Rainbow matchings in edge-colored complete split graphs

European Journal of Combinatorics 70 (2018) 297–316

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European Journal of Combinatorics journal homepage: www.elsevier.com/locate/ejc

Rainbow matchings in edge-colored complete split graphs Zemin Jin a , Kecai Ye a , Yuefang Sun b , He Chen c a b c

Department of Mathematics, Zhejiang Normal University, Jinhua 321004, PR China Department of Mathematics, Shaoxing University, Shaoxing, 312000, PR China Department of Mathematics, Southeast University, Nanjing, 210096, PR China

article

info

Article history: Received 22 March 2017 Accepted 24 January 2018

a b s t r a c t In 1973, Erdős et al. introduced the anti-Ramsey number for a graph G in Kn , which is defined to be the maximum number of colors in an edge-coloring of Kn which does not contain any rainbow G. This is always regarded as one of rainbow generalizations of the classic Ramsey theory. Since then the anti-Ramsey numbers for several special graph classes in complete graphs have been determined. Also, the researchers generalized the host graph for the anti-Ramsey number from the complete graph to general graphs, including bipartite graphs, complete split graphs, planar graphs, and so on. In this paper, we study the anti-Ramsey number of matchings in the complete split graph. Since the complete split graph contains the complete graph as a subclass, the results in this paper cover the previous results about the anti-Ramsey number of matchings in the complete graph. © 2018 Elsevier Ltd. All rights reserved.

1. Introduction An edge-colored graph is called rainbow if the colors on its edges are distinct. In 1973, Erdős et al. [5] introduced the anti-Ramsey number for a graph G in Kn , denoted by ar(Kn , G), which is defined to be the maximum number of colors in an edge-coloring of Kn which does not contain any rainbow G. It is the first rainbow generalization of the classic Ramsey theory. Note that some researchers use the notation rainbow number for the graph G in Kn , which is defined to be the minimum integer k such that any k-edge-coloring of Kn contains a rainbow G. Clearly, the rainbow number equals the anti-Ramsey number plus one. E-mail addresses: [email protected] (Z. Jin), [email protected] (Y. Sun), [email protected] (H. Chen). https://doi.org/10.1016/j.ejc.2018.01.010 0195-6698/© 2018 Elsevier Ltd. All rights reserved.

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Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

As shown by Erdős et al. [5], the anti-Ramsey number is closely related to the Turán number. Since then, the anti-Ramsey numbers for some special graph classes in the complete graph have been determined, see [1,2,4,5,9,12–15,18,27,28,30,31]. A natural variation of the anti-Ramsey number is to change the host graph from the complete graph to others, including complete bipartite graphs (see [3,13,17,21,24]), planar graphs, regular bipartite graphs, complete split graphs, etc. (see [7,8,10,11,16,19,20,23,25,29,32]). More results on anti-Ramsey numbers can be found in the comprehensive surveys [6,22]. Note that it is almost impossible to characterize precisely the value of an anti-Ramsey number in a general graph. Xu et al. [32] gave a bound for the anti-Ramsey number of cliques in a general graph. Among these host graphs, the complete split graph is one of the graph classes that has closely related to the complete graph. Gorgol [8] studied the anti-Ramsey numbers of short cycles in the complete split graph. In this paper, we consider the matching in the complete split graph and we determine the formula of its anti-Ramsey number. Note that the complete split graph contains the complete graph as a subclass and so the results in this paper cover the previous results about the anti-Ramsey number of matchings in the complete graph. The paper is organized as follows. In Section 2, we present some necessary preliminaries, including the Turán results and the canonical decomposition of graphs. In Section 3, we present the outline of our main results. And we complete the proof of our main results in the last section. 2. Preliminaries For two disjoint graphs G and H, the join of G and H, denoted by G + H, is defined to be the graph G ∪ H + {uv : u ∈ V (G), v ∈ V (H)}. A complete split graph Kn + K¯s is a join of a complete graph Kn and an empty graph K¯ s . Let V (Kn ) = N and V (K¯ s ) = S. Let G be a graph and kK2 be a matching of size k in G. For two disjoint subsets R, T ⊆ V (G), denote by [R, T ]G the set of all edges between R and T in G. We use G[R] to denote the subgraph induced by R in G when R ⊆ V (G) or R ⊆ E(G). Let M be a matching in G. A vertex of G is said to be saturated by M if it is incident with an edge of M. If each vertex of U ⊆ V (G) is saturated by M, then we say that U is saturated by M. Let c be an edge-coloring of G. We use c(W ) to denote the set of colors of W ⊆ E(G). When W = {e}, we use c(e) for short. Given H ⊆ G, the Turán number ext(G, H) for a subgraph H in G is defined to be the maximum number of edges of a subgraph of G which does not contain any subgraph isomorphic to H. First, we present the Turán result for matchings in bipartite graphs as follows. Lemma 2.1 ([24]). If m ≥ n ≥ k, then ext(Km,n , kK2 ) = m(k − 1). Given a graph G, denote by D(G) the set of all vertices in G which are not saturated by at least one maximum matching of G. Let A(G) be the set of vertices in V (G) − D(G) adjacent to at least one vertex in D(G). Finally, let C (G) = V (G) − A(G) − D(G). We denote by D(G), A(G) and C (G) the canonical decomposition of G. A near-perfect matching in G is a matching which saturates all but exactly one vertex of G. A graph G is said to be factor-critical if G − v has a perfect matching for each vertex v ∈ V (G). The following famous fact is attributed to Gallai and Edmonds. Lemma 2.2 ([26]). Let D(G), A(G) and C (G) be the canonical decomposition of G. Then we have that (a) each component of the graph G[D(G)] is factor-critical; (b) the graph G[C (G)] has a perfect matching; (c) the bipartite graph obtained from G − C (G) − E(G[A(G)]) by contracting each component of D(G) to a single vertex has positive surplus (viewed from A(G)); (d) each maximum matching M of G contains a near-perfect matching of each component of the graph G[D(G)], a perfect matching of the graph G[C (G)], and a maximum matching in EG (A(G), D(G)); (e) the size of a maximum matching M in G is 21 (|V (G)| − ω(G[D(G)]) + |A(G)|), where ω(G[D(G])) denotes the number of components of the graph G[D(G)]. We know that the anti-Ramsey number is closely related to the Turán number. In order to prove our main result, we present the Turán number for the matching in the complete split graph as follows.

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Lemma 2.3. Let n + s ≥ 2k. Then we have ext(Kn + K¯ s , kK2 ) = r where

(n) ⎧ ⎪ + (2k − n − 1)n, max{ ⎪ ⎪ 2 ⎪ ⎪ ( ) ⎪ ⎪ ⎪ k−1 ⎪ + (k − 1)(n + s − k + 1)}, if k ≤ n ≤ 2k − 2; ⎪ ⎪ ⎪ 2 ⎨ r = ) ( ⎪ 2k − 1 ⎪ ⎪ ⎪ , max { ⎪ ⎪ 2 ⎪ ⎪ ( ) ⎪ ⎪ ⎪ k−1 ⎪ ⎩ + (k − 1)(n + s − k + 1)}, if n ≥ 2k − 1. 2

Proof. First, we present three spanning subgraphs of Kn + K¯s which do not contain any k-matching.

(n)

(1) For k ≤ n ≤ 2k − 2, the graph (Kn + K¯ 2k−n−1 ) ∪ K¯ n+s−2k+1 has 2 + (2k − n − 1)n edges. ( 2k−1 ) (2) For n ≥ 2k − 1, the graph K2k−1 ∪ K¯ n+s−2k+1 has edges. 2 ( k−1 ) (3) The graph Kk−1 + K¯ n+s−k+1 has 2 + (k − 1)(n + s − k + 1) edges. So ext(Kn + K¯ s , kK2 ) ≥ r. Now we only need to show that ext(Kn + K¯ s , kK2 ) ≤ r. Let G be a spanning subgraph of Kn + K¯ s . Assume that the maximum matching of G is of size k − 1. To prove ext(Kn + K¯ s , kK2 ) ≤ r, we only need to prove that |E(G)| ≤ r. Let D(G), A(G) and C (G) be the canonical decomposition of G. Denote the components of G[D(G)] by D1 , D2 , . . . , Dq . Let |V (Di )| = 2li + 1 for 1 ≤ i ≤ q and l1 ≥ · · · ≥ lq and |A(G)| = p. Let

|V (Di ) ∩ S | = sdi , |C (G) ∩ S | = sc and 1) = n + s − q + p, ∑q |A(G) ∩ S | = sa . By Lemma 2.2, we have 2(k −∑ q i.e., q = n + s − 2k + 2 + p. Since i=1 (2li + 1) + |C (G)| + p = n + s, we have 1 + i=1 2li + |C (G)| = 1 + n + s − p − q = 2k − 2p − 1. Case 1. k ≤ n ≤ 2k − 2. Subcase 1.1. l1 = 0. Then we have l1 = l2 = · · · = lq = 0 and |C (G)| = n + s − q − p = 2k − 2 − 2p. So

|E(G)| ≤ =

(

(p)

+ p(n + s − p) +

2

+ p(n + s − p) +

2

) −

2

(

(p)

2k − 2 − 2p

2k − 2 − 2p

)

(s ) a

2

( −

2

−

(s )

sa + sc 2

)

− sa sc

.

