A Tight Bound for Conflict-free Coloring in terms of Distance to Cluster
aa r X i v : . [ m a t h . C O ] S e p A Tight Bound for Conflict-free Coloring in termsof Distance to Cluster
Sriram Bhyravarapu and Subrahmanyam KalyanasundaramDepartment of Computer Science and Engineering, IIT Hyderabad { cs16resch11001,subruk } @iith.ac.in October 2, 2020
Abstract
Given an undirected graph G = ( V, E ), a conflict-free coloring withrespect to open neighborhoods (CFON coloring) is a vertex coloring suchthat every vertex has a uniquely colored vertex in its open neighborhood.The minimum number of colors required for such a coloring is the CFONchromatic number of G , denoted by χ ON ( G ).In previous work [WG 2020], we showed the upper bound χ ON ( G ) ≤ dc ( G ) + 3, where dc ( G ) denotes the distance to cluster parameter of G . Inthis paper, we obtain the improved upper bound of χ ON ( G ) ≤ dc ( G ) + 1.We also exhibit a family of graphs for which χ ON ( G ) > dc ( G ), therebydemonstrating that our upper bound is tight. Given a graph G = ( V, E ), a conflict-free coloring is an assignment of colors toevery vertex of G such that there exists a uniquely colored vertex in the openneighborhood of each vertex. This problem was motivated by the frequencyassignment problem in cellular networks [5], where base stations and clientscommunicate with each other. It is required that there exists a base stationwith a unique frequency in the neighborhood of each client. We formally definethe problem as follows. Definition 1 (Conflict-Free Coloring) . A CFON 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. Thesmallest number of colors required for a CFON coloring of G is called the CFONchromatic number of G , denoted by χ ON ( G ).This problem has been studied from both algorithmic and structural per-spectives [1, 2, 3, 6, 7, 9]. Combinatorial bounds on this problem have been1tudied with respect to vertex cover, treewidth, pathwidth, feedback vertex setand neighborhood diversity [2, 3, 6]. In this paper, we study the relation be-tween CFON chromatic number and the distance to cluster parameter, whichis formally defined as follows. Definition 2 (Distance to Cluster) . Let G = ( V, E ) be a graph. The distanceto cluster of G , denoted dc ( G ), is the size of a smallest set X ⊆ V such that G [ V \ X ] is a disjoint union of cliques.Reddy [8] showed that 2 dc ( G )+1 colors are sufficient to CFON color a graph G . This bound has been improved to dc ( G ) + 3 in [2]. In this paper, we furtherimprove the bound to dc ( G ) + 1. Theorem 3.
For any graph G , we have χ ON ( G ) ≤ max { , dc ( G ) + 1 } .Further, we show graphs for which dc ( G ) colors are not sufficient, therebydemonstrating that the above bound is tight. Theorem 4.
For each value d ≥
1, there exist graphs G such that dc ( G ) = d and χ ON ( G ) > d .Theorem 4 is first proved in Section 2. The rest of the paper is devoted tothe proof of Theorem 3. In this paper, we consider only simple, finite, undirected and connected graphsthat have at least two vertices. If the graph has more than one connected com-ponent, Theorem 3 follows by its application to each component independently.Moreover, we assume that G does not have any isolated vertices as there is noCFON coloring for such graphs. We denote the set { , , · · · , d } by [ d ]. We usethe function C : V → [ d + 1] to denote the color assigned to a vertex. The openneighborhood of a vertex v , denoted by N ( v ), is the set of vertices adjacent to v . The degree of a vertex v , denoted deg( v ) is defined as | N ( v ) | . Sometimes,we use the notation deg A ( v ) = | N ( v ) ∩ A | , where A ⊆ V . We use the notation G [ A ] to denote the induced graph on the vertex set A . We use standard graphtheoretic terminology from the textbook by Diestel [4].During the coloring process, for each vertex v , we will designate a vertex asthe uniquely colored neighbor of v , denoted by U ( v ). The vertex U ( v ) is a vertex w ∈ N ( v ) such that C ( w ) = C ( x ), ∀ x ∈ N ( v ) \ { w } . We will also frequentlyrefer to a set X such that G [ V \ X ] is a disjoint union of cliques. For the sake ofbrevity, instead of referring to a component or maximal clique K of G [ V \ X ],we will say “ K is a clique in G [ V \ X ]”. In this section, we prove Theorem 4. We will see the existence of graphs G forwhich χ ON ( G ) > dc ( G ). 2 onstruction of graph G : Given a positive integer d , we construct the graph G such that dc ( G ) = d . It consists of three parts as described below. • The set X is an independent set of d vertices v , v , · · · , v d . Note that | X | = d , and G [ V \ X ] will be a disjoint union of cliques. • Singleton cliques K ( i,j ) , ∀ ≤ i < j ≤ d . For each K ( i,j ) , we have N ( K ( i,j ) ) = { v i , v j } . • A clique b K that has ( d +1)2 d vertices. The vertices of the clique b K consistof 2 d disjoint sets T W , one corresponding to each subset W ⊆ X . For each v ∈ T W , we have N ( v ) ∩ X = W . Moreover, we have | T W | = d + 1, foreach W ⊆ X . Proof of Theorem 4.
It can easily be noted that dc ( G ) = d . We will now seethat χ ON ( G ) > d .The singleton cliques K ( i,j ) force each vertex in X to be assigned a distinctcolor. WLOG, let C ( v i ) = i for each v i ∈ X . The colors 1 , , · · · , d are usedexactly once in X . Now, we prove that d colors are not sufficient to color theclique b K .We first consider a vertex u ∈ T ∅ . The vertex u does not have any neigh-bors in X and hence has its uniquely colored neighbor from b K . Let the vertex w ∈ b K be the uniquely colored neighbor of u . WLOG, let C ( w ) = 1. Now,consider the vertices in T { v } . At least d vertices in T { v } see both v and w astheir neighbors, and hence these vertices cannot have 1 as the unique color intheir neighborhood. Let u be such a vertex. WLOG, let w ∈ C be the vertexsuch that C ( w ) = 2 and w acts as the uniquely colored neighbor for u . Ofthe vertices in T { v ,v } , at least d − w , w in addition to v , v as neighbors. Hence these vertices cannot have the colors 1 or 2 as the uniquecolor in their neighborhood.We continue this reasoning and show that there exists at least one vertex,say u d +1 ∈ T X , that sees all the colors { , , . . . , d } at least twice in its neigh-borhood. Hence u d +1 cannot have any of the colors 1 , , . . . , d as the uniquecolor in its neighborhood. Hence we require a new color to CFON color G . In this section, we prove Theorem 3. Since it involves several cases and detailedanalyses, we first present an overview of the proof, before getting into the details.
Given a graph G = ( V, E ), and a set of vertices X ⊆ V such that | X | = d , wehave that G [ V \ X ] is a disjoint union of cliques. We require d colors, one foreach vertex in X . This is because G [ V \ X ] may contain (cid:0) d (cid:1) singleton cliques,such that each of these cliques has degree 2, and adjacent to a pair of verticesin X . Since a clique can be colored using at most 3 colors, it is easy to see that d + 3 colors are sufficient to CFON color G when G [ X ] is connected. Though3t is less straightforward, the bound of d + 3 can be extended to the case when G [ X ] is not connected as well [2]. It is a challenge to further improve the boundto d + 1. Our proof requires several cases and subcases since there does notseem to be a universal approach that leads to a desired coloring. The detailedcase analysis is necessary because of the different forms the induced graph G [ X ]can take.Except for some special cases, we will color each vertex in X with a distinctcolor from [ d ]. Our coloring algorithm consists of two phases, an initial phaseand a completion phase. In the initial phase, we color all the vertices of X , andidentify uniquely colored neighbors for some vertices in X . The key requirementof this phase to identify a free color f , which is a color in [ d ] that will not serve asa unique color in the neighborhood of any vertex in X . This is straightforwardin some cases, like when G [ X ] has a component of size at least 3. The caseswhere all the vertices of G [ X ] have degree 1 (Lemma 12), or all the vertices of G [ X ] have degree 0 or 1 (Lemma 13) prove to be particularly challenging. Insome of the cases, this is accomplished by coloring one or two of the cliques in G [ V \ X ]. The full set of conditions that are to be satisfied by the initial phaseis listed as the Rules of Lemma 9.After the initial phase, we are ready to run the completion phase, which isexecuted in Lemma 9. The goal of the completion phase is to color the rest ofthe graph while retaining the uniquely colored neighbors of those vertices thathad been identified in the intial phase. In the completion phase, we first identifyuniquely colored neighbors for those vertices in X , for which it has not beenidentified. This involves coloring some of the vertices in V \ X , and hence maypartially color some of the cliques in G [ V \ X ]. The cliques in G [ V \ X ] arecolored one by one. We have to use different approaches to color them, basedon the number of vertices that are already colored in the clique. The generalresults are presented for the case when d = | X | ≥
3, and the cases d = 1 and d = 2 need to be treated differently.The case d = 1 is straightforward, but the case d = 2 is somewhat involvedin itself. These cases are presented first in Section 3.2 as they serve as “warm-ups” for the pattern and the type of arguments that will be used in the general d ≥ | X | < We handle the cases | X | = 1 and | X | = 2 separately. Lemma 5.
Let G = ( V, E ) be a graph and X ⊆ V be a set of vertices suchthat | X | = 1 and G [ V \ X ] is a disjoint union of cliques. Then χ ON ( G ) ≤ Proof.
