On Minimal Energy of a Class of Bicyclic Graphs
with given Cycles’ Length and Pendent Vertices
∗
Yan Mi, Liancui Zuo
†and Chunhong Shang
Abstract
The energy ofGis defined as the sum of absolute value of all eigenvalues of the adjacency matrix A(G). Let B2(n, a, b, p)be the set of all bicyclic graphs on n ver-tices withppendent vertices and two cycles Ca and Cb which have unique common vertexu0,B2θ(n, a, b, p)the graph class obtained by coinciding the common vertexu0
of Ca and Cb with the center of the star Sn−(a+b−1)+1,
andB2
µ(n, a, b, p)the set obtained by coinciding the com-mon vertexu0 ofCa andCb with the center of the star Sp and connecting a pendent pathPn−(a+b−1)−(p−1) on
pointu0. In this paper, it is obtained thatBθ2(n, a, b, p) has the minimal energy in all graphs which have only pendent vertices except two cycles, andB2
µ(n, a, b, p)has the minimal energy in all graphs which have prescribed cycles’ length and pendent vertices.
Keywords: Bicyclic graphs; Quasi-Order method; Min-imal energy
1
Introduction
In this paper, all graphs are finite, connected, undirected and simple. In recent years, many parameters and classes of graphs were studied. For example, in [1], the restricted connectivity of Cartesian product graphs were obtained, and in [2, 12], some results on 3-equitable labeling and the n-dimensional cube-connected complete graph were gained. Let G be a graph with ordern and adjacency matrix A(G). The characteristic polynomial of G, de-noted byφ(G), is defined as
φ(G, x) = det(xI−A(G)) =
n
X
i=0
aixn−i,
where I is the identity matrix of order n. The roots of the equationφ(G) = 0, denoted byλ1, λ2,· · ·, λn, are the eigenvalues of A(G). It’s easy to see that eachλi is real sinceA(G)is real symmetric. The energy ofG, denoted byE(G), is defined as
E(G) =
n
X
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 code 201510065052.
†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 given in Sachs theorem,
ai(G) =
X
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 a
cycle) of Gthat contain i vertices, ω(S) is the number of connected components of S, and c(S) is the number of cycles contained inS.
It is well known that the Coulson integral formula of the energy is expressed as the following form
E(G) = 21πR−∞+∞x12ln[( bn
2c
P
i=0
(−1)ib2ix2i)2
+(
bn
2c
P
i=0 (−1)ib
2i+1x2i+1)2]dx.
(1)
Obviously, the formula (1) is a strictly monotonically increasing function of bi(G), that is to say, for any two graphs G1 andG2 with the same order, there exists the
following relationship:
bi(G1)≥bi(G2)hold f or 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 write
G1G2 or G2≺G1; if G1 G2 and there exists some
i0 such that bi0(G1)> bi0(G2), then we write G1G2. Combining with the formula (1), the increasing property G1G2⇒E(G1)> E(G2)of graph energy is obtained.
In theoretical chemistry, the energy of a molecular graph can be approximately used to represent π-electron en-ergy of the molecule, which is an important application of the energy of graphs in the chemical field, and has been widely studied by scholars. For a more detailed explanation can refer to the literature [3-4,10-11,13-15]. In addition, for some special class of graphs, searching for their extreme energy becomes an interesting topic. For the graphs with cycles, many researching conclu-sions have been obtained, for example, [7,9] give unicycle graphs with minimal energy and maximal energy, respec-tively. In [8,10], bicyclic graphs with minimal energy and maximal energy are gotten. On this basis, the relevant conclusions about the extreme energy of the graphs with
IAENG International Journal of Applied Mathematics, 47:1, IJAM_47_1_15
given parameters are also gained. For example, [13] ob-tains unicycle graphs with given diameter and minimal energy. In [5], the minimal energy of unicyclic graphs with prescribed girth and pendent vertices is given. In [6], the minimal energy of bicyclic graphs with a given diameter is obtained.
