Abstract
Shiu and Kwong (2008) studied the full friendly index set of , which only addressed the cases where or 1. In this paper, we significantly extend their work by determining the full index set for all values of . Our key approach is to utilize graph embedding and recursion methods to deduce for general . In particular, we embed small graphs like and into and apply recursive techniques to prove the main results. This work expands the scope of previous graph labeling studies and provides new insights into determining the full index set of product graphs. Given the broad range of applications for labeled graphs, this research can potentially impact fields like coding theory, communication network design, and more.
1. Introduction
A graph labeling is an assignment of integers to the vertices or edges, or both. Motivated by different settings and conditions, the problems of labeling various types of graphs were raised and studied. We refer the readers to the survey paper, A dynamic survey of graph labelings by Gallian [1], in which the sheer number of research papers regarding different graph labeling methods in graph theory has been reviewed.
Let be a simple connected graph with the vertex set and the edge set . Let be a function from to . For each edge , we assign the label , where . For , a vertex is called an vertex if . And we let denote the number of vertices in under labeling . The concepts of edge and can be defined similarly. Chartrand et al. [2] introduced the concept of friendly index set. And a vertex labeling is said to be friendly if .
Definition 1. The friendly index set of is defined as
Readers can refer to the literature [3, 4] for the friendly index set of graphs if interested. Shiu and Ling [5] extended this concept to the full friendly index set.
Definition 2. The set is called the full friendly index set of .
A friendly labeling of a graph is also known as a bisection of , which has been studied widely in the theory of graph partitions [6, 7]. According to the algorithm [8], it is NP-hard to find the maximum or minimum bisection (namely, friendly labeling with maximum or minimum 1 edge) of an arbitrary graph. There were many interesting results focusing on some specific graphs. For example, Sinha and Kaur [9] studied the full friendly index sets of , and . Shiu and his coauthors determined the full friendly index sets of cylinder graphs [10], some permutation Petersen graphs [11], slender and flat cylinder graphs [12], and [13]. Gao et al. [14] deduced of a family of cubic graphs, which are full vertices blow-up of with .
In this paper, we generalize the concept of the full friendly index set to the following, which we call the full index set.
Definition 3. The full index set of is defined as the set
According to the above definition, when or , the full index set of a graph is its full friendly index set.
Lemma 4. Let . Then, the full index set of is defined as
Proof. Note that the conditions and are symmetric, we find that is equal under both conditions. Then, . Then, we have . According to the definition of , we know that . □
Thus, we only need to compute the value of .
2. Preliminaries
We now discuss the full index set of for some specific values of . We name the vertices on the two paths as and , and Figure 1 represent the graph . We shall call the square the th square, the edge the th vertical edge.
We omit the subscripts in and for simplicity without causing confusion. In the remaining of the paper, if not specifically stated, only the vertices labeled 1 are listed and those not listed are labeled 0. For , we denote as .
Consider the graph . Let be any vertex labeling such that , since . Thus, we know . Note that is a positive integer, which implies that is an even integer.
First, we give some examples of in order to prove the following theorem.
Example 1. Consider the graph , where .(1)For (2)For (3)For , (4)For , For Example 1, we only give the specific notation of (4).
The labeled graph of Example 1 (4) is shown in the following, where the solid points represent vertices labeled 1, the hollow points represent vertices labeled 0, the thick line represents 1 edge, and the thin line represents 0 edge.
Thus, from Figure 1, it is clearly to see that .
Example 2. Consider the graph , where .(1)For , (2)For ,
Example 3. Consider the graph , where .(1)For
Example 4. Consider the graph , where .(1)For (2)For
Example 5. Consider the graph , where .(1)For (2)For (3)For (4)For (5)For
Example 6. Consider the graph , where .(1)For (2)For
Example 7. Consider the graph , where .(1)For (2)For (3)For (4)For (5)For Next, we introduce some relevant definitions.
