Saturation numbers for Ramsey-minimal graphs

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1 Saturation numbers for Ramsey-minimal graphs Martin Rolek and Zi-Xia Song Department of Mathematics University of Central Florida Orlando, FL 3816 August 17, 017 Abstract Given graphs H 1,..., H t, a graph G is (H 1,..., H t )-Ramsey-minimal if every t- coloring of the edges of G contains a monochromatic H i in color i for some i {1,..., t}, but any proper subgraph of G does not possess this property. We define R min (H 1,..., H t ) to be the family of (H 1,..., H t )-Ramsey-minimal graphs. A graph G is R min (H 1,..., H t )-saturated if no element of R min (H 1,..., H t ) is a subgraph of G, but for any edge e in G, some element of R min (H 1,..., H t ) is a subgraph of G + e. We define sat(n, R min (H 1,..., H t )) to be the minimum number of edges over all R min (H 1,..., H t )-saturated graphs on n vertices. In 1987, Hanson and Toft conjectured that sat(n, R min (K k1,..., K kt )) = (r )(n r + ) + ( ) r for n r, where r = r(k k1,..., K kt ) is the classical Ramsey number for complete graphs. The first non-trivial case of Hanson and Toft s conjecture for sufficiently large n was setteled in 011, and is so far the only settled case. Motivated by Hanson and Toft s conjecture, we study the minimum number of edges over all R min (K 3, T k )-saturated graphs on n vertices, where T k is the family of all trees on k vertices. We show that for n 18, sat(n, R min (K 3, T 4 )) = 5n/. For k 5 and n k +( k/ +1) k/, we obtain an asymptotic bound for sat(n, R min (K 3, T k )) by showing that ( k ) n c sat(n, Rmin (K 3, T k )) ( k ) n + C, where c = ( 1 k ) + 3 k and C = k 6k + 3 ( k k 1 k ) 1. AMS Classification: 05C55; 05C35. Keywords: Ramsey-minimal; saturation number; saturated graph 1 Introduction All graphs considered in this paper are finite and without loops or multiple edges. For a graph G, we will use V (G) to denote the vertex set, E(G) the edge set, G the number of Corresponding author. addresses: mrolek@knights.ucf.edu (M. Rolek), Zixia.Song@ucf.edu (Z-X. Song). 1

2 vertices, e(g) the number of edges, δ(g) the minimum degree, (G) the maximum degree, and G the complement of G. Given vertex sets A, B V (G), we say that A is complete to (resp. anti-complete to) B if for every a A and every b B, ab E(G) (resp. ab / E(G)). The subgraph of G induced by A, denoted G[A], is the graph with vertex set A and edge set {xy E(G) : x, y A}. We denote by B\A the set B A, e G (A, B) the number of edges between A and B in G, and G\A the subgraph of G induced on V (G)\A, respectively. If A = {a}, we simply write B\a, e G (a, B), and G\a, respectively. For any edge e E(G), we use G + e to denote the graph obtained from G by adding the new edge e. The join G + H (resp. union G H) of two vertex disjoint graphs G and H is the graph having vertex set V (G) V (H) and edge set E(G) E(H) {xy : x V (G), y V (H)} (resp. E(G) E(H)). Given two isomorphic graphs G and H, we may (with a slight but common abuse of notation) write G = H. For an integer t 1 and a graph H, we define th to be the union of t disjoint copies of H. We use K n, K 1,n 1, C n, P n and T n to denote the complete graph, star, cycle, path and a tree on n vertices, respectively. Given graphs G, H 1,..., H t, we write G (H 1,..., H t ) if every t-edge-coloring of G contains a monochromatic H i in color i for some i {1,,..., t}. The classical Ramsey number r(h 1,..., H t ) is the minimum positive integer n such that K n (H 1,..., H t ). A graph G is (H 1,..., H t )-Ramsey-minimal if G (H 1,..., H t ), but for any proper subgraph G of G, G (H 1,..., H t ). We define R min (H 1,..., H t ) to be the family of (H 1,..., H t )- Ramsey-minimal graphs. It is straightforward to prove by induction that a graph G satisfies G (H 1,..., H t ) if and only if there exists a subgraph G of G such that G is (H 1,..., H t )-Ramsey-minimal. Ramsey s theorem [18] implies that R min (H 1,..., H t ) for all integers t and all finite graphs H 1,..., H t. As pointed out in a recent paper of Fox, Grinshpun, Liebenau, Person, and Szabó [13], it is still widely open to classify the graphs in R min (H 1,..., H t ), or even to prove that these graphs have certain properties. Some properties of R min (H 1,..., H t ) have been studied, such as the minimum degree s(h 1,..., H t ) := min{δ(g) : G R min (H 1,..., H t )}, which was first introduced by Burr, Erdős, and Lovász [4]. Recent results on s(h 1,..., H t ) can be found in [1, 13]. For more information on Ramsey-related topics, the readers are referred to a very recent informative survey due to Conlon, Fox, and Sudakov [6]. In this paper, we study the following problem. A graph G is R min (H 1,..., H t )-saturated if no element of R min (H 1,..., H t ) is a subgraph of G, but for any edge e in G, some element of R min (H 1,..., H t ) is a subgraph of G + e. This notion was initiated by Nešetřil [16] in 1986 when he asked whether there are infinitely many R min (H 1,..., H t )-saturated graphs.

