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Communications in Mathematical
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MATCH Commun. Math. Comput. Chem. 66 (2011) 959-970
ISSN 0340 - 6253
On Biregular Graphs whose Energy
Exceeds the Number of Vertices∗
Liqiong Xu
School of Sciences, Jimei University,
Xiamen Fujian 361023, China
e-mail: [email protected]
(Received April 16, 2010)
Abstract
The energy E(G) of a graph G with n vertices is equal to the sum of absolute values of
the eigenvalues of G . Gutman (On graphs whose energy exceeds the number of vertices, Lin.
Algebra Appl. 429 (2008) 2670–2677) recently established sufficient conditions under which the
inequality E(G) ≥ n is obeyed for quadrangle–free biregular graphs and biregular graphs with
disjoint quadrangles. A new proof and sufficient conditions are provided. Moreover, the sufficient
condition for E(G) ≥ n in the case of biregular graphs with no three of their quadrangles
having a common vertex, quadrangle–free triregular graphs, and triregular graphs with disjoint
quadrangles are discussed.
∗
The Project Supported by the Science-technology Foundation for Young Scientists of Fujian Province
(No. 2007F3070).
-960Introduction
Let G be a simple graph of order n , and let λ1 , λ2 , · · · , λn be its eigenvalues (i. e., the
eigenvalues of the adjacency matrix of G) [3]. Then the energy of G is defined as [5]
E = E(G) =
n
|λi | .
i=1
The motivation for this definition comes from chemistry, where the first results on E were
obtained as early as the 1940s [2]; for details on this matter see the reviews [6, 7, 11], the recent
papers [14–17, 19, 20, 24], and the references cited therein.
The problem of characterizing (molecular) graphs for which the condition E(G) > n is
obeyed seems to be first time considered by England and Ruedenberg [4]. They asked: “Why is
the delocalization energy negative?”. Translated into the language of graph spectral theory, their
question reads: “Why does the energy exceed the number of vertices?”, understanding that the
graph in question is “molecular”. A molecular graph means a connected graph in which there
are no vertices of degree greater than three [12]. There are large classes of graphs were shown to
satisfy the condition E ≥ n . In an earlier work [13], the validity of this inequality was confirmed
for regular graphs. Later, the conditions under which biregular graphs and triregular graphs
satisfy the inequality were also determined [8, 16, 22].
The biregular graph and the triregular graph are defined as follows.
Let a , b be positive integers, 1 ≤ a < b . A graph G is said to be (a, b)-biregular if its vertices
have degree either a or b , and if it possesses vertices of degree a and b . The number of vertices
of G of degree a and b will be denoted by na and nb , respectively.
Let a , b , and c be integers, 1 ≤ a < b < c . A graph is said to be (a, b, c)-triregular if the
degrees of its vertices assume exactly three different values: a , b and c . The number of vertices
of G of degree a , b , and c will be denoted by na , nb , and nc , respectively.
For biregular graphs Gutman [8] proved the following results.
Theorem 1 [8]. Let G be a quadrangle–free (a, b)-biregular graph. Then for 2 ≤ a < b ≤ 2a−1 ,
E(G) ≥ n holds.
-961Theorem 2 [8]. Let G be an (a, b)-biregular graph with disjoint quadrangles. Then for 2 ≤
a < b ≤ 2a − 1 + 3/a , E(G) ≥ n holds.
In this paper we present a different proof of the above results and obtain another sufficient
conditions of the results. Moreover, we discuss the sufficient conditions for E(G) ≥ n in the case
of biregular graphs with no four of their quadrangles having a common vertex, quadrangle–free
triregular graphs, and triregular graphs with disjoint quadrangles.
