Extremal colorings and independent sets

Transcription

1 Exremal colorings and independen ses John Engbers Aysel Erey Ocober 17, 2017 Absrac We consider several exremal problems of maximizing he number of colorings and independen ses in some graph families wih fixed chromaic number and order. Firs, we address he problem of maximizing he number of colorings in he family of conneced graphs wih chromaic number k and order n where k 4. I was conjecured ha exremal graphs are hose which have clique number k and size ( k 2 n k. We affirm his conjecure for 4-chromaic claw-free graphs and for all k-chromaic line graphs wih k 4. We also reduce his exremal problem o a finie family of graphs when resriced o claw-free graphs. Secondly, we deermine he maximum number of independen ses of each size in he family of n-verex k-chromaic graphs (respecively conneced n-verex k-chromaic graphs and n-verex k-chromaic graphs wih c componens. We show ha he unique exremal graph is K k E n k, K 1 (K k 1 E n k and (K 1 (K k 1 E n k c1 E c 1 respecively. 1 Inroducion and Saemen of Resuls Le G = (V (G, E(G be a finie simple graph. For an ineger x 1, a proper x-coloring of G, or simply x-coloring of G, is a funcion f : V (G {1,..., x} so ha f(v 1 f(v 2 for every v 1 v 2 E(G. We le π(g, x denoe he chromaic polynomial of G and so for posiive inegers x, π(g, x is simply he number of x-colorings of G. The chromaic number of G, denoed χ(g, is he smalles posiive ineger so ha π(g, x 0, and we say ha G is k-chromaic if χ(g = k. A graph G is called criical if χ(g v < χ(g for every verex v of G. A k-chromaic criical graph is called k-criical. I is easy o see ha if G is a k-criical graph hen δ(g k 1 and k-criical graphs are 2-conneced. Much recen work has invesigaed he quesion of maximizing he number of x-colorings over various families of graphs, including n-verex m-edge graphs [19, 20], n-verex 2-conneced graphs [10], conneced graphs wih fixed minimum degree [8, 17], biparie regular graphs [16], and regular graphs [6, 14, 15]. One family ha we focus on in his noe is he family of n-verex k-chromaic graphs. In his family Tomescu [25] showed ha he disjoin union of he complee graph K k wih he empy graph E n k uniquely maximizes π(g, x for all x k. When resricing o he se of conneced n-verex k-chromaic graphs (which we denoe by C k (n, he problem of deermining he maximum value of π(g, x for G C k (n seems o be much more difficul. The answer is rivial for k = 2, where he exremal graphs are rees (when x 3, and is known for k = 3 (see [24, 26]. For k 4, we have he following conjecure [7, 23]. Le Ck (n be he se conaining all n-verex graphs obained from a k-clique by recursively aaching leaves. john.engbers@marquee.edu; Deparmen of Mahemaics, Saisics and Compuer Science, Marquee Universiy, Milwaukee, WI aysel.erey@gmail.com; Deparmen of Mahemaics, Universiy of Denver, Denver, CO

2 Conjecure 1.1. Le k 4 and G C k (n. Then for every ineger x k we have wih equaliy if and only if G C k (n. π(g, x (x k (x 1 n k A complee answer o Conjecure 1.1 is no ye known, alhough i has been verified for k 4 and x large [2] and graphs wih he addiional consrain of having independence number a mos 2 [11] (equivalenly, graphs ha are complemens of riangle-free graphs. For k = 4 he conjecure is reduced o undersanding a finie number of graphs [12] and is also known o hold when he graphs are required o be planar [26]. I is no difficul o see ha Conjecure 1.1 holds for graphs G wih χ(g = ω(g. Therefore, when