We explain how to assign C : V → { , , } such that C is a CFONcoloring of G . Let X = { v } . We assign C ( v ) = 1. Initial phase:
We first choose a clique K in G [ V \ X ]. Since G is connected,there exists v ∈ K such that v ∈ N ( v ). There are two cases.4 | K | = 1.That is, K = { v } . We assign C ( v ) = 2. We have U ( v ) = v and U ( v ) = v . • | K | ≥ v ′ ∈ K \ { v } . We assign C ( v ) = 2, C ( v ′ ) = 3, and thevertices (if any) in K \ { v, v ′ } are assigned 1. We have U ( v ) = U ( v ′ ) = v ,and for all y ∈ K \ { v ′ } , we have U ( y ) = v ′ . Completion phase:
The lone vertex v of X is already colored 1 and sees2 as the unique color in its neighborhood. We have also colored one clique in G [ V \ X ]. For all the uncolored cliques K in G [ V \ X ], we color K as per theapplicable case. • All the vertices in K see v as their neighbor.For all y ∈ K , assign C ( y ) = 3. The vertex v acts as the uniquely coloredneighbor for all y ∈ K . • There exists v ∈ K such that v / ∈ N ( v ).Notice that if | K | = 1, then the lone vertex in K has to necessarily see v as a neighbor. Hence in this case, we have | K | ≥
2. Choose a vertex v ′ ∈ K \ { v } . Assign C ( v ) = 2, C ( v ′ ) = 3, and the vertices (if any) in K \ { v, v ′ } the color 1.We have U ( v ′ ) = v and for all y ∈ K \ { v ′ } , we have U ( y ) = v ′ . Lemma 6.
Let G = ( V, E ) be a graph and X ⊆ V be a set of vertices suchthat | X | = 2 and G [ V \ X ] is a disjoint union of cliques. Then χ ON ( G ) ≤ Proof.
Let X = { v , v } . We explain how to assign C : V → { , , } to geta CFON coloring of G . There are two cases depending on whether X is anindependent set or not. Case 1: X is an independent set. That is, { v , v } / ∈ E ( G ) . We have twosubcases. • Every vertex in V \ X has at most one neighbor in X . Initial phase:
Since G is connected, there exists a clique K in G [ V \ X ]such that N ( v ) ∩ K = ∅ and N ( v ) ∩ K = ∅ . Let v ∈ N ( v ) ∩ K and v ′ ∈ N ( v ) ∩ K . Notice that v = v ′ , otherwise we would be violating thesubcase we are in.We assign C ( v ) = C ( v ) = 3, C ( v ) = 1, C ( v ′ ) = 2 and the remainingvertices (if any) in K \ { v, v ′ } the color 3.We get that U ( v ) = v and U ( v ) = v ′ . Also U ( v ) = v ′ and for each y ∈ K \ { v } , U ( y ) = v . Completion phase:
Now we color the remaining cliques in G [ V \ X ].Let K be an uncolored clique. There are two possibilities.5 Each vertex y ∈ K has deg X ( y ) = 1.For each vertex y ∈ K , if v ∈ N ( y ), assign C ( y ) = 2. Else, assign C ( y ) = 1. The uniquely colored neighbor of each vertex in K is itslone neighbor in X . – There exists v ∈ K such that deg X ( v ) = 0.Since G is connected, there exists v ′ ∈ K such that deg X ( v ′ ) = 0.WLOG let v ∈ N ( v ′ ). We assign C ( v ) = 1 , C ( v ′ ) = 2 and thevertices (if any) in K \ { v, v ′ } the color 3.We get that U ( v ) = v ′ and for each y ∈ K \ { v } , U ( y ) = v . • There exists a vertex v ∈ V \ X such that deg X ( v ) = 2. Initial phase:
Let v ∈ K , where K is a clique in G [ V \ X ]. We firstassign C ( v ) = 1 and C ( v ) = 2.If | K | = 1, we assign C ( v ) = 1 and we get that U ( v ) = U ( v ) = v and U ( v ) = v .Else | K | ≥ – There exists a vertex v ′ ∈ K \ { v } such that v / ∈ N ( v ′ ).We assign C ( v ) = 2, C ( v ′ ) = 3 and the vertices (if any) in K \ { v, v ′ } the color 1. We get that U ( v ) = U ( v ) = v , U ( v ′ ) = v and U ( y ) = v ′ for all y ∈ K \ { v ′ } . – Else, for each y ∈ K , we have v ∈ N ( y ).We assign C ( v ) = 1 and the remaining vertices in K \ { v } with thecolor 3. We get that U ( v ) = U ( v ) = v and U ( y ) = v for all y ∈ K .In each of the above case, both v and v have the same unique colorsfrom { , } . The other color in { , } does not serve as the unique colorof v and v and we refer to it as the free color . Completion phase:
WLOG let v and v have the color 1 as the uniquecolor, and 2 is the free color.Now, we extend this coloring to the cliques K ⊆ G [ V \ X ]. – There exists a vertex v ∈ K such that v ∈ N ( v ).Assign C ( v ) = 3 and the vertices (if any) in K \ { v } the color 2. Wehave U ( v ) = v and for all y ∈ K \ { v } , U ( y ) = v . – None of the vertices in K see v as a neighbor. ∗ All the vertices in K see v as a neighbor.Assign the color 3 to all the vertices in K . For each vertex y ∈ K ,we have U ( y ) = v . ∗ There exists v, v ′ ∈ K such that v ∈ N ( v ) and v / ∈ N ( v ′ ).Assign C ( v ) = 3 , C ( v ′ ) = 1 and the vertices (if any) in K \{ v, v ′ } the color 2. We get that U ( v ) = v ′ and y ∈ K \ { v } , U ( y ) = v . The notion of free color is used crucially in the proof of the general | X | ≥ ase 2: X is not an independent set. That is, { v , v } ∈ E ( G ) . Before we explain how we CFON color the graph, we need to set up notation.A clique K ⊆ G [ V \ X ] is called v -seeing ( v -seeing ) if for all vertices y ∈ K ,we have v ∈ N ( y ) ( v ∈ N ( y )). • Each vertex v ∈ V \ X that has deg X ( v ) = 1 appears in a clique K thatis v -seeing or v -seeing.In other words, each clique K in G [ V \ X ] satisfies one of the two condi-tions: (i) K is v -seeing or v -seeing, or (ii) each vertex v ∈ K has either deg X ( v ) = 0 or deg X ( v ) = 2. Initial phase:
There are two cases. – There is a vertex v ∈ V \ X such that deg X ( v ) = 2.Let v ∈ K , where K is a clique in G [ V \ X ]. We assign C ( v ) = 1, C ( v ) = 2, C ( v ) = 3 and the vertices (if any) in K \ { v } the color 2.We have U ( v ) = U ( v ) = v , U ( v ) = v and for the vertices (if any) y ∈ K \ { v } , U ( y ) = v . – For all y ∈ V \ X , we have deg X ( y ) < G is connected, there is a vertex v in each of the cliques in G [ V \ X ] such that deg X ( v ) = 1. By the case definition, we havethat each of the cliques must be v -seeing or v -seeing. If all thecliques were v -seeing, it follows that N ( v ) = { v } . This meansthat dc ( G ) = 1 and this case has been addressed in Lemma 5. Byan analogous argument, all the cliques cannot be v -seeing as well.Hence there are cliques K , K ⊆ G [ V \ X ] such that K is v -seeingand K is v -seeing.By the case definition, we have that for all vertices y ∈ K , deg X ( y ) <
2. Since K is v -seeing, it follows that for all y ∈ K , we have N ( y ) ∩ X = { v } . We choose a vertex v ∈ K . We assign C ( v ) = 1, C ( v ) = 3, and the vertices (if any) in K \ { v } , the color 2. We have U ( v ) = v , U ( v ) = v , and for the vertices (if any) y ∈ K \ { v } , U ( y ) = v .Similarly, for all y ∈ K , we have N ( y ) ∩ X = { v } . We choose avertex v ′ ∈ K and assign C ( v ) = 2, C ( v ′ ) = 3, and the vertices (ifany) in K \ { v } , the color 1. We have U ( v ) = v ′ , U ( v ′ ) = v , andfor the vertices (if any) y ∈ K \ { v ′ } , U ( y ) = v ′ .In each of the above cases, we have C ( v ) = 1, C ( v ) = 2, and the uniquecolor seen by v and v is 3. Completion phase:
Now we color the remaining cliques in G [ V \ X ].Let K be an uncolored clique. There are two possibilities. – For each y ∈ K , we have deg X ( y ) ≥ K is v -seeing, then for each vertex y ∈ K , assign C ( y ) = 2 and weget U ( y ) = v . If K is v -seeing, then for each vertex y ∈ K , assign7 ( y ) = 1 and we get U ( y ) = v . (If deg X ( y ) = 2 for each y ∈ K ,then either of the above assignments work.) – There exists a vertex v ∈ K such that deg X ( v ) = 0.Since G is connected, K has a vertex v ′ such that deg X ( v ′ ) >
0. Since K is not v -seeing or v -seeing, it must be the case that deg X ( v ′ ) = 2.We assign C ( v ) = 3, C ( v ′ ) = 1 and the vertices (if any) in K \ { v, v ′ } the color 2. We get that U ( v ) = v ′ and for each y ∈ K \{ v } , U ( y ) = v . • There exists a vertex v ∈ V \ X that has deg X ( v ) = 1 that appears in aclique K that is neither v -seeing nor v -seeing. Initial phase:
WLOG, let N ( v ) ∩ X = { v } . Since K is not v -seeing,there exists v ′ ∈ K such that v / ∈ N ( v ′ ).We assign C ( v ) = 1, C ( v ) = 2, C ( v ) = 1, C ( v ′ ) = 3, and the vertices (ifany) in K \ { v, v ′ } the color 2. We have U ( v ) = v , U ( v ) = v , U ( v ′ ) = v ,and for each y ∈ K \ { v ′ } , U ( y ) = v ′ . Note that the unique color seen byboth v and v is 1. Completion phase:
Now we color the remaining cliques in G [ V \ X ].Let K be an uncolored clique. We have the following cases. – There exists a vertex v ∈ K such that v ∈ N ( v ).Assign C ( v ) = 3, and the vertices (if any) in K \ { v } the color 2. Wehave U ( v ) = v , and for any y ∈ K \ { v } , U ( y ) = v . – None of the vertices in K see v as a neighbor. ∗ For all y ∈ K , we have N ( y ) ∩ X = { v } .Assign the color 3 to all the vertices in K . All the vertices in K see v as their uniquely colored neighbor. ∗ There is a vertex v ∈ K such that deg X ( v ) = 0.Since G is connected, | K | ≥
2. Choose a vertex v ′ ∈ K \ { v } .Assign C ( v ) = 1, C ( v ′ ) = 3, and the vertices (if any) in K \{ v, v ′ } the color 2. We have U ( v ′ ) = v , and for y ∈ K \ { v ′ } , U ( y ) = v ′ . | X | ≥ and X is an independent set We start handling the general case of | X | ≥
