Chromatic bounds for the subclasses of pK_2-free graphs
aa r X i v : . [ m a t h . C O ] F e b Chromatic bounds for the subclasses of pK -free graphs Athmakoori Prashant , S. Francis Raj and M. Gokulnath Department of Mathematics, Pondicherry University, Puducherry-605014, India. : [email protected] [email protected] [email protected] Abstract
In this paper, we study the chromatic number for graphs with forbidden induced sub-graphs. We improve the existing χ -binding functions for some subclasses of K -free graphs,namely { K , H } -free graphs where H ∈ { K − e, K + P , K + C } . In addition,for p ≥ , we find the polynomial χ -binding functions { pK , H } -free graphs where H ∈{ gem, diamond, HV N, K − e, K + P , butterf ly, dart, gem + , C , K + C , P } . Key Words:
Coloring, Chromatic number, χ -binding funtion, pK -free graphs. All graphs considered in this paper are simple, finite and undirected. Let G be a graph withvertex set V ( G ) and edge set E ( G ) . For any positive integer k , a proper k -coloring of a graph G is a mapping c : V ( G ) → { , , . . . , k } such that for any two adjacent vertices u, v ∈ V ( G ) , c ( u ) = c ( v ) . If a graph G admits a proper k -coloring then G is said to be k -colorable. Thechromatic number, χ ( G ) , of a graph G is the smallest k such that G is k -colorable. All coloringsconsidered in this paper are proper. In this paper, P n , C n and K n respectively denotes the path, thecycle and the complete graph on n vertices. Also for S, T ⊆ V ( G ) , we define N T ( S ) = N ( S ) ∩ T where N ( S ) denotes the set of neighbours of S in G . For S, T ⊆ V ( G ) , let h S i denote thesubgraph induced by S in G and let [ S, T ] denote the set of all edges with one end in S and theother end in T . If every vertex in S is adjacent with every vertex in T , then [ S, T ] is said to becomplete. For any graph G , let G denotes the complement of G .Let F be a family of graphs. We say that G is F -free if it contains no induced subgraph whichis isomorphic to a graph in F . For two vertex-disjoint graphs G and G , the join of G and G ,denoted by G + G , is the graph whose vertex set V ( G + G ) = V ( G ) ∪ V ( G ) and the edgeset E ( G + G ) = E ( G ) ∪ E ( G ) ∪ { xy : x ∈ V ( G ) , y ∈ V ( G ) } . In this paper, we write H ⊑ G whenever H is an induced subgraph of G . A clique (independent set) in a graph G is a set1f pairwise adjacent (non-adjacent) vertices. The size of a largest clique (independent set) in G iscalled the clique number (independence number) of G , and is denoted by ω ( G ) (cid:0) α ( G ) (cid:1) .A graph G is said to be perfect if χ ( H ) = ω ( H ) , for every induced subgraph H of G . A class G of graphs is said to be χ -bounded [9] if there is a function f (called a χ -binding function) suchthat χ ( G ) ≤ f ( ω ( G )) , for every G ∈ G . We say that the χ -binding function f is special linear if f ( x ) = x + c where c is a constant. If c = 1 , then this special upper bound is called the Vizingbound for the chromatic number. There has been extensive research done on χ -binding functionfor various graph classes. See for instance, [10, 12, 13]. The study of χ -binding functions and χ -bounded graphs was initiated by Gyárfás in [9]. Before we look at more results in χ -bindingfunctions. Let us look at a famous result given by Erd˝os. Theorem 1.1 [6] For any positive integers k, l ≥ , there exists a graph G with g ( G ) ≥ l and χ ( G ) ≥ k .By Theorem 1.1, Gyárfás was able to deduced that there exists no χ -binding function for G ( H ) whenever H contains a cycle. Hence, this led to the conclusion that the only possibility when G ( H ) has a χ -binding function is when H is acyclic and which gave rise to the following conjecture. Conjecture 1.2 [9] G ( H ) is χ -boundforeveryfixed forest H .Our work was motivated by the following(unsolved) problem posed by Gyárfás. Problem 1.3 [9]What istheorderofmagnitudeofthesmallest χ -bindingfunctionfor G (2 K ) ?In [15], Wagon showed that the class of pK -free graphs admit an O ( ω p − ) χ -binding functionfor any p ∈ N . In particular, he showed that the χ -binding function for K -free graphs is (cid:0) ω +12 (cid:1) and the best known lower bound is R ( C ,K ω ( G )+1 )3 , where R ( C , K ω ( G )+1 ) denotes the smallest k such that every graph on k vertices contains either a clique of size x + 1 or K . The lower boundis non-linear as Chung in [5] proved that R ( C , K t ) is atleast t ǫ for some ǫ > . It is extremelyinteresting to see that Brause.et.al. in [4] proved that the class of (2 K , H ) -free graphs, where H is a graph with independence number α ( H ) ≥ , does not admit a linear χ -binding function. Thisraised the question if there existed any linear subfamilies of K -free graphs.In [11], Karthick and Mishra proved that the families of { K , H } -free graphs, where H ∈{ HV N, diamond, gem, K + C , P , P ∪ P , K − e } admit special linear χ -binding functions. Inparticular they proved that { K , gem } -free, { K , HV N } -free, { K , K − e } -free and { K , K + C } -free admit to χ -binding functions ω ( G ) + 1 , ω ( G ) + 3 , ω ( G ) + 4 and ω ( G ) + 5 respectively.Brause.et.al [4] improved the χ -binding function for { K , gem } -free graphs to max { , ω ( G ) } .They also proved that for s = 1 or ω ( G ) = 2 the class of { K , ( K + K ) + K s } -free with ω ( G ) ≥ s is perfect and the class of { K , K + K r } -free graphs with ω ( G ) ≥ r for some2 b bb bb bbb bb bbb b bbb b b bbb b b b bb b b diamond dart gem butterfly HV N gem + Figure 1: Some special graphsinteger r ≥ is perfect. Clearly when s = 2 and r = 3 , ( K ∪ K ) + K s ∼ = HV N and K + K r ∼ = K − e which implies that the class of { K , HV N } -free and { K , K − e } -freegraphs are perfect for ω ( G ) ≥ and ω ( G ) ≥ respectively which improves the bounds given in[11].Our Motivation was to see how the bounds for these graph classes would change when { K , H } -free was