sudying his problem we only need o consider graphs whose chromaic number is differen from he clique number. An imporan family of such graphs is he family of claw-free graphs. So, in his paper we firs consider he graphs in C k (n which are addiionally claw-free. For k = 4 we obain a resul for all n. Theorem 1.2. Le G be a conneced n-verex claw-free 4-chromaic graph. Then for every ineger x 4 we have π(g, x (x 4 (x 1 n 4 wih equaliy if and only if G C 4 (n. For general k we have he following resul which reduces he problem o a finie family of graphs. Theorem 1.3. For every k 4, here exiss a finie family F of k-chromaic claw-free graphs such ha if every graph G in F saisfies π(g, x (x k (x 1 V (G k hen so does every conneced k-chromaic claw-free graph. We also consider line graphs or, equivalenly, edge colorings of graphs. A proper x-edge-coloring of G, or simply x-edge-coloring of G is a funcion f : E(G {1,..., x} so ha f(e 1 f(e 2 for all disinc edges e 1 and e 2 in E(G ha share an endverex. The chromaic index of G, denoed χ (G, is he smalles ineger x for which G has an x-edge-coloring. The line graph L(G of G is he graph whose verices represen he edges of G (i.e. V (L(G = E(G and ef is an edge of L(G if and only if e and f are adjacen edges of G. Observe ha χ (G = χ(l(g for every graph G. A graph G is called a line graph if here exiss a graph H such ha G = L(H. No every graph is a line graph and line graphs form a subfamily of claw-free graphs. We find he n-verex k-chromaic line graphs ha maximize he number of proper x-colorings for all n and k. Theorem 1.4. Le G be a conneced n-verex k-chromaic line graph wih k 4. Then for every ineger x k we have π(g, x (x k (x 1 n k wih equaliy if and only if G is obained from a K k by aaching pahs of sizes n 1,..., n k o he k verices where 0 n i n 4 for i = 1,..., k. Le k 3. A ree is called k-sarlike if i has exacly one verex wih degree k and all oher verices have degree a mos 2. An immediae consequence of he above heorem is he following exremal resul for proper edge-colorings. Corollary 1.5. For every ineger x k 4, k-sarlike rees maximize he number of x-edgecolorings in he family of conneced n-edge k-edge-chromaic graphs. 2

3 We also consider independen ses in his paper. A se I V (G is an independen se (or sable se if v 1, v 2 V (G implies ha v 1 v 2 / E(G. The size of an independen se I is I. Le i(g denoe he number of independen ses of G and i (G denoe he number of independen ses of size in G. The quaniy i(g has also been referred o as he Fibonacci number of G [22] (as hese values for he pah P n are Fibonacci numbers, or in he field of molecular chemisry, he Merrifield-Simmons index of G [21]. Noice ha each color class of a proper coloring of G is an independen se. There has also been a large amoun of work on invesigaing which graphs maximize i(g and i (G in various families of graphs, we refer he reader o wo surveys and he references herein [4, 29] for a summary of some of he resuls and conjecures in his area. In [18] he n-verex graph wih clique number ω conaining he maximum number of independen ses of each fixed size is found, along wih he characerizaion of uniqueness. Also, i was shown ha he Turan graph T n,k is he unique n-verex k-chromaic graph wih he minimum number of independen ses [28] and he minimum number of independen ses of each size [18] (where i is implici ha 2 n k. We nex find he n-verex k-chromaic graph ha has he maximum number of independen ses of each size. We remark