Page 1

Computers and Mathematics with Applications 60 (2010) 1706–1710

Contents lists available at ScienceDirect

Computers and Mathematics with Applications

journal homepage: www.elsevier.com/locate/camwa

Extremal graphs with given order and the rupture degree$

Yinkui Lia,∗, Shenggui Zhangb

aDepartment of Mathematics, Qinghai Nationalities College, Xining, Qinghai 810000, PR China

bDepartment of Applied Mathematics, Northwestern Polytechnical University, Xi’an, Shaanxi 710072, PR China

a r t i c l ei n f o

Article history:

Received 12 July 2009

Received in revised form 2 July 2010

Accepted 6 July 2010

Keywords:

The rupture degree

Extremal graphs

Nonlinear integer programming

a b s t r a c t

The rupture degree of an incomplete connected graph G is defined by r(G) = max{ω(G −

X) − |X| − τ(G − X) : X ⊂ V(G),ω(G − X) > 1}, where ω(G − X) is the number of

components of G−X andτ(G−X) is the order of a largest component of G−X. In Li and Li

[5]andLietal.(2005)[4],itwasshownthattherupturedegreecanbewellusedtomeasure

the vulnerability of networks. In this paper, the maximum and minimum networks with

prescribed order and the rupture degree are obtained. Finally, we determine the maximum

rupture degree graphs with given order and size.

Crown Copyright © 2010 Published by Elsevier Ltd. All rights reserved.

1. Introduction

The stability of a communication network, composed of processing nodes and communication links, is of prime

importance to network designers. As the network begins losing links or nodes, eventually there is a loss in its effectiveness.

Thus,communicationnetworksmustbeconstructedtobeasstableaspossible,notonlywithrespecttotheinitialdisruption,

but also with respect to the possible reconstruction of the network. Many graph theoretical parameters have been used in

measuring the vulnerability of networks, such as connectivity, toughness [1], integrity [2], tenacity [3] and the rupture

degree [4]. The rupture degree (also called the additive dual of the tenacity) is the newest and well used parameter, for it

represents a trade-off between the amount of work done to damage the network and how badly the network is damaged,

see [5,4].

Before we formally define the rupture degree of a graph, we recall some terminologies and notation from [4]. Let G be a

finite simple graph with vertex set V(G) and edge set E(G). A set X ⊂ V(G) is a cut set of G, if either G − X is disconnected

or G − X has only one vertex. For any cut set X of G, let ω(G − X) and τ(G − X) denote the number of components and the

order of a largest component in G − X, respectively.

The rupture degree of G is defined as

r(G) = max{ω(G − X) − |X| − τ(G − X) : X ⊂ V(G),ω(G − X) > 1}.

In particular, the rupture degree of a complete graph Knis defined to be 1 − n.

A vertex cut set X of graph G is called an r-set of G if r(G) = ω(G−X)−|X|−τ(G−X). A network is called the maximum

(minimum) network if it has maximum (minimum) number of edges with prescribed order and some properties.

In this paper, we mainly consider the construction of the extreme graphs with given number of vertices and rupture

degree. In Section 2, the maximum networks with prescribed order and the rupture degree are obtained. In Section 3, the

minimum networks will be discussed. In Section 4, the maximum rupture degree graphs are characterized with prescribed

order and size.

$Supported by NSFC (No. 10861009) and SNAC (No. 09QH02).

∗Corresponding author.

E-mail address: lyk463@yahoo.com.cn (Y. Li).

0898-1221/$ – see front matter Crown Copyright © 2010 Published by Elsevier Ltd. All rights reserved.

doi:10.1016/j.camwa.2010.07.001

Page 2

Y. Li, S. Zhang / Computers and Mathematics with Applications 60 (2010) 1706–1710

1707

Throughout this paper, all graphs are finite, undirected and simple. We use Bondy and Murty [6] for terminologies and

notation not defined here. Let Kn,Enand G[X] denote complete graph, the null graph with n vertices and the induced

subgraph of G on a nonempty subset X ⊂ V(G), respectively. A comet Ct,sis defined as the graph obtained by identifying

one end of the path Pt(t ≥ 2) with the center of the star K1,sand the center is also called the center of the comet.

The Join of two disjoint graphs G and H, denoted by G + H, is the graph with vertex set V(G) ∪ V(H) and edge set

E(G) ∪ E(H) ∪ {uv|u ∈ V(G),v ∈ V(H)}.