3. In this section, we prove theupper bound for the case when X is an independent set. Theorem 7.
Let G ( V, E ) be a graph and X ⊆ V be a set of vertices such that | X | = d ≥ G [ V \ X ] is a disjoint union of cliques. If X is an independentset, then χ ON ( G ) ≤ d + 1.In order to show the above theorem, we first prove Lemma 8, where wehandle the case when every vertex in V \ X has at most one neighbor in X .After this, we prove Lemma 10, where there is a vertex v ∈ V \ X that has8t least two neighbors in X . The proof of Lemma 10 uses Lemma 9, whichalso serves as the completion phase for all the remaining cases (including thosewhere X is not an independent set). Lemma 8.
Let G = ( V, E ) be a graph and X ⊆ V be a set of vertices suchthat | X | = d ≥ G [ V \ X ] is a disjoint union of cliques. If X is anindependent set and every vertex in V \ X has at most one neighbor in X , then χ ON ( G ) ≤ d + 1. Proof.
We explain how to assign C : V → [ d +1] such that C is a CFON coloringof G . Let X = { v , v , . . . , v d } . Initial phase:
For each v i ∈ X , assign C ( v i ) = d + 1. For each v i ∈ X , choosean arbitrary neighbor w i ∈ V \ X and assign C ( w i ) = i . We get that U ( v i ) = w i .Now, each vertex in X is colored and has a uniquely colored neighbor. Completion phase:
Each uncolored singleton clique in G [ V \ X ] is assignedthe color d + 1. Note that all the singleton cliques have exactly one neighborin X , and this neighbor is the uniquely colored neighbor. What remains to beaddressed are cliques of size at least 2. • Clique K ⊆ G [ V \ X ] with at least two colored vertices. Color theuncolored vertices with the color d +1. Let v, v ′ ∈ K be two of the verticesthat were colored prior to this step. Hence it follows that C ( v ) , C ( v ′ ) ∈ [ d ]and C ( v ) = C ( v ′ ).Since deg X ( y ) ≤ y ∈ G [ V \ X ], one of v and v ′ will be the uniquelycolored neighbor of all vertices in K . • Clique K ⊆ G [ V \ X ] with exactly 1 colored vertex v . Let C ( v ) = j and hence v j ∈ N ( v ). – If | K | = 2. Let K = { v, v ′ } . ∗ If v j / ∈ N ( v ′ ), we assign C ( v ′ ) = j . We get that U ( v ) = v ′ and U ( v ′ ) = v . ∗ Else, we have N ( v ) ∩ X = N ( v ′ ) ∩ X = { v j } . We assign C ( v ′ )arbitrarily from [ d ] \ { j } . We get that U ( v ) = v ′ and U ( v ′ ) = v . – Else if | K | ≥ v ′ be arbitrarily chosen from K \ { v } . Since | X | ≥ X ( v ′ ) ≤
1, there exists a vertex v ℓ ∈ X , v ℓ = v j such that v ℓ / ∈ N ( v ′ ) ∩ X . Assign C ( v ′ ) = ℓ and color the rest of the ver-tices in K \ { v, v ′ } with the color d + 1. We get that for all vertices w ∈ K , either v or v ′ is a uniquely colored neighbor. • Clique K ⊆ G [ V \ X ] with no colored vertices . We first select twovertices v, v ′ ∈ K .Since deg X ( v ) ≤
1, we can choose j ∈ [ d ] such that v j / ∈ N ( v ) ∩ X . Since deg X ( v ′ ) ≤ | X | ≥
3, we can choose ℓ ∈ [ d ] such that ℓ = j and v ℓ / ∈ N ( v ′ ) ∩ X . 9ssign C ( v ) = j , C ( v ′ ) = ℓ and the rest of the vertices in K \ { v, v ′ } thecolor d + 1. For all vertices w ∈ K , either v or v ′ is a uniquely coloredneighbor.We first state Lemma 9, which will serve as the completion phase for almostall the remaining cases (even for those where X is not an independent set).This lemma states that the graph can be CFON colored provided it has beenpartially colored satisfying certain rules. Lemma 9.
Let G = ( V, E ) be a graph with X = { v , v , . . . , v d } ⊆ V suchthat d ≥ G [ V \ X ] is a disjoint union of cliques. Further, Y = { v i ∈ X :deg X ( v i ) ≥ } and C : V → [ d + 1] be a partial coloring that satisfies the belowrules.Then, C can be extended to a CFON coloring b C : V → [ d + 1] of all thevertices in V . Rules (i) For all v i ∈ X , C ( v i ) = i .(ii) For some number of cliques K in G [ V \ X ], all the vertices in K arecolored using colors from [ d +1]. The remaining cliques are uncolored.(iii) All the vertices in Y and all the colored vertices in V \ X have auniquely colored neighbor. Moreover, some vertices in X \ Y have auniquely colored neighbor.(iv) The uniquely colored neighbor is identified for all the vertices in X whose entire neighborhood is colored.(v) There exists 1 ≤ f ≤ d , such that (a) v f ∈ X has a uniquely coloredneighbor and (b) for each vertex in X , the color f is not the uniquecolor in its neighborhood. We refer to f as the free color .(vi) If a vertex v i ∈ X \ Y does not have a uniquely colored neighbor, thenthe color i is not assigned to any vertex in V \ X yet.Before proving Lemma 9, we prove the upper bound when X is an indepen-dent set and there is a vertex v ∈ V \ X that has at least two neighbors in X . Lemma 10.
Let G ( V, E ) be a graph and X ⊆ V be a set of vertices suchthat | X | = d ≥ G [ V \ X ] is a disjoint union of cliques. If X is anindependent set and there exists a vertex v ∈ V \ X , such that deg X ( v ) ≥ χ ON ( G ) ≤ d + 1. Proof.
The goal here is to partially color some vertices of G so that the rulesof Lemma 9 are satisfied. We then use Lemma 9 to extend the partial coloringand obtain a CFON coloring of G . 10et X = { v , v , . . . , v d } . We explain how to assign C : V → [ d + 1] suchthat C is a partial coloring that satisfies the rules of Lemma 9. For each vertex v i ∈ X , we assign a distinct color C ( v i ) = i . There are two cases depending onthe neighborhood of vertices in V \ X . • There exists a singleton clique K = { v } such that deg X ( v ) ≥ N ( v ) ∩ X = { v i , v i , . . . , v i m } , with m ≥
2. We assign C ( v ) = i .We get that U ( v ) = v i and for all 1 ≤ j ≤ m , U ( v i j ) = v . The color i will not be the unique color of any vertex in X and will be the free color. • All singleton cliques have degree equal to 1.
By assumption, thereis a vertex v ∈ V \ X , such that deg X ( v ) ≥
2. It follows that v ∈ K where K is a clique in G [ V \ X ] and | K | ≥
2. Let N ( v ) ∩ X = { v i , v i , . . . , v i m } ,with m ≥ X whose neighbors belong to only K . We refer to these vertices as S K .Formally, S K = { v i ∈ X : N ( v i ) ⊆ K } \ N ( v ). The vertices in N ( v ) ∩ X rely on v for their uniquely colored neighbor and hence does not requirespecial attention. – S K = ∅ .First, we assign C ( v ) = i and choose i as the free color. For eachvertex v j ∈ S K , we choose a vertex w j ∈ N ( v j ). Note that by thedefinition of S K , it follows that w j ∈ K . We assign C ( w j ) = j ifit is not already assigned ( w j could have been the chosen neighborfor some other vertex v j ′ as well). Now all the vertices in S K have auniquely colored neighbor. ∗ If all vertices in K are colored because of the above coloring,every vertex in K has a uniquely colored neighbor. We get that U ( v ) = v i . Each vertex w ∈ K \ { v } is assigned a distinctcolor, say j , because it is adjacent to v j ∈ S K , which serves asits uniquely colored neighbor. ∗ There exists at least one uncolored vertex K . · If there exists a uniquely colored neighbor for each uncoloredvertex in K , assign the color d + 1 to all the uncolored ver-tices. We get that U ( v ) = v i . The vertices w j ∈ K \ { v } rely on the corresponding v j ’s as mentioned above. · Else, let v ′ ∈ K be an uncolored vertex that does not see auniquely colored neighbor. This implies that N ( v ′ ) ∩{ v i , v i , · · · , v i m } = { v i } . We reassign C ( v ) = i , assign C ( v ′ ) = d + 1, and des-ignate i as the free color instead of i . We assign the color i to the remaining uncolored vertices in K . We have U ( v ′ ) = v and U ( w ) = v ′ , for all w ∈ K \ { v ′ } .11 S K = ∅ .This implies that, there is no vertex in X \ { v i , v i , · · · , v i m } thatwas relying on K for its uniquely colored neighbor. We first assign C ( v ) = i and choose i as the free color. We have the followingcases. ∗ There exists a vertex v ′ ∈ K \ { v } such that v i / ∈ N ( v ′ ).Assign C ( v ′ ) = d + 1 and assign the remaining vertices of K \{ v, v ′ } the color i .For every vertex w ∈ K \ { v ′ } , U ( x ) = v ′ . Finally, we have U ( v ′ ) = v . ∗ Else, for every vertex w ∈ K , we have v i ∈ N ( w ) ∩ X .Reassign C ( v ) = i and assign the color d + 1 to all the verticesin K \ { v } . The color i is the redesignated free color.And for each w ∈ K , U ( w ) = v i .Now C is a partial color assignment satisfying all the conditions in the rulesof Lemma 9. Hence by Lemma 9, we can extend C to a full CFON coloring of G that uses at most d + 1 colors.We conclude this section with the proof of Lemma 9. Proof of Lemma 9.
