International Journal of CombinatoricsVolume 2011 (2011), Article ID 649687, 9 pagesdoi:10.1155/2011/649687

Research Article

Ramsey Numbers for Theta Graphs

M. M. M. Jaradat,1,2 M. S. A. Bataineh,¹ and S. M. E. Radaideh¹

¹Department of Mathematics, Yarmouk University, Irbid 21163, Jordan²Department of Mathematics, Statistics, and Physics, Qatar University, Doha, Qatar

Received 14 January 2011; Revised 27 March 2011; Accepted 31 March 2011

Academic Editor: Alois Panholzer

Copyright © 2011 M. M. M. Jaradat et al. M. M. M. Jaradat et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

The graph Ramsey number is the smallest integer with the property that any complete graph of at least vertices whose edges are colored with two colors (say, red and blue) contains either a subgraph isomorphic to all of whose edges are red or a subgraph isomorphic to all of whose edges are blue. In this paper, we consider the Ramsey numbers for theta graphs. We determine , for . More specifically, we establish that for . Furthermore, we determine for . In fact, we establish that if is even, if is odd.

1. Introduction and Preliminaries

The graphs considered in this paper are finite, undirected, and have no loops or multiple edges. For a given graph , we denote the vertex set of a graph by and the edge set by . The cardinalities of these sets are denoted by and , respectively. Throughout this paper a cycle on vertices will be denoted by , the complete graph on vertices by . Suppose that and is non-empty, the subgraph of whose vertex set is and whose edge set is the set of those edges of that have both ends in is called the subgraph of induced by , denoted by . Let be a cycle in a graph , an edge in is called a chord of . Further, a graph has a -graph if has a cycle that has a chord in . Let be a graph and . The degree of a vertex in , denoted by , is the number of edges of incident to . The neighbor-set of a vertex of in a subgraph of , denoted by , consists of the vertices of adjacent to . The circumference, of the graph is the length of the longest cycle of . For vertex-disjoint subgraphs and of we let and .

The graph Ramsey number is the smallest integer with the property that any complete graph of at least vertices whose edges are colored with two colors (say, red and blue) contains either a subgraph isomorphic to all of whose edges are red or a subgraph isomorphic to all of whose edges are blue.

It is well known that the problem of determining the Ramsey numbers for complete graphs is very difficult, and it is easier to deal with paths, trees, cycles, and theta graphs. See the updated bibliography by Radziszowski [1]. In this paper we study in the case when both and are theta graphs.

The results concerning Ramsey numbers for cycles were established by Chartrand and Schuster [2] (for ), by Bondy and Erdös [3] (for odd and for the case when is much smaller than ), and for all the remaining values by Rosta [4] and by Faudree and Schelp [5], independently. These results are summarized in the following theorem.

Theorem 1.1. Let be integers. Then

In order to prove our results, we need to state the following results.

Theorem 1.2 (see [6]). Let be a graph on vertices with no cycles of length greater than . Then where .

Theorem 1.3 (see [7]). Every non-bipartite graph on vertices with more than edges contains cycles of every length , where .

Theorem 1.4 (see [8]). Let be a non-bipartite graph on vertices and contains no -subgraph. Then .

Theorem 1.5 (see [9]). Let be a non-bipartite graph on vertices and contains no -subgraph. Then .

In this paper, we consider Ramsey numbers for theta graphs. We determine , for . More specifically, we establish that for . Furthermore, we determine for . In fact, we establish that

Throughout this paper (Figures 1–5), solid lines represent red edges and dashed lines represent blue edges.

Figure 1: This figure represents the red edges in .

Figure 2: This figure depicts the case when is adjacent to two nonconsecutive vertices of .

Figure 3: This figure represents the situation in Subcase 1.1.

Figure 4: This figure represents the situation in Subcase 1.2.

Figure 5: This figure represents the situation in Subcase 1.3.

2. Main Result

In the following lemma we determine the Ramsey number .

Lemma 2.1. The Ramsey number.

Proof. First we show that . Let be colored as follows: the vertex set is the disjoint union of two subsets and each of order 3 and completely colored red. All edges between and are colored blue. This coloring contains neither a red -graph nor a blue . So, we conclude that .It remains to show that . Let a red-blue coloring of be given. By Theorem 1.1, has a red or a blue . If has a blue , then the result is obtained. So we need to consider the case when has a red . Let be the vertices of and assume are the vertices of the red . Any vertex of the remaining vertices , and is adjacent to the red by at least 2 blue edges as otherwise a red -graph is produced. Assume has a blue edge, say , then has a blue . So, we need to consider that has no blue edge. Thus, a red -graph is produced. This completes the proof.