( ) ( ) + p(n + s − p) + 2k−22−2p − sa +2 sc . Since h1 (p) is a convex function of p and ( k−1 0) ≤ p ≤ k − 1, we obtain |E(G)| ≤ max{h1 (0), h1 (k − 1)}. If p = k − 1, then h1 (k − 1) ≤ + (k − 1)(n + s − k + 1) ≤ r. 2 If p (= 0,)then ( we have ) (sa =) 0 and ( |C (G) ) | = 2k − 2. Since sc ≥ 2k − 2 − n, we have 2k−2 h1 (0) ≤ − 2k−22−n < 2k2−1 − 2k−21−n ≤ r. 2 Therefore, |E(G)| ≤ max{h1 (0), h1 (k − 1)} ≤ r. Subcase 1.2. l1 > 0. Let t be the maximum integer such that lt > 0. So t ≥ 1. Then we have Let h1 (p) =

|E(G)| ≤

(p)

c

2

2

(p) 2

−

+ p(n + s − p) +

(s ) a

2

−

q ( ∑ sd ) i

i=1

2

) q ( ∑ 2li + 1 2

i=1

−

(s ) c

2

( − sa

( +

q ∑ i=1

|C (G)| 2

) sdi + sc

)

300

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

≤

( ∑q ) 2 i=1 li + 1 + |C (G)| + p(n + s − p) + 2 ( 2t ) ( s ) ( ∑t s + s ) ∑ d c a i=1 i − − sa sdi + sc −

(p)

2

2

=

i=1

(

(p)

+ p(n + s − p) +

2

2k − 1 − 2p

)

( −

2

sa +

∑t

i=1 sdi

+ sc

2

)

.

∑ ( ) s + t s +s + p(n + s − p) + 2k−12−2p − a i=21 di c . Since h2 (p) is a convex function of ( k−p1 and ) 0 ≤ p ≤ k − 1, we obtain |E(G)| ≤ max{h2 (0), h2 (k − 1)}. If p = k − 1, then h2 (k − 1) ≤ + (k − 1)(n + s − k + 1) ≤ r. ∑ 2 q If p = 0, then we have sa = 0. Since i=1 (2li + 1) +|C (G)|+|A(G)| = n + s and q = n + s − 2k + 2 + p,

(

(p)

Let h2 (p) =

)

2

we have t t ∑ ∑ |V (Di )| + |C (G)| = (2li + 1) + |C (G)| = n + s − (q − t) = 2k − 2 + t . i=1

i=1

∑t

(Son )sa + i=1 sdi + sc ≥ 2k − 2 + t − n ≥ 2k − 1 − n. Thus, we have h2 (0) ≤ + (2k − n − 1)n ≤ r. 2 Therefore, |E(G)| ≤ max{h2 (0), h2 (k − 1)} ≤ r. Case 2. n ≥ 2k − 1.

( 2k−1 ) 2

−

( 2k−1−n ) 2

=

Then we have

|E(G)| ≤

(p) 2

− ≤ =

+ p(n + s − p) +

2

i=1

(s ) a

2

(p) 2

(p) 2

) q ( ∑ 2li + 1

q

−

∑ (s ) di

2

i=1

−

(s ) c

2

+ p(n + s − p) +

)

2

)

∑

sdi + sc

i=1

∑q

(

1+

(

2k − 2p − 1

+ p(n + s − p) +

|C (G)|

q

( − sa

( +

i=1

2li + |C (G)|

)

2 2

)

.

( 2k−2p−1 )

(p)

. Since h3 (p) is a convex function of p and 0 ≤ p ≤ k − 1, we Let h3 (p) = 2 + p(n + s − p) + 2 ( k−1 ) have |E(G)| ≤ max{h3 (0), h3 (k−1)}. If p = k−1, then we have h3 (k−1) = 2 +(k−1)(n+s−k+1) ≤ ( 2k−1 ) r. If p = 0, then we have h3 (0) = ≤ r. Thus, |E(G)| ≤ max{h3 (0), h3 (k − 1)} ≤ r. 2 This completes the proof of the lemma. ■ 3. Outline of main results Theorem 3.1. ar K2 + K¯2 , 2K2 = 3.

(

)

Proof. If we color K2 + K¯2 with 4 colors, then there ( exist two )independent edges which receive distinct colors, a contradiction. Therefore, we have ar K2 + K¯2 , 2K2 ≤ 3. Let V (K2 ) = {x1 , x2 } and V (K¯2 ) = {y1 , y2 }. We color x1 y1 and x2 y2 with 1, color x(1 y2 and x2 y1 with ) 2, and color x1 x2 with 3. Clearly, K2 + K¯2 does not contain any rainbow 2K2 . Thus, ar K2 + K¯2 , 2K2 = 3. ■ Theorem 3.2. ar K2 + K¯s , 2K2 = 2 for s ⩾ 3.

(

)

Proof. If we color K2 + K¯s with 3 colors, then K2,s receives at least two distinct colors. ( So there exist ) two independent edges which receive distinct colors in K2,s , a contradiction. Thus, ar K2 + K¯2 , 2K2 ≤ 2.

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

301

Let V (K2 ) = {x1 , x2 } and V (K¯s ) = {y1 , . . . , ys }, s ⩾ 3. We color x1 x2 with 1 and (remaining edges ) receive additional one color, then K2 + K¯s does not contain any rainbow 2K2 . Thus, ar K2 + K¯s , 2K2 = 2 for s ⩾ 3. ■ We have the following main result. Theorem 3.3. Let n + s ≥ 2k, n ≥ k ≥ 3, s ≥ 2. Then ar(Kn + K¯s , kK2 ) = r, where

(n) ⎧ ⎪ max{ + (2k − n − 2)n, ⎪ ⎪ 2 ⎪( ) ⎪ ⎪ ⎪ ⎪ k − 2 + (k − 2)(n + s − k + 2)} + 1, if k + 1 ≤ n ≤ 2k − 2; ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ( ) ⎪ ⎨ k−2 + (k − 2)(n + s − k + 2) + 2, if n = k; r = 2 ⎪ ⎪ ⎪ ⎪ ⎪ ( ) ⎪ ⎪ 2k − 3 ⎪ ⎪ ⎪ max { , ⎪ ⎪ 2 ⎪ ⎪ ( ) ⎪ ⎪ ⎪ ⎪ ⎩ k − 2 + (k − 2)(n + s − k + 2)} + 1, if n ≥ 2k − 1. 2

Outline of the proof of Theorem 3.3 First, we present a r-edge-coloring of Kn + K¯s which does not contain any rainbow k-matching. (1) For n ≥ 2k − 1, color the edges of a subgraph K2k−3 in Kn + K¯s by distinct colors and the remaining edges of Kn + K¯s by a new color. (2) For n ≥ k + 1, color the edges of a subgraph Kk−2 in Kn + K¯s and all the edges from Kk−2 to Kn + K¯s − V (Kk−2 ) by distinct colors and the remaining edges of Kn + K¯s by a new color. (3) For k + 1 ≤ n ≤ 2k − 2, color the edges of a subgraph Kn + K¯ 2k−n−2 in Kn + K¯s by distinct colors and the remaining edges of Kn + K¯s by a new color. (4) For n = k, color the edges of a subgraph Kk−2 in Kn + K¯s , the edges of (Kn + K¯s )[N ] − V (Kk−2 ) and all the edges from Kk−2 to Kn + K¯s − V (Kk−2 ) by distinct colors and the remaining edges of Kn + K¯s by a new color. Then ar(Kn + K¯s , kK2 ) ≥ r. Now we show ar(Kn + K¯s , kK2 ) ≤ r. That is to say we only need to prove that each (r + 1)-edge-coloring(of)Kn + K¯s contains a rainbow 2). ( 2kkK ( ) n −2 If n ≤ 2k − 2, then we have 2 + (2k − n − 2)n + 1 = + 1 − 2k−22−n . 2 Given a (r + 1)-edge-coloring c of Kn + K¯s , let G be a rainbow spanning subgraph which contains all colors in Kn + K¯ s . Clearly, |E(G)| = r + 1. Suppose that Kn + K¯s does not contain any rainbow k-matching. Since r + 1 > ext(Kn + K¯ s , (k − 1)K2 ), G has a rainbow (k − 1)-matching. Let D(G), A(G) and C (G) be the canonical decomposition of G. Denote the components of G[D(G)] by D1 , D2 , . . . , Dq . Let |A(G)| = p, |V (Di )| = 2li + ∑1qfor 1 ≤ i ≤ q and l1 ≥ · · · ≥ lq . Let |V (Di ) ∩ S | = sdi , |C (G) ∩ S | = sc and |A(G) ∩ S | = sa . So sa + i=1 sdi + sc = s and by Lemma 2.2, 2(k − 1) = n + s − q + p, i.e., q = n + s − 2k + 2 + p. Now we distinguish the following 5 cases to complete the proof. Case 1. p = k − 1. Case 2. k ≤ n ≤ 2k − 2, 0 ≤ p ≤ k − 2 and l1 ≤ k − p − 2. Case 3. n ≥ 2k − 1, 0 ≤ p ≤ k − 2 and l1 ≤ k − p − 2. Case 4. k ≤ n ≤ 2k − 2, 0 ≤ p ≤ k − 2 and l1 = k − p − 1. Case 5. n ≥ 2k − 1, 0 ≤ p ≤ k − 2 and l1 = k − p − 1. 4. Proof of Theorem 3.3 First we present a useful lemma which helps us to simplify the proof of Theorem 3.3.