3 This was answered in the positive by Galluccio, Siminovits, and Simonyi [14]. We define sat(n, R min (H 1,..., H t )) to be the minimum number of edges over all R min (H 1,..., H t )- saturated graphs on n vertices. This notion was first discussed by Hanson and Toft [15] in 1987 when H 1,..., H t are complete graphs. They proposed the following conjecture. Conjecture 1.1 Let r = r(k k1,..., K kt ) be the classical Ramsey number for complete graphs. Then sat(n, R min (K k1,..., K kt )) = { ( n ) (r )(n r + ) + ( ) r n < r n r Chen, Ferrara, Gould, Magnant, and Schmitt [5] proved that sat(n, R min (K 3, K 3 )) = 4n 10 for n 56. This settles the first non-trivial case of Conjecture 1.1 for sufficiently large n, and is so far the only settled case. Ferrara, Kim, and Yeager [11] proved that sat(n, R min (m 1 K,..., m t K )) = 3(m m t t) for m 1,..., m t 1 and n > 3(m m t t). The problem of finding sat(n, R min (K 3, T k )) was also explored in [5]. Proposition 1. Let k and t be integers. Then ( ) (t )(k 1) sat(n, R min (K t, T k )) n(t )(k 1) (t ) (k 1) + ( ) ( ) n k 1 r + +, k 1 where r = n (mod k 1). It was conjectured in [5] that the upper bound in Proposition 1. is asymptotically correct. Note that there is only one tree on three vertices, namely, P 3. A slightly better result was obtained for R min (K 3, P 3 )-saturated graphs in [5]. Theorem 1.3 For n 11, sat(n, R min (K 3, P 3 )) = 5n 5. Motivated by Conjecture 1.1, we study the following problem. Let T k be the family of all trees on k vertices. Instead of fixing a tree on k vertices as in Proposition 1., we will investigate sat(n, R min (K 3, T k )), where a graph G is (K 3, T k )-Ramsey-minimal if for any - coloring c : E(G) {red, blue}, G has either a red K 3 or a blue tree T k T k, and we define R min (K 3, T k ) to be the family of (K 3, T k )-Ramsey-minimal graphs. By Theorem 1.3, we see that sat(n, R min (K 3, T 3 )) = 5n/ 5 for n 11. In this paper, we prove the following two main results. We first establish the exact bound for sat(n, R min (K 3, T 4 )) for 3

4 n 18, and then obtain an asymptotic bound for sat(n, R min (K 3, T k )) for all k 5 and n k + ( k/ + 1) k/ +. Theorem 1.4 For n 18, sat(n, R min (K 3, T 4 )) = 5n. Theorem 1.5 For any integers k 5 and n k + ( k/ + 1) k/, there exist constants c = ( 1 k ) + 3 k and C = k 6k + 3 ( k k 1 k ) 1 such that ( ) ( k 3 n c sat(n, R min (K 3, T k )) + 1 ) k n + C. The constants c and C in Theorem 1.5 are both quadratic in k. We believe that the true value of sat(n, R min (K 3, T k )) is closer to the upper bound in Theorem 1.5. To establish the desired lower and upper bounds for each of Theorem 1.4 and Theorem 1.5, we need to introduce more notation and prove a useful lemma (see Lemma 1.6 below). Given a graph H, a graph G is H-free if G does not contain H as a subgraph. For a graph G, let c : E(G) {red, blue} be a -edge-coloring of G and let E r and E b be the color classes of the coloring c. We use G r and G b to denote the spanning subgraphs of G with edge sets E r and E b, respectively. We define c to be a bad -coloring of G if G has neither a red K 3 nor a blue T k T k, that is, if G r is K 3 -free and G b is T k -free for any T k T k. For any v V (G), we use d r (v) and N r (v) to denote the degree and neighborhood of v in G r, respectively. Similarly, we define d b (v) and N b (v) to be the degree and neighborhood of v in G b, respectively. Remark. One can see that if G is R min (K 3, T k )-saturated, then G admits at least one bad -coloring but, for any edge e E(G), G + e admits no bad -coloring. We will utilize the following Lemma 1.6(a) to force a unique bad -coloring of certain graphs in order to establish an upper bound for sat(n, R min (K 3, T k )). Lemma 1.6(b) and Lemma 1.6(c) will be applied to establish a lower bound for sat(n, R min (K 3, T k )). Lemma 1.6 For any integer k 3, let c : E(G) {red, blue} be a bad -coloring of a graph G on n k + vertices. (a) If e E(G) belongs to at least k 3 triangles in G, then e E b. (b) If G is R min (K 3, T k )-saturated and D 1,..., D p are the components of G b with D i < k/ for all i {1,..., p}, then p. Moreover, if p =, then V (D 1 ) is complete to V (D ) in G r. (c) If G is R min (K 3, T k )-saturated, and among all bad -colorings of G, c is chosen so that E r is maximum, then (G r ) n 3 and G r is -connected. 4

5 Proof. To prove (a), suppose that there exists an edge e = uv E r such that e belongs to at least k 3 triangles in G. Since G r is K 3 -free, we see that either d b (u) k 1 or d b (v) k 1. In either case, G b contains K 1,k 1 as a subgraph, a contradiction. To prove (b), let D 1,..., D p be given as in (b). We next show that p. Since G is R min (K 3, T k )-saturated, we see that, for any edge e in G, G + e admits no bad -coloring. We next show that, for any i, j {1,..., p} with i j, V (D i ) is complete to V (D j ) in G r. Suppose that there exist vertices u V (D i ) and v V (D j ) such that uv / E r. Then uv / E(G) and so we obtain a bad -coloring of G + uv from c by coloring the edge uv blue, a contradiction. Thus V (D i ) is complete to V (D j ) in G r for any i, j {1,..., p} with i j. Since G r is K 3 -saturated, it follows that p. It remains to prove (c). By the choice of c, G r is K 3 -free but G r + e contains a K 3 for any e E(G r ), and G b is T k -free for any T k T k. Note that G b is disconnected and every component of G b contains at most k 1 vertices. Since G is R min (K 3, T k )-saturated, we see that, for any edge e in G, G + e admits no bad -coloring. Suppose that (G r ) n. Let x V (G) with d r (x) = (G r ) and let v be the unique non-neighbor of x in G r if d r (x) = n. Since G r is K 3 -free, we see that N r (x) is an independent set in G r. By the choice of c, v must be complete to N r (x) in G r. Since n k +, we have N r (x) k. Let u N r (x) and let H be the component of G b containing u. Then H k 1 and V (H) N r (x). Let w N r (x)\v (H). Clearly, uw / E(G). We obtain a bad -coloring of G + uw from c by coloring the edge uw red, and then recoloring all edges incident with u in G r blue and all edges incident with u in G b red, a contradiction. This proves that (G r ) n 3. Finally, we show that G r is -connected. Suppose that G r is not -connected. Since G r is K 3 -free but G r + e contains a K 3 for any e E(G r ), we see that G r is connected and must have a cut vertex, say u. Since (G r ) n 3, u has a non-neighbor, say v, in G r. Let G 1 and G be two components of G r \u with v V (G ). Let w V (G 1 ). By the choice of c, wv / E b, otherwise we obtain a bad -coloring of G from c by recoloring the blue edge wv red. Thus wv / E(G) and then we obtain a bad -coloring of G + wv from c by coloring the edge wv red, a contradiction. Therefore G r is -connected. This completes the proof of Lemma 1.6. The remainder of this paper is organized as follows. In Section, we discuss K 3 -saturated graphs with a specified minimum degree and prove a structural result which we shall use in the proof of Theorem 1.4. We then prove Theorem 1.4 in Section 3 and Theorem 1.5 in 5

