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ISRN Algebra
Volume 2011 (2011), Article ID 193560, 10 pages
doi:10.5402/2011/193560
Research Article

Graph Polynomials

Mehdi Alaeiyan and Saeid Mohammadian

Department of Mathematics, Iran University of Science and Technology, Narmak, Tehran 16844, Iran

Received 23 May 2011; Accepted 5 July 2011

Academic Editor: M. Przybylska

Copyright © 2011 Mehdi Alaeiyan and Saeid Mohammadian. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

One of the most important and applied concepts in graph theory is to find the edge cover, vertex cover, and dominating sets with minimum cardinal also to find independence and matching sets with maximum cardinal and their polynomials. Although there exist some algorithms for finding some of them (Kuhn and Wattenhofer, 2003; and Mihelic and Robic, 2005), but in this paper we want to study all of these concepts from viewpoint linear and binary programming and we compute the coefficients of the polynomials by solving a system of linear equations with { 0 , 1 } variables.

1. Introduction

All graphs in this note are simple, connected, finite, and undirected, though it is probable that some of the obtained results are extendable to general or directed graphs.

Let 𝐺 = ( 𝑉 , 𝐸 ) be a simple and connected graph with | 𝑉 | = 𝑛 and | 𝐸 | = 𝑚 ; then the edge cover and edge dominating polynomials are of degree 𝑚 , and the vertex cover and dominating polynomials are of degree 𝑛 , in which coefficient of 𝑥 𝑘 is the number of edge cover, edge dominating, vertex cover, and dominating sets with 𝑘 elements, respectively. Also the independence and matching polynomials are at most of degree 𝑛 such that coefficient of 𝑥 𝑘 is the number of independence and matching sets with 𝑘 elements, respectively, for some positive integer 𝑘 .

For some notation being not defined here we refer the reader to [1].

A set 𝐿 ⊆ 𝐸 is an edge cover if every vertex 𝑣 ∈ 𝑉 is incident to some edge of 𝐿 .

A set 𝑄 ⊆ 𝑉 is a vertex cover if every edge 𝑒 ∈ 𝐸 has at least one endpoint in 𝑄 .

A set 𝑆 ⊆ 𝑉 is an independent set if non of two vertex in 𝑆 are not adjacent.

The maximum size of an independent set is named independence number.

A matching in graph 𝐺 is a set 𝑀 ⊆ 𝐸 with no shared endpoints.

In graph 𝐺 a set 𝐷 ⊆ 𝑉 is a dominating set if every vertex number in 𝐷 has a neighbor in 𝐷 , and finally a set 𝑊 ⊆ 𝐸 is an edge dominating set if every edge number in 𝑊 has a neighbor in 𝑊 .

We set | | 𝐿 | | M i n = 𝛽 , | | 𝑆 | | | | 𝑄 | | | | 𝑀 | | M a x = 𝛼 , M i n = 𝛽 , M a x = 𝛼 , | | 𝐷 | | | | 𝑊 | | M i n = 𝛾 , M i n = 𝛾 . ( 1 . 1 ) By [1],(i) 𝛼 + 𝛽 = 𝑛 , (ii) 𝛼 + 𝛽 = 𝑛 .

Also in every bipartite graph(iii) 𝛼 = 𝛽 , (iv) 𝛼 = 𝛽 .

We denote the Adjacency matrix by 𝐴 and Incidence matrix by 𝑅 , in which 𝐴 = [ 𝑎 𝑖 𝑗 ] 𝑛 × 𝑛 such that 𝑎 𝑖 𝑗 = t h e n u m b e r s o f e d g e s w i t h e n d p o i n t s 𝑣 𝑖 a n d 𝑣 𝑗 , ( 1 . 2 ) and also 𝑅 = [ 𝑟 𝑖 𝑗 ] 𝑛 × 𝑚 in which 𝑟 𝑖 𝑗 = 1 , 𝑣 𝑖 i s a n e n d p o i n t o f 𝑒 𝑗 , 0 , o t h e r w i s e . ( 1 . 3 ) We also define an Edge adjacency matrix 𝐵 = [ 𝑏 𝑖 𝑗 ] 𝑚 × 𝑚 as follows: 𝑏 𝑖 𝑗 = 1 , 𝑒 𝑖 i s a d j a c e n t t o 𝑒 𝑗 , 0 , o t h e r w i s e , ( 1 . 4 ) and 𝑏 𝑖 𝑖 = 0 .