For each colored vertex w ∈ G , b C ( w ) = C ( w ). We explainhow to extend b C : V → [ d + 1] to all vertices such that b C is a CFON coloringof G . Let X = { v , v , . . . , v d } and Y = { v i ∈ X : deg X ( v i ) ≥ } . Process to identify uniquely colored neighbors for X \ Y : For every v j ∈ X \ Y , that does not have a uniquely colored neighbor, choose an uncoloredneighbor of v j , say w j ∈ V \ X and assign b C ( w j ) = j . Rules (iv) and (vi)of Lemma 9 allow us to do this. Since v j does not have a uniquely coloredneighbor, rule (iv) implies that v j has an uncolored neighbor, and as per rule(vi), no vertex in V \ X is assigned the color j . Observation:
It is possible that all the neighbors of v j ∈ X \ Y may be coloredby the above coloring process on other vertices v i ∈ X \ Y even before applyingthe process on v j . In such a case, we choose an arbitrary neighbor of v j that wasalready colored by this process and assign it as the uniquely colored neighborfor v j . This neighbor acts as the uniquely colored neighbor for at least 2 verticesin X \ Y . This fact will be useful later.Now, every vertex in X \ Y has a uniquely colored neighbor. We now lookat the previously uncolored cliques K in G [ V \ X ]. For each such clique K , wecolor K as per the applicable case below. Case 1: K has no colored vertices • | K | = 1. Let K = { w } .We assign b C ( w ) = d + 1. As all the neighbors of w are distinctly colored,we assign one of the neighbors as U ( w ). • | K | ≥
2. We have two subcases here.12
There exists a vertex w ∈ K , such that N ( w ) ∩ X = ∅ .Choose another vertex w ′ ∈ K \ { w } such that N ( w ′ ) ∩ X = ∅ . Wehave two subcases. ∗ N ( w ′ ) ∩ X = { v f } , where v f ∈ X is the vertex that correspondsto the free color f .Assign b C ( w ′ ) = d + 1 and b C ( w ) = c , where c ∈ [ d ] \ { f } , chosenarbitrarily. For all the vertices (if any) x ∈ K \ { w, w ′ } , assign b C ( x ) = f .We have U ( w ′ ) = w and for all vertices x ∈ K \ { w ′ } , we have U ( x ) = w ′ . ∗ There exists a vertex v i ∈ N ( w ′ ) ∩ X , where v i = v f .Assign b C ( w ′ ) = d + 1. For all the vertices x ∈ K \ { w ′ } , assign b C ( x ) = f .We have U ( w ′ ) = v i and for all vertices x ∈ K \ { w ′ } , we have U ( x ) = w ′ . – For all w ∈ K , N ( w ) ∩ X = ∅ .Assign all the vertices in K the color d + 1. For all the vertices in K , we assign one of the neighbors in X as the respective uniquelycolored neighbor. Case 2: K has exactly one colored vertex WLOG, let v ∈ K be such that b C ( v ) = j . This implies that v j ∈ N ( v ) ∩ X . • | K | = 1.We have U ( v ) = v j and U ( v j ) = v (as was already assigned). • | K | = 2. Let K = { v, v ′ } . – N ( v ′ ) ∩ X = ∅ .We assign b C ( v ′ ) = d + 1. We get that U ( v ) = v ′ , U ( v ′ ) = v and U ( v j ) = v . – v ′ has a neighbor other than v j in X . That is, ∃ v k ∈ N ( v ′ ) ∩ X , with v k = v j .We assign b C ( v ′ ) = d + 1. We get that U ( v ) = v ′ , U ( v ′ ) = v k and U ( v j ) = v . – N ( v ′ ) ∩ X = { v j } . ∗ There exists a vertex v ℓ ∈ X \ Y , v ℓ = v j such that U ( v ℓ ) = v .We reassign b C ( v ) = ℓ and b C ( v ′ ) = d +1. We have that U ( v ) = v ′ and U ( v ′ ) = v . Note that U ( v j ) = U ( v ℓ ) = v , as before. ∗ No vertex in X other than v j sees v as its uniquely colored neigh-bor and v has another neighbor in X besides v j , say v k .We reassign b C ( v ) = d + 1 and assign b C ( v ′ ) = j . We get that U ( v ′ ) = v , U ( v ) = v k and we reassign U ( v j ) = v ′ .13 N ( v ) ∩ X = N ( v ′ ) ∩ X = { v j } .We reassign b C ( v ) to an arbitrarily chosen value from [ d ] \ { j, f } .Note that such a value exists since d ≥
3. We assign b C ( v ′ ) = d + 1. We get that U ( v ) = v ′ , U ( v ′ ) = v and U ( v j ) = v , asbefore. • | K | ≥ – There exists a vertex v ′ ∈ K \ { v } such that v j / ∈ N ( v ′ ).We assign b C ( v ′ ) = d + 1 and the vertices in K \ { v, v ′ } are coloredwith the free color f .We get that U ( v ′ ) = v and for all w ∈ K \ { v ′ } , U ( w ) = v ′ . Asbefore, U ( v j ) = v . – Every vertex in K is adjacent to v j . ∗ There exists a vertex v ′ ∈ K \{ v } such that ( N ( v ′ ) ∩ X ) \ { v j , v f } 6 = ∅ . Let v k ∈ N ( v ′ ) ∩ X , where v k = v j and v k = v f .Assign b C ( v ′ ) = d + 1 and rest of the vertices in K \ { v, v ′ } arecolored with the free color f . We get that U ( v ′ ) = v k and for all w ∈ K \ { v ′ } , U ( w ) = v ′ . As before, U ( v j ) = v . ∗ For all w ∈ K \ { v } , N ( w ) ∩ X ⊆ { v j , v f } .We choose k ∈ [ d ] \ { j, f } arbitrarily. Note that such a valueexists since d ≥ v ′ , v ′′ arbitrarily from K \ { v } and assign b C ( v ′ ) = d + 1, b C ( v ′′ ) = k and the remaining vertices in K \{ v, v ′ , v ′′ } are colored with f .We get that U ( v ′ ) = v ′′ , and for all w ∈ K \ { v ′ } , U ( w ) = v ′ . Asbefore U ( v j ) = v . Case 3: K has at least two colored vertices and there exists a vertexin K that is a uniquely colored neighbor for at least two vertices in X \ Y Let v ′ ∈ K be such that U ( v j ) = U ( v k ) = v ′ , where v j , v k ∈ X \ Y . Thevertex v ′ was colored in the process to identify uniquely colored neighbors forvertices in X \ Y . WLOG, we may assume that v ′ was colored when assigninga unique colored neighbor for v j . That is, b C ( v ′ ) = j . This also implies that novertices in K are colored k . There are two cases here. • All vertices in K are colored.In this case, every vertex in K will have a uniquely colored neighbor. Thisis because every vertex in K would have been assigned a distinct color. If w ∈ K is such that b C ( w ) = ℓ , then U ( w ) = v ℓ ∈ X \ Y . • There exists an uncolored vertex v ∈ K . There are two subcases. – v is adjacent to v k . 14e assign b C ( v ) = d + 1 and the remaining uncolored vertices in K (if any) are assigned the free color f .We get that U ( v ) = v k , and for all w ∈ K \ { v } , U ( w ) = v . Notethat U ( v j ) = U ( v k ) = v ′ , as before. – v is not adjacent to v k .Reassign b C ( v ′ ) = k , assign b C ( v ) = d + 1 and the remaining uncoloredvertices in K (if any) are assigned the free color f .We get that U ( v ) = v ′ , and for all w ∈ K \ { v } , U ( w ) = v . Note that U ( v j ) = U ( v k ) = v ′ , as before. Case 4: K has at least two colored vertices and every colored vertexin K is the uniquely colored neighbor for exactly one vertex in X \ Y Let v, v ′ ∈ K be two colored vertices such that b C ( v ) = j and b C ( v ′ ) = k . Thecolors j and k are assigned because they are adjacent to v j and v k respectively,where v j , v k ∈ X \ Y . We have cases depending on the neighborhood of K . • There exists a colored vertex in K that is adjacent to both v j and v k . – At least one of v or v ′ is adjacent to both v j and v k .WLOG let that vertex be v . Reassign b C ( v ′ ) = d + 1 and assign thecolor f to the remaining uncolored vertices (if any).We have that U ( v ′ ) = v k , and for all vertices w ∈ K \{ v ′ } , U ( w ) = v ′ .We reassign U ( v k ) = v , while U ( v j ) = v as before. – There exists a colored vertex v ′′ ∈ K \ { v, v ′ } such that { v j , v k } ⊆ N ( v ′′ ).Let b C ( v ′′ ) = ℓ because it was the chosen neighbor for v ℓ ∈ X \ Y in the coloring process stated in the beginning of this proof. Wereassign b C ( v ) = d + 1 and b C ( v ′ ) = f . It is important to note thatbecause of the case definition, this reassignment does not affect theuniquely colored neighbors of vertices in X \ ( Y ∪ { v j , v k } ).The remaining uncolored vertices in K (if any) are assigned the color f .We have U ( v ) = v j , and for every vertex w ∈ K \ { v } , U ( w ) = v .We reassign U ( v j ) = U ( v k ) = v ′′ , while U ( v ℓ ) = v ′′ as before. • There exists an uncolored vertex v ′′ ∈ K such that { v j , v k } ⊆ N ( v ′′ ).We reassign b C ( v ) = d +1, b C ( v ′ ) = f and assign b C ( v ′′ ) = k . The remaininguncolored vertices in K (if any) are assigned the color f .We have U ( v ) = v j , and for every vertex w ∈ K \ { v } , U ( w ) = v . Wereassign U ( v j ) = U ( v k ) = v ′′ . • No vertex in K is adjacent to both v j and v k .Assign the color f to the remaining uncolored vertices (if any). Everyvertex in K \ { v, v ′ } will see either v or v ′ as its uniquely colored neighbor.Also U ( v ) = v j and U ( v ′ ) = v k , while U ( v j ) = v and U ( v k ) = v ′ , asbefore. 15 .4 The case when | X | ≥ and X is not an independentset In this section, we prove the upper bound when | X | ≥ X is not anindependent set. Theorem 11.
