neighborhoods) is an assignment of colors to the vertices of thegraph G such that every vertex in G has a uniquely colored vertex in itsopen neighborhood. The conflict-free coloring problem asks to find thesmallest number of colors required for a conflict-free coloring.The conflict-free coloring problem is NP-complete. From results in Abelet. al. [SODA 2017], it can be inferred that every planar graph has aconflict-free coloring with at most nine colors. As the best known lowerbound for planar graphs is four colors, it was asked in the same paper iffewer colors would suffice. We make progress in answering this question,by showing that every planar graph can be colored using at most sixcolors. The same proof idea is used to show that every outerplanar graphcan be colored using at most five colors. Using a different approach, wefurther show that every outerplanar graph can be colored using at mostfour colors.Finally, we study the problem on Kneser graphs. We show that k + 2colors are necessary and sufficient to color the Kneser graph K ( n, k )when n ≥ k ( k + 1) + 1. A proper coloring of a graph is an assignment of a color to every vertex of thegraph such that adjacent vertices are of distinct colors. Conflict-free coloring isa variant of the graph coloring problem. A conflict-free coloring of a graph G isa coloring such that for every vertex in G , there exists a uniquely colored vertexin its neighborhood. This problem was first introduced in 2002 by Even, Lotker,Ron and Smorodinsky [1]. This problem was originally motivated by wirelesscommunication systems consisting of base stations and clients. The clients andbase stations have to send each other information and hence they communicatewith each other. Each base station is assigned a frequency and if two base stationswith the same frequency try to communicate with a client, it leads to interference.So for each client, there has to be a base station with a unique frequency. Sinceeach frequency band is expensive, there is a need to minimize the number offrequencies used by the base stations. Over the past two decades, this problemhas been very well studied, see for instance the survey by Smorodinsky [2].The conflict-free coloring problem has been studied with respect to openneighborhoods as well as closed neighborhoods. In this paper, we study theconflict-free coloring problem with respect to open neighborhoods. efinition 1 (Conflict-Free Coloring). A complete conflict-free (CF) col-oring of a graph G = ( V, E ) using k colors is an assignment C : V ( G ) →{ , , . . . , k } such that for every v ∈ V ( G ) , there exists an i ∈ { , , . . . , k } suchthat | N ( v ) ∩ C − ( i ) | = 1 . The smallest number of colors required for a com-plete conflict-free coloring of G is called the conflict-free chromatic number of G ,denoted by χ CF ( G ) . The conflict-free coloring problem and many of its variants are known to beNP-complete [3,4]. It was further shown in [4] that the CF coloring problem onopen neighborhoods is hard to approximate within a factor of n / − ε , unless P =NP. Since the problem is NP-hard, the parameterized aspects of the problem havebeen studied. The problems are fixed parameter tractable when parameterizedby vertex cover number, neighborhood diversity [4], distance to cluster, distanceto threshold graphs [5], and more recently, tree-width [6,7].In this paper, we look at the conflict-free open neighborhood problem, whichis considered as the harder of the open and closed neighborhood variants, see forinstance, remarks in [8,9]. It is easy to construct examples of bipartite graphs G ,for which χ CF ( G ) is Θ ( √ n ). Since any proper coloring is also a valid conflict-freeclosed neighborhood coloring, these examples have a CF closed neighborhoodcoloring using two colors. Further, Cheilaris [10] showed that for every graph G ,we have χ CF ( G ) ≤ √ n . On the contrary, a graph with maximum degree ∆ hasa conflict-free closed neighborhood coloring with at most O (log ε ∆ ) colors [8].Restrictions of the conflict-free coloring problem to special classes of graphshave been studied extensively. Of these, graphs arising out of intersection ofgeometric objects have attracted special interest, see for instance, [9,11,12]. Theproblem has also been studied for structural classes of graphs such as bipartitegraphs and split graphs [5].In [3], Abel et. al. considered the partial coloring variant of the problemwhere not all vertices need to be assigned a color. Definition 2 (Partial Conflict-Free Coloring). A partial conflict-free (CF)coloring of a graph G = ( V, E ) using k colors is an assignment C : V ( G ) →{ , , , . . . , k } such that for every v ∈ V ( G ) , there exists an i ∈ { , , . . . , k } such that | N ( v ) ∩ C − ( i ) | = 1 . Let us refer to the variant of the problem that we originally stated in Definition1, where all the vertices have to be colored, as the complete coloring variant.The key difference between partial CF coloring and complete CF coloring is thatin the partial variant, we allow some vertices to be assigned the color 0. It isconvenient to think of the vertices 0 as uncolored vertices. However, the uniquelycolored neighbor is not allowed to be of color 0. If a graph can be colored using k colors in the partial coloring variant, then all the uncolored vertices can beassigned the color k + 1, and thus a k + 1 complete coloring for the same graphcan be obtained.For the partial coloring variant, eight colors suffice to color a planar graph [3].It is easy to construct a planar graph that requires four colors. Starting with K , each original edge is subdivided by introducing a degree-two vertex on this ig. 1. Graph that requires 4 col-ors for a partial CF coloring.
Fig. 2.
