An extremal problem for vertex partition of complete multipartite graphs

Discrete Mathematics 339 (2016) 1699–1705

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An extremal problem for vertex partition of complete multipartite graphs Tomoki Nakamigawa Department of Information Science, Shonan Institute of Technology, 1-1-25 Tsujido-Nishikaigan, Fujisawa, Kanagawa 251-8511, Japan

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Article history: Received 7 March 2014 Received in revised form 4 November 2015 Accepted 21 January 2016

Keywords: Extremal graph theory Graph Ramsey theory Graph decomposition

abstract For a graph G and a family H of graphs, a vertex partition of G is called an H -decomposition, if every part induces a graph isomorphic to one of H . For 1 ≤ a ≤ k, let A(k, a) denote the graph which is a join of an empty graph of order a and a complete graph of order k − a. Let Ak = {A(k, a) : 1 ≤ a ≤ k}. In this paper, extremal problems related to H -decomposition of a complete multipartite graph, where H ⊂ Ak , are studied. Among other results, it is proved that for every complete multipartite graph G of order kℓ, where ℓ ≥ k − 2 ≥ 2, there is a positive integer a such that G admits an {A(k, a), A(k, a + 1), A(k, a + 2)}-decomposition. © 2016 Elsevier B.V. All rights reserved.

1. Introduction A graph is finite and undirected with no multiple edges or loops. Let H be a family of graphs. For a graph G, we call a vertex partition V (G) = V1 ∪ · · · ∪ Vℓ an H -decomposition, if G[Vi ] ∈ H for all 1 ≤ i ≤ ℓ, where G[Vi ] is a subgraph of G induced by Vi . In the following, we mainly consider the case where G is a complete multipartite graph and H consists of graphs with a common number of vertices. Our aim is to find sufficient conditions for the existence of an H -decomposition having some nice properties. The problems raised in the paper can be considered as a coin problem. Let a set of piles of coins be given. A pile of coins corresponds to a partite set of complete multipartite graph, and a rearrangement of coins corresponds to a vertex partition of a complete multipartite graph. The next results were proved in [3,4]. Theorem A ([3]). For every complete multipartite graph G of order (k + 1)ℓ − 1, where ℓ ≥ k − 2, there is an induced subgraph G′ of order kℓ such that G′ admits an {H }-decomposition with some complete multipartite graph H of order k. Theorem B ([4]). For every complete multipartite graph G of order kℓ, where k ≥ 2 and ℓ ≥ 2, there is a pair of complete multipartite graphs H1 , H2 of order k such that G admits an {H1 , H2 }-decomposition. In the following, Kn1 ,n2 ,...,ns is denoted by (n1 , n2 , . . . , ns ). Furthermore, if t partite sets have a common order a, we write as (. . . , at , . . .) instead of (. . . , a, a, . . . , a, . . .). In particular, (n) denotes the empty graph of order n, and (1n ) denotes the complete graph of order n. Let Ak = {(a, 1k−a ) : 1 ≤ a ≤ k}. Ak is a family of graphs of order k which consists of a complete graph (1k ), and an empty graph (k) and joins of a complete graph and an empty graph. In this paper, we focus on an H -decomposition, where H ⊂ Ak .

E-mail address: [email protected]. http://dx.doi.org/10.1016/j.disc.2016.01.012 0012-365X/© 2016 Elsevier B.V. All rights reserved.

