Dr.Gangaboraiah, PhD

Department of Community Medicine

Kempegowda Institute of Medical Sciences

Banashankari 2nd Stage, Bangalore-70

Mobile: 98451 28875

E-mail: gbphd@kimsbangalore.edu.in

Cutset and Cutvertices

Cut-sets

Let G be a connected graph. A set S of

edges of G is said to be a cut-set of G

if the following conditions hold

(1) The removal of S from G leaves G

disconnected

(2) Removal of a proper subset of S

from G does not disconnect G

Removal of a cut-set a, c, d, f from the graph cuts it into two

f

There are many other cut-sets,

such as a, b, g, a, b, e, f and

d, h, f. The edge k alone is also

a cut-set. The set of edges

a, c, h, d on the other hand is not

a cut set, because one of its

proper subsets a, c, h is a cut-set

5

Thus, a cut-set of a connected graph

G is a minimal set of edges whose

removal from G will separate G into

exactly two parts. It should be noted

that when an edge is removed

(deleted) from a graph, the end

vertices of the edge will continue to

remain in the graph.

Find the cut-set in the following graph

v1

v5

e5

Let S=e1, e4, e7 be the set of edges. If we

remove these edges then the graph gets

disconnected as shown above. Other cut-sets

of G are e2, e3,, e1, e3, e6, e3, e4, e6, e7.

Find the cut-set in the following graph

Note that

(1) In a complete graph G, the set of all

edges incident on a vertex is a cut-set.

(2) The removal of any branch from a tree

breaks the tree into two parts. Hence,

every edge (branch) of a tree is a cut-

set.

Bridge

A cut-set consisting of a single edge is called a

bridge (or a cut-edge or an isthmus)

Connectivity

Let G be a connected graph. Every cut-

set of G contains certain number of

edges. Take a cut-set that contains the

fewest number of edges. Such a cut-set

is called a smallest cut-set of G. The

number of edges in a smallest cut-set is

called the edge-connectivity or line-

connectivity of G and denoted by λ(G).

This is equivalent to saying that the

edge connectivity of G is the minimum

number of edges that we need to

delete in order to disconnect G. The

edge connectivity of every (connected)

graph is at lest one.

Since, every edge is a cut-set (bridge)

in a tree the edge connectivity of a tree

is one.

Is there a single cut-set in the following graph?

Vertex Connectivity

Let G be a connected graph. The vertex

connectivity (or point-connectivity) is

defined as the minimum number of

vertices whose removal leaves the graph

disconnected.

Note that the removal of a vertex implies

removal of all the edges incident on that

vertex.

A graph G is said to have vertex

connectivity only when G is connected,

not complete and has three or more

vertices. The vertex connectivity of a

graph G is denoted by к(G). Obviously,

к(G)≥1 for every connected graph G.

A connected graph G is said to be k-

connected if к(G)=k.

In a tree, removal of a single vertex

breaks the tree. Therefore, the vertex

connectivity of a tree is one. That is

every tree is 1-connected. Find the vertex connectivity of this graph

Separable graphsThe graph for which к(G)=1 form an

important class of graphs called

separable graphs

A separable graph is defined as 1-

connected graph, i.e., a graph whose

connectivity is one.

Cut-vertex

In a separable graph, there exists a (at

least one) vertex whose removal leaves

the graph disconnected. Such a graph is

called a cut-vertex or articulation point.

It follows immediately that a vertex v of a

connected graph G is a cut vertex of G if

and only if G-v is a disconnected graph.

Theorem

The vertex-connectivity of a graph can

never exceed it sedge-connectivity.

Proof:

Let λ=λ(G) be the edge-connectivity of

a graph G. Then, there exists a cut-set

S in G with λ edges. Let V1 and V2 be

the parts into which G is broken by S.

Then every edge in S has one end vertex

in V1 and other end in V2. Therefore, the

graph G becomes disconnected if the

end vertices of the edges of S that

belongs to V1 (or V2) are removed. The

number of such vertices is λ and by the

definition of vertex-connectivity, this

number cannot be less than к(G), i.e.,

λ(G)≥ к(G).

Using the result of the theorem “ a vertex

v in a connected graph G is a cut-vertex

if and only if there exists two vertices x

and y in G such that every path between

x and y passes through v” and the result

of the previous theorem, we get what is

called Whitney inequality given by

к(G) ≤ λ(G) ≤ δ(G)

where δ(G) is the degree of G

Example 1

Find the degree, edge-connectivity, and vertex

connectivity of this graph

By examining all the vertices, we note

that at least three edges are incident on

every vertex. Therefore, the degree of

the graph is 3, i.e., δ(G)=3.

