MATCH Communications in Mathematical and in Computer Chemistry
MATCH Commun. Math. Comput. Chem. 61 (2009) 419-442
ISSN 0340 - 6253
BIPARTITE BICYCLIC GRAPHS WITH LARGE ENERGIES Yong Yang and Bo Zhou∗ Department of Mathematics, South China Normal University, Guangzhou 510631, P. R. China (Received May 14, 2008)
Abstract The energy of a graph is defined as the sum of the absolute values of its eigenvalues. Let B(n) be the class of bipartite bicyclic graphs on n ≥ 6 vertices containing at least one pendent vertex. We determine the graphs with maximal energies in the class of graphs in B(n) of exactly three cycles, in the class of graphs in B(n) of exactly two cycles with a common vertex, and in the class of graphs in B(n) of two vertex-disjoint cycles, respectively.
1. INTRODUCTION Let G be a simple graph on n vertices and A(G) its adjacency matrix. The characteristic polynomial of G is the characteristic polynomial of A(G), denoted by φ(G, λ). The roots of the equation φ(G, λ) = 0, denoted by λ1 , λ2 , . . . , λn , are called the eigenvalues of G [1]. The energy of G is defined as [2] E(G) =
n
|λi |.
i=1
In theoretical chemistry, the energy of a graph has been extensively studied since it can be used to approximate the total π-electron energy of the molecule [2–5]. ∗
Corresponding author. E-mail:
[email protected]
- 420 Let G be a bipartite graph (depicting an alternant structure) on n vertices. Then [1, 3] n/2
φ(G, λ) =
(−1)i b2i (G)λn−2i ,
i=0
where b2i (G) ≥ 0 for all 0 ≤ i ≤ n/2. A graph whose components are cycles and/or complete graphs with two vertices is called a Sachs graph. Sachs theorem [1, 3] states that for 1 ≤ i ≤ n/2, b2i (G) = (−1)i
(−1)p(S) 2c(S) ,
S∈L2i
where L2i denotes the set of Sachs graphs of G with 2i vertices, p(S) denotes the number of components of S and c(S) denotes the number of cycles contained in S. In addition, b0 (G) = 1. Obviously, b2 (G) equals the number of edges of G. For convenience, let b2i (G) = 0 if i < 0 or i > n/2. It is known that E(G) can be expressed as the Coulson integral formula (see [3]) ⎤ ⎡ n/2 2 +∞ 1 2i E(G) = log ⎣ b2i (G)x ⎦dx. π 0 x2 i=0
(1)
A quasi-order relation can be introduced in the class of all bipartite graphs: Let G1 and G2 be two bipartite graphs. If b2i (G1 ) ≤ b2i (G2 ) for all i ≥ 1, then we write G1 G2 . If G1 G2 and there exists a k such that bk (G1 ) < bk (G2 ), then we write G1 ≺ G2 . From (1), we have the following increasing property on energy: G1 ≺ G2 ⇒ E(G1 ) < E(G2 ).
(2)
Thus, the coefficients in φ(G, λ) can be used to establish an ordering of bipartite graphs according to their energies. A connected graph with n vertices and n edges is called a unicyclic graph, and a connected graph with n vertices and n + 1 edges is called a bicyclic graph. Gutman [6] determined the trees with minimal, second–minimal, third–minimal, and fourth–minimal energies, as well as the trees with maximal and second–maximal energies. Since then, graphs with minimal or maximal energies for many classes of graphs have been determined. For minimal energies, see [7–23]. Zhang et al. [24] determined hexagonal chains with maximal energy. Hou et al. [25] determined the
- 421 bipartite unicyclic graphs with maximal energy. On the basis of the work in [26], Gutman et al. [27] determined the bipartite unicyclic graphs with second–maximal and third–maximal energies. Li and Zhang [28] showed that the graph obtained by attaching hexagons to the end vertices of a path on n − 12 vertices has the maximal energy among all n-vertex bipartite bicyclic graphs on n ≥ 16 with the exception of the graphs obtained by connecting two vertex-disjoint cycles whose lengths are at least 10 and congruent 2 modulo 4 by an edge. By appropriate computer-based investigations, Furtula et al. [29] showed that this result holds for all n-vertex bipartite bicyclic graphs with n > 12. Related work on trees with third–maximal and fourth– maximal energies may be found in [30, 31]. Let B(n) be the class of bipartite bicyclic graphs on n vertices containing at least one pendent vertex (i.e., vertex of degree 1). Denote by B1 (n), B2 (n) and B3 (n) respectively the class of graphs in B(n) with exactly three cycles, the class of graphs in B(n) with exactly two cycles of a common vertex, and the class of graphs in B(n) with two vertex-disjoint cycles. Note that the extremal graph in [28, 29] has no pendent vertices. In this paper, we study maximal energies of bicyclic graphs with at least one pendent vertex. We show that in the three classes B1 (n) (n ≥ 9), B2 (n) (n ≥ 12) and B3 (n) (n ≥ 15), the graphs Bn1 , Bn2 and Bn3 are the unique graphs with maximal energies, respectively, 2 3 and B13 are where the graphs Bn1 , Bn2 and Bn3 are shown in Fig. 1. Obviously, B81 , B11
bipartite bicyclic graphs with no pendent vertices, and that Bn1 (n ≥ 9), Bn2 (n ≥ 12) and Bn3 (n ≥ 14) have exactly one pendent vertex. s s s@ @s s s q q q s s s s s @ @s
Bn1
s s @ @s s s q q q s s s @ @s s @s s @ s s @ @s Bn2
s s @ s @s s s q q q s s @ @s s s s @ @s s s @ @s Bn3
Figure 1: Graphs with maximal energies in B1 (n) (n ≥ 9), B2 (n) (n ≥ 12) and B3 (n) (n ≥ 15).
- 422 2. PRELIMINARIES For two graphs G and H, G = H means G and H are not isomorphic. For v ∈ V (G), ΓG (v) denotes the set of neighbors of v in G and the degree of v is degG (v) = |ΓG (v)|. Denote by Vp (G) the set of pendent vertices in G. Let |G| = |V (G)|. Let Pn and Cn be respectively the path and cycle on n vertices. The maximum subgraph with no pendent vertices of a graph G is called the base 3 If v ∈ V (G) 3 and degG (v) > deg 3 (v), then v is called a graph of G, denoted by G. G branching vertex of G. For a graph G, denote by dG (u, v) the distance (length of a shortest path) between vertices u and v in G. For a vertex u and a subgraph H in G, let dG (u, H) = min{dG (u, v)|v ∈ V (H)}. Obviously, if u ∈ V (H) then dG (u, H) = 0. Let C(uv) be the set of cycles containing edge uv in G. Then [1, 3] φ(G, λ) = " φ(G − uv, λ) − φ(G − u − v, λ) − 2 C∈C(uv) φ(G − C, λ), from which the following lemma follows easily. Lemma 2.1. Let G be a bipartite graph. (i) If uv is a bridge in G, then b2i (G) = b2i (G − uv) + b2i−2 (G − u − v), in particular, if u is a pendent vertex, adjacent to v, then b2i (G) = b2i (G − u) + b2i−2 (G − u − v). (ii) If uv is an edge on some cycle, then b2i (G) = b2i (G − uv) + b2i−2 (G − u − v) − 2
l
(−1) 2 b2i−l (G − Cl ).