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2. Main results In this section, we present the main results of the paper. The proofs will be given in Section 3. Firstly, we have the following result. Theorem 1. For every complete multipartite graph G of order kℓ, where ℓ ≥ k − 2 ≥ 2, there is a positive integer a such that G admits an {(a, 1k−a ), (a + 1, 1k−(a+1) ), (a + 2, 1k−(a+2) )}-decomposition. The next statement is an immediate consequence of Theorem 1. Corollary 2. Every complete multipartite graph G of order kℓ, where ℓ ≥ k − 2 ≥ 2, admits an Ak -decomposition. The bound for ℓ in Corollary 2 (and also in Theorem 1) is tight. To see this, consider the complete bipartite graph G = ((k − 1)ℓ − 1, k − 2) with ℓ = k − 3 and k ≥ 4. Then G has order kℓ, but G has no Ak -decomposition. Assume to the contrary that G has an Ak -decomposition. Then G has a {(k), (k − 1, 1)}-decomposition, since G has only two partite sets and contains no copy of (a, 1k−a ) for 1 ≤ a ≤ k − 2. However, this is impossible, since G has at most ℓ − 1 vertex disjoint copies of (k − 1). A related result of Theorem 1 is as follows. Theorem 3. Let G be a complete multipartite graph of order kℓ. Then the following statements hold: (a) If k = 3, then G has a {(2, 1), (3)}-decomposition or {(13 ), (2, 1)}-decomposition. (b) If k ≥ 4 and ℓ ≥ 2k − 6, then there is a positive integer a such that G admits an {(a, 1k−a ), (a + 1, 1k−(a+1) )}-decomposition. The bound for ℓ in Theorem 3(b) is tight. To see this, consider the complete multipartite graph G = ((k − 1) (k − 3)− 1, (k − 1)(k − 3)− 1, k − 4) with ℓ = 2k − 7 and k ≥ 4. Then G has order kℓ, but G has no {(a, 1k−a ), (a + 1, 1k−(a+1) )}decomposition. For k = 4, then G = (2, 2) has clearly no Ak -decomposition. For k ≥ 5, let P1 , P2 and P3 be three partite sets of G with |P1 | = |P2 | = (k − 3)(k − 1) − 1 and |P3 | = k − 4. Suppose to the contrary that G has an {(a, 1k−a ), (a + 1, 1k−(a+1) )}-decomposition, Since G has only three partite sets, we have a ≥ k − 2. Furthermore, since G contains at most ℓ − 1 vertex disjoint copies of (k − 1), we have a = k − 2. Without loss of generality, we may assume P1 is partitioned into k − 3 copies of (k − 2) and P2 is partitioned into k − 4 copies of (k − 2). However, this is impossible, since the number of the remaining vertices of P2 is (k − 1)(k − 3) − 1 − (k − 4)(k − 2) = 2k − 6, which is greater than ℓ. If every complete multipartite graph of order kℓ admits an H -decomposition, where H is a family of graphs each of which has order k, then we need {(k), (k − 1, 1)} ⊂ H for G = (kℓ − 1, 1) and also need (1k ) ∈ H for G = (1kℓ ). Conversely, these three graphs (k), (k − 1, 1), (1k ) suffice when ℓ is sufficiently large. Theorem 4. Every complete multipartite graph of order kℓ, where k ≥ 4 and ℓ ≥ (k − 2)2 , admits a {(k), (k − 1, 1),

(1k )}-decomposition.

The bound for ℓ in Theorem 4 is tight. To see this, consider the multipartite graph G = ((k − 1)ℓ − 1, (k − 2)k−2 ) with ℓ = (k − 2)2 − 1 and k ≥ 4. Then G has order kℓ, but G has no {(k), (k − 1, 1), (1k )}-decomposition. Assume to the contrary that G has a {(k), (k − 1, 1), (1k )}-decomposition. Then G has a {(k), (k − 1, 1)}-decomposition, since G has no copy of (1k ). However, this is impossible, since G has at most ℓ − 1 vertex disjoint copies of (k − 1). Theorem 5. Let G be a complete multipartite graph of order kℓ. Then the following statements hold: (a) If k = 3, then G has a {(3), (2, 1)}-decomposition or a {(3), (13 )}-decomposition. (b) If k ≥ 4 and ℓ ≥ 21 (3k2 − 9k + 4), then G has a {(k), (k − 1, 1)}-decomposition or a {(k), (1k )}-decomposition. The bound for ℓ in Theorem 5(b) is tight. To see this, consider the multipartite graph G = (k(k − 1)(k − 3)− 1, (k − 1)2 − 1,