Further, note that the removal of any one

edge will not disconnect the graph. But

the removal of the edges BP and CP will

disconnect the graph

These two edges constitute the cut-set.

Hence, the edge-connectivity of the graph

is 2, i.e., λ(G)=2.

Lastly, observe that the removal of the

vertex P from the graph will disconnect the

graph. Therefore the vertex-connectivity of

the graph is 1, i.e., к(G)=1. Thus, for the

given graph we have δ(G)=3, λ(G)=2 , and

к(G)=1 and hence, к(G) ≤ λ(G) ≤ δ(G).

Example 2

Show that a graph with n vertices and

vertex-connectivity k must have at least

kn/2 edges

Solution:

Let m be the number of edges in a graph

G with n vertices and with vertex-

connectivity k. For this graph δ(G) ≤ 2m/n

and the Whitney inequality yields k ≤ δ(G)

≤2m/n. Hence, m≥ kn/2.

Example 3

Prove that a connected (n, m) graph,

with m ≤ n-1, n>2, is separable.

Solution:

Let G be a (m, n) graph. Then δ(G) ≤

2m/n and Whitney inequality yields

k ≤ δ(G) ≤ 2m/n

≤ 2(n-1)/n, if m ≤ n-1

= 2(1-1/n) < 2 if n>2

If G is connected, then к(G)≥1. Thus,

under the condition m ≤ n-1 and n>2

we should have к(G)=1, hence, G is

separable.

Cut-point (Cut-vertex)

A cutpoint of a graph is one whose

removal increases the number of

components and a bridge is such a

line (edge). If v is a cut-point of a

connected graph G, then G-v is

disconnected. A non-separable graph

is connected, nontrivial, and no

cutpoints.

A block of a graph is a maximal non-

separable subgraph. If G is non-separable,

then G itself is called a block

In the figure

v is a cut-point while w is not.

x is a bridge while y is not.

B1 ,B2, B3, and B4 are four blocks.Each edge of a graph lies in

exactly one of its blocks

Each vertex of a graph lies in

exactly one of its blocks which is

not isolated or a cutpoint.

The edges of any cycle of G also

lies entirely in a single block.

Hence, the blocks of a graph partition its

edges and its cycles regarded as set of

edges.

Theorem:

Let v be a vertex of a connected graph G.

The following statements are equivalent.

(1) V is a cutpoint of G

(2) There exists a vertices of u and w

distinct from v such that v is a on

every u-w path

(3) There exists a partition of the set of

vertices V-v into subsets U and W

such that for any vertices u ε U and

w ε W, the vertex v is on every u-w

path.

Proof

(1) Implies (3):

Since v is a cutpoint of G, G-v is

disconnected and has at least two

components. Form a partition of V-v by

letting U consists of the vertices of one of

these components and W the vertices of the

others. Then any to two points u ε U and w

ε W lie in different components of G-v.

Therefore every u-v path in G contains v.

(3) Implies (2):

This is immediate consequence of (2).

(2) Implies (1):

If v is on every path in G joining u and

w, then there cannot be a path joining

these points in G-v. Thus G-v is

disconnected, so v is a cutpoint of G

34

Theorem

Let x be an edge of a connected graph G. The

following statements are equivalent

(1) x is a bridge of G

(2) x is not on any cycle of G

(3) There exists vertices u and v of G such that

the edge x is on every path joining u and v

(3) There exists partition of V into subsets U

and W such that for any vertices u ε U and

w ε W, the edge x is on every path joining u

and w.

Theorem

Let G be a connected graph with at least three

points. The following statements are

equivalent

(1) G is a block

(2) Every two points of G lie on a common

cycle

(3) Every vertex and edge of G lie on a

common cycle

(4) Every two edges of G lie on a common

cycle

(5) Given two vertices and one edge of

G, there is a path joining the

vertices which contains the edge.

(6) For every three distinct vertices of

G, there is a path joining any two of

them which contains the third

(7) For every three distinct vertices of

G, there is a path joining any two of

them which does not contains the

third

Proof

(1) Implies (2):

Let u and v be distinct vertices of G,

and let U be the set of vertices

different from u which lie on a cycle

containing u. Since G has at least

three vertices and no cutpoints, it has

no bridges; therefore every point to u

is in U, so U is not empty.

