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HandOut and Work Sheet
CS 309 GRAPH THEORY AND COMBINATORICS
Semester: Five
Academic Year: 2018-2019
Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
HandOut and Work Sheet
CS 309 GRAPH THEORY AND COMBINATORICS
Semester: Five
FACULTY NAME
DESIGNATION
DEPARTMENT
: KEERTHI A S PILLAI
: ASSISTANT PROFESSOR
: COMPUTER SCIENCE AND ENGINEERING
VERIFIED BY,
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Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor
Dept of CSE
3
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor
Dept of CSE
4
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor
Dept of CSE
5
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor
Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
MODULE I
Graph
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A linear graph (graph) G=(V,E) consists of a set of objects V={v1,v2,...} called
vertices, and another set E={e1,e2,...} called edges, such that each edge ek is
identified with an unordered pair (vi,vj) of vertices.
The vertices vi,vj associated with edge ek are called the end vertices of ek.
Vertex is also referred to as a node, a junction, a point, 0- cell, or an 0-simplex.
Edges are also referred to as a branch, a line, an element, a 1-cell , an arc and a 1simplex.
Graph can be represented by means of diagrams in which vertices are represented as
points and each edge as a line segment joining its end vertices
Self loop: An edge having the same vertex as both its end vertices is called a self
loop(loop). Such an edge is associated with vertex pair (vi,vi). In the above figure,
edge e1 is a self loop associated with vertex pair(v2,v2)
Parallel edges: More than one edge associated with a given pair of vertices are called
parallel edges. In the above figure e4 and e5 are parallel edges.
Simple Graph: A graph that has neither self loops nor parallel edges is called
simple graph. A graph is also called a linear complex, a 1-complex, or a one
dimensional complex. Examples of simple graph are shown below:
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Applications of Graph
a) Konigsberg Bridge Problem: Two islands C and D were connected to each other
and to the banks A and B with seven bridges as shown in figure. The problem was to
start at any land areas A, B, C or D , walk over each of the seven bridges exactly
once, and return to the starting point
Euler represented this problem by means of a graph. Vertices represent the land areas
and the edges represents the bridges
Euler proved that a solution for this problem does not exists
b) Utilities Problem: There are three houses H1 , H2 and H3, each to be connected to
each of the three utilities water(W), gas(G) and electricity(E) by means of conduits. Is
it possible to make such connections without any crossovers of conduits?
This problem also can be represented by a graph where edges represented the conduits
and verticesrepresented the houses and utility centres
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Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
From the graph it is clear that no solution is exists for ulitilty problem
c) Electrical Network Problem: Topology of a electrical network is studied by means
of graphs. Vertices represented the electrical network junctions and the edges
represented the branches
Electrical Network and its graph
d) Seating problem:
 Problem: Problem: Nine members of a new club meet each day for lunch at a round
table. They decide to sit such that every member has different neighbours at ach
lunch. How many days can this arrangement last?
Solution: Seating arrangement of nine members in a round table can be represented
by means of a graph. Each vertex represents a member and an edge joining two
vertices represents the relationship of sitting next to each other
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Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Finite and Infinite Graph

A graph with finite number of vertices and finite number of edges is called a finite
graph, otherwise it is an infinite graph
Finite Graph
Infinite Graph
Incidence and Degree

When a vertex vi is an end vertex of some ej, vi and ej are said to incident with each
other.Two non parallel edges said to be adjacent if they are incident on a common
vertex.

For example:
a) e3,e4,e5 incident on the vertex v1
b) e3,e1,e2 incident on the vertex v2
c) e4,e5,e6 incident on the vertex v3
d) e2,e6,e7 incident on the vertex v4
e) e7 incident on the vertex v5
Degree or valency of a vertex
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The number of edges incident on a vertex vi, with self loop counted twice, is called
the degree , d(vi), of vertex vi.
For example:
a) d(v1)= d(v3)= d(v4)=3
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Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
b) d(v2)=4
c) d(v5)=1
Sum of the degrees of all vertices in G is twice the number of edges in G
𝒏
𝒅 𝒗𝒊 = 𝟐𝒆
𝒊=𝟏

For example: sum of degrees of vertices=d(v1)+d(v2)+d(v3)+ d(v4)+d(v5)
= 3+4+3+3+1
=14 = twice the number of edges
Theorem 1.1 :The number of vertices of odd degree in a graph is always even
Proof:
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Sum of the degrees of all vertices in G = 𝑛𝑖=1 𝑑(𝑣𝑖 )
If we consider the vertices with odd and even degree separately, the above equation
can be expressed as the sum of vertices of even and odd degree as follows:
𝑛
𝑑(𝑣𝑖 ) =
𝑖=1

𝑑 𝑣𝑗 +
𝑒𝑣𝑒𝑛
𝑑 𝑣𝑘
𝑜𝑑𝑑
Since the left hand side of the equation is even, and the first expression on the right
hand side is even (being a sum of even numbers), the second expression must also be
even:
𝑑 𝑣𝑘 = 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟
𝑜𝑑𝑑

Because each 𝑑 𝑣𝑘 is odd, the total number of terms in the sum must be even to
make the sum an even number. Hence the theorem
Steps for checking the given degree sequence form a graph
Step 1: Sort the degrees in decending order
Step 2: If the first degree is 0, then graph can be formed
Step 3: If the first degree has value d ,the following d degrees must be greater than 0. If not,
no such graph can be formed
Step 4: Take away(remove) the first degree (value of d) and reduce the following d degrees
by one
Step 5: Repeat the steps until all degrees are zero
Steps for drawing a graph, if the degree sequence is given
Step 1: Sort the degrees in decending order
Step 2: Connect the highest degree d to the next d vertices
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Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Step 3: Take away(remove) the first degree (value of d) and reduce the following d degrees
by one
Step 4: Repeat the steps until all degrees are zero
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Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor
Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Regular Graph

A graph in which all vertices are of equal degree is called a regular graph
Isolated Vertex

A vertex having no incident edge is called isolated vertex. Isolated vertices are
vertices with zero degree
Pendent Vertex or End vertex