Cl ∈C(uv)
¿From Lemma 2.1 (i), we have Lemma 2.2. Let G be a bipartite graph. If H is obtained from G by deleting some bridges and/or pendent vertices, then H G. For 1 ≤ t < n, let Pn,k,t be the tree obtained by attaching a path Pt to vertex vk of the path Pn−t labeled consecutively by v1 , v2 , . . . , vn−t . Let Pnl be the unicyclic
- 423 graph obtained by attaching a path Pn−l to a vertex of the cycle Cl for l < n. Let l Pn,k,t be the unicyclic graph obtained by identifying a vertex of a cycle Cl and vertex
vn−l−t+1 of the tree Pn−l+1,k,t . For convenience, let Qn = Pn,3,2 = Pn,3,n−5 for n ≥ 6 ∗ and Pll = Cl . Let Bn,a,b be the graph obtained by connecting a vertex in Ca and a
vertex in Cb by a path when n ≥ a + b and the graph consisting of two cycles Ca and ∗ . Cb with a common vertex when n = a + b − 1. For convenience, let Bn∗ = Bn,6,6
Lemma 2.3. [3] Let T be an acyclic graph on n ≥ 6 vertices and T = Pn , Qn . Then T ≺ Qn ≺ Pn . Lemma 2.4. [25] Let G be a bipartite graph on n vertices containing a unique cycle of length l with l ≡ 2 (mod 4). Then G Pnl . Lemma 2.5. [25] Let G be a bipartite graph on n vertices containing a unique cycle and G = Cn . Then G Pn6 . Lemma 2.6. [3] Let n = 4k, 4k + 1, 4k + 2 or 4k + 3. Then P1 ∪ Pn−1 P3 ∪ Pn−3 · · · P2k−1 ∪ Pn−2k+1 P2k+1 ∪ Pn−2k−1 P2k ∪ Pn−2k · · · P4 ∪ Pn−4 P2 ∪ Pn−2 . 6 6 P4 ∪ Pn−4 . Lemma 2.7. (i) For n ≥ 11, 1 ≤ t ≤ n − 7 and t = 2, n − 8, Pt ∪ Pn−t 6 (ii) For n ≥ 8 and 1 ≤ t ≤ n − 7, Pt ∪ Pn−t ≺ Pn−6 ∪ C6 .
(iii) For n ≥ 8, Pn ≺ Pn−6 ∪ C6 . Proof. ¿From Lemma 2.1 (ii), 6 ) = b2i (Ps ∪ Pn−s ) + b2i−2 (P4 ∪ Ps ∪ Pn−s−6 ) + 2b2i−6 (Ps ∪ Pn−s−6 ), b2i (Ps ∪ Pn−s
where 1 ≤ s ≤ n−7. By Lemma 2.6, Pt ∪Pn−t P4 ∪Pn−4 and Pt ∪Pn−t−6 P4 ∪Pn−10 . 6 6 P4 ∪ Pn−4 . This proves (i) Thus Pt ∪ Pn−t 6 The result in (ii) is true for n = 8 by Sachs theorem. For n ≥ 9, since Pt ∪ Pn−t 6 6 , by similar arguments as above, we need only to show that P2 ∪ Pn−2 ≺ P2 ∪ Pn−2
Pn−6 ∪ C6 . By Lemma 2.1 (i), 6 b2i (P2 ∪ Pn−2 ) = b2i (P2 ∪ Pn−8 ∪ C6 ) + b2i−2 (P2 ∪ P5 ∪ Pn−9 ),
b2i (Pn−6 ∪ C6 ) = b2i (P2 ∪ Pn−8 ∪ C6 ) + b2i−2 (P1 ∪ Pn−9 ∪ C6 ).
- 424 6 By Sachs theorem, P2 ∪ P5 ≺ P1 ∪ C6 . Thus P2 ∪ Pn−2 ≺ Pn−6 ∪ C6 . This proves (ii).
The result in (iii) is true for n = 8, 9 by Sachs theorem. For n ≥ 10, by Lemmas 2.1 and 2.6, b2i (Pn ) = b2i (P6 ∪ Pn−6 ) + b2i−2 (P5 ∪ Pn−7 ) ≤ b2i (P6 ∪ Pn−6 ) + b2i−2 (P4 ∪ Pn−6 ) = b2i (Pn−6 ∪ C6 ) − 2b2i−6 (Pn−6 ). Thus b2i (Pn ) ≤ b2i (Pn−6 ∪ C6 ) and b6 (Pn ) < b6 (Pn−6 ∪ C6 ), implying Pn ≺ Pn−6 ∪ C6 . This proves (iii). By Lemmas 2.1 (ii) and 2.7 (ii), we have 6 6 Pn−6 ∪ C6 . Lemma 2.8. For 6 ≤ t ≤ n − 6, Pt6 ∪ Pn−t
By Lemma 2.1 (i), we have Lemma 2.9. Let G, G be two bipartite graphs and u (resp. u ) a pendent vertex, adjacent to v (resp. v ), of the graph G (resp. G ). If G − u G − u and G − u − v ≺ G − u − v , or G − u ≺ G − u and G − u − v G − u − v , then G ≺ G . 3. GRAPHS IN B1 (n) WITH MAXIMAL ENERGY By Lemma 2.1 (ii), it is easily seen that Pn ≺ Pn6 for n ≥ 6 and Pn6 ≺ Bn1 for n ≥ 8. 3 < 8 and n ≥ 8. Then G ≺ B 1 . Lemma 3.1. Let G ∈ B1 (n), where |G| n Proof. We prove the lemma by induction on n. 3 with deg 3 (z) = 3 and w ∈ Γ 3 (z) Suppose that n = 8. Take a vertex z ∈ V (G) G G such that G − z − w = P6 . Obviously, zw lies on two cycles, say Cb and Cc with b ≤ c. By Lemmas 2.1 (ii), 2.3 and 2.5, b2i (G) = b2i (G − zw) + b2i−2 (G − z − w) − 2
"
l
l∈{b,c}
(−1) 2 b2i−l (G − Cl )
≤ b2i (P86 ) + b2i−2 (Q6 ) + 4b2i−6 (P2 ) = b2i (B81 ), implying G B81 . Since b2 (G − Cb ) < b2 (P2 ) if b = 6 or −b2 (G − Cb ) < b2 (P2 ) if b = 4, we have b8 (G) < b8 (B81 ). Thus G ≺ B81 .