Bicyclic graphs are defined as connected graphs with n vertices and n+ 1 edges. According to the character-istics of the number of common vertices in two circles in the bicyclic graphs, they can be divided into three classes: two cycles without any common vertex, two cy-cles with common edges, and two cycy-cles with an unique common vertex. Here we discuss the last case, which is denoted by B2(n, a, b, p), where a, b are the two
cy-cles’ length, andpis the number of pendent vertices. In addition,B2
θ(n, a, b, p)is the graph class obtained by co-inciding the common vertex u0 of Ca and Cb with the center of the starSn−(a+b−1)+1. LetBµ2(n, a, b, p)be the graph class obtained by coinciding the common vertex u0ofCaandCbwith 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 Bθ2(n, a, b, p) and B2µ(n, a, b, p)asBθ2(n, p)andB
2
µ(n, p)simply (see Fig.1-2,20 and 28), subgraphs ofGwithout pendent vertices is called the base graph of G, and the base graph class of graph class mentioned above is represented byB2
∞ (see
Fig.3). Let V0(G)denote the set of all pendent vertices of G. dG(x, y) is defined as the distance between two verticesxandy of a graphG, 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 discuss the graphs with minimal energy inB2(n, a, b, p)that are
given length of two cycles with unique common point and pendent vertices by mathematical induction. It is obtained that B2
θ(n, a, b, p) has the minimal energy in all graphs which have only pendent vertices except two cycles, and B2µ(n, a, b, p) has the minimal energy in all graphs which have prescribed cycles’ length and pendent vertices (please see Fig.1-2).
Lemma 1 ([10]) LetG be a simple graph ande =uv be a pendent edge ofGwith pendent vertexv, then
bi(G) =bi(G−v) +bi−2(G−v−u).
Lemma 2 ([10]) Let G be a simple graph. If H is a subgraph (proper subgraph) ofG, thenGH(GH).
2
Graphs with only pendent vertices
ex-cept two cycles in
B
2(
n, a, b, p
)
Theorem 3. LetG∈B2(n, a, b, p)withn=a+b+p−1
and p≥1. Assume that the cycles’ length a and b are fixed, and u0 is the common vertex of two cycles. If
G6=B2
θ(n, a, b, p),thenGBθ2(n, a, b, p).
Proof. We will show the theorem by induction onp. (1) Suppose thatp= 1, i. e., there is only one pendent vertex. Assume thate=u0v is a pendent edge andv is
its pendent vertex, as shown in Fig.2. By Lemma 1, for B2
∞as in Fig.3, we can get
bi(B2θ(n,1)) =bi(G−v) +bi−2(G−u0−v) =bi(B∞2) +bi−2(Pa−1∪Pb−1).
&% '$ q
q q q q q q q
Ca u0 Cb
p v
Fig.1B2θ(n, a, b, p)
&% '$ q
q
Ca u0 Cb
v
Fig.2Bθ2(n, a, b,1)
&% '$ q
Ca u0 Cb
Fig.3B2 ∞
&% '$ q
q
Ca u0 Cb
q
ut
v1
Fig.4A11
&% '$ q
q q q
Ca Cb
v u0
ut
v1
Fig.5B11
&% '$ q
q
Ca u0 Cb
q q q
ut
v1 us
v
Fig.6B12
&% '$ q
q
q q
q
Ca Cb
v u0
ut
v1 ws
w0
Fig.7B13
&% '$ q
Ca u0 Cb
q
q q q q q q q q
ut
v1 l−1
Fig.8C11
&% '$ q
q q q
q q q
q
Ca u0 Cb
q q q q q q q
l−k wt
v1
Fig.9C12
q
Ca u0
q q q q q q q q q
q
q qut+1
ut−1
Fig.10H1
q
q
Ca u0
v
q q q qq q q q q q
q
q
ut+1
ut−1
Fig.11H2
q
Ca u0
q q q q q q q q q
q
q
ut+1
ut−1
q
q
v1 ut
Fig.12H3
q q q q q q q q q q q q q q
v1
u1u2 ut ut−1
Fig.13H4
&% '$ q
q q qq qq q q
qqq qq q
u0 Cb
ws+1
ws−1 q
q
ut
v1
Fig.14H5
Assume thatv is adjacent to ut butu0 belongs to cycle
Cb, as shown in Fig.4. By Lemma 1, for graphs A11 in
IAENG International Journal of Applied Mathematics, 47:1, IJAM_47_1_15
Fig.4 andH1in Fig.10, we obtain that
bi(A11) =bi(A11−v1) +bi−2(A11−ut−v1) =bi(B∞2 ) +bi−2(H1).
q
q
Ca u0
q q q q
l−1
q q q qq q q q q q
q
q
ut+1
ut−1
Fig.15H6
&% '$ q
Ca u0 Cb
q q q q q q q q q
ut
v1 l−k
q q q q
wt
Fig.16H7
&% '$ q
qqq qq q qwt+1
qqq qq q q
Cb
u0
wt−1 q
q q q q q q q q
ut
v1 l−k
q q q
Fig.17H8
&% '$ q
q q q q q q q
q qqqq
Ca u0 Cb
p
Fig.18H9
Obviously, Pa−1∪Pb−1 is a proper subgraph of H1, so
Pa−1∪Pb−1≺H1, and thenBθ2(n,1)≺A11.