Let G = ( V, E ) be a graph and X ⊆ V be a set of vertices suchthat | X | = d ≥ G [ V \ X ] is a disjoint union of cliques. If X is not anindependent set, then χ ON ( G ) ≤ d + 1.The proof of Theorem 11 involves a lot of cases. The cases when G [ X ] is1-regular and all the vertices in G [ X ] have degree 0 or 1 needs particular care.We state these two cases below. The proofs of Lemmas 12 and 13, are proved inSections 3.5 and 3.6 respectively. All the proofs in this section and subsequentsections will only deal with the initial phase, i.e., to achieve a partial coloringof the graph that satisfies the Rules of Lemma 9. The completion phase followsby Lemma 9. Lemma 12.
Let G = ( V, E ) be a graph and X ⊆ V be a set of vertices suchthat | X | = d ≥ G [ V \ X ] is a disjoint union of cliques. If G [ X ] is 1-regular(perfect matching), then χ ON ( G ) ≤ d + 1. Lemma 13.
Let G = ( V, E ) be a graph and X ⊆ V be a set of vertices, suchthat | X | = d ≥ G [ V \ X ] is a disjoint union of cliques. Moreover all thevertices in G [ X ] have degree at most 1 and at least one vertex has degree 0.Then χ ON ( G ) ≤ d + 1.We first prove Theorem 11 assuming the above lemmas. Proof of Theorem 11.
Let X = { v , v , . . . , v d } and Y = { v i ∈ X : deg X ( v i ) ≥ } . Also, let A be the set of connected components of G [ X ].For each vertex v i ∈ X , we assign a distinct color C ( v i ) = i . We have thefollowing cases depending on the components. • There exists a component A ∈ A such that | A | ≥
3, and ∃ v j ∈ A withdeg X ( v j ) = 1.Let N ( v j ) ∩ X = { v k } . Since | A | ≥
3, there exists v ℓ ∈ N ( v k ) ∩ X suchthat v ℓ = v j . Every vertex in A \ { v k } chooses an arbitrary neighbor in A as its uniquely colored neighbor while we assign U ( v k ) = v ℓ . The color j is not the unique color for any vertex in X . Hence we use j as the freecolor for the rest of the coloring.For all A ′ ∈ A \ A , where | A ′ | ≥
2, a vertex chooses one of its neighborsin A ′ as its uniquely colored neighbor. Thus all the vertices in Y have auniquely colored neighbor.At this point, we have partially colored G satisfying the rules of Lemma9. Using Lemma 9, we can extend this to a full CFON coloring of G thatuses d + 1 colors. 16 There exists a A ∈ A such that | A | ≥
3, and all the vertices in A havedegree at least 2 in G [ X ].Choose a vertex v j ∈ A . Since every vertex in A has degree at least 2,it can be ensured that every vertex in A is assigned a uniquely coloredneighbor other than v j . We use the color j as the free color.For all A ′ ∈ A \ A , where | A ′ | ≥
2, a vertex chooses one of its neighborsin A ′ as its uniquely colored neighbor. Thus all the vertices in Y have auniquely colored neighbor.Since the rules of Lemma 9 are satisfied, we can extend this coloring to aCFON coloring of G that uses d + 1 colors. • For all the components A ∈ A , we have | A | = 2.In this case, we have X \ Y = ∅ . All the vertices v i ∈ X have deg X ( v i ) = 1.We apply Lemma 12 to CFON color G with d + 1 colors. • For all the components A ∈ A , we have | A | ≤
2. Moreover there exists A ′ ∈ A such that | A ′ | = 1.That is X \ Y = ∅ . By assumption, X is not an independent set. Hence Y = ∅ as well. We apply Lemma 13 to CFON color G with d + 1 colors. Let X = { v , v , . . . , v d } . Since each vertex v i ∈ X has deg X ( v i ) = 1, we havethat d = | X | is an even number. This implies that d ≥
4. WLOG, we mayassume that the edges in G [ X ] are { v , v } , { v , v } , . . . , { v d − , v d } . We explainhow to assign C : V → [ d + 1] such that C is a partial coloring that satisfies therules of Lemma 9.For each vertex v i ∈ X , we assign the color C ( v i ) = i . We have the followingcases. Case 1:
There exists a vertex v ∈ V \ X such that deg X ( v ) = | X | .Let v ∈ K , where K is a clique in G [ V \ X ]. Subcase 1.1: K is the only clique in G [ V \ X ].For all v i ∈ X \ { v } , we reassign C ( v i ) = d + 1. Assign C ( v ) = 2 and theremaining vertices (if any) in K \ { v } are assigned the color d + 1.We get that U ( v ) = v and for all x ∈ V \ { v } , U ( x ) = v . Thus the entiregraph is CFON colored. Subcase 1.2:
There exists a clique K = K , such that K = { w } anddeg X ( w ) ≥ • N ( w ) ∩ X contains a pair of adjacent vertices. WLOG, let v , v ∈ N ( w ). We have cases based on the size of the clique K . – | K | = 1. 17e assign C ( w ) = 3 and C ( v ) = 1. We have that U ( v ) = U ( w ) = v .Also, U ( v ) = w and for each v i ∈ X \ { v } , U ( v i ) = v . We havecolor 4 as the free color. – | K | ≥ v ′ ∈ K \ { v } . We have three casesbased on the neighborhood of v ′ . ∗ v / ∈ N ( v ′ ).We assign C ( w ) = 3, C ( v ) = 1, C ( v ′ ) = d + 1 and the vertices(if any) in K \ { v, v ′ } the color 4.We have that U ( w ) = v , U ( v ) = w , and for each v i ∈ X \ { v } , U ( v i ) = v . Also U ( v ′ ) = v and for each y ∈ K \ { v ′ } , U ( y ) = v ′ .We have color 4 as the free color. ∗ v ∈ N ( v ′ ).We assign C ( w ) = 3, C ( v ) = 2, C ( v ′ ) = d + 1 and the vertices(if any) in K \ { v, v ′ } the color 4.We have that U ( w ) = v , U ( v ) = w and for each v i ∈ X \ { v } , U ( v i ) = v . Also U ( v ′ ) = v and for each y ∈ K \{ v ′ } , U ( y ) = v ′ .We have color 4 as the free color. • None of the vertices in N ( w ) ∩ X are adjacent to each other. WLOG, let v , v ∈ N ( w ). We have cases based on the size of the clique K . – | K | = 1.We assign C ( v ) = 4 and C ( w ) = 1. We have U ( w ) = U ( v ) = v .Also, U ( v ) = w , and for each v i ∈ X \ { v } , U ( v i ) = v . We havecolor 3 as the free color. – | K | ≥ v ′ ∈ K \ { v } . We have two casesbased on the neighborhood of v ′ . ∗ v / ∈ N ( v ′ ).We assign C ( w ) = 3, C ( v ) = 2, C ( v ′ ) = d + 1 and the vertices(if any) in K \ { v, v ′ } the color 4.We have that U ( w ) = v , U ( v ) = w , and for each v i ∈ X \ { v } , U ( v i ) = v . Also U ( v ′ ) = v and for each y ∈ K \ { v ′ } , U ( y ) = v ′ .Color 4 is the free color. ∗ v ∈ N ( v ′ ).We assign C ( w ) = 1, C ( v ) = 4, C ( v ′ ) = d + 1 and the vertices(if any) in K \ { v, v ′ } the color 3.We have that U ( w ) = v , U ( v ) = w , and for each v i ∈ X \ { v } , U ( v i ) = v . Also U ( v ′ ) = v and for each y ∈ K \{ v ′ } , U ( y ) = v ′ .Color 3 is the free color. Subcase 1.3:
There exists a clique K = K , such that K = { w } anddeg( w ) = 1. 18LOG let N ( w ) = { v } . We have cases based on the size of the clique K . • | K | = 1.We assign C ( v ) = 2 and C ( w ) = 3. We have U ( v ) = U ( w ) = v . Also, U ( v ) = w , and for each v i ∈ X \ { v } , U ( v i ) = v . Color 4 is the freecolor. • | K | ≥ – There exists a vertex v ′ ∈ K \ { v } , such that v / ∈ N ( v ′ ).We assign C ( v ) = 2, C ( w ) = 3, C ( v ′ ) = d + 1 and the vertices (ifany) in K \ { v, v ′ } the color 4.We have that U ( w ) = v , U ( v ) = w , and for each v i ∈ X \ { v } , U ( v i ) = v . Also U ( v ′ ) = v and for each y ∈ K \ { v ′ } , U ( y ) = v ′ .Color 4 is the free color. – All the vertices in K are adjacent to v .Choose a vertex v ′ ∈ K \{ v } . Assign C ( w ) = 4, C ( v ) = 1, C ( v ′ ) = 3and the vertices (if any) in K \ { v, v ′ } the color d + 1.We have that U ( w ) = v , U ( v ) = v ′ , and for each v i ∈ X \ { v } , U ( v i ) = v . Also for each y ∈ K , U ( y ) = v . Color 4 is the free color. Subcase 1.4:
There exists a clique K = K , such that | K | ≥ G is connected, there is an edge between X and K . WLOG, weassume that v is adjacent to w ∈ K . We now divide the cases based on thesize of the clique K . • | K | = 1. – There exists a vertex w ′ ∈ K \ { w } such that v / ∈ N ( w ′ ).We assign C ( v ) = 2, C ( w ) = 3, C ( w ′ ) = d + 1, and the vertices (ifany) in K \ { w, w ′ } the color 4.We have that U ( w ′ ) = w , and for each y ∈ K \ { w ′ } , U ( y ) = w ′ .Also U ( v ) = v , U ( v ) = w , and for each v i ∈ X \ { v } , U ( v i ) = v .Color 4 is the free color. – All the vertices in K \ { w } are adjacent to v .We assign C ( v ) = 2, C ( w ) = 4, and the vertices in K \ { w } the color d + 1.We have that U ( v ) = U ( w ) = v and for each y ∈ K \ { w } , U ( y ) = v . Also U ( v ) = w , and for each v i ∈ X \ { v } , U ( v i ) = v . Color 1is the free color. • | K | ≥ – There exists a vertex v ′ ∈ K \ { v } such that v / ∈ N ( v ′ ).There are two subcases based on the neighborhood of the vertices in K . 19 There exists a vertex w ′ ∈ K \ { w } such that v / ∈ N ( w ′ ).We assign C ( v ) = 2, C ( v ′ ) = d + 1, C ( w ) = 3, C ( w ′ ) = d + 1 andthe vertices (if any) in K \ { v, v ′ } and K \ { w, w ′ } the color 4.We have that U ( v ′ ) = v , and for each y ∈ K \ { v ′ } , U ( y ) = v ′ . Also U ( w ′ ) = w , and for each y ∈ K \ { w ′ } , U ( y ) = w ′ .Moreover, U ( v ) = w , and for each v i ∈ X \ { v } , U ( v i ) = v .Color 4 is the free color. ∗ All the vertices in K \ { w } are adjacent to v .We assign C ( v ) = 2, C ( v ′ ) = d + 1, C ( w ) = 4, the vertices (ifany) in K \ { v, v ′ } the color 1 and the vertices in K \ { w } thecolor d + 1.We have that U ( v ′ ) = v , and for each y ∈ K \ { v ′ } , U ( y ) = v ′ .Also U ( w ) = v , and for each y ∈ K \{ w } , U ( y ) = v . Moreover, U ( v ) = w , and for each v i ∈ X \ { v } , U ( v i ) = v . Color 1 is thefree color. – All the vertices in K are adjacent to v .We choose v ′ ∈ K \ { v } arbitrarily. We assign C ( v ) = 1, C ( v ′ ) = 3and the vertices (if any) in K \ { v, v ′ } the color 4. We leave theclique K uncolored for now.We have that for each y ∈ K , U ( y ) = v . Moreover, U ( v ) = v ′ ,and for each v i ∈ X \ { v } , U ( v i ) = v . Color 4 is the free color. Case 2:
For all y ∈ V \ X , we have deg X ( y ) < | X | . And there is a vertex v ∈ V \ X such that deg X ( v ) ≥ v ∈ K where K is a clique in G [ V \ X ]. Recall that since G [ X ] is 1-regular, it follows that | X | ≥
4. We have the following cases depending on theneighborhood of v . • N ( v ) ∩ X contains a pair of adjacent vertices. WLOG, let v , v ∈ N ( v ). Since deg X ( v ) < | X | , WLOG we assume v / ∈ N ( v ). – | K | = 1.We assign C ( v ) = 2. We get that U ( v ) = v and U ( v ) = U ( v ) = v .For each v i ∈ X \ { v , v } , the uniquely colored neighbor is the loneneighbor of v i in X . The color 3 is the free color. – | K | ≥ ∗ There exists a vertex v ′ ∈ K \ { v } such that v / ∈ N ( v ′ ).We assign C ( v ) = 2, C ( v ′ ) = d + 1 and the vertices (if any) in K \ { v, v ′ } the color 4.We have that U ( v ′ ) = v , and for all y ∈ K \{ v ′ } , U ( y ) = v ′ . Also U ( v ) = U ( v ) = v , and for each v i ∈ X \ { v , v } , the uniquelycolored neighbor is the lone neighbor of v i in X . We have color4 as the free color. 20 All vertices in K \ { v } are adjacent to v .Let v ′ be arbitrarily chosen from K \ { v } . We assign C ( v ) = 1, C ( v ′ ) = 3 and the vertices (if any) in K \ { v, v ′ } the color d + 1.We have that U ( v ) = v , and for all y ∈ K \{ v } , U ( y ) = v . Also U ( v ) = v ′ , U ( v ) = U ( v ) = v , and for all v i ∈ X \ { v , v , v } ,the uniquely colored neighbor is the lone neighbor of v i in X .We have color 4 as the free color. • None of the vertices in N ( v ) ∩ X are adjacent to each other. WLOG v , v ∈ N ( v ) which implies v , v / ∈ N ( v ). – | K | = 1.We assign C ( v ) = 2. We get that U ( v ) = v and U ( v ) = U ( v ) = v .For each v i ∈ X \ { v , v } , the uniquely colored neighbor is the loneneighbor of v i in X . The color 3 is the free color. – | K | ≥ ∗ There exists a v ′ ∈ K \ { v } such that v / ∈ N ( v ′ ).We assign C ( v ) = 2, C ( v ′ ) = d + 1 and the vertices (if any) in K \ { v, v ′ } the color 3.We get that U ( v ′ ) = v and for all y ∈ K \ { v ′ } , U ( y ) = v ′ .Moreover, U ( v ) = U ( v ) = v , and for each v i ∈ X \ { v , v } , theuniquely colored neighbor is the lone neighbor of v i in X . Thecolor 3 is the free color. ∗ Every vertex in K is adjacent to v .We assign C ( v ) = 4 and the vertices in K \ { v } the color d + 1.We get that for all y ∈ K , U ( y ) = v . Moreover, U ( v ) = U ( v ) = v , and for each v i ∈ X \ { v , v } , the uniquely coloredneighbor is the lone neighbor of v i in X . The color 3 is the freecolor. Case 3:
Every vertex v ∈ V \ X has deg X ( v ) ≤ G is connected, there is a v ∈ V \ X such that deg X ( v ) = 1. Let v ∈ K where K is a clique in G [ V \ X ]. WLOG let N ( v ) ∩ X = { v } . • | K | = 1.We assign C ( v ) = 1. We get that U ( v ) = v . Also, U ( v ) = v and foreach v i ∈ X \ { v } , the uniquely colored neighbor is the lone neighbor of v i in X . The color 2 is the free color. • | K | ≥ – There exists a vertex v ′ ∈ K \ { v } such that v / ∈ N ( v ′ ).We assign C ( v ) = 1, C ( v ′ ) = d + 1 and the vertices (if any) in K \ { v, v ′ } the color 2.We get that U ( v ′ ) = v and for all y ∈ K \ { v ′ } , U ( y ) = v ′ . Moreover, U ( v ) = v , and for each v i ∈ X \ { v } , the uniquely colored neighboris the lone neighbor of v i in X . The color 2 acts as the free color.21 All vertices in K are adjacent to v .We assign C ( v ) = 3 and the vertices in K \ { v } the color d + 1.We get that for all y ∈ K , U ( y ) = v . Moreover, U ( v ) = v , and foreach v i ∈ X \{ v } , the uniquely colored neighbor is the lone neighborof v i in X . The color 2 acts as the free color.In Subcase 1.1, G has been assigned a full CFON coloring using d + 1 col-ors. In all the other cases (and subcases therein), G has been partially coloredsatisfying the rules of Lemma 9. The uncolored cliques K ∈ G [ V \ X ] can becolored with the application of Lemma 9, yielding a full CFON coloring of G that uses d + 1 colors. Let X = { v , v , . . . , v d } and Y = { v i ∈ X : deg X ( v i ) ≥ } . By the conditionsin the statement of the lemma, we have X \ Y = ∅ and for each vertex v ∈ Y ,deg X ( v ) = deg Y ( v ) = 1. For each vertex v i ∈ X , we assign the color C ( v i ) = i . High Level Idea:
We have four cases depending on how the vertices in V \ X interact with X \ Y and Y . In each case, we choose a vertex v ∈ K forsome clique K ⊆ G [ V \ X ]. We assign colors to the vertices in K such thatall the vertices in K and N ( v ) ∩ X have a uniquely colored neighbor, whilesatisfying the rules of Lemma 9. In particular, we identify a free color from theabove partial coloring. We use Lemma 9 to color the remaining vertices andobtain a CFON coloring of G .The key obstacle here is that while coloring the clique K , we could end upassigning the free color or the color d + 1 to multiple vertices of K . There couldexist vertices v i ∈ X \ Y , such that N ( v i ) ⊆ K and all the vertices in N ( v i ) areassigned the free color and the color d + 1. This may leave the vertex v i withouta uniquely colored neighbor. Hence, while coloring K , we need to handle thesevertices separately. Let S K be the set of such vertices.Formally, S K = { v i ∈ X \ Y : N ( v i ) ⊆ K } \ N ( v ). The vertices in N ( v ) ∩ ( X \ Y ) rely on v for their uniquely colored neighbor and hence does not requirespecial attention.Lemma 14 shows that we can color K in such a way that all the vertices in S K have a uniquely colored neighbor, and satisfying all the rules of Lemma 9.Lemma 14 will be proved after completing the proof of Lemma 13. For now, weshall assume Lemma 14 and proceed. Lemma 14.