ha when = 0 and = 1, all n-verex graphs have he same number of independen ses of size, and for an n-verex k-chromaic graph G we have α(g n k 1 and hence i (G = 0 for n k 2. Theorem 1.6. Le G be an n-verex k-chromaic graph. Then we have ( ( n k n k i (G k 1 For 2 n k 1 we have equaliy if and only if G = K k E n k. We remark ha he graph K k E n k also uniquely maximizes he number of proper colorings in his family. Theorem 1.6 immediaely gives he following. Corollary 1.7. Le G be an n-verex k-chromaic graph. Then we have wih equaliy if and only if G = K k E n k. i(g i(k k E n k = (k 12 n k As wih proper colorings, we now consider he conneced n-verex k-chromaic graph ha has he mos number of independen ses. Resuls on independen ses of size in graphs wih ω(g = k appear in [18] (while no explicily saed in Theorem 1.8 of [18], he maximizing resuls are for conneced graphs wih clique number k. Observe ha an independen se of size 2 induces an edge in he complemen of he graph. Therefore, maximizing he number of independen ses of size 2 is equivalen o minimizing he number of edges. And he laer problem was already solved. Theorem 1.8 ([23]. Le G be a conneced n-verex k-chromaic graph. Then we have ( n k i 2 (G (k 1(n k. 2 Furhermore, for k = 3, exremal graphs are unicyclic graphs wih an odd cycle, while for k 3, exremal graphs belong o C k (n. 3

4 When k = 2, he conneced 2-chromaic graphs ha maximize he number of independen ses of size are rees (as deleing edges from G canno decrease he number of independen ses of size. The maximizaion (and minimizaion of he number of independen ses of size in rees was solved for all by Wingard [27]; see also [18]. We generalize his o all k. Theorem 1.9. Le k 2 and le G be a conneced k-chromaic graph of order n. Then ( ( ( n k n k 0 i (G (k For 3 n k 1 we have equaliy if and only if G = K 1 (K k 1 E n k. This gives he following corollary, whose proof is included in Secion 3. Corollary Le G be a conneced k-chromaic graph wih n verices. Then i(g k2 n k 1 wih equaliy if and only if G = K 1 (K k 1 E n k. We can refine he exremal graphs based on he number of componens. Theorem Le G be an n-verex k-chromaic graph wih c componens. Then we have ( ( ( n k n k c 1 i (G (k For 3 n k 1 we have equaliy if and only if G = (K 1 (K k 1 E n k c1 E c 1. Corollary Le G be an n-verex k-chromaic graph wih c componens. Then we have i(g k2 n k 2 c 1 wih equaliy if and only if G = (K 1 (K k 1 E n k c1 E c 1. In he res of he paper we provide he proofs of hese resuls. In Secion 2, we sudy exremal colorings and give proofs of Theorems 1.2, 1.3 and 1.4. Secion 3 deals wih independen ses and we give proofs for Theorems 1.6, 1.9 and Lasly, in Secion 4 we pose several quesions on exremal colorings and independen ses. 2 Exremal Colorings In his secion, we presen he proofs of he resuls abou proper colorings. We begin by recalling a few resuls ha will be frequenly used in our proofs. We le ω(g denoe he size of he larges clique in G. The firs resul bounds he number of colorings of a k-chromaic graph which conains a clique of size k. Proposiion 2.1 ([11]. Le G C k (n and ω(g = k. Then for all inegers x k we have wih equaliy if and only if G C k (n. π(g, x (x k (x 1 n k The nex resul allows us o focus on subgraphs ha have nice properies. Proposiion 2.2 ([12]. If H is a conneced subgraph of a conneced graph G, hen for all x N we have π(g, x π(h, x(x 1 V (G V (H. 4

