on minimal energy of a class of bicyclic graphs with given

On Minimal Energy of a Class of Bicyclic Graphswith given Cycles’ Length and Pendent Vertices∗

Yan Mi, Liancui Zuo† and Chunhong Shang

Abstract

The energy of G is defined as the sum of absolute valueof all eigenvalues of the adjacency matrix A(G). LetB2(n, a, b, p) be the set of all bicyclic graphs on n ver-tices with p pendent vertices and two cycles Ca and Cbwhich have unique common vertex u0, B2

θ (n, a, b, p) thegraph class obtained by coinciding the common vertex u0of Ca and Cb with the center of the star Sn−(a+b−1)+1,and B2

µ(n, a, b, p) the set obtained by coinciding the com-mon vertex u0 of Ca and Cb with the center of the starSp and connecting a pendent path Pn−(a+b−1)−(p−1) onpoint u0. In this paper, it is obtained that B2

θ (n, a, b, p)has the minimal energy in all graphs which have onlypendent vertices except two cycles, and B2

µ(n, a, b, p) hasthe minimal energy in all graphs which have prescribedcycles’ length and pendent vertices.

Keywords: Bicyclic graphs; Quasi-Order method; Min-imal energy

1 Introduction

In this paper, all graphs are finite, connected, undirectedand simple. In recent years, many parameters and classesof graphs were studied. For example, in [1], the restrictedconnectivity of Cartesian product graphs were obtained,and in [2, 12], some results on 3-equitable labeling andthe n-dimensional cube-connected complete graph weregained. Let G be a graph with order n and adjacencymatrix A(G). The characteristic polynomial of G, de-noted by φ(G), is defined as

φ(G, x) = det(xI −A(G)) =n∑i=0

aixn−i,

where I is the identity matrix of order n. The roots ofthe equation φ(G) = 0, denoted by λ1, λ2, · · ·, λn, are theeigenvalues of A(G). It’s easy to see that each λi is realsince A(G) is real symmetric. The energy of G, denotedby E(G), is defined as

E(G) =n∑i=1

|λi|.

∗Manuscript received August 23,2016. This work was sup-ported by NSF of Tianjin with code 16JCYBJC23600 and Tian-jin Training Program of Innovation for Undergraduates with code201510065052.†Yan Mi, Liancui Zuo and Chunhong Shang are with College of

Mathematical Science, Tianjin Normal University, Tianjin, 300387,China. Email: [email protected];[email protected]

For the coefficients ai(G) of φ(G), set bi(G) =|ai(G)| (i = 0, 1, · · ·, n). The following formula is givenin Sachs theorem,

ai(G) =∑S∈Li

(−1)ω(S)2c(S),

where Li denotes the set of Sachs subgraphs (the sub-graphs in which every component is either a K2 or acycle) of G that contain i vertices, ω(S) is the numberof connected components of S, and c(S) is the numberof cycles contained in S.

It is well known that the Coulson integral formula of theenergy is expressed as the following form

E(G) = 12π

∫ +∞−∞

1x2 ln[(

bn2 c∑i=0

(−1)ib2ix2i)2

+(bn

2 c∑i=0

(−1)ib2i+1x2i+1)2]dx.

(1)

Obviously, the formula (1) is a strictly monotonicallyincreasing function of bi(G), that is to say, for any twographs G1 and G2 with the same order, there exists thefollowing relationship:

bi(G1) ≥ bi(G2) hold for i ≥ 0

⇒ E(G1) ≥ E(G2).

In order to compare the energy of graphs with a bet-ter way, we need to define the following Quasi-Order:if bi(G1) ≥ bi(G2) hold for all i ≥ 0, we can writeG1�G2 or G2≺G1; if G1 � G2 and there exists somei0 such that bi0(G1) > bi0(G2), then we write G1 � G2.Combining with the formula (1), the increasing propertyG1 � G2 ⇒ E(G1) > E(G2) of graph energy is obtained.

In theoretical chemistry, the energy of a molecular graphcan be approximately used to represent π-electron en-ergy of the molecule, which is an important applicationof the energy of graphs in the chemical field, and hasbeen widely studied by scholars. For a more detailedexplanation can refer to the literature [3-4,10-11,13-15].In addition, for some special class of graphs, searchingfor their extreme energy becomes an interesting topic.For the graphs with cycles, many researching conclu-sions have been obtained, for example, [7,9] give unicyclegraphs with minimal energy and maximal energy, respec-tively. In [8,10], bicyclic graphs with minimal energy andmaximal energy are gotten. On this basis, the relevantconclusions about the extreme energy of the graphs with

IAENG International Journal of Applied Mathematics, 47:1, IJAM_47_1_15

(Advance online publication: 23 February 2017)

______________________________________________________________________________________

given parameters are also gained. For example, [13] ob-tains unicycle graphs with given diameter and minimalenergy. In [5], the minimal energy of unicyclic graphswith prescribed girth and pendent vertices is given. In[6], the minimal energy of bicyclic graphs with a givendiameter is obtained.

Bicyclic graphs are defined as connected graphs with nvertices and n + 1 edges. According to the character-istics of the number of common vertices in two circlesin the bicyclic graphs, they can be divided into threeclasses: two cycles without any common vertex, two cy-cles with common edges, and two cycles with an uniquecommon vertex. Here we discuss the last case, whichis denoted by B2(n, a, b, p), where a, b are the two cy-cles’ length, and p is the number of pendent vertices. Inaddition, B2

θ (n, a, b, p) is the graph class obtained by co-inciding the common vertex u0 of Ca and Cb with thecenter of the star Sn−(a+b−1)+1. Let B2

µ(n, a, b, p) be thegraph class obtained by coinciding the common vertexu0 of Ca and Cb with the center of the star Sp, and con-necting a pendent path Pn−(a+b−1)−(p−1) on point u0.For sake of the convenience, we write B2

θ (n, a, b, p) andB2µ(n, a, b, p) as B2

θ (n, p) and B2µ(n, p) simply (see Fig.1-

2,20 and 28), subgraphs of G without pendent vertices iscalled the base graph of G, and the base graph class ofgraph class mentioned above is represented by B2

∞ (seeFig.3). Let V ′(G) denote the set of all pendent verticesof G. dG(x, y) is defined as the distance between twovertices x and y of a graph G, and

dG(x,Ca,b) = min{dG(x, y)| y ∈ V (Ca,b),x /∈ V (Ca,b)}.