2. Maximum graphs with given number of vertices and rupture degree

In this section, we characterize the graphs with maximum number of edges and given number of vertices and rupture

degree. By the definition of rupture degree, if a graph G has n vertices and rupture degree 1−n, then G = Kn. In the following

we only consider graphs with n vertices and rupture degree r ?= 1− n. First we give two useful lemmas which were proved

in [4].

Lemma 2.1. Let G be an incomplete connected graph with order n. Then 3 − n ≤ r(G) ≤ n − 3.

Lemma 2.2. There exists no graph G with n vertices such that r(G) = n − 4. For any integer r with 3 − n ≤ r ≤ n − 5 or

r = n − 3, there exist graphs with n vertices and rupture degree r.

Theorem 2.1. Let n be an integer greater than 2, and r be an integer with 3−n ≤ r ≤ n−5 or r = n−3. If G is a graph which

has the maximum number of edges among all graphs of order n and rupture degree r, then

Proof. LetX beanr-set ofG,i.e.,ω(G−X)−|X|−τ(G−X) = r(G).SupposethatthecomponentsofG−X areG1,G2,...,Gk,

andlet|X| = x,|V(Gi)| = nifori = 1,2,...,k.ThenG−X = G1∪G2∪···∪Gkandr(G) = k−x−τ(G−X),?k

Since G has the maximum number of edges among all graphs with n vertices and rupture degree r, we have

(1) G[X] is a complete subgraph of G;

(2) All Gi(i = 1,2,...,k) are complete subgraphs of G;

(3) All vertices in X are adjacent to all vertices in Gi(i = 1,2,...,k).

Then G = Kx+

|E(G)| = f(x,n1,n2,...,nk)

=

2

i=1

?x

=

22

1≤i<j≤k

G =

K n−r−1

2

+

n + r + 1

2

?n + r − 4

K1,

if n + r is odd;

K n−r−4

2

+

2

K1∪ 2K2

?

,

if n + r is even.

i=1ni= n−x

and 1 ≤ ni≤ n − x − (k − 1) = n + 1 − x − k for i = 1,2,...,k.

??k

?

?

?

i=1Kni

?

. So we have

?x

+

k ?

1

2

?ni

?

i=1

2(n − x)2+

2

?

+ x

?2

k ?

i=1

ni

=

2

+

k ?

ni

+

?

?

x −

1

2

?

k ?

(n − x) −

i=1

ni−

?

?

1≤i<j≤k

ninj

?x

+

1

x −

1

?

ninj.

Case 1. r + n is odd.

First, let us determine the maximum value of f(x,n1,n2,...,nk) for a given x by solving the following nonlinear integer

programming

?

nonlinear integer programming and suppose that n0

ming(n1,n2,...,nk) =

1≤i<j≤k

ninj

s.t.

1 ≤ ni≤ n − x − (k − 1)

k ?

ni∈ Z.

We set N = (n1,n2,...,nk) for convenience. Let N0= (n0

for i = 1,2,...,k

i=1

ni= n − x

1,n0

2,...,n0

k) be an arbitrary feasible solution of the above

jis the first number larger than 1 among n0

1,n0

2,...,n0

k. Then construct

Page 3

1708

Y. Li, S. Zhang / Computers and Mathematics with Applications 60 (2010) 1706–1710

a new feasible solution N1= (1,1,...,1

????

j

,n0

j+1+ n0

j− 1,n0

j+2,...,n0

k). It is clear that g(N1) ≤ g(N0). Repeating the above

process, we can finally get a feasible solution N∗= (1,1,...,1

?

???

k−1

,n − x − k + 1). Since N0is an arbitrary feasible solution,

N∗is an optimal solution of the above nonlinear integer programming. Then the maximum value of f(x,n1,n2,...,nk) for

the given x is

?n − x − k + 1

Next let us determine the maximum value of p(x). By the above analysis, τ(G − X) = nk = n − x − k + 1. Thus

r = k − x − τ(G − X) = 2k − n − 1. This implies that k =

Furthermore, since p(x) is an increasing function when 1 ≤ x ≤

x =

p(x) = f(x,1,1,...,1,n − x − k + 1) =

2

?

+

?x

2

?

+ x(n − x).

n+r+1

2

. Then by τ(G − X) ≥ 1 we get x ≤ n − k =

n−r−1

2

n−r−1

2

.

, then, p(x) will meet the maximum value when

n−r−1

2

. Therefore, the maximum graph G = K n−r−1

2

+n+r+1

2

K1.