A vertex of degree one is called a pendant vertex or an end vertex
Example
Isolated vertex: v4,v7
Pendant vertex: v3
Null Graph
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A graph without any edges is called null graph
Every vertex in null graph is an isolated vertex
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Isomorphism
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Two graphs G and G’ are said to be isomorphic if there is a one to one
correspondence between their vertices and between their edges such that the incidence
relationship is preserved
The two isomorphic graph must have
a) same number of vertices
b) same number of edges
c) equal number of vertices with a given degree
QN: Check whether the following pair of graphs are isomorphic or not
1.
G1
G2
Vertices and
𝒂, 𝒃, 𝒄, 𝒅, 𝒆
𝒗𝟏, 𝒗𝟐, 𝒗𝟑, 𝒗𝟒, 𝒗𝟓
Total number of vertices
5
5
Vertices with degree (arranged in
decending order)
a, c, d, b, e
v1, v3, v4, v2, v5
Total number of edges= sum of
degrees of all vertices/2
(3+3+3+2+1)/2=6
(3+3+3+2+1)/2=6
Since G1 and G2 have same number of vertices and same number of edges, G1 and G2 are
isomorphic Graphs
2.
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Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
G1
𝑨𝟏, 𝑨𝟐, 𝑨𝟑, 𝑨𝟒, 𝑨𝟓, 𝑨𝟔, 𝑨𝟕, 𝑨𝟖
G2
𝑩𝟏, 𝑩𝟐, 𝑩𝟑, 𝑩𝟒, 𝑩𝟓, 𝑩𝟔, 𝑩𝟕, 𝑩𝟖
Vertices and
Total number of
vertices
8
8
Vertices with degree
(arranged in
𝑨𝟏, 𝑨𝟐, 𝑨𝟑, 𝑨𝟒, 𝑨𝟓, 𝑨𝟔, 𝑨𝟕, 𝑨𝟖
𝑩𝟏, 𝑩𝟐, 𝑩𝟑, 𝑩𝟒, 𝑩𝟓, 𝑩𝟔, 𝑩𝟕, 𝑩𝟖
decending order)
Total number of
edges= sum of
(3+3+3+3+3+3+3+3)/2=12
(3+3+3+3+3+3+3+3)/2=12
degrees of all
vertices/2
Since G1 and G2 have same number of vertices and same number of edges, G1 and G2 are
isomorphic Graphs
3.
Vertices and
G1
G2
𝑨𝟏, 𝑨𝟐, 𝑨𝟑, 𝑨𝟒, 𝑨𝟓
𝑩𝟏, 𝑩𝟐, 𝑩𝟑, 𝑩𝟒, 𝑩𝟓,
, 𝑨𝟔, 𝑨𝟕, 𝑨𝟖, 𝑨𝟗, 𝑨𝟏𝟎 𝑩𝟔, 𝑩𝟕, 𝑩𝟖, 𝑩𝟗, 𝑩𝟏𝟎
G3
𝑪𝟏, 𝑪𝟐, 𝑪𝟑, 𝑪𝟒, 𝑪𝟓,
𝑪𝟔, 𝑪𝟕, 𝑪𝟖, 𝑪𝟗, 𝑪𝟏𝟎
Total number of
vertices
10
10
10
Vertices with degree
(arranged in
𝑨𝟏, 𝑨𝟐, 𝑨𝟑, 𝑨𝟒, 𝑨𝟓,
𝑩𝟏, 𝑩𝟐, 𝑩𝟑, 𝑩𝟒, 𝑩𝟓,
𝑪𝟏, 𝑪𝟐, 𝑪𝟑, 𝑪𝟒, 𝑪𝟓,
decending order)
𝑨𝟔, 𝑨𝟕, 𝑨𝟖, 𝑨𝟗, 𝑨𝟏𝟎 𝑩𝟔, 𝑩𝟕, 𝑩𝟖, 𝑩𝟗,B10
𝑪𝟔, 𝑪𝟕, 𝑪𝟖, 𝑪𝟗,C10
Total number of
edges= sum of
(3+3+3+3+3+3+3+3 (3+3+3+3+3+3+3+3+ (3+3+3+3+3+3+3+3+
degrees of all
+3+3)/2=15
3+3)/2=15
3+3)/2=15
vertices/2
Since G1,G2 and G3 have same number of vertices and same number of edges, G1, G2 and
G3 are isomorphic Graphs
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
4)
G1
𝑨𝟏, 𝑨𝟐, 𝑨𝟑, 𝑨𝟒, 𝑨𝟓, 𝑨𝟔
G2
𝑩𝟏, 𝑩𝟐, 𝑩𝟑, 𝑩𝟒, 𝑩𝟓, 𝑩𝟔
Vertices and
Total number of
vertices
6
6
Vertices with degree
(arranged in
𝑨𝟒, 𝑨𝟐, 𝑨𝟑, 𝑨𝟏, 𝑨𝟓, 𝑨𝟔
𝑩𝟑, 𝑩𝟐, 𝑩𝟓, 𝑩𝟏, 𝑩𝟒, 𝑩𝟔
decending order)
Total number of
edges= sum of
(3+2+2+1+1+1)/2=5
(3+2+2+1+1+1)/2=5
degrees of all
vertices/2
Since G1 and G2 have same number of vertices and same number of edges, G1 and G2 are
isomorphic Graphs
Sub graphs

A graph g is said to be a sub graph of a graph G, if all the vertices and all the edges of
g are in G, and each edge of g has the same end vertices in g as in G. And it is denoted
by g G
graph G
graph g
Properties of sub graph
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Every graph is its own sub graph
A sub graph of a sub graph of G is a sub graph of G
A single vertex in a graph G is a sub graph of G
A single edge in G, together with its end vertices , is also a sub graph of G
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Edge Disjoint Sub Graphs

Two sub graphs g1 and g2 of a graph G are said to be edge disjoint if g1 and g2 do not
have any edge in common
A Puzzle with multi coloured cubes
Problem: We are given four cubes. The six faces of of every cube are various coloured blue,
green, red or white. Is it possible to stack the cubes one on top of another to form a column
such that no color appears twice on any of the four sides of theis column.
Solution:
Step 1: Draw a graph with four vertices B, G, R and W one for each color. Three opposite
faces of the cube are represented by three edges. ie., if a blue face in cube 1 has a white face
opposite to it, draw an edge between vertices B and W in the graph. Put label 1 on all the
three edges resulting from cube 1. Repeat the procedure for all the three cubes one by one
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Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
on the same graph until we have a graph with four vertices and 12 edges. A particular set of
four coloured cubes and their graphs are shown below.
Step 2: Consider the graph resulting from the representation. The degree of each vertex is the
total number of faces with the corresponding color. Four cubes can be arranged (to form a
coloumn such that no color appears more than once on any side) if and only if there
exists two edge disjoint sub graphs, each with four edges, each of the edges labelled
differently, and such that each vertex is of degree two. For the above set of cubes, this
condition is satisfied, and the two sub graphs are shown below.
Walks
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A walk is defined as a finite alternating sequence of vertices and edges, beginning and
ending with vertices, such that each edge is incident with the preceding and following
it
No edge appeared more than once in a walk. A vertex may appear more than once
A walk is also referred to as an edge train or chain.
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics

Vertices with which a walk begins and ends are called its terminal vertices

Closed Walk: If a walk begins and ends with same vertex, then it is called closed
walk
Open Walk: If a walk is not closed is called Open Walk
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Path
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An open walk in which no vertex appears more than once is called a path
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The number of edges in a path is called the length of a path
A self loop can be included in a walk but not in a path
The terminal vertices of a path are of degree one and the rest of the vertices are of
degree two.
Circuit
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A closed walk in which no vertex appears more than one is called a circuit
Circuit is a closed non intersecting walk
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Dept of CSE
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
If the circuit is a sub graph of another graph , one must count degrees contributed by
the edges in the circuit only
A circuit is also called a cycle, elementary cycle, circular path and polygon
Every self loop is a circuit, but not every circuit is a self loop
Connected and Disconnected Graph

A graph G is said to be connected if there is atleast one path between every pair of
vertices in G. Otherwise G is disconnected
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A null graph of more than one vertex is disconnected

A disconnected graph consists of two or more connected graph. Each of these
connected subgraph is called a component
Theorem 1.2: A graph G is disconnected if and only if its vertex set V can be partitioned
into two non empty, disjoint subsets V1 and V2 such that there exists no edge in G whose
one end vertex vertex is in subset V1 and other in subset V2
Proof:


Suppose V can be partitioned into two disjoint subsets V1 and V2
Consider two arbitrary vertices a and b of G, such that 𝑎 ∈ 𝑉1 and 𝑏 ∈ 𝑉2
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Dept of CSE
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
No path can exist between vertices a and b; otherwise, there would be atleast one edge
whose one end vertex would be in V1 and the other is in V2
Hence if a partition exists, G is not connected
 Conversely, let G be a disconnected graph. Consider a vertex a in G.
 Let V1 be the set of all vertices that are joined by paths to a.
 Since G is disconnected, V1 does not include all vertices of G. The remaining vertices
will form a set V2
 No vertex in V1 is joined to any in V2 by an edge. Hence the partition.
Theorem 1.3: If a graph (connected or disconnected) has exactly two vertices of odd
degree, there must be a path joining these two vertices
Proof:
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
Let G be a graph with all even vertices(vertice with even degree) except vertices v1
and v2, which are odd.
For every component of a disconnected graph , no graph can have an odd number of
odd vertices.
Therfore in graph G, v1 and v2 must belong to the same component , and hence must
have a path between them.
Theorem 1.4: A simple graph (ie., a graph without parallel edges or self loops) with n
vertices and k components can have at most 𝒏 − 𝒌 (𝒏 − 𝒌 + 𝟏)/𝟐 edges
Proof:

Let the number of vertices in each of the k components of a graph G be n1,n2,..., nk.
Thus we have
𝑛1 + 𝑛2 + ⋯ + 𝑛𝑘 = 𝑛
𝑛𝑖 ≥ 1

Maximum number of edges in the ith component of G is2 𝑛𝑖 (𝑛𝑖 − 1)

