THE RAMSEY NUMBERS OF LINEAR
FOREST VERSUS
3
K
3∪
2
K
4I W. Sudarsana, E. T. Baskoro, H. Assiyatun, and S.
Uttunggadewa
Abstract.For given two graphsGandH, a graphFis called a (G, H)-good graphifF contains noGandFcontains noH. Furthermore, any (G, H)-good graph onnvertices will be denoted by (G, H, n)-good graph. TheRamsey numberR(G, H) is defined as the smallest natural numbernsuch that no (G, H, n)-good graph exists. In this paper, we determine the Ramsey numbersR(G, H) for disconnected graphsGandH. In particular, G=Sk
i=1PniandH= 3K3∪2K4.
1. INTRODUCTION
We consider finite undirected graphs without loops and multiple edges. Let G(V, E) be a graph, the notationV(G) andE(G) (in shortV andE) stand for the vertex set and the edge set of the graphG, respectively. A graph H(V′, E′) is a
subgraph ofG if V′ ⊆V and E′ ⊆ E. For A⊆ V, G[A] represents the subgraph
inducedbyA inG.
For given two graphsGandH, a graphF is called a (G, H)-good graphifF contains no Gand F contains no H. Furthermore, any (G, H)-good graph onn vertices will be denoted by (G, H, n)-good graph. TheRamsey numberR(G, H) is defined as the smallest natural numbernsuch that no (G, H, n)-good graph exists. The Ramsey numbers R(G, H) for connected graphs G and H have been inten-sively studied since Chv´atal and Harary [4] established the general lower bound R(G, H)≥(c(G)−1)(h−1) + 1, wherehis the chromatic number ofH andc(G) is the number of vertices of the largest component of G. A connected graphGis
Received 20-05-2009, Accepted 05-08-2009.
2000 Mathematics Subject Classification: 05C55
Key words and Phrases: Ramsey number, union graph, good graph, forest, linear forest
calledH-goodifR(G, H) = (|G| −1)(h−1) +s, wheresis the chromatic surplus of H. The chromatic surplus ofH is the minimum cardinality of a color class taken over all properhcolorings of H.
LetGi be a graph with the vertex setVi and the edge setEi, i= 1,2, ..., k. to a tree for everyithen the union is called aforest. A forest is called linear forest, if all the components are a path.
Some recent results on the Ramsey number for a combination of disconnected (union) and connected graphs can be found in ([1], [2], [6], [7], [9], [10], [11]). Other results concerning graph Ramsey numbers can be seen in [8]. In this note, we determine the Ramsey numbersR(Sk
i=1Pni,3K3∪2K4).
2. MAIN RESULTS
Let us note firstly the previous theorems and lemmas used in the proof of our results. graph with the chromatic number h≥2 and the chromatic surplus s≥1. Let Gi be connected graphs and satisfies |Gk| ≥ |Gk−1| ≥ ... ≥ |G1| ≥
Now, we are ready to prove our main results in the following.
Proof. The inequality R(Pn,3K3) ≥2n+ 1 is derived from Lemma 1. We will show the reverse inequality R(Pn,3K3) ≤ 2n+ 1 by the following reason. Take an arbitrary graph Gon 2n+ 1 vertices and contains no Pn. We will show that G contains 3K3. Since |G| > 2n then by Lemma 2 we obtain G ⊇ 2K3. Let B ={a1, b1, c1} ∪ {a2, b2, c2} be the vertex set of 2K3 in G. We let D=V(G)\B and T =G[D]. Clearly, |T| = 2(n−2)−1. Theorem 1 gives that T ⊇Pn−2 or T ⊇K3. IfT ⊇K3 then we finish the proof. Now, consider T containsPn−2 and call Pn−2 = (p1p2...pn−3pn−2). LetK =V(T)\V(Pn−2) and hence |K|=n−3.
Now, we consider the connection of the end vertices ofPn−2 and the vertex setK. SinceGcontains noPn then we obtain the following facts.
Fact 1. The vertex p1 orpn−2 is adjacent to at least two vertices in K.
We let p1 adjacent to y1 and y2 in K. Since G does not contain Pn then
{pn−2, y1, y2} is an independent set inG. Therefore, the vertex set{pn−2, y1, y2}
forms aK3in T and together with B we haveG⊇3K3.
Fact 2. The vertex p1 orpn−2 is adjacent to exactly one vertex in K.
We letp1 adjacent to xin K. SinceG contains noPn then pn−2 must not adjacent to any vertex in K except the vertex x. If K\x contains two indepen-dent vertices, call x1 and x2, inT then the vertexpn−2 together withx1 and x2 induce a K3 in T and hence G ⊇ 3K3. Therefore, the vertex set K\x forms a Kn−4 inT. Now, if there exists one vertex, say y, inK\xthat is not adjacent to one vertex in B then the vertex set {pn−2, x, y} ∪B induces a 3K3 in G. Since
otherwise we will get that every vertex in K\xis adjacent to every vertex in B, which is impossible sinceGdoes not containPn withn≥7. Thus,Gcontains 3K3.
Fact 3. The vertex p1 andpn−2 do not adjacent to any vertex inK.
IfKcontains two independent vertices, call x1 andx2, inT then the vertex set {pn−2, x1, x2} induce a K3 in T. Therefore, we finish the proof since we have
G⊇3K3. Now, considerK shapes aKn−3 in T. Thus without loss of generality, one of the following conditions holds:
(i). The vertexp1 orpn−2 is adjacent to every vertices inB.