In this paper, using Quasi-Order method, we discussthe graphs with minimal energy in B2(n, a, b, p) that aregiven length of two cycles with unique common pointand pendent vertices by mathematical induction. It isobtained that B2

θ (n, a, b, p) has the minimal energy inall graphs which have only pendent vertices except twocycles, and B2

µ(n, a, b, p) has the minimal energy in allgraphs which have prescribed cycles’ length and pendentvertices (please see Fig.1-2).

Lemma 1 ([10]) Let G be a simple graph and e = uvbe a pendent edge of G with pendent vertex v, then

bi(G) = bi(G− v) + bi−2(G− v − u).

Lemma 2 ([10]) Let G be a simple graph. If H is asubgraph (proper subgraph) of G, then G�H(G � H).

2 Graphs with only pendent vertices ex-cept two cycles in B2(n, a, b, p)

Theorem 3. LetG ∈ B2(n, a, b, p) with n = a+b+p−1and p ≥ 1. Assume that the cycles’ length a and b arefixed, and u0 is the common vertex of two cycles. IfG 6= B2

θ (n, a, b, p), then G � B2θ (n, a, b, p).

Proof. We will show the theorem by induction on p.

(1) Suppose that p = 1, i. e., there is only one pendentvertex. Assume that e = u0v is a pendent edge and v is

its pendent vertex, as shown in Fig.2. By Lemma 1, forB2∞ as in Fig.3, we can get

bi(B2θ (n, 1)) = bi(G− v) + bi−2(G− u0 − v)

= bi(B2∞) + bi−2(Pa−1 ∪ Pb−1).

��&%'$qq q q q q q qCa Cbu0

p v

Fig.1 B2θ(n, a, b, p)

��&%'$qq

Ca Cbu0

v

Fig.2 B2θ(n, a, b, 1)

��&%'$qCa Cbu0

Fig.3 B2∞

��&%'$qq

Ca Cbu0 qut

v1

Fig.4 A11

��&%'$qqqq

Ca Cb

v

u0

ut

v1

Fig.5 B11

��&%'$qq

Ca Cbu0 q qqut

v1

us

v

Fig.6 B12

��&%'$qqq qq

Ca Cb

v

u0

ut

v1

wsw0

Fig.7 B13

��&%'$qCa Cbu0 qqq q q q q q qut

v1l − 1

Fig.8 C11

��&%'$q qq q qqqqCa Cbu0

q q q q q q ql − k

wt

v1

Fig.9 C12

��qCa u0

q q q qq q q q qqqqut+1

ut−1

Fig.10 H1

��qq

Ca u0

v

q q q q qq q q q qqq

ut+1

ut−1

Fig.11 H2

��qCa u0

q q q qq q q q qqqut+1

ut−1

q

qv1ut

Fig.12 H3

q q q q q q q q q q q q qqv1

u1u2 ut ut−1

Fig.13 H4

&%'$qqqqqqqqqqqqqqq

u0 Cb

ws+1

ws−1qqut

v1

Fig.14 H5

Assume that v is adjacent to ut but u0 belongs to cycleCb, as shown in Fig.4. By Lemma 1, for graphs A11 in



______________________________________________________________________________________

Fig.4 and H1 in Fig.10, we obtain that

bi(A11) = bi(A11 − v1) + bi−2(A11 − ut − v1)= bi(B

2∞) + bi−2(H1).

��qq

Ca u0

q qq ql − 1

q q q q qq q q q qqq

ut+1

ut−1

Fig.15 H6

��&%'$qCa Cbu0 qqq q q q q q q ut

v1l − k

q qqq

wt

Fig.16 H7

&%'$qqqqqqqqwt+1qqqqqqq Cbu0

wt−1 qqq q q q q q q ut

v1l − k

q qqFig.17 H8

��&%'$qq q q q q qqqqqqq

Ca Cbu0

p

Fig.18 H9

Obviously, Pa−1 ∪ Pb−1 is a proper subgraph of H1, soPa−1 ∪ Pb−1 ≺ H1, and then B2

θ (n, 1) ≺ A11.

(2) For p = 2 and B2θ (n, 2), by Lemma 1, we can get

bi(B2θ (n, 2)) = bi(B

2θ (n, 2)− v)

+bi−2(B2θ (n, 2)− u0 − v)

= bi(B2θ (n− 1, 1)) + bi−2(Pa−1 ∪ Pb−1).

If G 6∼= B2θ (n, 2), then there is at least one pendent vertex

which is not adjacent to u0. The problem can be dividedinto the following cases.

Case 1. There is at least one pendent point which isadjacent to u0. Suppose that one point is such one andthe other is adjacent to ut on cycle Cb, as B11 shown inFig.5. By Lemma 1, we have

bi(B11) = bi(B11 − v1) + bi−2(B11 − ut − v1)= bi(B

2θ (n− 1, 1)) +Bi(H2),

where H2 is shown in Fig.11. Since Pa−1 ∪ Pb−1 isa proper subgraph of H2, Pa−1 ∪ Pb−1 ≺ H2. Thus,B2θ (n, 2) ≺ B11.