Graph that requires 3 col-ors for a partial CF coloring. edge. In addition, a pendant vertex is attached to every original vertex of the K . This graph (see Figure 1) is planar and requires four colors. One of theopen questions asked in [3] was to close the gap between the upper bound ofeight and lower bound of four for the partial coloring variant of the conflict-free chromatic number of a planar graph. In this paper we reduce this gap, byshowing that five colors suffice for the partial CF coloring of a planar graph.Using the same proof idea, we show that four colors are enough for the partialcoloring of an outerplanar graph. The lower bound for the partial CF coloring ofan outerplanar graph is three, see Figure 2. As noted before, a partial coloringof an outerplanar graph using four colors implies a complete coloring using fivecolors. Using a different approach, we show that four colors are sufficient for acomplete CF coloring of an outerplanar graph.The last section in this paper studies the conflict-free coloring on Knesergraphs. The Kneser graph K ( n, k ) is the graph whose vertices are k -subsets of[ n ], and two such vertices are adjacent if and only if the corresponding sets aredisjoint. Several properties of Kneser graphs have been subject to study. Thechromatic number of the Kneser graph K ( n, k ) was conjectured by Kneser [13]in 1955 to be n − k +2. This remained open till Lov´asz proved [14] the conjecturein 1978. When n ≥ k ( k + 1) + 1, we determine the exact conflict-free chromaticnumber of the Kneser graph K ( n, k ).We summarize our results in this paper below:1. Five colors are sufficient for the partial conflict-free coloring of a planargraph. This improves the previous best known bound of [3] that requiredeight colors.Four colors are sufficient for the partial conflict-free coloring of an outerpla-nar graph. These two results are discussed in Section 3.2. Four colors suffice for the complete conflict-free coloring of an outerplanargraph. Moreover, three colors are sufficient and sometimes necessary for acomplete conflict-free coloring of cactus graphs. These results are shown inSection 4.3. In Section 5, we compute bounds on the conflict-free coloring of Knesergraphs. We also determine that the χ CF ( K ( n, k )) = k + 2 when n ≥ k ( k +1) + 1. Preliminaries
For any two vertices u, v ∈ G , we denote the shortest distance between themin G by dist ( u, v ). The open neighborhood of v , denoted by N ( v ), is the set ofvertices adjacent to v . We denote the graph induced by a set of vertices V ′ in G as G [ V ′ ].In this paper, we consider only connected graphs with at least two verticesbecause the colorings of the connected components can combine to give a coloringof the graph. Also, an isolated vertex does not have a conflict-free coloring (inthe open neighborhood setting) since there are no neighbors.A planar graph is a graph that can be drawn in R (a plane) such that theedges do not cross each other in the drawing. Each such drawing divides theplane into regions and each region is called a face . A planar drawing of a graphhas one face that is unbounded. This face is called the outer face . All the otherfaces are referred to as inner faces . An outerplanar graph is a planar graph thathas a drawing in a plane such that all the vertices of the graph belong to theouter face. Throughout the paper, we use terminology from the textbook “GraphTheory” by Diestel [15]. In [3] Abel et. al., showed that eight colors are sufficient for the partial CFcoloring of a planar graph. In this section, we improve the bound to five colors.We need the following definition:
Definition 3 (Maximal Distance-3 Set).
For a graph G = ( V, E ) , a maxi-mal distance-3 set is a set S ⊆ V ( G ) that satisfies the following:1. For every pair of vertices w, w ′ ∈ S , we have dist ( w, w ′ ) ≥ .2. For every vertex w ∈ S , ∃ w ′ ∈ S such that dist ( w, w ′ ) = 3 .3. For every vertex x / ∈ S , ∃ x ′ ∈ S such that dist ( x, x ′ ) < . The set S is constructed by initializing S = { v } where v is an arbitraryvertex. We proceed in iterations. In each iteration, we add a vertex w to S if(1) for every v already in S , dist ( v, w ) ≥
3, and (2) there exists a vertex w ′ ∈ S such that dist ( w, w ′ ) = 3. We repeat this until no more vertices can be added.The main component of the proof is the construction of an auxiliary graph G ′ from the given graph G . Construction of G ′ : The first step is to pick a maximal distance-3 set V . Noticethat any distance-3 set is an independent set by definition. We let V denote theneighborhood of V . More formally, V = { w | { w, w ′ } ∈ E ( G ) , w ′ ∈ V } . Let V denote the remaining vertices i.e., V = V \ ( V ∪ V ).We note the following properties satisfied by the above partitioning of V ( G ).1. The set V is an independent set.. For every vertex w ∈ V , there exists a unique vertex w ′ ∈ V such that { w, w ′ } ∈ E ( G ). This is because if there are two such vertices, this willviolate the distance-3 property of V .3. Every vertex in V has a neighbor in V . If there exists v ∈ V without aneighbor in V , then v is an isolated vertex. By assumption, G does not haveisolated vertices.4. There are no edges from V to V .5. Every vertex in V has a neighbor in V , and is hence at distance 2 fromsome vertex in V . This is due to the maximality of the distance-3 set V .Now we define A = V ∪ V . We first remove all the edges of G [ V ] making A an independent set. We then contract every vertex v ∈ A to a neighbor f ( v ) ∈ N ( v ) ⊆ V . The contraction process is as follows: we first identify vertex v with f ( v ). Then for every edge { v, v ′ } , we add an edge { f ( v ) , v ′ } . The resultinggraph is G ′ . Theorem 4.