5 2.1 4-chromaic claw-free graphs In his secion, we presen a number of resuls which, a he end, are used o prove Theorem 1.2. To begin, le F n, be he family of graphs G wih n verices and n 2 edges such ha G conains an induced odd cycle C n and riangles such ha every riangle overlaps he cycle C n in an edge and no wo riangles share a common edge. I is easy o see ha if G F n,, hen π(g, x = (x 2 (x 1((x 1 n 1 1. Le G and H be wo graphs wih clique number a leas r. We le G r H denoe a graph which is obained from G and H by gluing hem a an r-clique. Proposiion 2.3. Le G be an n-verex 4-criical claw-free graph wih ω(g 4. Then for every ineger x 4 we have π(g, x < (x 4 (x 1 n 4. Proof of Proposiion 2.3. We shall consider wo cases. Case 1: G conains an odd hole. Le C r be an odd hole wih verices v 1,..., v r in sandard order. Subcase 1: There exis a verex u / V (C r such ha u is adjacen o hree consecuive verices of C r. If u is adjacen o all verices of C r, hen G has a subgraph H = K 1 C r whose chromaic polynomial is x π(c r, x 1 = x((x 2 r (x 2. By Proposiion 2.2, i suffices o show ha x ((x 2 r (x 2 (x 1 n r 1 < (x 4 (x 1 n 4 which is equivalen o (x 2 r 1 1 < (x 3 (x 1 r 2. Since r 5, o prove he laer i would be sufficien o show ha (x 2 4 < (x 3 (x 1 3. Calculaions show ha (x 3 (x 1 3 (x 2 4 has a posiive leading coefficien and is larges real roo is Hence we are done. Now suppose ha here is a verex v j of C r which is no adjacen o u. We may assume ha u is adjacen o v 1, v 2, v 3 and ha j 4. Since G is 4-criical, δ(g 3. So v j has a neighbor u which is no in V (C r. As G is claw-free, u mus be adjacen o v j 1 or v j1 (v 1 if r = 5; denoe one such neighbor by v. Le H be he subgraph of G wih verex se V (C r {u, u } and edge se E(C r {uv 1, uv 2, uv 3, u v j, u v}. Observe ha H v 1 v 3 = K4 2 H r 1,1 and H/v 1 v 3 = K3 1 H r 2,1 where H p,q denoes a graph in he family F p,q. I is easy o see ha π(k 4 2 H r 1,1, x = (x 1(x 2 2 (x 3((x 1 r 2 1 and π(k 3 1 H r 2,1, x = (x 1 2 (x 2 2 ((x 1 r 3 1. So by he edge addiion-conracion formula, π(h, x = π(h v 1 v 3, x π(h/v 1 v 3, x < (x 1 r 1 (x 2 3. Now by Proposiion 2.2, i suffices o show ha (x 2 2 x(x 3, which clearly holds for x 4. Subcase 2: There is no verex in G which is adjacen o hree consecuive verices of C r. Since δ(g 3 and C r is an induced subgraph, every verex of C r has a neighbor ouside of C r. Le u 1 be a verex such ha v 1 u 1 E(G and u 1 / V (C r. Since G is claw-free, eiher u 1 v 2 E(G or u 1 v r E(G. Wihou loss, we assume ha u 1 v 2 E(G. By he assumpion, u 1 canno be adjacen o v 3. So here exiss a verex u 3 such ha u 3 u 1 and u 3 v 3 E(G. Since G is claw-free, eiher u 3 v 2 E(G or u 3 v 4 E(G. So we shall consider wo cases again. Firs assume ha u 3 v 2 E(G. If here exiss a verex v j V (C r wih j 4 such ha v j has a neighbor, say w j, which is no in {u 1, u 3 }, hen w j would be adjacen o a neighbor of v j in C r, as G is claw free. So we would have a subgraph H of G which belongs o he family F r,3 and π(h, x = (x 2 3 (x 1((x 1 r 1 1. Now by Proposiion 2.2 i suffices o show ha (x 2 3 ((x 1 r 1 1(x 1 n r 2 is less han (x 4 (x 1 n 4 which follows from (x 2 2 x(x 3 5