Case 2. There is no pendent vertex which is adjacent tou0, and both of them are adjacent to ut and us on onecycle Cb, respectively, as B12 shown in Fig. 6. Then byLemma 1, we can get

bi(B2θ (n, 2)) = bi(B

2θ (n, 2)− v)

+bi−2(B2θ (n, 2)− u0 − v)

= bi(B2θ (n− 1, 1)) + bi−2(Pa−1 ∪ Pb−1),

bi(B12) = bi(B12 − v) + bi−2(B12 − us − v)= bi(H3) + bi−2(H4),

where H3, H4 are shown in Fig. 12-13.

We can obtain that B2θ (n−1, 1) ≺ H3 from (1). In addi-

tion, it is obvious that Pa−1 ∪Pb−1 is a proper subgraphof H4, so Pa−1 ∪ Pb−1 ≺ H4. Hence, B2

θ (n, 2) ≺ B12.

Case 3. There is no pendent vertex that is adjacentto u0, and two points are adjacent to ut and ws on twocycles Cb and Ca, respectively, as B13 shown in Fig. 7.By Lemma 1, we obtain that

bi(B11) = bi(B11 − v) + bi−2(B11 − u0 − v)= bi(H3) + bi−2(Pa−1 ∪H5),

and

bi(B13) = bi(B13 − v) + bi−2(B13 − ws − v)= bi(H3) + bi−2(H6),

where H5, H6 are shown in Fig.14-15, respectively.

Comparing Pa−1 ∪ H5 with H6, since the former is aproper subgraph of the latter, Pa−1 ∪H5 ≺ H6, that is,B11 ≺ B13. Combining the conclusion B2

θ (n, 2) ≺ B11

from (1), we get B2θ (n, 2) ≺ B13.

Thus, the theorem holds for p = 2.

(3) Assume that the theorem holds for p = l − 1. In thesequel, we prove that the theorem still holds for p = l.For B2

θ (n, l), we have

bi(B2θ (n, l)) = bi(B

2θ (n, l)− v) + bi−2(B

2θ (n, l)− u0 − v)

= bi(B2θ (n− 1, l − 1)) + bi−2(Pa−1 ∪ Pb−1).

If there is at least one pendent vertex whose neighbor isnot u0, as C11 shown in Fig.8, then by Lemma 1, we canget

bi(C11) = bi(C11 − v1) + bi−2(C11 − ut − v1)= bi(B

2θ (n− 1, l − 1)) + bi−2(H7),

where H7 is shown in Fig.16. Since H7 contains Pa−1 ∪Pb−1 as its proper subgraph, Pa−1 ∪ Pb−1 ≺ H7, so wehave B2

θ (n, a, b, l) ≺ C11.

Now suppose that there are p− k pendent vertices thatare adjacent to u0 (l ≥ k ≥ 2), as C12 shown in Fig.9(the pendent vertices are not shown all). At this time,choose a pendent edge wtv1 with wt 6= u0, then

bi(C12) = bi(C12 − v1) + bi−2(C12 − wt − v1)= bi(H8) + bi−2(H9),

where H8, H9 are shown in Fig.17-18. Note that H8 andH9 display a part of the pendent vertices except the de-formation by deleting one or two vertices from C12. SinceB2θ (n− 1, l− 1) and H8 are graphs which have only pen-

dent vertices except two cycles with n − 1 vertices, byinduction hypothesis, we can get B2

θ (n− 1, l − 1) ≺ H8,and Pa−1 ∪ Pb−1 ≺ H9 because Pa−1 ∪ Pb−1 is a propersubgraph of H9. Thus B2

θ (n, p) ≺ C12, and the theoremholds for p = l.

In a word, we have shown that B2θ (n, p) is the graph with

minimal energy for 1 ≤ p ≤ l, l ≥ 2 and n = a+b+p−1.

3 Bicyclic graphs with prescribed cy-cles’ length and pendent vertices inB2(n, a, b, p)

On the basis of Theorem 3, we study the case of n >a+ b+ p− 1. Let Qa,b,pn be graphs shown in Fig. 19.



______________________________________________________________________________________

Theorem 4. Let G ∈ B2(n, a, b, p) with n ≥ a+b+p−1and p ≥ 1. Assume that the cycles’ length a and bare fixed, and u0 is the common vertex of two cycles.If G 6∼= B2

µ(n, a, b, p) and G 6∼= Qa,b,pn , then E(G) >E(B2

µ(n, a, b, p)).

Proof. For n = a + b − 1 + p, by Lemma 1, we haveB2µ(n, a, b, p)

∼= B2θ (n, a, b, p), the result holds for this

case. We show that the result is also true for n − (a +b− 1 + p) ≥ 1 in the following.

Assume that V ′(G) is the set of all pendent verticesof G, v is the point in V ′(G) such that dG(v, Ca,b) =max{dG(x,Ca,b), x ∈ V ′(G)}, and u is its unique neigh-bor.

��&%'$qCa Cbu0qqqqqqqqq qq qp

��&%'$qCa Cbu0qqqqqqqqqq qq qp

Fig.19 Qa,b,pn

��&%'$qqq q qqq q vv′

u′

Ca Cbu0

p

Fig.20 B2µ(n, a, b, p)

��&%'$qq q q q q qqqqqu

v

Ca Cbu0

Fig.21 D11

��&%'$qqq q qqq qv

u qqCa Cbu0

Fig.22 D′11

��&%'$qq q q q q qqqqquqqq q qv

m

Ca Cbu0

Fig.23 D12

��&%'$qqqq qqv qu q

mqqq

q Ca Cbu0

Fig.24 D′12

��&%'$qqqq q q quq

v

Ca Cbu0

Fig.25 D121

��&%'$qqqu qqv q q

Ca Cbu0

qqFig.26 D122

��&%'$qqqq q q qu′q

v′ qqCa Cbu0

Fig.27 D122′

��&%'$qqq q qqqqqqqqq qq

vuu1

v1

Ca Cbu0

p

Fig.28 B2µ(n, p)

��&%'$qqq q qqq q qv

uu1

Ca Cbu0

p

Fig.29 B2µ(a+ b+ p+ 1, p)

Based on the value of n− (a+ b− 1 + p), there are two

cases that need to be discussed.