Case 2. r + n is even.

Clearly,ifτ(G−X) = 1,thenn+r = n+k−x−τ(G−X) = 2(n−x)−1isodd,acontradiction.Ifτ(G−X) = n−x−k+1,

then n+r = 2k−1 is odd, again a contradiction. So in this case 2 ≤ τ(G−X) ≤ n−x−k. In the following let us determine

the maximum value of f(x,n1,n2,...,nk) for a given x by solving the following nonlinear integer programming

?

(1,1,...,1

?

q(x) = f(x,1,1,...,2,n − x − k) =

2

ming(n1,n2,...,nk) =

1≤i<j≤k

for i = 1,2,...,k

ninj

s.t.

1 ≤ ni≤ n − x − k

k ?

2 ≤ max

ni∈ Z.

i=1

ni= n − x

1≤i≤kni≤ n − x − k

Similar to case 1, we can obtain that the optimal solution of the above nonlinear integer programming is N∗∗

,2,n − x − k). Thus the maximum value of f(x,n1,n2,...,nk) for the given x is

=

???

k−2

?n − x − k

?

+

?x

2

?

+ x(n − x) + 1.

Clearly, τ(G − X) = n − x − k, and then r = k − x − τ(G − X) = 2k − n, this means k =

τ(G − X) ≥ 2 get x ≤

the maximum value when x =

The proof is completed.

?

n+r

n−r−4

2

2. At the same time, by

, then, q(x) will meet

K1∪ 2K2

n−r−4

2

. Furthermore, since q(x) is an increasing function when 1 ≤ x ≤

n−r−4

2

. Therefore, the maximum graph G = K n−r−4

2

+?n+r−4

2

?

.

Corollary 2.1. Let n be an integer greater than 2, and r an integer with 3 − n ≤ r ≤ n − 5 or r = n − 3. If G is a graph which

has the maximum number of edges among all graphs of order n and rupture degree r, then

3. Related discussion for minimum graphs with given number of vertices and rupture degree

|E(G)| =

1

8(3n2− r2− 2nr − 4n + 1),

1

8(3n2− r2− 2nr − 6r − 10n + 8),

if n + r is odd;

if n + r is even.

In this section, we consider the minimum network with given order and rupture degree. Clearly, if |V(G)| = n and

r(G) = 1 − n, then G = Knand thus min|E(G)| =

min|E(G)| = n − 1. As for the general cases, it seems complicated. In the following we discuss some special cases. We

denote T(n,∆) as the family of trees with order n(≥3) and maximum degree ∆(≥2).

Lemma 3.1. The rupture degree of the comet is r(Ct,s) =

n(n−1)

2

. If |V(G)| = n and r(G) = n − 3, then G = K1,n−1and

?

s − 1,

s − 2,

if t is even,

if t is odd.

Lemma 3.2. Let T be a tree and X be an r-set of T, then τ(T − X) ≤ 2.

Page 4

Y. Li, S. Zhang / Computers and Mathematics with Applications 60 (2010) 1706–1710

1709

Proof. Assume that τ(T − X) > 2. Without loss of generality, let C1,C2,...,Ckbe all the largest components of T − X. It is

clear that every Cjis a subtree of T and|V(C1)| = |V(C2)| = ··· = |V(Ck)| > 2. Then select one vertex ujin every V(Cj) such

that dCj(uj) ≥ 2, and let X1= X ∪{u1,u2,...,uk}. Clearly,ω(T −X)−|X| ≤ ω(T −X1)−|X1| andτ(T −X1) ≤ τ(T −X)−1.

Thus ω(T − X) − |X| − τ(T − X) < ω(T − X1) − |X1| − τ(T − X1). This is a contradiction to X is an r-set of T. Therefore

τ(T − X) ≤ 2.

Lemma 3.3. Let T be a tree with maximum degree ∆,X be the r-set of T, then ω(T − X) − |X| ≥ ∆ − 1.

Proof. Let X be an r-set of T. If |X| = 1, assume that X = {u}. By the definition of r-set we know that u must be the

maximum degree vertex of T. Thus ω(T − X) − |X| = ∆ − 1. If |X| > 1, by the definition of rupture degree there exists a

maximum degree vertex v ∈ X. Combining this with the fact T is a tree we have that ω(T − X) − |X| is increasing with |X|.

Thus ω(T − X) − |X| ≥ ω(T − {v}) − |{v}| = ∆ − 1. The proof is completed.

