Abstract
An edge labeling of graph with labels in is an injection from to , where is the edge set of , and is a subset of . A graph is called -antimagic if for each subset of with , there is an edge labeling with labels in such that the sums of the labels assigned to edges incident to distinct vertices are different. The main result of this paper is that the Cartesian products of complete graphs (except ) and cycles are -antimagic.
1. Introduction
All graphs considered in this paper are finite, simple, and without isolated vertices. As usual, let denote the set of real numbers. For a graph and a vertex in , , , and denote the vertex set of , the edge set of , and the set of edges incident to in , respectively. In this paper, the following terminologies and notations are used. Let be a graph. When is a subset of with and the function is injective, we say that is an edge labeling of with labels in ; in this case, for any vertex of , we use to denote . If is a subset of with such that for each subset of with , there is an edge labeling of with labels in such that is not equal to for any two distinct vertices , of , then we say that is B-antimagic.
In the literature, a graph is antimagic if is -antimagic. The concept of antimagic graphs was introduced by Hartsfield and Ringel [1] in 1990. They conjectured that every connected graph with at least two edges was antimagic. This conjecture has not been completely solved yet. Some partial results are listed below. The antimagicness for some special types of regular graphs is verified by Cranston [2], Cranston et al. [3], and Liang and Zhu [4]. Then, Chang et al. [5] proved that all regular graphs with degree are antimagic.
Some studies have addressed the antimagicness of Cartesian products. In 2008, Wang and Hsiao [6] introduced new classes of antimagic graphs constructed through Cartesian products, and Wang [7] proved that any Cartesian product of two or more cycles is antimagic. The antimagicness of the Cartesian products of two paths and the Cartesian products of two or more regular graphs are proved in [8, 9] by Cheng. Moreover, Zhang and Sun [10] proved that if a regular graph is antimagic, then for any connected graph , the Cartesian product is antimagic.
Let denote the set of real numbers. A graph is universal antimagic if is -antimagic. Matamala and Zamora [11] proved that paths, cycles, and graphs whose connected components are cycles or paths of odd lengths are universal antimagic in 2020. In this paper, we generalize further and define -antimagic graphs. The methods of labeling on Cartesian products of cycles used in this paper are similar in [7, 8]. In Section 2, we show that wheels, cycles, and complete graphs of order are -antimagic. In Section 3, we show that Cartesian products are -antimagic, where each is a complete graph of order or a cycle.
2. -Antimagic Graphs and Uniformly -Antimagic Graphs
Let be a path on vertices. In [11], it is shown that is -antimagic, but , , are not -antimagic. Shang et al. [12] investigated the antimagicness of star forests. We prove that stars are -antimagic, but not -antimagic.
Remark 1. Let denote the star with edges. Then, is -antimagic, but not -antimagic.
Proof. Let be the star with and .
Let be the arbitrarily given positive numbers. We define an edge labeling of with labels in by for . Then,for , andWe haveHence, is -antimagic.
Let be real numbers with . Let be an arbitrary edge labeling of with labels in . Without loss of generality, is defined by for . We see that . Accordingly, is not -antimagic, which results in not -antimagic.
Let denote the complete graph of order , and the cycle of order . A wheel is the graph obtained by connecting a single vertex to every vertex of the cycle . In this section, we prove that wheels, cycles, and complete graphs of order are -antimagic.
Theorem 1. Every wheel is -antimagic.
Proof. Let be the wheel with and . To prove the theorem, let be the arbitrarily given real numbers. We distinguish two cases: Case 1, , and Case 2, .
Case 1. .
We define an edge labeling of with labels in by , for , , and for (see Figure 1). Then, , for , and . Note thatfor ,Hence,
Case 2. .
We define an edge labeling of with labels in by , for , , and for (see Figure 2). Then, , for , and . Note thatfor ,Hence,This completes the proof.
To prove the results in Section 3, we need the concept of uniformly -antimagic graphs, which is defined below. Let be a graph. Suppose that all the vertices of can be listed as such that for every with , there is an edge labeling of with labels in such that . Then, we say that is uniformly -antimagic and that the sequence of vertices has the uniformly -antimagic property. Note that in this definition, the ordering of the vertices satisfying the property is independent of the choice of the subset of . Obviously, every uniformly -antimagic graph is -antimagic.
Before proving our main result, we first describe uniformly -antimagic property with cycles and complete graphs.
Theorem 2. Every cycle is uniformly -antimagic.
Proof. Let be the cycle with vertex set and edge set . Let be the arbitrarily given real numbers. We define an edge labeling of with labels in by , for , and (see Figure 3).
