A Note on Incidence graphs

Electronic Notes in Discrete Mathematics 33 (2009) 87–93 www.elsevier.com/locate/endm

A Note on Incidence graphs N.Venkataraman Lecturer,Department Of Mathematics, R.V.S. College of Arts and science, Coimbatore

R.Sundareswaran & V.Swaminathan 1 JRF,Ramanujan Research Center in Mathematics, Saraswathi Narayanan College,Madurai. Reader(Retd.),Ramanujan Research Center in Mathematics, Saraswathi Narayanan College,Madurai.

Abstract An incidence graph of a given graph G, denoted by I(G) , has its vertex set V (I(G)) = {(ve) : v ∈ V (G), e ∈ E(G) and v is incident to e in G} such that the pair (ue)(vf ) of vertices (ue), (vf ) ∈ V (I(G)) is an edge of I(G) if and only if there exists at least one case of u = v, e = f, uv = e or uv = f . Study of Incidence graphs was made by Zhagn Zhong-fu et.al in [3]. The origin of Incidence graphs can be traced to a paper titled Incidence and Strong edge colorings of graphs by R.A.Brualdi,et.al [7] . In this paper, we make a study of domination and related parameters in Incidence graphs. Keywords: Incidence graphs, Domination , Independence number

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Email: sulanesri @ yahoo.com

1571-0653/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2009.03.013

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Introduction

From a given graph, there are several ways of constructing new graphs. Total graphs,Semi-total graphs, Incidence graphs,etc. are some of them. Incidence were first defined and incidence chromatic number and Strong edge colorings of graphs were studied by R.A.Brualdi et al[7]. Later, Zhagn Zhong-fu et.al in [3] formly defined incidence graphs and found out basic results of these graphs. In this note, domination and related parameters in the case of incidence graphs are studied. Definition 1.1 An incidence graph of a given graph G, denoted by I(G) , has its vertex set V (I(G)) = {(ve) : v ∈ V (G), e ∈ E(G) and v is incident to e in G} such that the pair (ue)(vf ) of vertices (ue), (vf ) ∈ V (I(G)) is an edge of I(G) if and only if there exists at least one case of u = v, e = f, uv = e or uv = f . Example 1.2 ue

w

u

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G:

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w

2

wvf

w

f

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I(G): ug w

e

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ve

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wg

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Domination number and Independence number of Incidence graphs

Theorem 2.1 Let G be a graph. Then γ(I(G)) ≤



pnc(v).

v∈V (G)

Proof. Let D be a dominating set of G. Any isolate of G will belong to D. Let the non-isolates of G in D be not isolated in < D >. Therefore, pn(v, D) = φ for all v ∈ D. Let T = {ueu : u ∈ D, u not an isolate of G and eu joins u with a private neigbbour of u}. Claim: T is a dominating set for I(G). Let vfv ∈ V (I(G)) − T .If v = u (or) fv = eu for ueu ∈ T , then vfv is dominated by ueu . Suppose v = u and fv = eu . Suppose v ∈ D. Then vfv ∈ T . Then vfv dominates ueu . Suppose v ∈ / D. Then there existsw ∈ D such that v is the private neighbour of w. Therefore, wgw ∈ T , where gw = vw .

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Then vfv is dominated by wgw . Therefore, T is a dominating set of I(G).  pnc(v) . 2 Therefore, γ(I(G)) ≤ |T | ≤ v∈V (G)

