# Cartesian Products of Some Regular Graphs Admitting Antimagic Labeling for Arbitrary Sets of Real Numbers

Authors: Yi-Wu Chang; Shan-Pang Liu
Journal: Journal of Mathematics
 (2021)

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## Abstract

An edge labeling of graphG with labels in A is an injection from EG to A, where EG is the edge set of G, and A is a subset of ℝ. A graph G is called ℝ-antimagic if for each subset A of ℝ with A=EG, there is an edge labeling with labels in A 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 K1) and cycles are ℝ-antimagic.


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## Body

## 1. Introduction

All graphs considered in this paper are finite, simple, and without isolated vertices. As usual, letℝ denote the set of real numbers. For a graph G and a vertex v in G, VG, EG, and EGv denote the vertex set of G, the edge set of G, and the set of edges incident to v in G, respectively. In this paper, the following terminologies and notations are used. Let G be a graph. When A is a subset of ℝ with A=EG and the function f:EG⟶A is injective, we say that f is an edge labeling of G with labels in A; in this case, for any vertex v of G, we use f+v to denote ∑e∈EGvfe. If B is a subset of ℝ with B≥EG such that for each subset A of B with A=EG, there is an edge labeling of G with labels in A such that f+u is not equal to f+v for any two distinct vertices u, v of G, then we say that G is B-antimagic.In the literature, a graphG is antimagic if G is 1,2,…,EG-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 ≥2 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 G is antimagic, then for any connected graph H, the Cartesian product G□H is antimagic.Letℝ+ denote the set of real numbers. A graph G is universal antimagic if G is ℝ+-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 ℝ-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 ≥3 are ℝ-antimagic. In Section 3, we show that Cartesian products G1□G2□⋯□Gnn≥2 are ℝ-antimagic, where each Gi is a complete graph of order ≥2 or a cycle.


