A CONSTRUCTION OF BALANCED DEGREE-MAGIC GRAPHS

unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract. A graph G is called degree-magic if it admits a labelling of the edges by integers 1,2, ..., |E(G)| such that the sum of the labels of the edges incident with any vertex v is equal to (1+ |E(G)|)deg(v)/2. Degree-magic graphs extend supermagic regular graphs. In this paper, a new construction of balanced degree-magic graphs is introduced.


INTRODUCTION
The finite simple graphs and multigraphs without loops and isolated vertices are considered.
If G is a graph, then V (G) and E(G) stand for the vertex set and the edge set of G, respectively.
Cardinalities of these sets are called the order and the size of G. For any integers p and q, the set of all integers z satisfying p ≤ z ≤ q is indicated by [p, q]. A magic labelling f of G is called a supermagic labelling if the set { f (e) : e ∈ E(G)} consists of consecutive positive integers. A graph G is said to be supermagic (magic) whenever there exists a supermagic (magic) labelling of G.
A bijective mapping f from E(G) into [1, |E(G)|] is called a degree-magic labelling (or only d-magic labelling) of a graph G if its index-mapping f * satisfies A graph G is said to be degree-magic (balanced degree-magic) (or only d-magic) when a d- The concept of magic graphs was put forward by Sedláček [10]. Later, supermagic graphs were introduced by Stewart [11]. Besides, a new constructuion of supermagic complements of some graphs was recommended [9]. Moreover, the notion of degree-magic graphs was then suggested by Bezegová and Ivančo [1] as an extension of supermagic regular graphs. Recently, numerous papers are published on degree-magic and supermagic graphs, see [2,3,4,5,6,7,8] for more complehensive references.
Let one recall the basic properties of d-magic graphs that will be used in the next.

BALANCED DEGREE-MAGIC GRAPHS
An injective mapping f from E(G) into the set of positive integers is called a singleconsecutive labelling (SC-labelling) of a graph G if its index-mapping f * satisfies Let f i , i ∈ {1, 2}, be a SC-labelling of a graph G i . The labellings f 1 and f 2 are called The complementary labellings f 1 and f 2 are called balanced if all Now, one is able to prove the following Proposition.
Proposition 2.1. Let H 1 and H 2 be spanning subgraphs of a graph G which form its decom- · · · < f * (v n ) and let g be a SC-labelling of H 2 such that g * (v 1 ) > g * (v 2 ) > · · · > g * (v n ). If f and g are complementary, then G is a supermagic graph.
Proof. Since f is a SC-labelling of consider a mapping ϕ from E(G) into the set of positive integers defined by and the labellings f and g are complementary, ϕ is a supermagic labelling of G. Therefore, G is a desired graph.
If the graph G in Proposition 2.1 is regular, then G is d-magic by Therorem 1.1. For balanced d-magic graphs, one can show the following assertion.
Proposition 2.2. Let H 1 and H 2 be spanning subgraphs of a regular graph G which form its If f and g are (balanced) complementary, then G is a (balanced) d-magic graph.
Proof. By using the same proof as Proposition 2.1, G is a supermagic graph. Because G is regular, G is d-magic by Theorem 1.1. Since f and g are balanced complementary, for each Thus, ϕ is a balanced d-magic labelling of G. That is, G is an expected graph.
The above two Propositions describe methods to construct supermagic graphs and d-magic graphs by using SC-labellings respectively. In order to use Proposition 2.2, one needs reasonable SC-labellings of some graphs.
and a SC-labelling g of G such that g(E(G)) = {h, h + . Moreover, if k = 1 and h = 3, then the SC-labellings f and g are balanced complementary.
Proof. Consider a mapping f from E(G) into the set of positive integers given by It is easy to see that f (E(G)) = {k, k + 1, k + 4, k + 6} and Hence, f is a desired SC-labelling of G. Moreover, consider a mapping g from E(G) into the set of positive integers defined by One can see that g(E(G)) = {h, h + 1, h + 3, h + 5} and Thus, g is a required SC-labelling of G. Now, consider the case k = 1 and h = 3, one then has Clearly, f and g are balanced complementary labellings.
Proof. Consider a mapping f from E(G) into the set of positive integers given by One is able to check that f (E(G)) = [k, k + n − 1] and f * (v 1 ) < f * (v 2 ) < · · · < f * (v n ). Thus, f is a desired SC-labelling of G. Besides, consider a mapping g from E(G) into the set of positive integers defined by One can get that g(E(G)) = [h, h + n − 1] and g * (v 1 ) > g * (v 2 ) > · · · > g * (v n ). Hence, g is a required SC-labelling of G. Now, consider the case k = 1 and h = n + 1, one then gets 2n : e = v 2 v 1 .
Evidently, f and g are balanced complementary labellings.
In the next results, one is able to prove some sufficient conditions for balanced d-magic graphs.
Theorem 2.5. Let G be a graph which can be decomposable into two spanning cycle subgraphs of order 4. Then G is a balanced d-magic graph.
Proof. Suppose that two spanning cycle subgraphs of G have vertices v 1 , v 2 , v 3 , v 4 . Thus, by Lemma 2.3, there are two balanced complementary SC-labellings f , g of these cycles such that Theorem 2.8. Let G be a graph which can be decomposable into two spanning cycle subgraphs of odd order n ≥ 3. Then G is a balanced d-magic graph.
Proof. Assume that two spanning cycle subgraphs of G of odd order n ≥ 3 have vertices v 1 , v 2 , ..., v n . Hence by Lemma 2.4, there are two balanced complementary SC-labellings f , g of these cycles such that f * (v 1 ) < f * (v 2 ) < · · · < f * (v n ) and g * (v 1 ) > g * (v 2 ) > · · · > g * (v n ). It is clear that these two cycles are regular and they form its decomposition, so G is a regular graph.
Therefore, according to Proposition 2.2, G is a balanced d-magic graph.
Combining Theorem 1.2 and Theorem 2.8, one suddenly has Corollary 2.9. For any positive integer k, if a graph G can be decomposable into 2k spanning cycle subgraphs of odd order n ≥ 3, then G is a balanced d-magic graph.
Joining Theorem 1.1 and Corollary 2.9, one certainly has f One can prove that Furthermore, consider the case k = 1 and h = 5, one then 13 : e = v 7 v 8 ,