Spectral results on regular graphs with (k, τ)-regular sets

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1 Discrete Mathematics 307 (007) Spectral results on regular graphs with (k, τ)-regular sets Domingos M. Cardoso, Paula Rama Dep. de Matemática, Univ. Aveiro, Aveiro, Portugal Received 5 October 00; received in revised form 15 January 004; accepted 18 November 005 Available online 7 November 006 Abstract A set of vertices S V (G) is (k, τ)-regular if it induces a k-regular subgraph of G such that N G (v) S =τ v / S. Note that a connected graph with more than one edge has a perfect matching if and only if its line graph has a (0, )-regular set. In this paper, some spectral results on the adjacency matrix of graphs with (k, τ)-regular sets are presented. Relations between the combinatorial structure of a p-regular graph with a (k, τ)-regular set and the eigenspace corresponding to each eigenvalue λ / {p, k τ} are deduced. Finally, additional results on the effects of Seidel switching (with respect to a bipartition induced by S) of regular graphs are also introduced. 006 Elsevier B.V. All rights reserved. Keywords: Adjacency matrix; Graph eigenvalues 1. Introduction In this paper we deal with undirected simple graphs, G = (V (G), E(G)), where V (G) denotes the nonempty set of vertices and E(G) the set of edges. It is assumed that G is of order n, i.e., V (G) =n. An element of E(G), which has the vertices i and j as endvertices, is denoted by ij.ifv V (G), then we denote the neighborhood of v by N G (v), that is, N G (v) ={w : vw E(G)}. The number of neighbors of v V (G) will be denoted by d G (v) and called, as usually, degree of v. IfG is such that v V (G) d G (v) = p then we say that G is p-regular. Given a subset of vertices S of a graph G, the vector x R V with x v = 1ifv S and x v = 0ifv/ S is called the characteristic vector of S. Throughout this paper, A G will denote the adjacency matrix of the graph G of order n>1, that is, A G = (a ij ) n n is such that { 1 if ij E(G) a ij = 0 otherwise, and λ min (A G ) will denote the minimum eigenvalue of A G. Note that A G is a symmetric matrix and then has n real eigenvalues. It is well known that if G has at least one edge then λ min (A G ) 1. Actually, λ min (A G ) = 0 if and only if G has no edges, λ min (A G ) = 1 if and only if G has at least one edge and each of its components is complete and otherwise λ min (A G ) [6]. Given a matrix A, σ(a) will denote the set of its eigenvalues. Throughout the text ê will denote the all-ones vector with n components. Supported by Centre for Research in Optimization and Control (CEOC) from the Fundação para a Ciência e a Tecnologia FCT, cofinanced by the European Community Fund FEDER. Corresponding author. addresses: dcardoso@mat.ua.pt (D.M. Cardoso), prama@mat.ua.pt (P. Rama) X/$ - see front matter 006 Elsevier B.V. All rights reserved. doi: /j.disc

2 . Basic properties of graphs with (k, τ)-regular sets and D.M. Cardoso, P. Rama / Discrete Mathematics 307 (007) According to [18] (see also [8]), a (ρ, γ)-set of a graph G is a subset of vertices S V (G) such that N G (v) S ρ if v S N G (v) S γ if v/ S, where ρ and γ are nonempty subsets of {0, 1,...,n}. Herein, when the subsets ρ and γ are singletons, ρ ={k} and γ ={τ}, we say that S is (k, τ)-regular, that is, S V (G) is (k, τ)-regular if and only if it is a ({k}, {τ})-set. Therefore, a (k, τ)-regular set S of a graph G is a subset of vertices which induces in G a k-regular subgraph such that every vertex out of S has τ neighbors in it. Therefore, if G has a (k, τ)-regular set S then in its complement, Ḡ, S is ( S k 1, S τ)- regular. From the above definition it is immediate that for all τ > 0a(0, τ)-regular set of a graph is a maximal stable set. Fig. 1 depicts a graph G where the set of vertices S ={1, 3, 4, 6} is (1, 4)-regular. As an application of (k, τ)-regular sets, note that according to [3] a connected graph has a perfect matching if and only if its line graph has a (0, )-regular set (related results can be found in [1,]). The following proposition is a generalization of a result introduced in [3]. Proposition 1. Let G be a graph with a (k 1, τ 1 )-regular set S 1 and a (k, τ )-regular set S. Then τ S 1 \S τ 1 S \S 1 = S 1 S (k 1 k ). (1) Proof. The number of edges xy with x S 1 and y S is τ 1 S \S 1 +k 1 S 1 S =τ S 1 \S +k S S 1 and then the proof follows. Note that if we consider k 1 = k or S 1 S = then (1) becomes τ S 1 \S =τ 1 S \S 1 and thus if τ = 0 then S 1 \S =(τ 1 /τ ) S \S 1. On the other hand, for τ 1 =τ =τ = 0, (1) becomes S 1 \S S \S 1 =(k 1 k )/τ S 1 S. Therefore we have the following corollary. Corollary 1. Let G be a graph with a (k 1, τ 1 )-regular set S 1 and a (k, τ )-regular set S, such that τ > 0. (a) If k 1 = k or S 1 S = then S 1 \S =(τ 1 /τ ) S \S 1. (b) If τ 1 = τ = τ then S 1 \S = S \S 1 +(k 1 k )/τ S 1 S. Fig. represents a graph with the (0, 1)-regular set S 1 ={, 5}, the (0, )-regular set S ={1, 3, 5} and the (1, )-regular set S 3 ={1, 3, 4, 6}. Then, according to Corollary 1 (part (a)), {} = S 1 \S = τ 1 S \S 1 = 1 {1, 3}, τ Fig. 1. Graph with the (1, 4)-regular set {1, 3, 4, 6}.