## 2.ℝ-Antimagic Graphs and Uniformly ℝ-Antimagic Graphs

LetPn be a path on n vertices. In [11], it is shown that Pnn≥3 is ℝ+-antimagic, but P3, P4, P5 are not ℝ-antimagic. Shang et al. [12] investigated the antimagicness of star forests. We prove that stars are ℝ+-antimagic, but not ℝ-antimagic.Remark 1.
LetSn denote the star with n edges. Then, Snn≥2 is ℝ+-antimagic, but not ℝ-antimagic.Proof.
LetSn be the star with VSn=v1,v2,…,vn∪v and ESn=vvi|i=1,2,…,n.
Letr1<r2<r3<⋯<rn be the arbitrarily given positive numbers. We define an edge labeling f of Sn with labels in r1,r2,r3,…,rn by fvvi=ri for i=1,2,…,n. Then,(1)f+vi=ri<ri+1=f+vi+1,for i=1,2,…,n−1, and(2)f+vn=rn<r1+r2+⋯+rn=f+v.
We have(3)f+v1<f+v2<⋯<f+vn<f+v.
Hence,Sn is ℝ+-antimagic.
Letr1<r2<r3<⋯<rn be real numbers with r1+r2+⋯+rn−1=0. Let f be an arbitrary edge labeling of Sn with labels in r1,r2,r3,…,rn. Without loss of generality, f is defined by fvvi=ri for i=1,2,…,n. We see that f+vn=rn=r1+r2+⋯+rn−1+rn=f+v. Accordingly, Sn is not r1,r2,r3,…,rn-antimagic, which results in Sn not ℝ-antimagic.
LetKn denote the complete graph of order n, and Cn the cycle of order n. A wheel Wnn≥3 is the graph obtained by connecting a single vertex to every vertex of the cycle Cn. In this section, we prove that wheels, cycles, and complete graphs of order ≥3 are ℝ-antimagic.Theorem 1.
Every wheel isℝ-antimagic.Proof.
LetWn be the wheel with VWn=v1,v2,…,vn∪v and EWn=v1v2∪vivi+2|i=1,2,…,n−2∪vn−1vn∪vvi|i=1,2,…,n. To prove the theorem, let r1<r2<r3<⋯<r2n be the arbitrarily given real numbers. We distinguish two cases: Case 1, rn−1+rn<rn+1+rn+2+⋯+r2n−1, and Case 2, rn+1+rn+2+⋯+r2n−1≤rn−1+rn.Case 1.
rn−1+rn<rn+1+rn+2+⋯+r2n−1.
We define an edge labelingf of Wn with labels in r1,r2,r3,…,r2n by fv1v2=r1, fvivi+2=ri+1 for i=1,2,…,n−2, fvn−1vn=rn, and fvvi=rn+i for i=1,2,…,n (see Figure 1). Then, f+v1=r1+r2+rn+1, f+vi=ri−1+ri+1+rn+i for i=2,…,n−1, and f+vn=rn−1+rn+r2n. Note that(4)f+v1=r1+r2+rn+1<r1+r3+rn+2=f+v2,f+vi=ri−1+ri+1+rn+i<ri+ri+2+rn+i+1=f+vi+1,for i=2,…,n−2,(5)f+vn−1=rn−2+rn+r2n−1<rn−1+rn+r2n=f+vn,f+vn=rn−1+rn+r2n<rn+1+rn+2+⋯+r2n−1+r2n=f+v.
Hence,(6)f+v1<f+v2<⋯<f+vn<f+v.Figure 1
Edge labeling ofWn if rn−1+rn<rn+1+rn+2+⋯+r2n−1.Case 2.
rn+1+rn+2+⋯+r2n−1≤rn−1+rn.
We define an edge labelingf of Wn with labels in r1,r2,r3,…,r2n by fv1v2=rn+1, fvivi+2=rn+i+1 for i=1,2,…,n−2, fvn−1vn=r2n, and fvvi=ri for i=1,2,…,n (see Figure 2). Then, f+v1=rn+1+rn+2+r1, f+vi=rn+i−1+rn+i+1+ri for i=2,…,n−1, and f+vn=r2n−1+r2n+rn. Note that(7)f+v1=rn+1+rn+2+r1<rn+1+rn+3+r2=f+v2,f+vi=rn+i−1+rn+i+1+ri<rn+i+rn+i+2+ri+1=f+vi+1,for i=2,…,n−2,(8)f+vn−1=r2n−2+r2n+rn−1<r2n−1+r2n+rn=f+vn,f+v=r1+r2+r3+⋯+rn−1+rn<r1+rn+1+rn+2+⋯+r2n−2+r2n−1≤r1+rn−1+rn=f+v1.