If λ1 , λ2 , . . . , λn are the eigenvalues of the graph G , then the k-th spectral moment of G is
defined as
Mk =
n
λki .
i=1
As well know in spectral graph theory, for a graph G with n vertices, m edges, q quadrangles,
and vertex degree d1 , d2 , . . . , dn ,
M2 (G) = 2m
and
M4 (G) = 2
n
d2i − 2m + 8q .
i=1
Rada and Tineo [23] obtained the following lower bound for graph energy:
@
M2 (G)
.
E(G) ≥ M2 (G)
M4 (G)
Gutman [8] obtained that if G is (a, b)-biregular, then the above inequality becomes
@
E(G)
2m
2m
≥
n
n
(4a + 4b − 2)m − 2abn + 8q
where m is the number of edges, q the number of quadrangles, and di the degree of the i-th
vertex, i = 1, 2, · · · , n .
On the energy of quadrangle–free biregular graphs
In this section we consider quadrangle–free biregular graphs, that is, (a, b)-biregular graphs
for which q = 0 . Then the above inequality can be written in the form
@
E(G)
d
≥d
n
(2a + 2b − 1)d − 2ab
-962where d = 2m/n is the average vertex degree of the graph G . Therefore a < d < b .
Now we let
f (x) =
x3
(2a + 2b − 1)x − 2ab
for x ∈ (a, b) and find when it is greater than unity.
Since
f (x) = 2x2
(2a + 2b − 1)x − 3ab
((2a + 2b − 1)x − 2ab)2
the non-zero root of the equation f (x) = 0 is
x0 =
3ab
.
2a + 2b − 1
Since a + 1 ≤ b , x0 ≤ b , if x0 ≤ a , that is, 2a ≥ b + 1 , then f (x) > 0 for x ∈ (a, b) , and then
the function f (x) will increase monotonically in the interval (a, b) . The minimum of f (x) for
x ∈ [a, b] is f (a) . If f (a) ≥ 1 , then E(G)/n ≥ 1 . Since
f (a) =
a3
a3
= 2
≥1
(2a + 2b − 1)a − 2ab
2a − a
thus Theorem 1 holds.
If x0 > a , i. e., b > 2a − 1 , we have f (x) < 0 for x ∈ (a, x0 ) ; and f (x) > 0 for x ∈ (x0 , b) .
Thus the minimum of f (x) for x ∈ (a, b) is f (x0 ) . If f (x0 ) ≥ 1 , then E(G)/n ≥ 1 . Let
f (x0 ) ≥ 1 . Then we have that
*
3ab
2a+2b−1
ab
which is equivalent to
(ab)2 ≥
+3
≥1
2a + 2b − 1
3
3
.
We thus proved the following:
Theorem 3 Let G be a quadrangle–free (a, b)-biregular graph. Then for b > 2a − 1 and
(ab)2 ≥
E(G) ≥ n holds.
2a + 2b − 1
3
3
-963On the energy of biregular graphs with disjoint quadrangles
In this section we consider biregular graphs that may possess quadrangles, but require that
all such quadrangles be disjoint. (i. e., no two of them have a common vertex). If all quadrangles
are disjoint, then q ≤ n/4 . Therefore,
@
d
E(G)
≥d
.
n
(2a + 2b − 1)d − 2(ab − 1)
In a same manner as in the preceding section, the above relation will hold if the function
g(x) =
x3
(2a + 2b − 1)x − 2(ab − 1)
is greater than unity for x ∈ (a, b) .
Since
g (x) = 2x2
(2a + 2b − 1)x − 3(ab − 1)
((2a + 2b − 1)x − 2(ab − 1))2
the root of the equation g (x) = 0 is
x0 =
3(ab − 1)
.
2a + 2b − 1
Since a < b , x0 ≤ b , if x0 ≤ a , that is, b ≤ 2a − 1 + 3/a , then g (x) > 0 for x ∈ (a, b) , and
then the function g(x) will increase monotonically in the interval (a, b) . The minimum of g(x)
for x ∈ (a, b) is g(a) . If g(a) ≥ 1 , then E(G)/n ≥ 1 . Since g(a) ≥ 1 if and only if a ≥ 2 , thus
Theorem 2 holds.
