On the minimal energy of conjugated unicyclic graphs with maximum degree at most 3

Discrete Applied Mathematics 186 (2015) 186–198

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On the minimal energy of conjugated unicyclic graphs with maximum degree at most 3 Hongping Ma a , Yongqiang Bai a,∗ , Shengjin Ji b a

School of Mathematics and Statistics, Jiangsu Normal University, Xuzhou 221116, China

b

School of Science, Shangdong University of Technology, Zibo 255049, China

article

info

Article history: Received 22 July 2014 Received in revised form 14 January 2015 Accepted 26 January 2015 Available online 21 February 2015

abstract The energy of a graph G, denoted by E (G), is defined as the sum of the absolute values of all eigenvalues of G. Let n be an even number and Un be the set of all conjugated unicyclic n

graphs of order n with maximum degree at most 3. Let Sn2 be the radialene graph obtained by attaching a pendant edge to each vertex of the cycle C n . Cao et al. (2009) showed that n

2

n

if n ≥ 8, Sn2  G ∈ Un and the girth of G is not divisible by 4, then E (G) > E (Sn2 ). Let An be the unicyclic graph obtained by attaching a 4-cycle to one of the two leaf vertices of the path P n −1 and a pendent edge to each other vertices of P n −1 . In this paper, we prove that 2 2 An is the unique unicyclic graph in Un with minimal energy for n ≥ 8. © 2015 Elsevier B.V. All rights reserved.

Keywords: Minimal energy Unicyclic graph Perfect matching Characteristic polynomial Degree

1. Introduction Let G be a simple graph with n vertices and A(G) the adjacency matrix of G. The eigenvalues λ1 , λ2 , . . . , λn of A(G) are said to be the eigenvalues of the graph G. The energy of G is defined as E = E (G) =

n 

|λi |.

i=1

This concept was intensively studied in chemistry, since it can be used to approximate the total π -electron energy of a molecular. For further details on the mathematical properties and chemical applications of E (G), see the recent book [17], reviews [8,9], and papers [1,4–6,18,22,29]. One of the fundamental question that is encountered in the study of graph energy is which graphs (from a given class) have minimal and maximal energies. A large of number of papers were published on such extremal problems, especially for various subclasses of trees and unicyclic graphs, see Chapter 7 in [17]. A conjugated unicyclic graph is a connected graph with one unique cycle that has a perfect matching. The problem of determining the conjugated unicyclic graph with minimal energy has been considered in [19,25], and Li et al. [19] proved that the conjugated unicyclic graph of order (even) n with minimal energy is U1 or U2 , as shown in Fig. 1. It has been shown that E (U1 ) < E (U2 ) by Li and Li [16]. Recently, results on ordering of conjugated unicyclic graphs by minimal energies have been extended in [24,29]. In particular, U2 is the unique conjugated unicyclic graph of order n with second-minimal energy. The degree of a vertex v in a graph G is denoted by dG (v). Denote by ∆ the maximum degree of a graph. From now on, let n be an even number. Let Un be the set of all conjugated unicyclic graphs of order n with ∆ ≤ 3. Let G ∈ Un , the length

∗

Corresponding author. E-mail addresses: [email protected] (H. Ma), [email protected] (Y. Bai), [email protected] (S. Ji).

http://dx.doi.org/10.1016/j.dam.2015.01.033 0166-218X/© 2015 Elsevier B.V. All rights reserved.

H. Ma et al. / Discrete Applied Mathematics 186 (2015) 186–198

187

Fig. 1. The conjugated unicyclic graphs with minimal and second-minimal energy.

Fig. 2. Four graphs in Un . n

of the unique cycle of G is denoted by g (G), or simply g, and the unique cycle of G is denoted by Cg (G), or simply Cg . Let Sn2 be the radialene graph obtained by attaching a pendant edge to each vertex of the cycle C n . Wang et al. [26] showed the 2

n

following results: Assume that n ≥ 6 and Sn2  G ∈ Un . Then if one of the following conditions holds: (i) n , 2

n 2

n 2

2) and g ≤ (ii) g ̸≡ ≡ 0 (mod 4), (iii) ≡ g ≡ 2 (mod 4), and g ≤ above results by proving the following theorem.

n , 2

n 2

n 2

≡ g ≡ 1 (mod

then E (G) > E (Sn ). Y. Cao et al. [2] improved the

n

n

Theorem 1.1 ([2]). If n ≥ 8, Sn2  G ∈ Un with g ̸≡ 0 (mod 4), then E (G) > E (Sn2 ). Let An , Bn , Dn and En be the graphs shown in Fig. 2. In this paper, we completely characterize graphs with minimal energy in Un by showing the following result. Theorem 1.2. An is the unique unicyclic graph in Un with minimal energy for n ≥ 8. 2. Preliminaries In this section, we first introduce some notations and properties which are need in the sequel. Then we give some results n

on the energies of graphs An , Bn , Dn , En and Sn2 . Let G be a graph of order k. The characteristic polynomial of A(G) is also called the characteristic polynomial of G, denoted k k−i by φ(G, x) = det(xI − A(G)) = . Using these coefficients of φ(G, x), the energy of G can be expressed as the i=0 ai (G)x Coulson integral formula [10]: E (G) =

1



2π

+∞ −∞

1 x2

 2  k 2  ⌊ 2k ⌋ ⌊2⌋     ln  (−1)i a2i (G)x2i  +  (−1)i a2i+1 (G)x2i+1   dx. i=0

(1)

i=0

Write bi (G) = |ai (G)|. Clearly, b0 (G) = 1, b1 (G) = 0, and b2 (G) equals the number of edges of G. For unicyclic graphs or bipartite graphs, it had been shown [10,11] that E (G) =

1 2π



+∞ −∞

 2  k 2  ⌊ 2k ⌋ ⌊2⌋   1   ln  b2i (G)x2i  +  b2i+1 (G)x2i+1   dx. 2

x

i=0

(2)

i =0

For convenience, we introduce the following quasi-order relation on unicyclic or bipartite graphs based on Formula (2) which is originally defined in [23]. Let G1 and G2 be two unicyclic or bipartite graphs with k vertices. Define G1 ≽ G2 if bi (G1 ) ≥ bi (G2 ) for all i = 2, . . . , k. If G1 ≽ G2 and there exists some j such that bj (G1 ) > bj (G2 ), then we write G1 ≻ G2 . Clearly, G1 ≽ G2 ⇒ E (G1 ) ≥ E (G2 ), and G1 ≻ G2 ⇒ E (G1 ) > E (G2 ). It is known [3] that b2i+1 (G) = 0 for a bipartite graph G, and b2i (G) equals the number of i-matchings of G if G is a tree. Lemma 2.1 ([20]). Let G be a graph whose components are all trees except at most one being a unicyclic graph. (1) If G contains exactly one cycle Cg , and uv is an edge on this cycle, then bi (G) = bi (G − uv) + bi−2 (G − {u, v}) − 2bi−g (G − Cg ) if g ≡ 0 (mod 4), bi (G) = bi (G − uv) + bi−2 (G − {u, v}) + 2bi−g (G − Cg ) if g ̸≡ 0 (mod 4).

188

H. Ma et al. / Discrete Applied Mathematics 186 (2015) 186–198

Fig. 3. The graphs Fn and Hn+1 .

