Vol. 13, No. 1, 2017, 9-26
ISSN: 2279-087X (P), 2279-0888(online) Published on 10 January2017
www.researchmathsci.org
DOI: http://dx.doi.org/10.22457/apam.v13n1a2
9
Annals of
Size Multipartite Ramsey Numbers for K
4-e Versus all
Graphs up to 4 Vertices
Chula Jayawardeneand LilanthiSamarasekaraDepartment of Mathematics University of Colombo-3, Sri Lanka
Corresponding author. Email: c_jayawardene@yahoo.com
Received 14November 2016; accepted 30December 2016
Abstract.Let G and H be two simple subgraphs of
K
j×s. The smallest positive integers such that any red and blue colouring of
K
j×s has a copy of red G or a blue H is called the multipartite Ramsey number of G and H. It is denoted bym
j(
G
,
H
)
. This paper presents exact values form
j(
B
2,
G
)
where G is a isolate vertex free graph up to four vertices.Keywords: Graph Theory, Ramsey Theory
AMS Mathematics Subject Classification (2010): 05C55, 05D10 1. Introduction
The exact determination of the classical diagonal Ramsey number of r( ss, ) (see [6] for a survey) has not made any headway beyond r(5,5) (as of now known to be between 43 and 49) for a few decades. Researches are now trying to approach this problem by using new techniques like fuzzy logic, genetic algorithms to improve the lower bound (see [7], [8] and [9]). One of the main branches of classical Ramsey numbers, introduced by Burger and Vuuren (see [1]) was multipartite Ramsey numbers. It is defined considering
s j
K
× , which consist of j partite sets, each of size s.V
(
K
j×s)
=
{
v
mn|
m
∈
{1,2,...,
j
}
and n∈{1,2,...,s}} denotes the vertex set of
K
j×s. The set of vertices in the th10
fan Fn and a windmill M2n for j≥2, s∈{2,3} and n≥6. Jayawardena and et al (see [3,4]) also investigated special cases of the multipartite Ramsey number. They found
)
,
(
H
G
m
j whereH
∈
{
C
3,
C
4}
and G is any graph on four vertices and for j≥3 .This paper is also on a special case ofm
j(
B
2,
G
)
where G is any isolate vertex free graph up to four vertices where j≥3. These values are shown in the following table.)
,
(
B
2G
m
j values =j 3 4 5 6 7 8 9 10 j≥11
1 1 1 1 1 1 1
2
2 2 1 1 1 1 1 1 1
3 2 1 1 1 1 1 1 1
4 2 2 1 1 1 1 1 1
infinity infinity infinity 2 1 1 1 1 1 ,
4 3 2 2 1 1 1 1 1
infinity 3 2 2 1 1 1 1 1 ,+
infinity infinity infinity 2 1 1 1 1 1
infinity infinity infinity 2 2 2 2 1 1
infinity infinity infinity infinity infinity infinity 2 2 1 Table 1:
m
j(
B
2,
G
)
values2. Size Ramsey numbers
m
j(
B
2,
P
2)
Theorem 1.If j≥3, then
≥4 if 1
3 = if 2
= ) , ( 2 2
j j
11
Proof: Consider the K3×1 where all its edges are colored in red. Then K3×1 has neither a red B nor a blue 2 P . Therefore 2 m3(B2,P2)≥2.
Next consider any red-blue colouring of K3×2. If it has a blue P , then we are done. 2 Otherwise all the edges of K3×2 are red and hence it contains a red B . Therefore, 2
2 = ) , ( 2 2
3 B P
m .
Clearly
m
j(
B
2,
P
2)
=
1
when j≥4 as r(B2,P2)=4 (see [2]).3.Size Ramsey numbers
m
j(
B
2,
P
3)
Theorem 2. If j≥3, then
≥5 if 1
4 = if 2
3 = if 3
= ) , ( 2 3
j j j
P B mj
Proof: Consider the coloring K3×2 =HR⊕HB where HB = K3 2. Then K3×2 has neither a red B nor a blue 2 P3. Therefore m3(B2,P3)≥3.