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Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

Lemma 4.1. Let n ≥ 2k−1 and s ≥ 2. If Kn +K¯ s has a rainbow spanning subgraph I = K2k−3 ∪P3 ∪K¯ n+s−2k − − ¯ ¯ or I = K2k −3 ∪ K3 ∪ Kn+s−2k where K2k−3 is obtained from K2k−3 by deleting an edge, then Kn + Ks has a rainbow k-matching. Proof. Suppose that Kn + K¯ s does not have any rainbow k-matching. First we consider the case I = K2k−3 ∪ P3 ∪ K¯ n+s−2k . Let x ∈ V (K2k−3 ) ∩ N, P3 = u1 v1 w1 and y ∈ V (K¯ n+s−2k ). Let f ∈ E(G) with c(f ) = c(xy). We claim that k = 3 and f ∈ E(K2k−3 − x). Otherwise, the graph P3 ∪ [K2k−3 − x] − f contains a (k − 1)-matching, and the union of such a matching and the edge xy forms a rainbow k-matching, a contradiction. So k = 3 and f ∈ E(K2k−3 − x). Let V (K2k−3 ) = {x, w, z }. Furthermore, we can assume that each edge between x and V (K¯ n+s−2k ) is of the color c(f ). If n + s − 2k ≥ 2 and N ∩ V (Kn+s−2k ) ̸ = ∅, then there are two vertices u2 , v2 ∈ V (Kn+s−2k ) with u2 v2 ∈ E(Kn + K¯ s ). It is clear that the components K3 ∪ P3 of I contain a 2-matching which does not contain any edge of the color c(u2 v2 ). Then the union of such a 2-matching and u2 v2 is a rainbow 3-matching, a contradiction. So either n + s − 2k = 1 or n + s − 2k ≥ 2 and V (Kn+s−2k ) ⊆ S. Since n ≥ 2k − 1, {w, z } ∩ N ̸ = ∅. Let w ∈ N. By the same analysis as the vertex x, we can assume that each edge between w and V (K¯ n+s−2k ) is of the color c(xz). Hence, n + s − 2k = 1. Otherwise, let y′ ∈ V (Kn+s−2k ) − y. Then {xy, w y′ , u1 v1 } is a rainbow 3-matching, a contradiction. Thus we have that I = K3 ∪ P3 ∪ K1 . Now we consider the color of the edge xu1 . We claim that c(xu1 ) = c(v1 w1 ). Otherwise, {xu1 , v1 w1 , wz , wx} contains a rainbow 3-matching, a contradiction. By the same analysis, we have that c(w u1 ) = c(v1 w1 ), c(u1 v1 ) = c(xw1 ) = c(ww1 ). It is clear that v1 ∈ N. Then at least one of {v1 y, u1 x, w1 w}, {v1 y, u1 x, w z }, and {v1 y, w1 x, w z } is a rainbow 3-matching, a contradiction. − ¯ The case I = K2k −3 ∪ K3 ∪ Kn+s−2k can be proved in the similar way and here we omit the details. This completes the proof of the lemma. ■ Details of the proof of Theorem 3.3 . As done in the outline of the proof, we take a (r + 1)-edge-coloring c of Kn + K¯s and let G be a rainbow spanning subgraph which contains all colors in Kn + K¯ s . Clearly, |E(G)| = r + 1. Suppose that Kn + K¯s does not contain any rainbow k-matching. Since r + 1 > ext(Kn + K¯ s , (k − 1)K2 ), G has a rainbow (k − 1)-matching. Let D(G), A(G) and C (G) be the canonical decomposition of G. Denote the components of G[D(G)] by D1 , D2 , . . . , Dq . Let |A(G)| = p, |V (Di )| = 2li + ∑1qfor 1 ≤ i ≤ q and l1 ≥ · · · ≥ lq . Let |V (Di ) ∩ S | = sdi , |C (G) ∩ S | = sc and |A(G) ∩ S | = sa . So sa + i=1 sdi + sc = s and by Lemma 2.2, 2(k − 1) = n + s − q + p, i.e., q = n + s − 2k + 2 + p. Case 1. p = k − 1. Since p + q = p + n + s + p − 2k + 2 = n + s + 2p − 2(k − 1) = n + s, we have C (G) = ∅ and l1 = · · · = lq = 0. From p = k − 1 < n, we have that D(G) ∩ N ̸ = ∅. Suppose that there is at most one vertex of D(G) with degree less than k − 1 in G. Take a vertex u ∈ D(G) with degree as small as possible and a vertex v ∈ D(G) − {u} such that {u, v} ∩ N ̸ = ∅. Since p = k − 1, we have that the subgraph H induced by [A(G), D(G) −{u, v}]G is isomorphic to Kn+s−k−1,k−1 . Clearly, H has a rainbow (k − 1)-matching which does not contain any edge of the color c(uv ). Then the union of that matching and {uv} is a rainbow k-matching in Kn + K¯ s , a contradiction. So there are at least two vertices u, v of D(G) such that dG (u) < k − 1 and dG (v ) < k − 1. Then the subgraph H induced by the edges [A(G), D(G) − {u, v}]G − {e : c(e) = c(uv )} is a spanning subgraph of Kn+s−k−1,k−1 . Since Kn + K¯ s does not contain any rainbow k-matching, H does not contain any (k − 1)matching. So

|E(G)| ≤ 1 + ext(Kn+s−k−1,k−1 , (k − 1)K2 ) + 2(k − 2) ( ) ( ) q ∑ k−1 sa + − − sa sdi 2

2

i=1

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

(

k−1 ≤ 1 + (k − 2)(n + s − k − 1) + 2(k − 2) + 2 ( ) k−2 = + (k − 2)(n + s − k + 2) + 1 ≤ r ,

303

)

2

a contradiction to the fact |E(G)| = r + 1. Case 2. k ≤ n ≤ 2k − 2, 0 ≤ p ≤ k − 2 and l1 ≤ k − p − 2. Subcase 2.1. p = k − 2. Then l1 ≤ k − p − 2 = 0. So l1 = · · · = lq = 0 and |C (G)| = n + s − q − p = 2. Then

|E(G)| ≤

(p) 2

− ≤

+ p(n + s − p) + 1 −

(s ) c

2

( − sa

q ∑

(s ) a

2

−

q ( ∑ sd ) i

2

i=1

) sdi + sc

i=1

(p)

+ p(n + s − p) + 1 ) k−2 + (k − 2)(n + s − k + 2) + 1 ≤ r ,

(2

=

2

a contradiction to the fact |E(G)| = r + 1. Subcase 2.2. 0 ≤ p ≤ k − 3 and |C (G)| ≥ 2. Claim 1. l1 ≥ 1. Proof. Suppose that l1 = · · · = lq = 0. Then p∑ + q + |C (G)| = n + s. Since q = n + s − 2k + 2 + p, q we have |C (G)| = 2k − 2p − 2. Since sa + sc + i=1 sdi = s, we have

|E(G)| ≤ ≤

(p) 2

(p) 2

( + p(n + s − p) +

(p)

)

2

( 2k−2−2p ) ( sc )

(

k−3

≤

k−3

=

k−2

≤

k−2

− sa

−

q ∑

sdi

(s ) c

2

.

. Since f1 (p) is a convex function of p and 0 ≤ p ≤ k−3,

(s ) c

2

) + k − 3 + (k − 3)(n + s − k + 2) + 6

) + (k − 2)(n + s − k + 2) + 1 − (n + s − k) + 3

2

(

)

+ (k − 3)(n + s − k + 3) + 6 −

2

(

)

i=1

)

2

(

2

sa + sc 2

2k − 2 − 2p

Let f1 (p) = 2 +p(n+s−p)+ − 2 we obtain f1 (p) ≤ max{f1 (0), f1 (k − 3)}. Since n + s ≥ 2k, k ≥ 3, we have f1 (k − 3) =

( −

2

( + p(n + s − p) +

|C (G)|

)

2

+ (k − 2)(n + s − k + 2) + 1 ≤ r .

If p = 0, then |C (G)| = 2k − 2. So sc ≥ 2k − 2 − n. Then we have

( f1 (0) =

2k − 2 2

) −

(s ) c

2

( ≤

2k − 2 2

)

( −

2k − 2 − n 2

)

< r.

Therefore, we have |E(G)| ≤ r, a contradiction to the fact |E(G)| = r + 1. This completes the proof of the claim. ■

304

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

Let t be the maximum integer such that lt > 0. So t ≥ 1. Then we have

|E(G)| ≤

(p) 2

−

≤

+ p(n + s − p) +

2

i=1

(s ) a

2

(p) 2

−

2

−

−

c

2

− sa

(s ) a

2

i=1 li

i=1 sdi

2 2

−

sa +

)

2

) sdi + sc

+ 1 + |C (G)| − 2

+ sc − 1

2

(

)

2

)

− sa

q ∑

) sdi + sc

i=1

( ∑q

(

q ∑

|C (G)|

2

( ∑t −

( +

i=1

2

+ p(n + s − p) +

( ) +

2

(s )

(

( ∑q

2

(p)

i

+ p(n + s − p) +

2

+

q ( ∑ sd ) i=1

( )

≤

) q ( ∑ 2li + 1

i=1 li

+ 1 + |C (G)| − 2

)

2

∑t

i=1 sdi

+ sc − 1

)

2

.