Hence,(9)f+v<f+v1<f+v2<⋯<f+vn.
This completes the proof.
To prove the results in Section3, we need the concept of uniformly ℝ-antimagic graphs, which is defined below. Let G be a graph. Suppose that all the vertices of G can be listed as u1,u2,…,um such that for every A⊆ℝ with A=EG, there is an edge labeling f of G with labels in A such that f+u1<f+u2<⋯<f+um. Then, we say that G is uniformly ℝ-antimagic and that the sequence of vertices u1,u2,…,um has the uniformly ℝ-antimagic property. Note that in this definition, the ordering of the vertices u1,u2,…,um satisfying the property f+u1<f+u2<⋯<f+um is independent of the choice of the subset A of ℝ. Obviously, every uniformly ℝ-antimagic graph is ℝ-antimagic.
Before proving our main result, we first describe uniformlyℝ-antimagic property with cycles and complete graphs.Figure 2
Edge labeling ofWn if rn+1+rn+2+⋯+r2n−1≤rn−1+rn.Theorem 2.
Every cycle is uniformlyℝ-antimagic.Proof.
LetCn be the cycle with vertex set v1,v2,…,vn and edge set v1v2∪vivi+2|i=1,2,…,n−2∪vn−1vn. Let r1<r2<r3<⋯<rn be the arbitrarily given n real numbers. We define an edge labeling f of Cn with labels in r1,r2,…,rn by fv1v2=r1, fvivi+2=ri+1 for i=1,2,…,n−2, and fvn−1vn=rn (see Figure 3).
Then,f+v1=r1+r2, f+vi=ri−1+ri+1 for i=2,…,n−1, and f+vn=rn−1+rn. Since r1+r2<r1+r3<r2+r4<r3+r5<r4+r6<⋯<rn−3+rn−1<rn−2+rn<rn−1+rn, we have f+v1<f+v2<⋯<f+vn. We see that the listing of vertices v1,v2,…,vn with the property f+v1<f+v2<⋯<f+vn is independent of the arbitrarily given r1<r2<r3<⋯<rn. Thus, Cn is uniformly ℝ-antimagic.Figure 3
Edge labeling ofCn.Theorem 3.
The complete graphKnn≥3 is uniformly ℝ-antimagic.Proof.
LetKn be the complete graph with vertex set VKn=v1,v2,…,vn and edge set EKn=vivj|1≤i≤j≤n. Let r1<r2<r3<⋯<rn2 be the arbitrarily given real numbers.
Letf be an edge labeling of Kn with labels in r1,r2,r3,…,rn2 such that for i=1,2,…,n−2, fvivi+1<fvivi+2<fvivi+3<⋯<fvivn<fvi+1vi+2. Hence, fv1v2<fv1v3<⋯<fv1vn<fv2v3<fv2v4<⋯<fv2vn<fv3v4<⋯<fvn−1vn.
For1≤i≤n−1, we have(10)f+vi=∑1≤k<ifvkvi+fvivi+1+∑i+1<k≤nfvivk<∑1≤k<ifvkvi+1+fvivi+1+∑i+1<k≤nfvi+1vk=f+vi+1.
Hence,f+v1<f+v2<⋯<f+vn. We see that the listing of vertices v1,v2,…,vn with the property f+v1<f+v2<⋯<f+vn is independent of the arbitrarily given r1<r2<r3<⋯<rn2. Thus, Kn is uniformly ℝ-antimagic.


