A red-blue coloring of K 5 .  

Contexts in source publication

Context 1

... first that mr(C 3 , H 1 ) ≤ R(C 3 , C 3 ) = 6 while mr(C 3 , P 4 ), mr(H 1 , P 4 ) ≤ R(P 4 , P 4 ) = 5 by (1). In fact, mr(C 3 , H 1 ) = 6 as the red-blue coloring of K 5 in Figure 2 shows since neither C 3 nor H 1 is a subgraph of C 5 . Similarly, the red-blue coloring of K 4 inducing a red C 3 + K 1 (the union of C 3 and K 1 ) and a blue K 1,3 shows that mr(H 1 , P 4 ) = 5. ...

Context 2

... Observation 1(a)(b), we assume that G / ∈ {P 2 , P 3 }. If G is not a subgraph of C 5 , then the red-blue coloring of K 5 shown in Figure 2 shows that mr(G, C 3 ) ≥ 6. Thus the result is immediate by (1) since R(C 3 , C 3 ) = 6. ...

Context 3

... assume that G ⊆ C 5 . Since the coloring shown in Figure 2 is the only red-blue coloring of K 5 inducing no monochromatic triangles, it follows that mr(G, C 3 ) ≤ 5. On the other hand, since there are red-blue colorings of K 3 and K 4 resulting in no monochromatic triangles (see Figure 1), it follows that mr(G, C 3 ) ≥ |V (G)| when |V (G)| = 4, 5. Finally, since the red-blue coloring of K 4 avoids monochromatic triangles if and only if it induces either (i) a red (blue) C 4 and a blue (red) 2P 2 or (ii) a red (blue) P 4 and a blue (red) P 4 , it follows that mr(2P 2 , C 3 ) = mr(P 4 , C 3 ) = 4. ...

Context 4

... mr(G, C 4 ) ≤ R(C 4 , C 4 ) = 6 by (1). Note also that a red-blue coloring c of K 5 induces no monochromatic 4-cycles if and only if c is either the coloring in Figure 2 or the coloring in Figure 3(a). If c is the coloring in Figure 3(a), then c results in a red H 2 and a blue H 2 , where H 2 is the graph shown in Figure 3(b). ...

Context 5

... suppose that G ⊆ H 2 and G ⊆ C 5 . Then G ⊆ P 5 and so either C 3 ⊆ G or K 1,3 ⊆ G. Then the red-blue coloring of K 5 in Figure 2 shows that mr(G, C 4 ) = 6. Corollary 6. mr(C 3 , C 5 )+1 = mr(C 4 , C 5 ) = 6 while mr(C 3 , C n ) = mr(C 4 , C n ) = 6 for every integer n ≥ 3 and n = 5. ...