∑q

Since q = n + s − 2k + 2 + p and i=1 (2li + 1) + |C (G)| + p = n + s, we have n + s − p − q = 2k − 2p − 2. So we obtain

|E(G)| ≤

(

(p) 2

( −

2k − 2p − 3

+ p(n + s − p) + 2 ∑t ) sa + i=1 sdi + sc − 1 .

∑q

i=1 2li

+ |C (G)| =

) +1

2

( ∑t ) ( 2k−3−2p ) sa + i=1 sd +sc −1 i + p(n + s − p) + + 1 − . Since f2 (p) is a convex function 2 2 2 of p and 0 ≤ p ≤ k − 3, we obtain |E(G)| ≤ max{f2 (0), f2 (k − 3)}. If p = k − 3, then we have ( ) k−3 f2 (k − 3) ≤ + (k − 3)(n + s − k + 3) + 4 Let f2 (p) =

(p)

2

( = <

k−2

) + (k − 2)(n + s − k + 2) − (n + s − k) + 2

2

(

k−2 2

)

+ (k − 2)(n + s − k + 2) < r .

If p = 0, then we have sa = 0. Since we have

∑q

i=1 (2li

+ 1) +|C (G)|+|A(G)| = n + s and q = n + s − 2k + 2 + p,

t t ∑ ∑ |V (Di )| + |C (G)| = (2li + 1) + |C (G)| = n + s − (q − t) = 2k − 2 + t . i=1

i=1

∑t

Since n ≤ 2k − 2 and t(≥ 1, )we have(sa + )i=1 sd(i + sc) − 1 ≥ 2k ( − 2 +) t − n − 1 ≥ 2k − 2 − n ≥ 0. 2k−3 Thus, we have f2 (0) ≤ + 1 − 2k−22−n < 2k2−2 + 1 − 2k−22−n . 2 Therefore, we have |E(G)| ≤ max{f2 (0), f2 (k − 3)} ≤ r, a contradiction to the fact |E(G)| = r + 1. Subcase 2.3. p ≤ k − 3 and |C (G)| = 0. ∑q Since l1 ≤ k − p − 2 , we have 2l1 + 1 ≤ 2k − 2p − 3. Since i=1 (2li + 1) + p = n + s and q = n + s − 2k + 2 + p, we have q ∑ i=2

2li = n + s − p − 2l1 − 1 − (q − 1) ≥ 2.

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

305

So l2 ≥ 1. Let t be the maximum integer such that lt > 0. So t ≥ 2. Then we have

|E(G)| ≤

(p) 2

−

≤

≤

(s ) a

2

2

(

i

2

i=1 sdi

−

3 2

−1

( +

i=1 sdi

∑q

i=1

2li − 2

)

2

∑

sdi

i=1

( ∑t

1+ q

− sa

2

−1

)

2

i=1 (2li

+ p(n + s − p) +

1+

∑q

i=1

2li − 2

)

2

.

∑q

(

2

sdi

)

+ p(n + s − p) + 3 +

sa +

(p)

q ∑

− sa

i=1

( ∑t

Since q = n+s−2k+2+p and So we obtain

|E(G)| ≤

2

( )

2

2

q ( ∑ sd )

+ p(n + s − p) +

a

(p)

−

i=1

(s )

−

) q ( ∑ 2li + 1 i=1

(p)

−

+ p(n + s − p) +

+1)+p = n+s, we have

2k − 2p − 3

)

2

( +3−

∑q

i=1 2li

sa +

∑t

= n+s−p−q = 2k−2p−2.

i=1 sdi

−1

)

2

.

∑ ( ) s + t s −1 + p(n + s − p) + 2k−32−2p + 3 − a i=2 1 di . Since f3 (p) is a convex function of p and 0 ≤ p ≤ k − 3, we obtain |E(G)| ≤ max{f3 (0), f3 (k − 3)}. If p = k − 3, then we have ( ) k−3 f3 (k − 3) ≤ + (k − 3)(n + s − k + 3) + 6

Let f3 (p) =

(

(p)

)

2

2

( =

k−2

) + (k − 2)(n + s − k + 2) − (n + s − k) + 4

2

( ≤

k−2

)

2

+ (k − 2)(n + s − k + 2) + 1 ≤ r .

If p = 0, then we have sa = 0. Since we have

∑q

i=1 (2li

+ 1) +|C (G)|+|A(G)| = n + s and q = n + s − 2k + 2 + p,

t t ∑ ∑ |V (Di )| + |C (G)| = (2li + 1) + |C (G)| = n + s − (q − t) = 2k − 2 + t . i=1

i=1

∑t Since t (≥ 2,)we have sa + ) i=1(sdi −) 1 ≥ 2k ( ( 2k−−22−n+) t − n − 1 > 2k − 2 − n. Thus, we have 2k−3 2k−2−n 2k−2 f3 (0) < + 3 − ≤ + 1 − . 2 2 2 2 Therefore, we have |E(G)| ≤ max{f3 (0), f3 (k − 3)} ≤ r, a contradiction to the fact |E(G)| = r + 1. Case 3. n ≥ 2k − 1, 0 ≤ p ≤ k − 2 and l1 ≤ k − p − 2. Subcase 3.1. p = k − 2. Then l1 ≤ k − p − 2 = 0. So l1 = · · · = lq = 0 and |C (G)| = n + s − q − p = 2. Then q ( (p) (s ) ∑ sdi ) a |E(G)| ≤ + p(n + s − p) + 1 − − 2

−

2

(s ) c

2

( − sa

q

∑ i=1

) sdi + sc

i=1

2

306

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

≤

(p)

+ p(n + s − p) + 1 ) k−2 + (k − 2)(n + s − k + 2) + 1 ≤ r ,

(2

=

2

a contradiction to the fact |E(G)| = r + 1. Subcase 3.2. 0 ≤ p ≤ k − 3 and |C (G)| ≥ 2. Claim 2. l1 ≥ 1. Proof. Suppose that l1 = · · · = lq = 0. So each Di is an isolated vertex and sdi ≤ 1 for 1 ≤ i ≤ q. We have p + q + |C (G)| = n + s. Since q = n + s − 2k + 2 + p, we have |C (G)| = 2k − 2p − 2. Let H be the subgraph induced by [A(G), D(G)]G . From Lemma 2.2, H contains a p-matching M1 and G[C (G)] has a (k − p − 1)-matching M2 . Note that M1 ∪ M2 is a maximum matching in G. Since q = n + s − 2k + 2 + p ≥ 1 + s + p, there exists a vertex u ∈ D(G) ∩ N. By the definition of D(G), there is a maximum matching in G which does not saturate the vertex u. So we can take the matching M1 such that u is not saturated. Take a vertex v ∈ D(G) − u which is not saturated by M1 . Clearly, uv ∈ E(Kn + K¯ s ). Clearly, uv ̸ ∈ E(G) and there is an edge e ∈ M1 ∪ M2 such that c(e) = c(uv ). Case A. e ∈ M2 . If G[C (G)] − e still has a (k − p − 1)-matching, then the union of that matching and M1 ∪ {uv} is a rainbow k-matching in Kn + K¯ s , a contradiction. So the maximum matching of G[C (G)] − e is of size k − p − 2. We have

|E(G)| = |E(G[C (G)])| + |E(G[A(G)])| + |[A(G), D(G)]G |. Since G[C (G)] − e does not contain any (k − p − 1)-matching and |C (G) ∩ S | = sc , |E(G[C (G)])| ≤ ¯ , (k − p − 1)K2 ) + 1. From Lemma 2.3, we have ext(K2k−2p−2−sc + K¯ sc , (k − p − 1)K2 ) ≤ ext(K2k −2p−2−s)c + K ( 2k ( sscc − ) ( k−p−2 ) ( ) ( k−p−2 ) −2p−3 1 −3 max{ − , + (k − p − 2)(k − p)}. Since 2k(−2p −) 2 − (k − p − 2)(k − p) = 2 2 2 ( k−p−2 ) 2 2k−2p−3 ¯ ≥ 0, we have ext(K + K , (k − p − 1)K ) ≤ . 2k−2p−2−sc sc 2 2 2 Thus, we have

|E(G)| ≤ ext(K2k−2p−2−sc + K¯ sc , (k − p − 1)K2 ) + 1 ( q ) (p) (s ) ∑ a + + p(n + s − p) − − sa sdi + sc 2

( ≤

2

2k − 2p − 3 2

( 2k−2p−3 )

) +1+

(p) 2

i=1

+ p(n + s − p).