If x0 > a , i. e., b > 2a − 1 + 3/a , we have g (x) < 0 for x ∈ (a, x0 ) ; and g (x) > 0 for
x ∈ (x0 , b) . Thus the minimum of g(x) is g(x0 ) . If g(x0 ) ≥ 1 , then E(G)/n ≥ 1 . Let g(x0 ) ≥ 1 .
Then
*
3(ab−1)
2a+2b−1
ab − 1
from which
(ab − 1)2 ≥
We thus proved the following:
+3
≥1
2a + 2b − 1
3
3
.
-964Theorem 4 Let G be an (a, b)-biregular graph in which all quadrangles are disjoint. Then for
b > 2a − 1 + 3/a and
(ab − 1)2 ≥
2a + 2b − 1
3
3
E(G) ≥ n holds.
On the energy of biregular graphs with no three of their quadrangles having
a common vertex
In this section we consider biregular graphs that may possess joint quadrangles, but we
require that no three quadrangles have a common vertex. If no three quadrangles have a common
vertex, then q ≤ n/2 . Therefore, by substituting n/2 instead of q , the inequality yields
@
d
E(G)
≥d
.
n
(2a + 2b − 1)d − 2(ab − 2)
In a same manner as in the preceding section, the relation will hold if the function
g(x) =
x3
(2a + 2b − 1)x − 2(ab − 2)
is greater than unity for x ∈ (a, b) . Proceeding in the same way as in the previous two cases,
we arrive at:
Theorem 5 Let G be an (a, b)-biregular graph with no three of their quadrangles having a common vertex. Then for 2 < a < b < 2a − 1 + 6/a or b > 2a − 1 + 6/a and
(ab − 2)2 ≥
2a + 2b − 1
3
3
E(G) ≥ n holds.
On the energy of quadrangle–free triregular graphs
We first determine the validity of the inequality for (2, 3, 4)-triregular polyomino. For these,
q = (n − 2)/2 , m = (3n − 4)/2 and for n ≥ 8 ,
n
i=1
(di )2 = 22 + 22 + 22 + 22 + 32 + 32 + 22 + 42 + · · · + 22 + 42 ≤ 10n − 24 .
-965Calculation analogous to that used in the preceding sections shows that the (2, 3, 4)-triregular
polyomino chains satisfy E(G)/n ≥ 1 . By Theorem 1, we have that also the polyomino chains
which are (2, 3)-biregular satisfy E(G)/n ≥ 1 . Since the polyomino chains which are not (2, 3, 4)triregular are (2, 3)-biregular, we conclude that all polyomino chains satisfy E(G)/n ≥ 1 .
Consider now quadrangle–free triregular graphs in general.
For a (a, b, c)-triregular graph with n vertices and m edges we have
na + nb + nc = n
and
ana + bnb + cnc = 2m
where na , nb , and nc are the number of vertices of degree a , b , and c respectively. From the
above inequalities follows
nb =
na (a − c) + (cn − 2m)
c−b
Then
n
;
nc =
na (b − a) − (bn − 2m)
.
c−b
d2i = a2 na + b2 nb + c2 nc
i=1
which combined with the above equalities yields
n
d2i = na (b − a)(c − a) + 2m(b + c) − bcn .
i=1
Thus, for quadrangle–free triregular graphs,
@
E(G)
d
≥ d 2n (b−a)(c−a)
a
n
+
d(2b
+ 2c − 1) − 2bc
n
where d = 2m/n is the average vertex degree of the graph G . Therefore a < d < c .
Now we need to examine the function
f (x) =
2na
n (b
x3
− a)(c − a) − 2bc + (2b + 2c − 1)x
for x ∈ (a, c) and find when it is greater than unity.