(2) Forp= 2andB2
θ(n,2), by Lemma 1, we can get bi(Bθ2(n,2)) =bi(Bθ2(n,2)−v)
+bi−2(Bθ2(n,2)−u0−v)
=bi(Bθ2(n−1,1)) +bi−2(Pa−1∪Pb−1).
IfG6∼=B2
θ(n,2), then there is at least one pendent vertex which is not adjacent tou0. The problem can be divided
into the following cases.
Case 1. There is at least one pendent point which is adjacent to u0. Suppose that one point is such one and
the other is adjacent touton cycleCb, asB11 shown in
Fig.5. By Lemma 1, we have
bi(B11) =bi(B11−v1) +bi−2(B11−ut−v1) =bi(B2θ(n−1,1)) +Bi(H2),
where H2 is shown in Fig.11. Since Pa−1 ∪ Pb−1 is
a proper subgraph of H2, Pa−1∪Pb−1 ≺ H2. Thus,
B2
θ(n,2)≺B11.
Case 2. There is no pendent vertex which is adjacent to u0, and both of them are adjacent to ut andus on one cycleCb, respectively, as B12 shown in Fig. 6. Then by
Lemma 1, we can get
bi(Bθ2(n,2)) =bi(Bθ2(n,2)−v)
+bi−2(Bθ2(n,2)−u0−v)
=bi(Bθ2(n−1,1)) +bi−2(Pa−1∪Pb−1),
bi(B12) =bi(B12−v) +bi−2(B12−us−v)
=bi(H3) +bi−2(H4),
whereH3, H4are shown in Fig. 12-13.
We can obtain thatB2
θ(n−1,1)≺H3from (1). In
addi-tion, it is obvious thatPa−1∪Pb−1is a proper subgraph
ofH4, so Pa−1∪Pb−1≺H4. Hence,B2θ(n,2)≺B12.
Case 3. There is no pendent vertex that is adjacent to u0, and two points are adjacent to ut andwson two cycles Cb andCa, 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),
whereH5, H6are shown in Fig.14-15, respectively.
Comparing Pa−1∪H5 with H6, since the former is a
proper subgraph of the latter, Pa−1∪H5≺H6, that is,
B11 ≺ B13. Combining the conclusion B2θ(n,2) ≺ B11
from (1), we get B2
θ(n,2)≺B13.
Thus, the theorem holds forp= 2.
(3) Assume that the theorem holds forp=l−1. In the sequel, we prove that the theorem still holds for p =l. ForB2
θ(n, l), we have
bi(Bθ2(n, l)) =bi(Bθ2(n, l)−v) +bi−2(Bθ2(n, l)−u0−v) =bi(Bθ2(n−1, l−1)) +bi−2(Pa−1∪Pb−1).
If there is at least one pendent vertex whose neighbor is notu0, asC11shown in Fig.8, then by Lemma 1, we can
get
bi(C11) =bi(C11−v1) +bi−2(C11−ut−v1) =bi(B2θ(n−1, l−1)) +bi−2(H7),
where H7 is shown in Fig.16. SinceH7containsPa−1∪
Pb−1 as its proper subgraph, Pa−1∪Pb−1 ≺H7, so we
have Bθ2(n, a, b, l)≺C11.
Now suppose that there arep−k pendent vertices that are 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 edgewtv1withwt6=u0, then
bi(C12) =bi(C12−v1) +bi−2(C12−wt−v1) =bi(H8) +bi−2(H9),
whereH8, H9 are shown in Fig.17-18. Note thatH8 and
H9display a part of the pendent vertices except the
de-formation by deleting one or two vertices fromC12. Since
B2θ(n−1, l−1)andH8are graphs which have only
pen-dent vertices except two cycles with n−1 vertices, by induction hypothesis, we can getB2
θ(n−1, l−1)≺H8,
and Pa−1∪Pb−1≺H9 because Pa−1∪Pb−1 is a proper
subgraph ofH9. ThusBθ2(n, p)≺C12, and the theorem
holds forp=l.
In a word, we have shown thatB2
θ(n, p)is the graph with minimal energy for1≤p≤l, l≥2andn=a+b+p−1.
3
Bicyclic graphs with prescribed
cy-cles’ length and pendent vertices in
B
2(
n, a, b, p
)
On the basis of Theorem 3, we study the case of n > a+b+p−1. LetQa,b,p
n be graphs shown in Fig. 19.
IAENG International Journal of Applied Mathematics, 47:1, IJAM_47_1_15
Theorem 4. LetG∈B2(n, a, b, p)withn≥a+b+p−1
and p ≥ 1. Assume that the cycles’ length a and b are 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 have B2µ(n, a, b, p) ∼= Bθ2(n, a, b, p), the result holds for this case. We show that the result is also true for n−(a+
b−1 +p)≥1in the following.