(k − 2)(k − 1) − 1, (k − 3)(k − 1) − 1, . . . , 2(k − 1) − 1, 1(k −k−4)/2 ) with ℓ = (1/2)(3k2 − 9k + 2) and k ≥ 4. Then G has order kℓ, but G has neither {(k), (k − 1, 1)}-decomposition nor {(k), (1k )}-decomposition. Firstly, we will show that G has no {(k), (k − 1, 1)}-decomposition. The maximum number of vertex disjoint copies of (k − 1) in G is 2

k(k − 3) − 1 +

k−2 

i=

i=1

1 2

(3k2 − 9k) < ℓ.

Hence, G has no {(k), (k − 1, 1)}-decomposition. Next, we will show that G has no {(k), (1k )}-decomposition. Note that the maximum number of vertex disjoint copies of (k) in G is

(k − 1)(k − 3) − 1 +

k−2  i =1

i=

1 2

(3k2 − 11k + 6) = ℓ − k + 2.

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Let us take ℓ − x vertex disjoint copies of (k), and let us denote the maximum number of the remaining vertices in a partite set by r (x). In order to show that the set of the remaining vertices cannot be decomposed into copies of (1k ), it suffices to show that x < r (x) for any x with k − 2 ≤ x ≤ ℓ. We proceed by induction on x. For x = k − 2, since the number of the remaining vertices of the first partite set is k − 1, we have r (k − 2) = k − 1. Suppose that x > k − 2 and x − 1 < r (x − 1). Since the number of the vertices of partite sets having at least k vertices are all different modulo k, we have r (x) ≥ r (x − 1) + 1 > x. 3. Proofs of the main results In this section, we will show the proofs of the theorems in the previous section. In order to prove the theorems, we need a preliminary lemma. Lemma 6. For every complete multipartite graph G of order n in which the order of every partite set is at most ℓ, there exists a vertex partition V (G) = V1 ∪ · · · ∪ Vℓ such that G[Vi ] ∼ = (1k ) or (1k+1 ), where k = ⌊n/ℓ⌋, for all 1 ≤ i ≤ ℓ. Proof. Let s be the number of partite sets of G, and let Pj be the jth partite set of G for all 1 ≤ j ≤ s. Let s0 = 0. Put pj = |Pj | and sj = p1 + · · · + pj for 1 ≤ j ≤ s. Let us label the vertices of G such that Pj = {vx : sj−1 + 1 ≤ x ≤ sj }. Let us define Vi as {vx : x ≡ i (mod ℓ)} for all 1 ≤ i ≤ ℓ. We have |Vi | = k or k + 1 for all 1 ≤ i ≤ ℓ. Furthermore, since pj ≤ ℓ for all 1 ≤ j ≤ s, we have |Vi ∩ Pj | ≤ 1 for any i, j with 1 ≤ i ≤ ℓ, 1 ≤ j ≤ s. Hence, G[Vi ] is isomorphic to a complete graph for all 1 ≤ i ≤ ℓ. Therefore, a vertex partition V (G) = V1 ∪ · · · ∪ Vℓ is as desired.  Proof of Theorem 1. Let G be a complete multipartite graph of order kℓ with partite sets P1 , . . . , Ps . Let us define a as the maximum integer x such that G contains ℓ vertex disjoint copies of (x). If a = k, we have a {(k)}-decomposition of G. If a = k − 1, by combining ℓ singletons to ℓ copies of (k − 1) one by one, we have a {(k), (k − 1, 1)}-decomposition of G. Hence, we may assume a ≤ k − 2. Choose a vertex partition V (G) = W ∪ V1 ∪ · · · ∪ Vℓ of G so that (1) G[Vi ] ∼ = (a) or (a + 1) for all 1 ≤ i ≤ ℓ, (2) |W | is as small as possible subject to (1). Suppose that G[Vi ] ∼ = (a) for 1 ≤ i ≤ t and G[Vi ] ∼ = (a + 1) for t + 1 ≤ i ≤ ℓ. Note that we have t ≥ 1 by the choice of a. Define r as max{|W ∩ Pj | : 1 ≤ j ≤ s}. Claim. r ≤ ℓ. Suppose to a contradiction that r ≥ ℓ + 1. Since ℓ ≥ k − 2, we have r ≥ k − 1 ≥ a + 1. Let us take a partite set Pα satisfying |Pα ∩ W | = r. Let us take another partite set Pβ such that Pβ ⊃ Vf with G[Vf ] = (a). Then we modify the original partition to make a new partition V (G) = W ′ ∪ V1′ ∪ · · · ∪ Vℓ′ such that Vf′ ⊂ W ∩ Pα with |Vf′ | = a + 1, W ′ = (W \ Vf′ ) ∪ Vf , and Vi′ = Vi for 1 ≤ i ≤ ℓ and i ̸= f . Then we have |W ′ | < |W |. This contradicts to the choice of the original partition, as claimed. Now, we apply Lemma 6 to W . Since |W | = (k − a − 1)ℓ + t = (k − a)t + (k − (a + 1))(ℓ − t ), we have a partition W = S1 ∪ · · · ∪ Sℓ such that G[Si ] ∼ = (1k−a ) for 1 ≤ i ≤ t and G[Si ] ∼ = (1k−(a+1) ) for t + 1 ≤ i ≤ ℓ. Put Wi = Vi ∪ Si for all 1 ≤ i ≤ ℓ. By the minimality of |W |, the number of sets Vi with G[Vi ] = (a) is minimum in the initial partition. Hence, for 1 ≤ i ≤ t, since no partite set intersects both Vi and Si , we have G[Wi ] ∼ = (a, 1k−a ). For t + 1 ≤ i ≤ ℓ, if k−(a+2) ∼ some partite set intersects both Vi and Si , we have G[Wi ] = (a + 2, 1 ), otherwise we have G[Wi ] ∼ = (a + 1, 1k−(a+1) ). Therefore, the partition V (G) = W1 ∪ · · · ∪ Wℓ has a required property.  Proof of Theorem 3. (a) For a complete multipartite graph G of order 3ℓ, let us choose a partition V (G) = V1 ∪ V2 ∪ · · · ∪ Vℓ with |Vi | = 3 for 1 ≤ i ≤ ℓ such that the number of indices i with G[Vi ] ∼ = (2, 1) is as large as possible. Such a partition satisfies a desired condition, because if there exists a pair of indices α and β with G[Vα ] ∼ = (13 ) and G[Vβ ] ∼ = (3), we have a copy of (2, 1) in G[Vα ∪ Vβ ], a contradiction. (b) Let G be a complete multipartite graph of order kℓ with partite sets P1 , . . . , Ps . Put pj = |Pj | for all 1 ≤ j ≤ s. Let us define a as the maximum integer x such that G contains ℓ vertex disjoint copies of (x). As in the proof of Theorem 1, we may assume a ≤ k − 2. Choose a vertex partition V (G) = W ∪ V1 ∪ · · · ∪ Vℓ of G so that (1) G[Vi ] ∼ = (a) for all 1 ≤ i ≤ ℓ, (2) r := max{|W ∩ Pj | : 1 ≤ j ≤ s} is as small as possible subject to (1), and (3) the number of partite sets Pj satisfying r = |W ∩ Pj | is as small as possible subject to (1) and (2). We claim that r ≤ ℓ. If the claim is proved, in the same way as in the proof of Theorem 1, by using Lemma 6, we have a required partition. Hence, it suffices to show the claim. Suppose to a contradiction that r ≥ ℓ + 1. Since ℓ ≥ 2k − 6, we have r ≥ 2k − 5 ≥ k − 1 ≥ a + 1. For all 1 ≤ j ≤ s, let rj = |W ∩ Pj |. We may assume r1 = r. Let us define a set of indices J as J = {1} ∪ {2 ≤ j ≤ s : Vi ⊂ Pj for some i with 1 ≤ i ≤ ℓ}. Claim 1. rj ≥ r − a for all j ∈ J \ {1}. Indeed, assume that rd < r − a for some d ∈ J \ {1}. Let Vf ⊂ Pd . Moreover, we can take Vf′ ⊂ W ∩ P1 with |Vf′ | = a by r ≥ a + 1. Then we obtain a new partition from the partition by replacing W and Vf by W ′ = (W \ Vf′ ) ∪ Vf and Vf′ , respectively. Then |W ′ ∩Pd | = rd + a < r and thus this contradicts (2) or (3). Hence we have rj ≥ r − a for all j ∈ J \ {1}. Put t = |J | and n′ = j∈J pj .