Theorem 3.1.

?∆ − 2,

Proof. Clearly, if n = ∆+ 1,T is unique and T = K1,n−1, thus r(T) = n− 3 = ∆− 2. In the following we consider the case

n > ∆ + 1.

If n = ∆ + 2,T is also unique and T = K+

∆ − 3. If n ≥ ∆ + 3. By Lemmas 3.2 and 3.3 we easily have r(T) ≥ ∆ − 3.

In fact, the results can be achieved by trees T1and T2; while n − ∆ is even, T1= Cn−∆+1,∆−1and while n − ∆ is odd, T2

obtained by identifying one end of path P3with the center of the comet Cn−∆,∆−2.

By Theorem 3.1, we immediately get the following corollary.

?

?

min

T∈T[n,∆]r(T) =

if n = ∆ + 1

if n > ∆ + 1.

∆ − 3,

1,n−1obtained by subdividing one edge of the star K1,n−1, thus r(T) = n − 5 =

?

Corollary 3.1. If |V(G)| = n and −1 ≤ r(G) ≤ n − 5, then there exists a tree with order n and rupture degree r and thus

min|E(G)| = n − 1.

As for the case r(G) ≤ −2, it seems complicated to describe the minimum graph, maybe this problem needs some other

parameters to co-consider. But the following observation is clear.

The case r = −2 (n ≥ 5): If n is odd, the odd cycle Cnis the minimum graph. If n is an even, the graph G obtained by

connecting one vertex of an odd cycle with one vertex of another odd cycle maybe the minimum graph. Thus we claim that

min|E(G)| = 2?n

the minimum graph. Thus we claim that min|E(G)| = n + 3 when r = −3.

2

?+ 1 when r = −2.

The case r = −3 (n ≥ 6): The graph G obtained by adding edges u1u3,u3u5,u2u4in cycle C = u1u2u3···unu1may be

4. Graphs with maximum rupture degree and given order and size

In this section, we discuss the maximum rupture degree of graphs with prescribed order and size. Let G be a connected

graph with order n and size m, then n − 1 ≤ m ≤

induced subgraph G[S] is complete.

Theorem 4.1. Let G be a connected graph among all graphs with order n and size m. If

?

Proof. Let G be a connected graph on n vertices and m edges with the maximum rupture degree. Denote by Π the family of

all r-sets X in G. Let X∗be an element of Π with maximum order and G1,G2,...,Gpbe all components of G − X∗. Next we

prove that τ(G − X∗) = 1.

Suppose that there are q components G1,G2,...,Gqsuch that|Gi| ≥ 2 in G−X∗. Clearly, 0 ≤ q ≤ p. In the following we

distinguish two cases to prove that q = 0.

Case 1. Suppose that q ≥ 2. Since G is connected, we can select a vertex ui ∈ V(Gi) for i = 1,2,...,q such that ui

is adjacent to some vertex of X∗. Thus construct a new graph G?from graph G as G?= G − ∪q−1

E?

Let X?= X∗∪ {uq}, then X?is a cut set of graph G?and ω(G?− X?) ≥ ω(G − X∗) + q − 1 ≥ ω(G − X∗) + 1,|X?| =

|X∗| + 1,τ(G?− X?) ≤ τ(G − X∗) − 1. Thus

r(G?) ≥ ω(G?− X?) − |X?| − τ(G?− X?)

≥ ω(G − X∗) + 1 − |X∗| − 1 − τ(G − X∗) + 1

= ω(G − X∗) − |X∗| − τ(G − X∗) + 1

= r(G) + 1 > r(G).

1

2n(n − 1). A clique of a simple graph G is a subset S of V such that the

?

k

2

?

+ (n − k)(k − 1) < m ≤

k

2

?

+ (n − k)k, then the maximum rupture degree of G is r(G) = n − 2k − 1.

i=1E?

i+ ∪q−1

i=1E??

i, where

i= {uivj

i|vj

i∈ N(ui) ∩ V(Gi)},E??

i= {uqvj

i|vj

i∈ N(ui) ∩ V(Gi)}. Clearly, G?is a connected graph on n vertices and m edges.

Page 5

1710

Y. Li, S. Zhang / Computers and Mathematics with Applications 60 (2010) 1706–1710

But this contradicts the fact that G is a graph with the maximum rupture degree.

Case 2. Suppose that q = 1. Without loss of generality, assume that |V(G1)| ≥ 2. Now distinguish two cases to complete

the proof.

