Regular graphs are antimagic

In this note we prove - with a slight modification of an argument of Cranston et al. \cite{cranston} - that $k$-regular graphs are antimagic for $k\ge 2$.


Introduction
Throughout the note graphs are assumed to be simple. Given an undirected graph G = (V, E) and a subset of edges F ⊆ E, F (v) denotes the set of edges in F incident to node v ∈ V , and d F (v) := |F (v)| is the degree of v in F . A labeling is an injective function f : E → {1, 2, . . . , |E|}. Given a labeling f and a subset of edges F , let f (F ) = e∈F f (e). A labeling is antimagic if f (E(u)) = f (E(v)) for any pair of different nodes u, v ∈ V . A graph is said to be antimagic if it admits an antimagic labeling.
Hartsfield and Ringel conjectured [4] that all connected graphs on at least 3 nodes are antimagic. The conjecture has been verified for several classes of graphs (see e.g. [3]), but is widely open in general. In [2] Cranston et al. proved that every k-regular graph is antimagic if k ≥ 3 is odd. Note that 1-regular graphs are trivially not antimagic. We have observed that a slight modification of their argument also works for even regular graphs, hence we prove the following.
It is worth mentioning the following conjecture of Liang [5]. Let G = (S, T ; E) be a bipartite graph. A path P = {uv, vw} of length 2 with u, w ∈ S is called an S-link. Conjecture 2. Let G = (S, T ; E) be a bipartite graph such that each node in S has degree at most 4 and each node in T has degree at most 3. Then G has a matching M and a family P of node-disjoint S-links such that every node v ∈ T of degree 3 is incident to an edge in M ∪ ( P ∈P P ).
Liang showed that if the conjecture holds then it implies that every 4-regular graph is antimagic. The starting point of our investigations was proving Conjecture 2. As Theorem 1 provides a more general result, we leave the proof of Conjecture 2 for a forthcoming paper [1].

