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Journal of Xinjiang University(Natural Science Edition)
LIN Hui-qiu, MENG Ji-xiang

*†*, TIAN Ying-zhi
(

*College of Mathematics and Systems Science, Xinjiang University, Urumqi, Xinjiang 830046, China*)
A subset

*S ⊂ E*(

*G*) is called a 4-restricted-edge-cut of

*G*, if

*G − S *is disconnected and every
component contains at least 4 vertices. The minimum cardinality over all 4-restricted-edge-cut of

*G *is called
the 4-restricted-edge connectivity of

*G*, denoted by

*λ*4(

*G*). In this paper, we prove that

*λ*4(

*Qn*) = 4

*n − *8.

Similarly, a subset

*F ⊂ V *(

*G*) is called a

*Rg*-vertex cut of

*G*, if

*G − F *is disconnected and each vertex

*u ∈ V *(

*G*)

*− F *has at least

*g *neighbors in

*G − F *. The minimum cardinality of all

*Rg*-vertex-cut is called
the

*Rg*-vertex connectivity of

*G*, denoted by

*κg*(

*G*). In this paper, we prove that

*κ*1(

*L*(

*Qn*)) = 3

*n − *4,

*κ*2(

*L*(

*Qn*)) = 4

*n − *8, where

*L*(

*Qn*) is the line graph of

*Qn*.

Key words
line graph; hypercube; restricted-edge-cut; restricted-edge-connectivity

*κ*1(

*L*(

*Qn*)) = 3

*n − *4

*κ*2(

*L*(

*Qn*)) = 4

*n − *8
Let

*G *= (

*V, E*) be a finite graph without loops and parallel edges. It is well know that the underlying
topology of a computer interconnection network is modeled by a graph

*G*, and the connectivity

*κ*(

*G*) orthe edge connectivity

*λ*(

*G*) of

*G *is an important measure for fault-tolerance of the network. In general,the larger

*κ*(

*G*) or

*λ*(

*G*) is, the more reliable the network is. However,

*κ*(

*G*) or

*λ*(

*G*) is the worst measurethat underestimates the resilience of the network. To overcome such shortcoming, Harary [1] introduced theconcept of conditional connectivity by putting some requirements on the components of

*G − F *.

Hypercubes are used as fundamental models for computer networks. There are many research articles on
the hypercubes (see, for example[2,3,4,5]). An n-dimensional hypercube is an undirected graph

*Qn *= (

*V,E*)with

*|V | *= 2

*n *and

*|E| *=

*n*2

*n−*1. Each vertex can be represented by an

*n*-

*bit *binary string. There is an edgebetween two vertices whenever their binary string representation differ in only one bit position. It is well know
Foundation Item The research is supported by NSFC (No.10671165).

Biography LIN Hui-qiu(1983-), Male, Master.

*† *Corresponding author E-mail: mjx@xju.edu.cn
that

*λ*(

*Qn*) =

*n*. A subset

*S ⊆ E*(

*G*) is called a

*m*-restricted-edge-cut of

*G *if

*G−S *is disconnected and everycomponent contains at least

*m *vertices.

*λm *is the minimum cardinality over all

*m*-restricted-edge-cuts. It iswell known that

*λ*2(

*Qn*) = 2

*n−*2 and

*λ*3(

*Qn*) = 3

*n−*4 (see[2,3]). In this paper, we show that

*λ*4(

*Qn*) = 4

*n−*8,for

*n ≥ *4. With every graph

*G *which is nonempty, there is associated a graph

*L*(

*G*), called the line-graph of

*G*, whose point set can be put in one to one correspondence with the edge set of

*G *in such a way that twopoints of

*L*(

*G*) are adjacent if and only if the corresponding edges of

*G *are adjacent. A subset

*F ⊂ V *(

*G*) iscalled a

*Rg*-vertex cut of

*G*, if

*G − F *is disconnected and each vertex

*u ∈ V *(

*G*)

*− F *has at least

*g *neighborsin

*G − F *. The cardinality of a minimum

*Rg*-vertex-cut of

*G*, denoted by

*κg*(

*G*). There are many researcharticles on the connectivity of line graph (see, for example[6,7]). We follow Bondy [8] for terminologies notgiven here.