substituted with { pK , H } -free graphs where p ∈ N .Throughout this paper, we use a particular partition of the vertex set of a graph G as de-fined initially by Wagon in [15] and improved by Bharathi et.al., in [1] as follows. Let A = { v , v , . . . , v ω ( G )= ω } be a maximum clique of G . Let us define the lexicographic ordering onthe set L = { ( i, j ) : 1 ≤ i < j ≤ ω } in the following way. For two distinct elements ( i , j ) , ( i , j ) ∈ L , we say that ( i , j ) precedes ( i , j ) , denoted by ( i , j ) < L ( i , j ) if ei-ther i < i or i = i and j < j . For every ( i, j ) ∈ L , let C i,j = { v ∈ V ( G ) \ A : v / ∈ N ( v i ) ∪ N ( v j ) }\{ S ( i ′ ,j ′ ) < L ( i,j ) C i ′ ,j ′ } . Note that, for any k ∈ { , , . . . , j − }\{ i } , [ v k , C i,j ] iscomplete. Hence ω ( h C i,j i ) ≤ ω ( G ) − j + 2 .For ≤ i ≤ ω , let us define I i = { v ∈ V ( G ) \ A : v ∈ N ( a ) , for any a ∈ A \{ v i }} . Since A is a maximum clique, for ≤ i ≤ ω , I i is an independent set and for any x ∈ I i , xv i / ∈ E ( G ) .Clearly, each vertex in V ( G ) \ A is non-adjacent to at least one vertex in A . Hence those verticeswill be contained either in I i for some integer i , ≤ i ≤ ω , or in C i,j for some ( i, j ) ∈ L . Thus V ( G ) = A ∪ (cid:18) S ≤ i ≤ ω I i (cid:19) ∪ S ( i,j ) ∈ L C i,j ! .In this paper, we begin by giving simpler proofs for the class of (2 K , gem ) -free and (2 K , HV N ) -free graphs and improve the bounds for the class of { K , K + C } -free graphs to ω ( G ) + 1 - when ω ≥ and the class of { K , K − − e } -free graphs to ω ( G ) when ω ≥ . We also prove that thatthe class of { K , K + P } -free graphs is ω ( G ) + 2 -colorable. Further, we prove that the familiesof { pK , H } -free graphs, where H ∈ { gem, diamond, HV N, K − e, K + P } admit linear χ -binding functions and the family of { pK , H } -free graphs, where H ∈ { butterf ly, dart, gem + } admit quadratic χ -binding functions. In addition, we show that the families of { pK , H } -freegraphs, where H ∈ { C , K + C , P } admit O ( ω p − ) χ -binding functions. Some graphs that areconsidered as a forbidden induced subgraphs in this paper are shown in Figure 1.Notations and terminologies not mentioned here are as in [16].3 Coloring of some classes of K -free graphs and pK -freegraphs Let us start this section by recalling some χ -binding result due to Wagon in [15], Gaspers andHuang [8] and Brandt in [3] Theorem 2.1 [15]If G is a K -free graph, then χ ( G ) ≤ (cid:0) ω ( G )+12 (cid:1) . Theorem 2.2 [8] If G isa K -free graph such that ω ( G ) ≤ ,then χ ( G ) ≤ . Theorem 2.3 [3] For p ≥ , if G is a pK -free graph suchthat ω ( G ) = 2 , then χ ( G ) ≤ p − . Theorem 2.4
For p ≥ ,if G isa pK -free graphsuchthat ω ( G ) = 3 ,then χ ( G ) ≤ p − p + 4 . Proof.
Let G be a graph pK -free graph with ω ( G ) = 3 , then V ( G ) = A ∪ (cid:18) S ≤ i ≤ I i (cid:19) ∪ ( C , ∪ C , ∪ C , ) and each of C , , C , and C , are ( p − K -free. Also, by the definiton of C i,j ω ( C , ) ≤ , ω ( C , ) ≤ and ω ( C , ) ≤ . We shall prove the result by induction on p ≥ .Clearly, the vertices of A ∪ (cid:18) S ≤ i ≤ I i (cid:19) , can be colored with { , , } by assigning the color i to the vertices in { v i } ∪ I i .For p = 3 , by using Theorem 2.1 and Theorem 2.2, C , and C , can be colored with colorseach and C , can be colored with colors. Therefore, on the whole G can be colored with atmost . colors. For s ≥ , let us assume that if G ′ is an sK -free graph with ω ( G ′ ) = 3 ,then χ ( G ′ ) ≤ s − s + 4 .Let us assume that G is an ( s + 1) K -free graph and ω ( G ) = 3 . By using Theorem 2.3, C , and C , can be colored with s − colors each and by our assumption C , can be colored withatmost s − s + 4 colors. Therefore, V ( G ) can be colored with s −
2) + 2 s − s + 4 =2( s + 1) − s + 1) + 4 colors.Hence, if G is a pK -free graph with ω ( G ) = 3 , then χ ( G ) ≤ p − p + 4 .Let us prove some properties of gem -free graphs in Lemma 2.5. Lemma 2.5 If G isa gem -freegraphand V ( G ) = A ∪ (cid:18) S ≤ i ≤ ω I i (cid:19) ∪ S ( i,j ) ∈ L C i,j ! ,foranyintegers i, j such that i < j and j ≥ thefollowingholds.(i) h C i,j i is P -free and henceperfect.(ii) For any l ∈ { , , . . . , ω } , if H is a component in h C i,j i and let a ∈ V ( H ) such that av l ∈ E ( G ) , then [ V ( H ) , v l ] is complete.(iii) For a ∈ C i,j if av l / ∈ E ( G ) , where l ∈ { , , . . . , ω } then [ a, I l ] = ∅ .4iv) If H isa componentof C i,j , then ω ( H ) ≤ | A \ N A ( V ( H )) | . Proof. (i) Suppose there exists a P ⊑ h C i,j i , say P . Since j ≥ , then there exists an integer s ∈ { , , . . . , j }\{ i, j } such that N ( v s ) ⊇ V ( P ) and hence, h V ( P ) ∪ { v s }i ∼ = gem , acontradiction.(ii) On the contrary, let H be a component of h C i,j i and a, b ∈ V ( H ) such that ab, av l ∈ E ( G ) but bv l / ∈ E ( G ) for some l ∈ { , , . . . , ω } . Since j ≥ , there exists an integer s ∈{ , , . . . , j }\{ i, j } such that N ( v s ) ⊇ { a, b, v i , v j , v l } and thus h{ v s , b, a, v l , v j }i ∼ = gem , acontradiction.(iii) Suppose there exist vertices a ∈ C i,j and b ∈ I l such that av l / ∈ E ( G ) and ab ∈ E ( G ) .Let k ∈ { i, j }\{ l } . Since j ≥ , there exists an integer s ∈ { , , . . . , j }\{ i, j } such that N ( v s ) ⊇ { a, b, v k , v l } and hence h{ v s , a, b, v k , v l }i ∼ = gem , a contradiction.(iv) Suppose there exists a component H ⊑ h C i,j i such that ω ( H ) > | A \ N A ( V ( H )) | . By (ii) ofLemma 2.5, we have [ V ( H ) , N A ( V ( H ))] is complete. Hence ω ( h V ( H ) ∪ N A ( V ( H )) i ) = ω ( H ) + ω ( h N A ( V ( H )) i ) > | A \ N A ( V ( H )) | + | N A ( V ( H )) | = | A | = ω ( G ) , a contradiction.Let us now consider { K , gem } -free graphs. Theorem 2.6 has already been proved by C.Brauseet al [4]. We have given an alternative proof which is much simpler compared to the one given in[4]. Theorem 2.6
Let G bea { K , gem } -freegraph, then χ ( G ) ≤ max { , ω ( G ) } . Proof.