- 425 3 = 1 for some u ∈ Vp (G). Then Suppose that n = 9. Suppose that dG (u, G) G − u ∈ B1 (8) and so G − u ≺ B81 . Let v ∈ ΓG (u). If G − u − v is an acyclic graph, then by Lemma 2.3, G − u − v P7 ≺ P76 . Otherwise, G − u − v contains a unique cycle, and then by Lemma 2.5, G − u − v P76 . Thus, by Lemma 2.9, G ≺ B91 . 3 ≥ 2 for any u ∈ Vp (G). Obviously, G has at most two Suppose that dG (u, G) 3 = A6 or A7 , where A6 , A7 are the bicyclic graphs obtained branching vertices. If G respectively by identifying an edge of two quadrangles and by identifying two adjacent 3 edges of a quadrangle and a hexagon, then by Sachs theorem, G ≺ B91 . Otherwise, G is the complete bipartite graph K2,3 with bipartition ({v1 , v2 }, {v3 , v4 , v5 }), and one of v3 , v4 , v5 , say v3 , is not a branching vertex of G. Then by Lemmas 2.1, 2.2, 2.3, 2.5 and 2.6, b2i (G) = b2i (G − v1 v3 − v3 v2 ) + b2i−2 (G − v3 − v2 − v4 v1 ) +b2i−2 (G − v1 − v3 − v2 v4 ) − 2b2i−4 (G − v1 − v3 − v2 − v5 ) ≤ b2i (P86 ) + b2i−2 (P2 ∪ P5 ) + b2i−2 (P2 ∪ P5 ) ≤ b2i (P86 ) + b2i−2 (P1 ∪ C6 ) + b2i−2 (Q7 ) = b2i (B91 ) − 4b2i−6 (P2 ). Thus b2i (G) ≤ b2i (B91 ) and b6 (G) < b6 (B91 ), implying G ≺ B91 . Now suppose that n ≥ 10, the result holds for n − 1 and n − 2. Let G ∈ B1 (n). Let 3 = max{dG (x, G)|x 3 ∈ Vp (G)} and let v be the unique u ∈ Vp (G) such that dG (u, G) neighbor of u. Obviously, G − u ∈ B1 (n − 1) and thus by the induction hypothesis, 1 . G − u ≺ Bn−1
3 = 1. If G − u − v is an acyclic graph, then by Lemma 2.3, Case 1. dG (u, G) 1 G − u − v Pn−2 ≺ Bn−2 . Otherwise, G − u − v contains a unique cycle, and then 6 1 ≺ Bn−2 . by Lemma 2.5, G − u − v Pn−2
3 ≥ 2. If G − u − v is disconnected, then G − u − v is a subgraph Case 2. dG (u, G) of G , where G ∈ B1 (n − 2) is obtained from G − u − v by attaching all isolated 3 and by the induction hypothesis and Lemma vertices of G − u − v to a vertex of G, 1 . If G − u − v is connected, then G − u − v ∈ B1 (n − 2), 2.2, G − u − v ≺ G ≺ Bn−2 1 . and by the induction hypothesis, G − u − v ≺ Bn−2 1 . Now by Lemma 2.9, G ≺ Bn1 . Combining Cases 1 and 2, G − u − v ≺ Bn−2
- 426 3 + 2, |G| 3 ≥ 8, |Vp (G)| = 1 and G = B 1 . Lemma 3.2. Let G ∈ B1 (n), where n = |G| n Then G ≺ Bn1 . Proof. Let x be the unique pendent vertex of G. Let u be the neighbor of x and v the unique branching vertex of G. Obviously, n ≥ 10. Case 1. degG (v) = 4. If G is the bicyclic graph obtained by attaching a path P2 to a vertex of degree 3 in B81 , then n = 10 and it can be checked by Sachs theorem that 1 . Otherwise, we take w ∈ ΓG3 (v) such that vw lies on two cycles Cb and Cc G ≺ B10
with c = 6. By Lemmas 2.1, 2.3 and 2.5, b2i (G)
" l = b2i (G − vw) + b2i−2 (G − v − w) − 2 l∈{b,c} (−1) 2 b2i−l (G − Cl ) " l ≤ b2i (Pn6 ) + b2i−2 (P2 ∪ Pn−4 ) − 2 l∈{b,c} (−1) 2 b2i−l (P2 ∪ Pn−l−2 ) = b2i (Pn6 ) + b2i−2 (P2 ∪ P4 ∪ Pn−8 ) + b2i−4 (P2 ∪ Pn−9 ) " l +2b2i−6 (P2 ∪ Pn−9 ) − 2 l∈{b,c} (−1) 2 b2i−l (P2 ∪ Pn−l−2 ),
b2i (Bn1 ) = b2i (Pn6 ) + b2i−2 (Q6 ∪ Pn−8 ) + 4b2i−6 (P2 ∪ Pn−8 ) = b2i (Pn6 ) + b2i−2 (P2 ∪ P4 ∪ Pn−8 ) + b2i−4 (P2 ∪ Pn−8 ) +2b2i−6 (P2 ∪ Pn−9 ) + 2b2i−6 (P2 ∪ Pn−8 ) + 2b2i−8 (P2 ∪ Pn−10 ). By Lemma 2.2, b2i−4 (P2 ∪ Pn−9 ) ≤ b2i−4 (P2 ∪ Pn−8 ), and by Lemma 2.1 (i), 2b2i−c (P2 ∪ Pn−c−2 ) ≤ · · · ≤ 2b2i−10 (P2 ∪ Pn−12 ) ≤ 2b2i−8 (P2 ∪ Pn−10 ) if c ≡ 2 (mod 4) or b
−2b2i−c (P2 ∪Pn−c−2 ) ≤ 2b2i−8 (P2 ∪Pn−10 ) if c ≡ 0 ( mod 4). Note that −(−1) 2 2b2i−b (P2 ∪ Pn−b−2 ) ≤ 2b2i−6 (P2 ∪ Pn−8 ). Then b2i (G) ≤ b2i (Bn1 ). Since b2 (P2 ∪ Pn−9 ) < b2 (P2 ∪ Pn−8 ), we have b6 (G) < b6 (Bn1 ). Thus G ≺ Bn1 . Case 2. degG (v) = 3. Obviously, v lies outside some cycle, say Ca . Take a vertex 3 such that d 3 (w, Ca ) = 1. Let z be the neighbor of w in Ca . Then zw lies w ∈ V (G) G on the other two cycles, say Cb and Cc . Let G be the graph obtained from G by removing edge uv and adding edge uw. If a ≡ 2 ( mod 4), then either b = 6 or c = 6 and by similar arguments as those in Case 1, G ≺ Bn1 . So, to prove G ≺ Bn1 , it suffices to prove G G . By Lemmas 2.1 3 + b2i−2 (P1 ∪ P a ) = b2i (G ), implying G G . (i) and 2.4, b2i (G) ≤ b2i (P2 ∪ G) n−3 Suppose that a ≡ 0 (mod 4). Suppose that dG3 (v, Ca ) = 1. Then G = G . If b = 6 or c = 6, then by similar arguments as those in Case 1, G ≺ Bn1 . If b = c = 6,
- 427 8 then G must be the bicyclic graph obtained from P11 by adding edge wy between the
neighbor of z outside the cycle C8 and the vertex (in the cycle) of distance 4 from 8 1 , and by Sachs theorem, G ≺ B11 . z, where z is the unique vertex of degree 3 in P11
Now suppose that dG3 (v, Ca ) ≥ 2. Then we can take two vertices w and z as above such that G − z − w = Pn−2 . By Lemmas 2.1 (ii) and 2.3, " l b2i (G) = b2i (G − zw) + b2i−2 (G − z − w) − 2 l∈{b,c} (−1) 2 b2i−l (G − Cl ) " l a ) + b2i−2 (Qn−2 ) − 2 l∈{b,c} (−1) 2 b2i−l (P2 ∪ Pn−l−2 ), ≤ b2i (Pn,k,2 6 b2i (Bn1 ) = b2i (Pn,n−7,1 ) + b2i−2 (Qn−2 ) + 4b2i−6 (P2 ∪ Pn−8 ), a 6 ≺ Pn,n−7,1 . where 1 < k < n − a − 2. So to prove G ≺ Bn1 , we need only to prove Pn,k,2
If a = 4, then by Lemma 2.1, 4 b2i (Pn,k,2 ) = b2i (Pn−1,k,2 ) + b2i−2 (Pn−3,k,2 ) + b2i−2 (Pn−4,k,2 ), 6 b2i (Pn,n−7,1 ) = b2i (Pn−1 ) + b2i−2 (P6 ∪ Pn−8 ) + b2i−2 (P4 ∪ Pn−6 ) + 2b2i−6 (Pn−6 ).