Definition 5. Let be an integer, if under the vertex labeling of the graph , we denote the labeled graph as . For convenience, we use and to denote the labeled subgraphs and , respectively. Define as a square of , where .
Definition 6. . Given the labeled graph of , we can replace the two edges and with paths and of length 2 and joint and . This process is called a -embedding, which results in a new labeled graph .
Definition 7. . Given the labeled graph of , we can replace the two edges and with paths and of length 3 and jointing and . This process is called a -embedding, which results in a new labeled graph .
In the following discussion, all -embeddings or -embeddings on are embeddings for edges and of , where .
For the convenience of description, some substructures are given below.
Let denote a labeled -square, where . And let represent the number of 1 edge changes after the square is embedded. Write the different labels of as , , , and .
We make a -embedding in , denoted by , where . represents the number of 1 edge changes after embedding.
Thus, obtained after embedding is shown in Table 1.
From the results of Table 1, the Lemmas 8 and 9 are obtained.
Lemma 8. Given , there must exist some value such that .
Lemma 9. Given , there must exist some value such that .
Lemma 10. Let be the maximum value in , then we have
Proof. Let be any vertex labeling of such that . From , we obtain . Additionally, the maximum degree of the vertex of is 3 and , so .
Lemma 11. When and , we have .
Proof. Let be any vertex labeling of such that . Then, and , which implies that . Then, . Thus, we can know does not hold. When is the only possibility, note that a 0 vertex can lead to the number of 1 edge at most 3, and , so . □
Given Lemma 11, we shall always assume in the following discussion that in , when is an integer.
Lemmas 12 and 14 are some of the early results of Shiu and Kwong [13] on the cartesian product .
Lemma 12 (see [13]). Let be a labeling of a graph that contains a subgraph cycle . If contains at least a 1 edge, then the number of 1 edges in must be a positive even number.
Corollary 13. For , there exists no labeling of such that .
Proof. Let be any vertex labeling of such that . Since , . From Lemma 12, it is easy to get . The corollary follows immediately.
Lemma 14 (see [13]). We have
3. The Full Index Sets of
In this section, we give the full index set of by the embedding method and recursive method.
Theorem 15. In holds only when , and in all other cases, .
Proof. Suppose there exists a labeling of with , when and . In accordance with Lemma 12, the three 1 edges must occur in two adjacent squares, say the th and the squares, and at least one of the three 1 edges is vertical. All possibilities can be divided into the following three cases: Case 1. There is only one vertical 1 edge, which must be . If and are the other 1 edges, then and all other vertices are labeled 1. At this point, contradicts . If and are the other 1 edges, then for and for . At this point, contradicts . Neither of the subcases can exist. The other two cases are similar. Case 2. There are two vertical 1 edges, both of which belong to the or square. Due to symmetry, we may assume the vertical 1 edge is and and the other 1 edge is , and then and are labeled 0 and all other vertices are labeled 1. At this point, holds. The other two cases are similar. Case 3. All three vertical edges are 1 edges. This situation does not exist because .This completes the proof.
Theorem 16. In holds only when , and in all other cases, .
Proof. Suppose there exists a labeling of with , when . In accordance to Lemma 12, the two 1 edges must occur in a square of . Case 1. The two 1 edges are in a square . In this case, there must be . Then, there must be , for . At this point, contradicts . Case 2. Both 1 edges are either in square or . We assume that the two 1 edge are in square . Then, there must be or . In either of the two cases, we have . The other case is similar.This completes the proof.
Theorem 17. For ,.
Proof. From Lemma 14, we know . Under any friendly label of , there exists at least one square belonging type , , , or , in which can be embedded, where . That is, three of the embedded vertices are labeled 1 and one is labeled 0. Thus, we observe that in after making a -embedding in . By Lemma 8, we have .