Let G = ( V, E ) , X, Y be as above. Let v ∈ K where K is a cliquein G [ V \ X ] such that | K | ≥
2. Let C ( v i ) = i for all v i ∈ X and all the verticesin K \ { v } are uncolored. Suppose C ( v ) is assigned and the free color f isidentified, in such a way that v relies on a color other than f as the unique colorin its neighborhood. Then K can be colored in such a way that all the verticesin S K have a uniquely colored neighbor, and satisfying all the rules of Lemma9. 22t will be convenient to denote an application of Lemma 14 by the 4-tuple,( v, C ( v ) , f, K ). For example, we will say “applying Lemma 14 to ( v, , , K )”to denote an application of Lemma 14 where v ∈ K , C ( v ) = 1 and 3 is the freecolor.We have four cases based on the neighborhoods of the vertices in V \ X . Case 1: There exists a vertex v ∈ V \ X such that | N ( v ) ∩ ( X \ Y ) | ≥ . Let v ∈ K , where K is a clique in G [ V \ X ]. WLOG let v , v ∈ X \ Y suchthat v , v ∈ N ( v ). • | K | = 1.We assign C ( v ) = 1 and we get that U ( v ) = v , U ( v ) = U ( v ) = v . Weget the color 2 as the free color. • | K | ≥ – S K = ∅ .We assign C ( v ) = 1 and we get that U ( v ) = v and for all v i ∈ N ( v ) ∩ ( X \ Y ), we have U ( v i ) = v . We now apply Lemma 14 to( v, , , K ) ensuring that K is colored, while taking the remainingvertices of S K into account. Color 2 is the free color. – S K = ∅ . ∗ There exists a vertex v ′ ∈ K \ { v } such that v / ∈ N ( v ′ ).Assign C ( v ) = 1, C ( v ′ ) = d + 1 and the vertices (if any) in K \ { v, v ′ } the color 2.We get that U ( v ′ ) = v and for all y ∈ K \ { v ′ } , U ( y ) = v ′ . Also U ( v ) = U ( v ) = v . Color 2 is the free color. ∗ Every vertex in K is adjacent to v .Assign C ( v ) = 2 and the vertices in K \ { v } the color d + 1.We get that U ( y ) = v for all y ∈ K . Also U ( v ) = U ( v ) = v .Color 1 is the free color.In all the above cases, for each v i ∈ Y , the uniquely colored neighbor is the loneneighbor of v i in X . Case 2: There exists a vertex v ∈ V \ X such that | N ( v ) ∩ ( X \ Y ) | = 1 and N ( v ) ∩ Y contains a pair of vertices that are adjacent to each other. Since Case 1 is already addressed, we assume for all y ∈ V \ X , we have | N ( y ) ∩ ( X \ Y ) | ≤ v ∈ K , where K is a clique in G [ V \ X ]. WLOG let N ( v ) ∩ ( X \ Y ) = { v } ,and let v , v ∈ N ( v ) ∩ Y such that { v , v } ∈ E ( G ). • | K | = 1.We assign C ( v ) = 1. We get that U ( v ) = v and U ( v ) = U ( v ) = U ( v ) = v . We have the color 2 as the free color. • | K | = 2.Let K = { v, v ′ } . We have the following cases.23 S K = ∅ .We assign C ( v ) = 1 and we get that U ( v ) = v , U ( v ) = U ( v ) = U ( v ) = v . We apply Lemma 14 to ( v, , , K ). We have the color 2as the free color. – S K = ∅ . Subcase 1: v / ∈ N ( v ′ ).We assign C ( v ) = 1 and C ( v ′ ) = d + 1. We get that U ( v ) = v , U ( v ′ ) = v and U ( v ) = U ( v ) = U ( v ) = v . We have color 2 as thefree color. Subcase 2: v ∈ N ( v ′ ). That is, N ( v ′ ) ∩ ( X \ Y ) = { v } .We first check if there exists a clique b K ⊆ G [ V \ X ] such that N ( v ) ∩ b K = ∅ or N ( v ) ∩ b K = ∅ .If there is no such clique b K , we reassign C ( v ) = 2, assign C ( v ) = 3and C ( v ′ ) = d + 1. We get that U ( v ) = v and U ( v ) = U ( v ) = U ( v ) = U ( v ′ ) = v . Color 2 is the free color.Else, there exists a clique b K such that N ( v ) ∩ b K = ∅ or N ( v ) ∩ b K = ∅ . WLOG let N ( v ) ∩ b K = ∅ . We assign C ( v ) = 2 and C ( v ′ ) = d + 1.Now the vertex v does not have a uniquely colored neighbor. Let w ∈ N ( v ) ∩ b K .If S b K = ∅ , we assign C ( w ) = 3. We have U ( v ) = U ( v ′ ) = v , U ( v ) = U ( v ) = v , U ( v ) = w and U ( w ) = v . Due to the case definition, | N ( w ) ∩ ( X \ Y ) | ≤
1. For the lone vertex v i ∈ N ( w ) ∩ ( X \ Y ) (if itexists), we have U ( v i ) = w . We now apply Lemma 14 to ( w, , , b K )to color the remaining vertices b K taking care of the vertices in S b K .We have color 1 as the free color.Else if S b K = ∅ , we do the following . ∗ There exists a vertex w ′ ∈ b K \ { w } , such that v / ∈ N ( w ′ ).We assign C ( w ) = 3, C ( w ′ ) = d + 1 and the vertices (if any) in b K \ { w, w ′ } the color 1.We get that U ( v ) = U ( v ′ ) = v , U ( v ) = U ( v ) = v , U ( v ) = w , U ( w ′ ) = w and for all vertices x ∈ b K \ { w ′ } , U ( x ) = w ′ . Wehave the color 1 as the free color. ∗ For each x ∈ b K , we have v ∈ N ( x ).We assign C ( w ) = 1 and the rest of the vertices (if any) in b K \{ w } the color d + 1.We get that U ( v ) = U ( v ′ ) = v , U ( v ) = U ( v ) = v , U ( v ) = w ,and for all vertices x ∈ b K , U ( x ) = v . Color 3 is the free color. • | K | ≥ One may wonder about the possibility of vertices v i ∈ X \ Y such that N ( v i ) ⊆ K ∪ b K ,and be concerned that these vertices v i do not feature in S K or S b K . We note that there areno such vertices v i . This is because, we have N ( v ) ∩ ( X \ Y ) = N ( v ′ ) ∩ ( X \ Y ) = { v } inorder to be in Subcase 2. S K = ∅ .We assign C ( v ) = 1 and we get that U ( v ) = v , U ( v ) = U ( v ) = U ( v ) = v . We apply Lemma 14 to ( v, , , K ). We have the color 2as the free color. – S K = ∅ . ∗ There exists a vertex v ′ ∈ K \ { v } such that v / ∈ N ( v ′ ).We assign C ( v ) = 1, C ( v ′ ) = d + 1 and the vertices in K \ { v, v ′ } the color 3.We get that U ( v ′ ) = v and for all y ∈ K \ { v ′ } , U ( y ) = v ′ . Also U ( v ) = U ( v ) = U ( v ) = v . We have the color 3 as the freecolor. ∗ Every vertex in K is adjacent to v .Choose two vertices v ′ , v ′′ ∈ K \{ v } and assign C ( v ) = 1, C ( v ′ ) =2, C ( v ′′ ) = d + 1 and the vertices (if any) in K \ { v, v ′ , v ′′ } thecolor 3.We get that U ( v ′′ ) = v ′ and for all y ∈ K \ { v ′′ } , U ( y ) = v ′′ .Also U ( v ) = U ( v ) = U ( v ) = v . We have the color 3 as thefree color.In each of the above cases, for each v i ∈ Y \ { v , v } , the uniquely coloredneighbor is the lone neighbor of v i in X . Case 3: There exists a vertex v ∈ V \ X such that | N ( v ) ∩ ( X \ Y ) | = 1 and | N ( v ) ∩ Y | ≥ . Moreover, none of the vertices in N ( v ) ∩ Y areadjacent to each other. Let v ∈ K for a clique K ⊆ G [ V \ X ]. WLOG let v ∈ N ( v ) ∩ ( X \ Y )and v ∈ N ( v ) ∩ Y . Let v be the lone neighbor of v in Y . It follows that v / ∈ N ( v ). • | K | = 1.We assign C ( v ) = 1 and we get that U ( v ) = v , U ( v ) = U ( v ) = v and U ( v ) = v . Color 3 is the free color. • | K | ≥ – S K = ∅ .We assign C ( v ) = 1 and we get that U ( v ) = v , U ( v ) = U ( v ) = v and U ( v ) = v . We apply Lemma 14 to ( v, , , K ). Color 3 is thefree color. – S K = ∅ . ∗ There exists a vertex v ′ ∈ K \ { v } such that v ∈ N ( v ′ ).We assign C ( v ) = 1, C ( v ′ ) = d + 1 and the vertices (if any) in K \ { v, v ′ } the color 3.We get that U ( v ′ ) = v and for all y ∈ K \ { v ′ } , U ( y ) = v ′ . Also U ( v ) = U ( v ) = v and U ( v ) = v . We have color 3 as the freecolor. 25 None of the vertices in K \ { v } are adjacent to v .We assign C ( v ) = 2 and the vertices in K \ { v } the color d + 1.We get that U ( v ) = v and for all y ∈ K \ { v } , U ( y ) = v . Also U ( v ) = U ( v ) = v and U ( v ) = v . We have color 3 as the freecolor.In each of the above cases, for each v i ∈ Y \ { v , v } , the uniquely coloredneighbor is the lone neighbor of v i in X . Case 4: For each y ∈ V \ X such that | N ( y ) ∩ ( X \ Y ) | = 1 , we have | N ( y ) ∩ Y | = 0 . Since Case 1 is addressed, we assume that each vertex in V \ X has at most1 neighbor in X \ Y .Since G is connected and since | X \ Y | ≥