(2) If uv is a cut edge of G, then bi (G) = bi (G − uv) + bi−2 (G − {u, v}). In particular, if uv is a pendent edge with pendent vertex u, then bi (G) = bi (G − u) + bi−2 (G − {u, v}). Lemma 2.2 ([7]). Let T be a tree on n vertices. If T is different from the path Pn and the star Sn , then Pn ≻ T ≻ Sn . Let T be a tree of order n ≥ 3, e = uv be a non-pendent edge of T . Denote by T1 and T2 the two components of T − e with u ∈ T1 and v ∈ T2 . If T ′ is the tree obtained from T by contracting the edge e = uv and attaching a pendent vertex to the vertex u (=v ), we say that T ′ is obtained from T by edge-growing transformation (on edge e = uv ), or e.g.t (on edge e = uv ) for short [27,21]. Lemma 2.3 ([21]). If T ′ is obtained from T by one step of e.g.t, then T ≻ T ′ . Lemma 2.4 ([23]). Let G be a unicyclic or bipartite graph, uv be a cut edge of G. Then G ≻ G − uv . Lemma 2.5 ([23]). Let G be a unicyclic or bipartite graph, v be a non-isolated vertex in G, and K1 be the trivial graph of order 1. Then G ≻ (G − v) ∪ K1 . Let Fn and Hn+1 be the graphs shown in Fig. 3. Lemma 2.6 ([28]). Among conjugated trees of order n with ∆ ≤ 3, Fn has minimal energy. Lemma 2.7. For An , Bn , Dn and En , we have that (1)b2i (An ) = b2i (Fn ) + b2i−2 (Fn−2 ) − 2b2i−4 (Fn−4 ) = b2i (Hn−1 ) + 2b2i−2 (Hn−3 ). (2) b2i (Bn ) = b2i (An ) + 2b2i−6 (Fn−8 ). (3) b2i (An ) = b2i (An−2 ) + b2i−2 (An−2 ) + b2i−2 (An−4 )

(n ≥ 8),

(3)

b2i (Bn ) = b2i (Bn−2 ) + b2i−2 (Bn−2 ) + b2i−2 (Bn−4 )

(n ≥ 10),

(4)

b2i (Dn ) = b2i (Dn−2 ) + b2i−2 (Dn−2 ) + b2i−2 (Dn−4 ) b2i (En ) = b2i (En−2 ) + b2i−2 (En−2 ) + b2i−2 (En−4 )

(n ≥ 10), (n ≥ 12).

(5) (6)

Proof. (1) By picking the edge e1 in Fn (see Fig. 3), Lemma 2.1(2) shows that b2i (Fn ) = b2i (Hn−1 ) + b2i−2 (Fn−2 ).

(7)

From Lemma 2.1(1), by picking the edge e2 in An (see Fig. 2), we have b2i (An ) = b2i (Fn ) + b2i−2 (Fn−2 ) − 2b2i−4 (Fn−4 ) = [b2i (Fn ) − b2i−2 (Fn−2 )] + 2[b2i−2 (Fn−2 ) − b2i−4 (Fn−4 )] = b2i (Hn−1 ) + 2b2i−2 (Hn−3 ) (by Equality (7)). (2) By picking the edge e3 in Hn+1 (see Fig. 3), Lemma 2.1(2) shows that b2i (Hn+1 ) = b2i (Fn ) + b2i−2 (Fn−2 ). From Lemma 2.1(2), by picking the edge e4 in Bn (see Fig. 2), we have b2i (Bn ) = b2i (K2 ∪ An−2 ) + b2i−2 (Hn−3 ) = b2i (K2 ∪ An−2 ) + b2i−2 (Fn−4 ) + b2i−4 (Fn−6 ) (by Equality (8)). From Lemma 2.1(2), by picking the edge e5 in An (see Fig. 2), we have b2i (An ) = b2i (K2 ∪ An−2 ) + b2i−2 (An−4 ) = b2i (K2 ∪ An−2 ) + b2i−2 (Fn−4 ) + b2i−4 (Fn−6 ) − 2b2i−6 (Fn−8 ). Hence b2i (Bn ) = b2i (An ) + 2b2i−6 (Fn−8 ).

(8)

H. Ma et al. / Discrete Applied Mathematics 186 (2015) 186–198

189

(3) From Lemma 2.1(2), by picking the edges e6 , e5 in An in order (see Fig. 2), we have b2i (An ) = b2i (An − u) + b2i−2 (An−2 ) = b2i (An − {u, v}) + b2i−2 (An − {u, v, w}) + b2i−2 (An−2 ) = b2i (An−2 ) + b2i−2 (An−4 ) + b2i−2 (An−2 ). Hence Equality (3) holds. Similarly, we can prove that Equalities (4)–(6) hold.



Theorem 2.8. Bn ≻ An for n ≥ 8. Proof. It is easy to obtain that

φ(B8 , x) = x8 − 8x6 + 16x4 − 8x2 , φ(A8 , x) = x8 − 8x6 + 16x4 − 6x2 . So B8 ≻ A8 . Suppose n > 8. By Lemma 2.7, b2i (Bn ) − b2i (An ) = 2b2i−6 (Fn−8 ) ≥ 0 and b8 (Bn ) > b8 (An ). So Bn ≻ An for n > 8. The proof is thus complete.  Theorem 2.9. D8 ≻ E8 and En ≻ Dn for n ≥ 10. Proof. It is easy to obtain that

φ(E8 , x) = x8 − 8x6 + 14x4 − 8x2 + 1, φ(D8 , x) = x8 − 8x6 + 15x4 − 8x2 + 1, φ(E10 , x) = x10 − 10x8 + 29x6 − 31x4 + 12x2 − 1, φ(D10 , x) = x10 − 10x8 + 29x6 − 28x4 + 10x2 − 1, φ(E12 , x) = x12 − 12x10 + 47x8 − 74x6 + 51x4 − 14x2 + 1, φ(D12 , x) = x12 − 12x10 + 47x8 − 72x6 + 46x4 − 12x2 + 1. So D8 ≻ E8 , E10 ≻ D10 and E12 ≻ D12 . Suppose n > 12. By picking the edges e8 in En and e7 in Dn (see Fig. 2), respectively, Lemma 2.1(2) shows that b2i (En ) = b2i (E12 ∪ Fn−12 ) + b2i−2 (E10 ∪ Fn−14 ), b2i (Dn ) = b2i (D12 ∪ Fn−12 ) + b2i−2 (D10 ∪ EF −14 ). Hence we have En ≻ Dn for n > 12. The proof is thus complete.



n 2

Theorem 2.10. E (S84 ) > E (B8 ) and Sn ≻ Bn for n ≥ 10.

.

.

n

Proof. It is easy to obtain that 9.65685 = E (S84 ) > E (B8 ) = 9.15298. Suppose n ≥ 10. Since b2i+1 (Sn2 ) ≥ 0 = b2i+1 (Bn ), we n

only need to consider b2i (Sn2 ) and b2i (Bn ). By Lemmas 2.1 and 2.7, Equalities (7) and (8), we have

 n   n  K1 , b ( F ) + b ( F ) − 2b 2i n 2i − 2 n − 4 n 2i− 2 2 b2i (Sn2 ) = n   b2i (Fn ) + b2i−2 (Fn−4 ) + 2b2i− n K1 , 2 2

if g ≡ 0 (mod 4) if g ̸≡ 0 (mod 4),

and b2i (Bn ) = b2i (Fn ) + b2i−2 (Fn−2 ) − 2b2i−4 (Fn−4 ) + 2b2i−6 (Fn−8 ) = b2i (Fn ) + b2i−2 (Fn−4 ) + b2i−4 (Fn−6 ) − b2i−4 (Fn−4 ) + 2b2i−6 (Fn−8 ) = b2i (Fn ) + b2i−2 (Fn−4 ) − b2i−6 (Fn−6 ) + b2i−6 (Fn−8 ) = b2i (Fn ) + b2i−2 (Fn−4 ) − b2i−8 (Fn−8 ) − b2i−8 (Fn−10 ). Since b

n 2i− 2n

2 n 2

K1



 =

1, 0,

if 2i =

n

2 otherwise, n

5 we have b2i (Sn ) ≥ b2i (Bn ), b8 (S10 ) > b8 (B10 ), and b10 (Sn2 ) > b10 (Bn ) when n ≥ 12. The proof is thus complete.