Under some friendly label of , when the number of 1 edge is 2 or 5, there exist at least one of the squares , , and , where can be embedded. is obtained from Lemma 9. It follows from Corollary 13 that 0, 1, . Since , we find by Theorem 16. From Lemma 10, we know the maximum value of is . Combined with (2), (3), and (4) in Example 1, we conclude that .
Theorem 18. For ,.
Proof. From Theorem 17, we know . Under any label that satisfies in , there is at least one of the squares in which , , , , , and can be embedded, where . That is, 3 of the embedded 4 vertices are labeled with 1 and 1 with 0. Thus, in ; by Lemma 8, we have .
Under a label that satisfies in , when the number of 1 edge is 3 or 4, there must be a square , , and , where can be embedded. is obtained from Lemma 9. Since , we find by Theorem 15. Corollary 13 implies that . According to Lemma 10, we have . Combined with (1) and (2) in Example 2, we conclude that .
Theorem 19. For ,.
Proof. From Theorem 18, we know . Under any label that satisfies in , there is at least one of the squares , , , , , and can be embedded, where . That is, 3 out of the embedded 4 vertices are labeled with 1 and the other vertex is labeled with 0. By Lemma 8, we have .
Under a label that satisfies in , when the number of 1 edge is 2 or 5, there must be a square in , , and , embedded in . is obtained from Lemma 9. Since , we find by Theorem 16. Corollary 13 implies that . According to Lemma 10, we have . Combined with (7) in Example 3, we conclude that .
Theorem 20. For , we haveFor , we have
Proof. Case 1. . We will prove by induction. By Theorem 18, we have . Suppose for , in , then . Under any label that satisfies in , there is at least one of the squares , , , , , and can be embedded in this square, where . That is, 3 out of the 4 vertices of the embedding are labeled 1 and the other vertex is labeled 0. Thus, we observe that in after the square is embedded. By Lemma 8, we have . Under a label that satisfies in , when the number of 1 edge is 2 or 5, there must be a square , , , and can be embedded. is obtained from Lemma 9. For , let , and for , let . This implies that . Thus, we know that holds. Since , we find by Theorem 16. By Corollary 13, we have . By Lemma 10, we find the maximum value of is . Therefore, we have . In , . Under any labeling in which satisfies , there exists at least one of the squares , , , , and in which can be embedded, where . That is, 3 out of the 4 vertices of the embedding are labeled 1 and the other vertex is labeled 0. Thus, by Lemma 8, holds when in . Under any labeling in satisfying and when the number of its 1 edges is 3 or 4, there must exist a square similar as , , , and can be embedded. By Lemma 9, we have . In , for , let , and for , let . Clearly, . Thus, we know that holds. Since , we find by Theorem 15. By Corollary 13, we observe that . According to Lemma 10, the maximum value of is . Therefore, we have . To conclude, when , we have Case 2.. Similarly, we can obtain at .This completes the proof.
Theorem 21. For , .
Proof. According to Lemma 10, we have . Corollary 13 implies that . Combined with Example 5, we know .
Theorem 22.
Proof. According to Lemma 10, we have . Corollary 13 implies that . Combined with Example 4, we know . □
Therefore, according to Lemma 14 and Theorems 15–17 and 20–22, we get the full index set of as follows:(1)(2)For , (3)For , (4)For , (5)For , (6)For , (7)
4. The Full Index Sets of
In a similar way, we embed the labeled graph in to obtain the full index set of .
Lemma 23. For , we have .
Proof. Suppose that . In , let for , , then . Therefore, we have .
Lemma 24. For , we have .
Proof. Suppose that . In , let for , then . Therefore, we have .
Lemma 25. For , we have .
Proof. For , suppose that . In , let when ; when . Then, . Similarity, let when ; when . Then, . Therefore, we have .
For .
In , let when ; when . Then, . Let when ; when . Then, . Thus, .
Lemma 26. Given , there must exist such that .