Proof of Theorem 1
A trail in a graph G = (V, E) is an alternating sequence of nodes and edges v 0 , e 1 , v 1 , . . . , e t , v t such that e i is an edge connecting v i−1 and v i for i = 1, 2, . . . , t, and the edges are all distinct (but there might be repetitions among the nodes). The trail is open if v 0 = v t , and closed otherwise. The length of a trail is the number of edges in it. A closed trail containing every edge of the graph is called an Eulerian trail. It is well known that a graph has an Eulerian trail if and only if it is connected and every node has even degree.
Proof. Note that |T | is even. Arrange the nodes of T into pairs in an arbitrary manner and add a new edge between the members of every pair. Take an Eulerian trail of the resulting graph and delete the new edges to get the |T |/2 open trails.
The main advantage of Lemma 3 is that the edge set of the graph can be partitioned into open trails such that at most one trail starts at every node of V . Indeed, there is a trail starting at v if and only if v has odd degree in G. This is how we see the Helpful Lemma of [2].
Corollary 4 (Helpful Lemma of [2]). Given a bipartite graph G = (U, W ; E) with no isolated nodes in U , E can be partitioned into subsets E σ , T 1 , T 2 , . . . , T l such that d E σ (u) = 1 for every u ∈ U , T i is an open trail for every i = 1, 2, . . . , l, and the endpoints of T i and T j are different for every i = j.
Proof. Take an arbitrary E ⊆ E with the property d E (u) = 1 for every u ∈ U . A component of G − E containing more than one node is called nontrivial. If there exists a nontrivial component of G − E that only contains even degree nodes then let uw 1 ∈ E − E be an edge in this component with u ∈ U and w 1 ∈ W , and let uw 2 ∈ E . Replace uw 2 with uw 1 in E . After this modification, the component of G − E that contains u has an odd degree node, namely w 1 . Iterate this step until every nontrivial component of G − E has some odd degree nodes. Let E σ = E and apply Lemma 3 to get the decomposition of E − E σ into open trails.
In what follows we prove that regular graphs are antimagic: for sake of completeness we include the odd regular case, too. We emphasize the differences from the proof appearing in [2].
Proof of Theorem 1. Note that it suffices to prove the theorem for connected regular graphs. Let G = (V, E) be a connected k-regular graph and let v * ∈ V be an arbitrary node. Denote the set of nodes at distance exactly i from v * by V i and let q denote the largest distance from v * . We denote the edge-set of the antimagic labeling f of G as follows. We reserve the |E q | smallest labels for labeling E q , the next |E σ q | smallest labels for labeling E σ q , the next |E q | smallest labels for labeling E q , the next |E q−1 | smallest labels for labeling E q−1 , etc. There is an important difference here between our approach and that of [2] as we switched the order of labeling E σ i and E i , and we don't yet define the labels, we only reserve the intervals to label the edge sets. Next we prove a claim that tells us how to label the edges in E i .
Claim 5. Assume that we have to label the edges of E i from interval s, s + 1, . . . , (where |E i | = − s + 1), and that we are given a trail decomposition of E i into open trails. We can label E i so that successive labels (in a trail) incident to a node v i ∈ V i have sum at most s + , and successive labels (in a trail) incident to a node v i−1 ∈ V i−1 have sum at least s + .
Proof. Our proof of this claim is essentially the same as the proof in [2]: we merely restate it for selfcontainedness. Let T be the trail decomposition of E i into open trails. Take an arbitrary trail T = u 0 , e 1 , u 1 , . . . , e t , u t of length t from T and consider the following two cases (see Figure 1 for an illustration).
• Case A: If u 0 ∈ V i−1 then label e 1 , . . . , e t by s, , s + 1, − 1, . . . in this order. In this case the sum of 2 successive labels is s + at a node in V i , and it is s + + 1 at a node in V i−1 .
• Case B: If u 0 ∈ V i then label e 1 , . . . , e t by , s, − 1, s + 1, . . . in this order. In this case the sum of 2 successive labels is s + − 1 at a node in V i , and it is s + at a node in V i−1 .
We prove by induction on |T |. The proof is finished by the following cases.
1. If T contains a trail of even length, then let T be such a trail (and again t denotes the length of T ). If the endpoints of T fall in V i−1 then apply Case A. On the other hand, if the endpoints of T fall in V i then apply Case B. In both cases we use t 2 labels from the lower end of the interval, and t 2 labels from the upper end, therefore we can label the edges of the trails in T − T from the (remaining) interval s + t 2 , s + t 2 + 1, . . . , − t 2 , so that the lower bound s + t 2 + − t 2 = s + holds for the sum of two successive labels at every v i−1 ∈ V i−1 , and the same upper bound holds at each node v i ∈ V i .
2. Every trail in T has odd length. If T contains only one trail then label it using either of the two cases above and we are done. Otherwise let T 1 and T 2 be two trails from T , and let t i be the length of T i for both i = 1, 2. Label first the edges of T 1 using Case A (starting at the endpoint of T 1 that lies in V i−1 ). Note that the remaining labels form the interval s + t1+1 2 , . . . , − t1−1 2 . Next label the edges of T 2 using Case B (starting at the endpoint of T 2 that lies in V i ). Note that the sum of successive labels in the trail T 2 becomes s+ t1+1 2 +( − t1−1 2 )−1 = s+ at a node in V i , and it is s+ t1+1 2 +( − t1−1 2 ) = s+ +1 at a node in V i−1 , which is fine for us. Finally, the remaining labels form the interval s+ t1+1 2 + t2−1 2 , . . . , − t1−1 2 − t2+1 2 , therefore we can label the edges of the trails in T − {T 1 , T 2 } from the remaining interval so that the lower bound s + t1+1 2 + t2−1 2 + − t1−1 2 − t2+1 2 = s + holds for the sum of two successive labels at every node of V i−1 , and the same upper bound holds at every node of V i . Now we specify how the labels are determined to make sure f (E(u)) = f (E(v)) for every u = v. We label the edges of every E i arbitrarily from their dedicated intervals. Label the edges of every E i in the manner described by Claim 5. For any node We label the edges in E σ q , E σ q−1 , . . . , E σ 1 as in [2]: if we already labeled E σ q , E σ q−1 , . . . , E σ i+1 then p(v i ) is already determined for every v i ∈ V i . So we order the nodes of V i in an increasing order according to their p-value and assign the label to their σ edge in this order. This ensures that f (E(u)) = f (E(v)) for an arbitrary pair u, v ∈ V i .
We have fully described the labeling procedure. This labeling scheme ensures that f (E(v i )) < f (E(v j )) if v i ∈ V i , v j ∈ V j and i ≥ j + 2 since G is regular and the edges in E(v j ) get larger labels than those in E(v i ). Similarly, f (E(v * )) > f (E(v)) for every v ∈ V − v * for the same reason. It is only left is to show that f (E(v i )) = f (E(v i−1 )) for arbitrary v i ∈ V i , v i−1 ∈ V i−1 and i ≥ 2.