6 for x 4 and (x 1 r 1 1 < (x 1 r 1. Now we may assume ha N G (v j \V (C r {u 1, u 3 }. Since here is no verex adjacen o hree consecuive verices of C r, we ge u 1 v r, u 1 v 3, u 3 v 4, u 3 v 1 / E(G. Also, u 1 v 4, u 3 v r E(G by he assumpions. Then we mus have r 7, since if r = 5 hen he verices u 1, v 3, v 4, v 5 would induce a claw. As G is claw-free, u 3 v r 1 and u 1 v 5 are in E(G. Now, we have four edge disjoin riangles wih verex ses {u 1, v 1, v 2 }, {u 3, v 2, v 3 }, {u 3, v r, v r 1 } and {u 1, v 4, v 5 }. Le H be a minimal subgraph conaining hese four riangles. I is easy o see ha π(h, x = x(x 1 4 (x 2 4 < (x 4 (x 1 5 and he resul follows by Proposiion 2.2. Now le us assume ha u 3 v 4 E(G (and so u 3 v 2 / E(G. As in he previous case, we may assume ha N G (v j \ V (C r {u 1, u 3 }. Again i mus be ha r 7, as r = 5 implies ha he neighbor of v 5 is adjacen o hree consecuive verices of C 5. Furhermore, by he assumpions we have u 1 v r, u 3 v 5 / E(G and u 3 v r, u 1 v 5 E(G. Now, since G is claw free eiher u 3 v r 1 E(G or u 3 v 1 E(G (oherwise {u 3, v r 1, v r, v 1 } would induce a claw. If u 3 v r 1 E(G (resp. u 3 v 1 E(G hen G has a subgraph H 1 (resp. H 2 show in Figure 1. In each case, i is easy o check ha π(h i, x < (x 4 (x 1 V (H 4 holds and we are done by Proposiion 2.2. v r 1 v r u 3 v 1 v 2 v 3 v 4 v 5 u 1 v 1 v 2 v 3 v 4 v 5 u 1 v r u 3 H 1 H 2 Figure 1: The graphs H 1 and H 2. Case 2: G does no conain an odd hole. By assumpion, G is no a perfec graph. So by he srong perfec graph heorem [3], G mus conain an odd ani-hole. The graph G canno conain an ani-hole of order 5 because C 5 = C5. Also, G canno conain an odd ani-hole of order larger han 7 because oherwise i would conain a K 4. Hence, G mus conain a C 7. Calculaions show ha π(c 7, x = x (x 1 (x 2 (x 3 ( x 3 8 x 2 25 x 29. Hence, by Proposiion 2.2, i suffices o show ha (x 1 3 (x 3 8x 2 25x Bu (x 1 3 (x 3 8x 2 25x 29 is a quadraic wih posiive leading coefficien and no real roos. Thus, he resul follows. Proof of Theorem 1.2. Suppose ha G is a 4-chromaic claw-free graph. If ω(g = 4, hen by Proposiion 2.1 we have π(g, x (x 4 (x 1 n 4 wih equaliy if and only if G C4 (n. If ω(g < 4, hen we firs find a subgraph G of G ha is 4-criical and claw-free by removing some verices from G. Proposiions 2.2 and 2.3 hen show ha π(g, x < (x 4 (x 1 n 4, which finishes he proof of Theorem Claw-free graphs of large order We nex prove Theorem 1.3. To do so we use he following resul, which provides a large number of disjoin riangles in G. 6

7 Theorem 2.4 ([13]. If G is an n-verex claw-free graph, hen G conains a leas verex disjoin riangles. Theorem 2.4 allows us o analyze criical graphs. Proposiion 2.5. Le G be an n-verex k-criical claw-free graph where k 4 and n > 3k ( (k 2! k 3 log 1 (. (k 1 k 2 log k 2 k 1 Then for every ineger x k we have π(g, x < (x k (x 1 n k. ( δ(g 2 n δ(g 1 3 Proof of Proposiion 2.5. Since G is k-criical, δ(g k 1. By Theorem 2.4, G conains a leas (k 3n 3k verex disjoin riangles. Le H be a minimal conneced spanning subgraph conaining hese riangles. So H is a block graph and i is easy o see ha π(h, x = x(x 2 (x 1 n 1 where = (k 3n 3k. I suffices o show ha, for every x k, which is equivalen o ( (x 2 k 2 Now he laer follows as log assumpion on n. x(x 2 (x 1 n 1 < (x k (x 1 n k ( (x 2 k 2 > log (x 1 k 2 (x 1 k 2 1 log( x 2 x 1 1 (. log x 2 x 1 is a decreasing funcion on [k, and by he Proof of Theorem 1.3. The resul follows from Proposiions 2.5 and 2.2, as every k-chromaic claw-free graph conains a k-criical subgraph which is claw-free. 2.3 Line Graphs In his secion we prove Theorem 1.4. We begin wih a classic resul for edge-colorings. Theorem 2.6 (Vizing s Theorem. For every graph H, eiher χ (H = (H or χ (H = (H 1. A graph G is called chromaic index criical if G is conneced, χ (G = (G 1 and χ (G e < χ (G for every edge e of G. Theorem 2.7 (Vizing s Adjacency Lemma. Le H be a chromaic index criical graph. If v and w are wo adjacen verices of H wih deg H (v = (H, hen w is adjacen o a leas wo verices of degree (H. Lemma 2.8. Le G be an n-verex k-criical line graph wih ω(g = k 1 and k 4. Then for every ineger x k we have π(g, x < (x k (x 1 n k. 7