By Lemma 2.3, b2i (Pn−1,k,2 ) ≤ b2i (Pn−1 ), and by Lemma 2.6, b2i−2 (Pn−3,k,2 ) ≤ b2i−2 (P6 ∪ 4 6 ) < b6 (Pn,n−7,1 ). Pn−8 ). Note that b2i−2 (Pn−4,k,2 ) ≤ b2i−2 (P4 ∪ Pn−6 ) and that b6 (Pn,k,2 4 6 ≺ Pn,n−7,1 . We have Pn,k,2
If a ≥ 8, then by Lemmas 2.1 and 2.3, a b2i (Pn,k,2 )
= b2i (Pn,k,2 ) + b2i−2 (Pa−2 ∪ Pn−a,k,2 ) − 2b2i−a (Pn−a,k,2 ) = b2i (Pn−2 ) + b2i−2 (Pn−2 ) + b2i−2 (Pk−1 ∪ Pn−k−2 ) +b2i−2 (Pa−2 ∪ Pn−a,k,2 ) − 2b2i−a (Pn−a,k,2 ) ≤ b2i (Pn−2 ) + b2i−2 (P6 ∪ Pn−8 ) + b2i−2 (Pk−1 ∪ Pn−k−2 ) +b2i−2 (Pa−2 ∪ Pn−a ) + b2i−4 (P3 ∪ Pn−9 ) +b2i−6 (P2 ∪ Pn−9 ) + b2i−6 (P3 ∪ Pn−9 ),
6 ) = b2i (Pn−2 ) + b2i−2 (P6 ∪ Pn−8 ) + b2i−2 (Pk−1 ∪ Pn−k−2 ) b2i (Pn,n−7,1
+b2i−2 (P4 ∪ Pn−6 ) + b2i−4 (Pk−2 ∪ Pn−k−3 ) + 2b2i−6 (Pn−6 ). By Lemma 2.6, b2i−2 (Pa−2 ∪ Pn−a ) ≤ b2i−2 (P4 ∪ Pn−6 ), and by Lemma 2.2, b2i−6 (P2 ∪ Pn−9 ) + b2i−6 (P3 ∪ Pn−9 ) ≤ 2b2i−6 (Pn−6 ). Note that b2i−4 (P3 ∪ Pn−9 ) ≤ b2i−4 (Pk−2 ∪ a 6 ≺ Pn,n−7,1 . Pn−k−3 ) and that b2 (P2 ∪ Pn−9 ) < b2 (Pn−6 ). We have Pn,k,2
- 428 Similarly, we have 3 + 1, |G| 3 ≥ 8 and G = B 1 . Then G ≺ B 1 . Lemma 3.3. Let G ∈ B1 (n), where n = |G| n n 3 ≥ 8, G = Bn1 and n ≥ 9. Then G ≺ Bn1 . Lemma 3.4. Let G ∈ B1 (n), where |G| 3 Proof. We prove the lemma by induction on n − |G|. 3 = 1. Let h ≥ 2 and suppose that the By Lemma 3.3, the result is true for n − |G| 3 = h. 3 < h. Let G ∈ B1 (n) and n − |G| result holds for 1 ≤ n − |G| 3 = max{dG (x, G)|x 3 ∈ Vp (G)} and let v be the Let u ∈ Vp (G) such that dG (u, G) unique neighbor of u. Obviously, G − u ∈ B1 (n − 1) and thus by the induction 1 1 if G − u = Bn−1 . hypothesis, G − u ≺ Bn−1
3 = 1. If G − u − v is an acyclic graph, and by Lemma 2.3, Case 1. dG (u, G) 1 . Otherwise, G − u − v contains a unique cycle, and then G − u − v Pn−2 ≺ Bn−2 6 1 ≺ Bn−2 . So, by Lemma 2.9, G ≺ Bn1 . by Lemma 2.5, G − u − v Pn−2
3 ≥ 2. If G − u − v is disconnected, then G − u − v is a subgraph Case 2. dG (u, G) of G , where G ∈ B1 (n − 2) is obtained from G − u − v by attaching all isolated 3 and by the induction hypothesis and Lemma vertices of G − u − v to a vertex of G, 1 . Thus, by Lemma 2.9, G ≺ Bn1 . Suppose that G − u − v 2.2, G − u − v ≺ G Bn−2
3 ≥ 3, then 3 = 2, then by Lemma 3.2, G ≺ B 1 . If n − |G| is connected. If n − |G| n 1 , and thus by the induction hypothesis, G − u − v ∈ B1 (n − 2) and G − u = Bn−1 1 G − u − v Bn−2 . So, by Lemma 2.9, G ≺ Bn1 .
Combining Lemmas 3.1 and 3.4, and using the increasing property (2), we obtain the following theorem. Theorem 3.1. Let G ∈ B1 (n), where G = Bn1 and n ≥ 9. Then G ≺ Bn1 and E(G) < E(Bn1 ).
Remark. Let A16 , A17 , A17 and A18 be the graphs shown in Fig. 2. For n = 6, 7, 8, we have (i) if G ∈ B1 (n) with G = A1n , then G ≺ A1n and E(G) < E(A1n ) for n = 6, 8,
(ii) if G ∈ B1 (7) with G = A17 , A17 , then G ≺ A17 , A17 , and E(G) < E(A17 ) = 8.5702 < E(A17 ) = 8.6332.
- 429 s @s s s@ @ @s
A16
s
s s
s s s
s s
s s
A17
s s
s s
s s s
s
A17
s @ s @s s
s
A18
Figure 2: Graphs with maximal energies and a related graph in B1 (n), where n = 6, 7, 8. 4. GRAPHS IN B2 (n) WITH MAXIMAL ENERGY By Lemma 2.1 (ii), it is easily seen that Pn6 ≺ Bn2 for n ≥ 11. 3 < 11 and n ≥ 11. Then G ≺ B 2 . Lemma 4.1. Let G ∈ B2 (n), where |G| n Proof. We prove the lemma by induction on n. 3 with deg 3 (z) = 4 and w ∈ Γ 3 (z) Suppose that n = 11. Take the vertex z ∈ V (G) G G 6 , G − z − w = P2 ∪ P7 . Obviously, zw lies on one cycle, say such that G − zw = P11
Cb . By Lemmas 2.1 (ii), 2.3, 2.5 and 2.6, b
b2i (G) = b2i (G − zw) + b2i−2 (G − z − w) − (−1) 2 2b2i−b (G − Cb ) 6 2 ≤ b2i (P11 ) + b2i−2 (P4 ∪ P5 ) + 2b2i−6 (P5 ) = b2i (B11 ). 6 2 Note that G − zw ≺ P11 . We have G ≺ B11 .
3 = 1 for some u ∈ Vp (G). Then Suppose that n = 12. Suppose that dG (u, G) 2 . Let v ∈ ΓG (u). If G − u − v is an acyclic G − u ∈ B2 (11) and thus G − u ≺ B11 6 . If G − u − v contains a cycle, graph, then by Lemma 2.3, G − u − v P10 ≺ P10 6 2 . Therefore, by Lemma 2.9, G ≺ B12 . then by Lemma 2.5, G − u − v P10
3 ≥ 2 for any u ∈ Vp (G). Obviously, there is at most Now suppose that dG (u, G) one branching vertex in a quadrangle of G, denoted by v1 v2 v3 v4 v1 . Suppose without loss of generality that degG3 (v1 ) = 4 and v2 is not a branching vertex. By Lemmas 2.1, 2.3, 2.5 and 2.6, b2i (G) = b2i (G − v1 v2 − v2 v3 ) + b2i−2 (G − v2 − v3 − v4 v1 ) +b2i−2 (G − v1 − v2 − v3 v4 ) 6 6 ≤ b2i (P11 ) + b2i−2 (P10 ) + b2i−2 (P1 ∪ P4 ∪ P5 ) 2 = b2i (B12 ) − 2b2i−6 (P1 ∪ P5 ). 2 2 2 . Therefore G B12 and b6 (G) < b6 (B12 ), implying G ≺ B12
- 430 Suppose that n ≥ 13, the result holds for n − 1 and n − 2, and G ∈ B2 (n). By similar arguments as those in Lemma 3.1, we have G ≺ Bn2 . a ∗ Let Bn,b be the graph obtained from Ba+b−1,a,b by attaching a path Pn−a−b+1 to a ∗ ∗ ) with dBa+b−1,a,b (w, Ca ) = 1. vertex w ∈ V (Ba+b−1,a,b
3 + 2, |G| 3 ≥ 11, |Vp (G)| = 1 and G = Bn2 . Lemma 4.2. Let G ∈ B2 (n), where n = |G| Then G ≺ Bn2 . Proof. Let x be the unique pendent vertex of G, and let u be the neighbor of x and v the unique branching vertex in G. Let Ca and Cb be the two cycles of G. Suppose that v lies on Cb . Then n = a + b + 1. a a Claim 1. G Bn,b for G = Bn,b .
