Complete solution to a conjecture on the maximal energy of unicyclic graphs

Abstract

For a given simple graph G, the energy of G, denoted by E(G), is defined as the sum of the absolute values of all eigenvalues of its adjacency matrix. Let be the unicyclic graph obtained by connecting a vertex of Cℓ with a leaf of Pn−ℓ. In [G. Caporossi, D. Cvetković, I. Gutman, P. Hansen, Variable neighborhood search for extremal graphs. 2. Finding graphs with extremal energy, J. Chem. Inf. Comput. Sci. 39 (1999) 984–996], Caporossi et al. conjectured that the unicyclic graph with maximal energy is Cn if n≤7 and n=9,10,11,13,15, and for all other values of n. In this paper, by employing the Coulson integral formula and some knowledge of real analysis, especially by using certain combinatorial techniques, we completely solve this conjecture. However, it turns out that for n=4 the conjecture is not true, and should be the unicyclic graph with maximal energy.

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arXiv:1011.4658v1 [math.CO] 21 Nov 2010
Complete Solution to a Conjecture
on the Maximal Energy of Unicyclic Graphs∗
Bofeng Huo1,2, Xueliang Li1, Yongtang Shi1
1Center for Combinatorics and LPMC-TJKLC
Nankai University, Tianjin 300071, China
E-mail: huobofeng@mail.nankai.edu.cn; lxl@nankai.edu.cn; shi@nankai.edu.cn
2Department of Mathematics and Information Science
Qinghai Normal University, Xining 810008, China
Abstract
For a given simple graph G, the energy of G denoted by E(G), is defined as the
sum of the absolute values of all eigenvalues of its adjacency matrix. Let Pℓ
unicyclic graph obtained by connecting a vertex of Cℓwith a leaf of Pn−ℓ. In [G.
Caporossi, D. Cvetkovi´ c, I. Gutman, P. Hansen, Variable neighborhood search for
extremal graphs. 2. Finding graphs with extremal energy, J. Chem. Inf. Comput.
Sci. 39 (1999) 984–996], Caporossi et al. conjectured that the unicyclic graph with
maximal energy is Cnif n ≤ 7 and n = 9,10,11,13,15 , and P6
of n. In this paper, by employing the Coulson integral formula and some knowledge
of real analysis, especially by using certain combinatorial techniques, we completely
solve this conjecture. However, it turns out that for n = 4 the conjecture is not
true, and P3
4should be the unicyclic graph with maximal energy.
nbe the
nfor all other values
Keywords: graph energy; Coulson integral formula; unicyclic graph
AMS classification 2010: 05C50, 05C90, 15A18, 92E10.
1Introduction
For a given simple graph G of order n, denote by A(G) the adjacency matrix of G.
The characteristic polynomial of A(G) is usually called the characteristic polynomial of
G, denoted by
φ(G,x) = det(xI − A(G)) = xn+ a1xn−1+ ··· + an,
∗Supported by NSFC and “the Fundamental Research Funds for the Central Universities”.

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If G is a bipartite graph, the characteristic polynomial of G has the form
φ(G,x) =
⌊n/2⌋
?
k=0
a2kxn−2k=
⌊n/2⌋
?
k=0
(−1)kb2kxn−2k,
where b2k= (−1)ka2kand b2k≥ 0 for all k = 1,...,⌊n/2⌋, especially b0= a0= 1. In
particular, if G is a tree, the characteristic polynomial of G can be expressed as
φ(G,x) =
⌊n/2⌋
?
k=0
(−1)km(G,k)xn−2k,
where m(G,k) is the number of k-matchings of G.
For a graph G, let λ1,λ2,...,λndenote the eigenvalues of φ(G,x). The energy of G is
defined as
E(G) =
n
?
i=1
|λi|.
This definition was put forward by Gutman [5] in 1978. The following formula is also
well-known
E(G) =1
π
−∞
where i2= −1. Furthermore, in the book of Gutman and Polansky [8], the above equality
was converted into an explicit formula as follows:

k=0
?+∞
1
x2log|xnφ(G,i/x)|dx,
E(G) =
1
2π
?+∞
−∞
1
x2log



⌊n/2⌋
?
(−1)ka2kx2k


2
+


⌊n/2⌋
?
k=0
(−1)ka2k+1x2k+1


2
dx.
For more results about graph energy, we refer the reader to the survey of Gutman, Li and
Zhang [7].
For two given trees, or bipartite graphs, or unicyclic graphs G1and G2, according to
the corresponding coefficients of the characteristic polynomials, one can introduce a quasi
order to compare the values of E(G1) and E(G2). Actually, the quasi order method is
commonly used to compare the energies of pairs of such graphs. However, for general
graphs, it is difficult to define such a quasi order. If, for two trees, or bipartite graphs, or
unicyclic graphs, the above quantities m(T,k) or |ak(G)| can not be compared uniformly,
then the quasi order method is invalid, and this happened very occasionally. Recently,
for these quasi-order incomparable problems, we find an efficient approach to determine
which one attains the extremal value of the energy, such as our earlier papers [11]–[15].
Let Cnbe the cycle of order n, Pnthe path of order n, and Pℓ
obtained by connecting a vertex of Cℓwith a leaf of Pn−ℓ. In [2], Caporossi et al. proposed
a conjecture on the unicyclic graph with maximal energy.
nthe unicyclic graph
Conjecture 1 Among all unicyclic graphs on n vertices, the cycle Cnhas maximal energy
if n ≤ 7 and n = 9,10,11,13 and 15. For all other values of n, the unicyclic graph with
maximal energy is P6
n.

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In [10], the authors proved Theorem 1 that is weaker than the above conjecture,
namely that E(P6
differing from Cn. And they also claimed that the energy of Cnand P6
incomparable.
n) is maximal within the class of the unicyclic bipartite n-vertex graphs
nis quasi-order
Theorem 1 Let G be any connected, unicyclic and bipartite graph on n vertices and
G ≇ Cn. Then E(G) < E(P6
n).
Very recently, our another paper [14] and Andriantiana [1] independently proved that
E(Cn) < E(P6
the maximum value of the energy among all the unicyclic bipartite graphs for n = 8,12,14
and n ≥ 16, which partially solves the above conjecture.
n), and then completely determined that P6
nis the only graph which attains
Theorem 2 For n = 8,12,14 and n ≥ 16, E(P6
n) > E(Cn).
In this paper, by employing the Coulson integral formula and some knowledge of real
analysis, especially by using certain combinatorial techniques, we completely solve this
conjecture by proving the following theorem and corollary. However, we find that for
n = 4 the conjecture is not true, and P3
energy.
4should be the unicyclic graph with maximal
Theorem 3 Among all unicyclic graphs of order n ≥ 16, the unicyclic graph with maxi-
mal energy is P6
n.
Corollary 1 Among all unicyclic graphs on n vertices, the cycle Cnhas maximal energy
if n ≤ 7 but n ?= 4, and n = 9,10,11,13 and 15; P3
all other values of n, the unicyclic graph with maximal energy is P6
4has maximal energy if n = 4. For
n.
2Preliminaries
Let G(n,ℓ) be the set of all connected unicyclic graphs on n vertices, containing the
cycle Cℓas a subgraph. Denote by C(n,ℓ) the set of all unicyclic graphs obtained from
Cℓby adding to it n − ℓ pendent vertices. In the following, we list some results given in
[10] which will be useful in our proof.
Lemma 1 Let G ∈ G(n,ℓ) and n > ℓ. If G has maximal energy in G(n,ℓ), then G is
either Pℓ
nor, when ℓ = 4r, a graph from C(n,ℓ).
Lemma 2 Let G ∈ C(n,ℓ) and n > ℓ. If ℓ is even with ℓ ≥ 8 or ℓ = 4, then E(G) <
E(P6
n).
Lemma 3 Let ℓ be even and ℓ ≥ 8 or ℓ = 4. Then E(Pℓ
n) < E(P6
n).

