The Domination Polynomials of Cubic graphs of order 10
aa r X i v : . [ m a t h . C O ] M a y The Domination Polynomials of CubicGraphs of Order 10
Saieed Akbari a , Saeid Alikhani b,d, , Yee-hock Peng c,da Department of Mathematical Sciences, Sharif University of Technology11365-9415 Tehran, Iran b Department of Mathematics, Yazd University89195-741, Yazd, Iran c Department of Mathematics, and d Institute for Mathematical Research, University Putra Malaysia43400 UPM Serdang, Malaysia
ABSTRACT
Let G be a simple graph of order n . The domination polynomial of G is the polynomial D ( G, x ) = P ni = γ ( G ) d ( G, i ) x i , where d ( G, i ) is the number of dominating sets of G of size i , and γ ( G ) is thedomination number of G . In this paper we study the domination polynomials of cubic graphsof order . As a consequence, we show that the Petersen graph is determined uniquely by itsdomination polynomial. Mathematics Subject Classification:
Keywords:
Domination polynomial; Equivalence class; Petersen graph; Cubic graphs.
Let G = ( V, E ) be a simple graph. The order of G is the number of vertices of G . For anyvertex v ∈ V , the open neighborhood of v is the set N ( v ) = { u ∈ V | uv ∈ E } and the closedneighborhood of v is the set N [ v ] = N ( v ) ∪ { v } . For a set S ⊆ V , the open neighborhood of S is N ( S ) = S v ∈ S N ( v ) and the closed neighborhood of S is N [ S ] = N ( S ) ∪ S . A set S ⊆ V isa dominating set if N [ S ] = V , or equivalently, every vertex in V \ S is adjacent to at least onevertex in S . The domination number γ ( G ) is the minimum cardinality of a dominating set in G .A dominating set with cardinality γ ( G ) is called a γ -set . The family of all γ -sets of a graph G isdenoted by Γ( G ). For a detailed treatment of these parameters, the reader is referred to [2]. The Corresponding author. E-mail: [email protected] -subset of V ( G ) is a subset of V ( G ) of size i . Let D ( G, i ) be the family of dominating sets of agraph G with cardinality i and let d ( G, i ) = |D ( G, i ) | . The domination polynomial D ( G, x ) of G is defined as D ( G, x ) = P | V ( G ) | i = γ ( G ) d ( G, i ) x i , where γ ( G ) is the domination number of G (see[1]).We denote the family of all dominating sets of G with cardinality i and contain a vertex v by D v ( G, i ), and d v ( G, i ) = |D v ( G, i ) | . Two graphs G and H are said to be dominating equivalence ,or simply D -equivalent, written G ∼ H , if D ( G, x ) = D ( H, x ). It is evident that the relation ∼ of being D -equivalence is an equivalence relation on the family G of graphs, and thus G ispartitioned into equivalence classes, called the D -equivalence classes . Given G ∈ G , let[ G ] = { H ∈ G : H ∼ G } . We call [ G ] the equivalence class determined by G . A graph G is said to be dominating unique ,or simply D -unique , if [ G ] = { G } .The minimum degree of G is denoted by δ ( G ). A graph G is called k -regular if all vertices havethe same degree k . A vertex-transitive graph is a graph G such that for every pair of vertices v and w of G , there exists an automorphism θ such that θ ( v ) = w . One of the famous graphs isthe Petersen graph which is a symmetric non-planar cubic graph. In the study of dominationpolynomials, it is interesting to investigate the dominating sets and domination polynomial ofthis graph. We denote the Petersen graph by P .In this paper, we study the dominating sets and domination polynomials of cubic graphs oforder 10. As a consequence, we show that the Petersen graph is determined uniquely by itsdomination polynomial. In the next section, we obtain domination polynomial of the Petersengraph. In Section 3, we list all γ -sets of connected cubic graphs of order 10. This list will beused to study the D -equivalence of these graphs in the last section. In Section 4, we prove thatthe Petersen graph is D -unique. In the last section, we study the D -equivalence classes of somecubic graphs of order 10. 2 Domination Polynomial of the Petersen Graph
In this section we shall investigate the domination polynomial of the Petersen graph. First, westate and prove the following lemma:
Lemma 1.
Let G be a vertex transitive graph of order n and v ∈ V ( G ) . For any ≤ i ≤ n , d ( G, i ) = ni d v ( G, i ) . Proof.