Let G be a { K , gem } -free graph and V ( G ) = A ∪ (cid:18) S ≤ i ≤ ω I i (cid:19) ∪ S ( i,j ) ∈ L C i,j ! . Since, G is K -free, h C i,j i is an independent set for every ( i, j ) ∈ L . If ω ( G ) = 2 , by using Theorem2.1, the result follows. Now, let us consider ω ( G ) ≥ . For j ≥ , we claim that S j − i =1 C i,j is an independent set. If there exists vertices a ∈ C i,j and b ∈ C k,j such that ab ∈ E ( G ) and i < k , then we can find an integer s ∈ { , , . . . , j }\{ i, k, j } such that av s , bv s ∈ E ( G ) andhence h{ v s , a, b, v i , v j }i ∼ = gem , a contradiction. Next, let us consider C , . If a ∈ C , such that [ a, I ] = ∅ , then we shall show that for p = 1 , [ a, { v p } ∪ I p ] = ∅ and [ a, C i,j ] = ∅ for every ( i, j ) ∈ L , ( i, j ) = (1 , (i.e.) N ( a ) ∈ I . Let a ∈ C , and b ∈ I such that ab ∈ E ( G ) . If s ∈ { , . . . , ω }\{ , } such that av s ∈ E ( G ) , then h{ v s , a, b, v , v }i ∼ = gem , a contradiction.If there exists vertices a ∈ C , and c ∈ C i,j such that ac ∈ E ( G ) , then by using the fact that [ a, A ] = ∅ . h{ a, c, v i , v j }i ∼ = 2 K , a contradiction.Let the set of colors be { , , . . . , ω } . For ≤ i ≤ ω , let us assign the color i to the vertices of { v i } ∪ I i . By using (iii) of Lemma 2.5, for j ≥ , assign the color j to the vertices of S j − i =1 C i,j andassign the colors and to the vertices of C , and C , respectively. Finally for C , , if a ∈ C , [ a, I ] = ∅ , then assign the color 1 to a or else assign the color 2. Clearly this is a propercoloring of G and thus χ ( G ) ≤ ω ( G ) .For p ≥ , let us define a sequence of functions f p : N → N to serve as a χ -binding functionfor { pK , gem } -free graphs as follows. For p ≥ , t, s ≥ , f p (1) = 1 , f (2) = 3 , f ( t ) = t , f t (2) = 2 t − , f t ( s ) = f t − ( s ) + 2 s − . Theorem 2.7
For p ≥ , if G is a { pK , gem } -freegraph, then χ ( G ) ≤ f p ( ω ( G )) . Proof.
For ω = 1 , the result is obvious. For ω = 2 , the result follows from Theorem 2.1 andTheorem 2.3. Now for ω ≥ , we prove the results by induction on p . For p = 2 , by usingTheorem 2.6, the result holds. By induction hypothesis, for s ≥ , let us assume that if G ′ is an { sK , gem } -free graph, then χ ( G ′ ) ≤ f s ( ω ( G ′ )) .Let G be an { ( s + 1) K , gem } -free graph. Let V ( G ) = A ∪ (cid:18) S ≤ i ≤ ω I i (cid:19) ∪ S ( i,j ) ∈ L C i,j ! . For ≤ i ≤ ω , let us assign the color i to the vertex v i and to the vertices of I i . For j ≥ , if H is a component of h C i,j i , then by using (i) of Lemma 2.5, χ ( H ) = ω ( H ) . Also, for i = 2 and j ≥ , by using (iii) and (iv) of Lemma 2.5, every component of h C i,j i can be colored using thecolors given to the vertices in A . We claim this to be a proper coloring. Suppose there exists twoadjacent vertices a ∈ C i,j and b ∈ C k,l which receives the same color, say q . Without loss ofgenerality, assume ( i, j ) < L ( k, l ) , then either i < k or i = k and j < l . If i < k , then bv i ∈ E ( G ) .Since the vertices a and b have been assigned the same color q , we have av q , bv q / ∈ E ( G ) . Also,for i = 2 , k = 2 and j, l ≥ , N ( v ) ⊇ { a, b, v i , v q } and thus h{ v , a, b, v i , v q }i ∼ = gem , acontradiction. If j < l , then bv j ∈ E ( G ) and by similar arguments, we get a contradiction.For any l ≥ , we know that ω ( h C ,l i ) ≤ ω ( G ) − . For i = 2 and j ≥ , we have [ v , C i,j ] iscomplete. By using (iii) of Lemma 2.5, the color is available for all the vertices of ∪ ωl =3 C ,l . Forany r > s ≥ , we claim that [ C ,r , C ,s ] = ∅ . If there exist vertices a ∈ C ,r and b ∈ C ,s suchthat ab ∈ E ( G ) , then h{ v , b, a, v s , v }i ∼ = gem , a contradiction. By using (i) of Lemma 2.5, wecan color all the vertices of ∪ ωl =3 C ,l with the color together with at most ω − new set of colors.Thus the vertices of V ( G ) \ C , can be colored using ω − colors. Since h C , i is { sK , gem } -free, by induction hypothesis, it can be colored with f s ( ω ) colors. Therefore, V ( G ) can be coloredwith at most f s ( ω ) + 2 ω − f s +1 ( ω ) colors.Since diamond is an induced subgraph of gem , for p ≥ , f p ( ω ( G )) will be the χ -bindingfunction for { pK , diamond } -free graphs as well. Corollary 2.8
For p ≥ ,if G isa { pK , diamond } -freegraph, then χ ( G ) ≤ f p ( ω ( G )) .In [11], Karthick and Mishra have showed that if G is a { K , HV N } -free graph, then χ ( G ) ≤ ω ( G ) + 3 . In Theorem 2.9, we reduce the bound to ω ( G ) for all ω ≥ .6 heorem 2.9 If G isa { K , HV N } -free graph, suchthat ω ≥ ,then χ ( G ) = ω ( G ) . Proof.