In the following lemma we determine the Ramsey number .

Lemma 2.2. The Ramsey number .

Proof. First we show that . Let be colored as follows: The vertex set is the disjoint union of three subsets , and each of order 3 and completely colored red and the red edges between , , and are shown in Figure 1. The remaining edges are colored blue.This coloring contains neither a red nor a blue -graph. So, we conclude that . It remains to show that . Let a red-blue coloring of be given. By Theorem 1.1, contains a red or a blue . Without loss of generality we assume that has a red . Let be the vertices of and assume be the red in . Observe that has a red or a blue . So, we consider the following two cases.Case 1. Suppose has a red , say . Let , and be the remaining vertices in . Suppose has no blue edge, then has a red -graph. So, we need to consider the case when has a blue edge, say is the blue edge. Observe that any vertex in must be adjacent to each of and by two blue edges as otherwise a red -graph is produced. Thus and incident with two blue edges that have a common vertex in and incident with two blue edges that have a common vertex in and so a blue -graph is produced.Case 2. Now we need to consider the case when has a blue , say . Observe that every vertex in the red is adjacent to the blue by two red edges as otherwise a blue -graph is produced. Further, every vertex in the blue is adjacent to the red by two blue edges as otherwise a red -graph is produced.Thus, there are at least six red edges between the red and the blue and at least six blue edges between the blue and the red . We know that, . This is a contradiction.

Now, we begin with the following construction which will be used throughout our results. Let . Then, in color the edges of a complete bipartite graph with blue and all the remaining edges with red. Then this coloring contains neither a red -graph nor a blue -graph where . Thus, where . In the following lemma we determine the Ramsey number .

Lemma 2.3. The Ramsey number .

Proof. It is enough to show that . Let a red-blue coloring of be given that contains neither a red -graph nor a blue -graph. By Theorem 1.1, must contain a blue or a red . Without loss of generality we assume that has a red . Let be the vertices of and assume is the red . Define and . Now we have the following observations.(i) has no red chord as otherwise a red -graph is produced. Thus, contains a blue . (ii)Every vertex in is adjacent by at most 3 red (blue) edges to as otherwise a red (blue) -graph is produced. (iii)If a vertex in is adjacent to by 3 red (blue) edges, then it must be adjacent to 3 consecutive vertices in with red (blue) color as otherwise would have a red (blue) -graph ( are consecutive with red (blue) if is adjacent to with a red (blue) edge and is adjacent to with a red (blue) edge). (iv)Assume there are two vertices in , say and are adjacent to by 3red (blue) edges each. Then and are adjacent to 3 consecutive vertices in and as otherwise a red (blue) -graph is produced. (v)There are exactly two vertices in adjacent to vertices of by exactly 3 red edges and two blue edges each, say and , and so each of and is adjacent to vertices of by exactly 3 blue edges and two red edges. To this end, one can notice from the above observations that if is adjacent to two nonadjacent vertices of by the red edges, then a red -graph is produced (Figure 2 depicts the situation), a contradiction. If is adjacent to two adjacent vertices of by the red edges, then adjacent to three non-consecutive vertices (of the internal blue cycle) by blue edges. And so a blue -graph is produced, a contradiction. This completes the proof.

In the following lemma we determine the Ramsey number .

Lemma 2.4. The Ramsey number .

Proof. It is enough to show that . Let a red-blue coloring of be given. By Lemma 2.3, must contain a blue or a red -graph. If contains a blue -graph, then we are done. So, suppose has a red -graph. So, has a red triangle. Let be the red triangle. Let be the remaining vertices. By Theorem 1.1, has a red or a blue . We consider the following two cases.Case 1. contains a blue . Let be the blue . Every vertex in the blue is adjacent at least by two blue edges to , as otherwise would has a red -graph. Let denote the number of blue edges from to . We consider 3 subcases according to the number of blue edges of , and to .Subcase 1.1. , then a blue -graph is produced. Figure 3 depicts the situation.Subcase 1.2. and , then a blue -graph is produced. Figure 4 depicts the situation.Subcase 1.3. , then a blue -graph is produced. Figure 5 depicts the situation.Case 2. contains a red , say . Let be the remaining vertices of . Observe that each vertex of is adjacent to at least two vertices of each and which are colored by blue, as otherwise, contains a red -graph. Hence, if has a blue edge, say , then a blue is produced. Hence, by the above case a blue -graph is produced. So, we need to consider the case when has no blue edges. Observe that the number of blue edges in is at least 18. Further, the induced graph by blue edges is non-bipartite. By Theorem 1.5, would have a blue -graph. This completes the proof.

In the following lemma we determine the Ramsey number .

Lemma 2.5. The Ramsey number .