4-restricted-edge connectivity of hypercubes
We express

*Qn *as

*Qn *=

*L ⊕ R*, where

*L *and

*R *are (

*n − *1)-subcubes of

*Qn *induced by the vertices with
the leftmost coordinate 0 and 1, respectively, that is, all the vertices in

*L *are the form 0

*Xn−*1 and all thevertices in

*R *are of the form 1

*Xn−*1. Edges are also labelled from 0 to

*n − *1 such that any edge labelled

*i*connects two vertices whose labels differ in the

*i*th bit. Since the first bit position is the 0th position, soall the edges labelled 0 are cross edges. The neighbors of a vertex

*u *are called bordering vertices of

*u*. Thebordering vertex of

*u *across dimension

*i *is denote by

*ui*. The bordering vertex of

*ui *across dimension

*j *isdenoted by

*uij*. Similarly, the neighboring edges of a vertex of a node

*u *are called bordering edges

*u*, thebordering edges of

*u *across dimension

*i *is denoted by

*ei*(

*u*), since

*e*0(

*u*) is a cross edge, we also denote

*eo*(

*u*)by

*ec*(

*u*).

*ω*(

*A*) denote all the edges with one endpoint in A.

Theorem 2

*λ*4(

*Qn*) = 4

*n−*8, for

*n ≥ *4.

Proof Let

*S *=

*ω*(

*C*4). Then

*|S| *=

*|ω*(

*C*4)

*| *= 4

*n − *8. We now verify that

*S *is a 4-restricted-edge-cut of

*Qn*. It is easy to see that

*Qn − C*4 is connected(For any

*C*4

*⊆ Qn*, we can find an expression as

*Qn *=

*L ⊕ R*,such that

*C*4

*⊆ V *(

*L*) or

*C*4

*⊆ V *(

*R*). Since vertices in

*V *(

*L*) and

*V *(

*R*) are one-to-one corresponding, wehave

*Qn − C*4 is connected ). And for any

*n ≥ *4,

*|V *(

*Qn − C*4)

*| ≥ *4, so

*S *is a 4-restricted-edge-cut. Thus,

*λ*4

*≤ |S| *=

*|ω*(

*C*4)

*| *= 4

*n−*8. To show that

*λ*4(

*Qn*) = 4

*n−*8, it is sufficient to show that

*λ*4(

*Qn*)

*≥ *4

*n−*8, whichwill discussed by contradiction in the following paragraphs.

Suppose

*A ⊆ E*(

*G*) ,

*|A| ≤ *4

*n − *9 and every component of

*Qn − A *has at least four vertices. In the
following, we will prove that

*Qn − A *is connected. Let

*AL *=

*A ∩ E*(

*L*),

*AR *=

*A ∩ E*(

*R*). Since

*L ∩ R *= ∅,

*|AL| *+

*|AR| ≤ *4

*n − *9, then either

*|AL| ≤ *2

*n − *5 or

*|AR| ≤ *2

*n − *5. Without loss of generality, we suppose that

*|AR| ≤ *2

*n−*5. In the following, we will prove that the vertices in

*R−AR *is connected to each other in

*Qn−A*.

Case 1 If there is no isolated vertex in

*R − AR*, since

*λ *(

*R*) =

*λ *(

*Qn−*1) = 2

*n − *4

*> *2

*n − *5

*≥ |AR|*, we
have that

*R − AR *is connected.

Case 2 If there exists an isolated vertex

*uR *in

*R−AR*, then

*λ*(

*R−uR*)

*≥ κ*(

*R−uR*)

*≥ κ*(

*R*)

*−*1 =

*n−*2,
and

*|AR|−dR*(

*uR*)

*≤ n−*4. Thus

*R−uR *is connected. In the following we will prove that

*uR *is connected to thesubgraph

*R−AR−uR *in

*Qn−A*. Since there is no isolated vertex in

*Qn−A*, the cross edge

*ec*(

*uR*) = (

*uR,uL*)

*∈ A*.

For 1

*≤ i ≤ n − *1, if there exists an

*i *such that both

*ei*(

*uL*)

*∈ A *and

*ec*(

*ui *)

*∈ A*, then

*u*
to

*G *by the path:

*uRuLui ui *, we are done. So we may suppose that for each

*i *at least one of

*e*
*ec*(

*ui *)

*∈ A*. Let

*B *=

*{e*
)

*|i *= 1

*, *2

*, · · · , n − *1

*} ∩ A*, then

*|B| ≥ n − *1. Since there is no isolated edge

*i*(

*uL*)

*, ec*(

*uiL*
in

*Qn − A*, there exists a

*j *such that

*ej*(

*uL*) = (

*uL,vL*)

*∈ AL*. If both

*ei*(

*vL*) and

*ec*(

*vi *) do not belongs to

*A*,
then we are done. So we suppose that at least one edge from each of the above

*n−*2 pairs belongs to

*A*. Let

*C *=

*{ei*(

*vL*)

*,ec*(

*vi *)

*|i *= 1

*,*2

*,··· ,n − *1

*but i *=

*j}*. Then

*|C| ≥ n − *2. Since every component of

*Q*
vertices, so

*D *=

*ω*(

*ej*(

*uL*))

*− AR *= ∅. Suppose that

*wL *is an edge set of an element of

*D*, and

*wL ∈ {uL,vL}*.