1, we can choose a clique K ⊆ G [ V \ X ] with distinct vertices v, v ′ ∈ K such that N ( v ) ∩ ( X \ Y ) = ∅ and N ( v ′ ) ∩ Y = ∅ . WLOG let v ∈ N ( v ) ∩ ( X \ Y ) and v ∈ N ( v ′ ) ∩ Y . Let v bethe lone neighbor of v in Y . It follows that v , v / ∈ N ( v ). • S K = ∅ .We assign C ( v ) = 3, C ( v ′ ) = d + 1 and the vertices (if any) in K \ { v, v ′ } the color 1. We get that U ( v ′ ) = v and for all y ∈ K \ { v ′ } , U ( y ) = v ′ .Also U ( v ) = v , U ( v ) = v and U ( v ) = v . Color 1 is the free color. • S K = ∅ .We assign C ( v ) = 3 and we have U ( v ) = v , U ( v ) = v and U ( v ) = v .Recall that S K = { v i ∈ X \ Y : N ( v i ) ⊆ K } \ N ( v ). For each v i ∈ S K ,choose a vertex w i ∈ N ( v i ), assign C ( w i ) = i and let U ( v i ) = w i . Sinceeach vertex in V \ X has at most 1 neighbor in X \ Y , it also follows that N ( v i ) ∩ N ( v i ′ ) = ∅ for any two vertices v i , v i ′ ∈ S K .Because of the condition of Case 4, N ( v ′ ) ∩ ( X \ Y ) = ∅ . Assign C ( v ′ ) = d + 1, and assign the color 1 to all the remaining uncolored vertices (ifany) in K . We have U ( v ′ ) = v and for all y ∈ K \ { v ′ } , U ( y ) = v ′ . Color1 is the free color .In each of the above cases, for each v i ∈ Y \ { v , v } , the uniquely coloredneighbor is the lone neighbor of v i in X .We have concluded the four cases. In each of the cases we have a free color f . We use Lemma 9 to get a uniquely colored neighbor for remaining verticesin X \ Y , the cliques in G [ V \ X ] and thereby obtain a CFON coloring. Lemma 14 (Restated) . Let G = ( V, E ) be a graph and X = { v , v , . . . , v d } ⊆ V be a set of vertices such that G [ V \ X ] is a disjoint union of cliques. Let This is where we make use of the assumption that X \ Y is nonempty. Because of the definition of Case 4, it follows that v = v ′ . One may wonder why we did not apply Lemma 14 to ( v, , , K ) in this situation. This isbecause Lemma 14 requires v to rely on a color other than the free color as the unique colorin its neighborhood. There is no assignment that meets this requirement. = { v i ∈ X : deg X ( v i ) ≥ } . Let v ∈ K where K is a clique in G [ V \ X ] suchthat | K | ≥
2. Let C ( v i ) = i for all v i ∈ X and all the vertices in K \ { v } areuncolored. Suppose C ( v ) is assigned and the free color f is identified, in such away that v relies on a color other than f as the unique color in its neighborhood.Then K can be colored in such a way that all the vertices in S K have a uniquelycolored neighbor, and satisfying all the rules of Lemma 9. Proof.
Recall that S K = { v i ∈ X \ Y : N ( v i ) ⊆ K } \ N ( v ). Let S K = { v j , v j , · · · , v j m } , for some m ≥
1. For each v i ∈ S K , choose an uncoloredvertex w i ∈ N ( v i ), assign C ( w i ) = i and let U ( v i ) = w i . If all the vertices in N ( v i ) are colored, we arbitrarily choose a vertex in N ( v i ) as U ( v i ). WLOGlet the colors used in K because of the above process be { j , j , · · · , j m ′ } where m ′ ≤ m . Note that the vertex v is colored prior to the application of this lemma,and has a uniquely colored neighbor as well. Hence we do not talk about C ( v )and U ( v ) in this proof.We have the following cases based on the number of uncolored vertices in K . • All the vertices in K are colored.Each vertex in K \ { v } was colored because it was chosen as w i by some v i ∈ S K . Hence U ( w i ) = v i . • K contains exactly one uncolored vertex.Let the uncolored vertex in K be v ′ . If v ′ has a uniquely colored neighbor,we assign C ( v ′ ) = d + 1. Now, every vertex in K has a uniquely coloredneighbor.If v ′ does not have a uniquely colored neighbor, we have two cases de-pending on the number of colors used in K . – m = m ′ .This means that each vertex v i ∈ S K chose a neighbor w i ∈ K and as-signed the color i to it. So v ′ sees each of the the colors j , j , · · · , j m twice in its neighborhood. This means that v j , v j , · · · , v j m ∈ N ( v ′ ).Recall that f is the free color, where 1 ≤ f ≤ d , and hence v f ∈ X sees a color other than f as the unique color in its neighborhood. Wedo the following to obtain a uniquely colored neighbor for v ′ : ∗ There exists a vertex v ′′ ∈ K \ { v, v ′ } such that v f ∈ N ( v ′′ ) ∩ X .Let C ( v ′′ ) = k due to a vertex v k ∈ N ( v ′′ ) ∩ S K . Assign C ( v ′ ) = k and reassign C ( v ′′ ) = d +1. We have that U ( v ′ ) = v k , U ( v ′′ ) = v f and we reassign U ( v k ) = v ′ . ∗ None of the vertices in K \ { v, v ′ } is adjacent to v f .Note that v f / ∈ N ( v ′ ), else v f would have served as a uniquelycolored neighbor for v ′ .Choose a vertex v ′′ ∈ K \ { v, v ′ } . Suppose C ( v ′′ ) = k and thisimplies that v k ∈ N ( v ′′ ) ∩ S K . We reassign C ( v ′′ ) = f and assign27 ( v ′ ) = d + 1. We have U ( v ′ ) = v k and we reassign U ( v k ) = v ′ .The assignment of d + 1 as the unique color in the neighborhoodof v k is an exception. However, this is fine as N ( v k ) is containedin K , and does not interact with any other cliques in G [ V \ X ]. – m > m ′ .This implies that there exists a vertex v j ∈ S K such that the color j is not given to any vertex in K . So v j must be seeing a vertex v ′′ ∈ K \ { v ′ } as its uniquely colored neighbor.We claim that v j / ∈ N ( v ′ ). If v j ∈ N ( v ′ ), then v j is the lone vertex in N ( v ′ ) that is colored j , and hence is a uniquely colored neighbor for v ′ . As per the scope of this case, v ′ does not have a uniquely coloredneighbor. This is a contradiction.We reassign C ( v ′′ ) = j and assign C ( v ′ ) = d + 1. We get that U ( v ′ ) = v ′′ . • K contains at least two uncolored vertices.We first check if there exists an uncolored vertex v ′ in K such that v ′ hasa uniquely colored neighbor other than v f . If such a v ′ exists, then weassign C ( v ′ ) = d + 1 and the remaining uncolored vertices in K the freecolor f . For all w ∈ K such that C ( w ) = f , we have U ( w ) = v ′ .If such a vertex v ′ does not exist, we have the following cases based onthe relation between m and m ′ . – m = m ′ .Choose a colored vertex w ∈ K \ { v } and an uncolored vertex v ′ ∈ K .Suppose C ( w ) = j , which means that v j ∈ N ( w ). Since v ′ does notsee a uniquely colored neighbor other than v f , it is the case that v j ∈ N ( v ′ ).We reassign C ( w ) = f , assign C ( v ′ ) = d + 1 and the remaininguncolored vertices (if any) in K the free color f . We get that U ( v ′ ) = v j and U ( v j ) = v ′ . All the vertices in K \ { v ′ } will have v ′ as theiruniquely colored neighbor.The assignment of d + 1 as the unique color in the neighborhood of v j is an exception. However, this is fine as N ( v j ) is contained in K ,and does not interact with any other cliques in G [ V \ X ]. – m > m ′ .This implies that there exists a vertex v j ∈ S K such that the color j was not used in K . This also implies that none of the uncoloredvertices in K have v j in their neighborhood. This is because if v j had an uncolored neighbor in K , then that neighbor would havebeen colored j in the coloring process performed at the beginning ofthis proof.We choose two uncolored vertices v ′ , v ′′ ∈ K and assign C ( v ′ ) = d +1, C ( v ′′ ) = j and the remaining uncolored vertices (if any) the color f .28e get that U ( v ′ ) = v ′′ and for all other vertices w ∈ K \ { v ′ } willhave U ( w ) = v ′ . References [1] Zachary Abel, Victor Alvarez, Erik D. Demaine, Sndor P. Fekete, AmanGour, Adam Hesterberg, Phillip Keldenich, and Christian Scheffer. Conflict-free coloring of graphs.
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