Assume that V0(G) is the set of all pendent vertices
of G, v is the point in V0(G) such that d
G(v, Ca,b) =
max{dG(x, Ca,b), x∈V0(G)}, anduis its unique neigh-bor. &% '$ q
Ca u0 Cb
qqqqq q q qq q q q p &% '$ q
Ca u0 Cb
q qqqqq q q qq q q q p
Fig.19 Qa,b,pn
&% '$ q
qq q q q q q v v0 u0
Ca u0Cb
p
Fig.20B2
µ(n, a, b, p)
&% '$ q
q q q q q q q q q q u v
Ca u0 Cb
Fig.21D11 &% '$ q
qq q q q q q v u q q
Ca u0Cb
Fig.22D110
&% '$ q
q q q q q q q q q q u qq q q q
v m
Ca u0 Cb
Fig.23D12 &% '$ q q qq q q v q u q m q q q
qCa u0 Cb
Fig.24D120
&% '$ q q q
q q q q
u q
v
Ca u0 Cb
Fig.25D121
&% '$ q q q u q q
v q q
Ca u0 Cb
q q
Fig.26D122
&% '$ q q q
q q q q
u0 q
v0 q
q
Ca u0 Cb
Fig.27D1220
&% '$ q
qq q q q q q q q q q q qq vu u1 v1 Ca u0Cb
p
Fig.28B2µ(n, p)
&% '$ q
qq q q q qq q v u u1
Ca u0Cb
p
Fig.29Bµ2(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 getdG(v, Ca,b) = 2.
Subcase 1. 1. d(u) = 2. By Lemma 1, it is not difficult to verify that
bi(Bµ2(n, p)) =bi(Bµ2(n, p)−v0) +bi−2(Bµ2(n, p)−v0−u0).
Suppose thatG6∼=Bµ2(n, p), then we must haveG∼=D11
orG∼=D0
11(bothD11andD110 are graphs withnvertices
and ppendent vertices, and a part of pendent vertices which are adjacent to vertices on cycles are shown in Fig.21-22). TakeD11as an example, we have
bi(D11) =bi(D11−v) +bi−2(D11−v−u),
whereD11−v∈B2(n−1, p)andD11−v−u∈B2(n− 2, p−1). Since
B2µ(n, p)−v0 =∼Bθ2(n−1, p)
and
Bµ2(n, p)−v0−u0∼=Bθ2(n−2, p−1),
we have
Bθ2(n−1, p)≺B2(n−1, p), Bθ2(n−2, p−1)≺B2(n−2, p−1)
by Theorem 3. On account of G6∼=B2
µ(n, p), we can get Bµ2(n, p) −v0≺D11−v
and
Bµ2(n, p) −v0−u0≺D11−v−u.
Thus, Bµ2(n, p) ≺ D11. Similarly, we can obtain that
B2µ(n, p)≺D011.
Subcase 1. 2. d(u)≥3.
We get firstly
bi(Bµ2(n, p)) =bi(Bµ2(n, p)−v) +bi−2(Bµ2(n, p)−v−u0)
=bi(Bµ2(n−1, p−1)) +bi−2(Pa−1∪Pb−1∪P2), (2)
Assume that G 6∼= B2
µ(n, a, b, p), we must get G ∼= D12
or G∼=D012 (both D12 and D012 are graphs withn
ver-tices and ppendent vertices, and a part of pendent ver-tices that are adjacent to verver-tices on cycles are shown in Fig.23-24). TakeD12as 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. Suppose that there are m pendent vertices adjacent to u, then |N(u)| = m + 1. Since G 6∼= Qa,b,p
n , we have p ≥ m + 1. Set D12 −v − u = G0 ∪(m −1)P1, then
G0∈B2(n−m−1, p−m).
Now we show the result by induction n and p in the following. Assume that the result holds for smalln and p, then B2
µ(n−1, p−1) ≺ B2(n−1, p−1). Because
IAENG International Journal of Applied Mathematics, 47:1, IJAM_47_1_15
p≥m+ 1, d(u)≥3, anddG(v, Ca,b) = 2, there are at least four pendent vertices inG, son≥a+b+ 4. (1) Assume thatn−(a+b) = 4, i. e.,n=a+b+ 4, and G∈B2(a+b+ 4,4). Suppose thatG∼=D
121,|V(D121−
v)| −(a+b) = 3 and D121−v ∈ B2(a+b+ 3,3) (see
Fig.25). In the following, we use Quasi-Order method to compare Bµ2(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(B2(a+b,1)).
By Subcase 1.1, we haveB2
µ(a+b+2,2)≺B2(a+b+2,2), and
Pa−1∪Pb−1∪P2≺B2(a+b,1)
sincePa−1∪Pb−1∪P2is a proper subgraph ofB2(a+b,1),
so we get
Bµ2(a+b+ 3,3)≺D121−v
and
Bµ2(a+b+ 3,3)≺B2(a+b+ 3,3).