In the following, we will show that E (An ) < E (Dn ) by using the Coulson integral formula method, which had successfully been applied to compare the energy of two given graphs by Huo et al., see [12–15]. Before proving it, we prepare some results as follows.

190

H. Ma et al. / Discrete Applied Mathematics 186 (2015) 186–198

Lemma 2.11 ([3]). Let uv be an edge of G. Then

φ(G, x) = φ(G − uv, x) − φ(G − {u, v}, x) − 2



φ(G − C , x),

C ∈C (uv)

where C (uv) is the set of cycles containing uv . In particular, if uv is a pendent edge with pendent vertex v , then φ(G, x) = xφ(G − v, x) − φ(G − {u, v}, x). Lemma 2.12 ([30]). For any real number X > −1, we have X 1+X

≤ log(1 + X ) ≤ X .

In particular, log(1 + X ) < 0 if and only if X < 0. Lemma 2.13 ([8]). If G1 and G2 are two graphs with the same number of vertices, then E (G1 ) − E (G2 ) =

π

   φ(G1 , ix)   dx. φ(G , ix) 

+∞



1

−∞

log 

2

From Lemma 2.11, we can easily obtain the following lemma. Lemma 2.14. φ(An , x) = (x2 − 1)φ(An−2 , x) − x2 φ(An−4 , x) for n ≥ 8, and φ(Dn , x) = (x2 − 1)φ(Dn−2 , x) − x2 φ(Dn−4 , x) for n ≥ 10. By some easy calculations, we have φ(A6 , x) = x6 −6x4 +6x2 , φ(A8 , x) = x8 −8x6 +16x4 −6x2 , φ(D6 , x) = x6 −6x4 +5x2 −1 and φ(D8 , x) = x8 − 8x6 + 15x4 − 8x2 + 1. Now for convenience, we define some notations as follows: Y1 (x) = Z1 (x) = A1 (x) =

x2 − 1 +

√

x4 − 6x2 + 1

2

−x 2 − 1 +

√

,

x4 + 6x2 + 1

2 φ(A8 , x) − Y2 (x)φ(A6 , x)

Y2 (x) =

,

x2 − 1 −

√

x4 − 6x2 + 1

2

Z2 (x) =

−x 2 − 1 −

,

√

x4 + 6x2 + 1

2 φ(A8 , x) − Y1 (x)φ(A6 , x)

,

, A2 (x) = , (Y1 (x))4 − (Y1 (x))2 x2 (Y2 (x))4 − (Y2 (x))2 x2 φ(D8 , x) − Y2 (x)φ(D6 , x) φ(D8 , x) − Y1 (x)φ(D6 , x) B1 (x) = , B2 (x) = , 4 2 2 (Y1 (x)) − (Y1 (x)) x (Y2 (x))4 − (Y2 (x))2 x2 f6 (x) = x6 + 6x4 + 6x2 , f8 (x) = x8 + 8x6 + 16x4 + 6x2 , g6 (x) = x6 + 6x4 + 5x2 + 1, g8 (x) = x8 + 8x6 + 15x4 + 8x2 + 1.

It is easy to check that Y1 (ix) = Z1 (x), Y2 (ix) = Z2 (x), A1 (ix) =

f8 (x) + Z2 (x)f6 (x)

,

(Z1 (x)) + (Z1 (x)) g8 (x) + Z2 (x)g6 (x) B1 (ix) = , (Z1 (x))4 + (Z1 (x))2 x2 4

2 x2

A2 (ix) =

f8 (x) + Z1 (x)f6 (x)

, (Z2 (x))4 + (Z2 (x))2 x2 g8 (x) + Z1 (x)g6 (x) B2 (ix) = , (Z2 (x))4 + (Z2 (x))2 x2

Z1 (x) + Z2 (x) = −x2 − 1 and Z1 (x)Z2 (x) = −x2 . In addition, for x > 0, 0 <

Z1 (x) x

< 1; for x < 0, −1 <

Z1 (x) x

< 0.

Lemma 2.15. For n ≥ 6 and x ̸= 0, the characteristic polynomials of An and Dn have the following forms: n

n

n

n

φ(An , x) = A1 (x)(Y1 (x)) 2 + A2 (x)(Y2 (x)) 2 and

φ(Dn , x) = B1 (x)(Y1 (x)) 2 + B2 (x)(Y2 (x)) 2 . Proof. By Lemma 2.14, we have that φ(An , x), φ(Dn , x) satisfy the recursive formula f (n, x) = (x2 − 1)f (n − 2, x) − x2 f (n − n 4, x). Therefore, the form of the general solution of the linear homogeneous recursive relation is f (n, x) = C1 (x)(Y1 (x)) 2 + n C2 (x)(Y2 (x)) 2 . By some simple calculations, together with the initial values φ(A6 , x) and φ(A8 , x) (φ(D6 , x) and φ(D8 , x), respectively), we can get that Ci (x) = Ai (x) (Ci (x) = Bi (x), respectively), i = 1, 2.  Theorem 2.16. E (An ) < E (Dn ) for n ≥ 6.

H. Ma et al. / Discrete Applied Mathematics 186 (2015) 186–198

191

Proof. By Lemma 2.13, we have 1

E (An ) − E (Dn ) =



π

+∞

−∞

   φ(An , ix)    dx. log  φ(Dn , ix) 

From Lemma 2.15, we know that both φ(An , ix) and φ(Dn , ix) are polynomials of x with all real coefficients. For convenience, we abbreviate Ak (ix), Bk (ix) and Zk (x) to Ak , Bk and Ck for k = 1, 2, and abbreviate fk (x) and gk (x) to fk and gk for k = 6, 8, respectively. In the following, we assume that x ̸= 0. We distinguish two cases in terms of the parity of n/2. Case 1. n = 4k (k ≥ 2). Notice that Z1 Z2 = −x2 . When n → ∞, n

n

Z A2 + A1 ( x1 )n A1 Z12 + A2 Z22 A2 φ(An , ix) = → . n = n Z1 n φ(Dn , ix) B2 2 2 B2 + B1 ( x ) B1 Z1 + B2 Z2

We will show that

     φ(An , ix)   A2 (ix)     = log A2 .  log  < log  φ(D , ix)  B (ix)  B n

2

2

Assume that

       φ(An , ix)   A2  F1 (n, x)     2 log  − 2 log   = log 1 + . φ(Dn , ix)  B2 G1 (n, x) Then we get that G1 (n, x) = (A2 (ix)φ(Dn , ix))2 > 0 and n

n

n

n

F1 (n, x) = B22 (A1 Z12 + A2 Z22 )2 − A22 (B1 Z12 + B2 Z22 )2 n

= (A21 B22 − A22 B21 )Z1n + 2A2 B2 (A1 B2 − A2 B1 )(Z1 Z2 ) 2 . n

n

Notice that Z1n > 0, (Z1 Z2 ) 2 = (−x2 ) 2 > 0 and A2 B2 > 0. By some elementary calculations, we have

√ (−3x8 − 15x6 − 8x4 ) x4 + 6x2 + 1 A1 B2 − A2 B1 = < 0, x6 (x4 + 6x2 + 1) and A1 B2 + A2 B1 =

4x4 + 17x6 + 20x8 + 3x10 x6 (x4 + 6x2 + 1)

> 0.

Therefore F1 (n, x) < 0. Hence by Lemma 2.12, 1

π



+∞

log −∞

A2 (ix) B2 (ix)

dx ≤

1



π

+∞

−∞



A2 (ix)

 . 1 − 1 dx = (−0.8538292323) < 0. B2 (ix) π

Consequently, E (An ) − E (Dn ) ≤

1

π



+∞

log −∞

A2 (ix) B2 (ix)

dx < 0.