Proof. It is enough to show that . Let a red-blue coloring of be given that contains neither a red nor a blue . Suppose has a blue cycle of length 6. Let be the vertices of , and assume is the blue . Then has no blue chord as otherwise a blue -graph is produced. So, and are red triangles. Every vertex of the remaining vertices must be adjacent to by at least 4 blue edges as otherwise a red -graph is produced. Now, let be a vertex of the remaining vertices that is adjacent to by 4 blue edges. We consider three cases.Case 1. is adjacent to 4 consecutive vertices in . Assume is adjacent to , and , then a blue -graph is produced.Case 2. is adjacent to 3 consecutive vertices and a vertex separated in . Assume is adjacent to , and , then a blue -graph is produced.Case 3. is adjacent to a pair of 2 consecutive vertices separated from each other in . Assume is adjacent to and , then a blue -graph is produced.So, we need to consider that has no blue . We need to prove that has a red -graph. By contradiction, suppose has no red -graph. By Theorem 1.1, has a red . So, the subgraph induced by red edges is a non-bipartite graph. By Theorem 1.4, the number of red edges is at most 27. So, the number of blue edges is at least 28. By Lemma 2.1, has a blue . Hence, the subgraph induced by blue edges is a non-bipartite graph. By Theorem 1.3, there is a blue , this is a contradiction. This completes the proof.

In the following theorem we determine the Ramsey number , for .

Theorem 2.6. The Ramsey number.

Proof. It is enough to show that ,  . We prove it by contradiction. Let a red-blue coloring of be given. Suppose has a blue cycle of length . Let be the vertices of , and assume is the blue . Then has no blue chord as otherwise a blue -graph is produced. So, contains a red -graph. This is a contradiction.Now, we need to consider the case when has no blue cycle of length . By Theorem 1.1, contains a red . Let be the induced subgraph of the blue edges. Note that the subgraph induced by the red edges is a non-bipartite graph and contains no red . Hence, the number of red edges is at most . Thus, the number of blue edges is Observe that is a non-bipartite graph ( and does not contain a red -graph and so it contains a blue , Lemma 2.1). If , then by Theorem 1.3, has a blue , this is a contradiction. If , then by Theorem 1.2 which contradicts the inequality (2.1) for . This completes the proof.

In the following theorem we determine the Ramsey number , for .

Theorem 2.7. The Ramsey number .

Proof. It is enough to show that , . We prove it by contradiction. Let a red-blue coloring of be given. Suppose has a blue cycle of length . Let be the vertices of , and assume is the blue . Then has no blue chord as otherwise a blue -graph is produced. So, contains a red -graph. This a contradiction.Now, we consider the case when has no blue . Now, we have the following observations.(i),  . Thus, the induced graph on the red edges is a non-bipartite graph.(ii). Thus, the induced graph by the blue edges is a non-bipartite graph.Let be the graph induced by the blue edges. Since has no red -graph, by Theorem 1.5 the number of red edges is at most . Thus, as in the above theorem, edges. If , then by Theorem 1.3, there is a blue . This is a contradiction. If , then This contradicts the inequality (2.3) for .

Theorem 2.8. For ,

Proof. First we consider the case when is odd. In color the edges of two vertex disjoint complete graphs of order with a red color and the remaining edges with a blue color. This coloring contains neither a red nor a blue -graph. We conclude that .Let a red-blue coloring of be given that contains neither a red nor a blue . We know, by Theorem 1.1, that . Thus, contains either a red or a blue . Without loss of generality, we suppose that has a blue . Then there are no chords in as otherwise a blue is produced. So, contains a red . This is a contradiction.Now we consider the case when is even. Let be colored with two colors, say red and blue, as follows: the edges of vertex disjoint and are colored blue and the remaining edges are colored red. This coloring contains neither a red nor a blue . We conclude that .To show that , we follow, word by word, the above argument when is odd by taking into account that . The proof is complete.

We conclude this paper by highlighting an interesting open problem. We begin with the following constructions.

Let . If contains an odd cycle, then in color the edges of a complete bipartite graph with blue and all the remaining edges with red. This coloring contains neither a red -graph nor a blue -graph. Thus, .

Now, we consider the case when contains no an odd cycle. If is even, then in , color the edges of a complete bipartite graph with blue and all the remaining edges with red. This coloring contains neither a red -graph nor a blue -graph. Thus, . If is odd, then in , color the edges of a complete bipartite graph with red and all the remaining edges with blue and in , color the edges of a complete bipartite graph with blue and all the remaining edges with red. and , respectively, provide examples for lower bounds when is odd, respectively, when contains no an odd cycle. Thus, .

From the above construction we conjecture that for ,

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