6 Section 4. K 3 -saturated graphs In this section we list known results and establish new ones on K 3 -saturated graphs that we shall need to prove our main results. Given a graph H, a graph G is H-saturated if G is H-free but, for any edge e E(G), G + e contains a copy of H as a subgraph. We define sat(n, H) to be the minimum number of edges over all H-saturated graphs on n vertices. This notion was introduced by Erdős, Hajnal, and Moon [9] in Results on H-saturated graphs can be found in surveys by either Faudree, Faudree, and Schmitt [10] or Pikhurko [17]. In this section we are interested in the case when H = K t. Erdős, Hajnal, and Moon [9] showed that if G is a K t -saturated graph on n vertices, then e(g) (t )n ( ) t 1. Moreover, they showed that the graph K t + K n t+ is the unique K t -saturated graph with n vertices and (t )n ( ) t 1 edges. Notice that this extremal graph has minimum degree t. One may ask: what is the minimum number of edges in a K t -saturated graph with specified minimum degree? This was first studied by Duffus and Hanson [8] in They proved the following two results. Theorem.1 If G is a K 3 -saturated graph on n 5 vertices with δ(g) =, then e(g) n 5 edges. Moreover, if e(g) = n 5, then G can be obtained from C 5 by repeatedly duplicating vertices of degree. Theorem. If G is a K 3 -saturated graph on n 10 vertices with δ(g) = 3, then e(g) 3n 15. Moreover, if e(g) = 3n 15, then G contains the Petersen graph as a subgraph. Alon, Erdős, Holzman, and Krivelevich [1] showed that any K 4 -saturated graph on n 11 vertices with minimum degree 4 has at least 4n 19 edges. This has recently been generalized by Bosse, the second author, and Zhang [3] by showing that any K t -saturated graph on n t + 7 vertices with minimum degree t 3 has at least tn ( ) t+1 9 edges. Moreover, they showed that the graphs K t 3 + H are the only K t -saturated graphs with n vertices and tn ( ) t+1 9 edges, where H is a K3 -saturated graph on n t vertices with δ(h) = 3. Theorem.3 below is a result of Day [7] on K t -saturated graphs with prescribed minimum degree. It confirms a conjecture of Bollobás [] when t = 3. Theorem.3 For any integers p 1 and t 3, there exists a constant c = c(p) such that if G is a K t -saturated graph on n vertices with δ(g) p, then e(g) pn c. 6

7 For our proof of Theorem 1.4, we will need a structural result on K 3 -saturated graphs with minimum degree at most. The graph J depicted in Figure.1 is a K 3 -saturated graph with minimum degree, where A and either B = C = or B and C ; A, B and C are independent sets in J and pairwise disjoint; A is anti-complete to B C and B is complete to C; N J (y) = A B and N J (z) = A C; and A + B + C = J. It is straightforward to check that e(j) = ( J ) + B C B C J 5. Moreover, e(j) = J 5 when B = 1 or C = 1. That is, e(j) = J 5 when J is obtained from C 5 by repeatedly duplicating vertices of degree. Lemma.4 below yields a new proof of Theorem.1, and has been generalized for all K t -saturated graphs with minimum degree at most t 1 in [3]. Figure.1: The graph J Lemma.4 Let G be a K 3 -saturated graph with n vertices and δ(g) = δ. (a) If δ = 1, then G = K 1,n 1. (b) If δ =, then G = J, where the graph J is depicted in Figure.1. (c) If δ 3, then e(g) max{(δ + 1)n δ 1, (δ + )n δ(δ + t) }, where t := min{d(v) : v is adjacent to a vertex of degree δ in G}. Proof. Let x V (G) be a vertex with d(x) = δ. Since G is K 3 -saturated, we see that G is connected and K 3 -free. First assume that d(x) = 1. Let y be the neighbor of x. If there exists a vertex z V (G) such that yz / E(G), then G + xz is K 3 -free, contrary to the fact that G is K 3 -saturated. Thus y is complete to V (G) \ {y}. Clearly, N(y) is an independent set because G is K 3 -free. Thus G = K 1,n 1. This proves (a). 7

8 Next assume that d(x) =. Let N(x) = {y, z}. Then yz / E(G) because G is K 3 -free. If there exists a vertex w V (G) such that wy, wz / E(G), then G+xw is K 3 -free, contrary to the fact that G is K 3 -saturated. Hence N(y) N(z) = V (G)\{y, z}. Let A := N(y) N(z), B := N(y) \ N(z), and C := N(z) \ N(y). Then A + B + C = n, and A, B, C are pairwise disjoint. Clearly, x A, and either B = C = or B and C because δ(g) =. Since G is K 3 -free, we see that A, B, C are independent sets in G, and A is anti-complete to B C. We next show that B must be complete to C when B and C. Suppose there exist vertices b B and c C such that bc / E(G). Then G + bc is K 3 -free, a contradiction. Thus G = J, where J is depicted in Figure.1. It remains to prove (c). Let δ 3 and let t be given as in (c). We first show that e(g) (δ + 1)n δ 1. Since G is K 3 -saturated, every vertex in V (G)\N[x] has at least one neighbor in N(x), yielding v N(x) d(v) V (G)\N[x] + d(x) = n 1. Therefore e(g) = d(x) + d(v) + d(v) v N(x) δ + n 1 + δ(n δ 1) (δ + 1)n δ 1. v V (G)\N[x] We next show that e(g) (δ + )n δ(δ + t). We may assume that there exists a vertex y N(x) with d(y) = t. Notice that x and y have no common neighbor. Let M := V (G)\(N(x) N(y)). Then M = n δ t. Since G is K 3 -saturated, each vertex in M has at least one neighbor in N(x)\y and at least one neighbor in N(y)\x. v N(x)\y d(v) n t 1, and v N(y)\x d(v) n δ 1. Then e(g) = d(x) + d(y) + v N(x)\y d(v) + v N(y)\x d(v) + v M δ + t + (n t 1) + (n δ 1) + δ(n δ t) = (δ + )n δ(δ + t). This completes the proof of Lemma.4. d(v) Thus Corollary.5 Let G be a K 3 -saturated graph on n 5 vertices with δ(g) =. If e(g) = n k for some k {0, 1,, 3, 4, 5}, then G = J with B C B C = 4 k, where A, B, C, and J are as depicted in Figure.1 and the values of B and C are summarized in Table.1. 8