Therfore maximum number of edges in G is
1
1
2
𝑘
𝑖=1
1
𝑛𝑖 − 1 𝑛𝑖 =
2
=
𝑘
𝑖=1
𝑘
1
2
1
𝑛𝑖 2 −
2
𝑛𝑖 2 −
𝑖=1
𝑘
𝑛𝑖
𝑖=1
𝑛
2
1
≤ 𝑛2 − 𝑘 − 1 2𝑛 − 𝑘
2
1
= 𝑛−𝑘 𝑛−𝑘−1
2
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−
𝑛
2
Dept of CSE
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Points to Remember
A linear graph (graph) G=(V,E) consists of a set of vertices and set of edges
Applcation of Graphs: Konigsberge Bridge Problem, Utility Problem, Electrical
Network Problem, Seating Problem
A graph with finite number of vertices and finite number of edges is called a finite
graph, otherwise it is an infinite graph
The number of edges incident on a vertex vi, with self loop counted twice, is called
the degree , d(vi), of vertex vi.
Sum of the degrees of all vertices in G is twice the number of edges in G
𝒏
𝒅 𝒗𝒊 = 𝟐𝒆















𝒊=𝟏
The number of vertices of odd degree in a graph is always even
A graph in which all vertices are of equal degree is called a regular graph
Isolated vertices are vertices with zero degree
A vertex of degree one is called a pendant vertex or an end vertex
A graph without any edges is called null graph
The two isomorphic graph must have
a) same number of vertices
b) same number of edges
c) equal number of vertices with a given degree
A graph g is said to be a sub graph of a graph G, if all the vertices and all the edges
of g are in G, and each edge of g has the same end vertices in g as in G
Two sub graphs g1 and g2 of a graph G are said to be edge disjoint if g1 and g2 do not
have any edge in common
No edge appeared more than once in a walk. A vertex may appear more than once
Closed Walk: If a walk begins and ends with same vertex, then it is called closed
walk
Open Walk: If a walk is not closed is called Open Walk
An open walk in which no vertex appears more than once is called a path
A closed walk in which no vertex appears more than one is called a circuit
A graph G is said to be connected if there is atleast one path between every pair of
vertices in G. Otherwise G is disconnected
A simple graph (ie., a graph without parallel edges or self loops) with n vertices
and k components can have at most 𝒏 − 𝒌 (𝒏 − 𝒌 + 𝟏)/𝟐 edges
PROBLEMS
1. Prove that any two simple connected graph with n vertices , all of degree two are
isomorphic
2. Are the two graphs in the figure are isomorphic? Why?
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3. Show that two graphs in the figure are isomorphic.
4. Prove that a simple graph with n vertices must be connected if it has more than
[ 𝑛 − 1 𝑛 − 2 ]/2 edges (Hint: Use Theorem1.4)
5. Prove that if a connected graph G is decomposed into two sub graphs g 1 and g2, there
must be atleast one vertex common between g1 and g2
6. Prove that a connected graph G remains connected after removing an edge e i from G, if
and only if ei is in some circuit G
7. Draw a connected graph that becomes disconnected when any edge is removed from it
8. Prove that a graph with n vertices satisfying the condition –“ a connected graph that
becomes disconnected when any edge is removed from it “ is a)simple b) has exactly
n-1 edges
9. What is the length of the path from the entrance to the center of the maze in the
following figure.
10. List all different paths between vertices 5 and 6 in the figure. Give the length of each of
these paths?
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PREVIOUS YEAR UNIVERSITY QUESTIONS
1. Consider a graph G with 4 vertices: v1, v2, v3 and v4 and the degrees of vertices are 3, 5,
2 and 1 respectively. Is it possible to construct such a graph G? If not, why?
2. Draw a disconnected simple graph G1 with 10 vertices and 4 components and also
calculate the maximum number of edges possible in G1.
3. Write any two applications of graphs with sufficient explanation
4. Discuss the Konigsberg Bridge problem.Is there any solution to the problem? Justify your
answer.
5. Prove that a simple graph with n vertices must be connected, if it has more than (n-1)(n2)/2 edges.
6. Explain any two applications of graphs
7. Define the term a)simple graph b)Complete graph c) Null graph
8. Define Simple Graph, Multi Graph and Pseudo Graph with examples
9. Explain the terms:
a) Null Graph
b) Complete Graph and
c) Pendant Vertex with examples
10. Prove that the number of vertices of odd degree in a graph is always even
OR
Prove that in any undirected graph, the number of vertices of odd degree should be even
11. Differentiate a Walk, Path & Circuit in a Graph
12. What are the steps involved in proving that two graphs are isomorphic or not?
13. Prove that, if a graph has exactly two vertices of odd degree, then there must be a path
joining these two vertices
14. Prove that “ a simple graph with n vertices and k components can have atmost 𝑛 −
𝑘(𝑛−𝑘+1)/2 edges
15. Prove the following:
“A graph G is disconnected if and only if its vertex set V can be partitioned into two
non empty disjoint subsets V1 and V2 such that there exists no edge in G whose one
end vertex is in V1 and other in V2”
16. State four cube problem and find the solution to the four cube problem for the set of
cubes drawn below
23
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Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
WORKSHEET
1. Draw graphs representing problems of a)two houses and three utilities; b)four houses and
four utilities. say, water, gas, electricity, and telephone
2. Consider a graph G with 4 vertices: v1, v2, v3 and v4 and the degrees of vertices are 3, 5, 2
and 1 respectively. Is it possible to construct such a graph G? If not, why?
3. Draw a disconnected simple graph G1 with 10 vertices and 4 components and also calculate
the maximum number of edges possible in G1.
4. What are the basic conditions to be satisfied for two graphs to be isomorphic? Are the two
graphs below isomorphic? Explain with valid reasons
5. Define isomorphism between two graphs. Are the following graphs are isomorphic to each
other? Justify your answer.
24
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Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
6. Draw all simple graph of one, two, three and four vertices
7. Prove that the maximum number of edges in a simple graph with n vertices is
𝑛∗
𝑛−1
2
8. Determine whether the given pair of graph is isomorphic or not
9. Draw two isomorphic graphs with 6 vertices and 9 edges
10. State Seating problem and find all its solutions
11. Prove that a graph with n vertices, n-1 edges and no circuits is connected
12. Check whether the two graphs given below are isomorphic?
13. Prove that the ring sum of two circuits in a graph G is either a circuit or an edge disjoint
union of circuits
14. What is the largest possible number of vertices in a graph with 19 edges and all vertices are
of degree at least 3
15. Prove that the ring sum of two circuits in a graph G is either a circuit or an edge disjoint
union of circuits
16. Is it possible to have simple graph with the following degree sequences? If yes, draw the
graphs.
a. 2,3,3,3,3,3,4,5
b. 1,3,3,4,5,6,6
c. 1,2,3,3,4,5,6
17. Show that there is no graph G with 𝑉(𝐺) = 12 and 𝐸(𝐺) = 28 in which each vertex is
of degree 3 or 6
18. Does there exists a graph with 28 edges and 12 vertices each of degree 3 or 4?
25
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Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
MODULE II
Euler Graph


If some closed walk in a graph contains all the edges of the graph, then the walk is
called an Euler Line and a graph that contain Euler line is called Euler graph
Euler graph is always connected
Theorem 2.1: A given connected graph G is an Euler graph if and only if all vertices of
G are of even degree
Proof:










Suppose that G is and Euler graph. It therefore contains an Euler line( which is a
closed walk).
In tracing this walk, observe that every time the walk meets a vertex v it goes through
two “new” edges incident on v –with one we entered v and with the other “exited”.
This is true not only of all intermediate vertices of the walk but also of the terminal
vertex, because we “exited” and entered the same vertex at the beginning and end of
the walk, respectively.
Thus if G is an Euler graph, the degree of every vertex is even
Conversely, let us assume that each vertex of G has even degree. We need to show
that G is Eulerian.
Let us start with a vertex v0 ∈ 𝑉(𝐺). Assume G is connected, there exists a vertex v1
∈ 𝑉(𝐺) that is adjacent to v0. Since G is a simple graph and𝑑 𝑣𝑖 ≥ 2, for each vertex
vi ∈ 𝑉(𝐺), there exists a vertex v2 ∈ 𝑉(𝐺), that is adjacent that is adjacent to v1 with
𝑣2 ≠ 𝑣0 .Similarly, there exists a vertex v3 ∈ 𝑉(𝐺), that is adjacent that is adjacent to v2
with 𝑣3 ≠ 𝑣1
As the number of vertices is finite, the process of getting a new vertex will finally end
with a vertex 𝑣𝑖 being adjacent to a vertex 𝑣𝑘 for some 𝑖, 0 ≤ 𝑖 ≤ 𝑘 − 2. Hence,
𝑣𝑖 − 𝑣𝑖+1 − 𝑣𝑖+2 − ⋯ − 𝑣𝑘 − 𝑣𝑖 forms a circuit, say 𝐶, in 𝐺.
If 𝐶 contains every edge of 𝐺, then 𝐶 gives rise to a closed Eulerian trail and we are
done. Let us assume that 𝐸 𝐶 is a proper subset of 𝐸 𝐺 .
Now consider the graph 𝐺1 that is obtained by removing all the edges in 𝐶 from
𝐺. Then, 𝐺1 may be a disconnected graph but each vertex of 𝐺1 still has even degree.
Hence, we can do the same process explained above to 𝐺1 also to get a closed
Eulerian trail, say 𝐶1 .
If 𝑪𝟏 contains all edges of 𝑮𝟏 , then 𝑪 ∪ 𝑪𝟏 is a closed Euler trail in G. If not, let 𝐺2
be the graph obtained by removing the edges of 𝐶1 from 𝐺1 .
26
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Dept of CSE



HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Since G is a finite graph, we can proceed to find out a finite number of cycles only. Le
the process of finding cycles, as explained above, ends after a finite number of steps
say, 𝑟.
Then the reduced graph 𝐺𝑟 = 𝐺𝑟−1 − 𝐸 𝐶𝑟−1 = 𝐺 − 𝐸(𝐶 ∪ 𝐶1 ∪ 𝐶𝑟−1 ) will be an
empty graph(null graph). Then 𝐶 ∪ 𝐶1 ∪ 𝐶2 … ∪ 𝐶𝑟−1 is a closed Euler trail in G
Therefore, G is Eulerian, This completes the proof.
Konigsberg Bridge Problem:

Two islands C and D were connected to each other and to the banks A and B with
seven bridges as shown in figure. The problem was to start at any land areas A, B,
C or D , walk over each of the seven bridges exactly once, and return to the
starting point

It can be represented by means of the following graph and the problem can be solved
by checking whether the following graph is an Euler or not. If it is Euler graph then
we can solve the problem. Otherwise not.

From the figure we find that not all its vertices are of even degree . Hence it is not an
Euler graph. Thus it is not possible to walk over each of the seven bridges exactly
once and return to the starting point
Unicursal line or open Euler line:

27
An open walk that includes all edges of a graph without retracing any edge is a an
open Euler line or Unicursal line. A graph that has a unicursal line is called a
unicursal graph
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Dept of CSE

HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
A connected graph is unicursal if and only if it has exactly two vertices of odd degree
Theorem 2.2: In a connected graph G with exactly 2k odd vertices, there exist k
edge-disjoint subgraphs such that they together contain all edges of G and that each
is a unicursal graph
Proof:




Let the odd vertices of the given graph G be named as v1,v2,...,vk; w1,w2,...,wk in any
arbitrary order. Add k edges to G between the vertex pairs (v1,w1),( v2,w2),...,( wk,wk)
to form a new graph G’.
Since every vertex of G’ is of even degree, G’ consists of anEuler line 𝜌 .
Now if we remove 𝜌 the k edges we just added, 𝜌 will split into k walks, each of
which is a unicursal line
The first removal will leave a single unicursal line; the second removal will split that
into two unicursal line; and each successive removal will split a unicursal line into
two, until there are k of them. Thus the theorem.
Theorem2.3: A connected graph G is an
decomposed into circuits
Euler graph if and only if it can be
Proof:


Suppose graph G can be decomposed into circuits; that is , G is a union of edge
disjoint circuits.
Since the degree of every vertex in a circuit is two , the degree of every vertex in G is
even. Hence G is an Euler graph
Arbitrarily Traceable Graphs

A vertex v in an Euler graph have a property that an Euler line is always obtained
when one follows any walk from vertex v. Such a graph is called an arbitrarily
traceable graph from vertex v.
28
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Hamiltonian Circuits

A circuit in a connected graph G is said to be Hamiltonian if it includes every vertex
of G

In the figure a, starting at vertex v, if one traverse along the edges shown in heavy
lines- passing through each vertex exactly once- one gets a Hamiltonian circuit
Hamiltonian circuit in a graph of n vertices consists of exactly n edges

Hamiltonian Path


If we remove any one edge from a Hamiltonian circuit , we are left with a path. This
path is called a Hamiltonian Path
The length of a Hamiltonian path in a connected graph of n vertices is n-1
29
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
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Since Hamiltonian circuit traverses every vertex exactly once, it cannot include a
self loop or a set of parallel edges.
Complete Graph




A simple graph in which there exists an edge between every pair of vertices is called
complete graph.
A complete graph is also called universal graph or clique
Since every vertex is joined with every other vertex through one edge , the degree of
every vertex is n-1 in a completed graph G of n vertices
Total number of edges in G is 𝒏(𝒏 − 𝟏)/𝟐
Note: It is easy to construct a Hamiltonian circuit in a complete graph of n vertices. Let
the vertices be numbered v1, v2,..., vn. Since an edge exists between any two vertices, we
can start from v1 and traverse to v2, and v3 and so on to vn, and finally from vn and v1. This
is a Hamiltonian circuit
Theorem 2.4: In a complete graph with n vertices there are (𝒏 − 𝟏)/𝟐 edge –disjoint
Hamiltonian circuits, if n is an odd number ≥ 𝟑
Proof:

Note that a complete graph has
𝑛 (𝑛−1)
2
edges and a Hamiltonian cycle in 𝐾𝑛 contains
only n edges. Therefore, the maximum number of edge disjoint Hamiltonian cycle is
𝑛 −1
2


Now, assume that 𝑛 ≥ 3 and is odd. Construct a sub graph 𝐺 of 𝐾𝑛 as explained
below:
The vertex 𝑣1 is placed at the centre of a circle and the remaining 𝑛 − 1 vertices are
placed on the circle, at equal didtances along the circle such that the angle made at the
360
centre by two points is 𝑛 −1 degrees.


The vertices with odd suffixes are placed along the upper half of the circle and the
vertices with even suffixes are placed along the lower half circle.
Then draw edges 𝑣𝑖 𝑣𝑖+1 , where 1 ≤ 𝑖 ≤ 𝑛 , with the meaning that 𝑣𝑛+1 = 𝑣1 are
drawn as shown in the figure.
30
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Clearly the reduced graph 𝐺1 is a cycle covering all vertices of 𝐾𝑛 . That is if we rotate
360
the vertices along the curve for 𝑛 −1 degrees, we get another Hamiltonian sub graph
𝐺2 of 𝐾𝑛 , which has no common edges with 𝐺1

In a similar way, rotate the polygonal pattern clockwise by
360
𝑛 −1
degrees. After
𝑛 − 1 𝑡𝑕 rotation, all vertices will be exactly as in the figure 1. Therefore 𝑛 − 1
rotation are valid. But, it can be noted that the cycle 𝐺𝑖 obtained after the 𝑖th rotation
and the cycle 𝐺𝑛 −1 +𝑖 are isomorphic graphs, because all the vertices in the upper half
2
cycle in 𝐺𝑖 will be in the lower half cycle in 𝐺𝑛 −1+𝑖 and vice versa, in the same order.