Thus, forn=a+b+ 4andp= 4, we have Bµ2(a+b+ 3,3)≺B2(a+b+ 3,3),
that is,
Bµ2(n−1, p−1)≺B2(n−1, p−1).
(2) Suppose thatn−(a+b) = 5, i. e.,n=a+b+ 5. On the basis of (1), add one vertex to graphD121 such
that the number of pendent vertices increasing, then the added pendent vertex is adjacent touor other vertex on cycles, and we obtainD122, D1220∈B2(a+b+ 5,5)that
are shown in Fig.26-27.
Suppose that G∼=D122 or G∼=D122, then |V(D122−
v)|=|V(D1220−v0)|=a+b+4andD122−v, D1220−v0 ∈
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(B2(a+b,1)),
and
bi(D1220−v0) =bi(B2(a+b+3,3))+bi−2(B2(a+b+1,2)).
Combining with the result of (1), we have Bµ2(a+b+ 3,3)≺B2(a+b+ 3,3).
Since Pa−1∪Pb−1 ∪P2 is a proper subgraph of both
B2(a+b,1)andB2(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),
Bµ2(a+b+ 4,4)≺D122−v
and
Bµ2(a+b+ 4,4)≺D1220−v0,
so we have
Bµ2(a+b+ 4,4)≺B2(a+b+ 4,4).
Thus, forn=a+b+ 5andp= 5, we have Bµ2(a+b+ 4,4)≺B
2
(a+b+ 4,4),
that is,
Bµ2(n−1, p−1)≺B2(n−1, p−1).
(3) Assume that the theorem holds forn−(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 for n−(a+b) =p. Suppose thatG=∼D12orG∼=D120 (see
Fig. 23-24). TakeD12as an example. Clearly,|V(D12−
v)|=a+b+p−1andD12−v∈B2(a+b+p−1, p−1).
A comparison betweenBµ2(a+b+p−1, p−1)andD12−v
will be accomplished in the following. By Lemma 1, we have
bi(Bµ2(a+b+p−1, p−1)) =
bi(Bµ2(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).
Becausep−m≥1, it is not difficult to see thatPa−1∪
Pb−1∪P2 is a proper subgraph ofB2(a+b+p−m− 1, p−m), so
Pa−1∪Pb−1∪P2≺B2(a+b+p−m−1, p−m),
and
Bµ2(a+b+p−1, p−1)≺D12−v.
Thus, forn=a+b+p, we have
B2µ(a+b+p−1, p−1)≺B2(a+b+p−1, p−1),
that is,
Bµ2(n−1, p−1)≺B2(n−1, p−1).
By induction, we have
Bµ2(n−1, p−1)≺B2(n−1, p−1),
and the relationship between the first item on right of equations (2) and (3): Bµ2(n−1, p−1)≺D12−v. The
problem is turned into a proof of the relationship between the second:Pa−1∪Pb−1∪P2≺D12−v−u.
IAENG International Journal of Applied Mathematics, 47:1, IJAM_47_1_15
As what mentioned before,D12−v−u∈B2(n−m−1, p−
m). What remains is to prove thatPa−1∪Pb−1∪P2≺
B2(n−m−1, p−m). Sincep−m≥1, it is clear that
Pa−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 B2
µ(n, p) ≺ D12. Similarly, we can
ob-tain that B2
µ(n, p)≺D012. 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 neighbor
vertex ofu.
Assume that G6∼=Bµ2(n, a, b, p), we must have G∼=D13
which has n vertices and p pendent vertices, where a part of pendent vertices and pendent pathes are shown in Fig.31. &% '$ q q q qqq qq q
q q q q q
Ca u0Cb
v u u1
Fig.30B2µ(a+b+p+ 2, p+ 1)
&% '$ q q q q q q q q q q
Ca u0Cb
v0 u0 u
0 1 Fig.31D13 &% '$ q q q q q q q q q q
Ca u0Cb
q
v0 u0 u 0 1
Fig.32D131
&% '$ q q q q q q q q q q
Ca u0Cb
q
v0 u0 u
0 1
Fig.33D132
&% '$ q
qq q q q q
Ca u0Cb
p
Fig.34D1320
&% '$ q
qq q q q q
q
Ca u0Cb
p−1
Fig.35 D14 &% '$ q
qq q q q q q q q q q q v1 Ca u0Cb
p
Fig.36B2µ(a+b+p−1, p)
&% '$ q
qq q q q q q q q q q q qq v2u2
Ca u0Cb
p−1
Fig.37Bµ2(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 get B2
µ(n−1, p)≺D13−v andB2µ(n−2, p)≺D13−v−u, we
need to prove thatB2
µ(n−1, p)≺B2(n−1, p)andBµ2(n−
2, p) ≺ B2(n−2, p). We show the result by induction
on n. Assume that the result holds for small n, then B2
µ(n−1, p)≺B2(n−1, p)and Bµ2(n−2, p)≺B2(n−
2, p). Note thatdG(v, Ca,b)≥3, andd(u) = d(u1) = 2.