By Lemma 2.1 (i), b2i (G) = b2i (G − uv) + b2i−2 (G − u − v), a ∗ a ) = b2i (P2 ∪ Bn−2,a,b ) + b2i−2 (P1 ∪ Pn−3 ). b2i (Bn,b
If degG3 (v) = 4, then G−u−v = P1 ∪Pa−1 ∪Pb−1 , and by Lemma 2.1, b2i−2 (G−u−v) ≤ a a b2i−2 (P1 ∪Pn−3 ). If degG3 (v) = 2, then G−u−v = P1 ∪Pn−3,k,t , where 1 ≤ t ≤ n−a−4
and k = n − a − t − 2, and by Lemma 2.1, a ) + b2i−4 (Pt−1 ∪ Pa−1 ∪ Pn−a−t−3 ) b2i−2 (G − u − v) = b2i−2 (Pt ∪ Pn−t−3 a a ) + b2i−4 (Pt−1 ∪ Pn−t−4 ) ≤ b2i−2 (Pt ∪ Pn−t−3 a = b2i−2 (P1 ∪ Pn−3 ). ∗ a Clearly, b2i (G − uv) = b2i (P2 ∪ Bn−2,a,b ). Hence G Bn,b . a a = Bn2 , then Bn,b ≺ Bn2 . Claim 2. If Bn,b
Case 1. b = 4. Then n = a + 5 and a ≥ 8. By Lemma 2.1, a ) = b2i (Pna ) + b2i−2 (P2 ∪ P2 ∪ Pa−1 ) − 2b2i−4 (P2 ∪ Pa−1 ) b2i (Bn,4
= b2i (Pna ) + b2i−2 (P2 ∪ Pa−1 ) − b2i−4 (P2 ∪ Pa−1 ), b2i (Bn2 )
= b2i (Pn6 ) + b2i−2 (P4 ∪ P5 ∪ Pa−6 ) + 2b2i−6 (P5 ∪ Pa−6 ).
- 431 By Lemma 2.5, b2i (Pna ) ≤ b2i (Pn6 ), and by Lemmas 2.2 and 2.6, b2i−2 (P2 ∪ Pa−1 ) ≤ a a b2i−2 (Pa+1 ) ≤ b2i−2 (P4 ∪ P5 ∪ Pa−6 ). Thus Bn,4 Bn2 and b6 (Bn,4 ) < b6 (Bn2 ), implying a ≺ Bn2 . Bn,4
Case 2. b ≥ 6. By Lemma 2.1, b
a ) = b2i (Pna ) + b2i−2 (P2 ∪ Pb−2 ∪ Pa−1 ) − (−1) 2 2b2i−b (P2 ∪ Pa−1 ) b2i (Bn,b
= b2i (Pna ) + b2i−2 (P2 ∪ P2 ∪ Pb−4 ∪ Pa−1 ) + b2i−4 (Pb−5 ∪ Pa−1 ) b
+b2i−6 (Pb−5 ∪ Pa−1 ) − (−1) 2 2b2i−b (P2 ∪ Pa−1 ), a b2i (Bn,6 ) = b2i (Pna ) + b2i−2 (P4 ∪ Pb−4 ∪ Pa−1 ) + 2b2i−6 (Pb−4 ∪ Pa−1 )
= b2i (Pna ) + b2i−2 (P2 ∪ P2 ∪ Pb−4 ∪ Pa−1 ) + b2i−4 (Pb−4 ∪ Pa−1 ) +2b2i−6 (Pb−5 ∪ Pa−1 ) + 2b2i−8 (Pb−6 ∪ Pa−1 ). a a By Lemma 2.2, b2i−4 (Pb−5 ∪ Pa−1 ) ≤ b2i−4 (Pb−4 ∪ Pa−1 ). Thus Bn,b Bn,6 and a a b2 (Pb−5 ∪ Pa−1 ) < b2 (Pb−4 ∪ Pa−1 ), implying Bn,b ≺ Bn,6 for b ≥ 8.
By Lemma 2.1 (i), a ∗ a b2i (Bn,6 ) = b2i (Pn−a−5 ∪ Ba+5,a,6 ) + b2i−2 (Pn−a−6 ∪ Pa+4 ), 6 ∗ 6 ) = b2i (Pn−a−5 ∪ Ba+5,a,6 ) + b2i−2 (Pn−a−6 ∪ Pa+4 ). b2i (Bn,a a 6 a 6 By Lemma 2.5, b2i−2 (Pn−a−6 ∪ Pa+4 ) ≤ b2i−2 (Pn−a−6 ∪ Pa+4 ). Thus Bn,6 Bn,a .
By Lemma 2.1 (ii), b2i (Bn2 ) = b2i (Pn6 ) + b2i−2 (P4 ∪ P5 ∪ Pn−11 ) + 2b2i−6 (P5 ∪ Pn−11 ). If a = 4, then by Lemma 2.1 (ii), 6 ) = b2i (Pn6 ) + b2i−2 (P1 ∪ P1 ∪ P5 ∪ Pn−9 ) − b2i−4 (P5 ∪ Pn−9 ). b2i (Bn,4
By Lemmas 2.2 and 2.6, b2i−2 (P1 ∪ P1 ∪ P5 ∪ Pn−9 ) ≤ b2i−2 (P4 ∪ P5 ∪ Pn−11 ). Thus 6 6 6 Bn2 and b6 (Bn,4 ) < b6 (Bn2 ), implying Bn,4 ≺ Bn2 . Bn,4
Suppose that a ≥ 8. Then by Lemma 2.1 (ii), a
6 ) = b2i (Pn6 ) + b2i−2 (Pa−2 ∪ P5 ∪ Pn−a−5 ) − (−1) 2 2b2i−a (P5 ∪ Pn−a−5 ). b2i (Bn,a 6 ≺ Bn2 as above. If n = a + 7, then by Lemma 2.6, b2i−2 (Pa−2 ∪ If n = a + 7, then Bn,a 6 6 P5 ∪ Pn−a−5 ) ≤ b2i−2 (P4 ∪ P5 ∪ Pn−11 ). Thus Bn,a Bn2 and b6 (Bn,a ) < b6 (Bn2 ), 6 ≺ Bn2 . implying Bn,a
- 432 Similarly, we have 3 + 1, |G| 3 ≥ 11 and G = B 2 . Then Lemma 4.3. Let G ∈ B2 (n), where n = |G| n G ≺ Bn2 . By similar arguments as those in Lemma 3.4, we have 3 ≥ 11, G = B 2 and n ≥ 12. Then G ≺ B 2 . Lemma 4.4. Let G ∈ B2 (n), where |G| n n Combining Lemmas 4.1 and 4.4, and using the increasing property (2), we obtain the following theorem. Theorem 4.1. Let G ∈ B2 (n), where G = Bn2 and n ≥ 12. Then G ≺ Bn2 and E(G) < E(Bn2 ).
Remark. Let A28 , A29 , A29 , A210 and A211 be the graphs shown in Fig. 3. For n = 8, 9, 10, 11, we have (i) if G ∈ B2 (n) with G = A2n , then G ≺ A2n and E(G) < E(A2n ), for n = 8, 10, 11,
(ii) if G ∈ B2 (9) with G = A29 , A29 , then G ≺ A29 or A29 , E(A29 ) = 10.3376 < E(A29 ) = 10.9418, and so if G ∈ B2 (9) with G = A29 , then E(G) < E(A29 ). s s s @s @ s @ @s @ @s @ @s 4 A28 (= B8,4 )
s s s s s @s @ s @ @s @ @s @ @s
A29
s s s @s @ s @ @s @ @s @ @s
4 A29 (= B9,4 )
s s s @s s @ @ @s @s @
s s @ @s s
6 A210 (= B10,4 )
s s s @s s @ @ @s @s @
s @ @s s
6 A211 (= B11,4 )
Figure 3: Graphs with maximal energies and a related graph in B2 (n), where n = 8, 9, 10, 11. 5. GRAPHS IN B3 (n) WITH MAXIMAL ENERGY Let Bn,t for 1 ≤ t ≤ n − 13 (Bn,t for 1 ≤ t ≤ n − 12, resp.) be the graph obtained ∗ , where y and w are by attaching a path Pt to vertex y (vertex w, resp.) in Bn−t ∗ neighbors of a vertex of degree 3 in Bn−t such that y lies outside any cycle and w lies ∗ a . Denote by Un,t be the graph obtained by attaching a path on some cycle in Bn−t a , where Pt to a vertex on the cycle that is adjacent to the vertex of degree 3 in Pn−t
1 ≤ t ≤ n − a − t. Lemma 5.1. For n ≥ 16, Bn,2 ≺ Bn3 .
