Perfect Matching and Zero-Sum 3-Magic Labeling

: A mapping l : E ( G ) → A , where A is an abelian group written additively, is called an edge labeling of the graph G . For every positive integer h ⩾ 2, a graph G is said to be zero-sum h -magic if there is an edge labeling l from E ( G ) to Z h \{ 0 } such that s ( v ) = (cid:80) uv ∈ E ( G ) l ( uv ) = 0 for every vertex v ∈ V ( G ). In 2014, Saieed Akbari, Farhad Rahmati and Sanaz Zare proved that if r ( r (cid:44) 5) is odd and G is a 2-edge connected r -regular graph, G admits a zero-sum 3-magic labeling, and they also conjectured that every 5-regular graph admits a zero-sum 3-magic. In this paper, we first prove that every 5-regular graph with every edge contained in a triangle must have a perfect matching, and then we denote the edge set of the perfect maching by EM , and we make a labeling l : E ( EM ) → 2, and E ( E ( G ) − EM ) → 1. Thus we can easily see this labeling is a zero-sum 3-magic, confirming the above conjecture with a moderate condition.


Introduction and Basic Definitions
Graphs G considered here are all connected, finite and undirected with multiple-edges without loops.Finding a matching in a bipartite graph can be treated as a network flow.The study of matchings in a bipartite graph can be traced back to the early of the 20th century.In 1935, P. Hall proved a well-known result (Hall's Theorem) which provided a necessary and sufficient condition for an X, Y-bipartite graph to have a matching that saturates X [1, 3.1.11 Theorem].When |X| = |Y|, Hall's Theorem is the Marriage Theorem, proved originally by Frobenius in 1917 [2].Later in 1947, Tutte [3] investigated the problem concerning the existence of a perfect matching in a general graph and provided a necessary and sufficient condition (Tutte's Condition) for such a graph to have a perfect matching [ Lemma 1].If every vertex in a graph has the same degree r then this graph is referred to as a r-regular graph.In case of regular graphs, Peterson [4] proved that every 3-regular graph without a bridge (an edge whose deletion disconnects the graph) has a perfect matching.In 1981, Naddef and Pulleyblank considered K-regular graphs with specified edge connectivity and showed some classical theorems and some new results concerning the existence of matchings can be proved by using polyhedral characterization of Edmonds [5].It is not hard to see that a 5-regular 4-edge connected graph has a perfect matching (see [5] for detail).However very little is known about the existence of a perfect matching in a general 5-regular graph.In this paper we provide a sufficient condition for a 5-regular graph to have a perfect matching and our main result is as follows.
Theorem 1.Every 5-regular graph in which each edge is contained in a triangle has a perfect matching.
Our motivation for this work is towards resolving a conjecture of Saieed Akbari, Farhad Rahmati and Sanaz Zare [6].Before discussing this conjecture we require some more notation.A mapping l : E(G) → A, where A is an abelian group written additively, is called an edge labeling of the graph G. Given an edge labeling l of the graph G, the symbol s(v), which represents the sum of the labels of edges incident with v, is defined to be s(v) = uv∈E(G) l(uv), where v ∈ V(G).For every positive integer h ⩾ 2, a graph G is said to be zero-sum h-magic if there is an edge labeling from E(G) into Z h \{0} such that s(v) = 0 for every vertex v ∈ V(G).The null set of a graph G, denoted by N(G), is the set of all natural numbers h ∈ N such that G admits a zero-sum h-magic labeling.
Recently, Choi, Georges and Mauro [7] proved that if G is a 3-regular graph, then N(G) is N\{2} or N\{2, 4}.Saieed Akbari, Farhad Rahmati and Sanaz Zare [6] extend this result by showing that if G is an r-regular graph, then for even r (r > 2), N(G) = N and for odd r (r 5), N\{2, 4} ⊆ N(G).Moreover, they proved that if r (r 5) is odd and G is a 2-edge connected r-regular graph, then N(G) = N\{2}, implying G admits a zero-sum 3-magic labeling.Thus they proposed the following conjecture.
As a consequence of our main result, we obtain the following result, partially confirming Conjecture 1.
Corollary 1.Every 5-regular graph in which each edge is contained in a triangle admits a zero-sum 3-magic labeling.
By Theorem 1, every 5-regular graph in which each edge is contained in a triangle has a perfect matching.We denote the edge set of the perfect maching by EM, and we make a labeling l : E(EM) → 2, and E(E(G) − EM) → 1.It is easily to see this labeling is a zero-sum 3-magic.Thus every 5-regular graph in which each edge is contained in a triangle admits a zero-sum 3-magic labeling, proving Corollary 1.
We next introduce some basic definitions which will be used in the proof of our main result.If X and Y are disjoint vertex sets of G with x ∈ X and y ∈ Y, an edge{x, y} is usually written as xy (or yx), and such an edge xy is also called an X − Y edge.The set of all X − Y edges in the edge set E is denoted by E(X, Y).We denoted by G(X, Y) the subgraph of G with vertex set X Y and the edge set E(X, Y).An odd (or even) component of a graph is a connected component of odd (or even) number of vertices, and an odd (or even) vertex is a vertex with odd (or even) degree.
Definition 1.For a graph G, for any vertex set S of G, and any odd component Q of G − S , consider a connected component of G(S , V(Q)) which has odd number of edges in E(S , V(Q)).The vertices of this component induce a subgraph P in G, and we call it an odd component induced subgraph.
Definition 2. Let S be any vertex set and v ∈ S .The vertex v is called removable from S if q(G −S ′ ) = q(G − S ) − 1, where q(G − S ) denotes the number of odd components of G − S and S ′ = S − v. Definition 3. Let S be any vertex set, and let v, w ∈ S .The vertex pair (v, w) is called removable from S if q(G − S ′ ) = q(G − S ) − 2, where S ′ = S − v − w.Definition 4. Let S be any vertex set, Q be one of the odd components of G − S , we call Q the first type of odd component (with respect to S ) if there is an odd component induced subgraph P, such that there exists a removable vertex or a removable vertex pair in S ∩ V(P).Otherwise, Q is called the second type of odd component (with respect to S ).
Definition 5. Let S be any vertex set, Q be an odd component of G − S and P be an odd component induced subgraph.A connected subgraph K of P is called a 2-claw substructure, if the following conditions hold: 1. V(K) ∩ S has 3 vertices {v, v 1 , v 2 }, and each vertex is connected to exactly 3 odd components of G − S .2. The degree of v in P is 3.In this case the vertices {v, v 1 , v 2 } are called roots of K.