Theorem 2.2 Let G be a graph without isolates. If there exists a γ-set D of G such that pnc(v) = 1, v ∈ D and < D > has no isolates, then γ(I(G)) = γ(G). Proof. Let D be a minimum dominating set of G such that pnc(v) = 1, for all v ∈ D. Then for every v ∈ D, there exists uv ∈ V − D such that N (vv ) ∩ D = {v} and vv is unique. Let T = {ueu : eu = uuv ; u ∈ D}. By earlier theoerm, T is a dominating set of I(G). Therefore, γ(I(G) ≤ |T | = |D| = γ(G). Let T1 be a γ-set of I(G). Let D1 = {u : ueu ∈ T1 }. Let vfv ∈ V (I(G) − T1 . Claim: D1 is a dominating set of G. Let v ∈ V (G) − D1 . Since v is not an isolate, there exists vfv ∈ V (I(G)). Since T1 is a γ-set of I(G), there exists wgw ∈ T1 such that wgw dominates vfv . Therefore, w = v (or) fv = gw (or) vw = gw (or) fv . Since fv = gw , fv = gw = vw. Therefore, v and w aer adjacent. Therefore, w dominates v. Suppose vw = fv or gw . Then also v and w are adjacent. Therefore, w dominates v. Thus, in any case D1 dominates v. Therefore, D1 is a domninating set of G. Therefore, γ(G) ≤ |D1 | ≤ |T1 | = γ(I(G)). Therefore, γ(G) = γ(I(G)). 2 Remark 2.3 From the second part of the above theorem, it is evident that if G has no isolates, then γ(G) ≤ γ(I(G)). Also, if G has k isolates, γ(G) − k ≤ γ(I(G)). Remark 2.4 Let u = v . If ueu and vfv are adjacent in I(G), then u and v are adjacent . If ueu abnd vfv dominate each other, then u and v dominate each other. Remark 2.5 Let G be a graph without isolates. Let u, v ∈ V (G) . Suppose u and v are adjacent. Let ueu and vfv be two vertices in I(G).Then ueu and vfv need not be adjacent. In example : 1, u and w are adjcent in G. But ve and wg are not adajcent in I(G). Remark 2.6 Let S be an independent set of G, where G has no isolates. For each u ∈ S , choose exactly one eu from edges incident at u. Let S1 = {ueu : u ∈ S}. Sicne S is independent, S1 is independent in I(G). Therefore, |S| = |S1 | ≤ β0 (I(G)). Since this is true for any S, β0 (G) ≤ β0 (I(G)). In the above example 1, β0 (K3 ) = 1, β0 (I(K3 )) = 2. Remark 2.7 Let u and v be adjacent in G. Then ueu , vfv are independent if

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eu = uv, fv = uv. Remark 2.8 β0 (Kn ) = 1, β0 (I(Kn ) = n − 1. Theorem 2.9 Given any positive integer k, there exists a connected graph G such that β0 (I(G)) − β0 (G) = k. Proof. Let G = Kk+2 . Then , β0 (G) = 1. β0 (I(G)) = k + 1.

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Theorem 2.10 If G has a full degree vertex , then β0 (I(G)) = n − 1. Proof. Suppose G has a full degree vertex. Let V (G) = {u, x1 , x2 , · · · , xn−1 }. Let uxi = e1 , 1 ≤ i ≤ n − 1. Then x1 e1 , x2 e2 , · · · , xn−1 en−1 are independent in I(G). β0 (I(G)) ≥ n − 1. |V (G)| = 2|E(G)| ≥ 2(n − 1). Suppose S is independent set in I(G). Suppose uei ∈ S. Then xi fk ∈ / S, for any edge fk through xi . Therefore, |S| ≤ n−1. Therefore, uei ∈ / S, for any i, 1 ≤ i ≤ n−1. Let xi fxi ∈ S. Then S can contain at most n − 1 vertices, namely, x1 fk1 , x2 fk2 , · · · , xi−1 fki−1 , xi+1 fki+1 , · · · , xn−1 fkn−1 . Since any two vertices is the form xi gi , xi hi are adjacent, |S| ≤ n − 1. 2 Hence β0 (I(G)) = n − 1. Theorem 2.11 β0 (I(G)) ≤ n − 1, for all graph G, with |V (G)| = n. Proof. Let V (G) = {u1 , u2 , · · · , un }. Let S be an independent set in I(G).If ui e ∈ S, then ui f ∈ / S, for any edge f through ui . Therefore, S can contain at most one vertex of the form ui e , for any i. Therefore, |S| ≤ n. β0 (I(G)) ≤ n. Also, Suppose u1 e ∈ S. Then e = u1 ui , for some i, 2 ≤ i ≤ n. u1 e is adjacent to every vertex of the form ui f , where f is an edge incident at ui . Therefore, |S| ≤ n − 1. 2 Therefore, β0 (I(G)) ≤ n − 1. Remark 2.12 The bound is sharp. For: when G = K1,n . β0 (I(G)) = n = |V (G)| − 1. Theorem 2.13 K1,n with n odd (n ≥ 3) can not be an incidence graph. Proof. Consider K1,n with n = 2m + 1. Let K1,n = I(G) for some G. Then |V (I(G))| = 2m + 2 = 2|E(G)|. Therefore, |E(G)| = m + 1. Since I(G) is K1,n , I(G) is planar. Therefore, Δ(G) ≤ 2 ( by cor. 3, 2.2 [1]) Since I(G) is connected, G is connected.

