Abstract
A directed Toeplitz graph with vertices , , , is a directed graph whose adjacency matrix is a Toeplitz matrix. In this paper, we investigate the Hamiltonicity in directed Toeplitz graphs with and .
1. Introduction
Let be a finite vertex-labeled graph with vertex set and edge set . A graph is called a subgraph of if the vertex set and edge set of is a subset of the vertex set and edge set of , respectively. If , where , then is called a cycle. A cycle minus one edge is called a path. A subpath is a path making up part of a larger path. A cycle that visits each vertex of a graph is called Hamiltonian, and is then called a Hamiltonian graph. We consider here simple graphs as multiple edges and loops play no role in Hamiltonicity. The adjacency matrix of is the matrix in which if the vertex is adjacent to the vertex in , and otherwise. The main diagonal is zero, i.e., as has no loop.
A square matrix having constant values along all diagonals parallel to the main diagonal is called a Toeplitz matrix. A graph whose adjacency matrix is a Toeplitz matrix of order is called a Toeplitz graph and is denoted by , where and are the label of the diagonal, above and below the main diagonal, respectively, containing ones such that and , see Figure 1. We can calculate the length of an edge by . An edge is called an increasing (decreasing) edge if (). We have both increasing and decreasing edges in of length and , respectively, for some and . Note that if we reverse the direction of all edges of the graph , then the graph is obtained. If the Toeplitz matrix is symmetric, then the corresponding Toeplitz graph is undirected and can be denoted as . Hamiltonicity results obtained in the undirected case for a Toeplitz graph have a direct impact on the directed case. Hamiltonicity of means Hamiltonicity of .
Different properties of Toeplitz graphs have been studied in the literature, for example, colourability, bipartiteness, planarity, cycle discrepancy, metric dimension, decomposition, edge irregularity strength, and labeling. Hamiltonian properties of undirected Toeplitz graphs were first investigated by van Dal et al. in [1], and then these studies have been extended in [2–5]. The Hamiltonicity in directed Toeplitz graphs has also been studied in the literature.
The Hamiltonicity of the directed Toeplitz graphs and were investigated in the literature, which completes the Hamiltonicity investigation in the directed Toeplitz graphs with . In the literature, the Hamiltonicity of the directed Toeplitz graphs was also investigated. In this paper, we first prove a conjecture regarding stated in [6]. Then, we investigate the Hamiltonicity of the directed Toeplitz graphs with , , , and . We also investigate the Hamiltonicity of the directed Toeplitz graphs with , , and , i.e., . Thus, in this paper, we complete the Hamiltonicity investigation of the directed Toeplitz graphs with and .
Suppose that is a Hamiltonian cycle in . The Hamiltonian cycle is determined by two paths (from to ) and (from to ), i.e., . Since the path is Hamiltonian in the subgraph of induced by , the vertices which are not covered by would be covered by .
We denote a path of length , from vertex , as . By path , we mean a path from vertex to vertex .
2. Toeplitz Graphs
In this section, we first prove a conjecture regarding stated in [6]. Then, we investigate the Hamiltonicity of the directed Toeplitz graphs with , , , and .
Theorem 1 (see [7]). is Hamiltonian for .
In Theorem 1, it was proved that is Hamiltonian for , and in [6], it was stated as conjecture that is non-Hamiltonian for . Here, we prove this conjecture which allows us to restate Theorem 1 as follows:
Theorem 2. is Hamiltonian if and only if .
Proof. If , then Theorem 1 asserts that is Hamiltonian.
Conversely, we show that is non-Hamiltonian for . Assume to the contrary, that is Hamiltonian for . Let be a Hamiltonian cycle in . Since the path is Hamiltonian in the subgraph of induced by , the vertices which are not covered by would be covered by . Let , where each is a disjoint set of successive vertices. Since has increasing edges of length , , and only, for each , we have either or . But since we have decreasing edges of length only, , and one can observe that there is no such that , so for all . Since , then we have either or . If , then clearly, by keeping in mind that for all , and that the decreasing edges are of length only, the only possible subpath of is , but then it would be stuck at vertex , see Figure 2, this is a contradiction. If , then as in the previous case, the possible subpath of is but then it would be stuck at vertex , see Figure 3. This is a contradiction.
This completes the proof.
Theorem 3 (see [6]). is Hamiltonian if and only if .
Theorem 4 (see [6]). is Hamiltonian if and only if .
Theorem 5 (see [7]). is Hamiltonian if and only if .
Theorem 6 (see [6]). is Hamiltonian if and only if .