Now consider any red-blue colouring of K3×3. Assume it has no blue P3 or a red B . 2
Then it contains a red C3, say v11v21v31v11 (since m3(C3,P3)=2 by [3]). Due to the
absence of a red B , at least two of the vertices of2 { : {1,2,3}}
3
2
∈ =
i v j
ij
∪
must be adjacentto each other in blue(say u and v). But then in order to avoid a red B , 2 v must be adjacent to some vertex of the red C3 in blue. This forces a blue P3. A contradiction. Therefore, m3(B2,P3)=3.
Next consider the red-blue colouring of K4×1 having v11v21 and v31v41 as the only blue edges. Then K4×1 has no red B and has no blue 2 P3. Therefore, m4(B2,P3)≥2.
Consider any red-blue colouring of K4×2 which is blue P3 free. Then K4×2has a red C3
12
adjacent in red to at least two vertices of the red C3. ThusK4×2 has a red B . Hence 2
2 = ) , ( 2 3
4 B P
m .
Clearly
m
j(
B
2,
P
3)
=
1
when j≥5 as r(B2,P3)=5 (see [2]).4.Size Ramsey numbers
m
j(
B
2,2
K
2)
Theorem 3.If j≥3, then
≥ ∈ 5 if 1 {3,4} if 2 = ) ,2 ( 2 2
j j
K B mj
Proof: The graph K4×1=HR⊕HB where H consist only of the red R C3, v11v21v31v11 is red B free and blue 2 2K free. Therefore 2 m3(B2,2K2)≥m4(B2,2K2)≥2.
Consider the graph K3×2 =HR⊕HB which is red B free. Then it contains a blue 2 P 2
(say v11v21). If all the edges not incident to v or 11 v are red then 21 K3×2 has a red B2, a
contradiction. Therefore, there is a blue edge not incident to v or 11 v . Hence 21 K3×2 has a blue 2K . Therefore 2 m3(B2,2K2)=m4(B2,2K2)=2.
Clearly,
m
j(
B
2,2
K
2)
=
1
when j≥5 as r(B2,2K2)=5 (see [2]). 5.Size Ramsey numbersm
j(
B
2,
P
4)
Theorem 4.If j≥3, then
≥ ∈ 6 if 1 {4,5} if 2 3 = if 4 = ) , ( 2 4
j j j P B mj
Proof: Consider the coloring K3×3=HR⊕HB where H consist only of the three 3-B cycles in {v1iv2iv3iv1i:i∈{1,2,3}}. Then K3×3 has neither a red B nor a blue 2 P . 4
Therefore m3(B2,P4)≥4.
Now consider any red-blue colouring of K3×4. Assume it has no red B and no blue 2 P . 4
13 {1,2,3}} : { = )
(H2 ∪4=2 v i∈
V j ij has a blue P3. By relabeling we can assume them to be
31 21 11v v
v and v14v24v33. As K3×4 is blue P free, 4 v and 14 v31 are not incident to any blue edges other than v14v24 and v31v21 respectively. Then the red edges
} { {2,3}} : { {2,3}} :
{v14v2i i∈ ∪ v31v2i i∈ ∪ v14v31 forces a red B , a contradiction. 2
Therefore, m3(B2,P4)=4.
Next consider the coloring K5×1 =HR⊕HB. where H consist only of the edge B v21v31 and the 3-cycle v41v51v11v41. Then K5×1 has neither a blue P nor a red 4 B . Hence 2
2 ) , ( ) ,
( 2 4 5 2 4
4 B P ≥m B P ≥
m .Now consider any red-blue coloring of K4×2 which is blue
4
P free. Then it has a red C3, say v11v21v31v11 (as m4(C3,P4)=2 by [3]). Suppose each of v and 41 v are incident to two vertices of 42 V(C3) in blue. Then K4×2 has a blue P , 4 a contradiction. Therefore at least one of v or 41 v (say 42 v ) is incident to two vertices 41 (say v and 11 v ) of 21 V(C3) in red. Then the red edges v41v11,v41v21 together with the red
3
C forms a red B . 2 Therefore, 2≥m4(B2,P4). Thus, we get 2 = ) , ( = ) ,
( 2 4 5 2 4
4 B P m B P
m .