9 Proof. Since δ(g) =, by Lemma.4(b), G = J with e(g) = (n ) + B C B C and either B = C = or B, C, where A, B, C, and J are as depicted in Figure.1. We see that B C B C = 4 k because e(g) = n k, where k {0, 1,, 3, 4, 5}. Solving the resulting equation in each case of k yields explicit constructions of J, which are summarized in Table.1. k e(j) values of B and C with B C 5 n 5 B = 1 and C 1 4 n 4 B = C = or B = C = 0 3 n 3 B = and C = 3 n B = and C = 4 1 n 1 B = and C = 5 or B = C = 3 0 n B = and C = 6 Table.1: Construction of the graph J determined by k 3 Proof of Theorem 1.4 We are now ready to prove Theorem 1.4. We first establish the desired upper bound for sat(n, R min (K 3, T 4 )) by constructing an R min (K 3, T 4 )-saturated graph with the desired number of edges. Let n 8 be an integer and let H = ( n/ 4)K. When n 8 is even, let G even be the graph obtained from H by adding eight new vertices y, z, y 1, y, y 3, z 1, z, z 3, and then joining: y to all vertices in V (H) {y 1, y, y 3, z 1, z, z 3 }; z to all vertices in V (H) {y 1, y, y 3, z 1, z }; y 1 to all vertices in {y, z 1, z, z 3 }; y to all vertices in {z 1, z, z 3 }, z 1 to z ; and z 3 to y 3. When n is odd, let G odd be the graph obtained from H by adding nine new vertices y, z, y 1, y, y 3, y 4, z 1, z, z 3, and then joining: y to all vertices in V (H) {y 1, z 1, z, z 3 }; z to all vertices in V (H) {y 1, y, y 3, y 4, z 1, z, z 3 }; z 1 to all vertices in {y 1, y, y 3, y 4, z }; z to all vertices in {y 1, y, y 3, y 4 }, y to y 3 ; and y 4 to z 3. The graphs G odd and G even are depicted in Figure 3.1. It can be easily checked that e(g odd ) = (5n 1)/ and e(g even ) = 5n/. We next show that G odd and G even are R min (K 3, T 4 )-saturated. One can easily check that the coloring c : E(G) {red, blue} for each of G odd and G even given in Figure 3.1 is a bad -coloring. We next show that c is the unique bad -coloring for each of G odd and G even. To find a bad -coloring for G odd, by Lemma 1.6(a), the edges zz 1, zz, z 1 z must be colored blue and so all the other edges incident with z, z 1, z must be red. Then yy 1, y y 3, y 4 z 3 and all edges in E(H) must be blue and all the other edges 9

10 Figure 3.1: Two R min (K 3, T 4 )-saturated graphs with a unique bad -coloring, where dashed lines indicate blue and solid lines indicate red. incident with y must be red. This proves that G odd has a unique bad -coloring, as depicted in Figure 3.1. To find a bad -coloring for G even, by Lemma 1.6(a), y 1 y must be colored blue. We next show that z 1 z must be colored blue. Suppose that z 1 z is colored red. To avoid a red K 3, we may assume that yz 1 is colored blue. Then all edges z 1 y 1, z 1 y, yy 1, yy must be red, and so z y 1, z y must be blue, which then forces y 1 z to be red and z 1 z to be blue. Now the edges z 3 y and z 3 y 1 must be colored red, which yields a red K 3 with vertices y, z 3, y 1. This proves that z 1 z must be colored blue. Similar to the argument for G odd, one can see that the coloring of G even, depicted in Figure 3.1, is the unique bad -coloring of G even. It is straightforward to see that both G odd and G even are R min (K 3, T 4 )-saturated, and so sat(n, R min (K 3, T 4 )) 5n/. We next show that sat(n, R min (K 3, T 4 )) 5n/. Let G be an R min (K 3, T 4 )-saturated graph on n 18 vertices. e E(G), G + e has no bad -coloring. Then, for any edge Suppose that e(g) < 5n/ if n is even and e(g) < (5n 1)/ if n is odd. Among all bad -colorings of G, let c : E(G) {red, blue} be a bad -coloring of G with E r maximum. By the choice of c, G r is K 3 -saturated. Note that G b is disconnected and every component of G is isomorphic to K 1, K, P 3 or K 3. By Lemma 1.6(c), we have (1) (G r ) n 3 and G r is -connected. We next show that 10