- 433 Proof. By Lemma 2.1 (ii), 6 6 6 ) = b2i (Pn,7,2 ) + b2i−2 (P4 ∪ Pn−6 ) + 2b2i−6 (Pn−6 ), b2i (Bn,2 6 6 6 ) + b2i−2 (P4 ∪ Pn−6 ) + 2b2i−6 (Pn−6 ). b2i (Bn3 ) = b2i (Pn,7,n−13 6 6 So, to prove Bn,2 ≺ Bn3 , it suffices to prove Pn,7,2 ≺ Pn,7,n−13 . By Lemma 2.1 (i), 6 6 6 ) = b2i (P6 ∪ Pn−6 ) + b2i−2 (P5 ∪ P2 ∪ Pn−9 ), b2i (Pn,7,2 6 6 ) = b2i (P6 ∪ Pn−6 ) + b2i−2 (P5 ∪ Pn−13 ∪ C6 ). b2i (Pn,7,n−13 6 6 6 ≺ Pn−13 ∪ C6 and so Pn,7,2 ≺ Pn,7,n−13 . By Lemma 2.7 (ii), P2 ∪ Pn−9
Similarly, we have Lemma 5.2. For n ≥ 15, Bn,1 ≺ Bn3 . ≺ Bn3 . Lemma 5.3. For n ≥ 15, Bn,2 Proof. If 15 ≤ n ≤ 16, then by direct checking (see Table 1), we have Bn,2 ≺ Bn3 . ≺ Bn3 , we need only to prove Suppose that n ≥ 17. By Lemma 5.1, to prove Bn,2 Bn,2 . By Lemmas 2.1 and 2.7 (iii), Bn,2 ∗ 6 ) = b2i (P2 ∪ Bn−2 ) + b2i−2 (P1 ∪ Pn−3 ) b2i (Bn,2 ∗ ) + b2i−2 (Pn−3 ) + b2i−4 (P4 ∪ Pn−9 ) + 2b2i−8 (Pn−9 ) = b2i (P2 ∪ Bn−2 ∗ ≤ b2i (P2 ∪ Bn−2 ) + b2i−2 (Pn−9 ∪ C6 ) + b2i−4 (P4 ∪ Pn−15 ∪ C6 ) ), +2b2i−8 (Pn−15 ∪ C6 ) = b2i (Bn,2 Bn,2 . implying Bn,2
Similarly, we have ≺ Bn3 . Lemma 5.4. For n ≥ 14, Bn,1
By Lemma 2.1 (ii), we have Pn6 ≺ Bn3 for n = 13. By Lemmas 2.1 (ii) and 5.4, we ≺ Bn3 for n ≥ 14 . have Pn6 ≺ Bn,1
3 < 13, |G| 3 = B ∗ and n ≥ 13. Then G ≺ B 3 . Lemma 5.5. Let G ∈ B3 (n), where |G| 12 n
- 434 Proof. Obviously, G contains at least one quadrangle. We prove the lemma by induction on n. Suppose that n = 13. If each vertex of some quadrangle v1 v2 v3 v4 v1 is not a branching vertex of G, then let degG3 (v1 ) = 3 and by Lemmas 2.1, 2.5 and 2.7 (i), b2i (G) = b2i (G − v1 v2 − v2 v3 ) + b2i−2 (G − v2 − v3 − v4 v1 ) +b2i−2 (G − v1 − v2 − v3 v4 ) 6 6 ≤ b2i (P1 ∪ P12 ) + b2i−2 (P1 ∪ P10 ) + b2i−2 (P1 ∪ P1 ∪ P96 ) 6 6 ) + b2i−2 (P11 ) + b2i−2 (P4 ∪ P76 ) ≤ b2i (P12 3 = b2i (B13 ) − 2b2i−6 (P76 ). 3 3 3 and b6 (G) < b6 (B13 ), implying G ≺ B13 . Hence, G B13
Suppose that there exists at least one branching vertex in each quadrangle of G. If there exists exactly a branching vertex v on some quadrangle such that v has 3 then by similar arguments as above, we have exactly one pendent neighbor outside G, 3 . Otherwise, there exists some edge zv in G such that degG3 (z) = 3 and G−zv G ≺ B13
contains exactly two vertex-disjoint bipartite unicyclic graphs, the one containing z possesses 6 vertices and the other possesses 7 vertices. By Lemmas 2.1 (i), 2.3 and 2.5, b2i (G) = b2i (G − zv) + b2i−2 (G − z − v) 3 ≤ b2i (C6 ∪ P76 ) + b2i−2 (P5 ∪ C6 ) = b2i (B13 ), 3 3 3 . Moreover b4 (G) < b4 (B13 ). So G ≺ B13 . Thus G B13
3 = 1 for some u ∈ Vp (G). Then Suppose that n = 14. Suppose that dG (u, G) 3 . Let v ∈ ΓG (u). If v lies on some cycle, then G − u ∈ B3 (13) and thus G − u ≺ B13 6 . If v lies outside any cycle, then by Lemma 2.5, by Lemma 2.5, G − u − v P12 6 G − u − v P54 ∪ P76 , C4 ∪ P86 or C6 ∪ C6 , and by Sachs theorem, P12 , P54 ∪ P76 , C4 ∪ P86 3 C6 ∪ C6 . Therefore, by Lemma 2.9, G ≺ B14 .
3 ≥ 2 for any u ∈ Vp (G). Obviously, G have at most three Suppose that dG (u, G) branching vertices. If each vertex in some quadrangle is not a branching vertex of G, or there exist two quadrangles in which each contains exactly a branching vertex 3 3 = B∗ . Otherwise, G of G, then by similar arguments as above, G ≺ B14 10,4,6 , the 3 . quadrangle in G has two branching vertices, and thus by Sachs theorem, G ≺ B14
- 435 Suppose that n ≥ 15, the result holds for n − 1 and n − 2, and G ∈ B3 (n). By similar arguments as those in Lemma 3.1, we have G ≺ Bn3 . 3 = B ∗ and n ≥ 15. Then G ≺ B 3 . Lemma 5.6. Let G ∈ B3 (n), where |G| 12 n Proof. By similar arguments as those in Lemma 3.1, we have G Bn,n−12 for G ∈ 3 3 = B ∗ and n ≥ 13. Now we show that B B3 (n), where |G| 12 n,n−12 ≺ Bn for n ≥ 15. By
Lemma 2.1 (i), 6 b2i (Bn,n−12 ) = b2i (C6 ∪ Pn−6 ) + b2i−2 (P5 ∪ Pn−7 ), 6 b2i (Bn3 ) = b2i (C6 ∪ Pn−6 ) + b2i−2 (P5 ∪ Pn−13 ∪ C6 ). By Lemma 2.7 (iii), Pn−7 ≺ Pn−13 ∪ C6 . Thus Bn,n−12 ≺ Bn3 . ∗ Denote by Bn,a,b the bipartite bicyclic graph obtained from Bn−1,a,b by attaching ∗ ∗ ) with dBn−1,a,b (w, Ca ) = 1, where n ≥ a pendent vertex to a vertex w ∈ V (Bn−1,a,b
a + b + 2. For convenience, let b ≥ a if n = a + b + 2. Lemma 5.7. For a, b ≥ 6, Bn,a,b Bn,1 . Proof. First, we show that Bn,a,b Bn,a,6 for b ≥ 8. If n > a + b + 2, then by Lemmas
2.1 (i) and 2.5, 6 6 ) ≤ b2i (Ca ∪ Pn−a ) + b2i−2 (Pa−1 ∪ Pn−a−2 ) = b2i (Bn,a,6 ). b2i (Bn,a,b Hence Bn,a,b Bn,a,6 , in particular, Bn,6,a Bn,1 .