Subcase 2.1. If V(G1) is a clique of G, for any vertex u1 ∈ V(G1), let X?= X∗∪ V(G1)/{u1}, then ω(G − X?) =

ω(G−X∗),|X?| = |X∗|+τ(G−X∗)−1,τ(G−X?) = 1. Clearly,ω(G−X?)−|X?|−τ(G−X?) = ω(G−X∗)−|X∗|−τ(G−X∗).

Hence, X?is an r-set with |X?| > |X∗|, which is a contradiction to the choice of X∗.

Subcase 2.2. If V(G1) is not a clique of G, it is clear that there exist a cut set X0of G1, we let X?= X∗∪ X0, then

ω(G − X?) ≥ ω(G − X∗) + 1,|X?| = |X∗| + |X0|,τ(G − X?) ≤ τ(G − X∗) − |X0| − 1. Thus

ω(G − X?) − |X?| − τ(G − X?) ≥ ω(G − X∗) + 1 − |X∗| − |X0| − τ(G − X∗) + |X0| + 1

= ω(G − X∗) − |X∗| − τ(G − X∗) + 2

= r(G) + 2 > r(G).

We thus have a contradiction to the definition of rupture degree of G.

By the above analysis, we have that τ(G − X∗) = 1 for the r-set X∗. Denote |X∗| = x. Then

r(G) = ω(G − X∗) − |X∗| − τ(G − X∗)

= n − 2x − 1.

In the following we prove: If |E(G)| = m satisfies

then the rupture degree r(G) = n − 2x − 1 of G will meet the maximum value while x = k.

First, it is clear that x ≥ k. In fact, if x ≤ k − 1, then

|E(G)| ≤

2

?k − 1

=

2

which contradicts the value range of the number of edges of G. Thus we consider the case x ≥ k and easily get that

r(G) ≤ n − 2k − 1. At the same time, we let G0= Kk+ (n − k)K1, since |E(G0)| =

then r(G) ≥ r(G0) = n − 2k − 1. Hence, r(G) = n − 2k − 1.

The proof is completed.

?

?

k

2

?

+ (n − k)(k − 1) < m ≤

?

k

2

?

+ (n − k)k for a positive integer k,

?x

?

+ (n − x)x

?

?

≤

2

+ (n − k + 1)(k − 1)

?k

+ (n − k)(k − 1)

?

k

2

?

+ (n − k)k and r(G0) = n − 2k − 1,

Corollary 4.1. Let n be an integer greater than 4, and m be an integer with

an integer k. Then the maximum rupture degree graph among all graphs with order n and size m is

?

k

2

?

+ (n − k)(k − 1) < m ≤

?

k

2

?

+ (n − k)k for

G = [Kk+ (n − k)K1] \ {e1,e2,...,et}

where t =

connected.

?

k

2

?

+ k(n − k) − m and e1,e2,...,etare edges whose ends are respectively in V(Kk) and V(G \ Kk) and let G be

5. Conclusions

In this paper we discuss some extremal properties related to rupture degree. The maximum and minimum networks

with prescribed order and rupture degree are obtained, and determine the maximum rupture degree graphs with given

order and size. As for the problem of determining the minimum networks and the minimum rupture degree graphs is much

more complicated than that of the maximum case. Here we only discuss the minimum networks for −1 ≤ r(G) ≤ n− 5. In

fact, the results of the minimum networks are more interesting. And thus it needs further research in this aspect.

Acknowledgement

The authors are grateful to an anonymous referee for helpful comments on an earlier version of this article.

References

[1] V. Chvátal, Tough graphs and Hamiltonian circuits, Discrete Math. 5 (1973) 215–228.

[2] C.A. Barefoot, R. Entringer, H. Swart, Vulnerability in graphs-A comparative survey, J. Combin. Math. Combin. Comput. 1 (1987) 12–22.

[3] M. Cozzen, D. Moazzami, S. Stueckle, The tenacity of a graph, in: Proc. Seventh International Conference on the Theory and Applications of Graphs,

Wiley, New York, 1995, pp. 1111–1122.

[4] Y. Li, S. Zhang, X. Li, The rupture degree of graphs, Int. J. Comput. Math. 82 (7) (2005) 793–803.

[5] F. Li, X. Li, Comuting the rupture degrees of graphs, in: Proc. ISPAN’2004, Hong Kong, IEEE Computer Society, 2004, pp. 368–373.

[6] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, Macmillan London and Elsevier, New York, 1976.