11 () δ(g r ) = and so G r = J with A, B, and C, where J, A, B, C are depicted in Figure.1. Proof. By (1), δ(g r ). Suppose that δ(g r ) 3. We next show that e(g r ) (5n 17)/. This is trivially true if δ(g) 5. So we may assume that 3 δ(g r ) 4. By Theorem. applied to G r when δ(g r ) = 3 and Lemma.4(c) applied to G r when δ(g r ) = 4, we see that e(g r ) (5n 17)/ because n 18. By Lemma 1.6(b), e(g b ) (n )/. Thus e(g) = e(g r ) + e(g b ) 5n/, a contradiction. Hence δ(g r ) =. By Lemma.4(b), G r = J, where J, A, B, C are depicted in Figure.1. By (1), B and C. For the remainder of the proof, let J, A, B, C, and y, z be given as in Figure.1, where A, B, and C. By (), G r = J. We next show that (3) B and C. Proof. Suppose that B = 1 or C = 1, say the latter. Let u be the vertex in C. If yz, yu E b, then d b (u) = 1 because G b is T 4 -free. Now for any w A, we obtain a bad -coloring of G + uw from c by coloring the edge uw red, and then recoloring the edge zu blue. Thus either yz / E(G) or yu / E(G). We may assume that yz / E(G). Then yu E b, otherwise, we obtain a bad -coloring of G + yu from c by coloring the edge yu blue, and then recoloring the edge zu blue, and all the edges incident with z and u in G b red. Notice that d b (u) = 1, for otherwise let w A be the other neighbor of u in G b and v B. Then d b (w) = 1 and so we obtain a bad -coloring of G + wv from c by coloring the edge wv red, and then recoloring the edge yw blue. We next claim that B = N b (z). Suppose that B N b (z). Let w B \ N b (z), and let K be the component of G b containing w. If V (K) B, then for any v A, we obtain a bad -coloring of G + wv from c by coloring the edge wv red, and then recoloring the edges yw, uw blue and all edges incident with w in G b red, a contradiction. Thus V (K) A. Let v V (K) A. We claim that V (K) = {w, v}. Suppose that K = 3. Let v be the third vertex of K. Then K is isomorphic to K 3. If v A, then we obtain a bad -coloring of G from c by recoloring the edge yw blue, and then recoloring the edges wv, wv red, contrary to the choice of c. Thus v B, which again yields a bad -coloring of G from c by recoloring the edge yv blue, and then recoloring the edges vw, vv red, contrary to the choice of c. Thus V (K) = {w, v}, as claimed. For any v (A B)\({w, v} N b (z)), we obtain a bad -coloring of G + wv from c by coloring the edge wv red, and then recoloring the edge wv red, and the edges yw, uw blue. Thus B = N b (z), as claimed. 11

12 Since B = N b (z), we have B. Then yu E b, otherwise by a similar argument for showing B = N b (z), we have A = N b (u) and so n 7, a contradiction. Let v B. If B = {v}, then by a similar argument for showing d b (u) = 1, we have d b (v) = 1. But then we obtain a bad -coloring of G + yz from c by coloring the edge yz blue, and then recoloring the edge yu red, and the edge yv blue. Thus B =. Let v be the other vertex in B. Then vv E b, otherwise we obtain a bad -coloring of G + vv from c by coloring the edge vv blue. But now we obtain a bad -coloring of G + yz from c by coloring the edge yz blue, and then recoloring the edges yu, vv, zv red, and edges yv, uv blue, a contradiction. By Lemma 1.6(b), G b has at most two isolated vertices. Thus e(g b ) (n )/. Since e(g) < 5n/, we see that e(g r ) n. By (3), B and C. By Corollary.5, e(g r ) n 4 and B + C 8. Thus A n We next show that (4) If P 3 is a component of G b \{y, z} with vertices x 1, x, x 3 in order, then x A and {x 1, x 3 } B = {x 1, x 3 } C = 1. Proof. Clearly, {x 1, x, x 3 } A B or {x 1, x, x 3 } A C, otherwise x 1 x 3 / E(G) and we obtain a bad -coloring of G + x 1 x 3 from c by coloring the edge x 1 x 3 blue. Since y, z / {x 1, x, x 3 }, we see that x A. Then {x 1, x 3 } B = {x 1, x 3 } C = 1. (5) yz / E(G). Proof. Suppose that yz E(G). Then yz E b. Since G b does not contain a T 4, we see that either d b (y) = 1 or d b (z) = 1. We may assume that d b (z) = 1. We claim that d b (y) = 1 as well. Suppose that d b (y) =. Let w C be the other neighbor of y in G b. Then d b (w) = 1. Let v A. We obtain a bad -coloring of G + wv from c by coloring the edge wv red, and recoloring the edge zw blue. Thus d b (y) = d b (z) = 1. Since e(g r ) n and A n 10 8, by Corollary.5 and (4), G b contains a component, say K, such that V (K) A and V (K) A B or V (K) A C. Let u V (K) A and w A\V (K). We obtain a bad -coloring of G + uw from c by coloring the edge uw red, and then recoloring the edges yu, zu blue, and all the edges incident with u in G b red, a contradiction. (6) G b has no isolated vertex. Proof. Suppose for a contradiction that G b has an isolated vertex, say u. Then d(u) = d r (u). By (1), d(u) n 3. For any w V (G)\N[u], adding a blue edge uw to G must yield a blue T 4, because G is R min (K 3, T 4 )-saturated. Hence, 1

13 ( ) every vertex of V (G)\N[u] belongs to a P 3 or K 3 in G b. We next claim that every vertex of A\u belongs to a P 3 or K 3 in G b. By ( ), this is obvious if u A B C. So we may assume that u {y, z}. By symmetry, we may further assume that u = z. By (5), yz / E(G). Suppose that there exists a vertex v A such that v belongs to a component, say K, with K. Then V (K) A B or V (K) A C. Let w / V (K) be a vertex in C. This is possible because C by (3). We then obtain a bad -coloring of G + vw from c by coloring the edge vw red, and recoloring the edge vu blue, a contradiction. Thus every vertex of A\u belongs to a P 3 or K 3 in G b, as claimed. Since B + C 8 and A n 10 8, by (4) and Corollary.5, we see that G b [A] has at least two components isomorphic to K 3. By Lemma 1.6(b), G b has at most two isolated vertices and so e(g b ) 6 + (n 8)/. Since e(g) < 5n/, we have e(g r ) n 3. By (3), B and C. By Corollary.5, n 4 e(g r ) n 3 and max{ B, C } 3. Thus A n By (4) and Corollary.5 again, G b [A] has at least three components isomorphic to K 3. Thus e(g b ) 9+ (n 11)/ and so e(g) (n 4)+9+ (n 11)/ 5n/, a contradiction. (7) d b (y) = d b (z) =. Proof. Suppose that d b (y) 1 or d b (z) 1. By (6), d b (y), d b (z) 1. We may assume that d b (y) = 1. By (5), yz / E(G). Let y 1 C be the unique neighbor of y in G b, and let z 1 B be a neighbor of z in G b. We claim that d b (y 1 ) = 1. Suppose that d b (y 1 ) =. Let y1 A C be the other neighbor of y 1 in G b. Then y1 A, otherwise, we obtain a bad -coloring of G + yy1 from c by coloring the edge yy1 blue. Let w B. Then we obtain a bad -coloring of G + y1w from c by coloring the edge y1w red and recoloring the edge y1y blue. Thus d b (y 1 ) = 1, as claimed. By (3), B and C. We next claim that N b (z) = B. Suppose that there exists a vertex u B such that uz / E(G b ). Then uz 1 / E b, otherwise, we obtain a bad -coloring of G + uz from c by coloring the edge uz blue. This implies that B\N b (z) is anti-complete to N b (z) in G b. Let K be the component of G b containing u. By (6), K. Since G b is T 4 -free, we see that N b [z] is anti-complete to V (K) in G b. Suppose first that V (K) B. If K is isomorphic to K 3 or N b (z) =, then B 4 and G b contains at least one K 3 (K or G[N b [z]]). By Corollary.5, e(g r ) n. By (6), e(g b ) 3 + (n 3)/. Hence e(g) = e(g r ) + e(g b ) (n ) (n 3)/ 5n/, a contradiction. Thus K is isomorphic to K and d b (z) = 1. Using a similar argument to show that d b (y 1 ) = 1, we have d b (z 1 ) = 1. Let V (K) = {u, u }. If B = {u, u, z 1 }, then we obtain a bad -coloring 13