Then, , for , and . Since , we have . We see that the listing of vertices with the property is independent of the arbitrarily given . Thus, is uniformly -antimagic.
Theorem 3. The complete graph is uniformly -antimagic.
Proof. Let be the complete graph with vertex set and edge set . Let be the arbitrarily given real numbers.
Let be an edge labeling of with labels in such that for , . Hence, .
For , we haveHence, . We see that the listing of vertices with the property is independent of the arbitrarily given . Thus, is uniformly -antimagic.
3. Main Results
Let be a graph and be a subset of with . If is an edge labeling of with labels in and , are nonempty subsets of such that for all , , then we write under . It is easy to see that the relation is transitive (i.e., if , , are nonempty subsets of , and , , then ). The following trivial lemma will be used in the proofs of Theorems 4 and 5.
Lemma 1. Let be an arbitrary graph and be a subset of with . Let be an edge labeling of with labels in . Suppose that , , , are pairwise disjoint nonempty subsets of the edge set with , such that and under . Then,
Proof. Let and be an arbitrary edge in . Since under , we have . Since under and , we haveNote thatandCombining (12)–(14) and , we haveWe need the following notations. Let be a graph, and be a subset of with . If is an edge labeling of with labels in and being a nontrivial connected subgraph of which contains no isolated vertices, then we use to denote the restriction of to with range . Obviously, is an edge labeling of with labels in . Moreover, for a vertex , we use to denote . Recall that is the set of all edges incident to in . Thus, .
Let and be two graphs with and , respectively. The Cartesian product of and , denoted by , is the graph with vertex set such that is adjacent to if either and or and . For the convenience of the following discussions, we will use the following notations in the proofs of Theorems 4 and 5. In the graph , the vertex is represented by . For , we use to denote the subgraph of induced by the vertices ().
Note 1. The graphs , , , …, are isomorphic, and for each , the vertices , , , …, are the corresponding vertices under these isomorphisms.
Also, we use to denote ; i.e., is the set of all edges in . For and , we use to denote the set , i.e., is the set of all edges joining the vertices in and the vertices in . We see that is the disjoint union of () and (, ).
The notations for the vertices , the subgraphs , and the edge sets of will be used in the proofs of Theorems 4 and 5.
Theorem 4. Let be a regular and uniformly -antimagic graph. Then, is also regular and uniformly -antimagic.
Proof. Since both and are regular, it is trivial that is regular. Since is uniformly -antimagic, we assume that is the sequence of vertices of with the uniformly -antimagic property. We see that the edge set is the union of () and ().
Now, we prove that is uniformly -antimagic. Let with be arbitrarily given. Define to be an edge labeling of with labels in by the following three rules: Rule 1. For , . Rule 2. For , and for , (i.e., ). Rule 3. For , and for , (i.e., ).The edge labeling with labels in can have Rule 3 derived from the fact that the sequence of vertices has the uniformly -antimagic property in and the fact stated in Note 1.
Claim 1. For , .
Check of Claim 1. We need to show for .
Let . Note thatBy Rule 3, .
By Rule 2, for , , it impliesThus, , which completes the Check of Claim 1.
Claim 2. For , .
Check of Claim 2. Let . Note thatLet , , , . Thus,By Rule 1, . Since , , , we have and . Also, note , . Thus, by Lemma 1, . Hence,By Rule 1, if , and if , and we see that , . Thus, , which impliesCombining (20) and (21), we obtain . This completes the Check of Claim 2.
From Claims 1 and 2, we obtainWe also see that the order of the vertices , , , …, , , , , …, , , , , …, , , …, , , , , …, satisfying the aforementioned strict inequalities is independent of the chosen with . Thus, is uniformly -antimagic.
It has been proved that Cartesian product of two or more cycles is antimagic [7]. We further propose that is (uniformly) -antimagic where is a regular and uniformly -antimagic graph. In , the labels we use are in each subset of real numbers with and the labels used in [7, 8] are in . Because of the difference in labels, we have to modify the order of labelings that are different from those in [7, 8]. We use some strategies in the construction of labelings.
Theorem 5. Let be a regular and uniformly -antimagic graph. Then, is also regular and uniformly -antimagic.
Proof. Since both and are regular, it is trivial that is regular. Now, we show that is uniformly -antimagic. By Theorem 4, is uniformly -antimagic. Thus, is uniformly -antimagic. Using Theorem 4 twice, we see that is uniformly -antimagic. Thus, is uniformly -antimagic since is isomorphic to . We assume that .
Assume that the cycle has vertex set and the edge set . We use the notations for the vertices, subgraphs, and edge sets of which are defined in Theorem 4, where is now taken to be . We see that the edge set is the union of () and , (), .