Case 1. n− (a+ b− 1 + p) = 1.

By the definition, we can get dG(v, Ca,b) = 2.

Subcase 1. 1. d(u) = 2. By Lemma 1, it is not difficultto verify that

bi(B2µ(n, p)) = bi(B

2µ(n, p)−v′)+bi−2(B2

µ(n, p)−v′−u′).

Suppose that G 6∼= B2µ(n, p), then we must have G ∼= D11

orG ∼= D′11 (bothD11 andD′11 are graphs with n verticesand p pendent vertices, and a part of pendent verticeswhich are adjacent to vertices on cycles are shown inFig.21-22). Take D11 as an example, we have

bi(D11) = bi(D11 − v) + bi−2(D11 − v − u),

where D11 − v ∈ B2(n− 1, p) and D11 − v− u ∈ B2(n−2, p− 1). Since

B2µ(n, p)− v′ ∼= B2

θ (n− 1, p)

andB2µ(n, p)− v′ − u′ ∼= B2

θ (n− 2, p− 1),

we have

B2θ (n− 1, p) ≺ B2(n− 1, p),

B2θ (n− 2, p− 1) ≺ B2(n− 2, p− 1)

by Theorem 3. On account of G 6∼= B2µ(n, p), we can get

B2µ(n, p) − v′ ≺ D11 − v

andB2µ(n, p) − v′ − u′ ≺ D11 − v − u.

Thus, B2µ(n, p) ≺ D11. Similarly, we can obtain that

B2µ(n, p) ≺ D′11.

Subcase 1. 2. d(u) ≥ 3.

We get firstly


2µ(n, p)− v) + bi−2(B

2µ(n, p)− v− u0)

= bi(B2µ(n− 1, p− 1)) + bi−2(Pa−1 ∪ Pb−1 ∪ P2), (2)

Assume that G 6∼= B2µ(n, a, b, p), we must get G ∼= D12

or G ∼= D′12 (both D12 and D′12 are graphs with n ver-tices and p pendent vertices, and a part of pendent ver-tices that are adjacent to vertices on cycles are shown inFig.23-24). Take D12 as an example. Clearly,

bi(D12) = bi(D12 − v) + bi−2(D12 − v − u), (3)

and D12 − v ∈ B2(n − 1, p − 1). All points in N(u)are pendent vertices except one on a cycle. Supposethat there are m pendent vertices adjacent to u, then|N(u)| = m + 1. Since G 6∼= Qa,b,pn , we have p ≥m + 1. Set D12 − v − u = G′ ∪ (m − 1)P1, thenG′ ∈ B2(n−m− 1, p−m).

Now we show the result by induction n and p in thefollowing. Assume that the result holds for small n andp, then B2

µ(n − 1, p − 1) ≺ B2(n − 1, p − 1). Because



______________________________________________________________________________________

p ≥ m + 1, d(u) ≥ 3, and dG(v, Ca,b) = 2, there are atleast four pendent vertices in G, so n ≥ a+ b+ 4.

(1) Assume that n− (a+ b) = 4, i. e., n = a+ b+4, andG ∈ B2(a+b+4, 4). Suppose that G ∼= D121, |V (D121−v)| − (a + b) = 3 and D121 − v ∈ B2(a + b + 3, 3) (seeFig.25). In the following, we use Quasi-Order method tocompare B2

µ(a + b + 3, 3) and D121 − v. By Lemma 1,we get

bi(B2µ(a+ b+ 3, 3))

= bi(B2µ(a+ b+ 2, 2)) + bi−2(Pa−1 ∪ Pb−1 ∪ P2),

and

bi(D121− v) = bi(B2(a+ b+ 2, 2)) + bi−2(B

2(a+ b, 1)).

By Subcase 1.1, we have B2µ(a+b+2, 2) ≺ B2(a+b+2, 2),

andPa−1 ∪ Pb−1 ∪ P2 ≺ B2(a+ b, 1)

since Pa−1∪Pb−1∪P2 is a proper subgraph of B2(a+b, 1),so we get

B2µ(a+ b+ 3, 3) ≺ D121− v

andB2µ(a+ b+ 3, 3) ≺ B2(a+ b+ 3, 3).

Thus, for n = a+ b+ 4 and p = 4, we have

B2µ(a+ b+ 3, 3) ≺ B2(a+ b+ 3, 3),

that is,

B2µ(n− 1, p− 1) ≺ B2(n− 1, p− 1).

(2) Suppose that n− (a+ b) = 5, i. e., n = a+ b+ 5.

On the basis of (1), add one vertex to graph D121 suchthat the number of pendent vertices increasing, then theadded pendent vertex is adjacent to u or other vertex oncycles, and we obtain D122, D122

′ ∈ B2(a+ b+5, 5) thatare shown in Fig.26-27.

Suppose that G ∼= D122 or G ∼= D122, then |V (D122 −v)| = |V (D122

′−v′)| = a+b+4 andD122−v,D122′−v′ ∈

B2(a+ b+ 4, 4). By Lemma 1, we get

bi(B2µ(a+ b+ 4, 4))

= bi(B2µ(a+ b+ 3, 3)) + bi−2(Pa−1 ∪ Pb−1 ∪ P2),

bi(D122− v) = bi(B2(a+ b+ 3, 3)) + bi−2(B

2(a+ b, 1)),

and

bi(D122′−v′) = bi(B

2(a+b+3, 3))+bi−2(B2(a+b+1, 2)).