Let G be a { K , HV N } - free graph such that ω ( G ) ≥ . Since, G is K -free, h C i,j i is an independent set for every ( i , j ) ∈ L . We claim that [ I i , I j ] = complete , for ≤ i, j ≤ ω ( G ) . If there exists vertices a ∈ I i and b ∈ I j such that ab / ∈ E ( G ) , then there exists integers p, q ∈ { , , . . . , ω }\{ i, j } such that h{ a, b, v j , v p , v q }i ∼ = HV N . Now, for j ≥ , we shall firstshow that C i,j = ∅ . If there exists a vertex a ∈ C i,j , then there exist at least two integers s, q ∈{ , , . . . , j }\{ i, j } such that av s , av q ∈ E ( G ) and h{ a, v s , v q , v i , v j }i ∼ = HV N , a contradiction.Thus V ( G ) = A ∪ C , ∪ C , ∪ C , ∪ (cid:18) S ≤ i ≤ ω I i (cid:19) . By using similar arguments we can show that N A ( C , ) = { v } , N A ( C , ) = { v } and if a vertex a ∈ C , then | [ a, A ] | ≤ . Moreover, we have [ C , , I k ] = ∅ for k = 2 , [ C , , I l ] = ∅ for l = 1 and if a vertex a ∈ C , , such that [ a, I p ] = ∅ then [ a, I q ] = ∅ for p = q . In particular, if av s ∈ E ( G ) then [ a, I q ] = ∅ for s = q . Otherwise, for k = 2 ,there exists a ∈ C , and b ∈ I k such that ab ∈ E ( G ) . Since ω ( G ) ≥ and N A ( C , ) = { v } ,there exist at least two integers s, q ∈ { , , . . . , ω }\{ , k } such that av s , av q / ∈ E ( G ) and thus h{ a, v , b, v s , v q }i ∼ = HV N , a contradiction. Similarly, we can show that [ C , , I l ] = ∅ , for l = 1 and if a vertex a ∈ C , such that [ a, I p ] = ∅ then [ a, I q ] = ∅ for p = q and if av s ∈ E ( G ) then [ a, I q ] = ∅ for s = q .Let the set of colors be { , , . . . , ω } . For ≤ i ≤ ω , let us assign the color i to the vertex v i and to all the vertices of I i and assign the colors and to the vertices of C , and C , respectivelyand we can color the vertices of C , with the colors and respectively. Clearly, this is a propercoloring of G and thus χ ( G ) ≤ ω ( G ) .C.Brause et al in [4] have already proved that if p = 1 or ω ( G ) = 2 and G is a { K , ( K ∪ K ) + K p } -free graph with ω ( G ) ≥ p , then G is a perfect. Theorem 2.9 becomes a corollary tothis when p = 2 but the proof given by us is much simpler compared to the one given in [4].Note that, all those properties mentioned in Theorem 2.9 are due to the fact that G is a HV N -free and hence those properties are also valid in Theorem 2.10.For p ≥ , let us define a sequence of functions f p : N → N to serve as a χ -binding functionfor { pK , HV N } -free graphs as follows. For p ≥ , t ≥ , m, s ≥ , f p (1) = 1 , f (2) = 3 , f (3) = 4 , f t (2) = 2 t − , f ( s ) = s , f t (3) = 2 t − t + 4 , f t ( m ) = f t − ( m ) + 2 f t − ( m − . Theorem 2.10
For p ≥ ,if G isa { pK , HV N } -free graph,then χ ( G ) ≤ f p ( ω ( G )) . Proof.
For ω = 1 , the result is obvious. For ω = 2 , the result follows from Theorem 2.1 andTheorem 2.3. For ω = 3 , the result follows from Theorem 2.2 and Theorem 2.4. Now for ω ≥ ,let us prove the results by induction on p . For p = 2 , by using Theorem 2.9, the result holds.By induction hypothesis, for s ≥ , let us assume that if G ′ is an { sK , HV N } -free graph, then χ ( G ′ ) ≤ f s ( ω ( G ′ )) . 7et G be an { ( s + 1) K , HV N } -free graph. For ω ≥ , we have V ( G ) = A ∪ C , ∪ C , ∪ C , ∪ (cid:18) S ≤ i ≤ ω I i (cid:19) from Theorem 2.9. For ≤ i ≤ ω , assign the color i to the vertex v i and tothe vertices of I i . Hence A ∪ (cid:18) S ≤ i ≤ ω I i (cid:19) can be colored with ω ( G ) colors. Clearly, for ( i, j ) ∈ L ,each h C i,j i is { sK , HV N } -free and ω ( h C , i ) ≤ ω ( G ) − and ω ( h C , i ) ≤ ω ( G ) − .Since N A ( C , ) = { v } and N A ( C , ) = { v } , we can reuse the color for the vertices in C , and reuse the color for the vertices in C , . Since [ C , , I k ] = ∅ , for k = 2 and [ C , , I k ] = ∅ ,for k = 1 , we can reuse the colors { , , . . . , ω } for the vertices of C , ∪ C , . By using inductionhypothesis, the vertices in C , and C , ∪ C , can be colored with at most f s ( ω ) and f s ( ω − − ω colors respectively. Therefore, V ( G ) can be colored with at most ω + f s ( ω ) + 2 f s ( ω − − ω = f s ( ω ) + 2 f s ( ω −
1) = f s +1 ( ω ) colors.In [11], Karthick and Mishra have showed that if G is a { K , K − e } -free graph, then χ ( G ) ≤ ω ( G ) + 4 . In Theorem 2.11, we reduce the bound to for ω ( G ) = 4 and we prove that G is ω ( G ) -colorable, for all ω ( G ) ≥ . Theorem 2.11 If G is a { K , K − e } -free graph,then χ ( G ) ≤ ( for ω ( G ) = 4 ω ( G ) for ω ( G ) ≥ . Proof.
