Table Of Content

A formulation of a (q + 1,8)–cage M. Abreu1, G. Araujo-Pardo2, C. Balbuena3, D. Labbate1 5 1 0 1 2 Dipartimento di Matematica Informatica ed Economia n Universita` degli Studi della Basilicata a J 1 Viale dell’Ateneo Lucano 10, I-85100 Potenza, Italy. 1 [email protected], [email protected] ] O C . 2Instituto de Matema´ticas h t a Universidad Nacional Autońoma de México m [ México D. F., México. 1 [email protected] v 8 4 4 2 3Departament de Matema´tica Aplicada III 0 . 1 Universitat Politècnica de Catalunya 0 5 Campus Nord, Edifici C2, C/ Jordi Girona 1 i 3 E-08034 Barcelona, Spain. 1 : v [email protected] i X r a Abstract Let q 2 be a prime power. In this note we present a formulation for obtaining the ≥ known (q+1,8)-cages which has allowed us to construct small (k,g)–graphs for k =q 1,q − and g =7,8. Furthermore, we also obtain smaller (q,8)-graphs for even prime power q. Keywords: Cages, girth, Moore graphs, perfect dominating sets. MSC2010: 05C35, 05C69, 05B25 1 1 Introduction Throughout this paper, only undirected simple graphs without loops or multiple edges are considered. Unlessotherwisestated,wefollowthebookbyBondyandMurty[14]forterminology and notation. Let G be a graph with vertex set V = V(G) and edge set E = E(G). The girth of a graph G is the number g = g(G) of edges in a smallest cycle. For every v V, N (v) denotes the G ∈ neighbourhood of v, i.e. the set of all vertices adjacent to v. The degree of a vertex v V is ∈ the cardinality of N (v). Let S V(G), then we denote by N (S) = N (s) and by G G s∈S G−S ⊂ ∪ N [S] = S N (S). G G ∪ Agraphiscalledregular ifalltheverticeshavethesamedegree. A(k,g)-graph isak-regular graph with girth g. Erd˝os and Sachs [15] proved the existence of (k,g)-graphs for all values of k and g provided that k 2. Since then most work carried out has focused on constructing ≥ a smallest one (cf. e.g. [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 16, 18, 21, 24, 25]). A (k,g)-cage is a k-regular graph with girth g having the smallest possible number of vertices. Cages have been intensely studied since they were introduced by Tutte [28] in 1947. More details about constructions of cages can be found in the recent survey by Exoo and Jajcay [17]. Inthisnoteweareinterestedin(k,8)–cages. Countingthenumberofverticesinthedistance partition with respect to an edge yields the following lower bound on the order of a (k,8)-cage: n (k,8) = 2(1+(k 1)+(k 1)2+(k 1)3). (1) 0 − − − A (k,8)–cage with n (k,8) vertices is called a Moore (k,8)–graph (cf. [14]). These graphs 0 have been constructed as the incidence graphs of generalized quadrangles Q(4,q) and W(q) [12, 17, 27], which are known to exist for q a prime power and k = q +1 