3 1308 D.M. Cardoso, P. Rama / Discrete Mathematics 307 (007) Fig.. Graph with the (0, 1)-regular set S 1 ={, 5}, the (0, )-regular set S ={1, 3, 5} and the (1, )-regular set S 3 ={1, 3, 4, 6}. and according to Corollary 1 (part (b)), {5} = S \S 3 = S 3 \S + k 1 k τ S S 3 = {4, 6} {1, 3}. Given a graph G, each (0, 1)-regular set L(IM) of the line graph L(G) defines an induced matching IM of G for which every edge of G belongs to the induced matching IM or has a common vertex with one and only one edge of IM. We designate such induced matching by perfect induced matching. Thus, if G has a perfect matching M E(G) and a perfect induced matching IM E(G) then, applying Corollary 1 (part (a)) on L(G), we may conclude that n/ = M M\IM = IM\M and hence IM\M n/4. 3. Eigenvalues and eigenspaces of graphs with (k, τ)-regular sets From now on, it is assumed that every (k, τ)-regular set is such that τ > 0 and the kernel (or null space) of a linear map, defined by a matrix A, is denoted by Ker(A). From [11] we have the following proposition, where (as assumed in the beginning of the paper) the graph has order n. [ Proposition ] (Neumaier [11]). IfGisap-regular graph with a (k, τ)-regular set S then k τ σ(a G ) and v = (p k)ê1 τê is an eigenvector corresponding to k τ, with ê 1 and ê being the all-ones vectors of dimension S and n S, respectively. It is obvious that the eigenvector v of Proposition is not orthogonal to the characteristic vector of the (k, τ)-regular set S. In the next proposition we conclude that for an eigenvalue λ σ(a G ), such that k τ = λ = p, its eigenspace is orthogonal to the characteristic vector of S. Proposition 3. Let G be a p-regular graph with a (k, τ)-regular set S. Then λ σ(a G ), such that k τ = λ = p, and v Ker(A G λi n ), v i = 0 and v i = 0. i S i/ S Proof. Let x be the characteristic vector of a (k, τ)-regular set S of G and let u be an eigenvector of λ σ(a G ), such that k τ = λ = p. Then A G u = λu which leads to x A G u = λ x u N G (i) S u i = λ u i i S i V (G) i S N G (i) S u i + i/ S N G (i) S u i = λ i S u i k i S u i + τ i/ S u i = λ i S u i (λ k) i S u i = τ(ê u i S u i ) (λ k + τ) i S u i = 0. As λ = k τ we have i S u i = 0 and, by regularity of G, ê u = 0, so we also have i/ S u i = 0.