  φ(A



,ix) 

Case 2. n = 4k + 2 (k ≥ 1). We will show that log  φ(Dn ,ix)  is monotonically decreasing in n. Assume that n

       φ(An+4 , ix)   φ(An , ix)  F2 (n, x)     2 log  − 2 log  = log 1 + . φ(Dn+4 , ix)  φ(Dn , ix)  G2 (n, x) Then we can obtain that G2 (n, x) = (φ(An , ix)φ(Dn+4 , ix))2 > 0 and F2 (n, x) = (φ(An+4 , ix)φ(Bn , ix))2 − (φ(An , ix)φ(Dn+4 , ix))2 n

= (A1 Z12

+2

n

+ A2 Z22

+2 2

n

n

n

) (B1 Z12 + B2 Z22 )2 − (B1 Z12

+2

n

+ B2 Z22

+2 2

n

n

) (A1 Z12 + A2 Z22 )2

= −(A1 B2 − A2 B1 )xn (Z12 − Z22 ) · F3 (n, x), n

where F3 (n, x) = (A1 B2 + A2 B1 )(Z1 Z2 ) 2 (Z12 + Z22 ) + 2A1 B1 Z1n+2 + 2A2 B2 Z2n+2 . Since A1 B2 − A2 B1 < 0, Z12 − Z22 < 0 and xn > 0, to prove F2 (n, x) < 0, it suffices to show that F3 (n, x) > 0. By some elementary calculations, we can get that F3 (n, x) =

2f8 g8 F4 (n, x) + 2f6 g6 F5 (n, x) + (f8 g6 + f6 g8 )F6 (n, x) x4 + 6x2 + 1

,

192

H. Ma et al. / Discrete Applied Mathematics 186 (2015) 186–198

where n

F4 (n, x) = Z1n−4 + Z2n−4 − (Z1 Z2 ) 2 −3 (Z12 + Z22 ) n

= (Z12 F5 (n, x) =

−1

n

− Z22

Z22 Z1n−4

+

−1

F6 (n, x) =

n

= Z1 Z2 [(Z12

n

−3

− Z22

Z12 Z2n−4

− (Z1 Z2 )

n −2 2

n −2 2

= (Z1 Z2 )2 (Z1 2Z2 Z1n−4

n

)(Z12

− Z2

2Z1 Z2n−4

+ −1

n

− Z22

−1

Z12k − Z22k = Z12k 1 −



x Z1

),

n −2 2

n −4 2

)(Z1

(Z12 + Z22 ) n

− Z22

−4

), n

− (Z1 + Z2 )(Z1 Z2 ) 2 −3 (Z12 + Z22 ) n

)(Z12

Notice that Z1 > 0 and Z2 < 0. We have



−3

4k 

n

−4

Z1k

− Z22

−

Z2k

−4

n

) + (Z12

−3

n

− Z22

−3

n

)(Z12

−2

n

− Z22

−2

)].

> 0 when k is odd. On the other hand, for k ≥ 1, we have

< 0.

Therefore, we have F4 (n, x) ≥ 0, F5 (n, x) ≥ 0, F6 (n, x) > 0, and so F3 (n, x) > 0 and F2 (n, x) < 0. Hence,

.

E (An ) − E (Dn ) ≤ E (A6 ) − E (D6 ) = 6.60272 − 7.20775 < 0. The proof is thus complete.



3. Main results Let G be a graph in Un with vertex set V and a unique cycle Cg . We use MG to denote one arbitrary selected prefect matching of G. Let x, y ∈ V . Denote by dG (x, y) (dG (x, Cg ), respectively) the distance between vertex x and y (Cg , respectively). Define d = d(G) = maxx∈V (G) {dG (x, Cg )}, V1 = V1 (G) = {x ∈ V |dG (x, Cg ) = d(G)}, and t = t (G) = |V1 (G)|. Clearly, the vertices in V1 are pendent vertices when d(G) ≥ 1. This section is devoted to prove Theorem 1.2. We first outline the proof. It follows from Theorems 2.8 and 2.10 that n

E (Sn2 ) > E (Bn ) > E (An ) for n ≥ 8. Hence by Theorem 1.1, it suffices to show that for a graph An  G ∈ Un with g ≡ 0 (mod 4), E (G) > E (An ). To this end, the proof will be divided into two steps. First we show that if g ≥ 8, then G ≻ Bn (see Theorem 3.6). Next we prove that if g = 4, then E (G) > E (An ) (see Theorem 3.7). In each step, the proof is by induction on d. Lemmas 3.3–3.5 deal with the cases g ≥ 8 and d = 0, 1, 2, respectively. Let us begin with two useful lemmas. Lemma 3.1 will be used in the proofs of Lemmas 3.4 and 3.5. Lemma 3.2 will be repeatedly used in the proofs of next results. Lemma 3.1. Let T be a tree of order n − 1 (n − 1 is odd) obtained from a path u1 u2 . . . un−t −1 by adding t pendent edges vk1 uk1 , vk2 uk2 , . . . , vkt ukt . Suppose that t ≥ 1, 1 ≤ k1 < k2 < · · · < kt ≤ n − t − 1, and ki+1 − ki is odd for 1 ≤ i ≤ t − 1. Then T ≽ Hn−1 . Proof. Since n − 1 is odd, one and only one of the two numbers k1 − 1 and n − t − 1 − kt is odd. Without loss of generality, we assume that k1 − 1 is odd. Denote k0 = 1, and kt +1 = n − t − 2 if kt < n − t − 1 − kt . Then ki+1 − ki is odd for 0 ≤ i ≤ t. Notice that if ki+1 − ki = 2j + 1 > 1, then from T , we can obtain a different tree T ′ with ∆ ≤ 3 by carrying out j steps of e.g.t. Therefore we can finally get Hn−1 from T by carrying out e.g.t repeatedly, if necessary. By Lemma 2.3, we have T ≽ Hn−1 .  Lemma 3.2. Let G ∈ Un with a cycle Cg = z1 z2 . . . zg , g ≡ 0 (mod 4). Suppose that d(G) ≥ 3 and x1 x2 x3 . . . xd z1 is a path with length d in G. Then we have (1) b2i (G) = b2i (G1 ) + b2i−2 (G1 ) + b2i−2 (G2 ),

(9)

where G1 = G − {x1 , x2 } and G2 = G − {x1 , x2 , x3 }. (2) Suppose that G′n ∈ Un , and G′n satisfies the following recursive formula b2i (G′n ) = b2i (G′n−2 ) + b2i−2 (G′n−2 ) + b2i−2 (G′n−4 ). ′

′

′

(10) ′

′

If G1 ≻ Gn−2 and G2 ≽ Gn−4 ∪ K1 , or G1 ≽ Gn−2 and G2 ≻ Gn−4 ∪ K1 , then G ≻ Gn . Proof. (1) Note that dG (x1 ) = 1 and dG (x2 ) = 2. By Lemma 2.1(2), we have b2i (G) = b2i (G − x1 ) + b2i−2 (G − {x1 , x2 }) = b2i (G − {x1 , x2 }) + b2i−2 (G − {x1 , x2 , x3 }) + b2i−2 (G − {x1 , x2 }) = b2i (G1 ) + b2i−2 (G1 ) + b2i−2 (G2 ). (2) The result directly follows from Equalities (9), (10) and the definition of quasi-order relation.



H. Ma et al. / Discrete Applied Mathematics 186 (2015) 186–198

193

Fig. 4. The graphs T and T ′ in Operation I.