Let G be a { K , K − e } -free graph. Since, G is K -free, h C i,j i is an independentset for every ( i, j ) ∈ L For ω ≥ , we claim that I i = ∅ , for all ≤ i ≤ . For some i , ≤ i ≤ , if there exists a vertex a ∈ I i , then there exist integers p, q and r other than i suchthat h{ a, v p , v q , v r , v i }i ∼ = K − e , a contradiction. Let us partition the graph into six independentsets, namely, V = v ∪ C , , V = v ∪ C , , V = v ∪ C , , V = v ∪ C , , V = C , and V = C , . Hence G is -colorable for ω = 4 . Let us consider ω ≥ . For all j ≥ , we claimthat C i,j = ∅ . For some j , j ≥ , if there exists a vertex a ∈ C i,j , then there exist integers p, q, r ∈ { , , . . . , j }\{ i, j } such that h{ a, v p , v q , v r , v i }i ∼ = K − e , a contradiction.Thus V ( G ) = A ∪ C , ∪ C , ∪ C , ∪ C , ∪ C , ∪ C , . For ≤ i ≤ , we prove that N A ( C i, ) = { v , v , v }\{ v i } . If there exists an integer j ≥ and for some vertex a ∈ C i, suchthat av j ∈ E ( G ) , then h{ a, v , v , v , v j }i ∼ = K − e , a contradiction. Moreover, for ≤ i, k ≤ ,we can show that [ C i, , C k, ] = ∅ . On contrary, for some ≤ i, k ≤ , if there exist vertices a ∈ C i, , b ∈ C k, such that ab ∈ E ( G ) , then h{ a, b, v , v }i ∼ = 2 K , a contradiction. Now, let usus partition the graph into ω independent sets in the following way, V = v ∪ C , , V = v ∪ C , , V = v ∪ C , , V = v ∪ C , ∪ C , ∪ C , , and for ≤ i ≤ ω , V i = v i . Hence G is perfect forall ω ≥ .Next, for p ≥ , let us define a sequence of functions f p : N → N to serve as a χ -bindingfunction for { pK , K − e } -free graphs as follows. For p ≥ , t ≥ , m ≥ , s ≥ , f p (1) = 1 , f (2) = 3 , f (3) = 4 , f (4) = 6 , f ( s ) = s , f t (2) = 2 t − , f t (3) = 2 t − t + 4 , f t ( m ) = t − ( m ) + 2 f t − ( m −
1) + 3 m − . Theorem 2.12 If G is a { pK , K − e } -free graph, then χ ( G ) ≤ f p ( ω ( G )) . Proof.
For ω = 1 , the result is trivial. For ω = 2 , the result follows from Theorem 2.1 andTheorem 2.3. For ω = 3 , the reult follows from Theorem 2.2 and Theorem 2.4. Now for ω ≥ ,let us prove the results by induction on p . For p = 2 , by using Theorem 2.11, the result holds.By induction hypothesis, for s ≥ , let us assume that if G ′ is an { sK , K − e } -free graph, then χ ( G ′ ) ≤ f s ( ω ( G ′ )) .Let G be an { ( s + 1) K , K − e } -free graph. If ω ≥ , then as discussed in Theorem 2.11, wehave V ( G ) = A ∪ C , ∪ C , ∪ C , ∪ C , ∪ C , ∪ C , and for ≤ i ≤ , N A ( C i, ) = { , , }\{ i } .For ≤ i ≤ , we claim that C i, is a union of cliques. On contrary, for some integer i , ≤ i ≤ ,if h C i, i contains an induced P , say P , then h V ( P ) ∪ { v , v , v }\{ v i }i ∼ = K − e , a contradiction.Hence for ≤ i ≤ , each C i, is ( ω − -colorable. Next for ≤ i ≤ ω , let us assign thecolor i to the vertex v i . Since [ v , C , ] is complete and [ v , C , ] is complete, h C , i and h C , i are { sK , diamond } -free. While coloring the remaining vertices of G , we can reuse the colors , and for the vertices of C , , C , and C , respectively. Altogether, we can reuse the colors { , , . . . , ω } for the vertices of C , ∪ C , ∪ C , . Note that, h C , i is an { sK , K − e } -freesubgraph of G . Hence let us assign a new set of colors to the vertices of C , , C , ∪ C , and C , ∪ C , ∪ C , with at most f s ( ω ) , f s ( ω − and ω − − ω colors. Therefore, G can becolored with at most ω + f s ( ω )+2 f s ( ω − ω − − ω = f s ( ω )+2 f s ( ω − ω − f s +1 ( ω ) colors. Lemma 2.13 If G is a { K , K + P } -free graph and V ( G ) = A ∪ (cid:18) S ≤ i ≤ ω I i (cid:19) ∪ S ( i,j ) ∈ L C i,j ! ,thenthefollowingholdforevery C i,j except C , , C , and C , .(i) h C i,j i is P -free and henceit isperfect.(ii) For j < l , let H be a component in h C i,j i and let a ∈ V ( H ) such that av l ∈ E ( G ) , then [ V ( H ) , v l ] is complete.(iii) For l ≤ ω ,let a ∈ C i,j suchthat av l / ∈ E ( G ) ,then [ a, I l ] = ∅ .(iv) If H isa componentof C i,j , then ω ( H ) ≤ | A \ N A ( V ( H )) | . Proof.
Similar to the proofs of lemma 2.5.
Theorem 2.14 If G is a { K , K + P } -free graph, suchthat ω ≥ ,then χ ( G ) ≤ ω ( G ) + 2 . Proof.