That is, we have now that there are
𝑛 −1
2
2
distinct such non-isomorphic edge-disjoint
cycles in 𝐾𝑛 . Hence, the number of edge-disjoint Hamiltonian cycles is
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𝑛 −1
2
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Seating Arrangement Problem







Problem: Nine members of a new club meet each day for lunch at a round table.
They decide to sit such that every member has different neighbours at ach lunch. How
many days can this arrangement last?
Seating arrangement of nine members in a round table can be represented by means of
a graph. Each vertex represents a member and an edge joining two vertices represents
the relationship of sitting next to each other
We can solve the seating arrangement at round table problem using the Theorem 2.4,
as follows
Representing a member x by a vertex and the possibility of his sitting next to another
member y by an edge between x and y, we construct a graph G
Since every member is allowed to sit next to any other member, G is a complete graph
of nine vertices- nine being the number of people to be aeated around the table. Every
seating arrangement around the table is clearly a Hamiltonian circuit.
The first day of their meeting they can sit in any order, and it will be a Hamiltonian
circuit H1. The second day, if they are to sit such that every member must have
didderent neighbours, we have to find another Hamiltonian circuit H 2 in G, with an
enitirely different set of edges from those in H1 ; that is H1 and H2 are edge disjoint
Hamiltonian circuits.
From Theorem 2.4, the number of edge disjoint Hamiltonian circuit in G is four ;
therefore only four such arrangements exixts among people
Dirac’s Theorem for Hamiltonicity
Theorem: Every graph G with n ≥ 3 vertices and minimum degree d(G) ≥ n/2 has a
Hamiltonian cycle.
Proof:


Suppose that 𝐺 = (𝑉, 𝐸) satisfies the hypothesis of the theorem. Then G is connected,
since otherwise the degree of any vertex in a smallest component C of G would be
𝑛
𝑛
atmost 𝐶 − 1 < 2 , contradicting the hypothesis 𝛿 𝐺 ≥ 2
Let 𝑃 = 𝑥0 𝑥1 … 𝑥𝑘 be the longest path in G, as seen in the figure given below:
32
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
Note that the length of P is k. Since P cannot be extended to a longer path, all the
neighbours of 𝑥0 lie on P. Assume the contradictory. Let y be an adjacent vertex of
𝑥0 which is not in P

Then the path 𝑃′ = 𝑦𝑥0 𝑥1 … 𝑥𝑘 is a path of length 𝑘 + 1, contradicting the hypothesis
that P is the longest path in G

Similarly, we note that and all the neighbours of 𝑥𝑘 will also lie on P, unless we reach
at a contradiction as mentioned above

Hence, at least

vertices 𝑥1 𝑥2 … 𝑥𝑘 are adjacent to 𝑥0 .
Another way of saying the second part of the last sentence is: atleast least

𝑛
2
of the vertices 𝑥0 𝑥1 … 𝑥𝑘−1 are adjacent to 𝑥𝑘 , and at least
𝑛
2
𝑛
2
of the
of the
vertices 𝑥𝑖 ∈ 𝑥0 , … , 𝑥𝑘−1 are such that 𝑥0 𝑥𝑖+1 ∈ 𝐸. Combining both statements and
using the pigeon-hole principle,we see that there is some 𝑥𝑖 with 0 ≤ 𝐼 ≤ 𝐾 −
1, 𝑥𝑖 𝑥𝑘 ∈ 𝐸 and 𝑥0 𝑥𝑖+1 ∈ 𝐸
Consider the cycle 𝐶 = 𝑥0 𝑥𝑖+1 𝑥𝑖+2 … 𝑥𝑘−1 𝑥𝑘 𝑥𝑖 𝑥𝑖−1 … 𝑥1 𝑥0 as given in the following
graph

We claim that the above cycle C is a Hamiltonian cycle of G. Otherwise, since G is
connected, there would be some vertex 𝑥𝑗 of C adjacent to a vertex y not in C, so
that 𝑒 = 𝑥𝑖 𝑦 ∈ 𝐸. But then we could attach e to a path ending in 𝑥𝑗 containing k
edges of C, constructing a path in G longer than P( see the graph given below), which
is contradiction to the hypothesis that P is the longest path in G.

Therefore C must cover all vertices of G and hence it is a Hamiltonian cycle in G.
33
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Travelling Salesman problem




Problem: A salesman is required to visit a number of cities during a trip. Given the
distance between the cities, in what order should he travel so as to visit every city
precisely once and return home, with the minimum mileage travelled?
We can represent the cities by vertices and the roads between them by edges in a
graph. In this graph, with every edge 𝑒𝑖 there is associated a real number 𝑤(𝑒𝑖 ). Such
a graph is called a weighted graph; 𝑤 𝑒𝑖 being the weight of edge 𝑒𝑖
In this problem, if each of the cities has a road to every other city, we have a
completed weighted graph . This graph has numerous Hamiltonian circuits and we are
to pick the one that has the smallest sum of distances
Theoretically , the problem of travelling salesman can always be solved by
enumerating all 𝑛 − 1 !/2 Hamiltonian circuits, calculating the distance traveled in
each and then picking the shortest one.
Directed Graph







A directed graph or digraph G consists of a set of vertices 𝑉 = 𝑣1 , 𝑣2 , … , a set of
edges 𝐸 = 𝑒1 , 𝑒2 , … and a mapping 𝜑 that maps every edge onto some ordered pair
of vertices (𝑣𝑖 , 𝑣𝑗 )
In digraph, vertices are represented by points and edges by a line segment between vi
and vj with an arrow directed from vi to vj
The vertex vi, which edge ek is incident out of is called the initial vertex of ek
The vertex vj, which edge ek is incident into is called the terminal vertex of ek
A digraph is also referred to as oriented graph.
The number of edges incident out of a vertex vi is called the out-degree (outvalence
or outward demidegree) of vi and is written 𝑑 +(𝑣𝑖 )
The number of edges incident into vertex vi is called the in-degree (in valence or
inward demidegree) of vi and is written 𝑑 − (𝑣𝑖 )\
34
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
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
In any digraph, sum of all in degree is equal to the sum of all out-degree, each sum
being equal to the number of edges in G
𝑛
𝑛
𝑑 + 𝑣𝑖 =
𝑖=1


𝑑 − 𝑣𝑖 = 𝑒
𝑖=1
An isolated vertex is a vertex in which the in-degree and the out-degree are both
equal to zero
A vertex in a digraph is called pendant if its is of degree one
𝑑 + 𝑣𝑖 + 𝑑 − 𝑣𝑖 = 1
Types of Digraphs

Simple Digraphs: A digraph that has no self loop or parallel edges is called a simple
digraph

Asymmetric Digraphs: Digraphs that have atmost one directed edge between a pair
of vertices, but contains self loops are called asymmetric or antisymmetric

Symmetric Digraphs: Digraphs in which for every edge (a,b), there is also an edge
(b,a)
35
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics

A digraph that is both simple and symmetric is called a simple symmetric digraph.
Similarly, a digraph that is both simple and asymmetric is called a simple
asymmetric digraph

Complete Digraphs:
Two types: Complete symmetric and complete asymmetric
 A complete symmetric digraph is simple digraph in which there is exactly
one edge directed from every vertex to every other vertex
 A complete asymmetric digraph of n vertices contains 𝑛(𝑛 − 1) edges
 A complete asymmetric digraph is an asymmetric digraph in which there is
exactly one edge between every pair of vertices
 A complete asymmetric digraph of n vertices contains 𝑛(𝑛 − 1)/2 edges
 It is also called a tournament or a complete tournament



Balanced Digraph: A digraph is said to be balanced if for every vertex vi the in
degree equals the out degree. 𝑑 + 𝑣𝑖 = 𝑑 − 𝑣𝑖 .
Balanced Digraph is also referred to as pseudosymmetric digraph or iso graph
Balanced Digraph is said to be regular if every vertex has same in degree and out
degree as every other vertex
Digraphs and Binary Relations

Reflexive relation: A relation R on a set X that satisfies 𝑥𝑖 𝑅𝑥𝑖 for every 𝑥𝑖 ∈ 𝑋 is
called a reflexive relation. The digraph of a reflexive relation will have a self loop at
every vertex. Such a digraph representing a reflexive binary relation on its vertex set
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may be called a reflexive digraph. A digraph in which no vertex has a self loop is
called an irreflexive digraph

Symmetric Relation: For some relation R it may happen that for all 𝑥𝑖 and 𝑥𝑗 , if
𝑥𝑖 𝑅𝑥𝑗 holds then 𝑥𝑗 𝑅𝑥𝑖 also holds. Such a relation is called a symmetric relation. The
digraph of a symmetric relation is a symmetric digraph because for every directed
edge from vertex 𝑥𝑖 to 𝑥𝑗 there is a directed edge from 𝑥𝑗 to 𝑥𝑖

Transitive Relation: A relation R is said to be transitive if for any three
elements𝑥𝑦 , 𝑥𝑗 and 𝑥𝑘 in the set , 𝑥𝑖 𝑅𝑥𝑗 and 𝑥𝑗 𝑅𝑥𝑘 always imply 𝑥𝑖 𝑅𝑥𝑘 . A digraph
representing a transitive relation is called a transitive directed graph.