(p)

Let g1 (p) = + 1 + 2 + p(n + s − p). Since g1 (p) is a convex function of p and 2 0 ≤ (p ≤)k − 2, we obtain |E(G)| ≤ max{g1 (0), g1 (k − 2)}. If p = k( − 2,)then we have g1 (k − 2) ≤ k−2 2k−3 1 + 2 + (k − 2)(n + s − k + 2). If p = 0, then we have g1 (0) = + 1. Therefore, we obtain 2 |E(G)| ≤ max{g1 (0), g1 (k − 2)} ≤ r, a contradiction to the fact |E(G)| = r + 1. Case B. e ∈ M1 . Then we have p ≥ 1. If H − {u, v} − e still has a p-matching, then the union of that matching and M2 ∪{uv} is a rainbow k-matching in Kn +K¯ s , a contradiction. So the maximum matching of H −{u, v}−e is of size p − 1. Then it follows from Lemma 2.1 that |E(H − {u, v})| ≤ ext(Kp,n+s−2k+p , pK2 ) + 1 = (p − 1)(n + s − 2k + p) + 1. So we have

|E(G)| = |E(G[C (G)])| + |E(G[A(G)])| + |[A(G), C (G) ∪ {u, v}]G | + |E(H − {u, v})| ( ) ( ) ( ) ( ) 2k − 2p − 2 sc p sa ≤ − + − + p(2k − 2p) 2

2

2

2

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

( + (p − 1)(n + s − 2k + p) − sa

q ∑

307

) sdi + sc

+1

i=1

( ≤

2k − 2p − 2

) +

2

(p) 2

+ p(n + s − p) − (n + s − 2k + p) + 1.

Subcase B.1. 2 ≤ p ≤ k − 2. ( 2k−2p−2 ) ( p ) Let g2 (p) = + 2 + p(n + s − p) − (n + s − 2k + p) + 1. Since g2 (p) is a convex function 2 of p and 2 ≤ p ≤ k − 2, we obtain |E(G)| ≤ max{g2 (2), g2 (k − 2)}. If p = k − 2, then we have

( g2 (k − 2) = 2 +

<

(

k−2 2

k−2

)

2

) + (k − 2)(n + s − k + 2) − (n + s − k − 2)

+ (k − 2)(n + s − k + 2) + 1.

If p = 2, then we have

( g2 (2) = 2 +

( =

2k − 6 2

2k − 6

=

2k − 5

+ 2k − 6 + 2 + n + s

) + 2k − 5 + 2 + n + s − (2k − 5)

2

( =

2k − 4

) + 2k − 4 + 2 + n + s − (4k − 9)

2

( =

2k − 3 2

+ 2(n + s − 2) − (n + s − 2k + 2)

)

2

(

)

)

+ n + s − 4k + 11.

If g2 (2) ≤ g2 (k − 2), then we have |E(G)| ≤ g2 (k − 2) < r, a contradiction to the fact |E(G)| = r + 1. If g2 (2) > g2 (k − 2), then

(

2k − 3 2

)

+ n + s − 4k + 11 >

So we have (k − 4)(n + s) <

5k2 −27k+28 . 2

(

k−2

)

2

+ (k − 2)(n + s − k + 2) + 1 − (n + s − k) + 3.

Then we obtain n + s <

5k−7 2

and k > 4. Therefore, we have

( ) 2k − 3 |E(G)| ≤ g2 (2) = + n + s − 4k + 11 2 ( ) 2k − 3 15 − 3k < + 2 2 ( ) 2k − 3 ≤ < r, 2

a contradiction to the fact |E(G)| = r + 1. Subcase B.2. p = 1. Let A(G) = {w}. By |E(H −{u, v})| ≤ (p − 1)(n + s − 2k + p) + 1, it must hold that |E(H −{u, v})| = 1, i.e., E(H − {u, v}) = {e}. Let e = w z. Then we have

|E(G)| = |[A(G), C (G)]G | + |[A(G), {u, v}]G | + |E(G[C (G)])| + |E(H − {u, v})|

308

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

(

2k − 4

≤

) + 2k − 2 + 1

2

( =

2k − 3

)

2

+ 3.

Note that |E(G)| = r + 1 ≥ + 2. So |[w, {u, v, z }]G | ≥ 2, i.e., |[w, {u, v}]G | ≥ 1. Note that 2 n ≥ 2k − 1 and s ≥ 2. Then D(G) − {u, v, z } ̸ = ∅. Let x ∈ D(G) − {u, v, z }. Assume that wv ∈ E(G). By the analysis of Case A and Case B, we can assume that ux is of the color of an edge in M1 . Then either M2 ∪ {ux, w z } or M2 ∪ {ux, wv} is a rainbow kK2 , a contradiction. So wv ̸ ∈ E(G) and w u, w z ∈ E(G). Furthermore, we can assume that each edge between u and D(G) − {u, z } is of the color c(w z). If N ∩ [D(G) − {u, z }] ̸ = ∅, then there are two vertices u′ , v ′ ∈ D(G) − {u, z } with u′ v ′ ∈ E(Kn + K¯ s ). Hence either M2 ∪{u′ v ′ , w z } or M2 ∪{u′ v ′ , w u} is a rainbow kK2 , a contradiction. So N ∩[D(G)−{u, z }] = ∅. Note that n ≥ 2k − 1 and s ≥ 2. Then z ∈ N. By the same analysis of the vertex u, we can assume that each edge between z and D(G) − {u, z } is of the color c(w u). Clearly, M2 ∪ {uv, zx} is a rainbow kK2 , a contradiction. This completes the proof of the claim. ■

( 2k−3 )

∑q

Since p + i=1 2li + 1 +|C (G)| = n + s and q = n + s − 2k + 2 + p, we have 1 + n + s − p − q − 1 = 2k − 2p − 3. Then we have

|E(G)| ≤

(p) 2

− ≤ ≤ =

+ p(n + s − p) +

) q ( ∑ 2li + 1 i=1

(s ) a

2

(p) 2

(p) 2

(p) 2

−

q ( ∑ sd ) i

i=1

2

−

(s ) c

2

( + p(n + s − p) + + p(n + s − p) +

− sa

q ∑

|C (G)|

i=1 2l2

+|C (G)|− 2 =

)

2

) sdi + sc

i=1

1+

∑q

1+

∑q

i=1

2li

)

( +

|C (G)|

2k − 2p − 3 2

)

)

2

i=1 2li + |C (G)| − 2

2

(

(p)

(

+

2

(

+ p(n + s − p) +

2

(

∑q

)

( ) +

2 2

+ 1.

( 2k−2p−3 )

Let f4 (p) = 2 + p(n + s − p) + + 1. Since f4 (p) is( a convex 2 ) function of p and 0 ≤ p ≤ k − 3, 2k−3 we obtain |E(G)| ≤ max{f4 (0), f4 (k − 3)}. Note that f4 (0) = + 1 and 2

( f4 (k − 3) =

) + (k − 3)(n + s − k + 3) + 4

2

( =

k−3

) + k − 3 + (k − 3)(n + s − k + 2) + 4

2

( = <

k−3

k−2

) + (k − 2)(n + s − k + 2) + 1 − (n + s − k) + 1

2

(

k−2 2

)

+ (k − 2)(n + s − k + 2) < r .

Therefore, we have |E(G)| ≤ max{f4 (0), f4 (k − 3)} ≤ r, a contradiction to the fact |E(G)| = r + 1. Subcase 3.3. 0 ≤ p ≤ k − 3 and |C (G)| = 0. ∑q ( 2li +1 ) Since l1 ≤ k − p − 2 , we have 2l1 + 1 ≤ 2k − 2p − 3. Since + p = n + s and i=1 2 q = n + s − 2k + 2 + p, we have q ∑ i=2

2li = n + s − p − 2l1 − 1 − (q − 1) ≥ 2, i.e., l2 ≥ 1.

Since

∑q ( 2li +1 ) i=1

2

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

309

∑q ( 2li )

+1 = n+s−p−(q−1) =

+p = n+s and q = n+s−2k+2+p, we have

i=1

2

2k − 2p − 2. Then

|E(G)| ≤ ≤ =

(p) 2

+ p(n + s − p) +

) q ( ∑ 2li + 1 2

i=1

(p) 2

(p) 2

( + p(n + s − p) +

−

q ( ∑ sd ) i

2

i=1

2l1 + 1 + 2l2 − 2 +

− sa

+ p(n + s − p) +

2k − 2p − 3

)

2

sdi

i=1

∑q

i=3

2li

2

(

q ∑

( )

) +

3 2

+ 3.

(p) ( 2k−2p−3 ) ( ) + 3. Note that f5 (0) = 2k2−3 + 3, f5 (1) = 2 ( 2k−Let ) f5 (p) = 2 + p(n + s − p) + 3 + n + s − 4k + 11 and 2 ( f5 (k − 3) =

) + (k − 3)(n + s − k + 3) + 6

2

( =

k−3

) + k − 3 + (k − 3)(n + s − k + 2) + 6

2

( =

k−2

) + (k − 2)(n + s − k + 2) + 1 − (n + s − k) + 3

2

( ≤

k−3

k−2

)

2

+ (k − 2)(n + s − k + 2) + 1 ≤ r .

( 2k−3 )

If p = 0 and |E(G)| ≤ + 1 ≤ r, then it contradicts to the fact |E(G)| = r + 1. If p = 0 and 2 ( ) |E(G)| ≥ 2k2−3 + 2, then we can deduce that G contains a subgraph isomorphic to I. It follows from Lemma 4.1 that there is a rainbow k-matching in Kn + K¯ s , a contradiction. If 1 ≤ p ≤ k − 3 and f5 (1) ≤ f5 (k − 3), then we have |E(G)| ≤ f5 (k − 3) < r, a contradiction to the fact |E(G)| = r + 1. If 1 ≤ p ≤ k − 3 and f5 (1) > f5 (k − 3), then we have

(

2k − 3 2

)

+ n + s − 4k + 11 >

So we have (k − 4)(n + s) <

5k2 −27k+28 . 2

(

k−2 2

)

+ (k − 2)(n + s − k + 2) + 1 − (n + s − k) + 3.

