Antimagic orientation of biregular bipartite graphs

An antimagic labeling of a directed graph $D$ with $n$ vertices and $m$ arcs is a bijection from the set of arcs of $D$ to the integers $\{1, \cdots, m\}$ such that all $n$ oriented vertex sums are pairwise distinct, where an oriented vertex sum is the sum of labels of all arcs entering that vertex minus the sum of labels of all arcs leaving it. An undirected graph $G$ is said to have an antimagic orientation if $G$ has an orientation which admits an antimagic labeling. Hefetz, M{\"{u}}tze, and Schwartz conjectured that every connected undirected graph admits an antimagic orientation. In this paper, we support this conjecture by proving that every biregular bipartite graph admits an antimagic orientation.


Introduction
Unless otherwise stated explicitly, all graphs considered are simple and finite. A labeling of a graph G with m edges is a bijection from E(G) to a set S of m integers, and the vertex sum at a vertex v ∈ V (G) is the sum of labels on the edges incident to v. If there are two vertices have same vertex sums in G, then we call them conflict. A labeling of E(G) with no conflicting vertex is called a vertex distinguishable labeling. A labeling is antimagic if it is vertex distinguishable and S = {1, 2, · · · , m}. A graph is antimagic if it has an antimagic labeling.
Hartsfield and Ringel [8] introduced antimagic labelings in 1990 and conjectured that every connected graph other than K 2 is antimagic. There have been significant progresses