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Claim 2. n′ ≤ (ℓ − 1)(a + 1) + ta. Indeed, if n′ ≥ (ℓ − 1)(a + 1) + ta + 1, then we have

  pj   pj − a ≥ a+1 a+1 j∈J j∈J = ≥

1 a+1 1 a+1

(n′ − at ) ((ℓ − 1)(a + 1) + 1).

Hence, we have j∈J ⌊pj /(a + 1)⌋ ≥ ℓ, a contradiction to the definition of the integer a. Let w = n′ − ((ℓ − 1)(a + 1) + ta + 1). By Claim 2, we have w < 0. On the other hand, by Claim 1, we have



w= ≥ = ≥ =

n′ − ((ℓ − 1)(a + 1) + ta + 1) r + (t − 1)(r − a) + aℓ − (ℓ − 1)(a + 1) − ta − 1 t (r − 2a) + 2a − ℓ t (ℓ + 1 − 2a) + 2a − ℓ

(t − 1)(ℓ − 2a) + t .

Case 1. ℓ ≥ 2k − 5. Since ℓ ≥ 2k − 5 and a ≤ k − 2, w is at least −(t − 1) + t = 1, a contradiction. Case 2. ℓ = 2k − 6. Since w < 0, we have a = k − 2, ℓ − 2a = −2 and t ≥ 3. We claim that r = ℓ + 1. Indeed, if r ≥ ℓ + 2, we have

w = n′ − ((ℓ − 1)(a + 1) + ta + 1) ≥ t (ℓ + 2 − 2a) + 2a − ℓ = 2, a contradiction. Let us define a subset J1 , J2 ⊂ J as follows.

(mod k − 2)},

J1 = {j ∈ J : pj ≡ −1

J2 = {j ∈ J : pj ≡ −1 (mod k − 1)}. Put ti = |Ji | for i = 1, 2. Note that since a = k − 2, we have rj ≡ pj (mod k − 2) for 1 ≤ j ≤ s. Furthermore, since r = ℓ + 1 = 2k − 5 ≡ −1 (mod k − 2), we have 1 ∈ J1 . Then we have a refinement of lower bounds of rj for j ̸∈ J1 such that rj ≥ r − a + 1. Hence, we have n′ ≥ r + (t1 − 1)(r − a) + (t − t1 )(r − a + 1) + aℓ

= t (k − 2) − t1 + (k − 2)(2k − 5). If j ∈ J1 ∩ J2 , then we have pj ≥ (k − 2)(k − 1) − 1

= k2 − 3k + 1. Since n = (2k − 6)k < 2(k2 − 3k + 1), we have |J1 ∩ J2 | ≤ 1. Hence, we have t1 + t2 ≤ t + 1. Then we have

  pj   pj − a  pj − (a − 1) ≥ + a+1 a+1 a+1 j∈J j∈J j∈J \J 2

= ≥ =

1 a+1 1 a+1 1

2

(n − at2 − (a − 1)(t − t2 )) ′

(t (k − 2) − t1 + (k − 2)(2k − 5) − at + t − t2 ) ((2k − 7)(k − 1) + 3 + t − t1 − t2 )

k−1 > ℓ − 1,

a contradiction to the definition of a. Therefore, we have r ≤ ℓ, and this completes the proof.