LIN Hui-qiu,

*et al.*: Restricted Connectivity of the Line Graph of Hypercube
Since there is no triangle in

*Qn*, the vertex

*wL *can be adjacent to just one of the two vertices

*uL*,

*vL*. Let

*E *=

*{ei*(

*wL*)

*,ec*(

*wi *)

*|i *= 1

*,*2

*,··· ,n − *1 and

*e*
*iwL*is not adjacet to

*uL *or

*vL}*. Since

*B*,

*C *,

*D*,

*E *are disjoint to
each other, we have

*|E∩A| ≤ |A|−dR*(

*uR*)

*−|B|−|C| ≤ n−*4. Since there are

*n−*2 pairs of edges in

*E*, so thereexists an

*i*1 such that neither

*ei *(

*v*
*R *can be connected to

*G*[(

*R−AR*)

*−uR*]
through the path

*P *=

*uRuLvLvi*1

*vi*1, thus completing our proof that the vertices in

*R − A*
In the following paragraph, we will prove that any vertex of

*L−AL *is connected to the subgraph

*R−AR*.

Suppose that

*xL *is any vertex in

*L−AL*, if

*ec*(

*xL*)

*∈ A*, then we are done. So we suppose that

*ec*(

*xL*)

*∈ A*.

If there exists an

*i ∈ {*1

*, *2

*, · · · , n − *1

*} *such that both

*ei*(

*xL*)

*∈ A *and

*ec*(

*xi*(

*L*))

*∈ A*, then we are done. So wesuppose that at least one edge from each of the above

*n−*1 pairs belongs to

*A*. Let

*B *=

*{ei*(

*xL*)

*,ec*(

*xi*(

*L*))

*|i *=1

*, *2

*, · · · , n − *1

*} ∩ A*, then

*|B | ≥ n − *1. Since there is no isolated vertex in

*Qn − A*, there exists a

*j *=

*i *suchthat

*ej*(

*xL*)

*∈ A*. Suppose that

*ej*(

*xL*) = (

*xL,yL*). For all

*i ∈ {*1

*,*2

*,··· ,n − *1

*} *and

*i *=

*j*, if both

*ei*(

*yL*)

*∈ A *and

*ec*(

*yi *)

*∈ A*, then we are done. So we suppose that at least one edge from each of the above

*n−*2 pairs belongs
to

*A*. Let

*C *=

*{ei*(

*yL*)

*,ec*(

*yi *)

*|i *= 1

*,*2

*,···n − *1

*and i *=

*j} ∩ A*, then

*|C | ≥ n − *2. Since there is no isolated
edge in

*Qn −A *and

*Qn *contains no triangle, without loss of generality, we suppose that

*ek *= (

*xL,wL*)

*∈ A*. Ifboth

*ei*(

*wL*)

*∈ A *and

*ec*(

*wi *)

*∈ A*, then we are done. So we suppose that at least one edge from each above

*n − *2 pairs belongs to

*A*. Let

*D *=

*{ei*(

*wL*)

*,ec*(

*wi *)

*|i *= 1

*,*2

*,···n − *1

*and i *=

*k}*, then

*|D | ≥ n − *2. Since every
component of

*Qn−A *has at least four vertices, we have

*ω*(

*{xL,yL,wL}*)

*−AR *= ∅ and

*zL ∈ {xL,yL,wL}*. Sincethere is no triangle in

*Qn*, then the vertex

*zL *at most adjacent to two of the vertices

*{xL,yL,wL}*.

If

*zL *is adjacent to one of

*{xL,yL,wL}*, without loss of generality, let

*el *= (

*zL,yL*). If both

*ei*(

*wL*)

*∈ A*
and

*ec*(

*wi *)

*∈ A*, then we are done. So we suppose at least one edge from each of the above

*n−*2 pairs belongs
to

*A*. Let

*E *=

*{ei*(

*zL*)

*,ec*(

*zi *)

*|i *= 1

*,*2

*,··· ,n − *1

*and i *=

*j}*, then

*|E | ≥ n − *2. Since

*e*
disjoint to each other, so

*|E ∩ A| ≤ n − *5. Since there are

*n − *2 pairs of edges in

*E *, so there exists an

*i*1such that neither

*zi *(

*z*
) belonging to

*A*. Thus we are done. If

*z*
*L *are adjacent to

*yL *and

*wL*, let

*el *= (

*zL,xL*) and

*em *= (

*zL,wL*). If both

*ei*(

*zL*)

*∈ A *and

*ec*(

*zi *)

*∈ A*, then we are done. So at least one edge
from each of the above

*n − *3 pairs belongs to

*A*. Let

*E *=

*{ei*(

*zL*)

*,ec*(

*zi *)

*|i *= 1

*,*2

*,.,n − *1

*and i *=

*l and m}*.