Consider any red-blue coloring of K6×1 with no blue P . Then it has a red 4 C3, say
11 31 21 11v v v
v (as m6(C3,P4)=1 by [3]). Now considering v and 41 v51 and arguing as above, it can be proved that
m
j(
B
2,
P
4)
=
1
when j≥6.6.Size Ramsey numbers
m
j(
B
2,
K
1,3)
Theorem 5. If j≥3, then
≥ ∈ 7 if 1 {5,6} if 2 4 = if 3 3 = if 4 = ) , ( 2 1,3
j j j j K B mj
14
Figure 1: When H consists of blue a B C9.
Then K3×3 has neither a red B nor a blue 2
K
1,3. Therefore,m
3(
B
2,
K
1,3)
≥
4
.Now consider any colouring of K3×4 =HR⊕HB. Assume the graph has no blue
K
1,3. Then K3×4 has a red C3, say v11v21v31v11 (asm
3(
C
3,
K
1,3)
=
3
by [3]). Let} , , {
= v11 v21 v31
W . In order to avoid a red B each vertex of 2 V(K3×4)\W is adjacent in
blue to one vertex in W .
By pigeon hole principle this results in a blue
K
1,3, a contradiction. Therefore,4
)
,
(
2 1,33
B
K
≤
m
.Hence
m
3(
B
2,
K
1,3)
=
4
.Next as
m
4(
C
3,
K
1,3)
=
3
(see [3]) and C3 is a subgraph of B , we get 2.
3
)
,
(
2 1,34
B
K
≥
m
Now consider any colouring of K4×3 =HR⊕HB. Assume the graph has no blue
K
1,3. Then K4×3 has a red C3, say v11v21v31v11 (asm
4(
C
3,
K
1,3)
=
3
by [3]). In order to avoid a red B each vertex in 2 {v41,v42,v43} is adjacent in blue to two vertices in{1,2,3}} :
{vi1 i∈ and further v is adjacent in blue to one vertex in 12 {vi1:i∈{2,3}}. By pigeon hole principle this results in a blue
K
1,3, a contradiction. Therefore,3
)
,
(
2 1,34
B
K
≤
15
Next consider any colouring of K6×1=HR⊕HB where H consist only of the two B blue cycles, v11v21v31v11 and v41v51v61v41. Then K6×1 has niether a red B nor a blue 2
1,3
K
. Therefore,m
5(
B
2,
K
1,3)
≥
m
6(
B
2,
K
1,3)
≥
2
.Now consider any red-blue coloring of K5×2. Assume the graph has no blue
K
1,3. Then there is a red C3 say v11v21v31v11 (sincem
5(
C
3,
K
1,3)
=
2
by [3]). If every vertex in} , , ,
{v41 v42 v51 v52 is adjacent in blue to at least two vertices of V(C3) then the graph has a blue
K
1,3, a contradiction. Therefore there is a vertex in {v41,v42,v51,v52} which is adjacent in red to two vertices of V(C3). This forces a red B . Hence 2m
5(
B
2,
K
1,3)
≤
2
. Therefore,m
j(
B
2,
K
1,3)
=
2
when j∈{5,6}.Finally, we have
m
j(
B
2,
K
1,3)
=
1
for j≥7 asr
(
B
2,
K
1,3)
=
7
(see [2]).7.Size Ramsey numbers
m
j(
B
2,
G
)
for other graphs G Theorem 6.If j≥3, then
≥ ∈ ∞
+
7 if 1
6 = if 2
{3,4,5} if
= ) ,
( 2 1,3
j j j
x K B mj
Proof: As
m
j(
C
3,
K
1,3+
x
)
=
∞
when j∈{3,4,5} and C3 is a subgraph of B , we 2 havem
j(
B
2,
K
1,3+
x
)
=
∞
when j∈{3,4,5}. Asm
6(
C
3,
K
1,3+
x
)
=
2
(see [3]) and3
C is a subgraph of B , we have 2
m
6(
B
2,
K
1,3+
x
)
≥
2
.Next consider any red B free colouring of 2 K6×2 =HR⊕HB. Then the graph has a blue
3
C , say v11v21v31v11(as m6(B2,C3)=m6(C3,B2)=2 by [3]). If any vertex of V(C3) is
adjacent to any vertex in
∪
i6=4{
v
i1,
v
i2}
is blue, then the graph has a blueK
1,3+
x
and hence we are done with the proof. Therefore, assume that every vertex of V(C3) is adjacent to every vertex in∪
6i=4{
v
i1,
v
i2}
in red. Since the graph has no red B all edges 262 52 61 52 61 42 52
42v ,v v ,v v ,v v
v are blue. This result in a blue
K
1,3+
x
. Therefore,2
)
,
(
2 1,36
B
K
+
x
≤
16
1
=
)
,
(
B
2K
1,3x
m
j+
asr
(
B
2,
K
1,3+
x
)
=
7
(see [2]).Theorem 7. If j≥3, then
≥ ∈ ∞ 7 if 1 6 = if 2 {3,4,5} if = ) , ( 2 3
j j j C B mj
Proof: See [3].
