    Ramsey Numbers of Trails

We initiate the study of Ramsey numbers of trails. Let k ≥ 2 be a positive integer. The Ramsey number of trails with k vertices is defined as the the smallest number n such that for every graph H with n vertices, H or the complete H contains a trail with k vertices. We prove that the Ramsey number of trails with k vertices is at most k and at least 2√(k)+Θ(1). This improves the trivial upper bound of ⌊ 3k/2⌋ -1.

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1 Introduction

Ramsey theory is one of the topics in discrete mathematics that has been studied over the years [1, 3]. For graphs, the Ramsey number was first studied for complete graphs, and later it was studied for other classes of graphs such as paths, cycles, and trees. For graphs and , the Ramsey number of the pair is the smallest number such that for every graph with vertices, contains a copy of or the complement contains a copy of . It is known that for every pair of finite graphs, the Ramsey number of exists, and the determination of the Ramsey number is the ultimate goal. However, even for complete graphs, the exact Ramsey number is not known: When we only know that the Ramsey number lies between and .

In this paper, we initiate the study of Ramsey numbers for trails. Unlike paths, trails may have a repetition of vertices. To study the Ramsey number of trails, we first fix the number of vertices in a trail. Let and be integers. Then, the Ramsey number of trails with vertices and vertices is defined as the smallest number such that for every graph with vertices, contains a trail with vertices or contains a trail with vertices.

The ultimate goal is to determine the Ramsey number of trails. Unfortunately, we are unable to provide a definite answer. Nonetheless, we give a progress toward the ultimate goal. We concentrate on the diagonal case, i.e., the case where . Our main theorems give an improved upper bound of , and also a lower bound of roughly . We note here that a trivial upper bound is , which will be sketched in the next section.

2 Preliminaries

In this paper, all graphs are finite, simple and undirected. A graph is defined as a pair of a finite set and , where is the set of vertices of and is the set of edges of . The degree of a vertex is the number of edges incident to , i.e., .

A graph is a subgraph of a graph if , and for every . For a graph , the complement of , denoted by , is a graph with vertex set and edge set . Namely, . A pair of graphs is called complementary if .

A graph is A complete if each pair of vertices is joined by an edge. The complete graph with vertices is denote by . A graph is a path if , and . The path with vertices is denote by .

A walk is a sequence of vertices and edges such that for , the edge . Here, is the number of vertices of the walk, and are called endpoints. A trail is a walk in which all the edges are different from each other. A trail that satisfies is called a circuit. A graph is connected if it has a trail from any vertex to any other vertex.

Let be a connected graph. An Eulerian circuit of is a circuit of that passes every edge exactly once. If has an Eulerian circuit, then is called Eulerian. An Eulerian trail of is a trail of that passes every edge exactly once. If has an Eulerian trail but no Eulerian circuit, then is called a semi-Eulerian. It is well-known and easy to prove that a connected graph is Eulerian if and only if the degree of every vertex of is even, and

is semi-Eulerian if and only if the number of odd-degree vertices is two.

For , we denote by the set of connected graphs that have an Eulerian circuit or an Eulerian trail with vertices. Note that in our definitions, vertices in trails and circuits are counted multiple times if they are passed multiple times. Therefore, some graphs in may have less than vertices.

Let and be two graph classes, i.e., possibly infinite sets of graphs. Then, the Ramsey number of and is the smallest number such that for every graph with vertices, contains a graph in or contains a graph in . The Ramsey number of and is denoted by . If and are singletons (i.e., contain only one graph as and ), then the Ramsey number of and is denoted by .

Gerencsér and Gyárfás  determined the exact value of the Ramsey number of paths, as in the following theorem.

Lemma 1 ().

Let . Then, .

Since belongs to , . This is the trivial upper bound mentioned in the previous section.

To gain the first impression, we have conducted a computer search of the Ramsey number for small values of . This has been performed with the following procedure. For , we generate all graphs with vertices. For each such , we calculate , which is defined as the number of vertices in the longest trail in or . Then, we determine , which is defined as the minimum value of for all with vertices. If satisfies , then we know that is equal to .

The result of the computer search is summarized in Table 1. We may observe that the upper bound of should be improved.

3 Main Theorem: Lower Bound

We begin with a lower bound of .

Theorem 1.