Notation and Lemmas
Let G be a graph. If G is bipartite with partite sets X and Y , we denote G by G[X, Y ].
Given an orientation of G and a labeling on E(G), for a vertex v ∈ V (G) and a subgraph H of G, we use ω H (v) to denote the oriented sum at v in H, which is the sum of labels of all arcs entering v minus the sum of labels of all arcs leaving it in the graph H. If v is of degree 2 in G, we say the labels at edges incident to v the label at v and write it as a pair trail is the number of edges in it. Occasionally, a trail T is also treated as a graph whose vertex set is the set of distinct vertices in T and edge set is the set of edges in T . We use the terminology "trail" without distinguishing if it is a sequence or a graph, but the meaning will be clear from the context. For two integers a, b with a < b, let [a, b] := {a, a + 1, · · · , b}.
We need the result below which guarantees a matching in a bipartite graph. A simple proof of this result can be found in [4]. Lemma 2.1 ([4]). Let H be a bipartite graph with partite sets X and Y . If there is no isolated vertex in X and d H (x) ≥ d H (y) holds for every edge xy with x ∈ X and y ∈ Y , then H has a matching which saturates X.
For even regular graphs, Petersen proved that a 2-factor always exists. Lemma 2.2 ([13]). Every regular (multi)graph with positive even degree has a 2-factor.
Also we need the following result on decomposing edges in a graph into trails. Proof. Note that G is antimagic by Corollary 3 in [5]. Assume that φ : E(G) → [1, |E(G)|] is an antimagic labeling of G. Define another labeling ϕ : E(G) → [a, b] based on φ as follows.
Since G is regular and φ is antimagic, it is clear that ϕ is a vertex distinguishable labeling of G. Furthermore, the sums fall into the interval [2a + 1, 2b − 1].
be an open trail with all vertices in Y having degree 2 except precisely two having degree 1. Suppose T has 2m edges. Let y 1 and y m+1 be the two degree 1 vertices in Y such that T starts at y 1 and ends at y m+1 . Let a, b be two integers with such that each of the following holds.
Proof. Since |E(T )| = 2m, and except precisely two degree 1 vertices, all other vertices in Y have degree 2, we conclude that |Y | = m + 1. Let Y = {y 1 , y 2 , · · · , y m+1 }. Then there are precisely m edges of T incident to vertices in Y with even indices, and m edges of T incident to vertices in Y with odd indices. We treat T as an alternating sequence of vertices and edges starting at y 1 and ending at y m+1 .
If m ≡ 0 (mod 2), following the order of the appearances of edges in T , assign edges incident to vertices in Y of even indices with labels a, a + 1, · · · , a + m − 1, and assign edges incident to vertices in Y of odd indices with labels That is, the label at and not equal to 1 or m + 1.
If m ≡ 1 (mod 2), following the order of the appearances of edges in T , assign edges incident to vertices in Y of odd indices with labels a, a + 1, · · · , a + m − 1, and assign edges incident to vertices in Y of even indices with labels b, b − 1, · · · , b − m + 1.
Let x be a vertex in X. Suppose that one appearance of x is adjacent to y i and y i+1 in the sequence T . If m ≡ 0 (mod 2), for even i with 2 ≤ i ≤ m, the labels on the two edges xy i and xy i+1 contribute a value of (a + i − 1) + (b − (i + 1) + 2) = a + b to ω T (x); for odd i with 1 ≤ i ≤ m − 1, the labels on the two edges xy i and xy i+1 contribute a value of (b − i + 1) + (a + (i + 1) − 2) = a + b to ω T (x). Since x appears d T (x)/2 times in T , . If m ≡ 1 (mod 2), for even i with 2 ≤ i ≤ m, the labels on the two edges xy i and xy i+1 contribute a value of (b − i + 1) + (a + (i + 1) − 2) = a + b to ω T (x); for odd i with 1 ≤ i ≤ m − 1, the labels on the two edges xy i and xy i+1 contribute a value of (a + i − 1) (i) ω C (x) = a + b for any x ∈ X.
for all y ∈ Y , and the sums in Proof. Denote by C = x 1 y 1 x 2 y 2 · · · x m y m x 1 with x i ∈ X and y i ∈ Y . Following the order of the appearances of edges in C, assign edges incident to vertices in {y 1 , y 3 , · · · , y m−1 } with labels a + 1, a, a + 2, a + 3, · · · , a + m − 2, a + m − 1.
Note that the labels are increasing consecutive integers after exchanging the positions of the first two; assign edges incident to vertices in {y 2 , y 4 , · · · , y m } with labels Note that the labels are decreasing consecutive integers after inserting the last number between the first two labels.
For each y i ∈ Y with 1 ≤ i ≤ m, following the appearances of the edges in the sequence of C, we denote the labels on the edges incident to y i by an ordered pair. Particularly, by the assignment of the labels, we have that The sums on y 1 , y 3 , · · · , y m−1 starting at 2a+1 strictly increase to 2a+2m−3 and all of them are odd; and the sums on y 4 , y 6 , · · · , y m−2 starting at 2b − 7 strictly decrease to 2b − 2m + 5 and all of them are odd; the sums on y 2 , y m are even numbers 2b − 2 and 2b − m, respectively. Since This shows both (ii) and (iii). can be decomposed into edge-disjoint p + q open trails T 1 , · · · , T p , T p+1 , · · · , T p+q , and ℓ cycles C p+q+1 , · · · , C p+q+ℓ . Suppose further that these p + q + ℓ subgraphs have no common vertex in Y , and for each of the trail, its vertices in Y are all distinct and its endvertices are contained in Y . Let 2m := |E(H)|, and c, d be two integers with c = d − 2m + 1. If the length of T 1 , · · · , T p are congruent to 2 modulo 4, and the length of each of the remaining trails and cycles is congruent to 0 modulo 4, then there exists a bijection from E(H) to such that each of the following holds.
for any x ∈ X. Then Proof. Since T 1 , · · · , T p+q and C p+q+1 , · · · , C p+q+ℓ are edge-disjoint and pairwise have no common vertex in Y , and for each of the trail, its vertices in Y are all distinct and its endvertices are contained in Y , we conclude that in the graph H, all vertices in X have even degree, all the endvertices of T i are precisely the degree 1 vertices in Y , and all other vertices and apply Lemma 2.6 on each By Lemma 2.5 and Lemma 2.6, we get that for any x ∈ X.
By Lemma 2.5, the sums at y 1i , y m i i , respectively, are and all these sums are distinct and odd.
are all distinct and fall into the intervals and all the sums in Since for each i, j with 1 ≤ i < j ≤ p + q + ℓ, and 2b p+q+ℓ − 2m p+q+ℓ + 5 is the smallest value in the set Furthermore, Hence, all the vertex sums at By Equalities (1a) and (1b), and the assumptions on the parity of each , since each m j is even By the above analysis, for each i with 1 ≤ i ≤ p + q, both ω G (y 1i ) and ω G (y (m i +1)i )