Note that the Hamiltonicity in or implies the Hamiltonicity in . Theorems 2 and 3 show that and are non-Hamiltonian for infinitely many , respectively. Theorems 4, 5, and 6 show that , , and are non-Hamiltonian for only a finite number of , respectively. Here, we study the Hamiltonicity of these cases by adding one more diagonal (containing one) below the main diagonal, say , i.e., . Then, we will see in Theorem 11, is Hamiltonian for all , , and . We use the following some existing results in the proof of Theorem 11.
Theorem 7 (see [7]). For , is Hamiltonian for all .
Theorem 8 (see [6, 7]). For , is Hamiltonian for all .
We have the following theorems in the literature.
Theorem 9. For , is Hamiltonian for all .
Theorem 10. For even , is Hamiltonian if .
Theorem 11. is Hamiltonian for all , , and .
Proof. Case 1. Let or or or .
For or , by Theorem 7, is Hamiltonian for all . And for or , by Theorem 8, is Hamiltonian for all .
Case 2. Let and .
If , then . For , by using Theorem 9, is Hamiltonian for all , because the Hamiltonicity in means the Hamiltonicity in . By using Theorem 5, is Hamiltonian for all different from 7. A Hamiltonian cycle in is . Thus is Hamiltonian for all . By using Theorems 6 and 10, is Hamiltonian for all .
If , then . For , by using Theorem 3, is Hamiltonian if . Let , then the smallest such is . A Hamiltonian cycle in is which contains the edge , see Figure 4.
Let has a Hamiltonian cycle containing the edge , for some nonnegative integer . We transform this Hamiltonian cycle in to a Hamiltonian cycle in by replacing the edge with the path . This shows that enjoys the same property, thus is Hamiltonian for all . By using Theorems 3 and 4, is Hamiltonian for all . By using Theorems 3 and 5, is Hamiltonian for all . By using Theorem 6, is Hamiltonian for all different from . A Hamiltonian cycle in is , see Figure 5.
Thus, is Hamiltonian for all .
If , then . For , by using Theorems 2 and 4, is Hamiltonian for all . Hamiltonian cycles in and are and , respectively, see Figure 6.
Thus, is Hamiltonian for all . By using Theorems 2 and 4, is Hamiltonian for all . By using Theorem 6, is Hamiltonian for all . A Hamiltonian cycle in is , see Figure 7. Thus, is Hamiltonian for all .
If , then . By using Theorems 4 and 5, is Hamiltonian for all . By using Theorems 4 and 6, is Hamiltonian for all .
If , then . By using Theorems 4 and 6, is Hamiltonian for all . This completes the proof.
3. Toeplitz Graphs
In this section, we discuss the Hamiltonicity of Toeplitz graphs still with but and .
We need the following lemma in the proof of Theorem 14.
Lemma 12. For even , is non-Hamiltonian.
Proof. Assume, to the contrary, that has a Hamiltonian cycle . Let , where each is a disjoint set of successive vertices. Since has no increasing edge of length or of length greater than 4, for each , we have either or .
Since in , so . Then, . But there is no path because starting from vertex , it can only use increasing edges of length or (otherwise for some ), but then this will never end up to odd . This is a contradiction.
Now we will discuss the Hamiltonicity of for both even and odd . In Theorem 13, we discuss it for even .
Theorem 13. For even , is Hamiltonian if and only if for , and .
Proof. For even , by using a result in the literature, is Hamiltonian for all odd . Now for even , such that for , and , we prove that is Hamiltonian.
Case 1. If .
(i)Let .Assume . The smallest is . A Hamiltonian cycle in is , which contains the edge , see Figure 8.
Let has a Hamiltonian cycle containing the edge for some nonnegative integer . We transform this Hamiltonian cycle in to a Hamiltonian cycle in by replacing the edge with the path . This shows that enjoys the same property. Thus, for and , is Hamiltonian for all .
Assume and . The smallest , different from , is . Hamiltonian cycles in and are and , respectively, where both cycles contain the edge , see Figure 9.
Let has a Hamiltonian cycle containing the edge for some nonnegative integer . We transform this Hamiltonian cycle in to a Hamiltonian cycle in by replacing the edge with the path . This shows that enjoys the same property. Thus, for and , is Hamiltonian for all .
(ii)Let and .The smallest , different from , is . A Hamiltonian cycle in is , which contains the edge , see Figure 10.
By using the same technique as in the previous subcase, for , is Hamiltonian for all such that .
Case 2. If .