Let be a positive integer. Then,

 R(Tk,Tk)≥⎧⎪ ⎪⎨⎪ ⎪⎩kif k≤6,⌈1+√16k−72⌉if k≥7.

The rest of the section is devoted to the proof of Theorem 1. We first consider the case when .

Let . Then, there is no graph of in the complete graph . Therefore, .

Let . Then, the complete graph has only one edge. So, there is no graph of in . Thus, .

Let . Consider the complementary pair of graphs with three vertices as shown in Figure 2. Since those two graphs have at most two edges, no element of is contained in either graph. Therefore, .

Let . Consider the complementary pair of graphs with four vertices as shown in Figure 3. Since those two graphs have three edges, no element of is contained in either graph. Hence, .

Let . Consider the complementary pair of graphs with five vertices as shown in Figure 4. Those graphs have five edges. Hence, for an element of to be contained in either of the two graphs, one of the two graphs must be Eulerian or semi-Eulerian. However, each graph has four odd-degree vertices. Thus, these graphs are neither Eulerian nor semi-Eulerian, and have no elements of .

Next, we consider the case where . To complete the proof, we use the following two lemmas.

Lemma 2.

The number of vertices of a complete graph with edges is .

Proof.

Let be the number of vertices of a complete graph with edges. In this case, . Solving for , we have . ∎

Lemma 3.

Let and let be such that the complete graph has at most edges. Then, there exists a subgraph of such that and have no element of .

Before proving Lemma 3, we finish the proof of Theorem 1 using Lemmas 2 and 3.

Proof of Theorem 1 when k≥7..

Let and let be the number of vertices of a complete graph with at most edges. Then, from Lemma 3. Therefore, by Lemma 2

 R(Tk,Tk)≥⌈1+√1+8(2k−1)2⌉=⌈1+√16k−72⌉.\qed

Thus, it suffices to prove Lemma 3.

Proof of Lemma 3..

We distinguish the cases , and .

Case 1: |E|≤2k−4.

Choose as any subgraph with . Then, since , it follows that and have no element of .

Case 2: |E|=2k−3.

Consider a complete graph with edges. Then,

 n=1+√1+8(2k−3)2=1+√16k−232>5

from Lemma 2. Let be the vertices of , and let . Then, the graph is a cycle contained in . We construct an Eulerian or a semi-Eulerian graph with edges that contains .

Before constructing such , we observe that this is enough for our purpose. Since contains and , there exist two edges of such that has exactly four odd-degree vertices. Thus, is neither Eulerian nor semi-Eulerian. Since has only edges, include no element of . Further, has only edges, and includes no element of , either.

To find a subgraph with the desired properties, we further distinguish two cases according to the parity of .

Case 2-1: n is odd.

Let . Then, is Eulerian since the degree of each vertex of is even and is connected. Thus, contains a trail with vertices. Let be a subgraph of obtained by the first edges of such a trail, and let . Then, is Eulerian or semi-Eulerian with edges.

Case 2-2: n is even.

Let . Then, is Eulerian since the degree of each vertex of is even and is connected.

Thus, contains a trail with vertices since and . Let be a subgraph of obtained by the first edges of such a trail, and let . Then, is Eulerian or semi-Eulerian with edges.

Case 3: |E|=2k−2.

This case is analogous to Case 2 where . Note that for a complete graph with edges. we have from Lemma 2.

We have to take care of the argument after constructing because has edges and we need a different argument to show that includes no element of . Remind that contains edges from a trail of and the edges of .

We distinguish two cases according to the parity of . First, let be odd. Then, the degree of every vertex of is even. Since has four odd-degree vertices, has four odd-degree vertices, too. Thus, is neither Eulerian nor semi-Eulerian. Since has only edges, includes no element of .

Next, let be even. We first observe that . We already know that , but if , then the number of edges of is , which is not of the form : this is impossible. Therefore, has at least four odd-degree vertices since has even-degree vertices and . Thus, is neither Eulerian nor semi-Eulerian. Since has only edges, includes no element of . ∎

4 Main Theorem: Upper Bound

We already observed that as a trivial upper bound. Now, we improve the upper bound in the next theorem.

Theorem 2.

For every integer ,

 R(Tk,Tk)≤k.

To this end, for any graph with vertices, we prove either or its complement contains a trail with vertices.

We begin with the following lemma which will be used in the proof of the theorem.

Lemma 4.