Let G be a { K , K + P } -free graph. Since, G is K -free, h C i,j i is an independentset for every ( i, j ) ∈ L . If ω ( G ) ≥ , we claim that [ I i , I j ] = complete ,for ≤ i, j ≤ ω ( G ) .9f there exists vertices a ∈ I i and b ∈ I j such that ab / ∈ E ( G ) , then there exists integers p, q ∈{ , , , . . . , ω } such that h{ v p .v q , a, b, v i , v j }i ∼ = K + P . Further we also claim that, if thereexists a vertex a ∈ C , such that [ a, I ] = ∅ , then [ a, I p ] = ∅ for p ≥ , [ a, A \{ v } ] = ∅ and [ a, C , ] = ∅ . If there exists vertices a ∈ C , , b ∈ I , c ∈ I p for p ≥ and d ∈ C , such that ab, ac ∈ E ( G ) , then h{ v , c, a, b, v , v }i ∼ = K + P . Similarly, we can also show that [ a, A \{ v } ] = ∅ . If ab, ad ∈ E ( G ) , then by using the result stated above h{ a, d, v , v }i ∼ = 2 K .Let us consider ω ( G ) ≥ , then for j ≥ C i,j , C k,j ] = ∅ . If there exists vertices a ∈ C i,j and b ∈ C k,j such that ab ∈ E ( G ) , then there exists integers p, q ∈ { , , . . . , j }\{ i, k, j } such that h{ v p , v q , a, b, v i , v j }i ∼ = K + P .Let the set of colors be { , , . . . , ω +2 } . For ≤ i ≤ ω , let us assign the color i to the vertex v i and to the vertices of I i . For j ≥ , assign the color j to the vertices of C i,j for all i < j and assignthe colors , , , ω + 1 and ω + 2 to the vertices of C , , C , , C , , C , and C , respectively. Fora vertex a ∈ C , if [ a, I ] = ∅ , then assign it the color 3 or else assign it the color 2. Clearly thisis a proper coloring of G and thus χ ( G ) ≤ ω ( G ) + 2 .Next, for p ≥ , let us define a sequence of functions f p : N → N to serve as a χ -bindingfunction for { pK , K + P } -free graphs as follows. For p ≥ , m, t ≥ , s ≥ , f p (1) = 1 , f (2) = 3 , f (3) = 4 , f ( s ) = s + 2 , f t (2) = 2 t − , f t (3) = 2 t − t + 4 , f t ( m ) = f t − ( m ) + 2 f t − ( m −
1) + 3 m − Theorem 2.15 If G is a { pK , K + P } -freegraphs, then χ ( G ) ≤ f p ( ω ( G )) . Proof.
For ω = 1 , the result is trivial. For ω = 2 , the result follows from Theorem 2.1 andTheorem 2.3. For ω = 3 , the reult follows from Theorem 2.2 and Theorem 2.4. Now for ω ≥ ,let us prove the results by induction on p . For p = 2 , by using Theorem 2.14, the result holds. Byinduction hypothesis, for s ≥ , let us assume that if G ′ is an { sK , K + P } -free graph, then χ ( G ′ ) ≤ f s ( ω ( G ′ )) .Let G be an { ( s + 1) K , K + P } -free graph. Let V ( G ) = A ∪ (cid:18) S ≤ i ≤ ω I i (cid:19) ∪ S ( i,j ) ∈ L C i,j ! .For ≤ i ≤ ω , let us assign the color i to the vertex v i and to the vertices of I i . For j ≥ , if H is a component of h C i,j i , then by using (i) of Lemma 2.13, χ ( H ) = ω ( H ) . Also, for i ≥ ,by using (iii) and (iv) of Lemma 2.13, every component of h C i,j i can be colored using the colorsgiven to the vertices in A . We claim this to be a proper coloring. Suppose there exists two adjacentvertices a ∈ C i,j and b ∈ C k,l which receives the same color, say q . Without loss of generality,assume ( i, j ) < L ( k, l ) , then either i < k or j < l . If i < k , then bv i ∈ E ( G ) . Since thevertices a and b have been assigned the same color q , we have av q , bv q / ∈ E ( G ) . Also, for i, k ≥ , N ( v ) ∪ N ( v ) ⊇ { a, b, v i , v q } and thus h{ v , v , a, b, v i , v q }i ∼ = K + P , a contradiction. If j < l ,then bv j ∈ E ( G ) and by similar arguments, we get a contradiction.For any l ≥ , we know that ω ( h C ,l i ) ≤ ω ( G ) − . For i ≥ , we have [ { v , v } , C i,j ] is10omplete. By using (iii) of Lemma 2.13, the color is available for all the vertices of ∪ ωl =4 C ,l . Forany r > s ≥ , we claim that [ C ,r , C ,s ] = ∅ . If there exist vertices a ∈ C ,r and b ∈ C ,s suchthat ab ∈ E ( G ) , then h{ v , v , b, a, v s , v }i ∼ = K + P , a contradiction. By using (i) of Lemma2.13, we can color all the vertices of ∪ ωl =4 C ,l with the color together with at most ω − new setof colors. Similarly, for l ≥ , we can color all the vertices of ∪ ωl =4 C ,l with the color togetherwith another new set of atmost ω − colors.Thus the vertices of V ( G ) \ ( C , ∪ C , ∪ C , ) can be colored using ω − colors. Clearly h C i,j i is { sK , K + P } -free for every ( i, j ) ∈ L . Since, [ v , C , ] = complete and [ v , C , ] = complete both h C , i and h C , i are { sK , gem } -free and can be colored with f s ( ω − colors each. Also, h C , i is { sK , gem } -free, by induction hypothesis, it can be colored with f s ( ω ) colors. Therefore, V ( G ) can be colored with at most f s ( ω ) + 2 f s ( ω −
1) + 3 ω − f s +1 ( ω ) colors.Next, for p ≥ , let us define a sequence of functions f p : N → N to serve as a χ -bindingfunction for { pK , butterf ly } -free graphs as follows. For p ≥ , m, t ≥ , s ≥ , f p (1) = 1 , f (2) = 3 , f (3) = 4 , f ( s ) = (cid:0) s +12 (cid:1) , f t (2) = 2 t − , f t ( m ) = f t − ( m ) + (cid:0) m +12 (cid:1) − Theorem 2.16 If G is a { pK , butterf ly } -freegraph, then χ ( G ) ≤ f p ( ω ( G )) . Proof.