are even. As all the sums at vertices in
are odd, all of them are distinct from these 2(p + q) ω G sums on vertices in {y 1i , y (m i +1)i | 1 ≤ i ≤ p + q}. Hence, to show that all the vertex sums at vertices in Y are distinct, we are left to check that all these 2(p + q) ω G sums are distinct with the sums on vertices in p+q+ℓ , and all these 2(p + q) ω G sums at vertices in Thus, all the sums at vertices either in Furthermore, by (2a) and (2b), Thus, the ω G sums on vertices in {y 1i , y (m i +1)i | 1 ≤ i ≤ p} are all distinct.
By the definition of the parameters b i and easy calculations, By (2a), (2b),(3a), and (3b), the ω G sums at vertices in {y 1i , y (m i +1)i | 1 ≤ i ≤ p + q} fall into the intervals we then conclude that these 2(p + q) ω G sums at vertices in By all the arguments above, we have shown that ω G (y) = ω G (z) for any distinct y, z ∈ Y .
3 Proof of Theorem 1.1 Consequently m ≤ n and |E(G)| = ms = nt. Given an orientation of G, we will denote the orientation by − → G .
Orient edges of G from X to Y . Thus, the oriented vertex sums for vertices in X are negative, and the oriented vertex sums for vertices in Y are positive. Hence, no two vertices x and y conflict if x ∈ X and y ∈ Y . Also, it is routine to check that no two vertices in X conflicting and no two vertices in Y conflicting. Hence the labeling of − → G is antimagic. Thus we assume t ≥ 2. We distinguish three cases for finishing the proof.
Case 1: t ≥ 3 Orient edges of G from X to Y , and denote the orientation by − → G . By the orientation of G, the sums of vertices in X are negative while the sums at vertices in Y are positive.
Hence in the following, we just need to find a labeling of − → G using labels in [1, sm], which guarantees that the sums at vertices in X are all distinct and the sums at vertices in Y are all distinct. By Lemma 2.1, G has a matching M saturating X. Assume, w.l.o.g, that For each y i with 1 ≤ i ≤ m, assign arbitrarily the edges incident to y i with labels and for each y i with m + 1 ≤ i ≤ n, assign arbitrarily the edges incident to y i with labels t(i − m − 1) + tm + 1, t(i − m − 1) + tm + 2, · · · , t(i − m − 1) + tm + t.
Assume, w.l.o.g., that under the above assignment of labels, ω H (x 1 ) ≤ ω H (x 2 ) ≤ · · · ≤ ω H (x m ). Now for each edge x i y i ∈ M, 1 ≤ i ≤ m, assign the edge x i y i with the label i.
We verify now that the labeling of − → G given above is antimagic. For each Next for each y i , y j ∈ {y 1 , y 2 , · · · , y m } with i < j, since ω H (y i ) = ω H (y j ) = t−1 2 ((t + 1)m + 1), we have that ω G (y i ) = ω H (y i ) + i < ω H (y j ) + j = ω G (y j ). By the assignment of labels on edges incident to y i with m + 1 ≤ i ≤ n, the sums at y i are pairwise distinct. The smallest vertex sum among these values is ω G (y m+1 ) = t 2 m + t i=1 i. The largest vertex sum among values in {ω G (y 1 ), · · · , ω G (y m )} is ω G (y m ) = t−1 2 ((t + 1)m + 1) + m. It is easy to check that ω G (y m ) < ω G (y m+1 ). Hence, all the sums at vertices in Y are distinct.

Subcase 1.2: t ≥ 3 and t is even
Reserve labels in {2, 4, · · · , 2m} for edges in M, and use the labels in {1, 3, · · · , 2m − 1} ∪ {2m + 1, · · · , tn = sm} for edges in H. For each y i with 1 ≤ i ≤ m, assign arbitrarily the edges incident to y i with labels and for each y i with m + 1 ≤ i ≤ n, assign arbitrarily the edges incident to y i with labels Assume, w.l.o.g., that under the above assignment of labels, ω H (x 1 ) ≤ ω H (x 2 ) ≤ · · · ≤ ω H (x m ). Now for each edge x i y i ∈ M, 1 ≤ i ≤ m, assign the edge x i y i with the label 2i.
We verify now that the labeling of − → G given above is antimagic. Obviously, for each Next for each y i , y j with we have that ω G (y i ) = ω H (y i ) + 2i < ω H (y j ) + 2j = ω G (y j ). By the assignment of labels on edges incident to y i with m + 1 ≤ i ≤ n, the sums at y i are pairwise distinct. The It is easy to check that Hence, all the sums at vertices in Y are distinct. Consequently, T i has even length. For each i with 1 ≤ i ≤ m/2, let where x 1i , x 2i , · · · , x m i i ∈ X and y 1i , y 2i , · · · , y m i +1i ∈ Y . Note that x 1i , x 2i , · · · , x m i i maynot be distinct vertices in X, but y 1i , y 2i , · · · , y m i +1i are distinct vertices in Y because all vertices in Y have degree 2 in G. Assume further, w.l.o.g., that there are p trails T 1 , · · · , T p of length congruent to 2 modulo 4, and q trails T p+1 , · · · , T p+q of length congruent to 0 modulo 4. Set Thus, Applying Lemma 2.7 on H with c = 2p + 1 and d = sm − 2q defined as above, we get an assignment of labels on E(H) such that Orient all the edges of G from X to Y , and denote the orientation by − → G .
Claim 1: The labeling of − → G given above is antimagic.
Proof. We first show that all the set of labels used is the set [1, sm]. We then show that all the oriented sums on vertices in − → G are pairwise distinct. We first examine the sums on vertices in Y . For any y ∈ Y , since Y = V (H) ∩ Y , we know that all the sums on vertices in Y are pairwise distinct by (ii) preceding Claim 1.
Next we show that in − → G , the oriented sums on vertices in X are all distinct. According to (i) preceding Claim 1 and the orientation of G, we see that for any x ∈ X, Since all the labels on the edges in M are distinct, and for any x ∈ X, ω G (x) = ω H (x) − the label on e ∈ M which is incident to x, we know that the oriented sums on vertices in X are all distinct.
Finally, we show that for any x ∈ X and y ∈ Y the oriented sums at x and y are distinct in G. This is clear since all the oriented sums at vertices in Y are positive while that at vertices in X are negative.
Case 3: t = 2 and s is even We may assume that s ≥ 4. Otherwise G is 2-regular and |E(G)| = 2m. By Lemma 2.4, G has an antimatic labeling by taking a := 1 and b := 2m, and the labeling is also an antimagic labeling of − → G obtained by orienting all edges from X to Y .