Five colors are sufficient to partial CF color a planar graph.Proof.
Let G be a planar graph. We first construct the graph G ′ as above. Sincethe steps for constructing G ′ involve only edge deletion, and edge contraction, G ′ is also a planar graph. By the planar four-color theorem [16], every planargraph has a proper four coloring. That is, there is an assignment C : V ( G ′ ) →{ , , , } such that no two adjacent vertices of G ′ are assigned the same color.Notice that we have colored all the vertices of G ′ , that is the entire set V .Now, we extend C to get a CF coloring for G . For all vertices v ∈ V , weassign C ( v ) = 1. The vertices in V are assigned the color 0.We will show that C is indeed a partial CF coloring of G . Consider a vertex v ∈ A which is contracted to a neighbor f ( v ) = w ∈ V . The color assigned to w is distinct from all w ’s neighbors in G ′ . Hence the color assigned to w is theunique color among the neighbors of v in G .For each vertex w ∈ V , w is a neighbor of exactly one vertex v ∈ V . Everyvertex v ∈ V is colored 1, which is different from all the colors assigned to theneighbors of w in G ′ . ⊓⊔ Algorithmic Note:
The steps in the proof of Theorem 4 lead to an algorithm.The steps involved are construction of maximal distance-3 set, contraction ofvertices in A and the planar 4 coloring [16]. All these steps can be performed in O ( | V ( G ) | ) time. Thus we have an O ( | V | ) time algorithm, that given a planargraph G , determines a partial CF coloring for G that uses five colors.Outerplanar graphs have a proper coloring using three colors. By argumentanalogous to Theorem 4, we infer the following. Corollary 5.
Four colors are sufficient to partial CF color an outerplanar graph.
The famous Hadwiger’s conjecture states that if a graph G does not contain K k +1 as a minor, then G is k -colorable (in the sense of proper coloring). By ananalogous argument again, we obtain the following: Corollary 6.
Suppose the Hadwiger’s conjecture is true and that G has no K k +1 minor. Then G admits a partial CF coloring using k + 1 colors. Complete CF Coloring of Outerplanar Graphs
We saw in Corollary 5 that outerplanar graphs can be partially CF colored using4 colors. This implies a complete CF coloring using 5 colors. In this section, weshow an improved bound.
Theorem 7.
Four colors are sufficient for a complete CF coloring of an outer-planar graph G . Note that whenever we refer to an outerplanar graph G , we will also be implicitlyreferring to a planar drawing of G with all the vertices appearing in the outerface. We will abuse language and say “faces of G ” when we want to refer to facesof the above planar drawing.Theorem 7 is proved using a two-level induction process. The first level isusing a block decomposition of the graph. Any connected graph can be viewed asa tree of its constituent blocks. We color the blocks in order so that when we colora block, at most one of its vertices is previously colored. Each block is coloredwithout affecting the color of the already colored vertex. The second level of theinduction is required for coloring each of the blocks. We use ear decompositionon each block and color the faces of the block in sequence. However, the proofis quite technical and involves several cases of analysis at each step.We summarize the relevant aspects of block decomposition below. The readeris referred to a standard textbook in graph theory [15] for more details on this. – A block is a maximal connected subgraph without a cut vertex. – Blocks of a connected graph are either maximal 2-connected subgraphs, oredges (the edges which form a block will be bridges). – Two distinct blocks overlap in at most one vertex, which is a cut vertex. – Any connected graph can be viewed as tree of its constituent blocks.In the following discussion, we explain how to construct a coloring C : V ( G ) →{ , , , } for an outerplanar graph G . At any intermediate stage, the coloring C will satisfy the following invariants: Invariants of C – Every vertex v that has already been assigned a color C ( v ) has a neigh-bor w , such that C ( w ) = C ( x ), where x ∈ N ( v ) \ { w } . For v , thefunction U : V ( G ) → { , , , } denotes the color of w , its uniquelycolored neighbor. – ∀ v ∈ V ( G ), C ( v ) = U ( v ). – ∀{ v, w } ∈ E ( G ), C ( v ) = C ( w ) and |{ C ( v ) , U ( v ) , C ( w ) , U ( w ) }| = 3. ( ⋆ ) Theorem 7 is proved by using an induction on the block decomposition ofthe graph G and the below results. The condition marked ⋆ is violated in a few cases. In the exceptional cases where itis violated, we shall explain how the cases are handled. emma 8. If G is a 2-connected outerplanar graph such that all its inner facescontain exactly 5 vertices, then G has a complete CF coloring using 3 colors. Theorem 9.