## 3. Main Results

LetG be a graph and A be a subset of ℝ with A=EG. If g is an edge labeling of G with labels in A and K, L are nonempty subsets of EG such that gx<gy for all x∈K, y∈L, then we write K≺L under g. It is easy to see that the relation ≺ is transitive (i.e., if K, L, M are nonempty subsets of EG, and K≺L, L≺M, then K≺M). The following trivial lemma will be used in the proofs of Theorems 4 and 5.Lemma 1.
LetG be an arbitrary graph and A be a subset of ℝ with A=EG. Let g be an edge labeling of G with labels in A. Suppose that A1, A2, B1, B2 are pairwise disjoint nonempty subsets of the edge set EG with A1=B1, A2=B2=1 such that A1≺B1∪B2 and A2≺B1 under g. Then,(11)∑e∈A1∪A2ge<∑e∈B1∪B2ge.Proof.
LetA2=a and b be an arbitrary edge in B1. Since A2≺B1 under g, we have ga<gb. Since A1≺B1∪B2 under g and A1=B2∪B1−b, we have(12)∑e∈A1ge<∑e∈B2∪B1−bge.
Note that(13)∑e∈A1∪A2ge=ga+∑e∈A1ge,and(14)∑e∈B1∪B2ge=gb+∑e∈B2∪B1−bge.
Combining (12)–(14) and ga<gb, we have(15)∑e∈A1∪A2ge<∑e∈B1∪B2ge.
We need the following notations. LetG be a graph, and A be a subset of ℝ with A=EG. If f is an edge labeling of G with labels in A and D being a nontrivial connected subgraph of G which contains no isolated vertices, then we use fED to denote the restriction of f to ED with range fED. Obviously, fED is an edge labeling of D with labels in fED. Moreover, for a vertex v∈VD, we use fED+v to denote fED+v. Recall that EDv is the set of all edges incident to v in D. Thus, fED+v=∑e∈EDvfe.
LetG and H be two graphs with VG=u1,u2,…,um and VH=v1,v2,…,vn, respectively. The Cartesian product of G and H, denoted by G□H, is the graph with vertex set VG×VH such that ui,vj is adjacent to uk,vl if either ui=uk and vjvl∈EH or vj=vl and uiuk∈EG. For the convenience of the following discussions, we will use the following notations in the proofs of Theorems 4 and 5. In the graph G□H, the vertex ui,vj∈VG×VH is represented by wi,j. For j=1,2,…,n, we use Gj to denote the subgraph of G□H induced by the vertices wi,j (i=1,2,…,m).Note 1.
The graphsG, G1, G2, …, Gn are isomorphic, and for each ii=1,2,…,m, the vertices ui∈VG, wi,1∈VG1, wi,2∈VG2, …, wi,n∈VGn are the corresponding vertices under these isomorphisms.
Also, we useEj to denote EGj; i.e., Ej is the set of all edges in Gj. For 1≤j<l≤n and vjvl∈EH, we use Ej,l to denote the set wi,jwi,l|i=1,2,…,m, i.e., Ej,l is the set of all edges joining the vertices in Gj and the vertices in Gl. We see that EG□H is the disjoint union of Ej (j=1,2,…,n) and Ej,l (1≤j<l≤n, vjvl∈EH).
The notations for the verticeswi,j, the subgraphs Gj, and the edge sets Ej,Ej,l of G□H will be used in the proofs of Theorems 4 and 5.Theorem 4.
LetG be a regular and uniformly ℝ-antimagic graph. Then, G□Knn≥2 is also regular and uniformly ℝ-antimagic.Proof.
Since bothG and Kn are regular, it is trivial that G□Kn is regular. Since G is uniformly ℝ-antimagic, we assume that u1,u2,…,umm≥3 is the sequence of vertices of G with the uniformly ℝ-antimagic property. We see that the edge set EG□Kn is the union of Ej (j=1,2,…,n) and Ej,l (1≤j<l≤n).
Now, we prove thatG□Knn≥2 is uniformly ℝ-antimagic. Let A⊆ℝ with A=EG□Kn be arbitrarily given. Define g to be an edge labeling of G□Kn with labels in A by the following three rules:Rule 1. Forj=1,2,…,n−1, Ej≺Ej,j+1≺Ej,j+2≺⋯≺Ej,n≺Ej+1.Rule 2. For1≤j<l≤n, and for i=1,2,…,m−1, gwi,jwi,l<gwi+1,jwi+1,l (i.e., gw1,jw1,l<gw2,jw2,l<gw3,jw3,l<⋯<gwm,jwm,l).Rule 3. Forj=1,2,…,n, and for i=1,2,…,m−1, gEj+wi,j<gEj+wi+1,j (i.e., gEj+w1,j<gEj+w2,j<gEj+w3,j<⋯<gEj+wm,j).