Equivalence relation: A binary relation is called an equivalence relation if it is
reflexive, symmetric and transitive. The graph representing equivalence relation is
called equivalence graph
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
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
Points to Remember
Euler graph: If some closed walk in a graph contains all the edges of the graph, then
the walk is called an Euler Line and a graph that contain Euler line is called Euler
graph

Unicursal line or open Euler line: An open walk that includes all edges of a graph
without retracing any edge is a an open Euler line or Unicursal line. A graph that
has a unicursal line is called a unicursal graph

Arbitrarily Traceable Graphs: A vertex v in an Euler graph have a property that an
Euler line is always obtained when one follows any walk from vertex v. Such a graph
is called an arbitrarily traceable graph from vertex v.

Hamiltonian Circuits: A circuit in a connected graph G is said to be Hamiltonian if
it includes every vertex of G

Hamiltonian Path: If we remove any one edge from a Hamiltonian circuit , we are
left with a path. This path is called a Hamiltonian Path
The length of a Hamiltonian path in a connected graph of n vertices is n-1


Complete Graph: A simple graph in which there exists an edge between every pair
of vertices is called complete graph.

Dirac’s Theorem for Hamiltonicity: Every graph G with n ≥ 3 vertices and
minimum degree d(G) ≥ n/2 has a Hamiltonian cycle.

Directed Graphs: A directed graph or digraph G consists of a set of vertices
𝑉 = 𝑣1 , 𝑣2 , … , a set of edges 𝐸 = 𝑒1 , 𝑒2 , … and a mapping 𝜑 that maps every edge
onto some ordered pair of vertices (𝑣𝑖 , 𝑣𝑗 )

The number of edges incident out of a vertex vi is called the out-degree (outvalence
or outward demidegree) of vi and is written 𝑑 +(𝑣𝑖 )

The number of edges incident into vertex vi is called the in-degree (in valence or
inward demidegree) of vi and is written 𝑑 − (𝑣𝑖 )

Types of Digraphs: Simple graphs, Asymmetric digraphs, Symmetric digraphs,
Complete digraphs, Balance digraphs
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PROBLEMS
1. You are given a 10-piece domino set whos titles have the following set of dots: (1,2);
(1,3); (1,4); (1,5); (2,3); (2,4); (2,5); (3,4); (3,5); (4,5). Discuss the possibility of
arranging the tiles in a connected series such that one number on a title always
touches the same number on its neighbour
(Hint: Use a five-vertex complete graph and see if it is an Euler graph)
2. Is it possible to move a knight on a chess board such that it completes every
permissible move exactly once? A move between two squares is counted as one
regardless of the direction in which it is made
(Hint: Is the graph is unicursal)
3. A round robin tournament(when every player plays against every other) among n
players (n being an even number) can be represented by a complete graph of n
vertices. Discuss how you would schedule the tournaments to finish in the shortest
possible time.
4. Draw a graph that has a Hamiltonian path but does not have a Hamiltonian circuit.
5. Show that neither of the graphs in the figure has a Hamiltonian path
6. Show that the graph of a rhombicdodecahedron (with eight vertices of degree three
and six vertices of degree four) has no Hamiltonian path
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7. Draw a graph in which an Euler line is also a Hamiltonian circuit. What can you say
about such graphs in general?
8. Is it possible , starting from any of the 64 squares of the chess board, o move a knight
such that it occupies every square exactly once and return to the initial position? if so,
give one such tour.
(Hint: Look for a Hamiltonian circuit in the graph)
9. Prove that a graph G with n vertices always has a Hamiltonian path if the sum of the
degrees of every pair of vertices vi,vj in G satisfies the condition
𝑑 𝑣𝑖 + 𝑑 𝑣𝑗 ≥ 𝑛 − 1
(Hint: First show that G is connected. then use induction on path length in G)
10. Using the result of the above problem, show that in a dancing ring of n children it is
always possible to arrange the children so that everyone has a friend at each side if
every child enjoys friendship with atleast half the children
PREVIOUS YEAR UNIVERSITY QUESTIONS
1. With suitable example explain arbitrarily traceable graphs
2. State Dirac’s theorem for hamiltonicity and why it is not a necessary condition for a
simple graph to have a Hamiltonian circuit.
3. Differentiate between symmetric and asymmetric digraphs with examples and draw a
complete symmetric digraph of four vertices.
4. Define Euler line, Euler graph and Hamiltonian Circuit
5. Mention the difference between an Euler graph and Unicursal graph
6. Prove that in a complete graph with n vertices there are (𝑛 − 1)/2 edge –disjoint
Hamiltonian circuits, if n is an odd number ≥ 3
OR
7. Prove that “ In a complete graph with n vertices, there are (𝑛 − 1)/2 edge disjoint
Hamiltonian circuits, if n is an odd number≥ 3𝑛
8. Prove that a given connected graph G is an Euler graph if and only if all vertices of G
are of even degree
9. Define Hamiltonian graph.Find an example of a non-Hamiltonian graph with a
Hamiltonian path.
1. Define the following with example
i) Isomorphic digraph ii) Complete symmetric digraph
2. Explain the travelling salesman problem
OR
Explain the travelling salesman problem using the concept of Hamiltonian circuit
3. Define Euler graph.
4. Define connected graph and components of a graph
5. Prove that a given connected graph G is an Euler graph if and only if all vertices of G
are of even degree
6. Explain the travelling salesman problem
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
OR
Explain the travelling salesman problem using the concept of Hamiltonian circuit
7. Define Euler graph.
8. What is the necessary and sufficient condition for a graph to be Euler? And also prove
it.
9. Prove that the number of odd degree vertices in a graph is always even.
10. Discuss the Konigsberg Bridge problem.Is there any solution to the problem? Justify
your answer.
11. Define connected graph and components of a graph
12. For a Eulerian graph G, prove the following properties.
i) The degree of each vertex of G is even. ii) G is an edge-disjoint union of cycles.
13. State Travelling-Salesman Problem and how TSP solution is related with Hamiltonian
Circuits?
WORK SHEET
1. Draw the graphs for the following:
a. Graph that is Euler and Hamiltonian
b. Graph that is Euler but not Hamiltonian
c. Graph that is Hamiltonian but not Euler
2. Consider the graph G given below:
Define Euler graph. Is G an Euler? If yes, write an Euler line from G.
3. Define Hamiltonian circuits and paths with examples. Find out the number of edgedisjoint Hamiltonian circuits possible in a complete graph with five vertices.
4. For the following graph, find the shortest path between fromv1 to v4. Also find a
Euler circuit.
5. 19 students in a nursery school play a game each day, where they hold hands to form
a circle. For how many days can they do this, with no students holding hands with the
same playmates more than once? Substantiate your answer with graph theoretic
concepts.
6. Show that in any group of two or more people, there are always two with exactly
the same number of friends inside the group.
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MODULE III
Trees


A tree is a connected graph without any circuits.
A tree is a simple graph(without anu self loppor parallel edges)
Applications


The genealogy of a family can be represented by means of a tree(Family Tree)
The sorting of mail according to zip code and the sorting of punched cards are done
according to a tree(Decision tree or Sorting tree). Sorting or Decision Trees are
frequently used in Computer Applications and Switching Theory
Properties of Tree
A graph with n vertices is called a tree if