For ω = 1 , the result is trivial. For ω = 2 , the result follows from Theorem 2.1 andTheorem 2.3. Now for ω ≥ , let us prove the results by induction on p . For p = 2 , by usingTheorem 2.1 and Theorem 2.2, the result holds. By induction hypothesis, for s ≥ , let us assumethat if G ′ is an { sK , butterf ly } -free graph, then χ ( G ′ ) ≤ f s ( ω ( G ′ )) .Let G be an { ( s + 1) K , butterf ly } -free graph. For ≤ i ≤ ω , let us assign the color i to thevertex v i and to all the vertices of I i . Clearly, h C , i is { sK , butterf ly } -free. So let us assign anew set of colors to the vertices of C , with at most f s ( ω ) colors. For j ≥ , we claim that C i,j is an independent set. If there exists an edge ab ∈ E ( C i,j ) , j ≥ , then there exists an integer p ∈ { , , . . . , j }\{ i, j } such that h{ a, b, v p , v i , v j } ∼ = butterf ly .Therefore, G can be colored with at most ω + f s ( ω ) + ω P j =3 ( j −
1) = f s +1 ( ω ) + (cid:0) ω +12 (cid:1) − colors.Next, for p ≥ , let us define a sequence of functions f p : N → N to serve as a χ -bindingfunction for { pK , dart } -free graphs as follows. For p ≥ , m, t ≥ , s ≥ , f p (1) = 1 , f (2) = 3 , f (3) = 4 , f ( s ) = (cid:0) s +12 (cid:1) , f t (2) = 2 t − , f t ( m ) = f t − ( m ) + (cid:0) m +12 (cid:1) + m − . Theorem 2.17 If G is a { pK , dart } -freegraph, then χ ( G ) ≤ f p ( ω ( G )) . Proof.
Let us prove the result by induction on p . For ω = 1 , the result is trivial. For ω = 2 ,the result follows from Theorem 2.1 and Theorem 2.3. Now for ω ≥ , let us prove the resultsby induction on p . For p = 2 , by using Theorem 2.1 and Theorem 2.2, the result holds. By11nduction hypothesis, for s ≥ , let us assume that if G ′ is an { sK , dart } -free graph, then χ ( G ′ ) ≤ f s ( ω ( G ′ )) .Let G be an { ( s + 1) K , dart } -free graph. By using Theorem 2.1 and Theorem 2.3, the resultis true for ω ( G ) = 2 . Let us consider ω ≥ . For ≤ i ≤ ω , let us assign the color i to thevertex v i and to all the vertices of I i . Clearly, h C , i is { sK , dart } -free. So let us assign a newset of colors to the vertices of C , with at most f s ( ω ) colors. Since (cid:2) v , ∪ ωj =3 C ,j (cid:3) is complete, ω (cid:0) h∪ ωj =3 C ,j i (cid:1) ≤ ω ( G ) − .We claim that h∪ ωj =3 C ,j i is P -free. Suppose there exists a P ⊑ (cid:10) ∪ ωj =3 C ,j (cid:11) , say P . Then h V ( P ) ∪ { v , v }i ∼ = dart , a contradiction. Now, let us assign a new set of colors to the vertices of ∪ ωj =3 C ,j with at most ω − colors. Similarly, we can prove that for ≤ i ≤ ω − , h∪ ωj = i +1 C i,j i is P -free and ω (cid:0) h∪ ωj = i +1 C i,j i (cid:1) ≤ ω ( G ) − i + 1 . For all i , ≤ i ≤ ω − , let us assign a new setof colors to the vertices of ∪ ωj = i +1 C i,j with at most ω − i + 1 colors. Therefore, G can be coloredwith at most ω + f s ( ω ) + ω − ω − P i =2 i = f s ( ω ) + (cid:0) ω +12 (cid:1) + ω − f s +1 ( ω ) colors. Theorem 2.18 If G is a { pK , gem + } -freegraph, then χ ( G ) ≤ f p ( ω ( G )) . Proof.
The bound for the chromatic number of { pK , dart } -free graph holds for { pK , gem + } -free graphs too. By using techniques used in Theorem 2.17, we can show that h∪ ωj =3 C ,j i is P -freeand hence perfect, by a result of [14]. Hence, we can properly color h∪ ωj =3 C ,j i with atmost ω − colors. Similarly, we can prove that for ≤ i ≤ ω − , h∪ ωj = i +1 C i,j i is P -free and can be coloredusing atmost ω ( G ) − i + 1 colors and entire G with atmost f s +1 ( ω ) colors as in the proof ofTheorem 2.17.Next, let us recall a result due to Blazsik et.al., in [2]. Theorem 2.19 [2]If G is a { K , C } -free graph,then χ ( G ) ≤ ω ( G ) + 1 .For p ≥ , let us define a sequence of functions f p : N → N to serve as a χ -binding functionfor { pK , C } -free graphs as follows. For p, s ≥ , m, t ≥ , f p (1) = 1 , f ( s ) = s + 1 , f t (2) = 2 t − , f t ( m ) = m P i =2 f t − ( i ) . Theorem 2.20 If G is a { pK , C } -free graph,then χ ( G ) ≤ f p ( ω ( G )) . Proof.
For ω = 1 , the result is trivial. For ω = 2 , the result follows from Theorem 2.1 andTheorem 2.3. Now for ω ≥ , let us prove the results by induction on p . For p = 2 , by usingTheorem 2.19, the result holds. By induction hypothesis, for s ≥ , let us assume that if G ′ is an { sK , C } -free graph, then χ ( G ′ ) ≤ f s ( ω ( G ′ )) .Let G be an { ( s + 1) K , C } -free graph. For ω ≥ , assign the color i to the vertex v i for12 ≤ i ≤ ω . For all ( i, j ) ∈ L , h C i,j i is an { sK , C } -free and the colors and can be reusedwhile coloring the vertices of C , . Hence by using induction hypothesis, each vertex in C , canbe colored using at most f s ( ω ) − colors. For j ≥ , we claim that [ C i,j , C k,j ] = ∅ . Supposethere exist vertices a ∈ C i,j and b ∈ C k,j such that ab ∈ E ( G ) , then h{ a, v k , v i , b }i ∼ = C , acontradiction. Hence, for any h < j , the color j can be reused while coloring the vertices of C h,j . By induction hypothesis, for all j ≥ and i < j , we can color the vertices of C i,j withat most f s ( ω − j + 2) − new set of colors together with the color j . Also, for ≤ i, j ≤ ω ,we claim that [ I i , I j ] = ∅ . On contrary, if there exist integers i, j such that a ∈ I i , b ∈ I j and ab ∈ E ( G ) , then h{ a, v j , v i , b }i ∼ = C , a contradiction. Hence, for all i , ≤ i ≤ ω , let us assigna new color to the vertices of I i . Therefore, G can be colored with at most ω + f s ( ω ) − (cid:18) ω P i =3 [ f s ( ω − i + 2) − (cid:19) + 1 = (cid:18) ω P i =2 f s ( i ) (cid:19) + 1 = f s +1 ( ω ) colors.In [11], Karthick and Mishra have showed that if G is a { K , K + C } -free graph, then χ ( G ) ≤ ω ( G ) + 5 . In Theorem 2.21, we improve the bound to ω ( G ) + 1 . Theorem 2.21 If G is a { K , K + C } -free graph suchthat ω ( G ) ≥ ,then χ ( G ) ≤ ω ( G ) + 1 . Proof.