Since

*ec*(

*xL*)

*,B ,C ,D *are disjoint to each other, so

*|E ∩ A| ≤ n − *5. Since there are

*n − *3 pairs of edges in

*E *, so there exists an

*i*1 such that neither

*ei *(

*z*
) belonging to

*A*. Thus we are done.

In the following, we shall determine

*κ*1(

*L*(

*Qn*)) and

*κ*2(

*L*(

*Qn*)).

Lemma 1[9] Let

*X *=

*L*(

*Y *) be a connected graph. Then

*X −F *contains exactly two components, if

*F*
is the minimum restricted-vertex-cut of

*G*.

For a subset

*E ⊆ E*(

*G*), we use

*G*[

*E *] to denote the edge-induced subgraph of G by

*E *. Let

*L*1 be a
subgraph of

*L*(

*G*) and

*E*1 =

*V *(

*L*1). Define

*G*1 =

*G*[

*E*1].

Lemma 2[9] If

*L*1 is a connected subgraph of L with at least two vertices, then the subgraph

*G*1

*⊆ G*
is connected and

*|V *(

*G*1)

*| ≥ *3.

Lemma 3 For each

*u, v ∈ V *(

*L*(

*Qn*)), if

*u,v *is not adjacent, then

*|N*(

*u*)

*∩N*(

*v*)

*| ≤ *2.

Proof By contradiction, if

*|N *(

*u*)

*∩ N *(

*v*)

*| ≥ *3, then each pair of disjoint edges in

*Qn *has at least three
crossing edges, which will induce a triangle. It is impossible, so

*|N *(

*u*)
For any

*e ∈ E*(

*LQn*),

*|N*(

*e*)

*| *= 3

*n−*4. Next we prove that

*κ*1(

*L*(

*Qn*)) = 3

*n−*4.

Theorem 2

*κ*1(

*L*(

*Qn*)) = 3

*n−*4, for

*n ≥ *4.

Proof First, we verify that

*κ*1(

*L*(

*Qn*))

*≤ *3

*n−*4. Let

*F *be the set of all neighbors of some edge

*e*. Denote

*F *=

*N *(

*e*), then

*L*(

*Qn*)

*−F *is disconnected, and

*|F | *= 3

*n−*4. We now verify that

*F *is a

*R*1-restricted-vertex cut.

Set

*e *=

*uv*, and for each vertex

*z ∈ L*(

*Qn*)

*−F −e*. Since

*|N*(

*z*)

*N*(

*u*)

*| ≤ *2 and

*|N*(

*z*)

*N*(

*v*)

*| ≤ *2 by lemma 1.

Then

*dL*(

*Qn*)

*−F−e*(

*z*)

*≥ *2

*n−*2

*−*4 = 2

*n−*6

*> *1. Then

*F *is a restricted-vertex-cut. So

*κ*1(

*L*(

*Qn*))

*≤ |F | *= 3

*n−*4.

We now show that

*κ*1(

*L*(

*Qn*))

*≥ *3

*n − *4. By contradiction, suppose that

*F *is the minimum restricted-
vertex-cut, with

*F ⊆ R*1,

*|F | ≤ *3

*n−*5. Then

*L*(

*Qn*)

*−F *has exactly two components by Lemma 1. We denotethem by

*L*1 and

*L*2, and

*|V *(

*L*1)

*|,|V *(

*L*2)

*| ≥ *2, then by Lemma 2, we have

*|V *(

*G*1)

*| ≥ *3

*,|V *(

*G*2)

*| ≥ *3 and both

*G*1 and

*G*2 are connected. Thus

*S *is a 3-restricted edge cut with

*|S| *=

*|F | ≥ *3

*n − *4, it is a contradiction.

Therefore we get that

*κ*1(

*L*(

*Qn*)) = 3

*n−*4.

Lemma 4 For any vertex

*u, v ∈ V *(

*L*(

*Qn*)), if u is adjacent to v, then

*|N*(

*u*)

*N*(

*v*)

*| *=

*n−*2.