38

Fig. a Fig. b

Suppose v is not in U. Let w be a point

for which the distance d(w, v) is

minimum. Let P0 be a shortest w-v path,

and let P1 and P2 be the two u-v paths of

a cycle containing u and v

vu

(see fig (a)). Since w is not a cutpoint,

there is a u-v path P′ not containing w

(see fig. (b)). Let w′ be the vertex nearest

u in P′ which is also in P0, and let u′ be

the last vertex of the u-w′ subpath of p′ in

either P1 or P2. Without loss of

generality we assume u′ is in P1. Let Q1

be the u-w` path consisting of the u-u`

subpath of P1 and the

u`-w` subpath of P`. Let Q2 be the u-w`

path consisting of P2 followed by the

w-w` subpath of P0. then Q1 and Q2 are

disjoint u-w` paths. Together they

form a cycle, so w` is in U. since w` is

on a shortest

w-v path, d(w`, v) < d(w, v). This

contradicts our choice of w, proving

that u and v do lie on a cycle.

(2) Implies (3):

Let u be a vertex and vw an edge of G.

Let Z be a cycle containing u and v. A

cycle Z` containing u and vw can be

formed as follows. If w is on Z, then Z`

consists of vw together with the v-w

path of Z containing u. If w is not on Z,

there is a w-u path P not containing v,

since otherwise v

would be a cutpoint by Theorem 1..

Let u` be the first edge of P on Z. Then

Z` consists of vw followed by the w-u`

subpath of P and u`-v path in Z

containing u.

(3) Implies (4)

This proof is analogous to the

preceding one.

(4) Implies (5):

Any two vertices of G are incident with

one edge each, which lie on a cycle by

(4). Hence, any two vertices of G lie on

a cycle, and we have (2), so also (3).

Let u and v be distinct vertices and x

an edge of G. By statement (3), there

are cycles Z1 containing u and x and Z2

containing v and x. Thus, we

need only consider the case where v in

not a Z1 and u is not on Z2. Begin with

u and proceed along Z1 until reaching

the first point w of Z2, then take the

path on Z2 joining w and v which

contains x.

This walk constitutes a path joining u

and v that contains x.

(5) Implies (6):

Let u, v, and w be distinct vertices of

G, and let x be any edge incident with

w. By (5), there is a path joining u and

v which contains x, and hence must

contain w.

(6) Implies (7):

Let u, v, and w be dstinct vertices of G.

By statement (6), there is a

u-w path P containing v. The u-v

subpath of P does not contain w.

(7) Implies (1):

By statement (7), for any two vertices

u and v, no vertex lies on every u-v

path. Hence G must be a block.

Theorem

Every nontrivial connected graph has

at least two vertices which are not

circuits.

Proof

Let u and v be vertices at maximum

distances in G, and assume v is a

cutpoint. Then there is a point w in a

different component of G-v than u.

Hence, v is in every path joining u and

w, so d(u, w) >d(u, v), which is

impossible. There, v and similarly u

are not cutpoints of G.

Block graphs and cutpoint graphs

There are several intersection graphs

derived from a graph G which reflect

its structure. If we take the blocks of

G as the family F of sets, then the

intersection graph Ω (F) is the block

graph of G denoted by B (G). The

blocks of G correspond to the points

of B (G) and two of these vertices are

Graph G

Block Graph B(G)

Cutpoint Graph C(G)

adjacent whenever the corresponding

blocks contain a common cutpoint of

G. On the other hand to obtain a

graph whose vertices correspond to

the cutpoints of G, we can take the

sets Si to be the union of all blocks

which contain the cutpoint vi. The

resulting intersection graph Ω (F) is

called the cutgraph, (G).

Thus, two points of C (G) are adjacent

if the cutpoints of G to which they

correspond lie on a common block.

Note that C (G) is defined only for

graphs G which have at least one

cutpoint.

Line graphs

Line graphs

Given a graph G, its line graph L(G) is

a graph such that each vertex of L(G)

represents an edge of G; and two

vertices of L(G) are adjacent if and

only if their corresponding edges

share a common endpoint ("are

adjacent") in G.

That is, it is the intersection graph of

the edges of G, representing each

edge by the set of its two endpoints.

In graph theory, the line graph L(G) of

an undirected graph G is another

graph L(G) that represents the

adjacencies between edges of G.

