A graph labeling is an assignment of integers to the vertices or edges or both, subject to certain conditions. Varieties of graph labeling have been investigated by many authors [2], [3] [5] and they serve as useful models for broad range of applications

Chandrashekar Adiga,

Shivakumar Swamy C.S.

Shrikanth A. S.

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International J.Math. Combin. Vol.3(2012), 113-117A Note on 1-Edge Balance Index SetChandrashekar Adiga, Shrikanth A. S., Shivakumar Swamy C.S.(Department of Studies in Mathematics, University of Mysore, Manasagangothri, Mysore-570006, India)E-mail: c adiga@hotmail.com, shrikanth.ait@gmail.com, cskswamy@gmail.comAbstract: Let G be a graph with vertex set V and edge set E, and Z2 = {0, 1}. Let f bea labeling from E to Z2, so that the labels of the edges are 0 or 1. The edges labelled 1 arecalled 1-edges and edges labelled 0 are called 0-edges. The edge labeling f induces a vertexlabeling f∗ : V −→ Z2 defined byf∗(v) =1 if the number of 1-edges incident on v is odd,0 if the number of 1-edges incident on v is even.For i ∈ Z2 let ef (i) = e(i) = |{e ∈ E : f(e) = i}| and vf (i) = v(i) = |{v ∈ V : f∗(v) = i}|.A labeling f is said to be edge-friendly if | e(0) − e(1) |≤ 1. The 1- edge balance index set(OEBI) of a graph G is defined by {| vf (0)− vf (1) | : the edge labeling f is edge-friendly}.The main purpose of this paper is to completely determine the 1-edge balance index set ofwheel and Mycielskian graph of a path.Key Words: Mycielskian graph, edge labeling, edge-friendly, 1-edge balance index set,Smarandachely induced vertex labeling, Smarandachely edge-friendly graph.AMS(2010): 05C78§1. IntroductionA graph labeling is an assignment of integers to the vertices or edges or both, subject to certainconditions. Varieties of graph labeling have been investigated by many authors [2], [3] [5] andthey serve as useful models for broad range of applications.Let G be a graph with vertex set V (G) and edge set E(G) and Z2 = {0, 1}. Let f be alabeling from E(G) to Z2, so that the labels of the edges are 0 or 1. The edges labelled 1 arecalled 1-edges and edges labelled 0 are called 0-edges. The edge labeling f induces a vertexlabeling f∗ : V (G) −→ Z2, defined byf∗(v) =1 if the number of 1-edges incident on v is odd,0 if the number of 1-edges incident on v is even.For i ∈ Z2, let ef (i) = e(i) = |{e ∈ E(G) : f(e) = i}| and vf (i) = v(i) = |{v ∈ V (G) :f∗(v) = i}|. Generally, let f : E(G) → Zp be a labeling from E(G) to Zp for an integer1Received June 28, 2011. Accepted September 20, 2012.114 Chandrashekar Adiga, Shrikanth A. S., Shivakumar Swamy C.S.p ≥ 2. A Smarandachely induced vertex labeling on G is defined by fv = (l1, l2, · · · , lp) withnk(v) ≡ lk(modp), where nk(v) is the number of k-edges, i.e., edges labeled with an integer kincident on v. Letek(G) =12∑e∈E(G)nk(v)for an integer 1 ≤ k ≤ p. Then a Smarandachely edge-friendly