(i)Let and .The smallest , different from , is . A Hamiltonian cycle in is , which contains the edge , see Figure 11.
By using the same technique as in Case (i), for , is Hamiltonian for all such that .
(ii)Let .The smallest , is . A Hamiltonian cycle in is , which contains the edge , see Figure 12.
By using the same technique as in Case 1(i), for , is Hamiltonian for all .
Conversely, we show that if for and if , then is not Hamiltonian, i.e., , , and are non-Hamiltonian.
Claim 1. is non-Hamiltonian.
Assume, to the contrary, that is Hamiltonian, and let be a Hamiltonian cycle in . Since for every vertex in , so . Then . Clearly, or . If , then , but then terminates at vertex , for otherwise vertices and would be missed. If , then either or , but then terminates at vertices and , respectively. This is a contradiction.
Claim 2. is non-Hamiltonian.
Assume, to the contrary, that is Hamiltonian, and let be a Hamiltonian cycle in . Let , where each is a disjoint set of successive vertices. Since has no increasing edge of length or of length greater than . Clearly, for each , we have either or .
Let be the set of all decreasing edges in , i.e., , so . Let be the set of all decreasing edges in , then clearly . Since in , so . But cannot have only these two edges as its decreasing edges, because otherwise there must be a path in , but this is not possible as otherwise, for some , we have either or . Thus .
Case 1. If . Since , then two subcases arise.
(i) . Then . Here, . If , but then terminates at vertex . If , then , but then , for some , in , which is not possible.
(ii) . Then . Here, . If , but then terminates at vertex . If , then , but then , for some , in . If , but then terminates at vertex or , this is a contradiction.
Case 2. If , then , but then we have only two possible paths for .
(i) , then the only possible path for is but then there is no path as otherwise for some .
(ii) , then the only possible path for is but then there is no path as otherwise for some . So this is a contradiction.
By Lemma 12, is non-Hamiltonian. This together with Claim and Claim shows that is non-Hamiltonian if for , and if .
This finishes the proof.
Now, in Theorem 17, we will discuss the Hamiltonicity of for odd and we will be using the following known results of the literature in the proof of Theorem 17.
Theorem 14. is Hamiltonian if and only if or .
Theorem 15. is Hamiltonian if and only if .
Theorem 16. For odd , is Hamiltonian if and only if .
Theorem 17. (1) is Hamiltonian if and only if or .(2) is Hamiltonian if and only if .(3)For odd , is Hamiltonian if .
Proof. (1)If , then by using Theorem 14, is Hamiltonian. If , then a Hamiltonian cycle in is , where , see an example in Figure 13Conversely, suppose is a Hamiltonian cycle in . Since has edges of length , , and , only, uses either all decreasing edges of length , i.e., , or decreasing edges of length 3 along with increasing edges of length (a decreasing edge of length , then an increasing edge of length , and then again an increasing edge of length ), i.e., , where , see Figure 14. This implies that is either a multiple of or a multiple of . Thus or .
(2)First, we show that is Hamiltonian if . By using Theorem 15, is Hamiltonian if . Now we show that is Hamiltonian for . Let be a Hamiltonian cycle in , then we have,
,
,
,
,
, and
, see Figure 15.
Conversely, we show that is non-Hamiltonian for .
Claim 1. is non-Hamiltonian.
Assume, to the contrary, that is Hamiltonian and let be a Hamiltonian cycle in . Since for every vertex in , so . Then is either or , but in both cases, the path would be stuck at vertex 2. This is a contradiction.
Claim 2. is non-Hamiltonian.
Assume, to the contrary, that has a Hamiltonian cycle . Let , where each is a disjoint set of successive vertices. Clearly, for each , we have either or , because has no increasing edge of length or of length greater than .
Let be the set of all decreasing edges in , i.e., , so . Let be the set of all decreasing edges in , then clearly . Since in , so . But cannot have only these two edges as its decreasing edges, because otherwise, there must be a subpath in , but this is not possible here, so . We also observe that , as otherwise, for some . Thus .
Case 1. If . Since , then subcases arise.
(i) . Then, . By keeping in mind that there is no in such that , so here, . But in all of these subpaths, we have , for some , in , which is not possible.
(ii) . Then, . Here, and . But then, for some , (say ) in , which is a contradiction.
(iii) . Then, . Here, and . But then, for some , (say ) in , which is a contradiction.
(iv) . Then, . Here and . But then, for some , (say ) in , which is a contradiction.
(v) . Then . Here, . But then, for some , (say or ) in , which is a contradiction.