From now on we set 𝑣 𝑉 = 1 , 𝑣 2 , … , 𝑣 𝑛 𝑡 , 𝑒 𝐸 = 1 , 𝑒 2 , … , 𝑒 𝑚 𝑡 , 1 𝑛 = ( 1 , 1 , … , 1 ) 𝑡 1 × 𝑛 . ( 1 . 5 )

2. Edge Cover Set and Edge Cover Polynomial

As previous notations we have the following theorem for obtaining the minimum size of edge cover set.

Theorem 2.1. One has 𝛽 ′ = m i n 𝑚 𝑖 = 1 𝑒 𝑖 s u b j e c t t o 𝑅 𝐸 ≧ 1 𝑛 , 𝑒 𝑖 ∈ { 0 , 1 } , w h e r e 𝑖 = 1 , 2 , … , 𝑚 . ( 2 . 1 )

Proof. Since an edge cover set of 𝐺 is a set 𝐿 of edges such that every vertex of 𝐺 is incident to some edge of 𝐿 and we want to obtain the optimal size of the sets in covering problems, so we will have a minimize problem; that is, the object function is 𝛽 ∑ = m i n 𝑚 𝑖 = 1 𝑒 𝑖 ; on the other hand for each 𝑣 𝑖 ∈ 𝑉 at least one edge with endpoint 𝑣 𝑖 must belong to 𝐿 ; in other words from every row of matrix 𝑅 at least one entry ( 𝑒 𝑖 ) must be equal to 1. Therefore 𝑟 1 1 𝑒 1 + 𝑟 1 2 𝑒 2 + ⋯ + 𝑟 1 𝑚 𝑒 𝑚 𝑟 ≥ 1 , 2 1 𝑒 1 + 𝑟 2 2 𝑒 2 + ⋯ + 𝑟 2 𝑚 𝑒 𝑚 ⋮ 𝑟 ≥ 1 , 𝑛 1 𝑒 1 + 𝑟 𝑛 2 𝑒 2 + ⋯ + 𝑟 𝑛 𝑚 𝑒 𝑚 𝑒 ≥ 1 , 𝑖 ∈ { 0 , 1 } , w h e r e 𝑖 = 1 , 2 , … , 𝑚 . ( 2 . 2 )

Definition 2.2. An edge cover polynomial is as follows: 𝐿 ( 𝑥 ) = 𝑎 0 𝑥 𝛽 ′ + 𝑎 1 𝑥 𝛽 ′ + 1 + ⋯ + 𝑎 𝑚 − 𝛽 ′ 𝑥 𝑚 , ( 2 . 3 ) where 𝛽 is the same as in (2.1) and 𝑎 𝑖 ’s are the number of edge cover sets with 𝛽 + 𝑖 elements.

Theorem 2.3. The coefficients 𝑎 0 , 𝑎 1 , … , 𝑎 𝑚 − 𝛽 ′ in edge cover polynomial are all of solutions of the following system for 𝑖 = 0 , 𝑖 = 1 , … , 𝑖 = 𝑚 − 𝛽 , respectively, 𝑅 𝐸 ≥ 1 𝑛 , 𝑒 ( ∗ ) 1 + 𝑒 2 + ⋯ + 𝑒 𝑚 = 𝛽 𝑒 + 𝑖 , 𝑗 ∈ { 0 , 1 } , 𝑤 ℎ 𝑒 𝑟 𝑒 𝑗 = 1 , 2 , … , 𝑚 . ( ∗ ∗ )

Proof. The first inequality ∗ is the condition for a set to be an edge cover set and ∗ ∗ for each 𝑖 causes that we have the edge cover sets with cardinality 𝛽 , 𝛽 + 1 , … , 𝑚 , respectively, and with this process we can compute 𝑎 0 , 𝑎 1 , … , 𝑎 𝑚 − 𝛽 ′ . It is trivial that 𝑎 𝑚 − 𝛽 ′ = 1 and this completes the proof.