and no example is known when k 1 is not a prime power (cf. [11, 13, 19, 22]). Since they are incidence graphs, − these cages are bipartite and have diameter 4. In this note, we present in Definition 2.1 a formulation for obtaining the known (q +1,8)- cages with q 2 a prime power. Then we check in Theorem 2.1 that the graph Γ with such q ≥ a labelling is a (q +1,8)–cage, for each prime power q 2. Finally, we describe in Section 3 ≥ 2 the utility of this equivalent description for the known (q +1,8)–cages, for constructing small (q 1,8)–graphs [6] and (q+1,7)–graphs [4, 5], that give rise to new and better upper bounds. − Furthermore, we also obtain smaller (q,8)-graphs for even prime power q. 2 The formulation We start by presenting the following graph: Definition 2.1 Let F be a finite field with q 2 a prime power and (cid:37) a symbol not belonging q ≥ to F . Let Γ = Γ [W ,W ] be a bipartite graph with vertex sets W = F3 ((cid:37),b,c) ,((cid:37),(cid:37),c) : q q q 0 1 i q i i ∪{ b,c F ((cid:37),(cid:37),(cid:37)) , i = 0,1, and edge set defined as follows: q i ∈ }∪{ } For all a F (cid:37) and for all b,c F : q q ∈ ∪{ } ∈   {(w, aw+b, a2w+2ab+c)0 : w ∈ Fq}∪{((cid:37),a,c)0} if a ∈ Fq; N ((a,b,c) ) = Γq 1   (c,b,w) : w F ((cid:37),(cid:37),c) if a = (cid:37). 0 q 0 { ∈ }∪{ } N (((cid:37),(cid:37),c) ) = ((cid:37),c,w) : w F ((cid:37),(cid:37),(cid:37)) Γq 1 { 0 ∈ q}∪{ 0} N (((cid:37),(cid:37),(cid:37)) ) = ((cid:37),(cid:37),w) : w F ((cid:37),(cid:37),(cid:37)) . Γq 1 { 0 ∈ q}∪{ 0} Or equivalently For all i F (cid:37) and for all j,k F : q q ∈ ∪{ } ∈   {(w, j −wi, w2i−2wj +k)1 : w ∈ Fq}∪{((cid:37),j,i)1} if i ∈ Fq; N ((i,j,k) ) = Γq 0   (j,w,k) : w F ((cid:37),(cid:37),j) if i = (cid:37). 1 q 1 { ∈ }∪{ } N (((cid:37),(cid:37),k) ) = ((cid:37),w,k) : w F ((cid:37),(cid:37),(cid:37)) ; Γq 0 { 1 ∈ q}∪{ 1} N (((cid:37),(cid:37),(cid:37)) ) = ((cid:37),(cid:37),w) : w F ((cid:37),(cid:37),(cid:37)) . Γq 0 { 1 ∈ q}∪{ 1} Notethat(cid:37)isjustasymbolnotbelongingtoF andnoarithmeticaloperationwillbeperformed q with it. 3 Next, we will make use of the following induced subgraph B of Γ . q q Notation 2.1 Let B = B [V ,V ] be a bipartite graph with vertex set V = F3, i = 0,1, and q q 0 1 i q edge set E(B ) defined as follows: q For all a,b,c F : N ((a,b,c) ) = (j, aj +b, a2j +2ab+c) : j F . ∈ q Bq 1 { 0 ∈ q} In order to proceed, we need the following definition of a q-regular bipartite graph H q introduced by Lazebnik, Ustimenko and Woldar [21]. Definition 2.2 [21] Let F be a finite field with q 2. Let H = H [U ,U ] be a bipartite graph q q q 0 1 ≥ with vertex set U = F3, r = 0,1, and edge set E(H ) defined as follows: r q q For all a,b,c F : N ((a,b,c) ) = (w, aw+b, a2w+c) : w F . ∈ q Hq 1 { 0 ∈ q} Lazebnik, Ustimenko and Woldar proved that H given in Definition 2.2 is q-regular, bipar- q tite, of girth 8 and order 2q3. Lemma 2.1 The graph B is isomorphic to the graph H . q q Proof Let H be the bipartite graph from Definition 2.2. Since the map σ : B H defined q q q → by σ((a,b,c) ) = (a,b,2ab+c) and σ((x,y,z) ) = (x,y,z) is an isomorphism, the result holds. 