Combining with the result of (1), we have

B2µ(a+ b+ 3, 3) ≺ B2(a+ b+ 3, 3).

Since Pa−1 ∪ Pb−1 ∪ P2 is a proper subgraph of bothB2(a+ b, 1) and B2(a+ b+ 1, 2), we obtain that

Pa−1 ∪ Pb−1 ∪ P2 ≺ B2(a+ b, 1),

Pa−1 ∪ Pb−1 ∪ P2 ≺ B2(a+ b+ 1, 2),

B2µ(a+ b+ 4, 4) ≺ D122− v

andB2µ(a+ b+ 4, 4) ≺ D122

′ − v′,

so we have

B2µ(a+ b+ 4, 4) ≺ B2(a+ b+ 4, 4).

Thus, for n = a+ b+ 5 and p = 5, we have

B2µ(a+ b+ 4, 4) ≺ B2(a+ b+ 4, 4),

that is,

B2µ(n− 1, p− 1) ≺ B2(n− 1, p− 1).

(3) Assume that the theorem holds for n−(a+b) = p−1,i. e., B2

µ(a+ b+ p− 2, p− 2) ≺ B2(a+ b+ p− 2, p− 2).

In the sequel, we prove that the theorem still holds forn− (a+ b) = p. Suppose that G ∼= D12 or G ∼= D′12 (seeFig. 23-24). Take D12 as an example. Clearly, |V (D12−v)| = a+ b+p− 1 and D12− v ∈ B2(a+ b+p− 1, p− 1).

A comparison between B2µ(a+b+p−1, p−1) and D12−v

will be accomplished in the following. By Lemma 1, wehave

bi(B2µ(a+ b+ p− 1, p− 1)) =

bi(B2µ(a+ b+ p− 2, p− 2)) + bi−2(Pa−1 ∪ Pb−1 ∪ P2),

and

bi(D12 − v) = bi(B2(a+ b+ p− 2, p− 2))

+bi−2(B2(a+ b+ p−m− 1, p−m)).

By induction assumption, we obtain that

B2µ(a+ b+ p− 2, p− 2) ≺ B2(a+ b+ p− 2, p− 2).

Because p−m ≥ 1, it is not difficult to see that Pa−1 ∪Pb−1 ∪ P2 is a proper subgraph of B2(a + b + p −m −1, p−m), so

Pa−1 ∪ Pb−1 ∪ P2 ≺ B2(a+ b+ p−m− 1, p−m),

andB2µ(a+ b+ p− 1, p− 1) ≺ D12 − v.

Thus, for n = a+ b+ p, we have

B2µ(a+ b+ p− 1, p− 1) ≺ B2(a+ b+ p− 1, p− 1),

that is,

B2µ(n− 1, p− 1) ≺ B2(n− 1, p− 1).

By induction, we have

B2µ(n− 1, p− 1) ≺ B2(n− 1, p− 1),

and the relationship between the first item on right ofequations (2) and (3): B2

µ(n− 1, p− 1) ≺ D12 − v. Theproblem is turned into a proof of the relationship betweenthe second:Pa−1 ∪ Pb−1 ∪ P2 ≺ D12 − v − u.



______________________________________________________________________________________

As what mentioned before, D12−v−u ∈ B2(n−m−1, p−m). What remains is to prove that Pa−1 ∪ Pb−1 ∪ P2 ≺B2(n −m − 1, p −m). Since p −m ≥ 1, it is clear thatPa−1 ∪ Pb−1 ∪ P2 is a proper subgraph of B2(n − m −1, p−m), so we get

Pa−1 ∪ Pb−1 ∪ P2 ≺ B2(n−m− 1, p−m).

Thus, we have B2µ(n, p) ≺ D12. Similarly, we can ob-

tain that B2µ(n, p) ≺ D′12. Hence the result in Case 1 is

proved.

Case 2. n− (a+ b− 1 + p) ≥ 2.

In the following, we deal with the problem in two sub-cases.

Subcase 2.1. dG(v, Ca,b) ≥ 3.

Subcase 2.1.1. d(u) = 2.

Subcase 2.1.1.1. d(u1) = 2 and u1 is the neighborvertex of u.

Assume that G 6∼= B2µ(n, a, b, p), we must have G ∼= D13

which has n vertices and p pendent vertices, where apart of pendent vertices and pendent pathes are shownin Fig.31.

��&%'$qqqqqqqqqq q qqqCa Cbu0

vuu1

Fig.30 B2µ(a+ b+ p+ 2, p+ 1)

��&%'$qq qqqqqqqqCa Cbu0

v′u′

u′1

Fig.31 D13

��&%'$qq qqqqqqqqCa Cbu0qv′

u′u′1

Fig.32 D131

��&%'$qq qqqqqqqqCa Cbu0

qv′

u′u′1

Fig.33 D132

��&%'$qqq q qqq

Ca Cbu0

p

Fig.34 D132′

��&%'$qqq q qqq q

Ca Cbu0

p− 1

Fig.35 D14

��&%'$qqq q qqqqqqq q q

v1

Ca Cbu0

p

Fig.36 B2µ(a+ b+ p− 1, p)


v2u2

Ca Cbu0

p− 1

Fig.37 B2µ(a+ b+ p− 1, p− 1)

Obviously, D13 − v ∈ B2(n − 1, p) and D13 − v − u ∈B2(n−2, p). In the following, the two items on the right

of equal will be compared, respectively. If we want to getB2µ(n−1, p) ≺ D13−v and B2

µ(n−2, p) ≺ D13−v−u, weneed to prove that B2

µ(n−1, p) ≺ B2(n−1, p) and B2µ(n−

2, p) ≺ B2(n − 2, p). We show the result by inductionon n. Assume that the result holds for small n, thenB2µ(n − 1, p) ≺ B2(n − 1, p) and B2

µ(n − 2, p) ≺ B2(n −2, p). Note that dG(v, Ca,b) ≥ 3, and d(u) = d(u1) = 2.Assume that the number of pendent vertices as p willnot change even though the order of graph G changesconstantly. There are at least two vertices except twocycles and pendent vertices, so n ≥ a+ b+ p+ 1.