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Form Lemmas 1–3 and Theorem 2, we conclude that for any n-vertex unicyclic graph
G, when the length of the unique cycle of G is even and n = 8,12,14 and n ≥ 16,
E(G) < E(P6
E(G) < E(Pℓ
every odd ℓ and n ≥ 16.
In the remainder of this section, we will introduce some lemmas and notations. At
first, we recall some knowledge on real analysis, for which we refer to [17].
n); when the length of the unique cycle of G is odd, if G ∈ G(n,ℓ), then
n). For proving Theorem 3, we only need to show that E(Pℓ
n) < E(P6
n) for
Lemma 4 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.
The following lemma on the difference of the energy of two graphs is a well-known result
due to Gutman [6], which will be used in the sequel.
Lemma 5 If G1and G2are two graphs with the same number of vertices, then
E(G1) − E(G2) =1
π
?+∞
−∞
log
????
φ(G1,ix)
φ(G2,ix)
????dx.
Now we present one basic formula of the characteristic polynomial φ(G,x), which can
be found in [3].
Lemma 6 Let uv be an edge of G. Then
φ(G,x) = φ(G − uv,x) − φ(G − u − v,x) − 2
?
C∈C(uv)
φ(G − C,x)
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).
From Lemma 6, we can easily obtain the following lemma.
Lemma 7 For any positive integer t ≤ n − 2, φ(Pt
particular, φ(P6
n,x) = xφ(Pt
n−1,x) − φ(Pt
n−2,x). In
n,x) = xφ(P6
n−1,x) − φ(P6
n−2,x).
Now for convenience, we introduce some notions as follows, which will be well used in
this sequel.
Y1(x) =x +√x2− 4
2
,Y2(x) =x −√x2− 4
2
.
It is easy to verify that Y1(x) + Y2(x) = x, Y1(x)Y2(x) = 1, Y1(ix) =
Y2(ix) =x−√x2+4
2
i. We define
Z1(x) = −iY1(ix) =x +√x2+ 4
2
Observe that Z1(x) + Z2(x) = x and Z1(x)Z2(x) = −1. In addition, for x > 0, Z1(x) > 1
and −1 < Z2(x) < 0; for x < 0, 0 < Z1(x) < 1 and Z2(x) < −1. In the rest of this paper,
we abbreviate Zj(x) to Zjfor j = 1,2.
x+√x2+4
2
i and
, Z2(x) = −iY2(ix) =x −√x2+ 4
2
.

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3Main results
Now we define some notions, which will be used frequently later.
A1(x) =Y1(x)φ(P6
8,x) − φ(P6
7,x)
(Y1(x))9− (Y1(x))7
t+2,x) − φ(Pt
(Y1(x))t+3− (Y1(x))t+1
(Y1(x))3− Y1(x),
,A2(x) =Y2(x)φ(P6
8,x) − φ(P6
7,x)
(Y2(x))9− (Y2(x))7
t+2,x) − φ(Pt
(Y2(x))t+3− (Y2(x))t+1
C2(x) =Y2(x)(x2− 1) − x
(Y2(x))3− Y2(x).
8,x) = x8− 8x6+ 19x4− 16x2+ 4 and φ(P6
,
B1(x) =Y1(x)φ(Pt
t+1,x)
, B2(x) =Y2(x)φ(Pt
t+1,x)
,
C1(x) =Y1(x)(x2− 1) − x
By some calculations, we get that φ(P6
x7− 7x5+ 13x3− 7x, and then
A1(ix) = −Z1f8+ f7
7,x) =
Z2
1+ 1
Z7
2,A2(ix) = −Z2f8+ f7
Z2
2+ 1
Z7
1,
where f8= φ(P6
8,ix) = x8+8x6+19x4+16x2+4 and f7= iφ(P6
7,ix) = x7+7x5+13x3+7x.
Lemma 8 For n ≥ 7 and odd integer 3 ≤ t ≤ n, the characteristic polynomials of P6
and Pt
nhave the following form
n
φ(P6
n,x) = A1(x)(Y1(x))n+ A2(x)(Y2(x))n
and
φ(Pt
n,x) = B1(x)(Y1(x))n+ B2(x)(Y2(x))n
where x ?= ±2.
Proof. By Lemma 7, we notice that φ(P6
xf(n − 1,x) − f(n − 2,x). Therefore, the general solution of this linear homogeneous
recurrence relation is f(n,x) = D1(x)(Y1(x))n+ D2(x)(Y2(x))n. By some elementary
calculations, we can easily obtain that Di(x) = Ai(x) for φ(P6
initial values φ(P6
obtained by the analogous method.
n,x) satisfies the recursive formula f(n,x) =
n,x), i = 1,2, from the
8,x), φ(P6
7,x). Similarly, the required expression of φ(Pt
n,x) can be
Utilizing the similar method as the proof of Lemma 8, we can obtain
Lemma 9 For positive integer t ≥ 3, we have
φ(Pt
+?C2(x)(Y2(x))t−2((Y2(x))4− x2+ 1)?− 2(x2− 1);
φ(Pt
t+2,x) =?C1(x)(Y1(x))t−2((Y1(x))4− x2+ 1)?
t+1,x) =?C1(x)(Y1(x))t−2((Y1(x))3− x)?+?C2(x)(Y2(x))t−2((Y2(x))3− x)?− 2x.
Proof. By Lemma 6, we notice that φ(Pn,x) satisfies the recursive formula f(n,x) =
xf(n − 1,x) − f(n − 2,x). Therefore, the general solution of this linear homogeneous
recurrence relation is f(n,x) = D1(x)(Y1(x))n+ D2(x)(Y2(x))n. By some elementary