Clearly, if D is a dominating set of G with size i , and θ ∈ Aut ( G ), then θ ( D ) is also adominating set of G with size i . Since G is a vertex transitive graph, then for every two vertices v and w , d v ( G, i ) = d w ( G, i ). If D is a dominating set of size i , then there are exactly i vertices v j , . . . , v j i such that D counted in d v jr ( G, i ), for each 1 ≤ r ≤ i . Hence d ( G, i ) = ni d v ( G, i ), andthe proof is complete.
Theorem 1. ([2], p.48) If G is a connected graph of order n with δ ( G ) ≥ , then γ ( G ) ≤ n . We need the following theorem for finding the domination polynomial of the Petersen graph.
Theorem 2.
Let G be a graph of order n with domination polynomial D ( G, x ) = P ni =1 d ( G, i ) x i .If d ( G, j ) = (cid:0) nj (cid:1) for some j , then δ ( G ) ≥ n − j . More precisely, δ ( G ) = n − l , where l =min n j | d ( G, j ) = (cid:0) nj (cid:1)o , and there are at least (cid:0) nl − (cid:1) − d ( G, l − vertices of degree δ ( G ) in G .Furthermore, if for every two vertices of degree δ ( G ) , say u and v we have N [ u ] = N [ v ] , thenthere are exactly (cid:0) nl − (cid:1) − d ( G, l − vertices of degree δ ( G ) . Proof.
Since d ( G, j ) = (cid:0) nj (cid:1) for any r ≥ j , every r -subset of the vertices of G forms a domi-nating set for G . Suppose that there is a vertex v ∈ V ( G ) such that deg ( v ) < n − j . Consider V ( G ) \ N [ v ]. Clearly, (cid:12)(cid:12)(cid:12) V ( G ) \ N [ v ] (cid:12)(cid:12)(cid:12) ≥ j , and V ( G ) \ N [ v ] is not a dominating set for G , a con-tradiction. Now, assume that S is a ( l − V ( G ) which is not a dominating set.Thus there is a vertex u ∈ V ( G ) \ S which is not covered by S . Since δ ( G ) ≥ n − l , we have deg ( u ) = n − l . Let S ′ = S be a ( l − u ′ ∈ V ( G ) \ S ′ which is not covered by S ′ . As we did before deg ( u ′ ) = n − l . We claimthat u = u ′ . Since S = S ′ , there exists a vertex x ∈ S ∩ ( V ( G ) \ S ′ ). We know that u ′ x ∈ E ( G ).If u = u ′ , then ux ∈ E ( G ), a contradiction. Thus u = u ′ and the claim is proved. If u and v aretwo vertices of degree δ ( G ) and N [ u ] = N [ v ], then V ( G ) \ N [ u ] = V ( G ) \ N [ v ] is an ( l − G . Thus we have at least (cid:0) nl − (cid:1) − d ( G, l −
1) vertices of degree δ ( G ) in G . The last part of theorem is obvious. G G G G G G G G G G G G G G G G G G G G G Figure 1: Cubic graphs of order 10.4ndeed, by Theorem 2, we have the following theorem which relates the domination polynomialand the regularity of a graph G . Theorem 3.
Let H be a k -regular graph, where for every two vertices u, v ∈ V ( H ) , N [ u ] = N [ v ] .If D ( G, x ) = D ( H, x ) , then G is also a k -regular graph. There are exactly 21 cubic graphs of order 10 given in Figure 1 (see [3]). Using Theorem 1,the domination number of a connected cubic graph of order 10 is 3. Three are just two non-connected cubic graphs of order 10. Clearly, for these graphs, the domination number is also3. Note that the graph G is the Petersen graph. For the labeled graph G in Figure 1, weobtain all dominating sets of size 3 and 4 in the following lemma. Lemma 2.
For the Petersen graph P , d ( P,
3) = 10 and d ( P,
4) = 75 . Proof.
First, we list all dominating sets of P of cardinality 3, which are the γ -sets of the labeledPetersen graph (graph G ) given in Figure 1. D ( P,
3) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Now, we shall compute d ( P, D ( P,
4) = n { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } o .Therefore by Lemma 1, d ( P,
4) = × = 75.We need the following lemma: Lemma 3.