Proof. Let , we make a -embedding in , where . Then, . Therefore, we have . □
In , there exist a vertex labeling such that . From , we get . Thus, we shall always assume that in the following discussion.
Theorem 27. In holds only when , and in all other cases, .
Proof. Suppose there exists a labeling of with , when . In accordance to Lemma 12, the two 1 edges must occur in a square of . Case 1. Both 1 edges are in a square . In this case, there must be . Then, there must be , for . At this point, contradicts . Case 2. Both 1 edges are either in square or . We assume that both 1 edges are in square . Then, there must be or . In either of the two cases, we have . The other case is similar.This completes the proof.
Theorem 28. In holds only when , and in all other cases, .
Proof. Suppose there exists a labeling of with , when . In accordance with Lemma 12, the three 1 edges must occur in two adjacent squares, say the th and the squares, and at least one of the three 1 edges is vertical. All possibilities can be divided into the following three cases. Case 1. There is only one vertical 1 edge, which must be . If and are the other 1 edges, then and all other vertices are labeled 1. At this point, contradicts . If and are the other 1 edges, then for and for . At this point, contradicts . Neither of the subcases can exist. The other two cases are similar. Case 2. There are two vertical 1 edges, both of which belong to the or square. Due to symmetry, we may assume the vertical 1 edge are and , and the other 1 edge is , and then and are labeled 0 and all other vertices labeled 1. At this point, holds. The other two cases are similar. Case 3. All three vertical edges are 1 edge. This situation does not exist because .This completes the proof.
Theorem 29. The followings hold true:(1)For , we have (2)For (3)For (4)(5)
Proof. For (1), according to Theorem 17, we know . Under any friendly label of , there exist at least one square belonging type , or , in which can be embedded. That is, one of the embedded vertices is labeled 1 and other one is labeled 0. Thus, we note that does not change after is embedded. From Lemma 26, we know . By Lemma 24, we find . Suppose that , , then we have . Let , , then we have . From Corollary 13, we find . Since , we find by Theorem 28. Using Lemma 10, we know . Thus, we conclude that . For (2), from Theorem 20, we have . Under any friendly label of , there exist at least one square belonging type , or , in which can be embedded. That is, one of the embedded vertices is labeled 1 and other one is labeled 0. Thus, by Lemma 26, we get . It follows from Lemmas 23 and 25 that . Since , we find by Theorem 27. In accordance with Lemma 10, we find . Thus, we conclude that . For (3), from Theorem 20, we have . Under any friendly label of , there exists at least one square belonging type , or , in which can be embedded. That is, one of the embedded vertices is labeled 1 and other one is labeled 0. From Lemma 26, we get . It follows from Lemmas 23 and 25 that . Since , we find by Theorem 28. Thus, combined with Lemma 10, we conclude that . For (4), according to Lemma 10, we have . Corollary 13 implies that . Combined with Example 7, we know . For (5), according to Lemma 10, we have . Corollary 13 implies that . Combined with Example 6, we know . □This completes the proof.
5. Conclusions
In this paper, we obtained the full index set of by embedding and recursion methods. We can also use this method to consider the full index set of other graphs.
The characterization of full index sets for various graph families lays the mathematical groundwork for a wide range of applications involving labeled graphs. Labeled graphs have been applied across diverse fields including coding theory, circuit layout, network design, and more. By expanding theoretical knowledge on balanced labelings and index sets of key graph classes like , this work provides fundamental insights that can inform labeled graph models in any application domain. Though the specific results focus on index sets, the techniques like embedding and recursion have broad implications for constructing balanced graph partitions. Overall, this research on graph labelings and index sets furthers a mathematical foundation that enables diverse real-world applications. The methods and labeled graph constructions can be extended to other graph families, complementing existing literature and providing a springboard for future studies. By elucidating balanced labelings of modular graph units, the work broadly enhances our ability to design and analyze application-oriented labeled graph models.
Data Availability
No underlying data were collected or produced in this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.