Let G be a { K , K + C } -free graph and V ( G ) = A ∪ (cid:18) S ≤ i ≤ ω I i (cid:19) ∪ S ( i,j ) ∈ L C i,j ! .Since, G is K -free, h C i,j i is an independent set for every ( i, j ) ∈ L Let the set of colors be { , , . . . , ω ( G ) + 1 } . For ≤ i ≤ ω , assign the color i to the vertex v i . For ≤ i, j ≤ ω , weclaim that [ I i , I j ] = ∅ . On contrary, let us assume that for some integers i, j , there exist vertices a ∈ I i and b ∈ I j such that ab ∈ E ( G ) . Since ω ≥ , there exists an integer s = { i, j } such that N ( v s ) ⊇ { a, b, v i , v j } and thus h{ v s , a, v j , v i , b }i ∼ = K + C , a contradiction. For all i , ≤ i ≤ ω ,let us assign the color ω + 1 to the vertices of I i . For j ≥ , we show that [ C i,j , C k,j ] = ∅ . Supposethere exist vertices a ∈ C i,j and b ∈ C k,j such that ab ∈ E ( G ) . Since j ≥ , there exists an integer s ∈ { , , . . . , j }\{ i, k, j } such that N ( v s ) ⊇ { a, b, v i , v k } and thus h{ v s , a, v k , v i , b }i ∼ = K + C ,a contradiction. For all j ≥ and i < j , let us assign the color j to the vertices of C i,j andassign the colors , and to the vertices of C , , C , and C , respectively. Hence G is ( ω + 1) -colorable.For p ≥ , let us define a sequence of functions f p : N → N to serve as a χ -binding functionfor { pK , K + C } -free graphs as follows. For p, s ≥ , m, t ≥ , f p (1) = 1 , f ( s ) = s + 1 , f t (2) = 2 t − , f t ( m ) = (cid:18) m P i =2 f t − ( i ) (cid:19) + f t − ( m −
1) + 1 . Theorem 2.22 If G is a { pK , K + C } -freegraph, then χ ( G ) ≤ f p ( ω ( G )) . Proof.
For ω = 1 , the result is trivial. For ω = 2 , the result follows from Theorem 2.1 andTheorem 2.3. Now for ω ≥ , let us prove the results by induction on p . For p = 2 , by using13heorem 2.21, the result holds. By induction hypothesis, for s ≥ , let us assume that if G ′ is an { sK , K + C } -free graph, then χ ( G ′ ) ≤ f s ( ω ( G ′ )) .Let G be an { ( s + 1) K , K + C } -free graph. For ω ≥ , let us assign the color i to the vertex v i for ≤ i ≤ ω . Clearly, for ( i, j ) ∈ L , each h C i,j i is { sK , K + C } -free. Also, the color , , can be reused while coloring the vertices of C , , C , and C , . Thus the vertices of C , , C , and C , can be colored with at most f s ( ω ) − , f s ( ω − − and f s ( ω − − new set of colorsrespectively. As discussed in Theorem 2.21, we have for j ≥ , i = k , [ C i,j , C k,j ] = ∅ and ∪ ωi =1 I i is an independent set. Hence the color j can be reused while coloring the vertices of C h,j , for any h < j . For each j ≥ and i < j , let us assign a new set of colors to the vertices of C i,j with atmost f s ( ω − j + 2) − colors together with the color j . Finally, let us assign a new color to allthe vertices of ∪ ωi =1 I i . Therefore, G can be colored with at most ω + f s ( ω ) − f s ( ω − − f s ( ω − − (cid:18) ω P i =4 f s ( ω − i + 2) − (cid:19) + 1 = (cid:18) ω P i =2 f s ( i ) (cid:19) + f s ( ω −
1) + 1 = f s +1 ( ω ) colors.Next, let us recall a result due to Fouquet et.al., in [7]. Theorem 2.23 [7]If G is a { K , P } -free graph,then χ ( G ) ≤ ω ( G ) .Next, for p ≥ , let us define a sequence of functions f p : N → N to serve as a χ -bindingfunction for { pK , P } -free graphs as follows. For p, s ≥ , m, t ≥ , f p (1) = 1 , f ( s ) = s , f t (2) = 2 t − , f t ( m ) = m + m P i =2 f t − ( i ) . Theorem 2.24 If G is a { pK , P } -free graph,then χ ( G ) ≤ f p ( ω ( G )) . Proof.
For ω = 1 , the result is trivial. For ω = 2 , the result follows from Theorem 2.1 andTheorem 2.3. Now for ω ≥ , let us prove the results by induction on p . For p = 2 , by usingTheorem 2.23, the result holds. By induction hypothesis, for s ≥ , let us assume that if G ′ is an { sK , P } -free graph, then χ ( G ′ ) ≤ f s ( ω ( G ′ )) .Let G be an { ( s + 1) K , P } -free graph. For ω ≥ , let us assign the color i to the vertex v i and to all the vertices of I i for ≤ i ≤ ω Clearly, for ( i, j ) ∈ L , each h C i,j i is { sK , P } -free.Let us assign a new set of colors to the vertices of C , with at most f s ( ω ) colors. For j ≥ , weclaim that [ C i,j , C k,j ] = ∅ . Suppose there exist vertices a ∈ C i,j and b ∈ C k,j such that ab ∈ E ( G ) ,then h{ a, v k , v i , b, v j }i ∼ = P , a contradiction. For each j ≥ and i < j , let us assign a new set ofcolors to the vertices of C i,j with at most f s ( ω − j + 2) colors. Therefore, G can be colored withat most ω + f s ( ω ) + ω P i =3 f s ( ω − i + 2) = ω + ω P i =2 f s ( i ) = f s +1 ( ω ) colors.14 cknowledgment For the first author, this research was supported by the Council of Scientific and Industrial Research, Gov-ernment of India, File No: 09/559(0133)/2019-EMR-I. And for the second author, this research was sup-ported by SERB DST, Government of India, File no: EMR/2016/007339. Also, for the third author, thisresearch was supported by the UGC-Basic Scientific Research, Government of India, Student id: [email protected].
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