Assume that the number of pendent vertices as p will not change even though the order of graph G changes constantly. There are at least two vertices except two cycles and pendent vertices, son≥a+b+p+ 1.
&% '$ q
q q q q q q q q q q q q q q q q q q q m vu u1
Ca u0Cb
Fig.38D15 &% '$ q
q q q q q q q q q q q q q qq q qq q m vu u1
Ca u0Cb
Fig.39D150
&% '$ q
qq q q q q q q q q q q qq v Ca u0Cb
p−m+ 1
Fig.40B2
µ(n−m−1, p−m+ 1)
&% '$ q
q q q q q q q q q q q q q qq q qq q m vu u1 q q q q
Ca u0Cb
Fig.41D16 &% '$ q
qq q q q q q v u q q q q q
Ca u0Cb
Fig.42D17 &% '$ q
q q q q q q q q q q u qq q q q
v m
Ca u0 Cb
q q q
Fig.43D18
ForB2
µ(n, a, b, p)andD13, by Lemma 1, we have
bi(Bµ2(n, p)) =bi(Bµ2(n−1, p)) +bi−2(B2µ(n−2, p)),
bi(D13) =bi(D13−v) +bi−2(D13−v−u).
(1)Ifn−(a+b−1)−p= 2, thenn=a+b+p+ 1. Suppose that G ∼= D131 (the pendent edges are not
shown all in Fig.32) sinceG6∼=B2
µ(a+b+p+1, p). We can getV(D13−v0) =a+b+pandD13−v0 ∈B2(a+b+p, p).
We use Quasi-Order method to compareB2
µ(a+b+p, p) andD131−v0. By Lemma 1, we get
bi(Bµ2(a+b+p, p)) =bi(B2µ(a+b+p−1, p))
+bi−2(Bµ2(a+b+p−2, p−1))
and
bi(D131−v0) =bi(B2(a+b+p−1, p))
+bi−2(B2(a+b+p−2, p−1).
Since
Bµ2(a+b−1 +p, p)=∼Bθ2(a+b−1 +p, p),
B2µ(a+b+p−2, p−1)=∼Bθ2(a+b+p−2, p−1).
IAENG International Journal of Applied Mathematics, 47:1, IJAM_47_1_15
By Theorem 3, we can obtain that Bµ2(a+b−1 +p, p)≺B
2
(a+b−1 +p, p),
and
B2µ(a+b+p−2, p−1)≺B2(a+b+p−2, p−1),
so we haveBµ2(a+b+p, p)≺D131−v0, that isBµ2(a+ b+p, p)≺B2(a+b+p, p). Thus, forn=a+b+p+ 1, we have Bµ2(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 D131
but the number of pendent vertices will not change, there exist two methods: lengthening the pendent path which contains v0 and u0 on D131 or other pendent
path. Without loss of generality, assume that we add one vertex such that the graph has the same pendent vertices, then we obtainG∼=D132 orG∼=D1320, where
D132, D1320∈B2(a+b+p+ 2, p).
(2) Ifn−(a+b−1)−p= 3, thenn=a+b+p+ 2, so it is not difficult to see that
V(D132−v0) =a+b+p+ 1,
D132−v0∈B2(a+b+p+ 1, p),
V(D1320−v0) =a+b+p+ 1,
and
D1320−v0∈B2(a+b+p+ 1, p).
ComparingB2µ(a+b+p+ 1, p)andD132−v0 as well as
B2
µ(a+b+p+ 1, p)andD1320−v0, by Lemma 1, we get
bi(Bµ2(a+b+p+ 1, p)) =bi(Bµ2(a+b+p, p))
+bi−2(Bµ2(a+b+p−1, p)),
bi(D132−v0) =bi(B2(a+b+p, p))
+bi−2(B2(a+b+p−1, p),
and
bi(D1320−v0) =bi(B2(a+b+p, p))
+bi−2(B2(a+b+p−1, p−1)).
Comparing three equations above, combining with the proof of (1), the result of the comparison between the first term of the right side of the equal isB2
µ(a+b+p, p)≺ B2(a+b+p, p).
For the second term of the right side of the equal, since Bµ2(a+b−1 +p, p)=∼Bθ2(a+b−1 +p, p),
by Lemma 1, we getB2
µ(a+b−1 +p, p)≺B2(a+b+ p−1, p).