G is connected and is circuitless
G is connected and has n-1 edges
G is circuitless and has n-1 edges
There is exactly one path between every pair of vertices in G
G is a minimally connected graph
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Theorem 3.1:
There is one and only one path between every pair of vertices in a tree, T.
Proof:


Since T is a connected graph, there must exist atleast one path between every pair of
vertices in T. Now suppose that between two vertices a and b of T there are two
distinct paths.
The union of these paths will contain a circuit and T cannot be a tree
Theorem 3.2:
If in a graph G there is one and only one path between every pair of vertices, G is a tree
Proof:



Existance of a path between every pair of vertices assures that G is connected.
A circuit in a graph implies that there is atleast one pair of vertices a,b such that there
are two distinct paths between a and b.
Since G has one and only one path between every pair of vertices, G can have no
circuit. Therfore G is a tree.
Theorem 3.3:
A tree with n vertices have n-1 edges
Proof:
Theorem 3.4:
A graph is a tree if and only if it is minimally connected
Proof:

In a tree, all vertices are connected together with minimum number of edges. A
connected graph is said to be minimally connected if removal of any one edge from
it disconnects the graph.
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

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A minimally connected graph cannot have a circuit; otherwise we could remove one
of the edges in the circuit and still leave the graph connected. Thus a minimally
connected graph is a tree
Conversely, if a connected graph G is not minimally connected, there must exist an
edge 𝑒𝑖 in G such that 𝐺 − 𝑒𝑖 is connected. Therefore, 𝑒𝑖 is in some circuit, which
implies that G is not a tree. Hence the theorem
Theorem 3.5:
A graph with n vertices, n-1 edges, and no circuit is connected
Proof:


To interconnect n distinct points , the minimum number of line segment needed is
𝑛 − 1. The resulting structure, according to Theorem 3.3 is a tree
We know that a connected graph with n vertices and without any circuit has
𝑛 − 1 edges. We also know that a graph with n vertices which has no circuit and has
𝑛 − 1.edges is always connected(i.e., it is a tree). Hence the theorem
Pendant Vertices in a Tree


Pendant vertex is defined as a vertex of degree one
Concept of pendant vertex can be used in computer programmes as data tree
Theorem 3.6:
In any tree (with two or more vertices), there are atleast two pendant vertices
Proof:




44
Pendant vertices are vertex of degree one. For a tree of n vertices we have n-1 edges
and hence 2(𝑛 − 1) degrees to be divided among n vertices
Since no vertex can be of zero degree, we must have atleast two vertices of degree
one in a tree.
Hence the theorem.
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Distance and Centres in a Tree

Distance, d(vi , vj) between any two vertices vi and vj is the length of the shortest path
(i.e., number of edgesin the shortest path) between them.
Metric:
(Note:Funtion f(x,y) is used to compute the distance between two vertices x and y)
A function that satisifies the following three conditions is called a metric
 Nonnegativity: 𝑓 𝑥, 𝑦 > 0 𝑎𝑛𝑑 𝑓 𝑥, 𝑦 = 0 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑥 = 𝑦
 Symmetry: 𝑓 𝑥, 𝑦 = 𝑓 𝑦, 𝑥
 Triangle inequality: 𝑓 𝑥, 𝑦 ≤ 𝑓 𝑥, 𝑧 + 𝑓 𝑧, 𝑦 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑧
The distance between the vertices of a connected graph is a metric
Proof:



Distance d(vi , vj) between two vertices of a connected graph satisfies nonnegativity
and symmetry
Since d(vi , vj) is the length of the shortest path between vertices vi and vj, this path
cannot be longer than another path between vi and vj, which goes through a specified
vertex vk. Hence 𝒅 𝒗𝒊 , 𝒗𝒋 ≤ 𝒅 𝒗𝒊 , 𝒗𝒌 + 𝒅 𝒗𝒌 , 𝒗𝒋
Therfore the distance between the vertices of a connected graph is a metric
Eccentricity of a vertex 𝑬(𝒗):
The eccentricity of a vertex 𝑬(𝒗) in a graph G is the distance from 𝒗 to the vertex
farthest from 𝒗 in G
𝑬 𝒗 = 𝒎𝒂𝒙𝒗𝒊,,𝒗 𝒅(𝒗, 𝒗𝒊 )

Centre of G: A vertex with minimum eccentricity in graph G is called a centre of G
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Theorem 3.7:
Every tree has either one or two centres
Proof:





The maximum distance,𝑚𝑎𝑥 𝑑 𝑣𝑖 , 𝑣 from a given vertex 𝑣 to any other vertex 𝑣𝑖
occurs only when 𝑣𝑖 is a pendant vertex.
With this observation, let us startwith a tree T having more than one two vertices.
Tree T must have two or more pendant vertices.
Delete all the pendant vertices from T. The resulting graph 𝑇 ′ is still a tree.. Removal
of all pendant vertices from Tuniformly reduced the eccentricity of the remaining
vertices (i.e., vertices of 𝑇 ′ ) by one.
Therefore all vertices that T had as centers will still remain centres in 𝑇 ′ . From 𝑇 ′ we
can again remove all pendant vertices and get another tree 𝑇 ′ ′
We continue this process until there is left either a vertex (which is the centre of T) or
an edge (whose end vertices are the two centres of T). Thus the theorem.
If a tree has two centres , the two centres must be adjacent
Radius and Diameter of a Tree:

The eccentricity of a centre in a tree is defined as the radius of the tree. The diameter
of a tree T is defined as the length of the longest path in T
Rooted and Binary Trees
Rooted Trees:


A tree in which one vertex (called the root) is distinguished from all the other is
called a rooted tree. In a diagram of a rooted tree, the root is generally marked
distinctly.
All rooted trees with four vertices are shown in figure where the roots are enclosed in
a small triangle
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Binary Trees:



A binary tree is defined as a tree in which there is exactly one vertex (root) is of
degree two, and each of the remaining vertices is of degree one or three.
Every binary tree is a rooted tree since the vertex of degree two is distinct from all
other vertices, this vertex serves as a root.
Most straightforward application of binary tree is in search procedures
Properties of Binary Tree

The number of vertices n in a binary tree is always odd
(Proof: This is because there is exactly one vertex of even degree, and the
remaining n-1 vertices are of odd degree. Since the number of vertices of odd
degree is even in a graph, n-1 is even. Hence n is odd)

Number of pendant vertices in a binary tree T is 𝑝 =
𝑛 +1
2
Proof:
 Let p be the number of pendant vertices in a binary tree T.
Number of vertices of degree three = 𝑛 − 𝑝 − 1.
Number of edges in T
=
1
𝑝+3 𝑛−𝑝−1 +2
𝑛+1
=
2
2
Internal Vertex

A non pendant vertex in a tree is called an internal vertex of a tree
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The number of internal vertices in a binary tree is one less than the number of pendant
vertices
Level of a vertex in a tree



In a binary tree a vertex vi is said to be at level li if vi is at a distance of li from the root
Root is at level 0
A 13 vertex , four level binary tree is shown in figure. The number of vertices at
levels 1, 2, 3 and 4 are 2, 2, 4 and 4 respectively

Maximum number of vertices in a k-level binary tree is
20 + 21 + 22 + ⋯ + 2𝑘 ≥ 𝑛
Height of the tree


The maximum level, lmax, of any vertex in a binary tree is called the height of the tree
Minimum
possible
height
of
an
n
vertex
binary tree
is
𝑚𝑖𝑛 𝑙𝑚𝑎𝑥 = log 2 𝑛 + 1 − 1

Maximum possible height of an n vertex binary tree is 𝑚𝑎𝑥 𝑙𝑚𝑎𝑥 =
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𝑛 −1
2
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Construct a binary tree of given n such that the farthest vertex is as far as possible from
the root, we have exactly two vertices at each level except at level 0
Question: Construct a binary tree for 7 nodes such that the farthest vertex is as far as possible
from the root(binary tree with max height) and find the height of the tree
Question: Construct a binary tree with 15 nodes such that the tree must have a minimum
height. Also find the height of the tree
Path length (External path length) of a tree