Theorem 8. If j≥3, then
≥
∈
7
if
1
{5,6}
if
2
4
=
if
3
3
=
if
4
=
)
,
(
2 4j
j
j
j
C
B
m
jProof: See [4].
Theorem 9. If j≥3, then
≥ ∈ ≤ ∞ 10 if 1 {6,7,8,9} if 2 5 if = ) , ( 2 2
j j j B B mj
Proof:Consider the coloring of K5×s where V(K5×s)={vmn |m∈{1,2,...,5}and }}
{1,2,...,s
17
Here
v
ij is connected in blue tov
i′j′ if i=1, i′∈{2,5} and i∈{2,3,4}, i′=i±1 and 5=
i , i′∈{4,1} for any j, j′∈{1,2,...,s}.
Figure 2: The graph H . B
Clearly the coloring K5×s =HR⊕HB doesn’t contain a red B nor a blue 2 B . 2
Therefore, we get that m5(B2,B2)≥s. Since s is an arbitrary positive integer, we get
that m5(B2,B2)=∞. Thus,
m
j(
B
2,
B
2)
=
∞
for all j≤5.Next we will prove that m6(B2,B2)≤2. To prove this first we will use the following lemma.
Lemma 10. Suppose that v is any vertex of K6×2, and that V ={v1,v2,...,v5} is a set of five vertices in NB(v), belonging to at least 4 partite sets which v doesn’t belong.Then, V will induce a red B or V will induce a blue 2 P3. Thus K6×2 will contain either a red or a blue B . 2
18
V in blue are P or 2 2P . But then clearly by exhaustive search we see that the induced 2 subgraph of V in red will have a red B , as required. 2
Lemma 11.. Consider the coloring of K6×2 =HR ⊕HB such that H contains a red R
3
3K where two K3’s will belong to the same partite sets. Then either H contains a R red B or 2 H contains a blue B B . 2
Proof:Suppose that the three disjoint red three cycles in H are generated by the sets }
, , {
= 11 21 31
1 v v v
H , H2 ={v41,v51,v61} and H3={v42,v52,v62}. Assume that H is R
2
B -free. As there is no red B containing 2 v , without loss of generality we may assume 22 that e=(v11,v22) will be a blue edge. Again as there is no red B , the edge 2 e will have a common neighboring vertex in H and 2 H3 in blue. Thus H contains a blue B B , as 2
required in the lemma.
To prove that m6(B2,B2)≤2, consider any red B free and blue 2 B free colouring of 2
B R H
H
K6×2 = ⊕ . Since r(C3,C3)=6 by symmetry the induced red subgraph H generated by
{
v
p1|
p
∈
{1,...,6}}
will have a copy of a red C3. Without loss of generality let the red C3 in H be v11v21v31v11. Let H1={v11,v21,v31} and} , , {
= 12 22 32
2 v v v
H . Let Y =K6×2\(H1∪H2). Let B and R denote the blue and red induced subgraphs by Y. Consider any vertex of H . As it together with 2 H do not 1 induce a red B we get that without loss of generality one of the possible cases. 2
Case 1: If one vertex say x (say v ), of 21 H is adjacent to two vertices of 1 H in blue 2 and another vertex of H is adjacent in blue to a the vertex of 1 H . 2
Figure 3: The two possibilities for Case 1
19
1
H , there are at least 12 blue edges between Y and H . But by lemma 10, as 1 v can be 11 adjacent to at most 4 vertices of Y(note in the case its adjacent to exactly four vertices it must belong to exactly two partite sets of Y) and also v can be adjacent to at most 2 21 vertices of Y. Therefore, we get that v31 must be adjacent in blue to all the vertices of Y . But then if B has a blue P3 we are done as it will result in a blue B . Therefore, 2 B
can have at most one blue edge or two disjoint blue edges or 3 disjoint blue edges. In the first two options clearly R will contain a red B , a contradiction. The last option will 2 result in R containing a red 2K3 as in lemma 11. This will lead to a contradiction by the lemma.