Let G be an outerplanar graph.1. If B is a block of G that is either a bridge, or contains an inner face F with | V ( F ) | 6 = 5 , then B has a complete CF coloring using at most 4 colors.2. If B is a block of G , with exactly one vertex v precolored with color C ( v ) andunique color U ( v ) , then the rest of B has a complete CF coloring using atmost 4 colors, while retaining C ( v ) and U ( v ) .Proof (Proof of Theorem 7). Let G be an outerplanar graph. We apply blockdecomposition on G which results in blocks that are either maximal 2-connectedsubgraphs or single edges.If G is 2-connected and all its inner faces have exactly 5 vertices, then byLemma 8, G has a complete CF coloring using 3 colors.If G does not fit the above description, then G has a block B such that either B is an edge, or B has an inner face F with | V ( F ) | 6 = 5. In this case, by Theorem9.1, B has a complete CF coloring using at most 4 colors.Viewing G as a tree of its blocks, we can start coloring blocks that areadjacent to blocks that are already colored. Suppose the block B is alreadycolored, and let B ′ be a block adjacent to B . Let x be the cut-vertex betweenthe blocks B and B ′ . We use Theorem 9.2 to obtain a complete CF coloring of B ′ using at most 4 colors, while retaining C ( x ) and U ( x ). ⊓⊔ We now proceed towards proving Lemma 8 and Theorem 9. Lemma 8 andTheorem 9 discusses the coloring of blocks, which is accomplished by means ofinduction on the faces of the blocks. Towards this end, we use the following factabout ear decomposition of 2-connected outerplanar graphs. For a proof of thebelow lemma, we refer the reader to [17] where this is stated as Observation 2.
Lemma 10 (Ear Decomposition).
Let B be a 2-connected block in an outer-planar graph. Then B has an ear decomposition F , P , P , . . . , P q satisfying thefollowing: – F is an arbitrarily chosen inner face of B . – Every P i is a path with end points v, w such that { v, w } is an edge in F ∪ S ≤ j
Proof. (Proof of Lemma 8) Since G is 2-connected, the entire graph forms asingle block. Let F , P , . . . , P q be an ear decomposition of G . Recall that allthe faces have exactly five vertices. Let F = v − v − v − v − v − v . Weassign the following colors to the vertices in F : C ( v ) = 1 , C ( v ) = 1 , C ( v ) = The coloring assigned in this proof does not satisfy the condition marked ⋆ . However,this is not an issue since we are coloring the whole of G in this lemma. , C ( v ) = 2 , C ( v ) = 3. We also have U ( v ) = 3 , U ( v ) = 2 , U ( v ) = 1 , U ( v ) =3 , U ( v ) = 1.Let P i be any subsequent face P i = w − w − w − w − w − w with { w , w } being the pre-existing edge in F ∪ S ≤ j
Lemma 11.
An uncolored face F , such that | V ( F ) | 6 = 5 , can be CF coloredusing 4 colors satisfying the invariants.Proof. Let F = v − v − v − · · · − v k − − v k − v be a face with | V ( F ) | = k , k = 5. We assign C ( v ) = 1 , C ( v ) = 2 , C ( v ) = 3 and for the remaining vertices(if any), we set C ( v i ) = C ( v i − ). In order to satisfy the invariants, we need tomake the following changes: – k ≡ – k ≡ C ( v k ) = 4. – k ≡ C ( v k − ) = 4 , C ( v k − ) = 2 , C ( v k − ) = 3 , C ( v k ) =4. Notice that this coloring does not satisfy the invariants if k = 5. However,the smallest k that we consider in this case is k = 8.In each of the above cases the unique color for each vertex v i is provided byits cyclical successor i.e., U ( v i ) = C ( v i +1 ). ⊓⊔ Lemma 12.
Let F be a face (cycle) in G with one vertex v such that C ( v ) and U ( v ) are already assigned, with C ( v ) = U ( v ) . Then the rest of F can be completeCF colored using at most 4 colors, while retaining C ( v ) and U ( v ) , and satisfyingthe invariants.roof. Let v be the colored vertex in the cycle F . We may assume w.l.o.g. that C ( v ) = 1 and U ( v ) = 2. Now, we extend C to the remainder of F . – | V ( F ) | = 3 with F = v − v − v − v .We assign: C ( v ) = 3, C ( v ) = 4 and U ( v ) = 1, U ( v ) = 1. – | V ( F ) | ≥ F = v − v − v − · · · − v k − − v k − v .We first assign: C ( v ) = 3 and C ( v ) = 2. For the remaining vertices v i , weset C ( v i ) = C ( v i − ) for 4 ≤ i ≤ k . However, we need to make some changesto this in order to satisfy the invariants. We have the following subcases: • k ≡ C ( v k ) = 4. • k ≡ C ( v k − ) = 4.In each of the above cases the unique color for each vertex v i is providedby its cyclical successor i.e., U ( v i ) = C ( v i +1 ). Observe that U ( v ) is leftunchanged, by ensuring v and v k , the neighbors of v , are not assigned thecolor U ( v ). ⊓⊔ Lemma 13.
Let P be a path in G whose endpoints are v , v . Suppose { v , v } ∈ E ( G ) and that v , v are already assigned the functions C and U satisfying theinvariants. Then the rest of P can be CF colored using at most 4 colors, whileretaining C and U values of the endpoints, and satisfying the invariants. Since the proof of the above lemma is a bit long and involved, we first proveTheorem 9 using Lemmas 11, 12 and 13.
Proof (Proof of Theorem 9).