The Proof of Theorem 1
We begin with several important lemmas.The first one is well known as Tutte's Theorem [3] and is essential in the present paper.
Lemma 1. (Tutte's Theorem) A graph G has a perfect matching if and only if q(G − S ) ⩽ |S | for all S ⊆ V(G) , where q(G − S ) denotes the number of odd components of G − S .
If a graph G has no perfect matching, then there must exist S ⊆ V(G) with q(G − S ) > |S |.Such a set S is called an anti f actor set of G.
From now on we always assume that G is a 5-regular graph in which each edge is contained in a triangle.
Lemma 2. For any vertex set S , every odd component of G − S has at least one odd component induced subgraph.
Proof.For any vertex set S , and any odd component Q of G − S , we assume that |E(S , V(Q))| = d, and the number of vertices in Q is p.Since Q is an odd component, we know that p is odd.Let q be the sum of the degrees of each vertex contained in Q. Clearly q is even by the handshaking lemma.Since G is a 5 -regular graph, we have q = 5p − d.
Thus d is odd as q is even and p is odd.That is, the sum of the edges of all connected portions of G(S , V(Q)) is odd, so G(S , V(Q)) has at least one connected subgraph having an odd number of edges.The vertices of this connected subgraph induce a subgraph in G which is the desired odd component induced subgraph.□ Lemma 3.For every vertex set S and every odd component induced subgraph P, there exists at least one vertex v ∈ S ∩ V(P), such that d P (v) ⩾ 3.
Proof.Assume that for every vertex v in S ∩ V(P), we have d P (v) ⩽ 2. Since v has an edge in P and this edge is also in a triangle, the fact that the triangle is a connected graph implys that it is in P, so we must have d P (v) 1, and thus d P (v) = 2 for every vertex v in S ∩ V(P).Suppose that there are k vertices in S ∩ V(P), and s edges in E(S ) ∩ E(P).Then |E(S , V(Q)) ∩ E(P)| = 2k − 2s, which is even, a contradiction.Thus the lemma holds.□ Lemma 4. If a vertex v in S is connected to d odd components (where d ⩽ 2), then v is a removable vertex .
Proof.If v ∈ S is connected to only one odd component Q, then we remove v out of S , and denote the new vertex set by S 1 = S − v. Since Q is an odd component, after adding a vertex v to Q, it becomes an even component.Note that there is no change in parity for all other odd components.Thus q(G − S 1 ) = q(G − S ) − 1, so v is a removable vertex.If v is connected to 2 odd components Q i and Q j , then we also remove v out of S , and denote the new vertex set by S 1 = S − v. Since v is connected to Q j and Q i , at this time, Q i , Q j and v become a new connected component, and the number of its vertices is |Q i | + |Q j | + 1 which is odd.Note that the number of odd components is reduced by one, that is, q(G − S 1 ) = q(G − S ) − 1, so v is a removable vertex.□ We remark that Lemma 4 does not hold when d ⩾ 3.However, we may obtain a similar result for a vertex pair when d = 3.
Lemma 5.If a vertex pair (v, w) in S is connected to exactly 3 odd components, then (v, w) is a removable vertex pair.