Case 2: If (v11,v22),(v21,v32) and (v31,v12) are the only blue edges between H and 1
.
2
H
Figure 4: The Case 2.
However, as each vertex of Y has to be adjacent to at least two vertices of H in blue, 1 we get that each of the 3 vertices of H will be adjacent to exactly 4 vertices of 1 Y
belonging to exactly 2 partite sets as otherwise H will have a vertex satisfying Lemma 1 10, resulting in a contradiction. If we investigate the internal structure of B there are two possible subcases.
Subcase 2.1: If B has one blue edge or two disjoint blue edges or 3 disjoint blue edges. If B≅K2∪4K1 or B≅2K2∪2K1, clearly we see that R contains a B . If 2 B≅3K2, then we obtain a contradiction by lemma 11.
Subcase 2.2: If B contains a P3
There are two possibilities corresponding to this subcase. In the first possibility if B
20
exactly 4 vertices of Y belonging to exactly 2 partite sets. Further, as there is no red B , 2 the blue neighborhood in Y of any two vertices belonging to {v11,v21,v31} will be distinct. Therefore, {vi1,v41,v42,v51} for some i∈{1,2,3} will induce a blue B , a 2
contradiction.
In the second possibility if B contains a P3 incident to 3 partite sets and no P3 incident to 2 partite sets. Without loss of generality say this path is given by v41v52v61. Then in order to avoid the first possibility (v41,v51), (v51,v61), (v42,v52) and (v52,v62) will have to be red edges. But then, if (v41,v61) is blue we get that W ={v41,v52,v61,vi1} for some i∈{1,2,3} will induce a blue B , a contradiction. Next suppose that 2 (v41,v61) is red. But then as R contains no red B both 2 v and 42 v62 will have to be adjacent in blue to at least one vertex in W′={v41,v51,v61}. This will give rise to a similar situation as in case 1, with H and 1 H replaced by W2 ′ and Y \W′ respectively, resulting in the required contradiction.
Therefore by the above two cases we can conclude that, m6(B2,B2)≤2.
Consider the coloring of K9×1=HR⊕HB where H is given in the Figure 5. This B coloring K9×1 contains no red B or a blue 2 B . Therefore, we obtain that 2
. 2 ) , ( 2 2
9 B B ≥
m
21
Since
m
i(
B
2,
B
2)
≤
m
j(
B
2,
B
2)
for i≥ j, using the inequalities m9(B2,B2)≥2 and 2) , ( 2 2
6 B B ≤
m , we can conclude that mi(B2,B2)=2 for i∈{6,7,8,9}.
Clearly
m
j(
B
2,
B
2)
=
1
when j≥10 as r(B2,B2)=10 (see [2]). Hence the Theorem. Theorem 12. If j≥3, then
≥ ∈
≤ ∞
11 if
1
{9,10} if
2
8 if
= ) , ( 2 4
j j
j
K B mj
Proof: Let s be a positive integer. Consider the coloring of K8×s =HR⊕HB where
R
H is as follows.
Figure 6: In the case s=2 the partite set Vi consists of the two vertices vi1 and vi2
22 ,
{1,2,3,4} ∈
i the edges between the pair of set Vi, Vi+4 are red. Also, the edges between
1
V and V8 are red.