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Proof of Theorem 4. Let G be a complete multipartite graph of order kℓ such that G admits no {(k), (k − 1, 1), (1k )}-decomposition. If ℓ > (k − 2)2 , G has at least one copy of (k) or (1k ). Hence, by the induction on ℓ, it suffices to show the existence of a {(k), (k − 1, 1), (1k )}-decomposition for ℓ = (k − 2)2 . Let G have partite sets P1 , . . . , Ps . We take vertex disjoint copies of (k − 1) in G as many as possible. Suppose that we take ℓ1 vertex disjoint copies of (k − 1). If ℓ1 ≥ ℓ, then we have a {(k), (k − 1, 1)}-decomposition. Hence, we may assume ℓ1 ≤ ℓ − 1. Let ℓ2 = ℓ − ℓ1 . On the other hand, if k − 2 ≤ ℓ2 , since any partite set has at most k − 2 remaining vertices, by Lemma 6, all the remaining vertices can be partitioned into ℓ2 copies of (1k ) and some singletons. Then we have a {(k), (k − 1, 1), (1k )}-decomposition. Hence, we may assume ℓ2 ≤ k − 3. Let Ri be the set of the remaining vertices of Pi , and let ri = |Ri | for all 1 ≤ i ≤ s. Let R = R1 ∪ · · · ∪ Rs . Put x = |R|. Note that x = kℓ − (k − 1)ℓ1

= (k − 2)(k − 2 + ℓ2 ) + ℓ2 . s Let y = i=1 max{0, ri − ℓ2 }. Since ri ≤ k − 2 for all 1 ≤ i ≤ s and x ≡ ℓ2 (mod k − 2), we have y ≤ ( x − ℓ2 ) ·

k − 2 − ℓ2

k−2 = (k − 2 + ℓ2 )(k − 2 − ℓ2 ) = ℓ − ℓ22 ≤ ℓ1 .

For all 1 ≤ i ≤ s, let us consider a partition Ri = Si ∪ Ti such that |Si | = min{ℓ2 , ri }. Let S = S1 ∪ · · · ∪ Ss . Since |S | = x − y ≥ kℓ2 , by Lemma 6, S can be partitioned into ℓ2 vertex disjoint copies of (1k ) and some singletons. Therefore G has a {(k), (k − 1, 1), (1k )}-decomposition, as required.  Proof of Theorem 5. (a) For a complete multipartite graph of order 3ℓ, let us take a vertex partition V (G) = W ∪ V1 ∪· · ·∪ Vs so that (1) G[Vi ] ∼ = (3) for all 1 ≤ i ≤ s and (2) s is as large as possible subject to (1). If s ≥ ℓ − 1, then the partition is what we want, so we may assume s ≤ ℓ − 2. Since |W | ≥ 6 and the number of vertices of each partite set in W is at most 2, by Lemma 6, W can be decomposed into copies of (13 ), as required. (b) Suppose to a contradiction that G is a complete multipartite graph of order kℓ with ℓ ≥ (1/2)(3k2 − 9k + 4) such that G admits neither {(k), (k − 1, 1)}-decomposition nor {(k), (1k )}-decomposition. Let G have partite sets P1 , . . . , Ps with |Pi | = pi for all 1 ≤ i ≤ s. Let ℓ1 be the maximum number of vertex disjoint copies of (k) in G. For all 1 ≤ j ≤ s, let rj be the number of the remaining vertices in the jth partite set. Let ℓ2 = ℓ − ℓ1 and r = max{rj : 1 ≤ j ≤ s}. If r ≤ ℓ2 , by Lemma 6, the set of the remaining vertices admits a {(1k )}-decomposition, as required. Hence, we may assume r ≥ ℓ2 + 1. For all 1 ≤ i ≤ s, let us define a nonnegative integer

 fi =



pi k−1

,

and gi =

p  i

k

.