Then we obtain n + s <

5k−7 2

and k > 4. Therefore, we have

( ) 2k − 3 |E(G)| ≤ f5 (1) = + n + s − 4k + 11 2 ( ) 2k − 3 15 − 3k < + 2 2 ( ) 2k − 3 ≤ < r, 2

a contradiction to the fact |E(G)| = r + 1. Case 4. k ≤ n ≤ 2k − 2, 0 ≤ p ≤ k − 2 and l1 = k − p − 1. Claim 3. If Kn + K¯ s contains a rainbow spanning subgraph I1 = (Kn + K¯ 2k−2−n ) ∪ K¯ n+s−2k+2 + u1 u2 + u2 u3 where u2 ∈ V (Kn ) and u1 , u3 ∈ V (K¯ n+s−2k+2 ), then Kn + K¯ s has a rainbow k-matching.

310

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

Fig. 1. The graph I1 for the case k = 3.

Proof. Suppose that Kn + K¯ s does not contain any rainbow k-matching. Since n ≥ k ≥ 3 and 2k − 2 ≥ 4, we can take vertices x, z ∈ V (Kn − u2 ) and y ∈ V (Kn + K¯ 2k−2−n − u2 ). Clearly, we have zu1 , xu3 ̸ ∈ E(I1 ) and yu2 ∈ E(I1 ). Suppose that k = 3. Then 3 = k ≤ n ≤ 2k − 2 = 4. So Kn + K¯ 2k−2−n ∼ = K4 . The graph I1 is illustrated in Fig. 1. We claim that c(zu1 ) ∈ c({xy, u2 u3 }) and c(xu3 ) ∈ c({u1 u2 , yz }). Otherwise, either {zu1 , u2 u3 , xy} or {xu3 , u1 u2 , yz } is a rainbow 3-matching in Kn + K¯ s , a contradiction. Then, Kn + K¯ s has a rainbow k-matching {zu1 , xu3 , yu2 }, a contradiction. So k ≥ 4. Then Kn + K¯ 2k−2−n − z − u2 ∼ = Kn−2 + K¯ 2k−2−n and Kn + K¯ 2k−2−n − z − u2 contains a (k − 2)-matching M. Without loss of generality, let xy ∈ M. Clearly, M ∪ {u2 u3 } is a maximum matching in I1 and there is an edge e1 ∈ M ∪ {u2 u3 } such that c(e1 ) = c(zu1 ). Suppose that e1 ∈ M. Then Kn + K¯ 2k−2−n − z − u2 − e1 contains a (k − 2)-matching, the union of which and {zu1 , u2 u3 } is a rainbow k-matching, a contradiction. So we have e1 = u2 u3 , i.e., c(zu1 ) = c(u2 u3 ). For the same reason, we have c(xu3 ) = c(u1 u2 ). Therefore, (M − xy) ∪ {zu1 , xu3 , yu2 } is a rainbow k-matching in Kn + K¯ s , a contradiction. This completes the proof. ■ Since p + 2l1 + 1 + q − 1 = n + s, we have C (G) = ∅ and l2 = · · · = lq = 0. So sdi ≤ 1 for 2 ≤ i ≤ q. Let v ∈ V (D1 ) ∩ N and H be the subgraph induced by [A(G), D(G) − V (D1 )]G . From the definition of D(G), there exists a maximum matching of G which does not saturate v . From Lemma 2.2, H contains a p-matching M1 and D1 − v has a l1 -matching M2 . Clearly, M1 ∪ M2 is a maximum matching of G which does not saturate v . Since q = n + s − 2k + 2 + p ≥ p + 2, there exists a vertex u ∈ D(G) − V (D1 ) not saturated by M1 ∪ M2 . Clearly, uv ̸ ∈ E(G) and there is an edge e ∈ M1 ∪ M2 such that c(e) = c(uv ). Subcase 4.1. e ∈ M1 . Since e ∈ M1 , we have p ≥ 1. If H − u − e still has a p-matching, then the union of that matching and M2 ∪ {uv} is a rainbow k-matching in Kn + K¯ s , a contradiction. So the maximum matching of H − u − e is of size p − 1. Then it follows from Lemma 2.1 that |E(H − u)| ≤ ext(Kp,n+s−2k+p , pK2 ) + 1 = (p − 1)(n + s − 2k + p) + 1. So we have

|E(G)| = |E(D1 )| + |E(G[A(G)])| + |[A(G), V (D1 ) ∪ {u}]G | + |E(H − u)| ( ) ( ) ( ) ( ) sd1 p sa 2l1 + 1 ≤ − + − + p(2k − 2p) 2

2

2

2

+ (p − 1)(n + s − 2k + p) + 1 − sa

q ∑

sdi

i=1

( ≤

2k − 2p − 1

)

(p)

+ p(2k − 2p) ( ) sd1 + sa + (p − 1)(n + s − 2k + p) + 1 − . 2

+

2

2

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

Let f6 (p) =

( 2k−2p−1 ) 2

+

(p) 2

311

+ p(2k − 2p) + (p − 1)(n + s − 2k + p) + 1 −

(

sd +sa 1

2

)

. If p = k − 2,

then we have

( f6 (k − 2) ≤ 3 +

k−2 2

) + 4(k − 2) + (k − 3)(n + s − k − 2) + 1

) ( k−2 + (k − 2)(n + s − k + 2) − (n + s − k − 2) + 1 =3+ 2 ) ( k−2 + (k − 2)(n + s − k + 2) + 1 + 5 − (n + s − k). = 2

Then we show that f6 (k − 2) ≤ r if k ≥ 4. If n + s ≥ 2k + 1, then f6 (k − 2) ≤ k + 2) + 1 ≤ r. If n + s = 2k, then

( f6 (k − 2) ≤

=

k−2

+ (k − 2)(n + s − k + 2) + 1 + 5 − (n + s − k) + (k − 3)(n + s − k + 2) + 8

)

2 3k2 − 7k 2

+ (k − 2)(n + s −

)

2

(

=

k−2

2

)

2

(

=

k−2

( k−2 )

+ (k − 3)(2k − k + 2) + 8 + 5.

( ) 2 + 1 − 2k−22−n = − n2 + (2k − 52 )n + 1. The symmetry axis of g(n) is n = 2k − 52 . ( 2k−2 ) ( ) 2 2 So g(n) ≥ g(k) = 3k 2−5k + 1 ≥ 3k 2−7k + 5. Then we have f6 (k − 2) ≤ g(k) ≤ + 1 − 2k−22−n if 2 k ≥ 4. Therefore, f6 (k − 2) ≤ r if k ≥ 4. ( 2k−3 ) ( 2k−2−n ) = 2 ( 2k−If2p) = 1, then ( 2ks−d21−+n )sa ≥ 2k − 3 + 1 − n = 2k − 2 − n. So we have f6 (1) ≤ 2 + 2k − 1 − + 2 − . 2 2 If k > 4, then 1 < 2 < k − 2. Since f6 (p) is a quadratic ( 2k−2 ) function ( of p with ) a positive coefficient at p2 , we have f6 (2) < max{f6 (1), f6 (k − 2)} ≤ max{ 2 + 2 − 2k−22−n , r }. So f6 (2) ≤ r. If k = 4, then f6 (2) = f6 (k − 2) ≤ r. Hence, f6((2) ≤)r if k ≥ 4. ( ) 2k−2 Suppose that p = 1. If |E(G)| ≤ + 1 )− 2k−22−n ≤ r, then it is a contradiction to the fact 2( ( 2k−2 ) |E(G)| = r + 1. If |E(G)| = + 2 − 2k−2−n , then we can deduce that G ∼ = I1 . By Claim 3, Kn + K¯ s Let g(n) =