1 1 0 0 In what follows, we will obtain the graph Γ from the graph B adding some new vertices q q and edges. We need a preliminary lemma. Lemma 2.2 Let B be the graph from Notation 2.1. For any given a F , the vertices in the q q ∈ set (a,b,c) : b,c F are mutually at distance at least four. And, for any given i F , the 1 q q { ∈ } ∈ vertices in the set (i,j,k) : j,k F are mutually at distance at least four. 0 q { ∈ } Proof Suppose that there exists a path of length two (a,b,c) (w,j,k) (a,b(cid:48),c(cid:48)) in B . By 1 0 1 q Definition 2.1, j = aw+b = aw+b(cid:48) and k = a2w+2ab+c = a2w+2ab(cid:48)+c(cid:48). From both equation 4 we get b = b(cid:48) and c = c(cid:48) which implies that (a,b,c) = (a,b(cid:48),c(cid:48)) contradicting that the path has 1 1 length two. Similarly suppose that there exists a path of length two (i,j,k) (a,b,c) (i,j(cid:48),k(cid:48)) . 0 1 0 Reasoning similarly, we obtain j = ai+b = j(cid:48), and c = a2i 2aj +k = a2i 2aj(cid:48)+k(cid:48) yielding − − (i,j,k) = (i,j(cid:48),k(cid:48)) which is a contradiction. 0 0 Figure 1 shows a spanning tree of Γ with the vertices labelled according to Definition 2.1. q (̺,̺,̺)1 (̺,̺,̺)0 (̺,̺,0)0 ······ (̺,̺,i)0 (̺,̺,0)1 ······ (̺,̺,a)1 (̺,0,0)1 (̺,j,0)1 (̺,0,i)1 (̺,j,i)1 (̺,0,0)0 (̺,0,c)0 (̺,a,0)0 (̺,a,c)0 ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· (0,0,0)0(0,0,w)0(0,j,0)0(0,j,w)0(i,0,0)0(i,0,w)0(i,j,0)0(i,j,w)0(0,0,0)1(0,t,0)1 (0,0,c)1(0,t,c)1 (a,0,0)1(a,t,0)1(a,0,c)1(a,t,c)1 Figure 1: Spanning tree of Γ . q Theorem 2.1 The graph Γ given in Definition 2.1 is a Moore (q+1,8)-graph for each prime q power q 2. ≥ Proof As a consequence of Lemma 2.2, we obtain the following claim. Claim 1: For all x,y F , the q vertices of the set (x,y,j) : j F are mutually at distance q 0 q ∈ { ∈ } at least 6 in B . q Proof: By Lemma 2.2, the q vertices (x,y,j) : j F are mutually at distance at least 4. 0 q { ∈ } Suppose by contradiction that B contains the following path of length four: q (x,y,j) (a,b,c) (x(cid:48),y(cid:48),j(cid:48)) (a(cid:48),b(cid:48),c(cid:48)) (x,y,j(cid:48)(cid:48)) , for some j(cid:48)(cid:48) = j. 0 1 0 1 0 (cid:54) Then y = ax+b = a(cid:48)x+b(cid:48) and y(cid:48) = ax(cid:48) +b = a(cid:48)x(cid:48) +b(cid:48). It follows that (a a(cid:48))(x x(cid:48)) = 0, − − which is a contradiction since by Lemma 2.2 a = a(cid:48) and x = x(cid:48). (cid:3) (cid:54) (cid:54) Let B(cid:48) = B(cid:48)[V ,V(cid:48)] be the bipartite graph obtained from B = B [V ,V ] by adding q2 new q q 0 1 q q 0 1 vertices to V labeled ((cid:37),b,c) , b,c F (i.e., V(cid:48) = V ((cid:37),b,c) : b,c F ), and new edges 1 1 ∈ q 1 1 ∪{ 1 ∈ q} NBq(cid:48)(((cid:37),b,c)1) = {(c,b,j)0 : j ∈ Fq} (see Figure 1). Then Bq(cid:48) has |V1(cid:48)|+|V0| = 2q3+q2 vertices 5 such that every vertex of V has degree q +1 and every vertex of V(cid:48) has still degree q. Note 0 1 that the girth of B(cid:48) is 8 by Claim 1. Further, Lemma 2.2 partially holds in B(cid:48). We write this q q fact in the following claim. Claim 2: For any given a F (cid:37) , the vertices of the set (a,b,c) : b,c F are mutually q 1 q ∈ ∪{ } { ∈ } at distance at least four in B(cid:48). q Claim 3: For all a F (cid:37) and for all c F , the q vertices of the set (a,t,c) : t F are q q 1 q ∈ ∪{ } ∈ { ∈ } mutually at distance at least 6 in B(cid:48). q Proof: By Claim 2, for all a F (cid:37) the q vertices of (a,t,c) : t F are mutually at q 1 q ∈ ∪{ } { ∈ } distance at least 4 in B(cid:48). Suppose that there exists in B(cid:48) the following path of length four: q q (a,t,c) (x,y,z) (a(cid:48),t(cid:48),c(cid:48)) (x(cid:48),y(cid:48),z(cid:48)) (a,t(cid:48)(cid:48),c) , for some t(cid:48)(cid:48) = t. 1 0 1 0 1 (cid:54) If a = (cid:37), then x = x(cid:48) = c, y = t, y(cid:48) = t(cid:48)(cid:48) and a(cid:48) = (cid:37) by Claim 2. Then y = a(cid:48)x+t(cid:48) = a(cid:48)x(cid:48)+t(cid:48) = y(cid:48) (cid:54) yielding that t = t(cid:48)(cid:48) which is a contradiction. Therefore a = (cid:37). If a(cid:48) = (cid:37), then x = x(cid:48) = c(cid:48) and (cid:54) y = y(cid:48) = t(cid:48). Thus y = ax+t = ax(cid:48)+t(cid:48)(cid:48) = y(cid:48) yielding that t = t(cid:48)(cid:48) which is a contradiction. Hence we may assume that a(cid:48) = (cid:37) and a = a(cid:48) by Claim 2. In this case we have: (cid:54) (cid:54) y = ax+t = a(cid:48)x+t(cid:48); z = a2x+2at+c = a(cid:48)2x+2a(cid:48)t(cid:48)+c(cid:48); y(cid:48) = ax(cid:48)+t(cid:48)(cid:48) = a(cid:48)x(cid:48)+t(cid:48); z(cid:48) = a2x(cid:48)+2at(cid:48)(cid:48)+c = a(cid:48)2x(cid:48)+2a(cid:48)t(cid:48)+c(cid:48). Hence (a a(cid:48))(x x(cid:48)) = t(cid:48)(cid:48) t; (2) − − − (a2 a(cid:48)2)(x x(cid:48)) = 2a(t(cid:48)(cid:48) t). (3) − − − If q is even, (3) leads to x = x(cid:48) and (2) leads to t(cid:48)(cid:48) = t which is a contradiction with our assumption. Thus assume q is odd. If a+a(cid:48) = 0, then (3) gives 2a(t(cid:48)(cid:48) t) = 0, so that a = 0 − yieldingthata(cid:48) = 0(becausea+a(cid:48) = 0)whichisagainacontradiction. Ifa+a(cid:48) = 0, multiplying (cid:54) equation (2) by a+a(cid:48) and subtracting both equations we obtain (2a (a+a(cid:48)))(t(cid:48)(cid:48) t) = 0. − − Then a = a(cid:48) because t(cid:48)(cid:48) = t, which is a contradiction to Claim 2. Therefore, Claim 3 holds. (cid:3) (cid:54) 6 LetB(cid:48)(cid:48) = B(cid:48)(cid:48)[V(cid:48),V(cid:48)]bethegraphobtainedfromB(cid:48) = B(cid:48)[V ,V(cid:48)]byaddingq2+qnewvertices q q 0 1 q q 0 1 to V0 labeled ((cid:37),a,c)0, a Fq (cid:37) , c Fq, and new edges NB(cid:48)(cid:48)(((cid:37),a,c)0) = (a,t,c)1 : t Fq ∈ ∪{ } ∈ q { ∈ } (see Figure 1). Then B(cid:48)(cid:48) has V(cid:48) + V(cid:48) = 2q3 + 2q2 + q vertices such that every vertex has q | 1| | 0| degree q+1 except the new added vertices which have degree q. Moreover the girth of B(cid:48)(cid:48) is 8 q by Claim 3. Claim 4: For all a F (cid:37) , the q vertices of the set ((cid:37),a,j) : j F are mutually at q 0 q ∈ ∪{ } { ∈ } distance at least 6 in B(cid:48)(cid:48). q Proof: Clearly these q vertices are mutually at distance at least 4 in B(cid:48)(cid:48). Suppose that there q exists in B(cid:48)(cid:48) the following path of length four: q ((cid:37),a,j) (a,b,j) (x,y,z) (a,b(cid:48),j(cid:48)) ((cid:37),a,j(cid:48)) , for some j(cid:48) = j. 0 1 0 1 0 (cid:54) If a = (cid:37) then x = j = j(cid:48) which is a contradiction. Therefore a = (cid:37). In this case y = ax+b = (cid:54) ax+b(cid:48) which implies that b = b(cid:48). Hence z = a2x+2ab+j = a2x+2ab(cid:48)+j(cid:48) yielding that j = j(cid:48) which is again a contradiction. (cid:3) Let B(cid:48)(cid:48)(cid:48) = B(cid:48)(cid:48)(cid:48)[V(cid:48),V(cid:48)(cid:48)] be the graph obtained from B(cid:48)(cid:48) by adding q +1 new vertices to V(cid:48) q q 0 1 q 1 labeled ((cid:37),(cid:37),a)1, a Fq (cid:37) , and new edges NB(cid:48)(cid:48)(cid:48)((cid:37),(cid:37),a)1 = ((cid:37),a,c)0 : c Fq , see Figure ∈ ∪{ } q { ∈ } 1. Then B(cid:48)(cid:48)(cid:48) has V(cid:48)(cid:48) + V(cid:48) = 2q3 +2q2 +2q +1 vertices such that every vertex has degree q | 1 | | 0| q + 1 except the new added vertices which have degree q. Moreover the girth of B(cid:48)(cid:48)(cid:48) is 8 by q Claim 4 and clearly these q + 1 new vertices are mutually at distance 6. Finally, the Moore (q+1,8)-graph Γ is obtained by adding to B(cid:48)(cid:48)(cid:48) another new vertex labeled ((cid:37),(cid:37),(cid:37)) and edges q q 0 N (((cid:37),(cid:37),(cid:37)) ) = ((cid:37),(cid:37),i) : i F (cid:37) . Γq 0 { 1 ∈ q ∪{ }} Remark 2.1 A coordinatization of classical generalized quadrangles Q(4,q) and W(q) in four dimensions are discussed in [23, 26, 29]. The formulation of a Moore (q+1,8)-graph given in Theorem 2.1 in three dimensions is equivalent to this coordinatization. 