14 of G + yz from c by coloring the edge yz blue, and then recoloring the edges y 1 u, y 1 u, yz 1 blue, and the edge yy 1 red. Thus B 4. By Corollary.5, C =. Let C = {w, y 1 }. Let v A be such that v and w are not in the same component of G b. This is possible because A 8. Then we obtain a bad -coloring of G + vw from c by coloring the edge vw red, and then recoloring the edges z 1 w, zw blue, and all the edges incident with w in G b red. This proves that V (K) B and so V (K) A. Let v V (K) A. We next show that V (K) = {u, v}. Suppose that K = 3. Let v be the third vertex of K. Then K is isomorphic to K 3. If v A, then we obtain a bad -coloring of G from c by recoloring the edge uy blue, and then recoloring the edges uv, uv red, contrary to the choice of c. If v B, then we obtain a bad -coloring of G from c by recoloring the edge vy blue, and then recoloring the edges vu, vv red, contrary to the choice of c. Thus v C. Now for any w A\v, we obtain a bad -coloring of G+uw from c by coloring the edge uw red, and then recoloring the edges uy, uy 1 blue, and uv red. Hence V (K) = {u, v}. For any v A\v, we obtain a bad -coloring of G + uv from c by coloring the edge uv red, and then recoloring the edges uy blue and uv red. Thus N b (z) = B, as claimed. Since N b (z) = B and d b (z) B, we see that B =. Let B = {z 1, z }. Then z 1 z E(G b ), otherwise, we obtain a bad -coloring of G + z 1 z from c by coloring the edge z 1 z blue. Let C = {y 1,..., y t }, where t = C. Then y 1 y j / E(G b ) for all j {,..., t} because d b (y 1 ) = 1. If t 4, then by Corollary.5, e(g r ) n. By (6), e(g b ) 3 + (n 3)/. Thus e(g) (n ) (n 3)/ 5n/, a contradiction. Thus t 3. Let v A be such that vy j / E(G) for all j {1,,..., t}. This is possible because A 8 and t 3. We obtain a bad -coloring of G + y v from c by coloring the edge y v red, and then when t =, recoloring the edges yz 1, z 1 y 1, z y, y z blue, the edges z 1 z, z 1 z, and all the edges incident with y in G b red; when t = 3, recoloring the edges y 1 z 1, y 1 z, zy, zy 3 blue, the edges yy 1, zz 1, zz, and all the edges between A and {y, y 3 } in G b red. By (7), d b (y) = d b (z) =. By (5), yz / E(G). Let N b (y) = {y 1, y } C and N b (z) = {z 1, z } B. Then y 1 y, z 1 z E b, otherwise, we obtain a bad -coloring of G + e from c by coloring the edge e blue, where e {y 1 y, z 1 z }. By (6), e(g b ) 6 + (n 6)/. Since e(g) < 5n/, by Corollary.5, we see that n is even and B = C =. Let v A. We obtain a bad -coloring of G + vz 1 from c by coloring the edge vz 1 red, and then recoloring the edges yz 1, yz, zy 1, zy blue, and edges yy 1, yy, zz 1, zz red, a contradiction. This completes the proof of Theorem

15 4 Proof of Theorem 1.5 Finally, we prove Theorem 1.5. We will construct an R min (K 3, T k )-saturated graph on n k + ( k/ + 1) k/ vertices which yields the desired upper bound in Theorem 1.5. For positive integers k, n with k 5 and n k + ( k/ + 1) k/, let t be the remainder of n k k/ + when divided by k/, and let H = K k/ 1 K k sk k/ tk k/ +1, where s 0 is an integer satisfying s k/ + t( k/ + 1) = n k k/ +. Let H 1, H be the two disjoint copies of K k, and let H 3, H 4 be the two disjoint copies of K k/ 1 in H, respectively. Finally, let G be the graph obtained from H by adding four new vertices y, z, u, w, and then joining: every vertex in H 1 to all vertices in H ; y to all vertices in V (H) {w}; z to all vertices in V (H) {u}; u to all vertices in {w} V (H ) V (H 3 ); and w to all vertices in V (H 1 ) V (H 4 ), as depicted in Figure 4.1. Figure 4.1: An R min (K 3, T k )-saturated graph with a unique bad -coloring, where dashed lines indicate blue and solid lines indicate red. Clearly, the coloring c : E(G) {red, blue} given in Figure 4.1 is a bad -coloring of G. We next show that c is the unique bad -coloring of G. By Lemma 1.6(a), each edge e E(H 1 ) E(H ) must be colored blue because e belongs to k 3 triangles in G. Then all edges between V (H 1 ) and V (H ) in G must be colored red and the edge yv must be colored red for some v V (H 1 ) V (H ), because G b is T k -free. Additionally, y can only be joined by a blue edge to a vertex in either V (H 1 ) or V (H ) but not both. It follows that y is complete to one of V (H 1 ) or V (H ) in G r. We next show that y is complete to V (H ) in G r. Suppose that y is complete to V (H 1 ) in G r. Then y is complete to V (H ) in G b since G r is K 3 -free, and so yw E r since G b is T k -free. This implies that z must be complete to V (H 1 ) in G b. But now w must be complete to V (H 1 ) in G r, which yields a red K 3 on y, w, v for any v V (H 1 ), a contradiction. Hence y is complete to V (H ) in G r. Then y must be complete to V (H 1 ) in G b. Since G b is T k -free, y is complete to {w} (V (H)\V (H 1 )) in 15