In the following, we introduce four graph operations. Let T be a tree of the form in Fig. 4, where T1 and T2 are subtrees of T with at least one vertices. If T ′ is obtained from T by deleting the edge xy and adding one new edge xw , we say that T ′ is obtained from T by Operation I. Proposition 1. If T ′ is obtained from T by Operation I, then T ≻ T ′ . Proof. By Lemma 2.1(2), we have b2i (T ) = b2i (T − x) + b2i−2 (T − {x, y}), b2i (T ′ ) = b2i (T ′ − x) + b2i−2 (T ′ − {x, w}). Note that T − x = T ′ − x, and T ′ − {x, w} is isomorphic to a proper subgraph of T − {x, y}. Then T − {x, y} ≻ T ′ − {x, w} by Lemma 2.4. Therefore T ≻ T ′ .  Let G ∈ Un with a cycle Cg = z1 z2 . . . zg , g ≡ 0 (mod 4), g ≥ 8. Suppose that d(G) = 2, t (G) ≥ 2, x1 y1 z1 and x2 y2 z2 are two paths with length 2 in G. If G′ is the graph obtained from G by deleting the edge z2 y2 and adding one new edge y2 y1 , then we say that G′ is obtained from G by Operation II. Clearly, G′ ∈ Un , and d(G′ ) = 3. Proposition 2. If G′ is obtained from G by Operation II, then G ≻ G′ . Proof. By Lemma 2.1, we have b2i (G) = b2i (G − x2 ) + b2i−2 (G − {x2 , y2 }), b2i (G′ ) = b2i (G′ − x2 ) + b2i−2 (G′ − {x2 , y2 }), and b2i (G − x2 ) = b2i (G − {x2 , x1 }) + b2i−2 (G − {x2 , x1 , y1 }) = b2i (G − {x2 , x1 , y2 }) + b2i−2 (G − {x2 , x1 , y2 , z2 }) + b2i−2 (G − {x2 , x1 , y1 , y2 }) + b2i−4 (G − {x2 , x1 , y1 , y2 , z2 }), b2i (G′ − x2 ) = b2i (G′ − {x2 , x1 }) + b2i−2 (G′ − {x2 , x1 , y1 }) = b2i (G′ − {x2 , x1 , y2 }) + 2b2i−2 (G′ − {x2 , x1 , y2 , y1 }). Denote G3 = G − {x2 , x1 , y2 , z2 }, G4 = G − {x2 , x1 , y1 , y2 , z2 } and G5 = G − {x2 , x1 , y2 , y1 }. Since G − {x2 , y2 } = G′ − {x2 , y2 }, G − {x2 , x1 , y2 } = G′ − {x2 , x1 , y2 }, and G − {x2 , x1 , y1 , y2 } = G′ − {x2 , x1 , y2 , y1 }, we have b2i (G) − b2i (G′ ) = b2i−2 (G3 ) + b2i−4 (G4 ) − b2i−2 (G5 ). Furthermore, b2i−2 (G5 ) = b2i−2 (G5 − {z1 z2 }) + b2i−4 (G5 − {z1 , z2 }) − 2b2i−g (G5 − Cg ) = b2i−2 (G5 − {z1 z2 , z2 z3 }) + b2i−4 (G5 − {z1 z2 } − {z2 , z3 }) + b2i−4 (G5 − {z1 , z2 }) − 2b2i−g (G5 − Cg ) = b2i−2 (G5 − {z2 }) + b2i−4 (G5 − {z2 , z3 }) + b2i−4 (G5 − {z1 , z2 }) − 2b2i−g (G5 − Cg ), b2i−2 (G3 ) = b2i−2 (G3 − {y1 }) + b2i−4 (G3 − {y1 , z1 }) = b2i−2 (G5 − {z2 }) + b2i−4 (G5 − {z1 , z2 }). Note that G5 − {z2 , z3 } = G4 − z3 . Hence G4 ≻ [G5 − {z2 , z3 }] ∪ K1 by Lemma 2.5. Therefore b2i (G) − b2i (G′ ) = b2i−4 (G4 ) − b2i−4 (G5 − {z2 , z3 }) + 2b2i−g (G5 − Cg ) ≥ 0, and b6 (G) − b6 (G′ ) = b2 (G4 ) − b2 (G5 − {z2 , z3 }) > 0. Thus the result G ≻ G′ holds.



Let G ∈ Un with a cycle Cg = z1 z2 . . . zg , g ≡ 0 (mod 4). Suppose that d(G) ≥ 3, t (G) ≥ 2, x1 x2 x3 . . . xd z1 and y1 y2 x3 . . . xd z1 are two paths with length d in G. If G′ is the graph obtained from G by deleting two edges x3 y2 , y2 y1 and adding two new edges y1 x1 and y2 x2 , then we say that G′ is obtained from G by Operation III. Clearly, G′ ∈ Un , d(G′ ) = d + 1, and y1 x1 x2 . . . xd z1 is a path with length d + 1 in G′ .

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Proposition 3. If G′ is obtained from G by Operation III, then G ≻ G′ . Proof. By Lemma 3.2(1), we have b2i (G) = b2i (G − {y1 , y2 }) + b2i−2 (G − {y1 , y2 }) + b2i−2 (G − {y1 , y2 , x3 }), b2i (G′ ) = b2i (G′ − {y1 , x1 }) + b2i−2 (G′ − {y1 , x1 }) + b2i−2 (G′ − {y1 , x1 , x2 }). It is easy to see that G − {y1 , y2 } ∼ = G′ − {y1 , x1 }. Let G1 = G − {x1 , x2 , y1 , y2 }. Then b2i−2 (G − {y1 , y2 , x3 }) = b2i−2 ((G1 − x3 ) ∪ K2 ) = b2i−2 (G1 − x3 ) + b2i−4 (G1 − x3 ), b2i−2 (G′ − {y1 , x1 , x2 }) = b2i−2 (G1 ) = b2i−2 (G1 − x3 ) + b2i−4 (G1 − {x3 , x4 }). Since G1 − x3 ≻ (G1 − {x3 , x4 }) ∪ K1 by Lemma 2.5, we have b2i (G) ≥ b2i (G′ ) and b6 (G) > b6 (G′ ). Thus the result G ≻ G′ holds.  Let G ∈ Un with a cycle Cg = z1 z2 . . . zg , g ≡ 0 (mod 4). Suppose that d(G) = 3, t (G) ≥ 2, x1 x2 x3 z1 and y1 y2 y3 zi (2 ≤ i ≤ g) are two paths with length 3 in G, where dG (x3 ) = dG (y3 ) = 2. If G′ is the graph obtained from G by deleting two edges y1 y2 , y2 y3 and adding two new edges y1 x1 and y2 x2 , then we say that G′ is obtained from G by Operation IV. Clearly, G′ ∈ Un , d(G′ ) = 4, and y1 x1 x2 x3 z1 is a path with length 4 in G′ . Proposition 4. If G′ is obtained from G by Operation IV, then G ≻ G′ . Proof. By Lemma 3.2(1), we have b2i (G) = b2i (G − {y1 , y2 }) + b2i−2 (G − {y1 , y2 }) + b2i−2 (G − {y1 , y2 , y3 }), b2i (G′ ) = b2i (G′ − {y1 , x1 }) + b2i−2 (G′ − {y1 , x1 }) + b2i−2 (G′ − {y1 , x1 , x2 }). It is easy to see that G − {y1 , y2 } ∼ = G′ − {y1 , x1 }. Let G1 = G − {x1 , y1 , y2 , y3 }. Then b2i−2 (G − {y1 , y2 , y3 }) = b2i−2 (G1 ) + b2i−4 (G1 − x2 ) b2i−2 (G′ − {y1 , x1 , x2 }) = b2i−2 (G′ − {y1 , y2 , x2 , x1 }) = b2i−2 (G1 − x2 ) + b2i−4 (G1 − {x2 , zi }). It follows from Lemma 2.5 that G1 ≻ (G1 − x2 ) ∪ K1 and G1 − x2 ≻ (G1 − {x2 , zi }) ∪ K1 . Hence we have b2i (G) ≥ b2i (G′ ) and b4 (G) > b4 (G′ ). Thus the result G ≻ G′ holds.  The following three lemmas deal with the cases g ≥ 8 and d = 0, 1, 2, respectively. Lemma 3.3. Let Cn be a cycle with n ≥ 8 vertices such that n ≡ 0 (mod 4). Then Cn ≻ Bn . Proof. By Lemma 2.1(1), b2i (Cn ) =



b2i (Pn ) + b2i−2 (Pn−2 ), b2i (Pn ) + b2i−2 (Pn−2 ) − 2,

if 2i ̸= n if 2i = n.