8 Proof. Suppose ha G is he line graph of H, i.e. L(H = G. I is clear ha χ(g = χ (H and ω(g = (H. Since G is conneced we may assume ha H is also conneced (by ignoring isolaed verices, if any. Since G is a criical graph, he graph H is a chromaic index criical graph. Firs le us show ha G conains a leas wo edge disjoin (k 1-cliques. Le v and w be wo adjacen verices of H wih deg H (v = k 1. By Vizing s adjacency lemma, he verex w is adjacen o a leas wo verices of degree k 1. Le w be a neighbor of w in H such ha deg H (w = k 1 and w v. Le E v (resp. E w be he se of edges of H which are inciden o he verex v (resp. w. Observe ha E v E w 1. Le G 1 and G 2 be subgraphs of G induced by he verices of G which represen he edges in E v and E w respecively. I is clear ha G 1 = G2 = Kk 1 and G 1 and G 2 are edge disjoin, as E v E w 1. Case 1: k 6. Le G be a conneced minimal spanning subgraph of G which conains G 1 and G 2. Exacly wo blocks of G are k 1-cliques and all he res of he blocks are edges. Therefore, π(g, x = (x k 1 (x 1 k 2 (x 1 n 2k3. (1 Since π(g, x π(g, x, i suffices o show ha π(g, x < (x k (x 1 n k which is equivalen o (x 2(x 3 (x k 3(x k 2 < (x k 1(x 1 k 4. Subcase 1: k 7. I is clear ha (x i < (x 1 for i = 2,... k 4. So we only need o show ha (x k 3(x k 2 (x k 1(x 1 holds for x k. Observe ha (x k 1(x 1 (x k 3(x k 2 = (k 5x k 2 6k 7. Since x k, we ge (k 5x k 2 6k 7 (k 5k k 2 6k 7 = k 7 0. Subcase 2: k = 6. In his case i suffices o show ha (x 2(x 3(x 4 < (x 5(x 1 2 holds for x 6. Calculaions show ha he polynomial (x 5(x 1 2 (x 2(x 3(x 4 has a posiive leading coefficien and is larges real roo is Thus he resul follows. Case 2: k = 5. Consider a verex u / V (G 1 V (G 2. Since deg(u 4 and G is claw-free, here exis a leas wo riangles T 1 and T 2 conaining he verex u. Now i is sraighforward o check ha he number of x-colorings of a minimal conneced spanning subgraph conaining G 1, G 2, T 1 and T 2, combined wih Proposiion 2.2, gives an upper bound sricly less han (x 5 (x 1 n 5. Case 3: k = 4. Since line graphs are claw-free, he resul follows from Proposiion 2.3. Proof of Theorem 1.4. Suppose ha G is a n-verex k-chromaic conneced line graph. If ω(g = k, hen he inequaliy follows from Theorem 2.1. Since line graphs are claw-free, he only graphs in Ck (n (which are he only graphs ha can achieve equaliy are hose wih pendan pahs aached o he verices of he k-clique. This gives he equaliy saemen in Theorem 1.4. Now suppose ha ω(g < k. We can delee verices of G unil we reach a k-criical line graph G which is a subgraph of G. Then he resul follows from Lemma 2.8 and Proposiion Independen ses In his secion, we prove he resuls relaing o independen ses. 8