Let be a bipartite graph with partite sets and , i.e., , and each edge of joins a vertex of and a vertex of . If and the degree of every vertex of is two, then contains a trail such that both endpoints belong to and the number of edges is .

Proof.

Denote the three elements of by ,, and . We distinguish the following two cases according to the existence of an isolated vertex (i.e., a vertex of degree zero) in .

Case 1: A has an isolated vertex.

Without loss of generality, assume that is an isolated vertex. Since each vertex in has degree two, it is adjacent to and . Hence, the bipartite graph is connected. Furthermore, the number of odd-degree vertices in is zero or two since the degree of and is , and the degree of every vertex in is two. Thus, is Eulerian or semi-Eulerian and has edges. When is Eulerian, contains a trail with edges such that both endpoints coincide with . When is semi-Eulerian, contains a trail with edges such that one endpoint is and the other endpoint is .

Case 2: A has no isolated vertex.

Without loss of generality, assume that there exists a vertex adjacent to and . Since there is no isolated vertex, there is a vertex adjacent to . As the degree of is two, is adjacent to either or . Therefore, the three vertices of , and are connected by paths. This implies that is connected since every vertex in is adjacent to one of , and . Since is bipartite and the degree of every vertex in is two, the sum of the degrees of , and is even. If there are an odd number of odd-degree vertices in , then the sum of the degrees of , and is odd, contradicting the fact that the sum of the degrees of , and is even. Therefore, the number of odd-degree vertices is zero or two.

When there is no odd-degree vertex, then is Eulerian, and contains a trail with edges such that both endpoints coincide with . When there are two odd-degree vertices, let them be and . Then, is semi-Eulerian, and contains a trail with edges such that one endpoint is and the other endpoint is . ∎

We are now ready for the proof of Theorem 2.

Proof of Theorem 2..

The proof uses the induction on . When , holds from Table 1.

Now, fix an arbitrary integer and suppose that the statement is true for . Consider a graph with vertices. For a subgraph with vertices of , by induction hypothesis, either or contains a trail with vertices. If contains , then contains because is a subgraph of . If contains , then contains because is a subgraph of . Therefore, either or contains . Without loss of generality, suppose contains . Let where for all , be the set of vertices in , and . Note that the size of can be smaller than since some vertices can be identical.

If there exists a vertex such that , then contains the trail with vertices. Similarly, if there exists a vertex such that , then contains the trail with vertices. If there is a vertex such that is not included in , then contains the trail with vertices. Similarly, if there is a vertex such that is not included in , then contains the trail with vertices. In all of these cases, contains a trail with vertices and we are done.

Hence, we only need to consider the cases where the following two conditions are satisfied.

Condition 1.

For every , and . That is, and .

Condition 2.

For every , if is not included , then . That is, . If is not included , then . That is, .

We distinguish the cases according to the “shape” of .

Case 1: S is a path.

Since is a path, contains no repeated vertex. Therefore, and . Let be the only vertex in . Since contains no repeated vertex, for , the edges are not included in . Also, for , the edges are not included in . From Condition 2, and for , and . Consider a subgraph of , where and . Each vertex of except is adjacent to . Hence, is connected. Further, since the degree of each vertex in except and is two, and the degrees of and are , the number of odd-degree vertices in is zero or two. Therefore, is Eulerian or semi-Eulerian. Since has edges, contains a trail with vertices. Since is a subgraph of , we conclude that contains .

Case 2: S is a circuit.

When is a circuit, . Therefore, and . Denote the elements of by .

If there exist and such that , then we have a trail

 T=w{w,ui}uieiui+1…uk−1e1u2…ui

since . Note that has vertices. Therefore, contains a trail with vertices.

Hence, we only need to consider the situation where , i.e., for every and every . We distinguish two cases according to the comparison of and .

Case 2-1: |U|≥|W|.

Choose two vertices arbitrarily, and let and . Consider the subgraph of . Then, is connected since every vertex in is adjacent to and . The degree of every vertex in is two, and the degree of and are both . Hence, the number of odd-degree vertices in is zero or two. Therefore, is Eulerian or semi-Eulerian. Since has edges, contains a trail with vertices. Since is a subgraph of , we conclude that contains .

Case 2-2: |U|≤|W|.

