2 cutsets cut vertices and the line graphs
TRANSCRIPT
Dr.Gangaboraiah, PhD
Department of Community Medicine
Kempegowda Institute of Medical Sciences
Banashankari 2nd Stage, Bangalore-70
Mobile: 98451 28875
E-mail: [email protected]
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