The line graph is also sometimes

called the edge graph, the adjoint

graph, the interchange graph, or the

derived graph of G.

Example

The following figures show a graph

(left, with red vertices) and its line

graph (right, with green vertices). Each

vertex of the line graph is shown

labeled with the pair of endpoints of

the corresponding edge in the original

graph. For instance, the green vertex

on the right labeled 1,3

corresponds to the edge on the left

between the red vertices 1 and 3.

Green vertex 1,3 is adjacent to three

other green vertices: 1,4 and 1,2

(corresponding to edges sharing the

endpoint 1 in the red graph) and 4,3

(corresponding to an edge sharing the

endpoint 3 in the red graph).

Graph G Line graph L(G)

Properties

Consider the set X of edges of a graph G

with at least one edge as a family of 2-

vertex subsets of V(G). The line graph of G,

denoted by L(G), is the intersection of the

graph Ω(G). Thus the vertices of L(G) are

the edges of G, with two vertices of L(G)

adjacent whenever the corresponding

edges of G are.

If x=uv is an edge of G, then the degree of x

in L(G) is clearly deg u + deg v -2. Observe

these two graphs.

Note that G2 = L(G1), so that

L(G2) = L(L(G1)).

We write L1(G) = L(G), L2(G)=L(L(G))

and in general, the iterated line graph

is Ln(G) = L(L

n-1(G))

Every cut point of L(G) is a bridge of

G which is not an end line, and

conversely.

The line graph of a connected graph is

connected. If G is connected, it

contains a path connecting any two of

its edges, which translates into a path

in L(G) containing any two of the

vertices of L(G). However, a graph G

that has some isolated vertices, and is

therefore disconnected, may

nevertheless have a connected line

graph.

The edge chromatic number of a

graph G is equal to the vertex

chromatic number of its line graph

L(G).

The line graph of an edge-transitive

graph is vertex transitive.

If a graph G has an Eϋler graph, that

is, if G is connected and has an even

number of edges at each vertex, then

the line graph of G is Hamiltonian.

Thus, the existence of Hamiltonian

cycles in line graphs may be tested

efficiently, despite the hardness of

the problem for more general

families of graphs.

Theorem

If G is a (p, q) graph whose vertices

have degrees di, then L(G) has q

vertices and qL edges where qL = - q + ½ Σ d2

i

Proof: By the definition of line graph,

L(G) has q vertices. The di edges

incident with a vertex vi

Contribute di C

2to q

Lso

qL

= Σdi C

2= ½ Σdi (di-1)

= ½ Σd2i - ½ Σdi

= ½ Σd2i – q

Theorem

A connected graph is isomorphic to its

line graph if and only if it is a cycle.

Theorem

Let G and G′ be connected graphs with

isomorphic line graphs. Then G and

G′ are isomorphic unless one is K3 and

the other is K1, 3.

Theorem

If G is Eulerian, then L(G) is both

Eulerian and Hamiltonian. If G is

Hamiltonian, then L(G) is Hamiltonian.

Proof

It is easy to supply counter examples

to the converses of these statements

(See example given)

In figure 1 L(G) is Eulerian and

Hamiltonian while G is not Eulerian.

In figure 2 L(G) is Hamiltonian while G

is not.

Theorem

A sufficient condition of L2(G) to be

Hamiltonian is that G be Hamiltonian

and a necessary condition is that L(G)

be Hamiltonian

74

Theorem

A graph G is Eulerian if and only if

L3(G) is Hamiltonian (Ref Harary page

81 for counter examples)

Theorem

If G is non trivial connected graph with

p vertices which is not a path, then

Ln(G) is Hamiltonian for all

n ≥ p – 3 (Harary p 81).

Total Graph

The vertices and edges are called

elements. Two elements of a graph are

neighbours if they are either incident

or adjacent. The total graph T(G) has

vertex set

V(G) U X(G), and two vertices of T(G)

are adjacent whenever they are

neighbours in G.

It is easy to see that T(G) always

contains both G and L(G) as induced

subgraphs.

Theorem

The total graph T(G) is isomorphic to

the square of the subdivision graph

S(G).

Corollary 1

If v is a vertex of G, then the degree of

vertex v in T(G) is 2 deg v. If x=uv is a line

of G, then the degree of vertex x in T(G) is

deg u + deg v.

Corollary 2

If G is a (p, q) graph whose points have

degrees di, then the total graph T(G) has pT= p + q vertices and qT= 2q+ ½ Σ d2

i