Main results

3 Main results

Syafrizal et.al. have shown in Corollary 1 [9] the size multipartite Ramsey numbers m j (S m ,C n ) of stars and cycles for m = 2 and 3. In this section we calculate the size multipartite Ramsey numbers of stars and cycles for m ≥ 4.

Theorem 4. For integers j and n,  3 for j = 3, n = 6   

Proof: Case 1 . j = 3, n = 6.

We show first that m 3 (S 4 ,C 6 ) ≥ 3. Let G 1 ∼ =C 6 and let G 2 be the complement of G 1 relative to K 3×2 . This implies, G 1 6⊇ S 4 and

G 2 6⊇ C 6 , se Fig.1. Thus, m 3 (S 4 ,C 6 ) ≥ 3.

Fig. 1. G 1 and G 2 for j = 3 and n = 6.

Now, we show that m 3 (S 4 ,C 6 ) ≤ 3. Let F 1 ⊕F 2 be any factor- ization of F ∼ =K 3×3 and suppose that F 1 6⊇ S 4 . Then, ∆(F 1 ) ≤ 2, so δ(F 2 ) ≥ 4. If ∆(F 1 ) ≤ 1 then, by Lemma 1, F 2 is Pancylic and so

F 2 ⊇C 6 . Now, without lost of generality, we may assume d(x) = 2 for every x ∈ V (F 1 ), 2-F 1 . Then d(x) = 4 for every x ∈ V (F 2 ), 4-F 2 . By Lemma 2, F 2 ⊇P 8 . This implies, F 2 ⊇P 6 . Then, we can to con- struct P 6 in F 2 as follow. Let b 6∈ N (a) in F 1 . Clearly ab ∈ E(F 2 ). Again, since 4-F 2 then we have P 6 := a P b in F 2 . This implies, we have C 6 in F 2 . Now, let ∆(F 1 ) = 2. If there is a vertex of degree 1 in F 1 , then add new edges so that each vertex of F 1 has degree 2.

4 Syafrizal Sy, E.T. Baskoro, S. Uttunggadewa, H. Assiyatun

Therefore, m 3 (S 4 ,C 6 ) ≤ 3.

Case 2. j = 5, n = 3 and 4. We show first that m 5 (S 4 ,C n ) ≥ 2. Let G 1 ∼ =C 5 and let G 2 be the complement of G 1 relative to K 5 =K 5×1 , so F 2 ∼ =C 5 . Then, K 5 =G 1 ⊕G 2 is a factorization such that G 1 6⊇ S 4 and G 2 6⊇ C n for n = 3 and 5. Therefore, m 5 (S 4 ,C n ) ≥ 2 for n = 3 and 4.

Let F 1 ⊕F 2 be any factorization of F ∼ =K 5×2 . We show that S 4 ⊆F 1 or C n ⊆F 2 for n = 3 and 4. Suppose that S 4 6⊆ F 1 . Then, ∆(F 1 ) ≤ 2, so δ(F 2 ) ≥ 6. Since δ(F 2 ) ≥ 5 then, by Lemma 1, F 2 is pancyclic and so F 2 ⊇C n for n = 3 and 4. Therefore, m 5 (S 4 ,C n )≤2 for n < 5.

Case 3. 5 ≤ n ≤ j. Clearly, m j (S 4 ,C n ) ≥ 1. Now, we will show that m j (S 4 ,C n ) ≤ 1. Let K j ×1 ∼ =K j := F 1 ⊕F 2 be any factorization of K j and suppose that

F 1 6⊇ S 4 . Then, ∆(F 1 ) ≤ 2, so δ(F 2 ) ≥ (j − 1) − ∆(F 1 ). We consider the following two subcases.

Subcase 3.1. j = 5. masih kurang betul If ∆(F 5

1 ) ≤ 1 then δ(F 2 )≥3≥ 2 . Thus, by Theorem 1, F 2 is

a Hamiltonian cycle. Therefore, F 2 ⊇C 5 . Now, let δ(F 1 ) = 2. Consider V (K 5 ) = {x 1 ,x 2 ,x 3 ,x 4 ,x 5 }. If F 1 form a cycle C 5 := x 1 x 2 x 3 x 4 x 5 x 1 (see Fig.2.(i)), or C 5 := x 1 x 3 x 5 x 2 x 4 x 1 then F 2 con- tains C 5 := x 1 x 3 x 5 x 2 x 4 x 1 (see Fig.2.(ii)) or C 5 := x 1 x 2 x 3 x 4 x 5 x 1 , respectively. Thus, F 2 ⊇C 5 . Therefore, m 5 (S 4 ,C 5 ) ≤ 1.

x 1 x 2 x 1 x 2 (i)

(ii)

Fig. 2. Illustration for n = j = 5 of Theorem 1.