16 G r, and z is complete to V (H 1 ) in G r. Since G r is K 3 -free, we see that all edges in each component of H must be colored blue, and then z must be complete to V (H ) in G b and w must be complete to V (H 4 ) in G b. By symmetry of y and z, it follows that z is complete to {u} (V (H)\V (H )) in G r, and u is complete to V (H 3 ) in G b. This proves that c is the unique bad -coloring of G. It is straightforward to see that G is R min (K 3, T k )-saturated. Using the facts that s k/ + t( k/ + 1) = n k k/ + and t k/ 1, we see that ( ) ( ) ( ) k 4 k/ k/ + 1 e(g) = (n ) + + ((k ) + 1) + (s + ) + t = (n + k 7k + 3) + (s + ) k/ k/ 1 ( k/ 1) t( k/ + 1) = (n + k 7k + 3) + k/ 1 ((s + ) k/ + t( k/ + 1)) + t ( ) k + 1 = (n + k 7k + 3) + k/ 1 ((s k/ + t( k/ + 1)) + k/ ) + t ( k (n + k 7k + 3) + k/ 1 (n k k/ + + k/ ) + t ( ) k + 1 ( ) k k n + k 6k + (k 1) + k/ 1 ( ) k + 1 ( ) k n + k 6k + 3 ( k k 1 ) k 1 ( 3 = + 1 ) k n + C, ) + 1 ( k 1 k ) 1. Therefore sat(n, Rmin (K 3, T k )) e(g) where C = k 6k + 3 k ( k ) n + C. Let c = ( 1 k ) + 3 k. We next show that sat(n, Rmin (K 3, T k )) ( k ) n c. Let G be an R min (K 3, T k )-saturated graph on n k + ( k/ + 1) k/ vertices. Then G + e has no bad -coloring for any edge e E(G). Among all bad -colorings of G, let c : E(G) {red, blue} be a bad -coloring of G with E r maximum. By the choice of c, G r is K 3 -saturated and G b is T k -free for any T k T k. Note that G b is disconnected and every component of G b contains at most k 1 vertices. By Lemma 1.6(c), we have (1) (G r ) n 3 and G r is -connected. Let D 1, D,..., D p be the components of G b. Since n k + ( k/ + 1) k/, we have p 3. We next show that () G[V (D i )] = K Di for all i {1,,..., p}. 16

17 Proof. Suppose that there exists a component of G b, say D 1, such that G[V (D 1 )] K D1. Let u, v V (D 1 ) be such that uv / E(G). We obtain a bad -coloring of G + uv from c by coloring the edge uv blue, a contradiction. (3) p e(g[v (D i )]) i=1 ( 1 k 1 ) ( 1 n k 1 ) k Proof. By (), G[V (D i )] = K Di for all i {1,,..., p}. By Lemma 1.6(b), at most two components D i have less than k/ vertices. Let t be the remainder of n k when divided by k/, and let s 0 be an integer such that n k = s k/ + t( k/ + 1). It is p straightforward to see that e(g[v (D i )]) is minimized when: two of the components, say i=1 D 1, D, are such that D 1, D < k/; t of the components, say D 3,..., D t+, are such that D 3 = = D t+ = k/ + 1; and s of the components, say D t+3,..., D t+s+, are such that D t+3 = = D t+s+ = k/. It follows that p ( ) ( ) k/ k/ + 1 e(g[v (D i )]) > s + t i=1 = s k/ k/ 1 ( 1 k = 1 ) ( k s + t ( 1 k 1 ) (n k) ( 1 k 1 ) ( 1 n ( k/ 1) t( k/ + 1) )) ( k + 1 k 1 ) k. + t ( ) k + 1 Assume that G b [V (D i )] = K Di for all i {1,,..., p}. By (3), E b ( 1 k 1 ) n ( 1 k 1 ) k. By Lemma 1.6(b) and Theorem., Er n 5. Therefore e(g) = E r + E b ( k ) ( n 1 k ( 1 ) k k ) ( n c, where c = 1 k + 3 ) k, as desired. So we may assume that G b [V (D i )] K Di for some i {1,,..., p}, say i = 1. Let u 1, u V (D 1 ) be such that u 1 u / E b. By (), u 1 u E r. Since G r is K 3 -saturated, we have N r (u 1 ) N r (u ) =. We next show that (4) for any j {,..., p} and any w V (D j ), if wu i / E r for some i {1, }, then N r (w) N r (u i )\(V (D 1 ) V (D j )). 17