Algorithm 2.4 (For computation 𝑎 𝑖 ). One has the following.Step 1. Solve 𝛽 ′ = m i n 𝑚 𝑖 = 1 𝑒 𝑖 , 𝑅 𝐸 ≥ 1 𝑛 , 𝑒 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑚 , ( 2 . 4 ) and obtain 𝛽 .Step 2. For 𝑖 = 0 to 𝑚 − 𝛽 − 1 , compute all of solutions: 𝑅 𝐸 ≥ 1 𝑛 , 𝑒 1 + 𝑒 2 + ⋯ + 𝑒 𝑚 = 𝛽 𝑒 + 𝑖 , 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑚 . ( 2 . 5 ) Step 3. Set 𝑎 𝑖 to be equal to all solutions of Step 2.

3. Independence Set and Independence Polynomial

In an independence set 𝑆 from every two adjacent vertices at most one of them belongs to 𝑆 ; this means that for all 𝑒 𝑖 ∈ 𝐸 with end points 𝑣 𝑖 and 𝑣 𝑗 at most 𝑣 𝑖 or 𝑣 𝑗 belongs to 𝑆 . Therefore we have the following.

Theorem 3.1. One has 𝛼 = m a x 𝑛 𝑖 = 1 𝑣 𝑖 s u b j e c t t o 𝑅 𝑡 𝑉 ≤ 1 𝑚 , 𝑣 𝑖 ∈ { 0 , 1 } , w h e r e 𝑖 = 1 , 2 , … , 𝑛 . ( 3 . 1 )

Definition 3.2. An independence polynomial is as follows: 𝑆 ( 𝑥 ) = 𝑏 0 𝑥 𝛼 + 𝑏 1 𝑥 𝛼 − 1 + ⋯ + 𝑏 𝛼 − 1 𝑥 , ( 3 . 2 ) where 𝛼 is the same as in (3.1) and 𝑏 𝑖 ’s are the numbers of independence sets with 𝛼 − 𝑖 elements.

Theorem 3.3. The coefficients 𝑏 0 , 𝑏 1 , … , 𝑏 𝛼 − 1 are all of solutions of the following system 𝑖 = 0 , 𝑖 = 1 , … , 𝑖 = 𝛼 − 1 , respectively, 𝑅 𝑡 𝑉 ≤ 1 𝑚 , 𝑣 1 + 𝑣 2 + ⋯ + 𝑣 𝑛 𝑣 = 𝛼 − 𝑖 , 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑛 . ( 3 . 3 )

Algorithm 3.4 (For computation 𝑏 𝑖 ). One has the following.Step 1. Solve 𝛼 = m a x 𝑛 𝑖 = 1 𝑣 𝑖 , 𝑅 𝑡 𝑉 ≤ 1 𝑚 , 𝑣 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑛 , ( 3 . 4 ) and obtain 𝛼 .Step 2. For 𝑖 = 0 to 𝛼 − 2 , compute all of solutions: 𝑅 𝑡 𝑉 ≤ 1 𝑚 , 𝑣 1 + 𝑣 2 + ⋯ + 𝑣 𝑛 𝑣 = 𝛼 − 𝑖 , 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑛 . ( 3 . 5 ) Step 3. Set 𝑏 𝑖 to be equal to all solutions of Step 2 of course 𝑏 𝛼 − 1 = 𝑛 .

4. Vertex Cover Set and Vertex Cover Polynomial

We have the following theorem for vertex cover set.