On size multipartite Ramsey numbers m j (S m ,C n ) 5

Subcase 3.2. j ≥ 6. Since j ≥ 6 then δ(F j

2 )≥j−3≥ 2 . Thus, by Lemma 1, F 2 is a pancyclic. Thus, F 2 ⊇C n for 3 ≤ n ≤ j. Therefore, m j (S 4 ,C n )≤1 where 5 ≤ n ≤ j with j ≥ 6.⊓ ⊔

Theorem 5. For integers j ≥ 3 and j < n,  n

4≤m≤⌈ j ⌉( 2 − 1) + 2,  

for

j (S m ,C n )= ⌈ j ⌉ + 1 for n odd and either

    [m=( 2 − 1)(t − 1) + 2, j even ] or   

[m= 1 2 (j − 1)(t − 1) + 2, j odd ]. Proof:

Case 1 j .4≤m≤⌈

j ⌉( 2 − 1) + 2.

Let r = ⌈ n j ⌉. We show first that m j (S m ,C n ) ≥ r. Let G 1 ∼ = (r − 1)C j and let G 2 be the complement of G 1 relative to K j ×(r−1) . Thus,

∆(G 1 ) = 2, so that G 1 6⊃ S m , for m ≥ 4. Since |G 2 | = j(r − 1) < n then G 2 6⊃ C n . Now, we will show that m j (S m ,C n ) ≤ r. Let K j ×r := F 1 ⊕F 2 be any factorization of K j ×r and suppose that S m is not a subgraph of

1 . Thus, ∆(F 1 ) ≤ m − 2 = r( 2 − 1). Thus, δ(F 2 ) ≥ (j − 1)r − r( 2 −

2 . Since δ(F 2 )≥ 2 , by Lemma 1, F 2 ⊇C n where n ≤ jr. Thus m j (S m ,C n ) ≤ r for all 3 ≤ n ≤ jr.

1) = jr

jr

Case 2 . n is odd. Let t = ⌈ n

j ⌉ + 1. To show that m j (S m ,C n ) = t, we consider two possibilities of j.

Subcase 2.1 j .m=(

2 − 1)(t − 1) + 2 and j is even. We show first that m j (S m ,C n ) ≥ t. Let G 1 ∼ = 2K j 2 ×(t−1) and let

G 2 be the complement of G 1 relative to K j ×(t−1) . Thus, ∆(G 1 )=

(t − 1)( j

2 − 1). Since ∆(G 1 ) = (t − 1)( 2 − 1) < m − 1 then G 1 6⊃ S m . Furthermore, since G 2 is a bipartite graph then G 2 6⊃ C n for odd n.

Therefore, m j (S m ,C n ) ≥ t for all odd n. We will show that m j (S m ,C n ) ≤ t. Let F 1 ⊕F 2 be any factor- ization of K j ×t and suppose that S m is not a subgraph of F 1 . Thus,

∆(F j

jt

1 ) ≤ m−2. This implies, δ(F 2 ) ≥ (j −1)t−(m−2) = 2 + 2 −1 ≥

2 , since j ≥ 3. By Lemma 2, F 2 ⊇C n . Therefore, m j (S m ,C n )≤t for all odd n.

jt

Subcase 2.2 .m= 1

2 (j − 1)(t − 1) + 2 and j is odd.

6 Syafrizal Sy, E.T. Baskoro, S. Uttunggadewa, H. Assiyatun

Let V 1 ,V 2 ,···,V j −1 and V j

be the partite sets of K j ×t . Let V j =

A ∪ B with ||A| − |B|| ≤ 1. Partition the partite sets of K j ×(t−1) into two new partite sets, namely X 1 =V 1 ∪V 2 ∪···∪V ⌊ j

∪ A and

X 2 =V ⌈ j 2 ⌉ ∪V ⌈ j 2 ⌉+1 ∪···∪V j −1 ∪ B. Let F [X 1 ] and F [X 2 ] be two new complete multipartite graphs. Let G 1 = F [X 1 ] ∪ F [X 2 ] and G 2

be the complement of G 1 relative to K j ×(t−1) , see Fig.3. Thus, G 2 is a bipartite graph. This implies, ∆(G 1

1 )≤ 2 (j − 1)(t − 1). Since ∆(G 1 )≤ 1 2 (j − 1)(t − 1) < m − 1 then F 1 6⊃ S m . Furthermore,

since G 2 is a bipartite graph then G 2 6⊃ C n for odd n. Therefore, m j (S m ,C n ) ≥ t for all odd n.

Fig. 3. A bipartite graph G 2 of Subcase 2.2

We will show that m j (S m ,C n ) ≤ t. Let F 1 ⊕F 2 be any fac- torization of K j ×t and suppose that S m is not a subgraph of F 1 . Thus, ∆(F 1 ) ≤ m − 2. This implies, δ(F 2 ) ≥ (j − 1)t − (m − 2) =

(j − 1)t − t( jt

2 − 1)t ≥ 2 . By Lemma 2, F 2 ⊇C n where n = jt. Therefore, m j (S m ,C n ) ≤ t for all odd n.⊓ ⊔

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