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Therefore, G is a connected graph with maximum degree ≤ 2. Case(i): Suppose degree of every vertex is 2. Then G is a cycle, namely Cm+1 . I(Cm+1 )  K1,n Case(ii): Suppose, there are vertices of degree one. Let u be a vertex of degree one and v its support. Since degG (v) ≤ 2, either deg(v) = 1 (or) deg(v) = 2. Suppose degG (v) = 1. Then G = K2 = K1,1 . But n ≥ 3. Therefore, deg(v) = 2. Therefore, there exists a vertex w adjacent to v, w = u. If deg(w) = 1, then we get P3 . Otherwise, deg(w) = 2. Proceeding in this way , we terminate with a path say Pm+2 . But , I(Pm+2 )  K1,n . Hence the theorem. 2 Theorem 2.14 Cn can not be an incidence graph. Proof. Suppose Cn , n = 2m, m ≥ 2 be an incidence graph. Let Cn = I(G). Then |V (I(G)| = 2|E(G)| = 2n. Therefore, |E(G)| = n. Since Cn is planar, Δ(G) ≤ 2. Also,G is connected. Arguing as in the previous theorem, either G is a cycle (or) G is a path. In either, I(G) is not a cycle. Hence the theorem. 2 Theorem 2.15 Let G be a connected planar graph of even order. Then G is 2 . an incidence graph if and only if G = P2(n−1) Proof. Suppose G is an incidence graph. Let G = I(H). Then |V (I(H))| = . Since G is a connected planar graph , 2|E(H)|. Therefore, |E(H)| = |V (G)| 2 H is connected and Δ(H) ≤ 2. H is either a cycle (or) a path. 2 2 (or) P2(n−1) . I(H) = C2n 2 2 2 . But, C2n is not planar. Therefore, Therefore, G = C2n (or) P2(n−1) 2 2 G = P2(n−1) . The converse is obvious. Domination number for Incidence graph of standard graphs (i) γ(I(Pn )) = n5 (ii) γ(I(Cn )) =

n 5

(iii) γ(I(Wn )) = 1 (iv) γ(I(K1,n )) = 1

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(v) γ(I(Kn )) = 1 (vi) γ(I(Dr,s )) = 2

Conidtion for Minimal Dominating set of I(G) Theorem 2.16 A subset D of V (I(G)) is a minimal dominating set of I(G) if for every ueu ∈ D one of the following holds : (i) for any vfv ∈ D either u and v are not adjacent in G (or) xuvy is a path in G, where xu = eu and vy = fv . (ii) there exists wgw ∈ V (I(G)) − D such that wgw is adjacent to ueu and for any vfv ∈ D either v and w are not adjacent in G (or) xuvy is a path in G where xu = eu and wy = gw . 2

Proof. Routine.

Remark 2.17 Let G be connected. Let ueu , vfv be two vertices in I(G) . Then , using a path from u to v in G, say ue1 x1 e2 x2 · · · ek v. We get a path namely, ueu x1 e1 , · · · , uek vfv from ueu and vfv . Therefore, I(G) is connected. Remark 2.18 If G is a connected graph with at least three vertices, then I(G) has no pendent vertex. Proof. Let ueu be a vertex in I(G), where eu = uv. d(ueu ) = d(v) + 2(d(u) − 1) = 1 if and only if d(v) + 2d(u) = 3. That is , if and only if d(u) = d(v) = 1. That is , if and only if G contains K2 as a component, a contradiction. Therefore, I(G) has no pendent vertex. 2 Theorem 2.19 Let   G be a connected graph of order ≥ 3 . Then γ(I(G)) < (n−γ(G))(n−γ(G)+2) . 2 Proof. Since G is connected, I(G) is connected. Therefore, γ(I(G)) ≤ |V (I(G))| = 2 + |E(G)|. Suppose γ(I(G)) = |E(G)|. Then I(G) is either C4 or H , where H is a connected graph. I(G) = C4 , for any graph G. Therefore, I(G) = H + . Therefore, I(G) has pendent vertices. Since o(G) ≥ 4, o(I(G)) ≥ 4. + Therefore, I(G) has no pendentvertices. Therefore,  I(G) cannot be H . Therefore, γ(I(G)) ≤ |E(G)| ≤

(n−γ(G))(n−γ(G)+2) 2

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