Theorem 4.1. One has 𝛽 = m i n 𝑛 𝑖 = 1 𝑣 𝑖 s u b j e c t t o 𝑅 𝑡 𝑉 ≧ 1 𝑚 , 𝑣 𝑖 ∈ { 0 , 1 } , w h e r e 𝑖 = 1 , 2 , … , 𝑛 . ( 4 . 1 )

Proof. Since a vertex cover set of 𝐺 is a set 𝑄 of vertices such that every edge of 𝐺 is incident to some vertex of 𝑄 and we want to obtain the optimal size of the sets in covering problems, so we will have a minimize problem; that is, the object function is ∑ 𝛽 = m i n 𝑛 𝑖 = 1 𝑣 𝑖 ; on the other hand for each 𝑒 𝑖 ∈ 𝐸 with endpoint 𝑣 𝑖 and 𝑣 𝑗 at least one of them must belong to 𝑄 ; in other words from every row of matrix 𝑅 𝑡 at least one entry ( 𝑣 𝑖 ) must be equal to 1. Therefore 𝑟 1 1 𝑣 1 + 𝑟 2 1 𝑣 2 + ⋯ + 𝑟 𝑛 1 𝑣 𝑛 𝑟 ≥ 1 , 1 2 𝑣 1 + 𝑟 2 2 𝑣 2 + ⋯ + 𝑟 𝑛 2 𝑣 𝑛 ⋮ 𝑟 ≥ 1 , 1 𝑛 𝑣 1 + 𝑟 2 𝑛 𝑣 2 + ⋯ + 𝑟 𝑚 𝑛 𝑣 𝑛 𝑣 ≥ 1 , 𝑖 ∈ { 0 , 1 } , 𝑤 ℎ 𝑒 𝑟 𝑒 𝑖 = 1 , 2 , … , 𝑛 . ( 4 . 2 )

Definition 4.2. A vertex cover polynomial is as follows: 𝑄 ( 𝑥 ) = 𝑐 0 𝑥 𝛽 + 𝑐 1 𝑥 𝛽 + 1 + ⋯ + 𝑐 𝑛 − 𝛽 𝑥 𝑛 , ( 4 . 3 ) where 𝛽 is the same as in (4.1) and 𝑐 𝑖 ’s are the number of vertex cover sets with 𝛽 + 𝑖 elements.

Theorem 4.3. The coefficients 𝑐 0 , 𝑐 1 , … , 𝑐 𝑛 − 𝛽 are all of solutions of the following system for 𝑖 = 0 , 𝑖 = 1 , … , 𝑖 = 𝑛 − 𝛽 , respectively, 𝑅 𝑡 𝑣 𝑉 ≥ 1 , ( 𝚤 ) 1 + 𝑣 2 + ⋯ + 𝑣 𝑛 𝑣 = 𝛽 + 𝑖 , 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑛 . ( 𝚤 𝚤 )

Proof. The first inequality 𝚤 is the condition for a set to be a vertex cover set and 𝚤 𝚤 for each 𝑖 causes that we have the vertex cover sets with cardinality 𝛽 , 𝛽 + 1 , … , 𝑛 , respectively, and with this process we can compute 𝑐 0 , 𝑐 1 , … , 𝑐 𝑛 − 𝛽 . It is trivial that 𝑐 𝑛 − 𝛽 = 1 and this completes the proof.

Algorithm 4.4 (For computation 𝑐 𝑖 ). One has the following.Step 1. Solve 𝛽 = m i n 𝑛 𝑖 = 1 𝑣 𝑖 , 𝑅 𝑡 𝑉 ≥ 1 𝑚 , 𝑣 𝑖 ∈ { 0 , 1 } , w h e r e 𝑖 = 1 , 2 , … , 𝑛 , ( 4 . 4 ) and obtain 𝛽 .Step 2. For 𝑖 = 0 to 𝑛 − 𝛽 − 1 , compute all of solutions: 𝑅 𝑡 𝑉 ≥ 1 𝑚 , 𝑣 1 + 𝑣 2 + ⋯ + 𝑣 𝑛 𝑣 = 𝛽 + 𝑖 , 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑛 . ( 4 . 5 ) Step 3. Set 𝑐 𝑖 to be equal to all solutions of Step 2.

5. Matching Set and Matching Polynomial

In a matching set ( 𝑀 ) from every two adjacent edges at most one of them belongs to 𝑀 and this means that for all 𝑒 𝑖 , 𝑒 𝑗 ∈ 𝐸 with common endpoint 𝑣 𝑖 at most 𝑒 𝑖 or 𝑒 𝑗 belongs to 𝑀 . Therefore we have the following.