The edge labelingg with labels in A can have Rule 3 derived from the fact that the sequence of vertices u1,u2,…,um has the uniformly ℝ-antimagic property in G and the fact stated in Note 1.Claim 1.
Forj=1,2,…,n, g+w1,j<g+w2,j<g+w3,j<⋯<g+wm,j.
Check of Claim1. We need to show g+wi,j<g+wi+1,j for i=1,2,…,m−1.
LetJ=1,2,…,n. Note that(16)g+wi,j=gEj+wi,j+∑l∈J−jgwi,jwi,l,g+wi+1,j=gEj+wi+1,j+∑l∈J−jgwi+1,jwi+1,l.
By Rule 3,gEj+wi,j<gEj+wi+1,j.
By Rule 2, for1≤j<l≤n, gwi,jwi,l<gwi+1,jwi+1,l, it implies(17)∑l∈J−jgwi,jwi,l<∑l∈J−jgwi+1,jwi+1,l.
Thus,g+wi,j<g+wi+1,j, which completes the Check of Claim 1.Claim 2.
Forj=1,2,…,n−1, g+wm,j<g+w1,j+1.
Check of Claim2. Let J=1,2,…,n. Note that(18)g+wm,j=gEj+wm,j+∑k∈J−jgwm,jwm,k=gEj+wm,j+gwm,jwm,j+1+∑k∈J−j,j+1gwm,kwm,j,g+w1,j+1=gEj+1+w1,j+1+∑k∈J−j+1gw1,j+1w1,k=gEj+1+w1,j+1+gw1,jw1,j+1+∑k∈J−j,j+1gw1,kw1,j+1.
LetA1=EGjwm,j⊆Ej, A2=wm,jwm,j+1⊆Ej,j+1, B1=EGj+1w1,j+1⊆Ej+1, B2=w1,jw1,j+1⊆Ej,j+1. Thus,(19)∑e∈A1∪A2ge=gEj+wm,j+gwm,jwm,j+1,∑e∈B1∪B2ge=gEj+1+w1,j+1+gw1,jw1,j+1.
By Rule 1,Ej≺Ej,j+1≺Ej+1. Since A1⊆Ej, B1⊆Ej+1, A2,B2⊆Ej,j+1, we have A1≺B1∪B2 and A2≺B1. Also, note A1=B1, A2=B2=1. Thus, by Lemma 1, ∑e∈A1∪A2ge<∑e∈B1∪B2ge. Hence,(20)gEj+wm,j+gwm,jwm,j+1<gEj+1+w1,j+1+gw1,jw1,j+1.
By Rule 1,Ek,j≺Ek,j+1 if k<j, and Ej,k≺Ej+1,k if k>j+1, and we see that wm,kwm,j∈Ek,j, w1,kw1,j+1∈Ek,j+1. Thus, gwm,kwm,j<gw1,kw1,j+1, which implies(21)∑k∈J−j,j+1gwm,kwm,j<∑k∈J−j,j+1gw1,kw1,j+1.
Combining (20) and (21), we obtain g+wm,j<g+w1,j+1. This completes the Check of Claim 2.
From Claims1 and 2, we obtain(22)g+w1,1<g+w2,1<⋯<g+wm,1<g+w1,2<g+w2,2<⋯<g+wm,2<g+w1,3<g+w2,3<⋯<g+wm,3<⋯<⋯<⋯<g+w1,n<g+w2,n<⋯<g+wm,n.
We also see that the order of the verticesw1,1, w2,1, w3,1, …, wm,1, w1,2, w2,2, w3,2, …, wm,2, w1,3, w2,3, w3,3, …, wm,3, w1,4, …, wm,n−1, w1,n, w2,n, w3,n, …, wm,n satisfying the aforementioned strict inequalities is independent of the chosen A⊆ℝ with A=G□Kn. Thus, G□Knn≥2 is uniformly ℝ-antimagic.
It has been proved that Cartesian product of two or more cycles is antimagic [7]. We further propose that G□Cn is (uniformly) ℝ-antimagic where G is a regular and uniformly ℝ-antimagic graph. In G□Cn, the labels we use are in each subset A of real numbers with A=EG and the labels used in [7, 8] are in 1,2,…,EG. 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.
LetG be a regular and uniformly ℝ-antimagic graph. Then, G□Cn is also regular and uniformly ℝ-antimagic.Proof.
Since bothG and Cn are regular, it is trivial that G□Cn is regular. Now, we show that G□Cn is uniformly ℝ-antimagic. By Theorem 4, G□K3 is uniformly ℝ-antimagic. Thus, G□C3 is uniformly ℝ-antimagic. Using Theorem 4 twice, we see that G□K2□K2 is uniformly ℝ-antimagic. Thus, G□C4 is uniformly ℝ-antimagic since G□K2□K2 is isomorphic to G□C4. We assume that n≥5.
Assume that the cycleCn has vertex set VCn=v1,v2,…,vn and the edge set ECn=v1v2∪vivi+2|i=1,2,…,n−2∪vn−1vn. We use the notations for the vertices, subgraphs, and edge sets of G□H which are defined in Theorem 4, where H is now taken to be Cn. We see that the edge set EG□Cn is the union of Ej (j=1,2,…,n) and E1,2, Ej,j+2 (j=1,2,…,n−2), En−1,n.