1. If the block is a bridge, say { v, w } , then we color it C ( v ) = 1 , C ( w ) = 2 with U ( v ) = 2 , U ( w ) = 1. Note that the invariant marked ⋆ is violated in thiscase. However, this does not cause an issue since this edge is a bridge, andit does not appear in any inner face.If the block is not a bridge, then by assumption, it contains a face F suchthat | V ( F ) | 6 = 5. By Lemma 11, we have a coloring of F using 4 colors andsatisfying the invariants. By the Lemma 10 (Ear Decomposition), the blockhas an ear decomposition F, P , P , . . . with F as the starting inner face.Recall that for every path P i , the end points form an edge in F ∪ S ≤ j
Proof (Proof of Lemma 13).
Let v and v be the end points of P . We extendthe coloring C to the remainder of P . According to the invariants of C , we haveonly 2 cases possible. Case 1: C ( v ) = C ( v ) , U ( v ) = U ( v ). W.l.o.g. we may assume C ( v ) = 1, C ( v ) = 2 and U ( v ) = 2, U ( v ) = 3. – | V ( P ) | = 3, P = v − v − v . Assign C ( v ) = 4 with U ( v ) = 2. – | V ( P ) | = 4, P = v − v − v − v . Assign C ( v ) = 4, C ( v ) = 3 with U ( v ) = 3, U ( v ) = 1. – | V ( P ) | ≥ P = v − v − · · · − v k − − v k − v . We first assign C ( v ) =1 , C ( v ) = 3 , C ( v ) = 4. For the remaining vertices v i , we initially assign C ( v i ) = C ( v i − ) for 6 ≤ i ≤ k . However, we need to make some changes tosatisfy the invariants. We have the following subcases: • k ≡ C ( v k − ) = 2 and C ( v k ) = 4 • k ≡ C ( v k − ) = 2. • k ≡ v i is provided byits cyclical successor i.e., U ( v i ) = C ( v i +1 ). Case 2: U ( v ) = U ( v ). W.l.o.g., we may assume C ( v ) = 1, C ( v ) = 2 and U ( v ) = U ( v ) = 3. – Case 2(i): | V ( P ) | = 3 and P = v − v − v . • Case 2(i)(a):
Vertices v and v are the only neighbors of v . Assign C ( v ) = 4 and U ( v ) = 2. The invariant marked ⋆ is not satisfied, butthat does not matter as v does not participate in any further faces. • Case 2(i)(b):
One of the edges { v , v } or { v , v } does not feature inan another face. W.l.o.g., say { v , v } be that edge. Assign C ( v ) = 4with U ( v ) = 1. The ⋆ invariant is violated for { v , v } here but it doesnot affect the further coloring. • Case 2(i)(c):
One of the edges { v , v } or { v , v } features in an uncol-ored face F such that | V ( F ) | 6 = 3. W.l.o.g., say { v , v } is that edge.We assign C ( v ) = 4 with U ( v ) = 1. Let | V ( F ) | = k with F = v − w − w − · · · − w k − − v − v . We assign C ( w ) = 3, C ( w ) = 1 and C ( w ) = 4 (if w exists). For all 4 ≤ i ≤ k − C ( w i ) = C ( w i − ). If k ≡ C ( w k − ) = 2, C ( w k − ) = 1 and C ( w k − ) = 4.The unique colors U for the vertices are assigned as follows: ∗ For k = 6, U ( w ) = 4 , U ( w ) = 3 , U ( w ) = 2 and U ( w ) = 2. ∗ For k = 6, we have for 1 ≤ i ≤ k − U ( w i ) = C ( w i +1 ) and U ( w k − ) = C ( v ) = 2. • Case 2(i)(d):
The only remaining case is when both the edges { v , v } or { v , v } feature in uncolored triangular faces. Let { v , v } form a tri-angular face with x and { v , v } with y . We have two subcases: v v x yz Fig. 3.
Case 2(i)(d) ∗ The edge { x, v } forms a triangular face with another vertex z (seeFigure 3). Assign C ( v ) = 1 , C ( x ) = 2 , C ( y ) = 4 , C ( z ) = 3 and U ( v ) = 4 , U ( x ) = 3 , U ( y ) = 2 , U ( z ) = 1. Some edges violate theinvariant marked ⋆ , but these edges are already part of two faces,and hence do not feature in the further coloring. ∗ The edge { x, v } is not part of a triangular face with another vertex.In this case, we assign C ( v ) = 4 , C ( x ) = 4 , C ( y ) = 1 and U ( v ) =2 , U ( x ) = 1 , U ( y ) = 2. Out of the edges that violate the invariantmarked ⋆ , the only one that can participate in the further coloring isthe edge { x, v } . By assumption, { x, v } is not part of a triangularface. In Lemma 14, we explain how to color the uncolored face thatis { x, v } may be a part of. – Case 2(ii): | V ( P ) | = 4, P = v − v − v − v . • Case 2(ii)(a):
The edge { v , v } forms a triangular face with a vertex x . We assign C ( v ) = 1 , C ( v ) = 4 , C ( x ) = 3, with U ( v ) = 3 , U ( v ) =1 , U ( x ) = 4. • Case 2(ii)(b):
The edge { v , v } is not part of an uncolored triangularface. We assign C ( v ) = C ( v ) = 4, with U ( v ) = 2 , U ( v ) = 1. If theedge { v , v } is part of an uncolored face F , by assumption, we knowthat | V ( F ) | ≥ F satisfyingthe invariants. – Case 2(iii): | V ( P ) | = 5 with P = v − v − v − v − v . We assign C ( v ) =1 , C ( v ) = 3 , C ( v ) = 2, with U ( v ) = 3 , U ( v ) = 2 , U ( v ) = 1. – Case 2(iv): | V ( P ) | ≥
6, with P = v − v − · · · − v k − − v k − − v k − v .We first assign C ( v ) = 4 and C ( v ) = 3. For 5 ≤ i ≤ k , assign C ( v i ) = C ( v i − ). If k ≡ C ( v k − ) = 1 and C ( v k ) = 2. Foreach vertex v i , the unique color is provided by its cyclical successor i.e., U ( v i ) = C ( v i +1 ). ⊓⊔ Lemma 14.