Suppose that n = a + b + 2. By Lemma 2.1, b
a a a ) = b2i (Pn,b+1,1 ) + b2i−2 (Pb−2 ∪ Pa+2 ) − (−1) 2 2b2i−b (Pa+2 ) b2i (Bn,a,b a a = b2i (Pn,b+1,1 ) + b2i−2 (Pb−2 ∪ Pa+1 ) + b2i−4 (P4 ∪ Pb−6 ∪ Ca ) b
a ), +b2i−6 (P3 ∪ Pb−7 ∪ Ca ) − (−1) 2 2b2i−b (Pa+2 a a a b2i (Bn,a,6 ) = b2i (Pn,b+1,1 ) + b2i−2 (P4 ∪ Pn−6,b−5,1 ) + 2b2i−6 (Pn−6,b−5,1 ) a a ) + b2i−2 (P4 ∪ Pa+b−5 ) + b2i−4 (P4 ∪ Pb−6 ∪ Ca ) = b2i (Pn,b+1,1 a a +2b2i−6 (Pb−7 ∪ Pa+3,2,1 ) + 2b2i−8 (Pb−8 ∪ Pa+2 ).
- 436 a By Lemma 2.2, b2i−6 (P3 ∪ Pb−7 ∪ Ca ) ≤ b2i−6 (Pb−7 ∪ Pa+3,2,1 ). To prove Bn,a,b Bn,a,6 ,
we need only to prove a a a b2i−2 (Pb−2 ∪ Pa+1 ) ≤ b2i−2 (P4 ∪ Pa+b−5 ) + b2i−6 (Pb−7 ∪ Pa+3,2,1 ).
By Lemmas 2.1, and 2.2, a a ) = b2i−2 (P4 ∪ Pb−6 ∪ Pa+1 ) + b2i−4 (P3 ∪ Pb−7 ∪ Ca ) b2i−2 (Pb−2 ∪ Pa+1
+b2i−6 (P3 ∪ Pb−7 ∪ Pa−1 ) a ≤ b2i−2 (P4 ∪ Pb−6 ∪ Pa+1 ) + b2i−4 (P4 ∪ Pb−7 ∪ Ca )
+b2i−6 (P3 ∪ Pb−7 ∪ Ca ) a a ≤ b2i−2 (P4 ∪ Pa+b−5 ) + b2i−6 (Pb−7 ∪ Pa+3,2,1 ). Bn,6,a for a ≥ 8. By Lemma 2.1, Next, we show that Bn,a,6 ∗ ) = b2i (Bn−1,a,6 ) + b2i−2 (Pn−a−8 ∪ C6 ∪ Ca ) + b2i−4 (Pn−a−9 ∪ P5 ∪ Ca ), b2i (Bn,a,6 ∗ b2i (Bn,6,a ) = b2i (Bn−1,a,6 ) + b2i−2 (Pn−a−8 ∪ Ca ∪ C6 ) + b2i−4 (Pn−a−9 ∪ Pa−1 ∪ C6 ). So, to prove Bn,a,6 Bn,6,a , it suffices to prove P5 ∪ Ca Pa−1 ∪ C6 . By Lemma 2.1
(ii), a
b2i (P5 ∪ Ca ) = b2i (P5 ∪ Pa ) + b2i−2 (P5 ∪ Pa−2 ) − (−1) 2 2b2i−a (P5 ), b2i (Pa−1 ∪ C6 ) = b2i (P6 ∪ Pa−1 ) + b2i−2 (P4 ∪ Pa−1 ) + 2b2i−6 (Pa−1 ). Since b2i (P5 ∪ Pa ) ≤ b2i (P6 ∪ Pa−1 ) and b2i−2 (P5 ∪ Pa−2 ) ≤ b2i−2 (P4 ∪ Pa−1 ) by Lemma Bn,6,a . 2.6, we have Bn,a,6 Note that Bn,6,a Bn,1 . The result follows. ∗ Let G be a bipartite bicyclic graph on n ≥ 16 vertices and G = Bn,a,b , where
a, b ≥ 10, a ≡ b ≡ 2 (mod 4) and n = a + b. Then [28] G Bn∗ . By this result and by Sachs theorem for 13 ≤ n ≤ 15, we have Lemma 5.8. Let G be a bipartite bicyclic graph on n ≥ 13 vertices. If G contains ∗ , where a, b ≥ 10, a ≡ two vertex-disjoint cycles but no pendent vertices, G = Bn,a,b
b ≡ 2 (mod 4) and n = a + b, then G Bn∗ .
- 437 3 + 2, |G| 3 ≥ 13, |Vp (G)| = 1 and G = B 3 . Lemma 5.9. Let G ∈ B3 (n), where n = |G| n Then G ≺ Bn3 . Proof. Let x be the unique pendent vertex of G. Let u the neighbor of x and v the unique branching vertex of G. Let Ca and Cb be the two vertex-disjoint cycles of G. Obviously, n ≥ 15. 3 = B∗ Case 1. G n−2,a,b , where a, b ≥ 10, a ≡ b ≡ 2 (mod 4) and n − 2 = a + b. Obviously, v lies on some cycle of G, say Cb , and n = a + b + 2 ≥ 22. Take a ∗ vertex w ∈ V (Bn−2,a,b ) with dG3 (w, Ca ) = 2. It is not difficult to show that G G ,
where G is obtained from G by removing edge uv and adding edge uw. By Lemma . By Lemmas 2.1 (ii) and 2.5, 5.1, to prove G ≺ Bn3 , we need only to prove G Bn,2 b b b ) + b2i−2 (Pa−2 ∪ Pn−a ) + 2b2i−a (Pn−a ) b2i (G ) = b2i (Un,2 b 6 6 ≤ b2i (Un,2 ) + b2i−2 (Pa−2 ∪ Pn−a ) + 2b2i−a (Pn−a ), 6 6 6 ) = b2i (Pn,7,2 ) + b2i−2 (P4 ∪ Pn−6 ) + 2b2i−6 (Pn−6 ). b2i (Bn,2 6 6 By Lemma 2.7 (i), b2i−2 (Pa−2 ∪ Pn−a ) ≤ b2i−2 (P4 ∪ Pn−6 ), and by Lemma 2.1 (i), 6 6 b 6 b2i−a (Pn−a ) ≤ b2i−6 (Pn−6 ). To prove G Bn,2 , it suffices to prove Un,2 Pn,7,2 . By
Lemmas 2.1 and 2.5, b ) b2i (Un,2
= b2i (Pn ) + b2i−2 (P2 ∪ Pn−b−2 ∪ Pb−2 ) + 2b2i−b (P2 ∪ Pn−b−2 ) = b2i (P2 ∪ Pn−2 ) + b2i−2 (P6 ∪ Pn−9 ) + b2i−2 (P2 ∪ Pn−b−2 ∪ Pb−2 ) +b2i−4 (P4 ∪ P5 ∪ Pn−14 ) + b2i−6 (P3 ∪ P5 ∪ Pn−15 ) +2b2i−b (P2 ∪ Pn−b−2 ),
6 ) = b2i (Pn,7,2 ) + b2i−2 (P4 ∪ Pn−6,7,2 ) + 2b2i−6 (Pn−6,7,2 ), b2i (Pn,7,2
= b2i (P2 ∪ Pn−2 ) + b2i−2 (P6 ∪ Pn−9 ) + b2i−2 (P2 ∪ P4 ∪ Pn−8 ) +b2i−4 (P4 ∪ P6 ∪ Pn−15 ) + 2b2i−6 (Pn−7,7,2 ) + 2b2i−8 (Pn−8,7,2 ). By Lemma 2.6, b2i−2 (P2 ∪ Pn−b−2 ∪ Pb−2 ) ≤ b2i−2 (P2 ∪ P4 ∪ Pn−8 ) and b2i−4 (P4 ∪ P5 ∪ Pn−14 ) ≤ b2i−4 (P4 ∪ P6 ∪ Pn−15 ). By Lemmas 2.2 and 2.6, b2i−6 (P3 ∪ P5 ∪ Pn−15 ) ≤ b2i−6 (P2 ∪ Pn−9 ) ≤ b2i−6 (Pn−7,7,2 ). Note that b2i−b (P2 ∪ Pn−b−2 ) ≤ b2i−8 (Pn−8,7,2 ). We b 6 have Un,2 Pn,7,2 .