Proof.Assume that the vertex pair (v 1 , v 2 ) is connected to 3 odd components Q i , Q j and Q k .We move (v 1 , v 2 ) out of S , and denote the new vertex set by S 1 .Clearly |S 1 | = |S | − 2. As before, there is no change in parity for all odd components except Q i , Q j and Q k .Since the vertex pair (v 1 , v 2 ) is connected to Q i , Q j and Q k , the components Q i , Q j and Q k and the vertex pair (v 1 , v 2 ) become a new connected component, and the number of its vertices So the number of odd components is reduced by two, and we have q(G − S 1 ) = q(G − S ) − 2. Thus the vertex pair (v 1 , v 2 ) is a removable vertex pair.□ Lemma 6.Let Q be a second type of odd component in G − S and P j be the jth odd component induced subgraph of Q.Then for all j, P j has a 2-claw substructure.
Proof.According to Lemma 2, Q has an odd component induced subgraph P j .Let v be a vertex of V(P j ) ∩ S such that d P j (v) is maximum among all vertices in V(P j ) ∩ S .By Lemma 3, we have 3 ⩽ d P j (v) ⩽ 5.
We now divide the rest of the proof into 3 cases.
In this case, since G is a 5-regular graph, all the edges connected to v are in P j , so there is only one odd component adjacent to v. By Lemma 4, v is a removable vertex, and thus Q is a first type of odd component, yielding a contradiction.
In this case, since G is a 5-regular graph, only one of the edges of v is not in P j , so there are at most 2 odd components adjacent to v. By Lemma 4, v is a removable vertex, and thus Q is a first type of odd component, yielding a contradiction.
Since d p j (v) = 3 and G is 5-regular, we know that v is connected to at most two other odd components.We may always assume that v is connected to two other odd components Q s and Q t .For otherwise, by Lemma 4, v is removable, and thus Q is the first type, yielding a contradiction.Subcase 3.1.v has 0 edge in E(S ) ∩ E(P j ).
In this subcase, as mentioned before, v is connected to two other odd components Q t and Q s .We assume that e Q t (v) is connected to Q t , where e Q t (v) denotes the neighbors of v in Q t , and e Q s (v) is connected to Q s .Let us consider e Q t (v).By the assumption of G, e Q t (v) must be an edge of a triangle, so there must exist another edge e t connected to v in this triangle.If e t = e Q s (v), then we deduce that Q t is connected to Q s , yielding a contradiction, because Q t and Q s are different odd components.The same argument shows that e t E(P j ), so e t is the 6th edge of v, yielding a contradiction.Subcase 3.2.v has 1 edge in E(S ) ∩ E(P j ).
In this subcase, v has 2 edges in E(V(Q), S ).Let v 1 , v 2 ∈ Q be the 2 vertices, which are adjacent to v.And v 3 ∈ S be the vertex which is adjacent to v. Let v 4 ∈ Q t and v 5 ∈ Q s , which are adjacent to v.
Since the edge e(v 4 , v) is in a triangle, we may let v q be such that vv 4 v q is a triangle.Thus v q is a neighbor of v, and so v q ∈ {v i |i = 1, 2, 3, 4, 5}.As proved before v q {v i |i = 1, 2, 4, 5} (as Q, Q t , Q s are different odd components, so any two of Q, Q t , Q s are not connected), so v q = v 3 .For the same reason, v 3 is connected to v 5 in Q t .Since v 3 ∈ V(P j ), there exists a vertex v 6 ∈ Q adjacent to v 3 .If (v, v 3 ) is connected to 3 different components, by Lemma 5, (v, v 3 ) is a removable vertex pair, yielding a contradiction.
Otherwise, the last neighbor v p of v 3 is in another odd component.Since the edge e(v 3 , v p ) is in a triangle, v 3 must have another neighbor v q which is adjacency to v p .As before v q {v i |i = p, 4, 5, 6}.Thus v q = v, which implies that v has 6 neighbors, yielding a contradiction.Subcase 3.3.v has 2 edge in E(S ) ∩ E(P j ).

Figure 3 .
Figure 3. d P j (v) = 3 and v has 0 edge in E(S ) ∩ E(P j )

Figure 4 .
Figure 4. d P j (v) = 3 and v has 1 edge in E(S ) ∩ E(P j )