7 3 Applications In this section we overview results that we have obtained for girth 7 in [4, 5] and for girth 8 in [6] using the labelling from Definition 2.1. Moreover, we also use the labelling to construct small (q,8)–graphs, for q even, which we match the bound on the order obtained by [18]. For girth 7, we have obtained the following results: Theorem 3.1 ([4, Theorem 3.5], [5, Theorem 2.4]) Let q 4 be an even prime power. Then, ≥ there is a (q+1)-regular graph of girth 7 and order 2q3+q2+2q. Theorem 3.2 ([4, Theorem 4.7], [5, Theorem 3.4]) Let q 5 be an odd prime power. Then, ≥ there is a (q+1)-regular graph of girth 7 and order 2q3+2q2 q+1. − A subset U V(G) is said to be perfect dominating set of G if for each vertex x V(G) U, ⊂ ∈ \ N (x) U = 1 (cf. [20]). Note that if G is a k-regular graph and U is a perfect dominating G | ∩ | set of G then G U is clearly a (k 1)-regular graph. In [6], using the labelling from Definition − − 2.1, we have obtained perfect dominating sets of (q+1,8)–cages and (q,8)–graphs, for q a prime power, that give rise to the following two theorems: Theorem 3.3 Let q 4 be a prime power. Then, there is a q-regular graph of girth 8 and order ≥ 2q(q2 2). − Theorem 3.4 Let q 4 be a prime power. Then, there is a (q 1)-regular graph of girth 8 and ≥ − order 2q(q 1)2. − The labelling from Definition 2.1 also allows us to improve the result in Theorem 3.3 for even q 4. ≥ Proposition 3.1 Let q 4 be an even prime power and Γ = Γ [V ,V ] the Moore (q +1,8)- q q 0 1 ≥ graph given in Definition 2.1. Let Q = ((cid:37),j,0) : j F ((cid:37),(cid:37),0) and S = (u,u,1+u+ 0 q 0 { ∈ }∪{ } { u2) : u F ((cid:37),1,1) . Then the set 1 q 1 ∈ }∪{ }   (cid:32) (cid:33) (cid:92) (cid:92) NΓq[Q]∪ NΓ2q(a)∪NΓq[S]∪ NΓ2q(b) a∈Q b∈S 8 is a perfect dominating set of Γ of cardinality 2(q2+4q+3) (see Figure 2). q Q (̺,0,0)0 (̺,1,0)0 ... (̺,̺,0)0 (̺,1,1)1 (0,0,1)1 ... (u,u,p(u))1 S ... ... ... ... ... ... ... ... (̺,̺,̺)0 (0,0,0)0 (0,t,0)0 (̺,0,1)1 (0,1,1)1 (u,u+1,p(u))1 Figure 2: Perfect dominating set of Theorem 3.1. Proof First assume that q 8. By Definition 2.1, we have for all element of Q: ≥ N (((cid:37),j,0) ) = (j,t,0) : t F ((cid:37),(cid:37),j) ; Γq 0 { 1 ∈ q}∪{ 1} N (((cid:37),(cid:37),0) ) = ((cid:37),t,0) : t F ((cid:37),(cid:37),(cid:37)) . Γq 0 { 1 ∈ q}∪{ 1} Then ((cid:37),(cid:37),(cid:37)) N2 (((cid:37),j,0) )) N2 (((cid:37),(cid:37),0) )) for all j F . Moreover, since q is even, 0 ∈ Γq 0 ∩ Γq 0 ∈ q 2tj = 0 and N ((j,t,0) ) = (w,jw+t,j2w) : w F ((cid:37),j,0) . Thus, for all j ,j F , Γq 1 { 0 ∈ q}∪{ 0} 1 2 ∈ q j = j , we have (w,j w + t ,j2w) = (w,j w + t ,j2w) if and only if w = 0 and t = t . 