Let F be a face with | V ( F ) | ≥ with such that the edge { v , v } ∈ E ( F ) and v and v already colored such that C ( v ) = C ( v ) and U ( v ) = U ( v ) .Then the rest of F can be CF colored using 4 colors satisfying the invariants.roof. W.l.o.g., we may assume C ( v ) = C ( v ) = 4, U ( v ) = 1 and U ( v ) = 2.We have the following cases: – | V ( F ) | = 4 with F = v − v − v − v − v . We assign: C ( v ) = 1, C ( v ) = 3and U ( v ) = 4 and U ( v ) = 4. – If | V ( F ) | = 5 with F = v − v − v − v − v − v . We assign: C ( v ) = 1, C ( v ) = 2, C ( v ) = 3 and U ( v ) = 2, U ( v ) = 3 and U ( v ) = 4. – If | V ( F ) | ≥ F = v − v − · · · − v k − − v k − v − v . We assign: C ( v ) = 3 and C ( v ) = 2. For all 5 ≤ i ≤ k , C ( v i ) = C ( v i − ). • k ≡ C ( v k − ) = 1. • k ≡ • k ≡ C ( v k − ) = 1 and C ( v k ) = 2.The unique color of each vertex v i is provided its cyclical successor i.e., U ( v i ) = C ( v i +1 ). ⊓⊔ Algorithmic Note:
The steps in the proof of Theorem 7 leads to an algorithm.Block decomposition, outerplanarity testing and embedding outerplanar graphs[18] can all be done in linear time, i.e., O ( | V ( G ) | ). Thus we have an O ( | V ( G ) | )time algorithm, that given an outerplanar graph G , determines a complete CFcoloring for G that uses four colors. Now, we show that a cactus graph can be complete CF colored using 3 colors.This is a tight bound.
Definition 15.
A cactus graph is a connected graph in which any two cycleshave at most one vertex in common.
Theorem 16.
Three colors are sufficient and sometimes necessary to completeCF color cactus graphs.Proof.
Cactus graphs are outerplanar and by definition 15, any two cycles haveat most one vertex in common. We apply the block decomposition on the cactusgraph G . Note that each block is a cycle or a bridge. Throughout the coloring,we maintain the invariant that for each vertex v , the unique color seen by v , U ( v ) = C ( v ).Let B be the first block considered. If B is a bridge { v, w } , we assign C ( v ) = 1, C ( w ) = 2 and U ( v ) = 2, U ( w ) = 1. Else B is a cycle with F = v − v − · · · − v k − − v k − v . Initially we assign C ( v ) = 1, C ( v ) = 2 and C ( v ) = 3. For all 4 ≤ i ≤ k , C ( v i ) = C ( v i − ). If k ≡ C ( v k − ) = 3. Ineach of the cases, for each vertex v , we can identify U ( v ) such that U ( v ) = C ( v ).Now, we choose a block that is adjacent to an already colored block. Such ablock has exactly one colored vertex. Let v be the that vertex with C ( v ) = 1and U ( v ) = 2. If the block is a bridge, say { v , w } , then assign C ( w ) = 3 with U ( w ) = 1. Else the block is an inner face F = v − v − · · · − v k − − v k − v with colored vertex v . We initially assign C ( v ) = 3, C ( v ) = 2. For all 4 ≤ i ≤ k , C ( v i ) = C ( v i − ). In the case when k ≡ C ( v k ) = 3, andin the case when k ≡ C ( v k − ) = 2. In this case too, wemaintain the invariant that C ( v ) = U ( v ) for each v .To see that the bound is tight, observe that Figure 2 is a cactus graph thatrequires three colors. ⊓⊔ In this section, we study the CF coloring of Kneser graphs. Throughout thissection, we use [ n ] to denote the set { , , .., n } . The Kneser graph K ( n, k ) isformally defined as follows: Definition 17 (Kneser graph).