By picking the edge e9 in Bn (see Fig. 2), Lemma 2.1(1) shows that b2i (Bn ) = b2i (G1 ) + b2i−2 (K2 ∪ Fn−4 ) − 2b2i−4 (K2 ∪ Fn−6 ), where G1 is the tree of order n obtained by attaching a path with 4 edges to one of the two vertices of degree 2 of Fn−4 . By Lemmas 2.2 and 2.4, Pn ≻ G1 , and Pn−2 ≻ Fn−2 ≻ K2 ∪ Fn−4 . On the other hand, we have bn (Cn ) = bn (Bn ) and b4 (Cn ) > b4 (Bn ). Hence Cn ≻ Bn .  Lemma 3.4. Let G ∈ Un , g (G) ≡ 0 (mod 4), g (G) ≥ 8, and d(G) = 1. Then G ≻ Bn . n

Proof. It is easy to see that n ≥ 10 and t (G) is even. If t = g, then G ∼ = Sn2 , and so G ≻ Bn by Theorem 2.10. So in the following, we suppose 2 ≤ t ≤ g − 2. Let Cg = y1 y2 . . . yg , and xk1 yk1 , xk2 yk2 , . . . , xkt ykt (1 = k1 < k2 < · · · < kt ) be all the edges outside Cg . Then there must exist an index ki such that ki + 1, ki + 2 (mod g) are not in the set {k1 , k2 , . . . , kt }, that is yki +1 yki +2 ∈ MG . Without loss of generality, we assume that ki = 1. Since G has a perfect matching, we have that 4 ≤ k2 < k3 < · · · < kt ≤ g, k2 − 3 is odd, ki+1 − ki is odd for 2 ≤ i ≤ t − 1, and g − kt is even. By Lemma 2.1, we have b2i (G) = b2i (G − x1 ) + b2i−2 (G − {x1 , y1 }), and b2i (G − x1 ) =



b2i (G − x1 − y1 y2 ) + b2i−2 (G − {x1 , y1 , y2 }), b2i (G − x1 − y1 y2 ) + b2i−2 (G − {x1 , y1 , y2 }) − 2,

if 2i ̸= g if 2i = g .

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195

Note that G − {x1 , y1 } is a conjugated tree of order n − 2 with ∆ ≤ 3, by Lemma 2.6, G − {x1 , y1 } ≽ Fn−2 . Denote T1 = G − x1 − y1 y2 and T2 = G − {x1 , y1 , y2 }. It follows from Lemma 3.1 that T1 ≽ Hn−1 and T2 ≽ Hn−3 . By Lemma 2.7, we have b2i (Bn ) = b2i (Hn−1 ) + 2b2i−2 (Hn−3 ) + 2b2i−6 (Fn−8 ). Hence for 2i ̸= g, we have b2i (G) ≥ b2i (Hn−1 ) + b2i−2 (Fn−2 ) + b2i−2 (Hn−3 ) = b2i (Hn−1 ) + 2b2i−2 (Hn−3 ) + b2i−4 (Fn−4 ) = b2i (Bn ) + b2i−4 (Fn−4 ) − 2b2i−6 (Fn−8 ) = b2i (Bn ) + b2i−4 (Fn−6 ) + b2i−8 (Fn−8 ) + b2i−8 (Fn−10 ) ≥ b2i (Bn ), and bg (G) ≥ bg (Bn ) + bg −4 (Fn−6 ) + bg −8 (Fn−8 ) + bg −8 (Fn−10 ) − 2 ≥ bg (Bn ), since g ≤ n − 2, bg −4 (Fn−6 ) ≥ 1 and bg −8 (Fn−8 ) ≥ 1. On the other hand, it is obvious that b4 (G) > b4 (Bn ). Thus G ≻ Bn .



Lemma 3.5. Let G ∈ Un , g (G) ≡ 0 (mod 4), g (G) ≥ 8, and d(G) = 2. Then G ≻ Bn . Proof. Let Cg = z1 z2 . . . zg . We apply induction on t. Suppose t = 1. Assume that dG (x1 , z1 ) = 2 and x1 y1 z1 is a path with length 2 in G. Then dG (y1 ) = 2 and x1 y1 ∈ MG . Since G has a perfect matching, either z1 z2 ∈ MG or z1 zg ∈ MG . We may assume that z1 z2 ∈ MG . Let E1 = {zi1 yi1 , zi2 yi2 , . . . , zik yik } (i1 < i2 < · · · < ik ) be the set of all edges in MG \ x1 y1 outside Cg . If E1 is not empty, then we have that 3 ≤ i1 < i2 < · · · < ik ≤ g, i1 − 3 is even, ij+1 − ij is odd for 1 ≤ j ≤ k − 1, and g − ik is even. By Lemma 2.1, we have b2i (G) = b2i (G − x1 ) + b2i−2 (G − {x1 , y1 }) = b2i (G − {x1 , y1 }) + b2i−2 (G − {x1 , y1 , z1 }) + b2i−2 (G − {x1 , y1 }). Denote G1 = G − {x1 , y1 } and G2 = G − {x1 , y1 , z1 }. Then G1 ∈ Un−2 with d(G1 ) ≤ 1. By Lemmas 3.3 and 3.4, we have G1 ≻ Bn−2 . It follows from Lemma 3.1 that G2 ≽ Hn−3 . Therefore by Lemmas 2.1 and 2.7, b2i (G) ≥ b2i (Bn−2 ) + b2i−2 (Bn−2 ) + b2i−2 (Hn−3 ) = b2i (Bn−2 ) + b2i−2 (Bn−2 ) + b2i−2 (Hn−5 ) + 2b2i−4 (Hn−7 ) + 2b2i−6 (Fn−8 ) = b2i (Bn−2 ) + b2i−2 (Bn−2 ) + b2i−2 (An−4 ) + 2b2i−6 (Fn−8 ) = b2i (Bn−2 ) + b2i−2 (Bn−2 ) + b2i−2 (Bn−4 ) + 2b2i−6 (Fn−8 ) − 2b2i−8 (Fn−12 ) = b2i (Bn−2 ) + b2i−2 (Bn−2 ) + b2i−2 (Bn−4 ) + 2b2i−6 (Fn−10 ) + 2b2i−8 (Fn−10 ) = b2i (Bn ) + 2b2i−6 (Fn−10 ) + 2b2i−8 (Fn−10 ) ≥ b2i (Bn ). Since G1 ≻ Bn−2 , there exists some i such that b2i (G) > b2i (Bn ). Hence G ≻ Bn for t = 1. Assume now that t ≥ 2 and the assertion holds for smaller values of t. Let V1 (G) = {x1 , x2 , . . . , xt }, and for each i ∈ {1, 2, . . . , t }, xi yi zki (1 = k1 < k2 < · · · < kt ≤ g) be a path with length 2 in G. Then dG (yi ) = 2 and xi yi ∈ MG . For convenience, let kt +1 = k1 . We consider the following two cases: Case 1. There exist two indices ki and ki+1 such that zki zki+1 is an edge on Cg . Without loss of generality, we assume that ki = 1, ki+1 = 2. Now let G′ be the graph obtained from G by Operation II. It follows from Proposition 2 that G ≻ G′ . Hence it suffices to prove that G′ ≽ Bn . Since d(G′ ) = 3 and x2 y2 y1 z1 is a path with length 3 in G′ , by Lemma 3.2(1), we have b2i (G′ ) = b2i (G′ − {x2 , y2 }) + b2i−2 (G′ − {x2 , y2 }) + b2i−2 (G′ − {x2 , y2 , y1 }). Denote G6 = G′ − {x2 , y2 } and G7 = G′ − {x2 , y2 , y1 }. Then it is easy to see that G6 ∈ Un−2 , G7 − x1 ∈ Un−4 , and for i = 6, 7, we have d(Gi ) ≤ 1 or d(Gi ) = 2 and t (Gi ) < t (G). Therefore by Lemmas 3.3 and 3.4 and the induction hypothesis, we have G6 ≻ Bn−2 and G7 − x1 ≻ Bn−4 . Note that dG7 (x1 ) = 0, and Bn satisfies Formula (10) by Lemma 2.7(3). Hence G′ ≻ Bn follows from Lemma 3.2(2). Case 2. For the indices 1 = k1 < k2 < · · · < kt ≤ g, we have ki+1 − ki ≥ 2 for 1 ≤ i ≤ t − 1 and kt ≤ g − 2. Since G has a perfect matching, we may assume that z1 z2 ∈ MG , and zki wki ∈ MG for 2 ≤ i ≤ t, where wki ∈ {zki −1 , zki +1 }. Similar to the above case t = 1, we have b2i (G) = b2i (G − {x1 , y1 }) + b2i−2 (G − {x1 , y1 , z1 }) + b2i−2 (G − {x1 , y1 }), and G − {x1 , y1 } ≻ Bn−2 . Denote T = G − {x1 , y1 , z1 }. Let T ′ be the tree obtained from T by deleting t − 1 edges x2 y2 , x3 y3 , . . . , xt yt and adding t new edges x2 wk2 , x3 wk3 , . . . , xt wkt . Then from T we can obtain T ′ by applying Operation I t − 1 times. By Proposition 1, we have T ≻ T ′ . Clearly, T ′ is a tree with ∆ ≤ 3. Now we can assume that zj1 , zj2 , . . . , zjl (2 < j1 < j2 < · · · < jl ≤ g) are all vertices

