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A comment to: Two classes of edge domination in graphs
Discrete Applied Mathematics 157 (2009) 400–401
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Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam
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A comment to: Two classes of edge domination in graphs S. Jahanbekam ∗ Department of Mathematical Sciences, Sharif University of Technology, Iran
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a b s t r a c t
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Article history: Received 30 January 2008 Received in revised form 30 April 2008 Accepted 11 June 2008 Available online 23 July 2008 Keywords: Signed edge domination function Signed edge domination number
Let G be a graph of order n and γs0 (G) denote the signed edge domination number of G. In [B. Xu, Two classes of edge domination in graphs, Discrete Appl. Math. 154 (2006) 1541–1546] it was proved that for any graph G of order n, γs0 (G) ≤ b 11 n − 1c. But the method given 6 in the proof is not correct. In this paper we give an example for which the method of proof given in [1] does not work. © 2008 Elsevier B.V. All rights reserved.
Let P G = (V , E ) be a nonempty graph. A function f : E → {+1, −1} is called the signed edge domination function P (SEDF) of G if e0 ∈N [e] f (e0 ) ≥ 1 for every e ∈ E (G). The signed edge domination number of G is defined as γs0 (G) = min{ e∈E f (e) | f is an SEDF of G}, and define γs0 (Kn ) = 0 for all totally disconnected graphs Kn . For each v ∈ V (G), let E (v) be the set of all edges adjacent to v . In [1], the following theorem has been proved: Theorem (Theorem 3 of [1]). For any graph G of order n, γs0 (G) ≤ b 11 n − 1c. 6 The proof of the above theorem that has been given is not true. In the following we present an example for which the proof does not work. In order to show that the method given in the proof of Theorem 3 in [1] fails, let G be the graph shown in Fig. 1. According to the Line 3 of the proof of Theorem 3, when the graph contains no Hamilton cycle, the author defines a spanning tree T of G whose the number of pendant vertices is as small as possible. Since G is not Hamiltonian, the path, u1 , u2 , . . . , u14 can be considered as an eligible spanning tree T of G, because the number of its pendant vertices is 2. Now, according to the proof of Theorem 3, the function f on the edges of this tree is defined as follows: f (ui ui+1 ) = +1,
for each i, 1 ≤ i ≤ 13.
Now, we have A = {v ∈ V (T ) | dT (v) = 1} = {u1 , u14 } and T0 = T \ A is the path, u2 , u3 , . . . , u13 . So A0 = {v ∈ V (T0 ) | dT0 (v) = 1} = {u2 , u13 }, but there is no edge e0 ∈ E (u0 ) \ E (T ) for the vertex u0 = u2 or u13 and so M = ∅. S St Thus G0 = G \ (E (T ) M ) = G \ E (T ). Now, the author defines G1 = G0 \ i=1 E (Hi ) where H1 , H2 , . . . , Ht are all even circuits of G0 . Since G0 contains no even circuit so G1 = G0 . On the other hand, G1 contains only one odd cycle, named Cr1 = u1 u6 u8 u10 u12 u1 and M1 = {u6 u8 , u12 u1 } is a maximum matching for this cycle. Define: f (u6 u8 ) = f (u1 u12 ) = −1,
f (u1 u6 ) = f (u8 u10 ) = f (u10 u12 ) = +1.
∗ Corresponding address: Department of Mathematical Sciences, Sharif University of Technology, Alley no.4, Mahmoodieh St Chamran Blv. 7194965118, Shiraz, Fars, Iran. Tel.: +98 9177067423; fax: +98 7116252353. E-mail addresses: [email protected], [email protected]. 0166-218X/$ – see front matter © 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.dam.2008.06.016
S. Jahanbekam / Discrete Applied Mathematics 157 (2009) 400–401
401
Fig. 1. The counterexample.
Now, let F = G1 \ E (Cr1 ). In fact V (F ) = {u1 , u3 , u4 , u6 } and E (F ) = {u1 u3 , u4 u6 }, thus in the last step of the definition of f we have f (u1 u3 ) = f (u4 u6 ) = −1. We claim that the function P f , introduced according to the proof of Theorem 3, is not an SEDF of G. To see this we note that for the edge u1 u6 we have e∈N [u1 u6 ] f (e) = f (u1 u6 )+ f (u1 u2 )+ f (u1 u3 )+ f (u1 u12 )+ f (u4 u6 )+ f (u5 u6 )+ f (u6 u7 )+ f (u6 u8 ) = 1 + 1 − 1 − 1 − 1 + 1 + 1 − 1 = 0. So f is not an SEDF of G. Remark. Recently, H. Karami, S.M. Sheikholeslami and A. Khodkar proved that for every simple graph G of order n, γs0 (G) ≤ d 3n e. 2 Acknowledgment I am very grateful to professor S. Akbari for his useful comments. References [1] B. Xu, Two classes of edge domination in graphs, Discrete Appl. Math. 154 (2006) 1541–1546.