Now, we prove thatG□Cn is uniformly ℝ-antimagic. Since G is uniformly ℝ-antimagic, we assume that u1,u2,…,um (m≥3) is the sequence of vertices of G with the uniformly ℝ-antimagic property. Let A⊆ℝ with A=EG□Cn be arbitrarily given. Define g to be an edge labeling of G□Cn with labels in A by the following three rules:Rule 4. Rules of≺ on G□Cn.(a)
E1≺E1,2≺E2,(b)
forj=2,3,…,n−2, Ej≺Ej−1,j+1≺Ej+1,(c)
En−1≺En−2,n≺En−1,n≺En (hence E1≺E1,2≺E2≺E1,3≺E3≺E2,4≺E4≺E3,5≺E5≺⋯≺En−3≺En−4,n−2≺En−2≺En−3,n−1≺En−1≺En−2,n≺En−1,n≺En).Rule 5. Forvjvl∈ECn,gw1,jw1,l<gw2,jw2,l<gw3,jw3,l<⋯<gwm,jwm,l.Rule 6. Forj=1,2,…,n, we have gEj+w1,j<gEj+w2,j<gEj+w3,j<⋯<gEj+wm,j.
The edge labelingg with labels in A can have Rule 6 derived from the fact that the sequence of vertices u1,u2,…,um has the uniformly ℝ-antimagic property in G and the fact stated in Note 1.Claim 3.
Forj=1,2,…,n, g+w1,j<g+w2,j<g+w3,j<⋯<g+wm,j.
Check of Claim3.
We need to showg+wi,j<g+wi+1,j for i=1,2,…,m−1. Note that(23)g+wi,j=gEj+wi,j+∑vjvl∈ECngwi,jwi,l,g+wi+1,j=gEj+wi+1,j+∑vjvl∈ECngwi+1,jwi+1,l.
By Rule 6,gEj+wi,j<gEj+wi+1,j.
From Rule 5, we obtain that for fixedi, i=1,2,…,m−1,(24)∑vjvl∈ECngwi,jwi,l<∑vjvl∈ECngwi+1,jwi+1,l.
Thus,g+wi,j<g+wi+1,j. This completes the Check of Claim 3.Claim 4.
Forj=1,2,…,n−1, g+wm,j<g+w1,j+1.
Check of Claim4. We distinguish five cases: Case 3, j=1; Case 4, j=2; Case 5, j=3,4,…,n−3; Case 6, j=n−2; and Case 7, j=n−1.Case 3.
j=1.
We need to show thatg+wm,1<g+w1,2. Let A1=EG1wm,1 and A2=wm,1wm,2. Then,(25)g+wm,1=gwm,1wm,3+∑e∈A1∪A2ge.
LetB1=EG2w1,2 and B2=w1,1w1,2. Then,(26)g+w1,2=gw1,2w1,4+∑e∈B1∪B2ge.
From Rule 4,E1≺E1,2≺E2≺E1,3≺E2,4. Since E1,3≺E2,4, we have(27)gwm,1wm,3<gw1,2w1,4.
SinceE1≺E1,2≺E2, A1⊆E1, A2,B2⊆E1,2, B1⊆E2, we have A1≺B1∪B2, A2≺B1. Since G is regular, we have A1=B1. Trivially, A2=B2=1. Thus, by Lemma 1,(28)∑e∈A1∪A2ge<∑e∈B1∪B2ge.
From the aforementioned, we obtaing+wm,1<g+w1,2.Case 4.
j=2.
We need to show thatg+wm,2<g+w1,3. Note that(29)g+wm,2=gE2+wm,2+gwm,1wm,2+gwm,2wm,4,g+w1,3=gE3+w1,3+gw1,1w1,3+gw1,3w1,5.
SinceE2≺E3 and G2 and G3 are regular with the same degree, we have(30)gE2+wm,2<gE3+w1,3.
SinceE1,2≺E1,3, we have(31)gwm,1wm,2<gw1,1w1,3.
SinceE2,4≺E3,5, we have(32)gwm,2wm,4<gw1,3w1,5.
Thus, we obtaing+wm,2<g+w1,3.Case 5.
j=3,4,…,n−3.
We need to show thatg+wm,j<g+w1,j+1. For n=5, we do not need to consider this case. Assume that n≥6. Note that(33)g+wm,j=gEj+wm,j+gwm,j−2wm,j+gwm,jwm,j+2,g+w1,j+1=gEj+1+w1,j+1+gw1,j−1w1,j+1+gw1,j+1w1,j+3.
From Rule 4(b), we haveEj≺Ej−1,j+1≺Ej+1≺Ej,j+2 for 2≤j≤n−3. Since Ej≺Ej+1 and Gj and Gj+1 are regular with the same degree, we have(34)gEj+wm,j<gEj+1+w1,j+1.
SinceEj−2,j≺Ej−1,j+1, we have(35)gwm,j−2wm,j<gw1,j−1w1,j+1.
SinceEj,j+2≺Ej+1,j+3, we have(36)gwm,jwm,j+2<gw1,j+1w1,j+3.
Accordingly, we obtaing+wm,j<g+w1,j+1.Case 6.
j=n−2.
We need to show thatg+wm,n−2<g+w1,n−1. Note that(37)g+wm,n−2=gEn−2+wm,n−2+gwm,n−4wm,n−2+gwm,n−2wm,n,g+w1,n−1=gEn−1+w1,n−1+gw1,n−3w1,n−1+gw1,n−1w1,n.
Also note thatEn−4,n−2≺En−2≺En−3,n−1≺En−1. Since wm,n−4wm,n−2∈En−4,n−2, w1,n−3w1,n−1∈En−3,n−1, we have(38)gwm,n−4wm,n−2<gw1,n−3w1,n−1.
SinceEGn−2wm,n−2⊆En−2, EGn−1w1,n−1⊆En−1, we have(39)gEn−2+wm,n−2<gEn−1+w1,n−1.
Furthermore,En−2,n≺En−1,n, this implies(40)gwm,n−2wm,n<gw1,n−1w1,n.
Hence, we obtaing+wm,n−2<g+w1,n−1.Case 7.