If , then the number of vertices in is less than 1, which contradicts the fact that the number of vertices in is . Therefore, . Choose two vertices arbitrarily, and let and . Consider the subgraph of . Since every vertex in is adjacent to and , is connected. The degree of every vertex in is two, and the degree of and are both . Hence, the number of odd-degree vertices in is zero or two. Therefore, is Eulerian or semi-Eulerian. Since has edges, contains a trail with vertices. Since is a subgraph of , we conclude that contains .

Case 3: S is not a path or a circuit.

Since is not a path, . Since is not a circuit, . We distinguish cases according to the size of .

Case 3-1: |U|=k−2.

Since is a trail with vertices and , there is only one vertex that is used more than once in . If and , then there are at most two vertices adjacent to either or in . If is or , then there are at most four vertices adjacent to either or in .

Let be the set of elements of that are not adjacent to either or in . Every vertex satisfies and , i.e. and . Further, . From Condition 1, for each vertex , we have and . Let , and consider the subgraph of . Since every vertex in is adjacent to and , is connected. The degree of every vertex in is two, and the degree of and are both . Hence, the number of odd-degree vertices in is zero or two. Therefore, is Eulerian or semi-Eulerian. Since has edges, contains a trail with vertices. Since is a subgraph of , we conclude that contains .

Case 3-2: |U|≤⌊k/2⌋.

From Condition 1, for each vertex , we have and . Let , and consider the subgraph of . Since every vertex in is adjacent to and , is connected. The degree of every vertex in is two, and the degree of and are both . Hence, the number of odd-degree vertices in is zero or two. Therefore, is Eulerian or semi-Eulerian. Since has edges, contains a trail with vertices. Since is a subgraph of , we conclude that contains .

Case 3-3: ⌊k/2⌋<|U|≤k−3.

By the induction hypothesis, in or , there exists a trail with vertices such that every vertex in is an element of . Let with and be the set of vertices used in .

We further distinguish two cases according to the containment of in or .

Case 3-3-1: T is included in G.

Assume that there exists a vertex adjacent to two vertices where . Then, we have a trail

 S′=u1e1u2e2…ui{ui,wx}wx…wy{wy,ui}uiei…ek−2uk−1.

The number of vertices of is at least . Thus, we only need to consider the case where, for every vertex , there exists at most one element of adjacent to in .

Let , and be any three different vertices of . For every vertex , there exists at most one element of adjacent to in . Then, is adjacent to at least two vertices of ,, and in . Therefore, has the following bipartite graph as a subgraph:

• The partite sets of are and ;

• The degree of each vertex in is two.

From Lemma 4, has a trail with edges, i.e., vertices. Since is a subgraph of , also has .

Case 3-3-2: T is included in ¯¯¯¯G.

Let , and be any three different vertices of . First, assume that in there exist three vertices that are adjacent to at least two of the vertices , and . Then, one of the graphs in Figure 5 always appears as a subgraph of . In both cases, there exists a cycle that has a vertex in . Therefore, we have a trail , and the number of vertices of is at least .

Second, assume there are at most two elements of that are adjacent to at least two vertices of , and in . Let and be those two vertices of . Then, there is a subbipartite graph in :

• The partite sets of are and ;

• The degree of each vertex in is two.

From Lemma 4, has a trail with vertices. Let be the endpoints of . We now construct a trail with vertices such that it only consists of the vertices and edges used in , and does not start at . If , then we have . If and is a circuit, then we have since and . If and is not a circuit, then we have since . Therefore, can be constructed.

Since is included in , is also included in . Let be the endpoints of . Then, we have a trail with edges. Hence, is a trail with vertices. ∎

5 Conclusion

From Theorems 1 and 2, we conclude that when and when .

Future work is to find stricter upper and lower bounds. Another challenge is to find upper and lower bounds of for any and .

References

•  R. L. Graham, B. L. Rothschild, J. H. Spencer, Ramsey Theory, Wiley, 1990.
•  L. Gerencsér, A. Gyárfás, On Ramsey-type problems, Annales Universitatis Scientiarum Budapestinensis, Eötvös Sect. Math., 10 (1967): pp. 167–170.
•  M. Katz, J. Reimann, An Introduction to Ramsey Theory: Fast Functions, Infinity, and Metamathematics. AMS, 2018.
•  S. Radziszowski, Small Ramsey numbers, The Electronic Journal of Combinatorics, DS1, Version 16 (2021).