4 D.M. Cardoso, P. Rama / Discrete Mathematics 307 (007) Fig. 3. A 4-regular graph G with (0, )-regular, (, 4)-regular and (, )-regular sets. Fig. 4. A 3-regular graph G with the (1, 1)-regular set {3, 6}. According to the above proposition we may conclude that when G is a p-regular graph, λ σ(a G ), such that k τ = λ = p, every eigenvector in Ker(A G λi n ) is orthogonal to the characteristic vector of any (k, τ)-regular set. Therefore, it provides information which could lead us to find (k, τ)-regular sets. Fig. 3 shows a 4-regular graph G, for which the spectrum is σ(a G )={[ ], [0] 3, [4] 1 }, 1 the corresponding columns of eigenvectors are and all (0, )-regular sets are maximum stable sets. Notice that from Proposition 3 and the fact that ê is an eigenvector of A G corresponding to the eigenvalue p (when G is p-regular) also follows the statement that k τ σ(a G ) and that its eigenspace is not orthogonal to the characteristic vector x of a (k, τ)-regular set, otherwise x would be orthogonal to all eigenvectors of A G but ê and thus x = μê, with μ = 0, which is a contradiction. As a direct consequence of Propositions and 3, considering that G is a p-regular graph with a (k, τ)-regular set S, we may conclude the following: k τ and p are the only eigenvalues with eigenspaces not orthogonal to the characteristic vector of S. If k = τ then 0 σ(a G ). If k τ then λ min (A G ) = k τ and thus the eigenspace of λ min (A G ) is orthogonal to the characteristic vector of the (k, τ)-regular set S. If G is strongly regular then its three distinct eigenvalues are integers (a similar conclusion was obtained in [11]). Fig. 4 represents a 3-regular graph with a (1, 1)-regular set S ={3, 6}, but with λ min (A G ) = = k τ. Note that from Proposition we may conclude that 0 σ(a G ) = {[ ], [0], [1] 1, [3] 1 } and the eigenspace corresponding to λ min (A G ) = is orthogonal to the characteristic vector of S. Fig. 5 illustrates the construction of an infinite. 1 Notice that [λ i ] m i means that the eigenvalue λ i has multiplicity m i. The eigenspace corresponding to is spanned by {( 1, 1, 1, 0, 0, 1), (1, 1, 1,,, 1)}

5 1310 D.M. Cardoso, P. Rama / Discrete Mathematics 307 (007) Fig. 5. Graphs (a) G with, 0, σ(a G ), (b) G 3 with 3, 1, 1, 3 σ(a G3 ) and (c) G 4 with 4,,, 4 σ(a G4 ). family of p-regular graphs of order p with (0,p)-regular and (p 1, 1)-regular sets. Considering the graphs of Fig. 5, the -regular graph G has the (0, )-regular set {, 3} and the (1, 1)-regular set {1, } (and then, 0, σ(a G )). The 3-regular graph G 3 has the (, 1)-regular set {1,, 3, 4} and the (0, 3)-regular set {, 3, 6, 7} (and then 3, 1, 1, 3 σ(a G3 )). The 4-regular graph G 4 has the (3, 1)-regular set {1,, 3, 4, 5, 6, 7, 8} and the (0, 4)-regular set {, 3, 6, 7, 10, 11, 14, 15} (and then 4,,, 4 σ(a G4 )). Proposition 4. Let G be a graph with two subsets of vertices S and S (S = S) such that S = S and both subsets induce k-regular subgraphs. Then λ min (A G ) S\ S k i S\ S N G(i) S. () S\ S Proof. Let x and x be the characteristic vectors of the two subsets of vertices S and S, respectively. Denoting the symmetric difference between S and S by S S, that is, S S = (S\ S) ( S\S), then as S and S have the same cardinality we may easily deduce that S S = S\ S = S\S. On the other hand, we know that and also { y } A G y λ min (A G ) = inf y : y = 0, (3) (x x) A G (x x) x x Furthermore, { k, i S, (A G x) i = N G (i) S, i / S = x A G x + x A G x x A G x. (4) S S and therefore x A G x = k S and x A G x = k S =k S. Finally, x A G x = k S S + N G (i) S =k S S + N G (i) S. i S\ S i S\S