j=n−1.
We need to show thatg+wm,n−1<g+w1,n. Let A1=EGn−1wm,n−1 and A2=wm,n−1wm,n. Then,(41)g+wm,n−1=gwm,n−3wm,n−1+∑e∈A1∪A2ge.
LetB1=EGnw1,n and B2=w1,n−1w1,n. Then,(42)g+w1,n=gw1,n−2w1,n+∑e∈B1∪B2ge.
Note thatEn−3,n−1≺En−1≺En−2,n≺En−1,n≺En. From En−3,n−1≺En−2,n and wm,n−3wm,n−1∈En−3,n−1, w1,n−2w1,n∈En−2,n, we have(43)gwm,n−3wm,n−1<gw1,n−2w1,n.
FromEn−1≺En−1,n≺En and A1⊆En−1, A2⊆En−1,n, B1⊆En, B2⊆En−1,n, we have A1≺B1∪B2 and A2≺B1. Since G is regular, we have A1=B1. Trivially, A2=B2=1. Thus, by Lemma 1,(44)∑e∈A1∪A2ge<∑e∈B1∪B2ge.
Therefore, we obtaing+wm,n−1<g+w1,n.
These complete the Check of Claim4.
From Claims3 and 4, we obtain(45)g+w1,1<g+w2,1<⋯<g+wm,1<g+w1,2<g+w2,2<⋯<g+wm,2<g+w1,3<g+w2,3<⋯<g+wm,3<⋯<⋯<⋯<g+w1,n<g+w2,n<⋯<g+wm,n.
We also see that the order of the verticesw1,1, w2,1, w3,1, …, wm,1, w1,2, w2,2, w3,2, …, wm,2, w1,3, w2,3, w3,3, …, wm,3, w1,4, …, wm,n−1, w1,n, w2,n, w3,n, …, wm,n satisfying the aforementioned strict inequalities is independent of the chosen A⊆ℝ with A=EG□Cn. Thus, G□Cn is uniformly ℝ-antimagic. This completes the proof of the theorem.
The following corollaries derive directly from Theorems4 and 5.Corollary 1.
The graphG1□G2□⋯□Gnn≥2 is uniformly ℝ-antimagic, where G1 is regular and uniformly ℝ-antimagic, and for i≥2, each Gi is a complete graph of order ≥2 or a cycle.Corollary 2.
The graphG1□G2□⋯□Gnn≥2 is uniformly ℝ-antimagic, where each Gi is a complete graph of order ≥2 or a cycle.Proof.
EachGi is a complete graph of order ≥2 or a cycle.Case 8.
SomeGi≠K2.
Without loss of generality, assumeG1≠K2. Then, G1 is a cycle or a complete graph of order ≥3. By Theorems 2 and 3, G1 is uniformly ℝ-antimagic. Then, the corollary derives from Corollary 1.Case 9.
Gi=K2 for i=1,2,…,n.
SinceK2□K2≅C4, by Theorem 2, G1□G2 is uniformly ℝ-antimagic. Again, the corollary derives from Corollary 1.
Note that the hypercubeQn is isomorphic to G1□G2□⋯□Gn, where each Gi=K2 for i=1,2,…,n. The following corollary derives from Corollary 2.Corollary 3.
HypercubeQnn≥2 is uniformly ℝ-antimagic.


## 4. Conclusions

In this paper, we propose the notion ofℝ-antimagic graph. This is a generalization of ℝ+-antimagic graph. Every ℝ-antimagic graph is ℝ+-antimagic, and every ℝ+-antimagic is antimagic. Not all ℝ+-antimagic graphs (e.g., stars and Pn,n=3,4,5) are ℝ-antimagic.In Section2, we show that wheels, cycles, and complete graphs of order ≥3 are ℝ-antimagic. Let G be a complete graph (except K1) or a cycle with VG=u1,u2,…,un. We have found that all the vertices of G can be listed as u1,u2,…,un such that for every A⊆ℝ with A=EG, there is an edge labeling f of G with labels in A such that f+u1<f+u2<⋯<f+un. The property we call uniformly ℝ-antimagic property is independent of the choice of the subset A of ℝ. We have found some graphs with uniformly ℝ-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 G1□G2□⋯□Gnn≥2 of G1, G2, …, Gn are (uniformly) ℝ-antimagic if each Gi is either a complete graph (except K1) or a cycle in Section 3.We construct some classes of uniformlyℝ-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: everyℝ+-antimagic graph is also ℝ-antimagic if the graph is regular. In further studies, we will propose ℝ-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.



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