Since  G admits no {(k), (k − 1, 1)}-decomposition, G has no set of ℓ vertex disjoint copies of (k − 1). Hence, we have 1≤i≤s fi ≤ ℓ − 1. For all 1 ≤ i ≤ s, let εi be the minimum integer x such that pi ≤ ri + k(k − 2 − ri ) + k(k − 1)x. Claim 1. For all 1 ≤ i ≤ s, εi = fi − gi . Furthermore,

s

i=1

εi ≤ ℓ2 − 1.

For the first statement, note that there exists a positive integer ci with ci ≤ k − 1 such that pi = ri + k(k − 2 − ri ) + k(k − 1)(εi − 1) + ci k. Hence, we have

 fi =

k−1

 =



pi

k(k − 2 + ci )



k−1

− ri + k(εi − 1)

= k − 1 + ci − ri + k(εi − 1). On the other hand, we have gi =

p  i

k

= k − 2 − ri + (k − 1)(εi − 1) + ci . Hence, we have fi − gi = εi .

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The second statement of the claim follows from the inequality s 

εi =

s 

i=1

fi −

i=1

s 

gi

i=1

≤ ℓ − 1 − ℓ1 = ℓ2 − 1, as claimed. For all 0 ≤ c ≤ k − 1, let Hc (or H≤c ) be the set of indices i with 1 ≤ i ≤ s such that ri = c (or ri ≤ c) and pi ≥ k. Put hc = |Hc | and h≤c = |H≤c |. Claim 2. h≤c ≤ k − ℓ2 − 1 + c for 0 ≤ c ≤ k − 1. Suppose to a contradiction that h≤c ≥ k − ℓ2 + c. Let us define U2 ⊂ V (G) such that for 1 ≤ i ≤ s, if i ∈ H≤c then |U2 ∩ Pi | = ri + k, and otherwise, |U2 ∩ Pi | = ri . Let U1 = V (G) \ U2 . Then we have U1 is a set of ℓ1 − h≤c vertex disjoint copies of (k). Since |U2 ∩ Pi | is at most c + k ≤ ℓ2 + h≤c for 1 ≤ i ≤ s, by Lemma 6, U2 has a {(1k )}-decomposition, a contradiction. We consider two cases depending on r. Case 1. r ≤ k − 2. In this case, we have kℓ =

s 

pi

i=1

≤

k−2 

hc k(k − 2 − c ) + k(k − 1)

c =0

s 

εi +

i=1

≤ (k − ℓ2 − 1)k(k − 2) + k

k−3 

s 

ri

i =1

i + k(k − 1)

s 

εi +

i=1

i=0

s 

ri

i=1

1

≤ (k − ℓ2 − 1)k(k − 2) + k(k − 3)(k − 2) + k(k − 1)(ℓ2 − 1) + kℓ2 2   1 ≤ k 2ℓ2 + (k − 3)(3k − 4) . 2

Since ℓ2 ≤ r − 1 ≤ k − 3, we have kℓ ≤ k · (1/2)(3k2 − 9k), a contradiction. Case 2. r = k − 1. In this case, we have kℓ =

s 

pi

i =1

≤

k−1 

hc k(k − 2 − c ) + k(k − 1)

c =0

s  i =1

k−3 

≤ (k − ℓ2 − 1)k(k − 2) + k

i=−1

εi +

s 

ri

i=1

i + k(k − 1)

s  i =1

εi +

s 

ri

i=1

1

≤ (k − ℓ2 − 1)k(k − 2) + k((k − 3)(k − 2) − 2) + k(k − 1)(ℓ2 − 1) + kℓ2 2   1 ≤ k 2ℓ2 + (k − 3)(3k − 4) − 1 . 2

Since ℓ2 ≤ r − 1 = k − 2, we have kℓ ≤ k · (1/2)(3k2 − 9k + 2), a contradiction.



4. Vertex partition of graphs In the previous sections, we only consider the problem for the decomposition of complete multipartite graphs. How about graphs instead of complete multipartite graphs? Hell, Manoussakis and Tuza noted the following theorem [2].