graph is defined as follows.Definition 1.1 A graph G is said to be Smarandachely edge-friendly if | ek(G)− ek+1(G) |≤ 1for integers 1 ≤ k ≤ p. Particularly, if p = 2, such a Smarandachely edge-friendly graph isabbreviated to an edge-friendly graph.Definition 1.2 The 1-edge balance index set of a graph G, denoted by OEBI(G), is definedas {| vf (1)− vf (0) |: f is edge-friendly}.For convenience, a vertex is called 0-vertex if its induced vertex label is 0 and 1-vertex, ifits induced vertex label is 1.In the mid 20th century there was a question regarding the construction of triangle-freek-chromatic graphs, where k ≤ 3. In this search Mycielski [4] developed an interesting graphtransformation known as the Mycielskian which is defined as follows:Definition 1.3 For a graph G = (V,E), the Mycielskian of G is the graph µ(G) with vertexset consisting of the disjoint union V ∪ V ′ ∪ {v0}, where V ′ = {x′ : x ∈ V } and edge setE ∪ {x′y : xy ∈ E} ∪ {x′v0 : x′ ∈ V ′}.u1u2u3u4un−2un−1unv1v2v3v4vn−2vn−1vnv0Figure 1 Mycielskian graph of the path PnRecently Chandrashekar Adiga et al. [1] have introduced and studied the 1-edge bal-ance index set of several classes of graphs. In Section 2, we completely determine the 1-edge balance index set of the Mycielskian graph of path Pn. In Section 3, we establish thatOEBI(Wn) = {0, 4, 8 . . . , n} if n ≡ 0(mod 4), OEBI(Wn) = {2, 6, 10 . . . , n} if n ≡ 2(mod 4)and OEBI(Wn) = {1, 2, 5 . . . , n} if n is odd.A Note on 1-Edge Balance Index Set 115§2. The 1-Edge Balance Index Set of µ(Pn)In this section we consider the Mycielskian graph of the path Pn (n ≥ 2), which consists of2n+1 vertices and 4n−3 edges. To determine the OEBI(µ(Pn)) we need the following theorem,whose proof is similar to the proof of the Theorem 1 in [6].Theorem 2.1 If the number of vertices in a graph G is even(odd) then the 1-edge balance indexset contains only even(odd)numbers.Now we divide the problem of finding OEBI(µ(Pn)) into two cases, viz,n ≡ 0(mod 2) and n ≡ 1(mod 2),Denoted by max{OEBI(µ(Pn))} the largest number in the 1-edge balance index set ofµ(Pn). Then we get the follpowing result.Theorem 2.2 If n ≡ 0(mod 2) i.e, n = 2k(k ∈ N), then OEBI(µ(Pn)) = {1, 3, 5, . . . , 2n+1}.Proof Let f be an edge-friendly labeling on µ(Pn). Since the graph contains 2n+1 = 4k+1vertices, 4n − 3 = 8k − 3 edges, we have two possibilities: i) e(0) = 4k − 1, e(1) = 4k − 2ii) e(0) = 4k − 2, e(1) = 4k − 1. Now we consider the first case namely e(0) = 4k − 1 ande(1) = 4k−2. Denote the vertices of µ(Pn) as in the Figure 1. Now we label the edges u2q−1v2q,u2q+1v2q for 1 ≤ q ≤ k − 1, uquq+1 for 1 ≤ q ≤ 2k − 3, u2k−2v2k−1, u2kv2k−1 and u2k−1u2k by1 and label the remaining edges by 0. Then it is easy to observe that v(0) = 4k + 1 and thereis no 1-vertex in the graph. Thus | v(1)− v(0) |= 4k + 1 = 2n+ 1 = max{OEBI(µ(Pn))}.Now we interchange the labels of the edges to get the remaining 1-edge balance indexnumbers. By