The Kneser graph K ( n, k ) is the graph whosevertices are (cid:0) [ n ] k (cid:1) , the k -sized subsets of [ n ] , and the vertices x and y are adjacentif and only if x ∩ y = ∅ (when x and y are viewed as sets). During the discussion, we shall use the words k -set or k -subset to refer to aset of size k . We shall sometimes refer to the k -subsets of [ n ] and the vertices of K ( n, k ) in an interchangeable manner. Lemma 18. k + 2 colors are sufficient to complete CF color a K ( n, k ) when n ≥ k − .Proof. We first assign a coloring to the vertices of K ( n, k ) and then argue thatthis coloring is a complete CF coloring. – For any vertex ( k -set) v that is a subset of { , , . . . , k − } , we assign C ( v ) = max ℓ ∈ v ℓ − ( k − – The set { k, k + 1 , . . . , k − } is assigned the color k + 1. – All the remaining vertices are assigned the color k + 2.For example, for the Kneser graph K ( n, { , , } , color 2 to the vertices { , , } , { , , } , { , , } , color 3 to the ver-tices { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , color 4 to the vertex { , , } and color 5 to all the remaining vertices.Now, we prove that in the above coloring, every vertex has a uniquely coloredneighbor. Let C i be the set of all vertices assigned the color i , i.e., the color classof the color i . Notice that C ∪ C ∪ · · · ∪ C k = (cid:0) [2 k − k (cid:1) . Let w k +1 denote the k -set { k, k + 1 , . . . , k − } . Any vertex v ∈ C ∪ C ∪ · · · ∪ C k is a neighborof w k +1 . Since w k +1 is the lone vertex colored k + 1, it serves as the uniquelycolored neighbor for any v ∈ C ∪ C ∪ · · · ∪ C k .Now we have to show the presence of uniquely colored neighbors for verticesthat have some elements from outside { , , . . . , k − } . Let v be the vertex In this coloring, the uniquely colored neighbor is not colored k + 2 for any of thevertices. Thus, by recoloring the color class k + 2 with the color 0, we get a partialcoloring that uses k + 1 colors. uch that it has some elements from outside { , , . . . , k − } . That is, v ∩{ , , . . . , k − } 6 = v . Let t = t ( v ) be the smallest nonnegative integer suchthat |{ , , . . . , k + t } \ v | = k . Since v has at least one element from outside { , , . . . , k − } , t is at most k − v has a lone neighbor colored t + 1, and this neighbor is givenby the set { , , . . . , k + t } \ v . By the choice of t , this is the only neighbor of v that is colored t + 1. It can be further observed that there are no neighbors of v that are assigned a color smaller than t + 1. ⊓⊔ Now we show that k + 2 colors are necessary, when n is large enough. Theorem 19. k + 2 colors are necessary to complete CF color K ( n, k ) when n ≥ k ( k + 1) + 1 .Proof. We prove this by contradiction. Suppose that K ( n, k ) can be coloredusing the k + 1 colors 1 , , , . . . , k + 1. For each 1 ≤ i ≤ k + 1, let C i denote thecolor class corresponding to the color i , i.e., the set of all vertices colored withthe color i . Let q = (cid:0) nk (cid:1) / ( k + 1).If for all i , | C i | < q , this implies that the total number of vertices is strictlyless than q ( k + 1) = (cid:0) nk (cid:1) . This is a contradiction. Hence there is at least one i ,such that | C i | ≥ q . For any vertex v , let d i ( v ) denote the number of neighborsof v in C i .Our strategy is to find one vertex, say x , which does not have a uniquelycolored neighbor. More formally, we want x to satisfy d i ( x ) = 1, for all 1 ≤ i ≤ k + 1. We construct the vertex ( k -set) x , by choosing elements in it as follows.Suppose there are C i ’s that are singleton, i.e., | C i | = 1. For all the singleton C i ’s we choose a hitting set. In other words, we choose entries in x so as toensure that x intersects with the vertices in all the singleton C i ’s. This partiallyconstructed x may also intersect with vertices in other C i ’s. Some of the other C i ’s might become “effectively singleton”, that is x may intersect with all thevertices in those C i ’s except one. We now choose further entries in x so as to hitthese effectively singleton C i ’s too. Finally, we terminate this process when allthe remaining C i ’s are not singleton.At this stage, x can have potentially k + 1 entries, one each to hit the k + 1color classes. However, the below claim shows that not all the color classes needto be hit. Claim:
There exists an i for which C i does not become singleton/effectivelysingleton. Proof of claim.
We have already seen that there is at least one C i for which | C i | ≥ q = ( nk ) k +1 . We show that this C i does not become effectively singleton.Let t be the number of entries in x when the above process terminates. Noticethat each entry in x can cause x to intersect with at most (cid:0) n − k − (cid:1) other vertices.We have t ≤ k + 1 entries in x , so x can intersect with at most ( k + 1) (cid:0) n − k − (cid:1) vertices. When n ≥ k ( k + 1) + 1, it can be verified that ( k + 1) (cid:0) n − k − (cid:1) < q − C i that do not intersect with x . ⊓⊔ ue to the above claim, the number of entries in x is t ≤ k . To fill up theremaining entries of x (if any), we consider the set(s) C j that have not becomeeffectively singleton. For each of these sets C j , we choose two distinct vertices,say y j , y ′ j ∈ C j . We choose the remaining entries of x so that x ∩ y j = ∅ and x ∩ y ′ j = ∅ . The number of such sets C j is at most k + 1. So for choosing theremaining entries of x , we have at least n − t − k ( k + 1) choices. Because n > k ,we can choose such entries. ⊓⊔ It is worth noting that the above proof technique cannot be applied forshowing a lower bound of k + 3. For such a proof, we would start with a k + 2coloring, and try for a contradiction. In this case, we could have k + 1 singletonsand effective singletons, which could require k +1 elements of [ n ] to hit. However, x can hold at most k elements. This is where the proof breaks down. In this section, we see some results on the CF closed neighborhood coloringof Kneser graphs. We abbreviate CF closed neighborhood coloring as CF-CNcoloring and denote by χ CF − CN ( G ), the conflict-free closed neighborhood chro-matic number of G . It is easy to see that a proper coloring of a graph G isalso a CF-CN coloring. That is, χ CF − CN ( G ) ≤ χ ( G ) for all graphs G . Since χ ( K ( n, k )) ≤ n − k + 2 [14], we have that χ CF − CN ( K ( n, k )) ≤ n − k + 2. Lemma 20.