ForBµ2(a+b+p−1, p)andB2(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
Bµ2(a+b+p−1, p)≺Bµ2(a+b+p−1, p−1).
From Lemma 1,we can get bi(Bµ2(a+b+p−1, p))
=bi(B2θ(a+b+p−2, p−1)) +bi−2(Pa−1∪Pb−1)
and
bi(B2(a+b+p−1, p−1))
=bi(Bθ2(a+b+p−2, p−1))
+bi−2(B2(a+b+p−3, p−2)).
SincePa−1∪Pb−1is a proper subgraph ofB2(a+b+p− 3, p−2), the relationship on the second term of the right side of the equal is
Pa−1∪Pb−1≺B2(a+b+p−3, p−2),
so
Bµ2(a+b+p−1, p)≺B2(a+b+p−1, p−1).
Thus, forn=a+b+p+ 2, we obtain that Bµ2(a+b+p+ 1, p)≺B2(a+b+p+ 1, p),
i. e.,
B2µ(n−1, p)≺B2(n−1, p).
(3) Ifn−(a+b−1)−p=N−1, thenn=a+b+p+N−2. Setn1=n, assume that the theorem holds forn1, i. e.,
we have
Bµ2(n1−2, p)≺B2(n1−2, p),
and
Bµ2(n1−1, p)≺B2(n1−1, p).
Suppose thatn−(a+b−1)−p=N, thenn=a+b+p+
N−1. Setn2=n. SinceV(G−v) =n2−1, without loss
of generality, assume thatG∼=D13 (the pendent pathes
are not shown all in the Fig.31). Obviously, D13−v ∈
B2(n
2−1, p). ComparingBµ2(n2−1, p)withD13−v, by
Lemma 1, we have bi(Bµ2(n2−1, p))
=bi(Bµ2(n2−2, p)) +bi−2(Bµ2(n2−3, p)),
and
bi(D13−v)
=bi(B2(n2−2, p)) +bi−2(B2(n2−3, p)).
Due ton2=n1+ 1, we can get
B2
µ(n2−2, p)∼=Bµ2(n1−1, p),
B2(n
2−2, p)∼=B2(n1−1, p),
B2
µ(n2−3, p)∼=Bµ2(n1−2, p),
and
B2(n2−3, p)=∼B2(n1−2, p).
By induction hypothesis, we have
Bµ2(n1−1, p)≺B2(n1−1, p),
and
Bµ2(n1−2, p)≺B2(n1−2, p).
IAENG International Journal of Applied Mathematics, 47:1, IJAM_47_1_15
Therefore,
Bµ2(n2−2, p)≺B2(n2−2, p),
and
Bµ2(n2−3, p)≺B2(n2−3, p)),
that is to say,
B2µ(n−2, p)≺B2(n−2, p),
and
B2µ(n−1, p)≺B2(n−1, p).
So
Bµ2(n−1, p)≺D13−v.
According to the induction above, we have Bµ2(n−1, p)≺D13−v,
and
Bµ2(n−2, p)≺D13−v−u,
that is,
bi(Bµ2(n−1, p))< bi(D13−v),
and
bi(Bµ2(n−2, p)< bi(D13−v−u).
Thus,B2µ(n, p)≺D13and then the result is obtained.
Subcase 2.1.1.2. d(u1)≥3.
Suppose that G∼=D14, where D14 is the graph with n
vertices and p pendent vertices and a part of pendent vertices and a pendent path are shown in Fig.35. For B2
µ(n, p)andD14, we have
bi(Bµ2(n, p)) =bi(Bµ2(n−1, p)) +bi−2(B2µ(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 and
subcase 1.2, we have B2
µ(n−1, p) ≺ B2(n−1, p), and B2
µ(n−2, p−1) ≺B2(n−2, p−1). SoBµ2(n−1, p)≺ D14−v,Bµ2(n−2, p−1)≺D14−v−u. Hence the problem
is turned into showing B2
µ(n−2, p)≺B2µ(n−2, p−1). (i)Forn−2−(a+b−1 +p) = 0, then B2
µ(n−2, p)∼= B2θ(n−2, p).
By Lemma 1, we get
B2θ(n−2, p) =bi(B2θ(n−3, p−1)) +bi−2(Pa−1∪Pb−1),
B2µ(n−2, p−1) =bi(Bθ2(n−3, p−1))+bi−2(B2θ(n−4, p−2)).
It is not difficult to find that Pa−1 ∪Pb−1 is a proper
subgraph ofB2
θ(n−4, p−2), so
B2µ(n−2, p)≺Bµ2(n−2, p−1).