interchanging the labels of edges u2qu2q+1 and u2qv2q+1 for 1 ≤ q ≤ k − 2,we get, | v(0) − v(1) |= 4k + 1 − 4q. Further interchanging u2k−1u2k and u2k−1v2k, we get| v(0)− v(1) |= 5.In the next four steps we interchange two pairs of edges as follows to see that 1, 3, 7, 11 ∈OEBI(µ(Pn))u1v2 and v1v0, u2v3 and v2v0.u3v2 and v3v0, u3v4 and v4v0.u4v5 and v5v0, u5v4 and v6v0.u5v6 and v7v0, u6v7 and v8v0.Now we interchange u2⌊ q−12⌋+7 v2⌈ q−12⌉+6 and v2q+7 v0, u2q+6 v2q+7 and v2q+8 v0 for 1 ≤ q ≤k − 5 to obtain | v(0) − v(1) |= 4q + 11. Finally by interchanging the labels of the edgesu2⌊ k−52⌋+7 v2⌈ k−52⌉+6 and u2k−2 u2k−1 we get | v(0) − v(1) |= 4k − 5 and u2⌊ k−42⌋+7 v2⌈ k−42⌉+6and u2k−1 v0 we get | v(0)− v(1) |= 4k − 1.Proof of the second case follows similarly. ThusOEBI(µ(Pn)) = {1, 3, 5, · · · , 2n+ 1}. Theorem 2.3 If n ≡ 1(mod 2) i.e, n = 2k + 1(k ∈ N), then OEBI(µ(Pn)) = {1, 3, 5, . . . ,2n+ 1}.116 Chandrashekar Adiga, Shrikanth A. S., Shivakumar Swamy C.S.Proof Let f be an edge-friendly labeling on µ(Pn). Since the graph contains 2n+1 = 4k+3vertices, 4n − 3 = 8k + 1 edges, we have two possibilities: i) e(0) = 4k + 1, e(1) = 4kii) e(0) = 4k, e(1) = 4k + 1. Now we consider the first case namely e(0) = 4k + 1 ande(1) = 4k. Denote the vertices of µ(Pn) as in the Figure 1. Now we label the edges u2q−1v2q,u2q+1v2q for 1 ≤ q ≤ k and uquq+1 for 1 ≤ q ≤ 2k by 1 and label the remaining edges by0. Then it is easy to observe that v(0) = 4k + 3 and there is no 1-vertex in the graph. Thus| v(1)− v(0) |= 4k + 3 = 2n+ 1 = max{OEBI(µ(Pn))}.Now we interchange the labels of the edges to get the remaining 1-edge balance indexnumbers. By interchanging the labels of edges u2qu2q+1 and u2qv2q+1 for 1 ≤ q ≤ k we get| v(0)−v(1) |= 4k+3−4q. Further interchanging u2kv2k+1 and v2k+1v0 we get | v(0)−v(1) |= 1.In the next four steps we interchange two pairs of edges as follows to see that 5, 9, 13.17 ∈OEBI(µ(Pn))u1v2 and v1v0, u2v3 and v2v0.u3v2 and v3v0, u3v4 and v4v0.u4v5 and v5v0, u5v4 and v6v0.u5v6 and v7v0, u6v7 and v8v0.And finally by interchanging the labels of edges u2⌊ q−12⌋+7 v2⌈ q−12⌉+6 and v2q+7 v0, u2q+6 v2q+7and v2q+8 v0 for 1 ≤ q ≤ k − 4, we Obtain | v(0)− v(1) |= 4q + 17.Proof of the second case follows similarly. ThusOEBI(µ(Pn)) = {1, 3, 5, . . . , 2n+ 1}. §3. The 1-Edge Balance Index Set of WheelIn this section we consider the wheel, denoted by Wn which consists of n vertices and 2n − 2edges. To determine the OEBI(Wn) we consider four cases, namely,n ≡ 0(mod 4), n ≡ 1(mod 4),n ≡ 2(mod 4), n ≡ 3(mod 4).Theorem 3.1 If n ≡ 0(mod 4) i.e, n = 4k(k ∈ N), then OEBI(Wn) = {0, 4, 8, . . . , n}.Proof Let f be an edge-friendly labeling on Wn. Since the graph contains n = 4k vertices,2n− 2 = 8k − 2 edges, we must have e(0) = e(1) = 4k − 1. Denote the vertices on the rim ofthe wheel by v0, v1, v2, · · · , v4k−1 and denote the center by v0. Now we label the edges vqvq+1for 1 ≤ q ≤ 4k − 2 and v4k−1v1 by 1 and