��&%'$qqq qq qq qqqqqqq q q qq qq

mvuu1

Ca Cbu0

Fig.38 D15

��&%'$qqq qq qqqqqqq qqq q qqq q

mvuu1

Ca Cbu0

Fig.39 D′15


v

Ca Cbu0

p−m+ 1

Fig.40 B2µ(n−m− 1, p−m+ 1)

��&%'$qqq qq qqqqqqq qqq q qqq q

mvuu1 qq qqCa Cbu0

Fig.41 D16

��&%'$qqq q qqq qv

u qqqqqCa Cbu0

Fig.42 D17

��&%'$qq q q q q qqqqquqqq q qv

m

Ca Cbu0

qqq

Fig.43 D18

For B2µ(n, a, b, p) and D13, by Lemma 1, we have


2µ(n− 1, p)) + bi−2(B

2µ(n− 2, p)),

bi(D13) = bi(D13 − v) + bi−2(D13 − v − u).

(1)If n− (a+ b− 1)− p = 2, then n = a+ b+ p+ 1.

Suppose that G ∼= D131 (the pendent edges are notshown all in Fig.32) since G 6∼= B2

µ(a+b+p+1, p). We canget V (D13−v′) = a+b+p and D13−v′ ∈ B2(a+b+p, p).We use Quasi-Order method to compare B2

µ(a+ b+p, p)and D131− v′. By Lemma 1, we get

bi(B2µ(a+ b+ p, p)) = bi(B

2µ(a+ b+ p− 1, p))

+bi−2(B2µ(a+ b+ p− 2, p− 1))

and

bi(D131− v′) = bi(B2(a+ b+ p− 1, p))

+bi−2(B2(a+ b+ p− 2, p− 1).

Since

B2µ(a+ b− 1 + p, p) ∼= B2

θ (a+ b− 1 + p, p),

B2µ(a+ b+ p− 2, p− 1) ∼= B2

θ (a+ b+ p− 2, p− 1).



______________________________________________________________________________________

By Theorem 3, we can obtain that

B2µ(a+ b− 1 + p, p) ≺ B2(a+ b− 1 + p, p),

and

B2µ(a+ b+ p− 2, p− 1) ≺ B2(a+ b+ p− 2, p− 1),

so we have B2µ(a+ b+ p, p) ≺ D131− v′, that is B2

µ(a+b+ p, p) ≺ B2(a+ b+ p, p). Thus, for n = a+ b+ p+ 1,we have B2

µ(a+ b+ p, p) ≺ B2(a+ b+ p, p), i. e.,

B2µ(n− 1, p) ≺ B2(n− 1, p).

On the basis above, to add one vertex to graph D131but the number of pendent vertices will not change,there exist two methods: lengthening the pendent pathwhich contains v′ and u′ on D131 or other pendentpath. Without loss of generality, assume that we addone vertex such that the graph has the same pendentvertices, then we obtain G ∼= D132 or G ∼= D132

′, whereD132, D132

′ ∈ B2(a+ b+ p+ 2, p).

(2) If n− (a+ b− 1)− p = 3, then n = a+ b+ p+ 2, soit is not difficult to see that

V (D132− v′) = a+ b+ p+ 1,

D132− v′ ∈ B2(a+ b+ p+ 1, p),

V (D132′ − v′) = a+ b+ p+ 1,

andD132

′ − v′ ∈ B2(a+ b+ p+ 1, p).

Comparing B2µ(a+ b+ p+ 1, p) and D132− v′ as well as

B2µ(a+ b+ p+1, p) and D132

′ − v′, by Lemma 1, we get

bi(B2µ(a+ b+ p+ 1, p)) = bi(B

2µ(a+ b+ p, p))

+bi−2(B2µ(a+ b+ p− 1, p)),

bi(D132− v′) = bi(B2(a+ b+ p, p))

+bi−2(B2(a+ b+ p− 1, p),

andbi(D132

′ − v′) = bi(B2(a+ b+ p, p))

+bi−2(B2(a+ b+ p− 1, p− 1)).

Comparing three equations above, combining with theproof of (1), the result of the comparison between thefirst term of the right side of the equal is B2

µ(a+b+p, p) ≺B2(a+ b+ p, p).

For the second term of the right side of the equal, since

B2µ(a+ b− 1 + p, p) ∼= B2

θ (a+ b− 1 + p, p),

by Lemma 1, we get B2µ(a + b − 1 + p, p) ≺ B2(a + b +

p− 1, p).

For B2µ(a+ b+ p− 1, p) and B2(a+ b+ p− 1, p− 1), by

Theorem 1, we have

B2µ(a+ b+ p− 1, p− 1) ≺ B2(a+ b+ p− 1, p− 1).

Now, we need only to prove that

B2µ(a+ b+ p− 1, p) ≺ B2

µ(a+ b+ p− 1, p− 1).

From Lemma 1,we can get

bi(B2µ(a+ b+ p− 1, p))

= bi(B2θ (a+ b+ p− 2, p− 1)) + bi−2(Pa−1 ∪ Pb−1)

andbi(B

2(a+ b+ p− 1, p− 1))= bi(B

2θ (a+ b+ p− 2, p− 1))

+bi−2(B2(a+ b+ p− 3, p− 2)).