(ii)Forn−2−(a+b−1 +p)≥1. By Lemma 1, we get bi(B2µ(n−2, p)) =bi(Bµ2(n−2, p)−v1)
+bi−2(B2µ(n−2, p)−v1−u1),
bi(Bµ2(n−2, p−1)) =bi(Bµ2(n−2, p−1)−v2)
+bi−2(B2µ(n−2, p−1)−v2−u2).
Note thatB2
µ(n−2, p)−v1∼=Bµ2(n−2, p−1)−v2. We
have
bi(B2µ(n−2, p)−v1) =bi(B2µ(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 of B2
µ(n−2, p−1)−v2−u2, we obtain that
Bµ2(n−2, p)−v1−u1≺Bµ2(n−2, p−1)−v2−u2,
thus
B2µ(n−2, p)≺Bµ2(n−2, p−1).
In summary, we getB2
µ(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 n
vertices andppendent vertices that are adjacent tou1,
and a part of pendent vertices and pendent path are shown in Fig.38.
ForB2
µ(n, a, b, p)andD15, by Lemma 1, we have
bi(Bµ2(n, p)) =bi(Bµ2(n, p)−v1)+bi−2(Bµ2(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 vertices
are pendent vertices except one pointu1 in N(u).
Suppose that there are m pendent vertices adjacent to u, then|N(u)|=m+ 1(please seeD0
15 in Fig.39). Let
D015−v−u=G0∪(m−1)P1. ThenG0 ∈B2(n−m− 1, p−m+ 1). Similar as subcase 1.2, we have
Bµ2(n−1, p−1)≺B2(n−1, p−1),
and
Bµ2(n−m−1, p−m+ 1)≺B2(n−m−1, p−m+ 1),
so we obtain thatBµ2(n−1, p−1)≺D150 −vandBµ2(n− m−1, p−m+1)≺D015−v−u. What remains is to prove
thatPa−1∪Pb−1∪Pn−a−b−p+2≺B2µ(n−m−1, p−m+1). SinceG6∼=Qa,b,p
n , p−m≥1.
(i) Forp=m+1,B2
µ(n−m−1, p−m+1)∼=B2µ(n−p,2), then by Lemma 1, we can get
bi(B2µ(n−m−1, p−m+ 1))
=bi(Bµ2(n−p,2))
=bi(Bµ2(n−p−1,1))
+bi−2(Pa−1∪Pb−1∪Pn−p−a−b),
and
bi(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).
IAENG International Journal of Applied Mathematics, 47:1, IJAM_47_1_15
Comparing the first term of the right side of the equal, it is not difficult to find thatPa−1∪Pb−1∪Pn−a−b−p+1
is a proper subgraph ofB2
µ(n−p−1,1), so we have Pa−1∪Pb−1∪Pn−a−b−p+1≺Bµ2(n−p−1,1).
Therefore,
Pa−1∪Pb−1∪Pn−a−b−p+2≺B2µ(n−m−1, p−m+ 1)
holds.
(ii) Forp≥m+ 2,by Lemma 1, we get bi(B2µ(n−m−1, p−m+ 1))
=bi(Bµ2(n−m−1, p−m+ 1)−v)
+bi−2(Bµ2(n−m−1, p−m+ 1)−v−u0) =bi(Bµ2(n−m−2, p−m))
+bi−2(Pa−1∪Pb−1∪Pn−a−b−p),
and
bi(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 alli≥0. Therefore,
Pa−1∪Pb−1∪Pn−a−b−p+2 ≺B2
µ(n−m−1, p−m+ 1).
Subcase 2.1.2.2. d(u1)≥3.
Without loss of generality, assume that G∼=D16 where
D16is the graph withnvertices andppendent vertices,
and a part of pendent vertices and pendent path are shown in Fig.41. Since d(u1) ≥ 3, p ≥ m+ 1.
Fur-thermore,D16−v∈B2(n−1, p−1) andD16−v−u∈
B2(n−m−1, p−m).The problem is transformed into the
similar situation above, so the corresponding conclusion is 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 where
D17is the graph withnvertices andppendent vertices,
and a part of pendent vertices and pendent path are shown in Fig.42.
Obviously, D17−v ∈ B2(n−1, p) and D17−v−u ∈
B2(n−2, p−1). The problem is transformed into the
above similar situation. Similar as Subcase 2.1.1.2, we have the corresponding conclusion.
Subcase 2.2.2. d(u)≥3.
Without loss of generality, assume that G∼=D18 where
D18is the graph withnvertices andppendent vertices,
and a part of pendent vertices and pendent paths are shown in Fig.43. Assume that G0 is discussed as above. Due toG6∼=Qa,b,pn ,p≥m+ 1. Obviously, we have
D18−v∈B2(n−1, p−1),
and
D18−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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