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with degree 3 in T ′ . Then we have that j1 − 2 is odd, ji+1 − ji is odd for 1 ≤ i ≤ l − 1, and g − jl is even. It follows from Lemma 3.1 that T ′ ≽ Hn−3 . Therefore we have T ≻ Hn−3 , and so similar to the above case t = 1, we can finally obtain G ≻ Bn . The proof is thus complete.  Theorem 3.6. Let G ∈ Un , g (G) ≡ 0 (mod 4), g (G) ≥ 8. Then G ≻ Bn . Proof. We apply induction on d. As the case d ≤ 2 was proved by Lemmas 3.3–3.5, we now suppose that d ≥ 3 and the assertion holds for smaller values of d. Let Cg = z1 z2 . . . zg . Assume that dG (x1 , z1 ) = d and x1 x2 . . . xd z1 is a path with length d in G. Then dG (x2 ) = 2 and x1 x2 ∈ MG . For convenience, denote xd+1 = z1 . Let G1 = G − {x1 , x2 } and G2 = G − {x1 , x2 , x3 }. By Lemmas 3.2(1) and 2.7(3), we have b2i (G) = b2i (G1 ) + b2i−2 (G1 ) + b2i−2 (G2 ), b2i (Bn ) = b2i (Bn−2 ) + b2i−2 (Bn−2 ) + b2i−2 (Bn−4 )

(n ≥ 10).

Now we prove the result for a given d by induction on t. Case 1. Suppose t = 1. It is easy to see that G1 ∈ Un−2 , and d(G1 ) < d. Hence by the induction hypothesis, G1 ≽ Bn−2 . To prove G ≻ Bn , by Lemma 3.2(2), we only need to show that G2 ≻ Bn−4 ∪ K1 . To this end, we will repeatedly make use of the following claim: Claim 1. If there exists a vertex u in G2 such that dG2 (u) = 1 (dG2 (u) = 0, respectively), G2 − u ∈ Un−4 and d(G2 − u) < d, then we have G2 ≻ Bn−4 ∪ K1 (G2 ≽ Bn−4 ∪ K1 , respectively). Proof. Since G2 − u ∈ Un−4 and d(G2 − u) < d, by the induction hypothesis, we have G2 − u ≽ Bn−4 . If dG2 (u) = 0, it is clear that G2 ≽ Bn−4 ∪ K1 . Otherwise, by Lemma 2.5, we obtain G2 ≻ (G2 − u) ∪ K1 , and so G2 ≻ Bn−4 ∪ K1 .  Subcase 1.1. dG (x3 ) = dG (x4 ) = 2. Note that x3 x4 ∈ MG . It is easy to see that G2 − x4 ∈ Un−4 and d(G2 − x4 ) < d. Hence by Claim 1, we have G2 ≻ Bn−4 ∪ K1 , and so we are done. Subcase 1.2. dG (x3 ) = 2, dG (x4 ) = 3 and d > 3. Note that x3 x4 ∈ MG . Suppose y2 ̸∈ {x3 , x5 } is a neighbor of x4 , and y1 y2 ∈ MG . Since t = 1, we have dG (y1 ) = 1 and dG (y2 ) = 2. Then we have G2 − y1 ∈ Un−4 , since (MG \ {x1 x2 , x3 x4 , y1 y2 }) ∪ {y2 x4 } is a perfect matching of G2 − y1 . Clearly, d(G2 − y1 ) < d. Hence by Claim 1, we have G2 ≻ Bn−4 ∪ K1 , and so we are done. Subcase 1.3. dG (x3 ) = 2, dG (x4 ) = 3 and d = 3, i.e., x4 = z1 . Note that x3 x4 ∈ MG . Since G has a perfect matching, g (G) ≡ 0 (mod 4), there exist k ≥ 1 pendent edges zi1 yi1 , . . . , zik yik (2 ≤ i1 < · · · < ik ≤ g) such that k is odd and zik yik ∈ MG . Then we have G2 − yi1 ∈ Un−4 , since