Since Pa−1∪Pb−1 is a proper subgraph of B2(a+ b+p−3, p−2), the relationship on the second term of the rightside of the equal is

Pa−1 ∪ Pb−1 ≺ B2(a+ b+ p− 3, p− 2),

so

B2µ(a+ b+ p− 1, p) ≺ B2(a+ b+ p− 1, p− 1).

Thus, for n = a+ b+ p+ 2, we obtain that

B2µ(a+ b+ p+ 1, p) ≺ B2(a+ b+ p+ 1, p),

i. e.,B2µ(n− 1, p) ≺ B2(n− 1, p).

(3) If n−(a+b−1)−p = N−1, then n = a+b+p+N−2.Set n1 = n, assume that the theorem holds for n1, i. e.,we have

B2µ(n1 − 2, p) ≺ B2(n1 − 2, p),

andB2µ(n1 − 1, p) ≺ B2(n1 − 1, p).

Suppose that n−(a+b−1)−p = N , then n = a+b+p+N−1. Set n2 = n. Since V (G−v) = n2−1, without lossof generality, assume that G ∼= D13 (the pendent pathesare not shown all in the Fig.31). Obviously, D13 − v ∈B2(n2−1, p). Comparing B2

µ(n2−1, p) with D13−v, byLemma 1, we have

bi(B2µ(n2 − 1, p))

= bi(B2µ(n2 − 2, p)) + bi−2(B

2µ(n2 − 3, p)),

and

bi(D13 − v)= bi(B

2(n2 − 2, p)) + bi−2(B2(n2 − 3, p)).

Due to n2 = n1 + 1, we can get

B2µ(n2 − 2, p) ∼= B2

µ(n1 − 1, p),B2(n2 − 2, p) ∼= B2(n1 − 1, p),B2µ(n2 − 3, p) ∼= B2

µ(n1 − 2, p),

andB2(n2 − 3, p) ∼= B2(n1 − 2, p).

By induction hypothesis, we have

B2µ(n1 − 1, p) ≺ B2(n1 − 1, p),

andB2µ(n1 − 2, p) ≺ B2(n1 − 2, p).



______________________________________________________________________________________

Therefore,

B2µ(n2 − 2, p) ≺ B2(n2 − 2, p),

andB2µ(n2 − 3, p) ≺ B2(n2 − 3, p)),

that is to say,

B2µ(n− 2, p) ≺ B2(n− 2, p),

andB2µ(n− 1, p) ≺ B2(n− 1, p).

SoB2µ(n− 1, p) ≺ D13 − v.

According to the induction above, we have

B2µ(n− 1, p) ≺ D13 − v,

andB2µ(n− 2, p) ≺ D13 − v − u,

that is,bi(B

2µ(n− 1, p)) < bi(D13 − v),

andbi(B

2µ(n− 2, p) < bi(D13 − v − u).

Thus, B2µ(n, p) ≺ D13 and then the result is obtained.

Subcase 2.1.1.2. d(u1) ≥ 3.

Suppose that G ∼= D14, where D14 is the graph with nvertices and p pendent vertices and a part of pendentvertices and a pendent path are shown in Fig.35. ForB2µ(n, p) and D14, we have


2µ(n− 1, p)) + bi−2(B

2µ(n− 2, p)),

bi(D14) = bi(D14 − v) + bi−2(D14 − v − u).

Obviously, D14 − v ∈ B2(n − 1, p) and D14 − v − u ∈B2(n − 2, p − 1). Similar as the subcase 2.1.1.1 andsubcase 1.2, we have B2

µ(n − 1, p) ≺ B2(n − 1, p), andB2µ(n − 2, p − 1) ≺ B2(n − 2, p − 1). So B2

µ(n − 1, p) ≺D14−v, B2

µ(n−2, p−1) ≺ D14−v−u. Hence the problemis turned into showing B2

µ(n− 2, p) ≺ B2µ(n− 2, p− 1).

(i)For n − 2 − (a + b − 1 + p) = 0, then B2µ(n − 2, p) ∼=

B2θ (n− 2, p).

By Lemma 1, we get

B2θ (n− 2, p) = bi(B

2θ (n− 3, p− 1)) + bi−2(Pa−1 ∪ Pb−1),

B2µ(n−2, p−1) = bi(B

2θ (n−3, p−1))+bi−2(B2

θ (n−4, p−2)).

It is not difficult to find that Pa−1 ∪ Pb−1 is a propersubgraph of B2

θ (n− 4, p− 2), so

B2µ(n− 2, p) ≺ B2

µ(n− 2, p− 1).

(ii)For n− 2− (a+ b− 1 + p) ≥ 1. By Lemma 1, we get

bi(B2µ(n− 2, p)) = bi(B

2µ(n− 2, p)− v1)

+bi−2(B2µ(n− 2, p)− v1 − u1),

bi(B2µ(n− 2, p− 1)) = bi(B

2µ(n− 2, p− 1)− v2)

+bi−2(B2µ(n− 2, p− 1)− v2 − u2).

Note that B2µ(n− 2, p)− v1 ∼= B2

µ(n− 2, p− 1)− v2. Wehave

bi(B2µ(n− 2, p)− v1) = bi(B

2µ(n− 2, p− 1)− v2).

Since

B2µ(n− 2, p)− v1 − u1 = Pa−1 ∪ Pb−1 ∪ Pn−a−b−p,

and Pa−1 ∪ Pb−1 ∪ Pn−a−b−p is a proper subgraph ofB2µ(n− 2, p− 1)− v2 − u2, we obtain that

B2µ(n− 2, p)− v1 − u1 ≺ B2

µ(n− 2, p− 1)− v2 − u2,

thusB2µ(n− 2, p) ≺ B2

µ(n− 2, p− 1).