( 2k−2 ) 2

2

2

contains a rainbow k-matching, a contradiction. So we have 2 ≤ p ≤ k − 2. Then k ≥ 4. Therefore, |E(G)| ≤ max{f6 (k − 2), f6 (2)} ≤ r, a contradiction to the fact |E(G)| = r + 1. Subcase 4.2. e ∈ M2 . Claim 4. Let n = k. If Kn + K¯ s contains a rainbow spanning subgraph I2 = Kk−2 + K¯ n+s−k+2 + u1 u2 + u2 u3 + u1 u3 where u1 , u2 ∈ V (K¯ n+s−k+2 ) ∩ N and u3 ∈ S, then Kn + K¯ s has a rainbow k-matching. Proof. Suppose that Kn + K¯ s does not contain any rainbow k-matching. By the definition of I2 , we have V (Kk−2 ) ⊆ N. Since n = k ≥ 3 and n + s ≥ 2k, we can take vertices x ∈ V (Kk−2 ) and y, z ∈ V (Kn+s−k+2 − u1 − u2 − u3 ). Clearly, we have xu3 ∈ E(I2 ) and u1 y, u2 z ̸ ∈ E(I2 ). Suppose that k = s = 3. Then I2 − {u1 , u2 , u3 } ∼ = P3 where P3 = yxz. The graph I2 is illustrated in Fig. 2. We claim that c(u2 z) ∈ c({xy, u1 u3 }) and c(u1 y) ∈ c({xz , u2 u3 }). Otherwise, either {u2 z , xy, u1 u3 } or {u1 y, xz , u2 u3 } is a rainbow 3-matching in Kn + K¯ s , a contradiction. Then Kn + K¯ s contains a rainbow 3-matching {u1 y, u2 z , xu3 }, a contradiction. So we have k ≥ 4 or s ≥ 4. Obviously, we have I2 − {u1 , u2 , u3 , z } ∼ = Kk−2 + K¯ n+s−k−2 . Then I2 − {u1 , u2 , u3 , z } contains a rainbow (k − 2)-matching M. Without loss of generality, let xy ∈ M. It is easy to see that M ∪ {u1 u3 } is a maximum matching in I2 . Clearly, there is an edge e1 ∈ M ∪ {u1 u3 } such that c(e1 ) = c(u2 z). Suppose that e1 ∈ M. Then I2 − {u1 , u2 , u3 , z } − e1 contains a (k − 2)matching, the union of which and {u1 u3 , u2 z } is a rainbow k-matching in Kn + K¯ s , a contradiction. So

312

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

Fig. 2. The graph I2 for the case k = 3, s = 3.

e1 = u1 u3 and c(u2 z) = c(u1 u3 ). For the same reason, we can prove c(u1 y) = c(u2 u3 ). Therefore, (M − xy) ∪ {u1 y, xu3 , u2 z } is a rainbow k-matching in Kn + K¯ s , a contradiction. This completes the proof. ■ Claim 5. Let n ≥ k + 1. If Kn + K¯ s contains a rainbow spanning subgraph I3 = Kk−2 + K¯ n+s−k+2 + u1 u2 + u2 u3 where u1 , u2 , u3 ∈ V (Kn+s−k+2 ), then Kn + K¯ s has a rainbow k-matching. Proof. Suppose that Kn + K¯ s does not have any rainbow k-matching. By the definition of I3 , we have V (Kk−2 ) ⊆ N. Since k ≥ 3 and n + s ≥ 2k, we can take vertices x ∈ V (Kk−2 ) and y, z ∈ V (Kn+s−k+2 − u1 − u2 − u3 ). Case A. {u1 , u2 , u3 } ⊆ N. Clearly, we have xy, xz , xu2 ∈ E(I3 ) and u1 y, u3 z ̸ ∈ E(I3 ). Suppose that k = 3 and s = 2. Since k + 1 ≤ n ≤ 2k − 2, we have n = 4. Clearly, I3 − {u1 , u2 , u3 } ∼ = P3 where P3 = yxz. The graph I3 is illustrated in Fig. 3. We claim that c(u3 z) ∈ c({xy, u1 u2 }) and c(u1 y) ∈ c({xz , u2 u3 }). Otherwise, either {u3 z , xy, u1 u2 } or {u1 y, xz , u2 u3 } is a rainbow 3-matching in Kn + K¯ s , a contradiction. Then Kn + K¯ s contains a rainbow 3-matching {u1 y, u3 z , xu2 }, a contradiction. So we have k ≥ 4 or s ≥ 3. Obviously, we have I3 − {u1 , u2 , u3 , z } ∼ = Kk−2 + K¯ n+s−k−2 . Then I3 − {u1 , u2 , u3 , z } contains a rainbow (k − 2)-matching M. Without loss of generality, let xy ∈ M. It is easy to see that M ∪{u1 u2 } is a maximum matching in I3 . Clearly, there is an edge e1 ∈ M ∪{u1 u2 } such that c(e1 ) = c(u3 z). Suppose that e1 ∈ M. Then I3 −{u1 , u2 , u3 , z }− e1 contains a (k − 2)-matching, the union of which and {u1 u2 , u3 z } is a rainbow k-matching in Kn + K¯ s , a contradiction. So e1 = u1 u2 and c(u3 z) = c(u1 u2 ). For the same reason, we have c(u1 y) = c(u2 u3 ). Therefore, (M − xy) ∪ {u1 y, xu2 , u3 z } is a rainbow k-matching in Kn + K¯ s , a contradiction. Case B. |N ∩ {u1 , u2 , u3 }| = 2. Without loss of generality, we assume that {u2 , u3 } ⊆ N and u1 ∈ S. Then |N | − |V (Kk−2 )| − |{u2 , u3 }| = n − (k − 2) − 2 ≥ 1. So there exists a vertex in V (I3 − {u1 , u2 , u3 }) ∩ N. Without loss of generality, let z be such a vertex. Clearly, we have xy, xz , xu2 ∈ E(I3 ) and u3 y, u1 z ̸ ∈ E(I3 ). Using the similar way as the proof of Case A, we can check the correctness for the case |N ∩ {u1 , u2 , u3 }| = 2 and here we omit the detailed proof. Case C. |N ∩ {u1 , u2 , u3 }| = 1. Then u2 ∈ N and {u1 , u3 } ⊆ S. Notice that |N | − |V (Kk−2 )| − |{u2 }| = n − (k − 2) − 1 ≥ 2. So there exist two vertices in V (I3 − {u1 , u2 , u3 }) ∩ N. Without loss of generality, let y, z be such vertices. Clearly, we have xy, xz , xu2 ∈ E(I3 ) and u1 y, u3 z ̸ ∈ E(I3 ). Using the similar way as the proof of Case A, we can check the correctness for the case |N ∩ {u1 , u2 , u3 }| = 1 and here we omit the detailed proof. This completes the proof of the claim. ■

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

313

Fig. 3. The graph I3 for the Case A.

If D1 −v−e still has a (k−p−1)-matching, then the union of that matching and M1 ∪{uv} is a rainbow k-matching in Kn + K¯ s , a contradiction. So the maximum matching of D1 −v − e is of size k − p − 2. Note that |(V (D1 − v )) ∩ S | = sd1 . We have |E(D1 − v )| ≤ ext(K2k−2p−2−sd + K¯ sd , (k − p − 1)K2 ) + 1. From 1

( 2k−2p1−3 )

Lemma 2.3, we have ext(K2k−2p−2−sd + K¯ sd , (k − p − 1)K2 ) ≤ max{ 1

2

1

−

(k − p − 2)(k − p)}. Note that |[V (D1 − v ), v]G | ≤ 2k − 2p − 2. Then we have

(

sd −1 1

2

) ( ) , k−p2−2 +

|E(G)| = |E(D1 − v )| + |[V (D1 − v ), v]G | + |E(G[A(G)])| + |[A(G), D(G)]G | ≤ ext(K2k−2p−2−s + K¯ s , (k − p − 1)K2 ) + 1 d1

+ 2k − 2p − 2 +

d1

(p) 2

+ p(n + s − p) −

(s ) a

2

− sa

q ∑

sdi .

i=1

Note that |V (D1 )| = 2l1 + 1 = 2k − 2p − 1. By Lemma 2.2, D1 is factor-critical. Suppose that sd1 ≥ k − p. Then |V (D1 ) ∩ N | ≤ k − p − 1. Let w ′ ∈ V (D1 ) ∩ N. Clearly, D1 − w ′ does not contain any perfect i.e., D1 is not factor-critical, a contradiction. So sd1 ≤ k − p − 1. ( matching, ) Then

( 2k−2p−3 ) 2

ext(K2k−2p−2−sd

1

( ) ( ) − k−p2−2 = k−p2−2 + (k − p − 2)(k − p). Thus, we have ( ) ( ) s −1 −3 + K¯ sd1 , (k − p − 1)K2 ) ≤ 2k−2p − d12 and 2

(

|E(G)| ≤

sd −1

−

1

2

≥

( 2k−2p−3 ) 2

2k − 2p − 3

)

2k − 2p − 3

)

(

)

(p) + 1 + 2k − 2p − 2 + 2 2 2 ( ) q (s ) ∑ a + p(n + s − p) − − sa sd1 − 1 + 1 + sdi −

sd1 − 1

2

( ≤

i=2

(p)

+ 2k − 2p − 1 + 2 ( ) sa + sd1 − 1 + p(n + s − p) − . 2

2

Let f7 (p) =

( 2k−2p−3 ) 2

( f7 (k − 2) ≤ 3 +

+ 2k − 2p − 1 + k−2 2

)

(p) 2

+ p(n + s − p) −

+ (k − 2)(n + s − k + 2).

(

sa +sd −1 1

2

)

. If p = k − 2, then we have

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Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

If p = 0, then sa = 0. So we have sa + sd1 − 1 ≥ 2k − 2 − n. Thus,

( f7 (0) ≤

2k − 3

)

( + 2k − 1 −

2

2k − 2 − n

)

2

( =

2k − 2 2

)

( +2−

2k − 2 − n 2

)

.