When n ≥ k + 1 , we have χ CF − CN ( K ( n, k )) ≤ k + 1 .Proof. We assign the following coloring to the vertices of K ( n, k ): – For any vertex ( k -set) v that is a subset of { , , . . . , k − } , we assign C ( v ) = max ℓ ∈ v ℓ − ( k − – All the uncolored vertices are assigned color k + 1.For 1 ≤ i ≤ k + 1, let C i be the color class of the color i . Notice that C ∪ C ∪ · · · ∪ C k = (cid:0) [2 k − k (cid:1) . Since any two k -subsets of { , , . . . , k − } intersect, it follows that (cid:0) [2 k − k (cid:1) is an independent set. Hence each of the colorclasses C , C , . . . , C k are independent sets. So if v is colored with color i , where1 ≤ i ≤ k , it has no neighbors of its own color. Hence, it serves as its ownuniquely colored neighbor.If v is colored k + 1, then v [2 k − v has some elements fromoutside [2 k −
1] = { , , . . . , k − } . Let t = t ( v ) be the smallest nonnegativeinteger such that |{ , , . . . , k + t } \ v | = k . Since v has at least one element fromoutside { , , . . . , k − } , t is at most k −
1. It is easy to verify that the vertexcorresponding to the set { , , . . . , k + t } \ v is the lone neighbor of v that iscolored t + 1, and thus serves as the uniquely colored neighbor of v . ⊓⊔ Lemma 21. χ CF − CN ( K (2 k + 1 , k )) = 2 , for all k ≥ .roof. Consider a vertex v ∈ V ( K (2 k + 1 , k )). If v ∩ { , } 6 = ∅ , we assign color1 to v . Else, we assign color 2 to v .Let C and C be the sets of vertices colored 1 and 2 respectively. Below, wediscuss the unique colors for every vertex of K ( n, k ). – If v ∈ C and { , } ⊆ v , then v is the uniquely colored neighbor of itself.This is because all the vertices in C contain either 1 or 2 and hence v hasno neighbors in C . – Let v ∈ C and | v ∩ { , }| = 1. W.l.o.g., let 1 ∈ v and 2 / ∈ v . In thiscase, v has a uniquely colored neighbor w ∈ C . The vertex w is the k -set w = [2 k + 1] \ ( v ∪ { } ). – If w ∈ C , w is the unique color neighbor of itself. This is because C isan independent set. For two vertices w, w ′ ∈ C to be adjacent, we need | w ∪ w ′ | = 2 k , but vertices in C are subsets of { , , , . . . , k + 1 } , whichhas cardinality 2 k − ⊓⊔ Lemma 22. χ CF − CN ( K (2 k + d, k )) ≤ d + 1 , for all k ≥ .Proof. We prove this by induction on d . The base case is when d = 1 which istrue from Lemma 21. Suppose K (2 k + d, k ) has a CF-CN coloring with d + 1colors. Let us consider K (2 k + d + 1 , k ). For all the vertices of K (2 k + d + 1 , k )that appear in K (2 k + d, k ) we use the same coloring as in K (2 k + d, k ). Thenew vertices (the vertices that contain 2 k + d + 1) are assigned the new color d + 2. As all the new vertices contain 2 k + d + 1, they form an independent set.Hence each of the new vertices serve as their own uniquely colored neighbor.The vertices of K (2 k + d + 1 , k ) already present in K (2 k + d, k ) get newneighbors, but all the new neighbors are colored with the new color d + 2. Hencethe unique color of the existing vertices are retained. ⊓⊔ So, from Lemma 20 and Lemma 22 we get the following. χ CF − CN [ K ( n, k )] ≤ (cid:26) n − k + 1 , for 2 k + 1 ≤ n ≤ kk + 1 , for n ≥ k + 1 (cid:27) . We note a few directions that are left open by this paper: – We showed that a planar graph has a partial CF coloring that uses at most5 colors. The best known lower bound is 4 colors. – Along similar lines, an outerplanar graph can be partial CF colored using 4colors, while the lower bound is 3. – We showed that the complete CF chromatic number of K ( n, k ) is k + 2 when n ≥ k ( k + 1) + 1. We believe this requirement on n can be relaxed. Acknowledgments:
We would like to thank I. Vinod Reddy for suggesting theproblem, Karthik R. for initial discussions and Rogers Mathew for the proof ofTheorem 19. eferences
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