label the remaining edges by 0. Then it is easy toobserve that v(0) = 4k and there is no 1-vertex in the graph. Thus | v(1) − v(0) |= 4k = n =max{OEBI(Wn)}.Now we interchange the labels of the edges to get the remaining 1-edge balance indexnumbers. By interchanging the labels of edges v2q−1v2q and v2q−1v0, v2qv2q+1 and v2qv0 for1 ≤ q ≤ k we get | v(0)− v(1) |= 4k − 4q. Thus 0, 4, 8, · · · , n are elements of OEBI(Wn).A Note on 1-Edge Balance Index Set 117Let ai = card{v ∈ V | number of 1-edges incident on v is equal to i }, i = 1, 2, 3, . . . , 4k−1.Then we have4k−1∑i=1iai = a1 + 2a2 + 3a3+, . . . ,+(4k − 1)a4k−1 = 8k − 2implies that a1 + 3a3 + 5a5+, . . . ,+(4k − 1)a4k−1 is even, which is possible if and only if,a1 + a3 + a5+, . . . ,+a4k−1 is even, that is, the number of 1-vertices is even and hence thenumber of 0-vertices is also even. Therefore, the numbers 2, 6, 10, . . . , n− 2 are not elements ofOEBI(Wn). Theorem 3.2 If n ≡ 1(mod 4) i.e, n = 4k + 1(k ∈ N), then OEBI(Wn) = {1, 3, 5, . . . , n}.Proof Let f be an edge-friendly labeling on Wn. Since the graph contains n = 4k + 1vertices, 2n− 2 = 8k edges, we must have e(0) = e(1) = 4k. Denote the vertices on the rim ofthe wheel by v0, v1, v2, · · · , v4k and denote the center by v0. Now we label the edges vqvq+1 for1 ≤ q ≤ 4k − 1 and v4kv1 by 1 and label the remaining edges by 0. Then it is easy to observethat v(0) = 4k + 1 and there is no 1-vertex in the graph. Thus | v(1) − v(0) |= 4k + 1 = n =max{OEBI(Wn)}.Now we interchange the labels of the edges to get the remaining 1-edge balance indexnumbers. By interchanging the labels of edges v2q−1v2q and v2q−1v0, v2qv2q+1 and v2qv0 for1 ≤ q ≤ 2k − 1, we get | v(0)− v(1) |=| 4k + 1 − 4q | and by interchanging the labels of edgesv4k−1v4k and v4k−1v0, v4kv1 and v4kv0, we get | v(0)− v(1) |= 4k − 1. ThusOEBI(Wn) = {1, 3, 5, . . . , n}. Similarly one can prove the following results.Theorem 3.3 If n ≡ 2(mod 4) i.e, n = 4k + 2(k ∈ N), then OEBI(Wn) = {2, 6, 10, . . . , n}.Theorem 3.4 If n ≡ 3(mod 4) i.e, n = 4k + 3(k ∈ N), then OEBI(Wn) = {1, 3, 5, . . . , n}.References[1] Chandrashekar Adiga, C.K.Subbaraya, Shrikanth A. S., Sriraj M. A., On 1-edge balanceindex set of some graphs, Proc. Jang. Math. Soc., 14(2011), 319-331.[2] L.W.Beineke and S.M.Hegde, Strongly multiplicative graphs, Discuss. Math. Graph The-ory, 21(2001), 63-76.[3] I.Cahit, Cordial graphs: A weaker version of graceful and harmonious graphs, Ars Com-bin., 23(1987), 201-207.[4] J.Mycielski, Sur le coloriage des graphs, Colloq. Math,, 3(1955), 161-162.[5] A.Rosa, On certain valuations of vertices of a graph, Internat. symposium, Rome, July1976, Gordon, New York,Breach and Dunod, Paris 1967.[6] Yuge Zheng and Ying Wang, On the Edge-balance Index sets of Cn×P2, 2010 InternationalConference on Networking and Digital Society, 2(2010)360-363.

A Note on 1-Edge Balance Index Set

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