If p = k − 3, then we have

( f7 (k − 3) ≤ 3 + 2k − 2(k − 3) − 1 +

( =8+

k−2 2

k−3 2

) + (k − 3)(n + s − k + 3)

) − (k − 3) + (k − 3)(n + s − k + 3)

( ) k−2 =8+ + (k − 3)(n + s − k + 2) 2 ( ) k−2 = + (k − 2)(n + s − k + 2) + 1 + 5 − (n + s − k). 2

Then we show that f7 (k − 3) ≤ r if k ≥ 4. If n + s ≥ 2k + 1, then f7 (k − 3) ≤ k + 2) + 1 ≤ r. If n + s = 2k, then

( f7 (k − 3) ≤

=

k−2

+ (k − 2)(n + s − k + 2) + 1 + 5 − (n + s − k) + (k − 3)(n + s − k + 2) + 8

)

2 3k2 − 7k 2

+ (k − 2)(n + s −

)

2

(

=

k−2

2

)

2

(

=

k−2

( k−2 )

+ (k − 3)(2k − k + 2) + 8 + 5.

( ) 2 + 1 − 2k−22−n = − n2 + (2k − 52 )n + 1. The symmetry axis of g(n) is n = 2k − 52 . So ( 2k−2 ) ( ) 2 2 g(n) ≥ g(k) = 3k 2−5k + 1 ≥ 3k 2−7k + 5. Then we have f7 (k − 3) ≤ + 1 − 2k−22−n . Therefore, 2 f7 (k − 3) ≤ r if k ≥ 4. If k ≥ 5, then we have 0 < 1 < k − 3. Since f7 (p) is a quadratic ( 2k−2 ) function ( of p) with a positive coefficient at p2 , we have f7 (1) < max{f7 (0), f7 (k − 3)} ≤ max{ 2 + 2 − 2k−22−n , r }. So we have f7 (1) ≤ r. If k = 4, then f7 (1) = f7 (k − ≤ we obtain ( 2k3) ) r. Thus,( 2k ) f7 (1) ≤ r if k ≥ 4. −2 −2−n Suppose that p = 0. If |E(G)| ≤ + 1 − ≤ r, then it is a contradiction to the fact 2 ( ) ( ) 2 |E(G)| = r + 1. If |E(G)| = 2k2−2 + 2 − 2k−22−n , then we can deduce that G − e + uv ∼ = I1 . From Claim 3, Kn + K¯ s contains a rainbow k-matching, a contradiction. So p ≥ 1. Suppose that 1 ≤ p ≤ k − 3. Then k ≥ 4. Hence, we have |E(G)| ≤ max{f7 (1), f7 (k − 3)} ≤ r, a contradiction to the fact |E(G)| = r( + 1.) k−2 So p = k − 2. If n = k and |E(G)| ≤ 2 + (k − 2)(n + s − k + 2) + 2 ≤ r, then it is a contradiction ( k−2 ) to the fact |E(G)| = r + 1. If n = k and |E(G)| = + (k − 2)(n + s − k + 2) + 3, then we can 2 ∼ ¯ s has a rainbow k-matching by Claim 4, a contradiction. If n > k and deduce that G I . So K + K = 2 n ( ) |E(G)| ≤ k−2 2 + (k −( 2)(n) + s − k + 2) + 1 ≤ r, then it is a contradiction to the fact |E(G)| = r + 1. k−2 If n > k and |E(G)| ≥ 2 + (k − 2)(n + s − k + 2) + 2, then we can deduce that either G ∼ = I3 or G contains a spanning subgraph isomorphic to I3 . Then Kn + K¯ s has a rainbow k-matching by Claim 5, a Let g(n) =

( 2k−2 ) 2

contradiction. Case 5. 0 ≤ p ≤ k − 2, l1 = k − p − 1 and n ≥ 2k − 1. Since p + 2l1 + 1 + q − 1 = n + s, we have C (G) = ∅ and l2 = · · · = lq = 0. So sdi ≤ 1 for 2 ≤ i ≤ q. Let H be the subgraph induced by [A(G), D(G) − V (D1 )]G . From Lemma 2.2, H contains a p-matching M1 and D1 has a l1 -matching M2 . Moreover, M1 ∪ M2 is a maximum matching of size k − 1 in G. Since n ≥ 2k − 1 and |V (D1 )| = 2k − 2p − 1 ≤ |N |, we can distinguish the following subcases to consider it.

Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

315

Subcase 5.1. V (D1 ) ̸ = N. Since q = n + s − 2k + 2 + p ≥ p + s + 1 and V (D1 ) ̸ = N, there exists a vertex u ∈ (D(G) − V (D1 )) ∩ N and one vertex v ∈ D(G) − V (D1 ) − u which are not saturated by M1 ∪ M2 . Clearly, uv ̸ ∈ E(G) and there is an edge xy ∈ M1 ∪ M2 such that c(xy) = c(uv ). Suppose that xy ∈ M2 . By Lemma 2.2, D1 − x contains a l1 -matching, the union of which and M1 ∪ {uv} is a rainbow k-matching in Kn + K¯ s , a contradiction. So xy ∈ M1 . Then we have p ≥ 1. Let y ∈ A(G). Suppose that there exists a vertex z ∈ V (D1 ) such that yz ∈ E(G). From Lemma 2.2, D1 − z contains a l1 -matching, the union of which and (M1 − {xy}) ∪ {uv, yz } is a rainbow k-matching in Kn + K¯ s , a contradiction. So [y, V (D1 )]G = ∅. Moreover, H − xy − {u, v} does not have any pmatching. Otherwise, H − xy − {u, v} has a p-matching, the union of which and M2 ∪ {uv} is a rainbow k-matching in Kn + K¯ s , a contradiction. So |E(H − {u, v})| ≤ ext(Kp,q−3 , pK2 ) + 1 ≤ (p − 1)(q − 3) + 1 = (p − 1)(n + s − 2k + p + 1) + 1. Then

|E(G)| = |E(D1 )| + |E(G[A(G)])| + |E(H − {u, v})| + |[A(G), V (D1 )]G | + |[A(G), {u, v}]G | ( ) ( ) 2k − 2p − 1 p ≤ + + 1 + (p − 1)(n + s − 2k + p + 1) 2

2

+ (2k − 2p − 1)(p − 1) + 2p. ( 2k−2p−1 ) ( p ) Let f8 (p) = + 2 +( 1 + )(p − 1)(n + s − 2k + p + 1) + (2k − 2p − 1)(p − 1) + 2p. Note ( 2k−3 )2 2k−3 that f8 (1) = + 3, f + n + s − 4k + 11 and 8 (2) = 2 2 ( ) k−2 f8 (k − 2) = + 3 + 1 + (k − 3)(n + s − 2k + k − 2 − 1) 2

+ (2k − 2(k − 2) − 1)(k − 3) + 2(k − 2) ( ) k−2 = + 3 + 1 + (k − 3)(n + s − k) + 2(k − 2) 2 ( ) k−2 = + (k − 2)(n + s − k + 2) + 1 − (n + s − k) + 3 2 ( ) k−2 ≤ + (k − 2)(n + s − k + 2) + 1 ≤ r . 2

( 2k−3 )

If p = 1 and |E(G)| ≤ + 1 ≤ r, then it contradicts to the fact |E(G)| = r + 1. If p = 1 and 2 ( ) |E(G)| ≥ 2k2−3 + 2, then G − xy + uv contains a spanning subgraph which is isomorphic to I. By Lemma 4.1, there is a rainbow k-matching in Kn + K¯ s , a contradiction. If 2 ≤ p ≤ k − 2 and f8 (2) ≤ f8 (k − 2), then we have |E(G)| ≤ f8 (k − 2) ≤ r, a contradiction to the fact |E(G)| = r + 1. If 2 ≤ p ≤ k − 2 and f8 (2) > f8 (k − 2), then

(

2k − 3 2

)

+ n + s − 4k + 11 >

So we have (k − 4)(n + s) <

(

5k2 −27k+28 . 2

(

k−2 2

)

+ (k − 2)(n + s − k + 2) + 1 − (n + s − k) + 3.

Then we obtain n + s <

)

2k − 3 |E(G)| ≤ f8 (2) = + n + s − 4k + 11 2 ( ) 2k − 3 15 − 3k < + 2 2 ( ) 2k − 3 ≤ < r, 2

a contradiction to the fact |E(G)| = r + 1.

5k−7 2

and k > 4. Therefore, we have

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Z. Jin et al. / European Journal of Combinatorics 70 (2018) 297–316

Subcase 5.2. V (D1 ) = N. Then p = 0, |V (D1 )| = 2k − 1 and q = s + 1. Let u ∈ D1 and v ∈ D(G) − V (D1 ). Since Kn + K¯ s does not have any rainbow k-matching, each (k − 1)-matching of D1 − u has an edge with the color c(uv ). Let xy ∈ E(D1 − u) with c(xy) = c(uv ). So G − {u, v} − xy does not contain any (k − 1)-matching. We consider the graph H = G − xy + uv . Clearly, H has a maximum matching of size k − 1 and each vertex of S ∪ {x, y} is not saturated by some maximum matchings of H. So S ∪ {x, y} ⊆ D(H) and the u ∈ A(H), i.e., |A(H)| ≥ 1. Applying similar technique in the proofs of Cases 1, 3 and Subcase 5.1, we can obtain contradictions. This completes the proof. ■ Acknowledgments This work was supported by National Natural Science Foundation of China (11571320, 11671366 and 11401389) and Zhejiang Provincial Natural Science Foundation (LY14A010009 and LY17A010017). The authors are very grateful to the referees for helpful comments and suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32]

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