Contents lists available at ScienceDirect
Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam
Note
A comment to: Two classes of edge domination in graphs S. Jahanbekam ∗ Department of Mathematical Sciences, Sharif University of Technology, Iran
article
a b s t r a c t
info
Article history: Received 30 January 2008 Received in revised form 30 April 2008 Accepted 11 June 2008 Available online 23 July 2008 Keywords: Signed edge domination function Signed edge domination number
Let G be a graph of order n and γs0 (G) denote the signed edge domination number of G. In [B. Xu, Two classes of edge domination in graphs, Discrete Appl. Math. 154 (2006) 1541–1546] it was proved that for any graph G of order n, γs0 (G) ≤ b 11 n − 1c. But the method given 6 in the proof is not correct. In this paper we give an example for which the method of proof given in [1] does not work. © 2008 Elsevier B.V. All rights reserved.
Let P G = (V , E ) be a nonempty graph. A function f : E → {+1, −1} is called the signed edge domination function P (SEDF) of G if e0 ∈N [e] f (e0 ) ≥ 1 for every e ∈ E (G). The signed edge domination number of G is defined as γs0 (G) = min{ e∈E f (e) | f is an SEDF of G}, and define γs0 (Kn ) = 0 for all totally disconnected graphs Kn . For each v ∈ V (G), let E (v) be the set of all edges adjacent to v . In [1], the following theorem has been proved: Theorem (Theorem 3 of [1]). For any graph G of order n, γs0 (G) ≤ b 11 n − 1c. 6 The proof of the above theorem that has been given is not true. In the following we present an example for which the proof does not work. In order to show that the method given in the proof of Theorem 3 in [1] fails, let G be the graph shown in Fig. 1. According to the Line 3 of the proof of Theorem 3, when the graph contains no Hamilton cycle, the author defines a spanning tree T of G whose the number of pendant vertices is as small as possible. Since G is not Hamiltonian, the path, u1 , u2 , . . . , u14 can be considered as an eligible spanning tree T of G, because the number of its pendant vertices is 2. Now, according to the proof of Theorem 3, the function f on the edges of this tree is defined as follows: f (ui ui+1 ) = +1,
for each i, 1 ≤ i ≤ 13.
Now, we have A = {v ∈ V (T ) | dT (v) = 1} = {u1 , u14 } and T0 = T \ A is the path, u2 , u3 , . . . , u13 . So A0 = {v ∈ V (T0 ) | dT0 (v) = 1} = {u2 , u13 }, but there is no edge e0 ∈ E (u0 ) \ E (T ) for the vertex u0 = u2 or u13 and so M = ∅. S St Thus G0 = G \ (E (T ) M ) = G \ E (T ). Now, the author defines G1 = G0 \ i=1 E (Hi ) where H1 , H2 , . . . , Ht are all even circuits of G0 . Since G0 contains no even circuit so G1 = G0 . On the other hand, G1 contains only one odd cycle, named Cr1 = u1 u6 u8 u10 u12 u1 and M1 = {u6 u8 , u12 u1 } is a maximum matching for this cycle. Define: f (u6 u8 ) = f (u1 u12 ) = −1,
f (u1 u6 ) = f (u8 u10 ) = f (u10 u12 ) = +1.
∗ Corresponding address: Department of Mathematical Sciences, Sharif University of Technology, Alley no.4, Mahmoodieh St Chamran Blv. 7194965118, Shiraz, Fars, Iran. Tel.: +98 9177067423; fax: +98 7116252353. E-mail addresses: [email protected], [email protected]. 0166-218X/$ – see front matter © 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.dam.2008.06.016
S. Jahanbekam / Discrete Applied Mathematics 157 (2009) 400–401
401
Fig. 1. The counterexample.
Now, let F = G1 \ E (Cr1 ). In fact V (F ) = {u1 , u3 , u4 , u6 } and E (F ) = {u1 u3 , u4 u6 }, thus in the last step of the definition of f we have f (u1 u3 ) = f (u4 u6 ) = −1. We claim that the function P f , introduced according to the proof of Theorem 3, is not an SEDF of G. To see this we note that for the edge u1 u6 we have e∈N [u1 u6 ] f (e) = f (u1 u6 )+ f (u1 u2 )+ f (u1 u3 )+ f (u1 u12 )+ f (u4 u6 )+ f (u5 u6 )+ f (u6 u7 )+ f (u6 u8 ) = 1 + 1 − 1 − 1 − 1 + 1 + 1 − 1 = 0. So f is not an SEDF of G. Remark. Recently, H. Karami, S.M. Sheikholeslami and A. Khodkar proved that for every simple graph G of order n, γs0 (G) ≤ d 3n e. 2 Acknowledgment I am very grateful to professor S. Akbari for his useful comments. References [1] B. Xu, Two classes of edge domination in graphs, Discrete Appl. Math. 154 (2006) 1541–1546.