6 D.M. Cardoso, P. Rama / Discrete Mathematics 307 (007) Proceeding with the development started in (4), (x x) A G (x x) x x = S k S S k i S\ S N G(i) S S\ S = S\ S k i S\ S N G(i) S. (5) S\ S Finally, from (3) and (5), λ min (A G ) S\ S k i S\ S N G(i) S. S\ S It was referred above that for a regular graph G with a (0, τ)-regular set, τ σ(a G ). In the next corollary, assuming the existence of a (0, τ)-regular set not equal to some stable set with the same cardinality, in the absence of regularity, we state that τ is still an upper bound for λ min (A G ). Corollary. Let G be a graph with two stable sets S and S (S = S) such that S = S and S is a (0, τ)-regular set. Then i S\ S λ min (A G ) N G(i) S = τ. S\ S Proof. Since S is a (0, τ)-regular set and S is another stable set of G such that S = S, then in () we have N G (v) S =τ, v S\ S and k = 0 which yields i S\ S λ min (A G ) N G(i) S = S\ S τ S\ S S\ S = τ. 4. Bipartitions with (k, τ)-regular sets and Seidel switching Given the graph G, a partition π=(v 1,...,V r ) of V (G) is such that its elements V j, for j =1,...,r, are nonempty, pairwise disjoint and r j=1 V j =V (G).A partition π=(v 1,...,V r ) of V (G) is equitable if for any pair i, j {1,...,r} and v V i the numbers m ij = N G (v) V j depend only on i and j, i.e, the number of neighbors which a vertex in V i has in V j is independent of the choice of the vertex in V i. For our purpose, we will consider equitable partitions of G with two subsets, π = (V 1,V ), which will be called equitable bipartitions. Equitable partitions were introduced in [13 15]. In the later one equitable bipartitions are used to obtain information about eigenvalues and eigenvectors of graphs. Equitable partitions appear related with automorphism groups of graphs [9], walk partitions and colorations [1], distance-regular graphs and covering graphs (see [7] for details and further references). Recent applications of equitable partitions may be found in [10] applied to the study of graphs with three eigenvalues and in [17] applied to the study of correlation structure of landscapes. A digraph D is a graph where the set of edges is replaced by a set of arcs, that is, a set of ordered pairs of vertices. If a digraph D has multiple arcs or loops then we say that D is a multi-digraph. The quotient graph G/π of G with respect to the equitable bipartition π is a multi-digraph with the subsets of π (V 1 and V ) as its vertices and with m ij arcs going from V i to V j. Thus G/π has, in general, both loops and multiple arcs. The adjacency matrix A G/π is the matrix with ij entries equal to m ij, [ ] m11 m A G/π = 1. m 1 m From the above definitions we may conclude that a bipartition of a graph H, π = (S, V (H )\S), is equitable if and only if S is (m 11,m 1 )-regular and V(H)\S is (m,m 1 )-regular.

7 131 D.M. Cardoso, P. Rama / Discrete Mathematics 307 (007) Fig. 6. Graphs G and G π with π = ({, 5}, {1, 3, 4, 6}). If G is[ a p-regular ] graph and S V (G) is (k, τ)-regular then π = (S, V (G)\S) is an equitable bipartition such that A G/π = k p k τ p τ. In particular, if S V (G) is a stable set such that N G (v) S =τ v / S, i.e., S is a (0, τ)-regular set, then G has an equitable bipartition π = (S, V (G)\S) with [ ] 0 p A G/π =. τ p τ Conversely, let G be a graph with an equitable bipartition π = (V 1,V ) such that V 1 is (k 1, τ 1 )-regular and V is (k, τ )-regular. Since d G (v 1 ) = k 1 + τ for v 1 V 1 and d G (v ) = k + τ 1 for v V then G is a regular graph if and only if k 1 k = τ 1 τ. Furthermore, from π = (V 1,V ) being a partition, we have that V 1 V =, V 1 + V =n and (1) turns into τ V 1 τ 1 V =0 τ (n V ) = τ 1 V V = nτ τ 1 + τ and hence V 1 =nτ 1 /(τ 1 + τ ). Next we will present the operation of the switching of edges which was introduced by Seidel [16]. Let us consider that each edge uv of a graph G corresponds to the set of arcs {(u, v), (v, u)}. Then, considering the bipartition π ={V 1,V } of the set of vertices of G, the graph obtained by switching in G with respect to the bipartition π, which will be denoted by G π, may be defined such that E(G π ) = E 11 ((V 1 V )\E 1 ) ((V V 1 )\E 1 ) E, and E ij = E(G) (V i V j ). The graph G π will be called the switching graph of G relative to the bipartition π. According to the above definition, it is obvious that π is an equitable bipartition of G if and only if π is an equitable bipartition of G π. Fig. 6 represents a pair of graphs which are the switching of each other with respect to an equitable bipartition. The following proposition includes a generalization of a result introduced in [3]. It must be noted that, as stated above (in the beginning of section 3), every (k, τ)-regular set considered here is such that τ > 0. Proposition 5. Let G be a p-regular graph with a (k, τ)-regular set S and π=(s, V (G)\S) the corresponding equitable bipartition. Then nτ S = (6) p k + τ and n p k + τ. (7) Furthermore, G π is a regular graph if and only if n = (p + S k τ), (8) and if p = k + τ, G π is a regular graph if and only if S =τ. (9) Proof. Proof of (6): If G is a p-regular graph and S is a (k, τ)-regular set of G then S (p k) = (n S )τ nτ = S (p k + τ) S = nτ p k + τ.