Let G be a cubic graph of order . Then the following hold:(i) d ( G, i ) = (cid:0) ni (cid:1) , for i = 7 , , , . ii) if t and s are the number of subgraphs isomorphic to K \{ e } ( e is an edge) and K in G ,respectively, then d ( G,
6) = (cid:0) (cid:1) − (10 − t − s ) .(iii) if G has no subgraph isomorphic to graphs given in Figure 2, then d ( G,
5) = (cid:0) (cid:1) − . Figure 2: Graphs illustrated in Lemma 3.
Proof. (i) It follows from Theorem 2.(ii) If G is a cubic graph of order 10, then for every v ∈ V ( G ), V ( G ) \ N [ v ] is not a dominatingset. Also, if S ⊂ V ( G ), | S | = 6 and S is not a dominating set, then S = V ( G ) \ N [ v ], for some v ∈ V ( G ). Note that if G has K \{ e } as a subgraph, then there are two vertices u and u such that G \ N [ u ] = G \ N [ u ]. Also if G has K as its subgraph, then there are fourvertices u i , 1 ≤ i ≤ G \ N [ u i ] = G \ N [ u j ], for 1 ≤ i = j ≤
4. Hence we have d ( G,
6) = (cid:0) (cid:1) − (10 − t − s ).(iii) It suffices to determine the number of 5-subsets which are not dominating set. Supposethat S ⊆ V ( G ), | S | = 5, and S is not a dominating set for G . Thus there exists v ∈ V ( G ) suchthat N [ v ] ∩ S = ∅ . Now, note that for every x ∈ V ( G ), V ( G ) \ ( N [ x ] ∪{ y } ), where y ∈ V ( G ) \ N [ x ]is a 5-subset which is not a dominating set for G . Also since none of the graphs given in Figure 2is a subgraph of G , for every two distinct vertices x and x ′ and any two arbitrary vertices y ∈ V ( G ) \ N [ x ] and y ′ ∈ V ( G ) \ N [ x ′ ], we have V ( G ) \ ( N [ x ] ∪ { y } ) = V ( G ) \ ( N [ x ′ ] ∪ { y ′ } ). Thisimplies that the number of 5-subsets of V ( G ) which are not dominating sets is 10 × d ( G,
5) = (cid:0) (cid:1) − Corollary 1.
For cubic graphs of order , the following hold: If G ∈ { G , G } , then d ( G,
6) = (cid:0) (cid:1) − . (ii) If G ∈ { G , G } , then d ( G,
6) = (cid:0) (cid:1) − . (iii) If G ∈ { G , G } , then d ( G,
6) = (cid:0) (cid:1) − . (iv) For each i, ≤ i ≤ , if i
6∈ { , , , , , } , then d ( G i ,
6) = (cid:0) (cid:1) − . (v) If G ∈ { G , G , G , G , G } , then d ( G,
5) = (cid:0) (cid:1) − . Theorem 4.
The domination polynomial of the Petersen graph P is: D ( P, x ) = x + (cid:18) (cid:19) x + (cid:18) (cid:19) x + (cid:18) (cid:19) x + h(cid:18) (cid:19) − i x + h(cid:18) (cid:19) − i x + 75 x + 10 x . Proof.
The result follows from Lemma 2 and Corollary 1. γ -Sets of cubic graphs of order . In this section, we present all γ -sets of connected cubic graphs G , G , . . . , G , G shown inFigure 1. The results here will be useful in studying the D -equivalence of these graphs in thelast section.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore, d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 22.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 12.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 17.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 15.7( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 24.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 10.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 6Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o .Therefore d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 6 . Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 10 . Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 10 . Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 12.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 15.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . There-fore d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 8.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore d ( G ,
3) = (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 22.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , }{ , , } , { , , } , { , , } , , , } , { , , } , { , , } o . Therefore (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 12.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore d ( G ,
3) = 6.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore, d ( G ,
3) = 10.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore, d ( G ,
3) = 16.Γ( G ) = n { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } , { , , } o . Therefore (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 13. D -Equivalence class of the Petersen Graph In this section we show that the Petersen graph is D -unique. Theorem 5.
The Petersen graph P is D -unique. Proof.