9 3.1 Fixed chromaic number Proof of Theorem 1.6. We proceed by inducion on n for all k wih 1 k n. The resul is clear when n = k or k = 1. So suppose ha k > 1 and n > k. Suppose ha here exiss a verex v such ha G v has chromaic number k. The number of independen ses of G wih size which do no conain v is equal o he number of independen ses ( of G v wih size. Then by inducion on number of verices, he graph G v has a mos n 1 k ( k n 1 k 1 independen ses wih size. The number of independen ses of G of size ha include v is a mos he number of independen ses of size 1 in G v. Again by inducion i 1 (G v ( ( n 1 k 1 k n 1 k 2. (Noe ha his bound sill holds when = 2. Therefore, i (G i (G v i 1 (G v ( ( ( ( n 1 k n 1 k n 1 k n 1 k k k ( ( n k n k = k 1 where he las equaliy follows from Pascal s ideniies ( ( n k = n k 1 ( n k 1 ( 1 and n k 1 = ( n 1 k ( 1 n 1 k 2. Suppose hen ha v is a verex such ha χ(g v = k 1. Then we know ha d(v k 1. As before, an upper bound on he number of independen ses of size ha do no include v, by inducion, is ( ( n 1 (k 1 (k 1 n 1 (k 1 1. The number of independen ses of size ha include v is a mos ( n k 1 (he number of 1 ses in he a mos n k remaining verices. In all cases for, summing he wo bounds gives he desired upper bound. As saed in he Inroducion, he ranslaion o he oal coun of all independen ses is rivial. 3.2 Conneced wih fixed chromaic number In his secion, we focus on he conneced graphs wih fixed chromaic number. Proof of Theorem 1.9. Noice ha for = 0 and = 1 he inequaliy in Theorem 1.9 is rue as he value of i (G is consan over all n-verex graphs. For = 2 he inequaliy in Theorem 1.9 is rue by Lemma 1.8. We proceed by inducion on he number of verices. The resul is clear if n = k or n = k 1, so we may assume ha n k 2. Furhermore, by he remarks in he previous paragraph, we assume ha 3. Le v be a verex of G such ha G v is conneced. Observe ha i (G = i (G v i 1 (G v N G (v i (G v i 1 (G v and for 3 n k 1 he equaliy can be achieved only if N G (v has no verex which belongs o an independen se of size 1 of G v. We consider wo cases. Firs suppose ha G v is k-chromaic. By inducion, i (G v ( ( n 1 k (k 1 n 1 k ( and i 1 (G v ( ( n 1 k 1 (k 1 n 1 k ( 2 0 2, and boh inequalies can be equaliies a he same ime only if G v = K 1 (K k 1 E n 1 k. Adding he righ sides of hese inequaliies and using Pascal s ideniiy gives he desired inequaliy. In he exremal case, le v be he dominaing verex of G v. So v is he only verex of G v which canno belong o any independen se of size 2. Therefore, v mus be adjacen o v only and G = K 1 (K k 1 E n k in he exremal case. Now suppose ha G v is k 1 chromaic. By inducion, we have ha i (G v ( n 1 (k 1 (k 2 ( ( n 1 (k Noe ha v has a leas k 1 neighbors as G v is k 1 chromaic. 9

10 Choosing any 1 verices from he n k remaining verices gives i 1 (G v N G (v ( n k 1. Summing hese bounds gives he inequaliy. In he exremal case, le v be he dominaing verex of G v, and noe ha v mus have exacly k 1 neighbors. The fac ha G v is k 1 chromaic while G is k-chromaic implies ha G = K 1 (K k 1 E n k. We now prove Corollary Proof of Corollary If n = k, hen G = K n and he resul holds. If n = k 1, hen α(g 2 and so he characerizaion of equaliy follows from Lemma 1.8 (noe ha when k = 3, he only unicyclic graph wih an odd cycle is K 1 (K 2 E 1, and for oher k he only graph in Ck is K 1 (K k 1 E 1. For n k 2, again noice ha K 1 (K k 1 E n k Ck (n, and so he resul follows from Lemma 1.8 and Theorem 1.9. The nex resuls inerpolae beween he resuls for fixed chromaic number and hose for conneced graphs wih fixed chromaic number in ha hey also fix a number of componens. Proof of Theorem Since removing edges does no decrease he number of independen ses, we may assume ha c 1 componens are each rees. For a fores on a fixed number of verices and edges, he disjoin union of a sar and isolaed verices maximizes he number of independen ses of any fixed size [5, Theorem 2.2]. Noice ha if G is he disjoin union of G 1 and G 2, hen i (G = k i k(g 