- 438 ∗ 3= Case 2. G Bn−2,a,b , where a, b ≥ 10, a ≡ b ≡ 2 (mod 4) and n − 2 = a + b.
Subcase 2.1. v lies on some cycle of G. By Lemma 5.3, to prove G ≺ Bn3 , we need only to prove G Bn,2 . Clearly,
G − x − u is a bicyclic graph containing no pendent vertices and G − x − u − v is a bipartite graph on n − 3 vertices containing a unique cycle. Then by Lemma 5.8, ∗ 6 , and by Lemma 2.5, G − x − u − v Pn−3 . Thus, by Lemma 2.9, G − x − u Bn−2 . G Bn,2
Subcase 2.2. v lies outside any cycle of G. Suppose that there exists some quadrangle v1 v2 v3 v4 v1 in G with degG3 (v1 ) = 3. Then by Lemmas 2.1, 2.5 and 2.7 (i) and (ii), b2i (G) = b2i (G − v1 v2 − v2 v3 ) + b2i−2 (G − v2 − v3 − v4 v1 ) +b2i−2 (G − v1 − v2 − v3 v4 ) 6 6 6 ) + b2i−2 (P1 ∪ Pn−3 ) + b2i−2 (P1 ∪ P1 ∪ Pn−4 ) ≤ b2i (P1 ∪ Pn−1 6 6 ≤ b2i (Pn−1 ) + b2i−2 (Pn−8 ∪ C6 ) + b2i−2 (P4 ∪ Pn−7 ) 6 ) − b2i−4 (P4 ∪ Pn−14 ∪ C6 ) − 2b2i−6 (Pn−6,n−13,1 ). = b2i (Bn,1 Hence G Bn,1 . By Lemma 5.2, Bn,1 ≺ Bn3 . Thus G ≺ Bn3 .
Suppose that G contains no quadrangles. Since G = Bn3 , we have n ≥ 16. By . Clearly, G−x ∈ B3 (n−1) Lemma 5.1, to prove G ≺ Bn3 , it suffices to prove G Bn,2
contains exactly one pendent vertex and G − x − u is a bicyclic graph containing ∗ . By Lemma 2.9, we no pendent vertices, and by Lemma 5.8, G − x − u Bn−2 . If G − x = Bn−1,a,b , then by Lemma 5.7, need only to prove G − x Bn−1,1 G − x Bn−1,1 . Otherwise, all neighbors of v lie outside any cycle of G. Then
G − x − u − v = Pta ∪ Psb , where t > a, s > b and n = t + s + 3, and thus by Lemmas 6 ∗ . Note that G − x − u Bn−2 . Hence, by 2.5 and 2.8, G − x − u − v C6 ∪ Pn−9 . Lemma 2.9, G − x Bn−1,1
Similarly, we have 3 + 1, |G| 3 ≥ 13 and G = Bn3 . Then Lemma 5.10. Let G ∈ B3 (n), where n = |G| G ≺ Bn3 .
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By similar arguments as those in Lemma 3.4, we have 3 ≥ 13, G = Bn3 and n ≥ 14. Then G ≺ Bn3 . Lemma 5.11. Let G ∈ B3 (n), where |G| Combining Lemmas 5.5, 5.6 and 5.11, and using the increasing property (2), we have the following theorem. Theorem 5.1. Let G ∈ B3 (n), where G = Bn3 and n ≥ 15. Then G ≺ Bn3 and E(G) < E(Bn3 ).
Remark. Let A39 , A310 , A310 , A310 , A311 , A312 and A312 be the graphs shown in Fig. 4. Let and A314 = B14,2 . For n = 9, 10, 11, 12, 13, 14, we have A313 = B13,1
(i) if G ∈ B3 (n) with G = A3n , then G ≺ A3n and E(G) < E(A3n ), for n = 9, 11, 13, 14,
(ii) if G ∈ B3 (10) with G = A310 , A310 , A310 , then G ≺ A310 , A310 or A310 , E(A310 ) =
11.8641 < E(A310 ) = 11.9315 < E(A310 ) = 12.4709 and so if G ∈ B3 (10) with G = A310 , then E(G) < E(A310 ),
(iii) if G ∈ B3 (12) with G = A312 , A312 , then G ≺ A312 or A312 , E(A312 ) = 15.1349 < E(A312 ) = 15.6722, and so if G ∈ B3 (12) with G = A312 , then E(G) < E(A312 ). s s s @ @s @ @s
s @s s @ @ @s
s s s @ @s @ @s
A39
s s @s s @ @ @s
s
s s s @ @s @ @s
A310
s s @s s @ @ @s
s s @ @s
A311
s @ @s s
s s @s s @ @ @s
s @s s @ @ @s
s s s @ @s @ @s
A310 s
s s @ @s
s @ @s s
A312
s
s @s s @ @ @s
A310 s
s s @s s @ @ @s
s s @ @s
s @ @s s
A312
Figure 4: Graphs with maximal energies and related graphs in B3 (n), where n = 9, 10, 11, 12. 6. COMMENTS We have determined the graphs in Bi (n) for i = 1, 2, 3 with maximal energies above. Now we determine the graphs with maximal energy in the class of bipartite bicyclic graphs with exactly two cycles and at least one pendent vertex. Theorem 6.1. Let G ∈ B2 (n) ∪ B3 (n), where G = Bn3 and n ≥ 15. Then G ≺ Bn3 and E(G) < E(Bn3 ).
- 440 Proof. By Theorems 4.1 and 5.1, and using the increasing property (2), it suffices to show Bn2 ≺ Bn3 . We prove this by induction on n. By direct checking (see Table 1), the result holds for n = 15 and n = 16. Suppose that n ≥ 17, the result holds for n − 1 and n − 2, and G ∈ B(n)\B1 (n). k k By Lemma 2.1 (i), b2i (Bnk ) = b2i (Bn−1 ) + b2i−2 (Bn−2 ), where k = 2, 3. Hence, by the
induction hypothesis, Bn2 ≺ Bn3 . Remark. For 15 ≤ n ≤ 18, by direct calculation of the eigenvalues, we find E(Bn1 ) < E(Bn3 ), which can not be deduced by use of the relation “≺” from the b2i -values. This may be seen from Table 1 for n = 15, 16. Table 1: b2i -values and energies of some graphs graph
coefficients
energy
b4
b6
b8
b10
b12
b14
b16
1 B15
100
317
550
520
245
43
19.7780
2 B15
99
308
519
470
207
32
19.6188
3 B15
100
316
549
532
270
56
19.8698
B15,2
100
315
541
507
240
44
19.7684
1 B16
115
403
794
887
531
143
9
21.2109
2 B16
114
393
755
813
460
111
4
20.9728
3 B16
115
402
791
893
557
168
16
21.3495
B16,2
115
401
784
872
530
156
16
21.3219
Acknowledgement. This work was supported by the National Natural Science Foundation of China (Grant No. 10671076).
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