1 (cid:54) 2 1 1 1 0 2 2 2 0 1 2 Let I = (cid:84) N2 (a). Then I = (0,t,0) : t F ((cid:37),(cid:37),(cid:37)) (see left side of Figure 2) Q a∈Q Γq Q { 0 ∈ q}∪{ 0} implying that N [Q] + I = (q+1)2+2(q+1). | Γq | | Q| Let p(u) = 1+u+u2 and observe that j = j for all j F since q is even. By Definition q − ∈ 2.1, we have for all element of S: N ((u,u,p(u)) = (a,ua+u,u2a+p(u)) : a F ((cid:37),u,p(u)) ; Γq 1} { 0 ∈ q}∪{ 0} N (((cid:37),1,1) ) = (1,1,a) : a F ((cid:37),(cid:37),1) ; Γq 1 { 0 ∈ q}∪{ 0} N ((a,ua+u,u2a+p(u)) ) = (w,ua+u+wa,w2a+u2a+p(u)) : w F ((cid:37),ua+u,a) . Γq 0 { 1 ∈ q}∪{ 1} N (((cid:37),u,p(u)) ) = (u,y,p(u)) : y F ((cid:37),(cid:37),u) ; Γq 0 { 1 ∈ q}∪{ 1} Then ((cid:37),0,1) N (((cid:37),(cid:37),1) )) N ((1,0,1+u) )) for all u F . Let I = (cid:84) N2 (b), 1 ∈ Γq 0 ∩ Γq 0 ∈ q S b∈S Γq we have ((cid:37),0,1) I (see right side of Figure 2). Moreover, two vertices (a ,a u +u ,a u2+ 1 ∈ S 1 1 1 1 1 1 p(u )) and (a ,a u +u ,a u2+p(u )) for u ,u ,a ,a F such that u = u and a ,a = 1 1 0 2 2 2 2 2 2 2 0 1 2 1 2 ∈ q 1 (cid:54) 2 1 2 (cid:54) have a common neighbor (w,x,y) for some w,x,y F if and only if 1 q ∈ x = u a +u +wa = u a +u +wa (4) 1 1 1 1 2 2 2 2 9 1 and y = u2a +w2a +p(u ) = u2a +w2a +p(u ). (5) 1 1 1 1 2 2 2 2 But (4) holds if and only if x+w = (u +w)(a +1) = (u +w)(a +1). (6) 1 1 2 2 Moreover, (5) holds if and only if y+w2+1 = (u2+w2)(a +1)+u = (u2+w2)(a +1)+u . (7) 1 1 1 2 2 2 Since q is even (u2 +w2) = (u +w)2. Therefore multiplying (6) by u +w and combining it i i 1 with (7) we get u = (u +w)(a +1)(u +u )+u . But u = u gives that (u +w)(a +1) = 1 1 2 2 1 2 2 1 2 2 2 (cid:54) as well as (u +w)(a +1) = 1 by (6). Hence, x = 1+w by (6) and y = 1+w +w2 = p(w) 1 1 by (7). We conclude that I = (u,1+u,p(u)) : u F ((cid:37),0,1) since (u,1+u,p(u)) S 1 q 1 1 { ∈ }∪{ } ∈ N (((cid:37),u,p(u)) ). Thus, N [S] + I = (q+1)2+2(q+1). Γq 0 | Γq | | S| Let D = N [Q] I and D = N [S] I . We get that D D = 2(q+1)2+4(q+1) = Q Γq ∪ Q S Γq ∪ S | Q∪ S| 2(q2+4q+3) since D and D are vertex disjoint for q 8. Q S ≥ Let us show that D D is a perfect dominating set of Γ . Q S q ∪ First, we check that by Definition 2.1, there is a matching joining each vertex of D V Q 1 ∩ with one vertex in D V . We have: S 0 ∩ - For all u F , ((cid:37),(cid:37),u) D is adjacent to ((cid:37),u,p(u)) D , and ((cid:37),(cid:37),(cid:37)) D is q 1 Q 0 S 1 Q ∈ ∈ ∈ ∈ adjacent to ((cid:37),(cid:37),1) D . 0 S ∈ - For all t F , ((cid:37),t,0) D is adjacent to (0,t,p(t)) D . q 1 Q 0 S ∈ ∈ ∈ - For all a F , (a,a,0) D is adjacent to (1,0,a2) N ((cid:37),0,1) D . ∈ q 1 ∈ Q 0 ∈ Γq 1 ⊂ S - For all a F , (a,a+1,0) D is adjacent to (1,1,a2) N ((cid:37),1,1) D . ∈ q 1 ∈ Q 0 ∈ Γq 1 ⊂ S - For all a,t F , t = 0,1, (a,a + t,0) D is adjacent to (x,ax + a + t,a2x) q 1 Q 0 ∈ (cid:54) ∈ ∈ 10