18 Proof. We may assume that wu 1 / E r. Since G r is K 3 -saturated, we see that N r (w) N r (u 1 ). Note that wu 1 / E(G). If N r (w) N r (u 1 )\(V (D 1 ) V (D j )) =, then we obtain a bad -coloring of G + wu 1 from c by coloring wu 1 red, and then recoloring all red edges incident with u 1 in D 1 blue and all red edges incident with w in D j blue, a contradiction. (5) For any j {,..., p} and any w V (D j ), N r (w)\v (D j ). Proof. This is obvious when wu 1, wu E r. So we may assume that wu 1 / E r. Since N r (u 1 ) N r (u ) =, it follows from (4) that either N r (w)\(v (D 1 ) V (D j )) when wu / E(G) or N r (w)\v (D j ) = N r (w)\(v (D 1 ) V (D j )) + N r (w) V (D 1 ) = when wu E(G). In both cases, N r (w)\v (D j ), as desired. For each vertex w V (G)\V (D 1 ), since G r is K 3 -saturated, we see that either wu 1 / E r or wu / E r. Let P := {w V (G) \ V (D 1 ) : wu 1, wu / E r }, Q := {w V (G) \ V (D 1 ) : wu 1 / E r, wu E r }, and R := {w V (G) \ V (D 1 ) : wu 1 E r, wu / E r }. Further, let Q 1 denote the set of vertices w Q such that N r (w) V (D 1 ) = {u }, and let R 1 denote the set of vertices w R such that N r (w) V (D 1 ) = {u 1 }. Let Q := Q\Q 1 and R := R\R 1. By definition, P, Q 1, Q, R 1, R are pairwise disjoint and P + Q + R = n V (D 1 ) n k+1. Let H be obtained from G\V (D 1 ) by deleting all edges in G[V (D i )] for all i {, 3,..., p}. Then E(H) E r and for each edge e in H, e is not in G[V (D i )] for any i {, 3,..., p}. For any w Q 1 R 1, by (4), N H (w)\p. We next show that (6) for any w Q 1, if w is adjacent to exactly one vertex, say v, in H\P, then v R. Proof. We may assume that w V (D ). Since w Q 1, we have N r (w) V (D 1 ) = {u }. By (4), vu 1 E r, and we may further assume that v V (D 3 ). Then vu / E r because G r is K 3 -free. Since D 1 is a component of G b, there must exist a vertex, say u V (D 1 ), such that uu E b. Then wu / E r (and so wu / E(G)) because N r (w) V (D 1 ) = {u }. Hence uv E r, otherwise, we obtain a bad -coloring of G + wu from c by coloring wu red and then recoloring all edges incident with w in D blue. Therefore v R. By symmetry, for any w R 1, if w is adjacent to exactly one vertex, say v, in H\P, then v Q. We next count the number of edges in H. Since N r (u 1 ) N r (u ) =, it follows from (4) that for each w P, e H (w, Q R) and so e H (P, Q R) P. Let Q 1 be the set of vertices w Q 1 such that w is adjacent to exactly one vertex in H\P. Similarly, let R 1 be the set of vertices w R 1 such that w is adjacent to exactly one vertex in H\P. By (6), e H (Q 1, R ) Q 1 and e H (R 1, Q ) R 1. Notice that for 18

19 any w (Q 1 R 1 )\(Q 1 R 1), w is adjacent to at least two vertices in H\(P Q 1 R 1) and so e(h\(p Q 1 R 1)) Q 1 \Q 1 + R 1 \R 1 = Q 1 + R 1 Q 1 R 1. Therefore e(h) = e H (P, Q R) + e H (Q 1, R ) + e H (R 1, Q ) + e(h\(p Q 1 R 1)) P + Q 1 + R 1 + Q 1 + R 1 Q 1 R 1 = P + Q 1 + R 1. Note that e G (V (D 1 ), Q R) Q 1 + Q + R 1 + R = Q + R + Q + R. We see that e(h) + e G (V (D 1 ), Q R) ( P + Q 1 + R 1 ) + ( Q + R + Q + R ) = ( P + Q + R ) n k +. By (3), where c = ( 1 k + 3 ) k. e(g) e(h) + e G (V (D 1 ), Q R) + p e(g[v (D i )]) i=1 ( 1 k (n k + ) + 1 ) n ( 3 = + 1 ) ( k 1 k n + 3 ( 3 = + 1 ) k n c This completes the proof of Theorem 1.5. ( 1 ) k + k 1 ) k Conclusion. For the graphs G odd and G even in the proof of Theorem 1.4, we want to point out here that we found the graph G odd when d b (y) = 1, d b (z) =, and G r = J with B = and C = 4; and the graph G even when d b (y) = d b (z) =, and G r = J with B = 3 and C =. We believe that the method we developed in this paper can be applied to determine sat(n, R min (K p, T k )) for any given tree T k and any p 3. Acknowledgment The authors would like to thank Christian Bosse, Michael Ferrara, and Jingmei Zhang for their helpful discussion. References [1] N. Alon, P. Erdős, R. Holzman and M. Krivelevich, On k-saturated graphs with restrictions on the degrees, J. Graph Theory 3 (1996) 1 0. [] B. Bollobás, On generalized graphs, Acta Math. Acad. Sci. Hungar. 16 (1965) [3] C. Bosse, Z-X. Song and J. Zhang, On R min (K 3, K 4 )-saturated graphs, in preparation. 19

20 [4] S. A. Burr, P. Erdős and L. Lovász, On graphs of Ramsey type, Ars Combin. 1 (1976) [5] G. Chen, M. Ferrara, R. J. Gould, C. Magnant and J. Schmitt, Saturation numbers for families of Ramsey-minimal graphs, J. Combinatorics (011) [6] D. Conlon, J. Fox and B. Sudakov, Recent developments in graph Ramsey theory, Surveys in Combinatorics 44 (015) [7] A. N. Day, Saturated graphs of prescribed minimum degree, Combinatorics, Probability and Computing 6 (017) [8] D. A. Duffus and D. Hanson, Minimal k-saturated and color critical graphs of prescribed minimum degree, J. Graph Theory 10 (1986) [9] P. Erdős, A. Hajnal and J. W. Moon, A problem in graph theory, Amer. Math. Monthly. 71 (1964) [10] J. R. Faudree, R. J. Faudree and J. R. Schmitt, A survey of minimum saturated graphs and hypergraphs, Electron. J. Combin. 18 (011) DS19. [11] M. Ferrara, J. Kim and E. Yeager, Ramsey-minimal saturation numbers for matchings, Discrete Math. 3 (014) [1] J. Fox and K. Lin, The minimum degree of Ramsey-minimal graphs, J. Graph Theory 54 (007) [13] J. Fox, A. Grinshpun, A. Liebenau, Y. Person and T. Szabó, On the minimum degree of minimal Ramsey graphs for multiple colors, J. Combin. Theory, Ser. B. 10 (016) [14] A. Galluccio, M. Simonovits and G. Simonyi, On the structure of co-critical graphs, Graph theory, combinatorics, and algorithms, Vol. 1, (Kalamazoo, MI, 199), , Wiley-Intersci. Publ., Wiley, New York, [15] D. Hanson and B. Toft, Edge-colored saturated graphs, J. Graph Theory, 11 (1987) [16] J. Nešetřil, Problem, in Irregularities of Partitions, (eds G. Halász and V. T. Sós), Springer Verlag, Series Algorithms and Combinatorics, vol 8, (1989) P164. (Proc. Coll. held at Fertőd, Hungary 1986). [17] O. Pikhurko, Results and open problems on minimum saturated graphs. Ars Combin. 7 (004) [18] F. P. Ramsey, On a problem of formal logic, Proc. Lond. Math. Soc. (3) 30 (1930)

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