(MG \ {x1 x2 , x3 x4 , zi1 yi1 , z2 z3 , z4 z5 , . . . , zi1 −2 zi1 −1 }) ∪ {z1 z2 , z3 z4 , . . . , zi1 −1 zi1 } is a perfect matching of G2 − yi1 . Clearly, d(G2 − yi1 ) < d. Hence by Claim 1, we have G2 ≻ Bn−4 ∪ K1 , and so we are done. Subcase 1.4. dG (x3 ) = 3. Since t = 1, x3 x4 ̸∈ MG , we assume that x3 y1 ∈ MG . Then dG (y1 ) = 1, dG2 (y1 ) = 0, d(G2 − y1 ) < d and G2 − y1 ∈ Un−4 . Hence by Claim 1, G2 ≽ Bn−4 ∪ K1 . Note that in this case we have G1 ≻ Bn−2 . Therefore we can obtain that G ≻ Bn by Lemma 3.2(2). Case 2. Assume now that t ≥ 2 and the assertion holds for smaller values of t. Note that G1 ∈ Un−2 with d(G1 ) = d and t (G1 ) < t. By the induction hypothesis, G1 ≽ Bn−2 . To prove G ≻ Bn , by Lemma 3.2(2), we only need to show that G2 ≻ Bn−4 ∪ K1 . Similar to Case 1, we can obtain the following claim: Claim 2. If there exists a vertex u in G2 such that dG2 (u) = 1 (dG2 (u) = 0, respectively), G2 − u ∈ Un−4 , and either d(G2 − u) = d, t (G2 − u) < t or d(G2 − u) < d, then we have G2 ≻ Bn−4 ∪ K1 (G2 ≽ Bn−4 ∪ K1 , respectively). Subcase 2.1. dG (x3 ) = dG (x4 ) = 2. It is easy to see that x3 x4 ∈ MG , G2 − x4 ∈ Un−4 , d(G2 − x4 ) = d and t (G2 − x4 ) < t. Hence by Claim 2, we have G2 ≻ Bn−4 ∪ K1 , and so we are done. Subcase 2.2. dG (x3 ) = 3 and x3 x4 ̸∈ MG . Since x3 x4 ̸∈ MG , we assume that x3 y1 ∈ MG . Then dG (y1 ) = 1, dG2 (y1 ) = 0, d(G2 − y1 ) = d, t (G2 − y1 ) < t and G2 − y1 ∈ Un−4 . Hence by Claim 2, G2 ≽ Bn−4 ∪ K1 . Note that in this case we have G1 ≻ Bn−2 . Therefore we can obtain that G ≻ Bn by Lemma 3.2(2). Subcase 2.3. dG (x3 ) = 2, dG (x4 ) = 3, and d > 3. Suppose y2 ̸∈ {x3 , x5 } is a neighbor of x4 , and y1 y2 ∈ MG . If dG (y2 ) = 3, let y3 ̸∈ {x4 , y1 } be a neighbor of y2 , and y4 y3 ∈ MG . Then we have dG (y1 ) = dG (y4 ) = 1, dG (y3 ) = 2, and G2 − y1 ∈ Un−4 , since (MG \ {x1 x2 , x3 x4 , y1 y2 } ∪ {y2 x4 }) is a perfect matching of G2 − y1 . Since d(G2 − y1 ) = d and t (G2 − y1 ) < t, by Claim 2, we have G2 ≻ Bn−4 ∪ K1 , and so we are done.

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Fig. 5. The graphs considered in Theorem 3.7.

Subcase 2.4. dG (x3 ) = 3 and x3 x4 ∈ MG . Suppose y2 ̸∈ {x2 , x4 } is a neighbor of x3 , and y1 y2 ∈ MG . Then we have dG (y1 ) = 1 and dG (y2 ) = 2. Let G′ be the graph obtained from G by Operation III. It follows from Proposition 3 that G ≻ G′ . Clearly, d(G′ ) = d + 1, t (G′ ) = 1, and y1 x1 . . . xd z1 is a path of length d + 1 in G′ . By Lemma 3.2(1), we have b2i (G′ ) = b2i (G′1 ) + b2i−2 (G′1 ) + b2i−2 (G′2 ), where G′1 = G′ − {y1 , x1 } and G′2 = G′ − {y1 , x1 , x2 , y2 }. It is easy to see that G′1 ∈ Un−2 , G′2 ∈ Un−4 , d(G′1 ) = d, t (G′1 ) < t, and either d(G′2 ) = d, t (G′2 ) < t or d(G′2 ) < d. Hence by the induction hypothesis, G′1 ≽ Bn−2 and G′2 ≽ Bn−4 . Therefore G′ ≽ Bn , and so G ≻ Bn . Subcase 2.5. dG (x3 ) = 2 and d = 3. Now x4 = z1 . Since t ≥ 2, suppose that y1 ∈ V1 (G) and y1 y2 y3 zi (i ̸= 2) is a path with length 3 in G. Then by the above subcases, we may assume that dG (y3 ) = 2. Let G′ be the graph obtained from G by Operation IV. Then we have G ≻ G′ by Proposition 4. Clearly, d(G′ ) = 4, t (G′ ) = 1, and y1 x1 x2 x3 z1 is a path of length 4 in G′ . By the same proof as that of Subcase 2.4, we can obtain G′ ≽ Bn . Hence G ≻ Bn . The proof is thus complete.  Theorem 3.7. Let G ∈ Un (n ≥ 6) with a cycle C4 . If G  An , then E (G) > E (An ). Proof. Let An  G ∈ Un and E1 be the set of C4 . We distinguish the following three cases. Case 1. |MG ∩ E1 | = 2. In this case, it suffices to prove that G ≻ An . We apply induction on d. For d ≤ 1, there is nothing to prove. Suppose d = 2. Then G is isomorphic to one of the following graphs B8 , I1 , I2 and I3 , as shown in Fig. 5. By Theorem 2.8, we have B8 ≻ A8 . It is easy to obtain that

φ(I1 , x) = x8 − 8x6 + 16x4 − 9x2 , φ(I2 , x) = x10 − 10x8 + 30x6 − 34x4 + 12x2 , φ(I3 , x) = x12 − 12x10 + 48x8 − 84x6 + 64x4 − 16x2 , φ(A8 , x) = x8 − 8x6 + 16x4 − 6x2 , φ(A10 , x) = x10 − 10x8 + 30x6 − 28x4 + 6x2 , φ(A12 , x) = x12 − 12x10 + 48x8 − 74x6 + 40x4 − 6x2 . Hence I1 ≻ A8 , I2 ≻ A10 , and I3 ≻ A12 . Now suppose that d ≥ 3 and the assertion holds for smaller values of d. Note that An satisfies Formula (10) by Lemma 2.7(3). By an argument similar to the proof of Theorem 3.6, we can obtain G ≻ An . Case 2. |MG ∩ E1 | = 1. Note that E (Dn ) > E (An ) by Theorem 2.16. Suppose that G  Dn , we will prove that G ≻ Dn . We apply induction on d. For d ≤ 1, there is nothing to prove. Suppose d = 2. Then G is isomorphic to I4 , as shown in Fig. 5. It is easy to obtain that

φ(I4 , x) = x10 − 10x8 + 29x6 − 32x4 + 12x2 − 1, φ(D10 , x) = x10 − 10x8 + 29x6 − 28x4 + 10x2 − 1. Hence I4 ≻ D10 . Now suppose that d ≥ 3 and the assertion holds for smaller values of d. Note that Dn satisfies Formula (10) by Lemma 2.7(3). We use the same notations as in Theorem 3.6. Then the proof is similar to that of Theorem 3.6. We can divide two subcases xd z1 ∈ Mg and xd z1 ̸∈ Mg to proceed. The difference is that we need to prove the result G ≻ Dn for the case: d = 3, t = 1, dG (x3 ) = 2 and z2 z3 ∈ M4 . Since G  Dn , G is isomorphic to one of the following graphs I5 , I6 and I7 , as shown in Fig. 5. It is easy to obtain that

φ(I5 , x) = x10 − 10x8 + 30x6 − 33x4 + 11x2 − 1, φ(I6 , x) = x10 − 10x8 + 30x6 − 33x4 + 12x2 − 1, φ(I7 , x) = x12 − 12x10 + 48x8 − 83x6 + 62x4 − 16x2 + 1, φ(D12 , x) = x12 − 12x10 + 47x8 − 72x6 + 46x4 − 12x2 + 1. Hence I5 ≻ D10 , I6 ≻ D10 and I7 ≻ D12 .

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Case 3. |MG ∩ E1 | = 0. Then n ≥ 8. Note that E8 ∼ = S84 . Hence it follows from Theorems 2.8–2.10 and 2.16 that E (En ) > E (An ). Suppose that G  En , we will prove that G ≻ En . We apply induction on d. For d ≤ 2, there is nothing to prove. Now suppose that d ≥ 3 and the assertion holds for smaller values of d. Note that En satisfies Formula (10) by Lemma 2.7(3). By an argument similar to the proof of Theorem 3.6, we can obtain G ≻ En . The proof is thus complete.  Proof of Theorem 1.2. The proof follows directly from Theorems 1.1, 2.8, 2.10, 3.6 and 3.7



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