Theorem 5.1. One has 𝛼 = m a x 𝑚 𝑖 = 1 𝑒 𝑖 s u b j e c t t o 𝑅 𝐸 ≤ 1 𝑛 , 𝑒 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑚 . ( 5 . 1 )

Definition 5.2. A matching polynomial is as follows: 𝑀 ( 𝑥 ) = 𝑑 0 𝑥 𝛼 ′ + 𝑑 1 𝑥 𝛼 ′ − 1 + ⋯ + 𝑑 𝛼 ′ − 1 𝑥 , ( 5 . 2 ) where 𝛼 is the same as in (5.1) and 𝑑 𝑖 ’s are the number of matching sets with 𝛼 − 𝑖 elements.

Theorem 5.3. The coefficients 𝑑 0 , 𝑑 1 , … , 𝑑 𝛼 ′ − 1 are all of solutions of the following system, respectively, 𝑖 = 0 , 𝑖 = 1 , … , 𝑖 = 𝛼 − 1 , 𝑅 𝐸 ≤ 1 𝑛 , 𝑒 1 + 𝑒 2 + ⋯ + 𝑒 𝑚 = 𝛼 𝑒 − 𝑖 , 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑚 . ( 5 . 3 )

Algorithm 5.4 (For computation 𝑑 𝑖 ). One has the following.Step 1. Solve 𝛼 = m a x 𝑚 𝑖 = 1 𝑒 𝑖 , 𝑅 𝐸 ≤ 1 𝑛 , 𝑒 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑚 , ( 5 . 4 ) and obtain 𝛼 .Step 2. For 𝑖 = 0 to 𝛼 − 2 , compute all of solutions: 𝑅 𝐸 ≤ 1 𝑛 , 𝑒 1 + 𝑒 2 + ⋯ + 𝑒 𝑚 = 𝛼 𝑒 − 𝑖 , 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑚 . ( 5 . 5 ) Step 3. Set 𝑑 𝑖 to be equal to all solutions of Step 2, 𝑖 = 0 , 1 , … , 𝛼 − 2 , of course 𝑑 𝛼 ′ − 1 = 1 .

6. Dominating Set and Dominating Polynomial

With the same argument in the previous sections we have the following theorem.

Theorem 6.1. One has 𝛾 = m i n 𝑛 𝑖 = 1 𝑣 𝑖 , s u b j e c t t o 𝐴 + 𝐼 𝑛 𝑉 ≧ 1 𝑛 , 𝑣 𝑖 ∈ { 0 , 1 } , w h e r e 𝑖 = 1 , 2 , … , 𝑛 . ( 6 . 1 )

Definition 6.2. A dominating polynomial is as follows: 𝐷 ( 𝑥 ) = 𝑓 0 𝑥 𝛾 + 𝑓 1 𝑥 𝛾 + 1 + ⋯ + 𝑓 𝑛 − 𝛾 𝑥 𝑛 , ( 6 . 2 ) where 𝛾 is the same as in (6.1) and 𝑓 𝑖 ’s are the number of dominating sets with 𝛾 + 𝑖 elements.

Theorem 6.3. The coefficients 𝑓 0 , 𝑓 1 , … , 𝑓 𝑛 − 𝛾 are all of solutions of the following system, respectively, 𝑖 = 0 , 𝑖 = 1 , … , 𝑖 = 𝑛 − 𝛾 , 𝐴 + 𝐼 𝑛 𝑉 ≥ 1 𝑛 𝑣 , ( † ) 1 + 𝑣 2 + ⋯ + 𝑣 𝑛 𝑣 = 𝛾 + 𝑖 , 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑛 . ( † † )

Proof. The first inequality † is the condition for a set to be a dominating set and † † for each 𝑖 causes that we have the dominating sets with cardinality 𝛾 , 𝛾 + 1 , … , 𝑛 , respectively, and with this process we can compute 𝑓 0 , 𝑓 1 , … , 𝑓 𝑛 − 𝛾 . It is trivial that 𝑓 𝑛 − 𝛾 = 1 and this completes the proof.