Since
f (x) = 2x2
(2b + 2c − 1)x − 3bc + 3nna (b − a)(c − a)
( 2nna (b − a)(c − a) − 2bc + (2b + 2c − 1)x)2
-966the root of the equation f (x) = 0 is
x0 =
3bc −
3na
n (b
− a)(c − a)
.
2b + 2c − 1
Obviously x0 ≤ c . If x0 ≤ a , that is
na
3bc − 2ab − 2ac + a
≥
n
3(b − a)(c − a)
then f (x) > 0 for x ∈ (a, c) , and then the function f (x) will increase monotonically in the
interval (a, c) . The minimum of f (x) will then be f (a) . If f (a) ≥ 1 , then E(G)/n ≥ 1 . When
x = a , we have a = b = c , and then
f (a) =
a2
≥1.
2a − 1
Because the parameter d is strictly greater than a , we have E(G)/n > 1 . If x0 ≥ a , that is
na
3bc − 2ab − 2ac + a
≤
n
3(b − a)(c − a)
then f (x) < 0 for x ∈ (a, x0 ) and f (x) > 0 for x ∈ (x0 , c) , and then the minimum of the
function f (x) for x ∈ (a, c) is f (x0 ) . If f (x0 ) ≥ 1 , then E(G)/n ≥ 1 . For f (x0 ) ≥ 1 , we have
that
*
bc −
+2
na
(b − a)(c − a) ≥
n
2a + 2b − 1
3
3
.
We thus prove the following:
Theorem 6 Let G be a quadrangle–free (a, b, c)-triregular graph. Then for
na
3bc − 2ab − 2ac + a
≥
n
3(b − a)(c − a)
or
3bc − 2ab − 2ac + a
na
<
n
3(b − a)(c − a)
and
*
bc −
E(G) ≥ n holds.
+2
na
(b − a)(c − a) ≥
n
2a + 2b − 1
3
3
-967On the energy of triregular graphs with disjoint quadrangles or with no
three of their quadrangles having a common vertex
If G is an (a, b, c)-triregular graph with disjoint quadrangles, then
@
d
E(G)
≥d
.
n
2 na (b−a)(c−a) + d(2b + 2c − 1) − 2(bc − 1)
n
The relation will hold if the function
f (x) =
2na
n (b
x3
− a)(c − a) − 2(bc − 1) + (2b + 2c − 1)x
is greater than unity for x ∈ (a, c) .
If G is an (a, b, c)-triregular graph with no three of its quadrangles having a common vertex,
then
E(G)
≥d
n
@
d
2 na (b−a)(c−a)
+ d(2b + 2c − 1) − 2(bc − 2)
n
.
The relation will hold if the function
f (x) =
2na
n (b
x3
− a)(c − a) − 2(bc − 2) + (2b + 2c − 1)x
is greater than unity for x ∈ (a, c) .
In a same manner as in the preceding sections, we can prove:
Theorem 7 Let G be an (a, b, c)-triregular graph in which all quadrangles are disjoint. Then
for
na
3bc − 2ab − 2ac + a − 3
≥
n
3(b − a)(c − a)
or
3bc − 2ab − 2ac + a − 3
na
<
n
3(b − a)(c − a)
and
*
bc − 1 −
+2
na
(b − a)(c − a) ≥
n
2a + 2b − 1
3
3
E(G) ≥ n holds.
Theorem 8 Let G be an (a, b, c)-triregular graph with no three of its quadrangles having a
common vertex. Then for
na
3bc − 2ab − 2ac + a − 6
≥
n
3(b − a)(c − a)
-968or
3bc − 2ab − 2ac + a − 6
na
<
n
3(b − a)(c − a)
and
*
bc − 2 −
+2
na
(b − a)(c − a) ≥
n
2a + 2b − 1
3
3
E(G) ≥ n holds.
Acknowledgement. The author gratefully acknowledges the suggestions from Prof. Weigen Yan
that helped in improving this article.
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