Now, we prove that is uniformly -antimagic. Since is uniformly -antimagic, we assume that () is the sequence of vertices of with the uniformly -antimagic property. Let with be arbitrarily given. Define to be an edge labeling of with labels in by the following three rules: Rule 4. Rules of on .(a),(b)for , ,(c) (hence ). Rule 5. For . Rule 6. For , we have .The edge labeling with labels in can have Rule 6 derived from the fact that the sequence of vertices has the uniformly -antimagic property in and the fact stated in Note 1.
Claim 3. For , .
Check of Claim 3.
We need to show for . Note thatBy Rule 6, .
From Rule 5, we obtain that for fixed , ,Thus, . This completes the Check of Claim 3.
Claim 4. For , .
Check of Claim 4. We distinguish five cases: Case 3, ; Case 4, ; Case 5, ; Case 6, ; and Case 7, .
Case 3. .
We need to show that . Let and . Then,Let and . Then,From Rule 4, . Since , we haveSince , , , , we have , . Since is regular, we have . Trivially, . Thus, by Lemma 1,From the aforementioned, we obtain .
Case 4. .
We need to show that . Note thatSince and and are regular with the same degree, we haveSince , we haveSince , we haveThus, we obtain .
Case 5. .
We need to show that . For , we do not need to consider this case. Assume that . Note thatFrom Rule 4(b), we have for . Since and and are regular with the same degree, we haveSince , we haveSince , we haveAccordingly, we obtain .
Case 6. .
We need to show that . Note thatAlso note that . Since , , we haveSince , , we haveFurthermore, , this impliesHence, we obtain .
Case 7. .
We need to show that . Let and . Then,Let and . Then,Note that . From and , , we haveFrom and , , , , we have and . Since is regular, we have . Trivially, . Thus, by Lemma 1,Therefore, we obtain .
These complete the Check of Claim 4.
From Claims 3 and 4, we obtainWe also see that the order of the vertices , , , …, , , , , …, , , , , …, , , …, , , , , …, satisfying the aforementioned strict inequalities is independent of the chosen with . Thus, is uniformly -antimagic. This completes the proof of the theorem.
The following corollaries derive directly from Theorems 4 and 5.
Corollary 1. The graph is uniformly -antimagic, where is regular and uniformly -antimagic, and for , each is a complete graph of order or a cycle.
Corollary 2. The graph is uniformly -antimagic, where each is a complete graph of order or a cycle.
Proof. Each is a complete graph of order or a cycle.
Case 8. Some .
Without loss of generality, assume . Then, is a cycle or a complete graph of order . By Theorems 2 and 3, is uniformly -antimagic. Then, the corollary derives from Corollary 1.
Case 9. for .
Since , by Theorem 2, is uniformly -antimagic. Again, the corollary derives from Corollary 1.
Note that the hypercube is isomorphic to , where each for . The following corollary derives from Corollary 2.
Corollary 3. Hypercube is uniformly -antimagic.
4. Conclusions
In this paper, we propose the notion of -antimagic graph. This is a generalization of -antimagic graph. Every -antimagic graph is -antimagic, and every -antimagic is antimagic. Not all -antimagic graphs (e.g., stars and ) are -antimagic.
In Section 2, we show that wheels, cycles, and complete graphs of order are -antimagic. Let be a complete graph (except ) or a cycle with . We have found that all the vertices of can be listed as such that for every with , there is an edge labeling of with labels in such that . The property we call uniformly -antimagic property is independent of the choice of the subset of . We have found some graphs with uniformly -antimagic property.
We use labelings modified from those in [7, 8] and make them more systematic in this paper. The proofs in this paper provide efficient algorithms for finding edge labelings of Cartesian products of cycles and complete graphs. Our contribution is to quickly find the edge labelings of Cartesian products of cycles and complete graphs through the algorithms we constructed. It has been proved the Cartesian products of , , …, are (uniformly) -antimagic if each is either a complete graph (except ) or a cycle in Section 3.
We construct some classes of uniformly -antimagic graphs through Cartesian products. Some join graphs which are antimagic have been proved in [13, 14]. In [13], they use the way of listing edges in [9] to show that a class of join graphs are antimagic. It makes the method of labelings in this paper more plausible.
We end this paper with the following observation: every -antimagic graph is also -antimagic if the graph is regular. In further studies, we will propose -antimagicness of more regular graphs (e.g., Petersen graph). Also, we will generalize the research results in this paper in the proposals of Cartesian product of some other regular graphs.
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare no conflicts of interest.