1 i k (G 2. This implies ha we may assume ha our graph G has a componen which is k-chromaic, a componen ha is a (possibly rivial, i.e. 1-verex sar, and c 2 isolaed verices. Le he he k-chromaic componen and he sar have x := n c 2 oal verices. We now show ha o maximizes he number of independen ses of size, he sar is an isolaed verex and he k-chromaic conneced graph is K 1 (K k 1 E x k 1. Suppose he sar has a verices and so he k-chromaic componen has x a verices. Then and ( a 1 i (K 1,a 1 = ( 0 1 ( x a k i (K 1 (K k 1 E x a k = (k 1 Now, for any fixed, we have (( x a k (k 1 i i=0 which simplifies o ( ( x k 1 x a k (k 1 1 ( x a k i 1 (( x k 1 1 ( 0 i 1 ( x a k 1 (( a 1 i ( x a k 2 ( ( i 1 ( ( a Comparing his o K 1 (K k 1 E n c1 k E 1, which has ( ( (( ( ( ( x k 1 x k 1 x k 1 x k (k independen ses of size and using ( ( y k z ( k yz k for all values of k, we see ha he maximizing graph in his family consiss of c 1 isolaed verices and a maximizing conneced k-chromaic graph on n c 1 verices. 10

11 Now we consider he cases of equaliy. Fpr c < n k, he k-chromaic componen has size a leas k 2, and so he characerizaion of equaliy follows from he equaliy characerizaion in Theorem 1.9. For c = n k, he k-chromaic componen has size k 1, and so he characerizaion of equaliy comes from Lemma 1.8. Bu here n = k 1, and so for any k he k-chromaic componen is K 1 (K k 1 E 1. Finally, for c = n k 1, he graph mus be K k E n k. We nex give he proof of Corollary Proof of Corollary The inequaliy is clear, as he exremal graphs for 3 n k 1 also have equaliy in he upper bound for = 0, 1, 2. When n > k 1 (and so n k 1 3 equaliy follows from he characerizaion of equaliy in Theorem When n = k we have c = 1 and G = K k. When n = k 1 and c = 1 he resul follows from Corollary When n = k 1 and c = 2 hen G = K k E 1. 4 Concluding Remarks We end his paper wih a few quesions and conjecures. While Conjecure 1.1 is sill open, i would also be ineresing o consider he class of n-verex k-chromaic claw-free conneced graphs for k > 4, and o show ha he analogous saemen o Theorem 1.2 holds for hese k. I would also be ineresing o invesigae he maximizing graphs for n-verex k-chromaic l-conneced graphs for oher values of l. There are also pleny of quesions relaed o independen ses. We have given resuls for maximizing i (G for graphs G which are k-chromaic and l-conneced for l = 0, 1. The following is a naural quesion. Quesion 4.1. Le l < k and le G be an l-conneced k-chromaic graph wih n verices. Fix 3. Is i rue ha i (G i (K l (K k l E n k? In our paper, we answer Quesion 4.1 in he affirmaive when l = 0 (Theorem 1.6 and l = 1 (Theorem 1.9. Wha abou oher values of k and l? Consider k = 2 and l 1. Noice ha l-conneced graphs have minimum degree a leas l. Maximizing in his larger family of fixed minimum degree graphs is already known. Theorem 4.2 ([1]. Le n, δ, and 3 be posiive inegers wih n 2δ. If G is a biparie graph on n verices wih minimum degree a leas δ, hen wih equaliy if and only if G = K δ,n δ. i (G i (K δ,n δ In oher words, his shows ha for n 2l we have K l,n l is he unique 2-chromaic l-conneced graph ha maximizes i (G for 3. This leads o a naural quesion. Quesion 4.3. Le l k and le G be a k-chromaic l-conneced graph wih n 2l verices. Fix 3. Is i rue ha i (G i ((K k 1 E l k1 E n l? We remark ha he quesion of minimizing he number of independen ses of size over graphs wih n verices and fixed conneciviy are sudied in [18]. 11

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