Algorithm 6.4 (For computation 𝑓 𝑖 ). One has the following.Step 1. Solve 𝛾 = m i n 𝑛 𝑖 = 1 𝑣 𝑖 , 𝐴 + 𝐼 𝑛 𝑉 ≥ 1 𝑛 , 𝑣 𝑖 ∈ { 0 , 1 } , w h e r e 𝑖 = 1 , 2 , … , 𝑛 , ( 6 . 3 ) and obtain 𝛾 .Step 2. For 𝑖 = 0 to 𝑛 − 𝛾 − 1 , compute all of solutions: 𝐴 + 𝐼 𝑛 𝑉 ≥ 1 𝑛 , 𝑣 1 + 𝑣 2 + ⋯ + 𝑣 𝑛 𝑣 = 𝛾 + 𝑖 , 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑛 . ( 6 . 4 ) Step 3. Set 𝑓 𝑖 to be equal to all solutions of Step 2.

7. Edge Dominating Set and Edge Dominating Polynomial

With the same argument in previous sections we have Theorems 7.1 and 7.3.

Theorem 7.1. One has 𝛾 = m i n 𝑚 𝑖 = 1 𝑒 𝑖 s u b j e c t t o 𝐵 + 𝐼 𝑚 𝐸 ≧ 1 𝑚 , 𝑒 𝑖 ∈ { 0 , 1 } , w h e r e 𝑖 = 1 , 2 , … , 𝑚 . ( 7 . 1 )

Definition 7.2. An edge dominating polynomial is a polynomial such as 𝑊 ( 𝑥 ) = 𝑓 0 𝑥 𝛾 ′ + 𝑓 1 𝑥 𝛾 ′ + 1 + ⋯ + 𝑓 𝑚 − 𝛾 ′ 𝑥 𝑚 , ( 7 . 2 ) where 𝛾 is the same as in (7.1) and 𝑓 𝑖 ’s are the number of edge dominating sets with 𝛾 + 𝑖 elements.

Theorem 7.3. The coefficients 𝑓 0 , 𝑓 1 , … , 𝑓 𝑚 − 𝛾 ′ are all of solutions of the following system, respectively, 𝑖 = 0 , 𝑖 = 1 , … , 𝑖 = 𝑚 − 𝛾 , 𝐵 + 𝐼 𝑚 𝐸 ≥ 1 𝑚 , 𝑒 1 + 𝑒 2 + ⋯ + 𝑒 𝑚 = 𝛾 𝑒 + 𝑖 , 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑚 . ( 7 . 3 )

Algorithm 7.4 (For computation 𝑓 𝑖 ). One has the following.Step 1. Solve 𝛾 = m i n 𝑚 𝑖 = 1 𝑒 𝑖 , 𝐵 + 𝐼 𝑚 𝐸 ≥ 1 𝑚 , 𝑒 𝑖 ∈ { 0 , 1 } , w h e r e 𝑖 = 1 , 2 , … , 𝑚 , ( 7 . 4 ) and obtain 𝛾 .Step 2. For 𝑖 = 0 to 𝑚 − 𝛾 − 1 , compute all of solutions: 𝐵 + 𝐼 𝑚 𝐸 ≥ 1 𝑚 , 𝑒 1 + 𝑒 2 + ⋯ + 𝑒 𝑚 = 𝛾 𝑒 + 𝑖 , 𝑗 ∈ { 0 , 1 } , w h e r e 𝑗 = 1 , 2 , … , 𝑚 . ( 7 . 5 ) Step 3. Set 𝑓 𝑖 to be equal to all solutions of Step 2.

References

  1. D. B. West, Introduction to Graph Theory, Prentice Hall Inc., Upper Saddle River, NJ, USA, 1996.
  2. F. Kuhn and R. Wattenhofer, “Distributed combinatorial optimization,” Tech. Rep. 426, Department of computer Science, ETH Zurich, Zurich, Switzerland, 2003.
  3. J. Mihelic and B. Robic, “Solving the k-center problem efficiently with a dominating set,” Algorithm Journal of Computing and Information Technology, vol. 13, no. 3, pp. 225–233, 2005.