Then this coloring K8×s contains no red B or a blue 2 K . Therefore, we obtain that 4
s K B
m8( 2, 4)≥ . Since s is arbitrary, we get that m8(B2,K4)=∞. Therefore we could conclude that
m
j(
B
2,
K
4)
=
∞
for all j≤8.Lemma 13. Suppose that v is any vertex of K9×2, adjacent in blue to V ={v1,v2,...,v9} belonging to 6 partite sets V1,V2,...,V6 such that v7∈V1, v8∈V2, v9∈V3 and vi∈Vi
for all i where 1≤i≤6. Then, K9×2 will induce a red B or a blue 2 K . Also if 4 v is adjacent in blue to seven vertices belonging to 7 partite sets then K9×2 will induce a red
2
B or a blue K . 4
Proof: Assume that K9×2 is red B free. 2
Remark: Assume that there is a red C3 in V such that the vertices of the red C3 belong to the partite sets Vi,
V
j and Vk where 3≤i, j,k≤6. Then any two vertices of V belonging to distinct partite sets outside ofV
i∪
V
j∪
V
k will be adjacent to each other in red. This is clear, as if there is such a blue edge ( wu, ) belonging to distinct partite sets of V outside ofV
i∪
V
j∪
V
k, then in order to avoid a red B , 2 u and w will have to be adjacent in blue to a common vertex of V(C3). That is, it will contain a blue C3 in V . Thus, K9×2 will induce a blue K (containing 4 v), as required.As r(C3,C3)=6(see [2]), under the conditions stated in the lemma, there is a red C3 induced by any 6 vertices of V belonging to 6 partite sets. By the above remark in order to avoid a red B , the vertices of 2 V(C3) must be contained in the first three partite sets. But even in this case this will force a red C3 contained in V4∪V5∪V6. However, in such a situation, by the repeated use of the remark on this red C3 will result in blue K , 4 as required.
The later part of the lemma follows directly as r(B2,C3)=7(see [2]).
23
Proof.Assume K9×2 has no red B . Clearly, V cannot have any red 2 P3 as it would force a red B in 2 K9×2. Thus the red edges of V must consist of a P , 2 2P or a 2 3P . 2 However, exhaustive search shows that in each of these possibilities a vertex not incident to a red edge in V will be contained in a blue induced subgraph of V isomorphic to a
4
K . Hence the lemma.
Lemma 15. Suppose that v is any vertex of K9×2 adjacent in blue to V ={v1,v2,...,v11} belonging to 6 partite sets V1,V2,...,V6 such that v6∈V6 and vi,vi+6∈Vi for all i such that 1≤i≤5. Then, K9×2 will induce a red B or a blue 2 C3.
Proof. Suppose that V doesn’t induce a red B nor a blue 2 C3. As r(C3,C3)=6(see [2]), there is a red C3 induced by any 6 vertices of V belonging to 3 partite sets. Let the vertices of this red C3 be {x,y,z}. In order to avoid a red B each vertex in 2
} , , { \ x y z
V , not belonging to the partite sets which x,y and
z
belong to, must be adjacent to at least 2 vertices of {x,y,z} in blue. Also in order to avoid a red B each 2 vertex in V\{x,y,z}, belonging to the partite sets which x,y andz
belong to, must be adjacent to at least 1 vertex of {x,y,z} in blue. By pigeon hole principle, without loss of generality we may assume that x is adjacent in blue to 5 vertices of V . Since these 5 vertices belong to at least 3 partite sets and V doesn’t contain a red B , these 5 vertices 2 must induce a blue edge say (p,q). But then we get the required contradiction asq p
x, , will induce a blue C3. Hence the lemma.
To prove that m9(B2,K4)≤2, consider any red B free and blue 2 K free colouring of 4
B R H
H
K9×2 = ⊕ . Since r(C3,K4)=9(see [2]) the induced subgraph H of H R generated by
{
v
p2|
p
∈
{1,...,9}}
will have a copy of a red C3, say v42,v52,v62,v42. Let} , , {
= v42 v52 v62
W . Since m6(B2,C3)=2, we may assume the induced subgraph of
B
24
1
W will be adjacent in blue to a vertex of W . Thus we get that without loss of generality there are two possible cases.
Remark: Note that we will be using the fact that any vertex outside of W ∪W1 will be adjacent to at least two vertices of W in blue. Also given any six vertices belonging to 3 partite sets outside W∪W1 will contain at least 4 vertices belonging to exactly 3 partite sets adjacent to some vertex of W in blue or else all three vertices of W will be adjacent in blue to exactly 4 vertices belonging to exactly two partite sets.