Claim 2:
The graph G[X, Y ] contains a subgraph F such that (1) F is a set of vertex disjoint cycles; and Proof. Suppressing all degree 2 vertices in Y , we obtain an s-regular (multi)graph G ′ . Since s is even, by applying Lemma 2.2, we find a 2-factor of G ′ . Subdivide each edge in the 2-factor of G ′ , we get the desired graph F .
the i-th cycle of length congruent to 2 modulo 4.
We pre-label edges in  label on e = Assume that there are q paths with positive length after deleting the vertices x 1i , x 2i , x m i i , Assume, w.l.o.g., that these paths are obtained from C h−q+1 , · · · , C h . For each 1 ≤ i ≤ q, denote these paths by and assume that P i starts at y 2(h−q+i) and ends at y (m h−q+i −1)(h−q+i) . Then each P i has length 2(m h−q+i − 3) ≡ 0(mod 4). Under this assumption, we know that C 1 , C 2 , · · · , C h−q are 6-cycles.

It is clear that
Then by Equations (6e) and (6f), the labels on the other edges not in H incident to y 2j and y (m j −1)j , respectively, are If s ≥ 6, then let c := 4h + 1 and d := (s − 2)m − 2h.
Then by Equations (7e) and (7f), the labels on the other edges not in H incident to y 2j and y (m j −1)j , respectively, are Thus, Apply If s = 4, orient the edges in {x 1i y 1i , x 1i y m i i | 1 ≤ i ≤ h} from Y to X, and orient all the remaining edges from X to Y . If s ≥ 6, orient the edges in {x 1i y 1i | 1 ≤ i ≤ h} from Y to X, and orient all the remaining edges from X to Y . Denote the orientation of G by − → G .

Claim 3:
The labeling of − → G given above is antimagic.
Proof. We first show that the set of labels used is the set [1, sm]. The labels used on edges in F are exactly numbers in the set [(s − 2)m + 1, sm]. If s = 4, then the set of labels used  The union of these sets is the set [1, sm].
We then show that the oriented sums on vertices in − → G are all distinct. We separate the proof according to if s = 4 or s ≥ 6.

Case s = 4:
For each i with 1 ≤ i ≤ h, by (6a)-(6d) and the orientation of G, the labels at y 1i , y m i i , respectively, are Thus, All these 2h values are pairwise distinct and fall into the interval [−2m + h + 2, 2h − 1].
For each i with 1 ≤ i ≤ h − q, by (6e) and (6f), the label at y 2i is, Thus, All these h − q values are pairwise distinct. are pairwise disjoint, we see that the oriented sums on vertices in Y are all distinct.
Next we show that in − → G , the oriented sums on vertices in X are all distinct. For each i Hence, For the graph F , by Lemma 2.4, the sums on vertices in V (F ) ∩ X are pairwise distinct.
Since the set of labels used on E(F ) is [2m + 1, 4m] and F is 2-regular, it follows that in and any two of the sums at vertices in V (F ) ∩ X differ an absolute value of at most 4m − 4. Because of ω G (x) = ω G−E(F ) (x) + ω F (x) for x ∈ X and the fact in (8), we conclude that the total oriented vertex sums at vertices in X are all distinct.
Finally, we show that for any x ∈ X and y ∈ Y the oriented vertex sums at x and y are distinct in − → G . By the analysis above, ω G (y) ∈ [−2m + h + 2, 8m − 1] for any y ∈ Y .
All these 2h values are pairwise distinct and fall into the interval [2h + 2, 6h].
For the graph F , by Lemma 2.4, the sums on vertices in V (F ) ∩ X are pairwise distinct, any two of the sums at vertices in V (F ) ∩ X differ an absolute value of at most 4m − 4. Because of ω G (x) = ω G−E(F ) (x) + ω F (x) for x ∈ X and the fact in (9), we conclude that the total oriented sums at vertices in X are all distinct.
Finally, we show that for any x ∈ X and y ∈ Y the oriented sums at x and y are distinct in − → G . By the analysis above, for any y ∈ Y , ω G (y) ∈ [2h + 2, 2sm − 1] is a positive integer.
The proof of Theorem 1.1 is now complete.