8 Proof of (7): From S τ > 0 and using (6) we get D.M. Cardoso, P. Rama / Discrete Mathematics 307 (007) n( S τ) = n S +nτ = n S + S (p k + τ) = S (p k + τ n) which yields (7), n p k + τ. Proof of (8): Consider graph G π and the adjacency matrix of G/π, [ ] k p k A G/π =. τ p τ For v S, N G π(v) S =k and N G π(v) (V (G)\S) =n S p + k, so d G π(v) = n S p + k. For v V (G)\S, N G π(v) S = S τ and N G π(v) (V (G)\S) =p τ, hence it follows that d G π(v) = S +p τ. Consequently, G π is regular if and only if n S p + k = S +p τ n = (p + S k τ). (10) Proof of (9): If G is regular and p = k + τ, then taking into account (6) and (8) it follows that S = (p + S k τ)τ p k + τ S (p k) + S τ = (p k τ)τ + S τ S (p k τ) = (p k τ)τ S =τ. Conversely, suppose that S =τ. Then, from (6) S (p k + τ) = nτ n = (p k + τ) n = (p k + S τ) and from (8) G π is regular. Theorem.4.9 of [4] states that if p is the degree of regularity of G π then p n/ σ(a G ). This statement can also be obtained from (10) taking into account that p = n S p + k = S +p τ, which leads to k = (p n + S +p)/ and τ = (p S p)/. Therefore, from k τ σ(a G ), it follows that p n/ = k τ is an eigenvalue of A G. From (8) we may conclude, as in [5, Corollary.4.10], that a graph with an odd number of vertices cannot be switched into a regular graph. Considering the graphs G and G π, by [5] σ(a G/π ) σ(a G ) and by [10] σ(a G π) = { σ(a G ) σ(a G π /π) } \σ(a G/π ), where the union and difference set operations are done taking into account the multiplicities of the eigenvalues. Then we may conclude that σ(a G π) = (σ(a G )\σ(a G/π )) σ(a G π /π). (11) For the next proposition we need some information about the spectra of G/π and G π /π in order to use (11). Lemma 1. LetGbeap-regular graph with a (k, τ)-regular set S and π = (S, V (G)\S) the corresponding equitable bipartition. Then σ(a G/π ) ={k τ,p} and p + k τ ± σ(a G π /π) = (p k + τ) + 4 S (n S p + k + τ) 4nτ.

9 1314 D.M. Cardoso, P. Rama / Discrete Mathematics 307 (007) Proof. The proof comes directly from the computation of the spectra of the matrices [ ] [ ] k p k k n S p + k A G/π = and A τ p τ G π /π =. S τ p τ The statement of Proposition that k τ σ(a G ) could be obtained applying Lemma 1 and the inclusion σ(a G/π ) σ(a G ) [5]. Proposition 6. LetGbeap-regular graph with a (k, τ)-regular set S and π = (S, V (G)\S) an equitable bipartition. Then λ min (A G π /π) k τ S n p + k n (p k + τ) (1) and if λ min (A G π /π)>k τ, p k + τ n<(p k + τ). (13) Furthermore, if λ min (A G π /π) k τ and λ min (A G ) = k τ then λ min (A G π) = λ min (A G π /π). Proof. Take into account Proposition, Lemma 1, (11) and the hypothesis σ(a G π) = ({..., k τ,...,p}\{k τ,p}) p + k τ ± (p k + τ) + 4 S (n S p + k + τ) 4nτ. (14) Let us prove (1). If λ min (A G π /π) k τ, then from Lemma 1, p + k τ (p k + τ) + 4 S (n S p + k + τ) 4nτ k τ, that is, (p k + τ) + 4 S (n S p + k + τ) 4nτ p k + τ (p k + τ) + 4 S (n S p + k + τ) 4nτ (p k + τ) S (n S p + k + τ) nτ, which by (6) is equivalent to S (n S p + k + τ) S (p k + τ) S n p + k. By (6) and the last inequality, nτ n p + k p k + τ nτ n(p k + τ) (p k)(p k + τ) n(p k) (p k)(p k + τ) 0 n (p k + τ). Let us prove (13). From the proof of (1), we have λ min (A G π /π)>k τ S >n p + k and n<(p k + τ). Then, from the last inequality and from (7), we conclude that p k + τ n<(p k + τ). Finally, if λ min (A G π /π) k τ and λ min (A G ) = k τ then, from (14), λ min (A G π) = λ min (A G π /π).