Case 2. If . Since , then subcases arise.
(i) . Then, . By keeping in mind that there is no in such that , so , but then the path would be stuck at vertex .
(ii) . Then, . Due to the same reason, here but then would be stuck at vertex .
(iii) . Then, . Here, , but then terminates at vertex .
(iv) . Then, . Here, , but in all of these subpaths, for some , (say or ) in .
(v) . Then, . Here, , but terminates at vertex .
(vi) . Then, . Here, , but terminates at vertex .
(vii) . Then, . Here, , but then would be stuck at vertex .
(viii)–(x) If or or , then clearly for some , say , or , or , respectively. This is a contradiction.
Case 3. If . Since , then subcases arise.
(i) . By keeping in mind that there is no in such that , so . But the subpath would be stuck at vertex or .
(ii) . Here, , but terminates at vertex .
(iii) . Here, , but terminates at vertex .
(iv) . Here, , but would be stuck at vertex .
(v)–(x) If or or or or , or , then clearly , for some , this is a contradiction.
(3)If . By using Theorem 16, for odd , is Hamiltonian if . Now we show that is Hamiltonian for . A Hamiltonian cycle in is , see Figure 16.A Hamiltonian cycle in is , see Figure 17.
A Hamiltonian cycle in is , see Figure 18. And a Hamiltonian cycle in is , see Figure 19.
This finished the proof.
In Theorem 17, we saw that for odd , is Hamiltonian if . In the following theorem, we study the Hamiltonicity of for and odd , with some restriction on .
Theorem 18. For odd , is Hamiltonian if and .
Proof. For odd , let and . We show that is Hamiltonian.
Case 1. If , then a Hamiltonian cycle in is , see an example in Figure 20.
Case 2. If . For , a Hamiltonian cycle in is , see an example in Figure 21.
A Hamiltonian cycle in is , see Figure 22.
This completes the proof.
4. Toeplitz Graphs
Theorem 19 (see [8]). Let . (1)If and are both even, then is Hamiltonian if and only if is odd.(2)If and are of opposite parity, then is Hamiltonian for all .(3)If and are both odd, and(a)if , then is Hamiltonian for all .(b)if , then is Hamiltonian if .In Theorem 19, the Hamiltonicity of have been studied. Now we will see, what happens, if we add one more diagonal (containing one) above the main diagonal, say . So here we discuss the Hamiltonicity of .
Theorem 20. For even and even , is Hamiltonian if and only if is odd. Otherwise it is Hamiltonian for all .
Proof. Case 1. If and are both even, then by using Theorem 19, is Hamiltonian for all odd . Conversely, suppose that has a Hamiltonian cycle . Now, we show that is odd. Assume, to the contrary, that is even. Since cannot use an increasing edge of length or of length greater than , cannot use any increasing edges of the types or . Hence, uses only edges of even length, and therefore vertices of the same parity, and this implies that is odd, because otherwise would never end up to vertex . This is a contradiction; thus, is odd.
Case 2. If and are of opposite parity, then by using Theorem 19, is Hamiltonian for all .
Case 3. If and are both odd and , then by using Theorem 19, is Hamiltonian for all . Now assume , where both and are odd, then by using Theorem 19, is Hamiltonian for .
Now we prove that, for both odd and , is Hamiltonian for . Clearly, here, is an even integer such that and (as if , then ). If , then , and therefore (because and ). A Hamiltonian cycle in is , see Figure 23.
If , then and then . By using Theorem 17, is Hamiltonian for . Since here (because , as ), we need to consider only . The Hamiltonian cycles in and are and , respectively, see Figure 24.
If , then by using Theorem 17, is Hamiltonian for , and by using Theorem 18, is Hamiltonian for and . Now, we need to show that is Hamiltonian for and . Let and . For , a Hamiltonian cycle in is , see Figure 25. And for , a Hamiltonian cycle in is , see Figure 26. Thus, if and are both odd, is Hamiltonian for all .
This completes the proof.
Conjecture: for odd , is non-Hamiltonian if and .
5. Concluding Remark
An affirmative resolution of the conjectures for will complete the study of the Hamiltonicity of the Toeplitz graph with and . The next task in our opinion is to investigate the Hamiltonicity of the Toeplitz graph with and , which will then complete the Hamiltonicity investigation in the Toeplitz graph with . To make this paper not very long, we have not discussed that case here.
Data Availability
No data has been used for producing the result of this paper.
Conflicts of Interest
The author declares that she has no conflicts of interest.