In summary, we get B2µ(n, p) ≺ D14.

Subcase 2.1.2. d(u) ≥ 3.

Subcase 2.1.2.1. d(u1) = 2.

Suppose that G ∼= D15 where D15 is the graph with nvertices and p pendent vertices that are adjacent to u1,and a part of pendent vertices and pendent path areshown in Fig.38.

For B2µ(n, a, b, p) and D15, by Lemma 1, we have


2µ(n, p)−v1)+bi−2(B2

µ(n, p)−v1−u0)

= bi(B2µ(n−1, p−1))+bi−2(Pa−1∪Pb−1∪Pn−a−b−p+2),

bi(D15) = bi(D15 − v) + bi−2(D15 − v − u).Obviously, D15 − v ∈ B2(n − 1, p − 1), and all verticesare pendent vertices except one point u1 in N(u).

Suppose that there are m pendent vertices adjacent tou, then |N(u)| = m + 1 (please see D′15 in Fig.39). LetD′15 − v − u = G′ ∪ (m− 1)P1. Then G′ ∈ B2(n−m−1, p−m+ 1). Similar as subcase 1.2, we have

B2µ(n− 1, p− 1) ≺ B2(n− 1, p− 1),

and

B2µ(n−m− 1, p−m+ 1) ≺ B2(n−m− 1, p−m+ 1),

so we obtain that B2µ(n−1, p−1) ≺ D′15−v and B2

µ(n−m−1, p−m+1) ≺ D′15−v−u. What remains is to provethat Pa−1∪Pb−1∪Pn−a−b−p+2 ≺ B2

µ(n−m−1, p−m+1).Since G 6∼= Qa,b,pn , p−m ≥ 1.

(i) For p = m+1, B2µ(n−m−1, p−m+1) ∼= B2

µ(n−p, 2),then by Lemma 1, we can get

bi(B2µ(n−m− 1, p−m+ 1))

= bi(B2µ(n− p, 2))

= bi(B2µ(n− p− 1, 1))

+bi−2(Pa−1 ∪ Pb−1 ∪ Pn−p−a−b),

andbi(Pa−1 ∪ pb−1 ∪ Pn−a−b−p+2)= bi(Pa−1 ∪ Pb−1 ∪ Pn−a−b−p+1)+bi−2(Pa−1 ∪ Pb−1 ∪ Pn−a−b−p).



______________________________________________________________________________________

Comparing the first term of the right side of the equal,it is not difficult to find that Pa−1 ∪ Pb−1 ∪ Pn−a−b−p+1

is a proper subgraph of B2µ(n− p− 1, 1), so we have

Pa−1 ∪ Pb−1 ∪ Pn−a−b−p+1 ≺ B2µ(n− p− 1, 1).

Therefore,

Pa−1 ∪ Pb−1 ∪ Pn−a−b−p+2 ≺ B2µ(n−m− 1, p−m+ 1)

holds.

(ii) For p ≥ m+ 2, by Lemma 1, we get

bi(B2µ(n−m− 1, p−m+ 1))

= bi(B2µ(n−m− 1, p−m+ 1)− v)

+bi−2(B2µ(n−m− 1, p−m+ 1)− v − u0)

= bi(B2µ(n−m− 2, p−m))

+bi−2(Pa−1 ∪ Pb−1 ∪ Pn−a−b−p),

andbi(Pa−1 ∪ pb−1 ∪ Pn−a−b−p+2)= bi(Pa−1 ∪ Pb−1 ∪ Pn−a−b−p+1)+ bi−2(Pa−1 ∪ Pb−1 ∪ Pn−a−b−p)

holds for all i ≥ 0. Therefore,

Pa−1 ∪ Pb−1 ∪ Pn−a−b−p+2

≺ B2µ(n−m− 1, p−m+ 1).

Subcase 2.1.2.2. d(u1) ≥ 3.

Without loss of generality, assume that G ∼= D16 whereD16 is the graph with n vertices and p pendent vertices,and a part of pendent vertices and pendent path areshown in Fig.41. Since d(u1) ≥ 3, p ≥ m + 1. Fur-thermore, D16 − v ∈ B2(n− 1, p− 1) and D16 − v − u ∈B2(n−m−1, p−m).The problem is transformed into thesimilar situation above, so the corresponding conclusionis still established.

In summary, the result in Subcase 2.1 is proved com-pletely.

Subcase 2.2. dG(v, Ca,b) = 2.

Subcase 2.2.1. d(u) = 2.

Without loss of generality, assume that G ∼= D17 whereD17 is the graph with n vertices and p pendent vertices,and a part of pendent vertices and pendent path areshown in Fig.42.

Obviously, D17 − v ∈ B2(n − 1, p) and D17 − v − u ∈B2(n − 2, p − 1). The problem is transformed into theabove similar situation. Similar as Subcase 2.1.1.2, wehave the corresponding conclusion.

Subcase 2.2.2. d(u) ≥ 3.

Without loss of generality, assume that G ∼= D18 whereD18 is the graph with n vertices and p pendent vertices,and a part of pendent vertices and pendent paths areshown in Fig.43. Assume that G′ is discussed as above.Due to G 6∼= Qa,b,pn , p ≥ m+ 1. Obviously, we have

D18 − v ∈ B2(n− 1, p− 1),

andD18 − v − u ∈ B2(n−m− 1, p−m),

the problem is transformed into the subcases above. Sim-ilar as Subcase 1.2, we can get the corresponding conclu-sion.

The proof is completed.

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on minimal energy of a class of bicyclic graphs with given

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