10 D.M. Cardoso, P. Rama / Discrete Mathematics 307 (007) Proposition 7. LetGbeap-regular graph with a (k, τ)-regular set S, such that p = k + τ and π = (S, V (G)\S) the corresponding equitable bipartition. If G π is a regular graph then (1) Sisa(k, τ)-regular set of G π. () σ(a G π) = σ(a G ). Proof. (1) If S is a (k, τ)-regular set of G then S is a (k, S τ)-regular set of G π. In addition, if G π is regular, then from Proposition 5, S =τ and hence S is a (k, τ)-regular set of G π. () From the regularity of G and k τ σ(a G ), it follows that σ(a G ) ={...,k τ,...,p}. On the other hand, from (11) and Lemma 1 we have σ(a G π) = (σ(a G )\σ(a G/π )) σ(a G π /π), σ(a G/π ) ={k τ,p}, p + k τ ± (p k + τ) + 4 S (n S p + k + τ) 4nτ σ(a G π /π) =. Taking into account the regularity of G π, then from Proposition 5, S =τ and n = (p + S k τ) = (p k + τ). Thus, accordingly, replacing S and n in σ(a G π /π) leads to σ(a G π /π) = = p + k τ ± p + k τ ± ={k τ,p}. (p k + τ) + 8τ(p k + τ) 8τ(p k + τ) (p k + τ) { } p + k τ ± (p k + τ) = Consequently, we conclude that σ(a G π) = σ(a G ). As an immediate consequence of this proposition, we may conclude that given a p-regular graph G, with a (k, τ)- regular set S such that p = k + τ and π = (S, V (G)\S) the corresponding equitable bipartition, if the switching graph G π is regular, then it is p-regular. It must be noted that combining the corollary of Lemma 6.6 in [4] (two regular connected graphs of the same degree are cospectral if and only if their ( 1, 1, 0)-adjacency matrices are cospectral) with Corollary 3.3 in [16] (switching equivalent graphs have the same ( 1, 1, 0)-adjacency eigenvalues), we may conclude that if a connected regular graph has a connected regular switching graph with the same degree, then they have the same spectrum. The difference between this conclusion and Proposition 7 (part ()) is that in the Proposition 7 we do not need to impose the connectivity on both graphs. On the other hand, the statement that both graphs have the same degree is a consequence of Proposition 7 (part ()). However, we need to impose that p = k + τ. For instance, using the notation of [4], we may apply Proposition 7 to the 11-regular graph G=(G 1 G 3 )+(G G 4 ), with the (3, 6)-regular set S = V(G 3 ) V(G 4 ), where G 1, G, G 3 and G 4 are the graphs depicted in Fig. 7. It must be noted that, as in [4], given the graphs H 1 and H, V(H 1 H ) = V(H 1 + H ) = V(H 1 ) V(H ), E(H 1 + H ) = E(H 1 ) E(H ) and E(H 1 H ) = E(H 1 ) E(H ) {ij : i V(H 1 ), j V(H )}. Consider the above referred graph G and the equitable bipartition π = (S, V (G)\S). Therefore, since 11 = p = k + τ = 9 and G π is regular, by Proposition 7, S is a (3, 6)-regular set of G π and σ(a G π) = σ(a G ). Note that the graph G is nonisomorphic to G π = (G 1 G 4 ) + (G G 3 ), σ(a G ) = σ(a G1 G 3 ) σ(a G G 4 ), with σ(a G1 G 3 ) = {[ 3] 1, [.4], [ ], [ 1], [0], [0.4], [1], [11] 1 } and σ(a G G 4 ) = {[ 3] 3, [ 1.6], [0] 6, [0.6], [11] 1 }, and σ(a G π) = σ(a G1 G 4 ) σ(a G G 3 ), with σ(a G1 G 4 ) = {[ 3], [.4], [ 1], [0] 4, [0.4], [1] 1, [11] 1 }and σ(a G G 3 ) = {[ 3], [ ], [ 1.6], [0] 4, [0.6], [1] 1, [11] 1 }.

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