We letp1 adjacent to every vertex inB. SinceG does not containPn then
{pn−2} ∪Bis an independent set inT which each element does not adjacent to any vertex inK. Therefore, it can be verified that the set {k1, k2, pn−2} ∪B induce a 3K3 inT, for anyk1, k2 in K. So,G⊇3K3.
We letp1adjacent to every vertex inB\a2. SinceGcontains noPn then it is not difficult to verify that the sets{pn−2, a2, c1}, {a1, c2, k2}and {b1, b2, k1} form
a 3K3in G, for anyk1, k2in K.
(iii). The vertexp1 orpn−2 is adjacent to four vertices in B.
We letp1adjacent to every vertex inB\{a2, c2}. Again, sinceGcontains no Pn then it can be verified that the sets {pn−2, a2, c2}, {a1, c1, k2} and{b1, b2, k1}
form a 3K3 inG, for anyk1, k2 inK.
(iv). The vertex p1 orpn−2 is adjacent to three vertices inB.
Without loss of generality, we distinguish the following two cases.
Case 1. The vertex p1 is adjacent toa1,b1 andc1 inB
Thus the set{p1, a2, b2, c2} is an independent set in T. Since Gcontain no Pnthen it is easy verify that the sets{pn−2, a1, c2},{a2, b2, p1}and{b1, c1, k}form
a 3K3in G, for everyk∈K.
Case 2. The vertex p1 is adjacent toa1,b1 andc2 inB
Therefore, the set{p1, a2, b2}is an independent set inT. SinceGcontain no Pnthen it is easy verify that the sets{p1, a2, b2},{a1, c1, pn−2}and{b1, c2, k}form a 3K3in G, for everyk∈K.
(v). The vertexp1 or pn−2 is adjacent to two vertices inB.
Without loss of generality, we distinguish the following two cases.
Case 1. The vertex p1 is adjacent toa1 andb1 inB
Thus the set{p1, a2, b2, c2}is an independent set inT. Now, if there exists one vertex, cally, inKthat is not adjacent to one vertex, sayc2, inB\{a1, b1}then the vertex sets{p1, a2, b2},{b1, c2, y}and{a1, c1, pn−2}form a 3K3 inG. Since other-wise we get that vertex setK∪{a2, b2, c2, c1}induces a graphKn−3+(K3∪K1) inT, which is impossible sinceGdoes not containPn withn≥7. Thus,Gcontains 3K3.
Case 2. The vertex p1 is adjacent toa1 anda2 in B
Since G contains no Pn then it is easy verify that the sets {pn−2, b1, c1},
(vi). The vertex p1 orpn−2 is adjacent to one vertex inB.
We letp1adjacent toa1inB. Now, consider the vertex set{b1, c1, a2, b2, c2}. If there exists a vertex, sayk1, inKthat is not adjacent to one vertex in{b1, c1, a2, b2, c2}
then the vertex set{p1, pn−2, k1}∪Binduces a 3K3inG. Since otherwise we obtain
that the vertex setK∪ {b1, c1, a2, b2, c2}induces a graphKn−3+ (K3∪K2) inT, which is impossible sinceGdoes not containPn withn≥7. Thus,Gcontains 3K3.
(vii). The vertexp1 orpn−2 does not adjacent to any vertices in B.
If there exists a vertex, say k, in K that is not adjacent to one vertex in B then the vertex set {p1, pn−2, k} ∪B induces a 3K3 in G. Since otherwise we
Suppose that the maximum of the right side of the equation (3) is achieved fori0. Writet0=Pk
j=i0nj andt=ni0+t0. The lower boundR(G,3K3)≥t+ 1 can be obtained by using Lemma 1. We will proveR(G,3K3)≤t+ 1.
LetF be a graph of ordert+ 1 and suppose that F contains no 3K3. We shall show thatF containsG. We prove this by induction oni. For i=k, we get G=Pnk. Sincet+ 1≥2nk+ 1 andF +3K3 then the theorem holds by Lemma
U ⊇ 2K4. Let L = V(U)\V(2K4) and Q = U[L]. Clearly, |Q| = l−9. By
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I W. Sudarsana: Combinatorial Mathematics Research Division, Faculty of Mathe-matics and Natural Sciences, Institut Teknologi Bandung (ITB), Jalan Ganesha No. 10 Bandung 40132, Indonesia dan Jurusan Matematika, FMIPA, Universitas Tadulako (UNTAD), Jl. Soekarno-Hatta Km. 8 Palu 94118, Indonesia. E-mail: sudarsanai-wayan@yahoo.co.id
E. T. Baskoro: Combinatorial Mathematics Research Division, Faculty of Mathematics and Natural Sciences, Institut Teknologi Bandung (ITB), Jalan Ganesha No. 10 Bandung 40132, Indonesia.
E-mail: ebaskoro@math.itb.ac.id
H. Assiyatun: Combinatorial Mathematics Research Division, Faculty of Mathematics and Natural Sciences, Institut Teknologi Bandung (ITB), Jalan Ganesha No. 10 Bandung 40132, Indonesia.
E-mail: hilda@math.itb.ac.id
S. Uttunggadewa: Combinatorial Mathematics Research Division, Faculty of Mathe-matics and Natural Sciences, Institut Teknologi Bandung (ITB), Jalan Ganesha No. 10 Bandung 40132, Indonesia.