T. Nakamigawa / Discrete Mathematics 339 (2016) 1699–1705

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Theorem C ([2]). For any integers k and t, and sufficiently large multiple n of k, every t-edge colored complete graph of order n has a vertex partition into monochromatic stars on k vertices such that the endpoints of each star induce a monochromatic complete graph of order k − 1. We focus on the case t = 2 in Theorem C. Let us define f (k) as the minimum integer ℓ such that every graph of order kℓ admits a {Kk , Kk , K1,k−1 , K1,k−1 }-decomposition. By Theorem C, it follows that f (k) is finite. Theorem 7. There exist some positive integers c1 and c2 such that c1 2k/2 ≤ f (k) ≤ c2 k−3/2 24k for any positive integer k. Proof. Both lower and upper bounds of f (k) come from the known bounds of the diagonal Ramsey number R(k), which is defined as the minimum integer n such that every graph of order n contains Kk or Kk as a subgraph. It is known that there exist some positive integers d1 and d2 such that for any positive integer k, we have d1 k2k/2 ≤ R(k) ≤ d2 k−1/2 22k (for example, see [1]). First, we will show the lower bound of f (k). Claim. If (k − 1)ℓ + 1 < R(k − 1), then we have ℓ < f (k). Let us take a graph G of order kℓ with (k − 1)ℓ + 1 < R(k − 1) such that G contains an induced subgraph H of order (k − 1)ℓ + 1 in which there exists no copy of Kk−1 or Kk−1 . For any partition V (G) = V1 ∪ V2 ∪ · · · ∪ Vℓ such that |Vi | = k for all 1 ≤ i ≤ ℓ, we have some index α such that Vα ⊂ V (H ). Since G[Vα ] ⊂ H has neither Kk−1 nor Kk−1 as a subgraph, we have G[Vα ] is isomorphic to none of Kk , Kk , K1,k−1 , K1,k−1 . Hence, G has no {Kk , Kk , K1,k−1 , K1,k−1 }-decomposition. Therefore, we have ℓ < f (k). By the Claim and the lower bound of R(k), if ℓ < (d1 (k − 1)2(k−1)/2 − 1)/(k − 1), we have ℓ < f (k), as desired. Next, we will show the upper bound of f (k). Let s = R(k) − k. Let G be a graph of order n = kℓ such that n ≥ max{2ks, R(2k − 1) + (s − 1)(2k − 1)}. Since n ≥ R(2k − 1) + (s − 1)(2k − 1), we can reserve s mutually disjoint sets of vertices A1 , A2 , . . . , As such that G[Ai ] is isomorphic to K2k−1 or K2k−1 for 1 ≤ i ≤ s, which may be mixed, from G. From the remaining vertices, remove vertex disjoint copies of Kk or Kk one by one, until s the number of the remaining vertices W is less than R(k). Since n ≡ 0 (mod k) and the number of reserved vertices is i=1 |Ai | ≡ −s (mod k), we have |W | is exactly s. Let W = {w1 , w2 , . . . , ws }. For all 1 ≤ i ≤ s, let us combine Ai and wi . By the pigeonhole principle, there is a set of k edges from wi to Ai , or a set of k nonedges from wi to Ai . Hence, if G[Ai ] ∼ = K2k−1 , we have a partition Ai ∪ {wi } = Bi ∪ Ci with wi ∈ Ci such that G[Bi ] ∼ = Kk or K1,k−1 , otherwise we have a partition Ai ∪ {wi } = Bi ∪ Ci with wi ∈ Ci such = Kk and G[Ci ] ∼ that G[Bi ] ∼ = Kk and G[Ci ] ∼ = Kk or K1,k−1 . Therefore, we have a {Kk , Kk , K1,k−1 , K1,k−1 }-decomposition. By the upper bound of R(k), if ℓ ≥ (d2 (2k − 1)−1/2 22(2k−1) + (d2 k−1/2 22k (2k − 1)))/k, we have f (k) ≤ ℓ, as desired.  Acknowledgments I am indebted to anonymous reviewers for their valuable comments, which greatly improved the presentation of the paper. This work was supported by KAKENHI (23540168). References [1] [2] [3] [4]

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