Assume that G is a graph such that D ( G, x ) = D ( P, x ). Since for every two vertices x, y ∈ V ( P ), N [ x ] = N [ y ], by Theorem 3, G is a 3-regular graph of order 10. Using the (cid:12)(cid:12)(cid:12) Γ( G i ) (cid:12)(cid:12)(cid:12) for i = 1 , . . . ,
21 in Section 3 we reject some graphs from [ P ]. Since d ( G ,
3) = d ( G ,
3) = d ( G ,
3) = d ( G ,
3) = 10, we compare the cardinality of the families of dominating sets ofthese four graphs of size 4. D ( G ,
4) = n { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } o . Therefore, by Lemma 1, d ( G ,
4) = × = 85 > d ( P,
4) = 75.9ow, we obtain the family of all dominating sets of G of size 4. D ( G ,
4) = n { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } o .Therefore, by Lemma 1, d ( G ,
4) = × = 85 > d ( P,
4) = 75.Now, for graph G we have, D ( G ,
4) = n { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , }{ , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } o .Therefore d ( G ,
4) = 91 > d ( P, . Hence [ P ] = { P } , and so the Petersen graph is D -unique.By the arguments in the proof of Theorem 5, we have the following corollary. Corollary 2. (i)
The graph G is D -unique, (ii) [ G ] = n G , G o with the following domination polynomial: x + (cid:18) (cid:19) x + (cid:18) (cid:19) x + (cid:18) (cid:19) x + ( (cid:18) (cid:19) − x + ( (cid:18) (cid:19) − x + 85 x + 10 x . D -equivalence class of cubic graphs of order 10 In this section, we shall study the D -equivalence classes of other cubic graphs of order 10.We need the following theorem: Theorem 6. ([1])
If a graph G has m components G , . . . , G m , then D ( G, x ) = D ( G , x ) · · · D ( G m , x ) . Corollary 3.
Two graphs G and G are D -equivalence, with the following domination poly-nomial: D ( G , x ) = D ( G , x ) = x + 10 x + 45 x + 120 x + 203 x + 216 x + 134 x + 36 x . Proof.
Two graphs G and G are disconnected with two components. In other words G = H ∪ K and G = H ′ ∪ K , where H and H ′ are graphs with 6 vertices. It is not hardto see that D ( H, x ) = D ( H ′ , x ) = x + 6 x + 15 x + 20 x + 9 x . On the other hand, D ( K , x ) = x + 4 x + 6 x + 4 x . By Theorem 6, we have the result. Theorem 7.
The graphs G , G , G , G , and G in Figure 1 are D -unique. Proof.
Using γ -sets in Section 3, (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 15 , (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 7 , (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 22, and (cid:12)(cid:12) Γ( G ) (cid:12)(cid:12) = 13.By comparing these numbers with the cardinality of γ -sets of other 3-regular graphs, we havethe result. Now, we consider graph G . Since d ( G ,
3) = d ( G ,
3) = d ( G ,
3) = 6, we shallobtain d ( G i ,
4) for i = 7 , , D ( G ,
4) = n { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } o . Therefore, by Lemma 1, d ( G ,
4) = × = 80.11 G G G D ( G , x ) = D ( G , x ) = x
10 + 10 x x x x x x x G G D ( G , x ) = D ( G , x ) = x
10 + 10 x x x x x x x D ( G , x ) = D ( G , x ) = x
10 + 10 x x x x x x x Figure 3: Cubic graphs of order 10 with identical domination polynomial. D ( G ,
4) = n { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } o . Therefore, by Lemma 1, d ( G ,
4) = × = 80.Now, we obtain d ( G , D ( G ,
4) = n { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , { , , , } , , , , } , { , , , } , { , , , } o . Therefore, by Lemma 1, d ( G ,
4) = × = 85. Hence[ G ] = { G } .By the arguments in the proof of Theorem 7, we have the following corollary. Corollary 4.
Two graphs G and G are D -equivalence. In summary, in this paper we showed that the Petersen graph is D -unique. Also, we provedthat the graphs G , G , G , G , G , G , G , G , G , and G are D -unique, and [ G ] = { G , G } , [ G ] = { G , G } , [ G ] = { G , G } (see Figure 3). We are not able to determinethe D -equivalence of G , G , G , G , and G , but we think that they are D -unique. Acknowledgement.
The first author is indebted to the Institute for Mathematical Research(INSPEM) at University Putra Malaysia (UPM) for the partial support and hospitality duringhis visit.
References [1] S. Alikhani, Y. H. Peng, Introduction to Domination polynomial of a graphs, Ars Combi-natoria, to appear.[2] T.W. Haynes, S.T. Hedetniemi, P.J. Slater, Fundamentals of Domination in Graphs, MarcelDekker, NewYork, 1998.[3] G. B. Khosrovshahi, Ch. Maysoori, Tayfeh-Rezaie, A Note on 3-Factorizations of K10