Case 1: If (v41,v62), (v51,v62) and (v61,v52) are blue edges between W and W . 1
Let U =V(K9×2). In order to avoid a red B each vertex in 2 U \(W ∪W1), must be
adjacent to at least 2 vertices of {v42,v52,v62} in blue. Also in order to avoid a red B 2
each vertex in W , must be must be adjacent to at least 1 vertex of W in blue. By pigeon 1 hole principleand lemma 13, without loss of generality we may assume that at least one vertex say
z
of {v42,v52,v62} is adjacent in blue to 9or more vertices of U \(W ∪W1) (by lemma 13, all three vertices cannot be adjacent to exactly 8 vertices of) (
\ W W1
U ∪ ).Futher, by lemma 13,
z
can not be incident in blue to 6 or more than 6 partite sets. Therefore,z
must be incident to 9 vertices belonging to 5 partite sets. Since62
v is incident to v and 41 v51we get z≠v62. If z= v52then there are two possibilities, namely (v41,v52) is blue or (v41,v52) is red. If (v41,v52) is blue, lemma 13 will give us the required contradiction. If (v41,v52) is red, applying lemma 14 with v= v52 or lemma 13 with v= v52 will give us a contradiction. Therefore, z =v42.Further, as shown above
since v52 and v62 can be adjacent to at most 8 vertices of U in blue, v will have to be 42
adjacent to 11 vertices of U in blue. Now applying lemma 15 to the 11 vertices v is 42 adjacent in blue, we get a blue C3 induced by these 11 vertices unless these 11 vertices belong to at least 7 partite sets. But in this case too, since r(B2,C3)=7(see [2]), we will get a blue C3 induced by these vertices. Thus, v together with these 3 vertices will 42 induce a blue K , a contradiction. 4
Case 2: If (v41,v62), (v51,v42) and (v61,v52) are the only blue edges between W and 1 .
W
25 62
= v
v will give us the required contradiction, unless v62 is adjacent to both v and 12 v 22
in red and v62 is adjacent in blue to at most 2 partite sets out of V7, V8 and V9. But in order to avoid a red B , this will force all edges between 2 {v12,v22} and {v42,v52} to be blue. But then by pigeon hole principle without loss of generality we may assume that
52
v is adjacent to at least 4 vertices of {vmn |m∈{7,8,9} andn∈{1,2}} in blue. Irrespective of whether these 4 vertices belong to two or three partite sets of V7, V8 and
9
V , applying lemma 13 with v= v52 will give us the required contradiction.
In the second scenario by symmetry we may assume that (v32,v62), (v22,v42) and
) ,
(v12 v42 are red. According to whether v62 is adjacent to two or three partite sets in blue of {vmn |m∈{7,8,9} and n∈{1,2}} we will obtain a required contradiction by applying lemma 14 and lemma 13 respectively to v= v62.Therefore we can conclude that, m9(B2,K4)≤2
Consider the coloring of K10×1 =HR ⊕HB where H is given in the Figure 7. This R coloring of K10×1 contains no red B or a blue 2 K . Therefore, we obtain that 4
2 ) , ( 2 4
10 B K ≥
m .
Figure 7: The graph H . R
Since
m
i(
B
2,
K
4)
≤
m
j(
B
2,
K
4)
for i≥ j, using m10(B2,K4)≥2 and m9(B2,K4)≤2, we can conclude that mj(B2,K4)=2 for j∈{9,10}.Clearly
m
j(
B
2,
K
4)
=
1
when j≥11 as r(B2,K4)=11 (see [2]). Hence the Theorem.REFERENCES
26
2. V.Chavatal, F.Harary, Generalized Ramsey theory for graphs, III. small off-diagonal numbers, Pacific Journal of Mathematics. 41-2 (1972) 335-345.
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4. C.J.Jayawardena and T.Hewage, Multipartite Ramsey numbers for C4 versus all graphs on 4 vertices, in progress.
5. V.Kanovei, M.Sabok and J.Zapletal, Canonical Ramsey theory on Polish spaces, Cambridge University Press. Cambridge Tracks in Mathematics 202.
6. V.Kavitha and R.Govindarajan, A study on Ramsey numbers and its bounds, Annals of Pure and Applied Mathematics, 8(2) (2014) 227-236.
7. Masulli, An introduction to fuzzy sets and systems, DISI - Dept. Computer and Information Sciences, University of Genova, Italy, (2010).
8. T.Pathinathan and J.Jesintha Rosline, Matrix representation of double layered fuzzy graph and its properties, Annals of Pure and Applied Mathematics, 8(2) (2014) 51-58. 9. M. S. Sunitha and Sunil Mathew, Fuzzy graph theory: a survey, Annals of Pure and
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