Proof Let

*e *=

*uv *= (

*xy, yz*), then

*u *=

*xy *has neighbors of (

*xx*1

*,xx*2

*,··· ,xxn−*1

*,yy*1

*,yy*2

*,··· ,yyn−*1),

*v *=

*yz *has neighbors of (

*yy*1

*,yy*2

*,··· ,yyn−*1

*,zz*1

*,zz*2

*,··· ,zzn−*1). Therefore

*|N*(

*u*)

*N*(

*v*)

*| *=

*n−*2.

Clearly, if

*uv ∈ E*(

*L*(

*Qn*)), then

*|N*(

*u*)

*N*(

*v*)

*| *= 0. By Lemma 4, we know that for any

*C*4

*⊆ V *(

*LQn*),

*|N *(

*C*4)

*| *= 4

*n−*8. Next we proof that

*κ*2(

*L*(

*Qn*)) = 4

*n−*8.

Theorem 3

*κ*2(

*L*(

*Qn*)) = 4

*n−*8, for

*n ≥ *4.

Proof First, we verify that

*κ*2(

*L*(

*Qn*))

*≤ *4

*n − *8. Let

*F *be the set of all neighbors of

*C*4 in

*L*(

*Qn*).

Denoted by

*F *=

*N *(

*C*4). Then

*L*(

*Qn*)

*−F *is disconnected, and

*|F | *=

*|N*(

*C*4)

*| *= 4

*n−*8. We now verify that

*F *isa

*R*2-restricted-vertex-cut. By Theorem 1, we know that

*Qn−ω*(

*C*4)

*−C*4 is connected. Then

*L*(

*Qn*)

*−F −C*4 isalso connected, for any

*n ≥ *4. Now we have to verify that for any vertex

*u ∈ L*(

*Qn*)

*−F−C*4,

*dL*(

*Q*
By contradiction, if there exist a vertex

*u ∈ L*(

*Qn*)

*−F −C*4, such that

*dL*(

*Q*
vertex

*x ∈ V *(

*Qn*)

*− C*4 such that

*dV *(

*Q*
(

*x*) = 1, thus,

*d *(

*x*) =

*n − *1

*≥ *3, which is impossible. Therefore,
(

*u*)

*≥ *2, then

*F ⊆ R*2. So

*κ*2(

*L*(

*Q*
*n*))

*≤ *4

*n − *8.

On the other hand, we show that

*κ*2(

*L*(

*Qn*))

*≥ *4

*n − *8. For this purpose, by contradiction,

*L*1 is a
connected subset of

*L*(

*Qn*), and

*F *=

*N*(

*L*1) is a

*R*2-restricted-vertex-cut with

*|F | < *4

*n − *8,

*dL *(

*u*)

*≥ *2 and
(

*v*)

*≥ *2, for any

*u ∈ L*
*n*)

*−N *(

*L*1)

*−L*1
1,

*v ∈ V *(

*L*(

*Qn*))

*− V *(

*N *(

*L*1))

*− V *(

*L*1). Then

*|V *(

*L*1)

*| ≥ *3 and

*|V *(

*L*2)

*| ≥ *3.

Since

*Qn *is triangle-free, we have

*|V *(

*G*1)

*| ≥ *4 and

*|V *(

*G*2)

*| ≥ *4, then

*S *is a 4-restricted edge cut of

*Qn*. ByTheorem 1, we have

*|S| ≥ *4

*n − *8, then

*|F | *=

*|S| ≥ *4

*n − *8, a contradiction. Therefore,

*κ*2(

*L*(

*Qn*))

*≥ *4

*n−*8.

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Source: http://xdxb.xju.edu.cn/qkbj/journal/rleasepdf/B/201001/B2010121100004.pdf

Dr. Mohammad Reza Abbaspour PhD of pharmaceutics Assistant professor Department of Pharmaceutics, School of Pharmacy & Nanotechnology Research Centre Ahvaz Jundishapur University of Medical Sciences P.O. Box: 61357-33184, Ahvaz, Iran Tel / fax: +98 611 3738381 E-mail: abbaspourmr@ajums.ac.ir; abbaspourmr@yahoo.com Education: - Ph.D of pharmaceutics, School of Pharmac

GLI INVESTIMENTI ITALIANI IN ROMANIA Considerando il numero di nuove imprese a partecipazione estera (dati del Registro del Commercio), nel 2012 sono state registrate in Romania 6.384 nuove aziende, portando a 185.791 il numero totale di imprese estere dal 1991. Al 31 dicembre 2012, secondo i dati dell'Ufficio del Registro Nazionale del Commercio, erano registrate complessivamente 34.185 imp