It can be defined as the sum of the path lengths from root to all pendant vertices (or
sum of the levels of all pendant vertices)
Path length related to the execution time of an algorithm
The path length of the binary tree in the following figure is 1 + 3 + 3 + 4 + 4 + 4 +
4 = 23
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.
Qn: Find the path length of trees for the following figure:
𝒑𝒂𝒕𝒉 𝒍𝒆𝒏𝒈𝒕𝒉 = 𝟐 + 𝟐 + 𝟑 + 𝟑 + 𝟑 + 𝟑 = 𝟏𝟔
𝒑𝒂𝒕𝒉 𝒍𝒆𝒏𝒈𝒕𝒉 = 𝟏 + 𝟐 + 𝟑 + 𝟒 = 𝟓 + 𝟓 = 𝟐𝟎
On counting trees


Arthur Cayley used a connected graph to represent and count the structural isomers
of the saturated hydrocarbons𝐶𝑘 𝐻2𝑘 +2 . Corresponding to their chemical valencies, a
carbon atom was represented by a vertex of degree four and a hydrogen atom by a
vertex of degree one
The total number of vertices in such a graph is 𝑛 = 3𝑘 + 2
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And total number of edges is
1
𝑒 = (𝑠𝑢𝑚 𝑜𝑓 𝑑𝑒𝑔𝑟𝑒𝑒𝑠)
2
1
= 4𝑘 + 2𝑘 + 2
2
= 3𝑘 + 1
Since the graph is connected and the number of edges is one less than number of
vertices, it is a tree. Thus the problem of counting structural isomers of a given
hydrocarbon becomesthe problem of counting trees
Labeled Graph

A graph in which each vertex is assigned a unique label or name (no two vertices have
the same label) is called labeled graph
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Prufer Code

We can represent tree using prufer code or prufer sequence
Algorithm for generating Prufer sequence
1. Take any tree, 𝑇 ∈ 𝑇𝑛 whose vertices are labelled from 1 to n
2. Take the vertex with the smallest label whose degree is equal to 1, delete it from the
tree and write down the value of its only neighbour
3. Repeat the process with the new, smaller tree. Continue until only one edge remains
Algorithm for constructing tree using prufer sequence(Decoding)
1. Find the smallest number from 1 to n that is not in the sequence P and attach the
vertex with that number to the vertex with the first number in P.(n=2+number of
elements in P)
2. Remove the first number of P from the sequence. Repeat this process considering
only the numbers whose vertex have not yet attained their correct degree
3. Do this until there are no numbers left in P. Remember to attach last number in P to
vertex n
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Cayley’s Theorem: The number of labeled trees with n vertices 𝒏 ≥ 𝟐 𝒊𝒔 𝒏𝒏−𝟐
Proof:
Algorithm for generating Prufer sequence
1. Take any tree, 𝑇 ∈ 𝑇𝑛 whose vertices are labelled from 1 to n
2. Take the vertex with the smallest label whose degree is equal to 1, delete it from the
tree and write down the value of its only neighbour
3. Repeat the process with the new, smaller tree. Continue until only one edge remains
Spanning Trees




A tree T is said to be a spanning tree of a connected graph G if Tis a sub graph of G
and T contains all vertices of G
Spanning tree is also referred to as a skeleton or scaffolding of G
Since spanning tree are the largest trees(with maximum number of edges) among all
trees in G, it is also called maximal tree subgraph or maximal tree of G
A disconnected graph with k components has a spanning forest consists of k
spanning tree.(Forest: Collection of trees)
Theorem: Every connected graph has atleast one spanning tree
Proof:


If G has no circuit, it is its own spanning tree
If G has a circuit , delete an edge from the circuit. This will still leave the graph
connected.
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If there are more circuits, repeat the operation till an edge from the last circuit is
deleted- leaving a connected, circuit free graph that contains all the vertices of G.
Thus we have Every connected graph has atleast one spanning tree.
Branch and Chord of a spanning tree


An edge in a spanning tree T is called a branch of T
An edge of G that is not in a given spanning tree T is called a chord of T
Theorem: With respect to any of the spanning trees, a connected graph of n vertices and
e edges has n-1 branches and e-n+1 chords
Rank and Nullity of a graph






The rank and nullity of a graph G with n verticres and e edges is defined as
𝑟𝑎𝑛𝑘,
𝑟 =𝑛−𝑘
𝑛𝑢𝑙𝑙𝑖𝑡𝑦,
𝜇 =𝑒−𝑛+𝑘
For a connected graph or a spanning tree, k=1. Therefore
𝐫𝐚𝐧𝐤 𝐨𝐟 𝐆 = 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐛𝐫𝐚𝐧𝐜𝐡𝐞𝐬 𝐢𝐧 𝐚𝐧𝐲 𝐬𝐩𝐚𝐧𝐧𝐢𝐧𝐠 𝐭𝐫𝐞𝐞 𝐨𝐟 𝐆 = 𝑛 − 1
𝐧𝐮𝐥𝐥𝐢𝐭𝐲 𝐨𝐟 𝐆 = 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐜𝐡𝐨𝐫𝐝𝐬 = 𝑒 − 𝑛 + 1
𝐫𝐚𝐧𝐤 + 𝐧𝐮𝐥𝐥𝐢𝐭𝐲 = 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐞𝐝𝐠𝐞𝐬 𝐢𝐧 𝐆
Nullity of a graph is also referred to as its cyclomatic number or first Betti number
Fundamental Circuits

A circuit formed by adding a chord to a spanning tree is called a fundamnetal circuit
Theorem: A connected graph G is a tree if and only if adding an edge between any two
vertices in G creates exactly one circuit
Proof:
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HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
If we add an edge between any two vertices of a tree G, a circuit is created. This is
because there already exists one path between any two vertices of a tree; adding an
edge between them creates an additional path , and hence a circuit.
let us now consider a spanning tree T in a connected graph G. Adding any one chord
to T will create exactly one circuit.(Fundamental circuit)
Therefore A connected graph G is a tree if and only if adding an edge between
any two vertices in G creates exactly one circuit.
Finding all spanning tree of a graph

We can generate one spanning tree from another, through addition of chords and
deletion of an appropriate branch, is called a cyclic interchange or elementary tree
transformation
Distance between two spanning trees



Distance between two spanning trees Ti and Tj of a graph G is defined as the number
of edges of G present in one tree but not in other.
Ring sum of two spanning trees Ti and Tj of G is the sub graph of G containing all
edges of G that are either in Ti or Tj but not in both
𝟏
𝒅 𝑻𝒊 , 𝑻𝒋 = 𝑵(𝑻𝒊 ⨁𝑻𝒋 )
𝟐
𝒘𝒉𝒆𝒓𝒆 𝑵 𝒈 𝒅𝒆𝒏𝒐𝒕𝒆𝒔 𝒕𝒉𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒆𝒅𝒈𝒆𝒔 𝒊𝒏 𝒈𝒓𝒂𝒑𝒉 𝒈
𝒅 𝑻𝒊 , 𝑻𝒋 is the minimumnumber of cyclic interchanges involved in going from 𝑻𝒊 to
𝑻𝒋
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Theorem:The distance between the spanning tree of a graph is metric. That is it satisfies
 𝒅 𝑻𝒊 , 𝑻𝒋 ≥ 𝟎 𝒂𝒏𝒅 𝒅 𝑻𝒊 , 𝑻𝒋 = 𝟎 𝒊𝒇 𝒂𝒏𝒅 𝒐𝒏𝒍𝒚 𝒊𝒇 𝑻𝒊 = 𝑻𝒋
 𝒅 𝑻𝒊 , 𝑻𝒋 = 𝒅 𝑻𝒋 , 𝑻𝒊
 𝒅 𝑻𝒊 , 𝑻𝒋 ≤ 𝒅 𝑻𝒊 , 𝑻𝒌 + 𝒅 𝑻𝒌 , 𝑻𝒋
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Points to Remember
A tree is a connected graph without any circuits.
There is one and only one path between every pair of vertices in a tree, T.
A tree with n vertices have n-1 edges
A graph is a tree if and only if it is minimally connected
A graph with n vertices, n-1 edges, and no circuit is connected
Pendant vertex is defined as a vertex of degree one
In any tree (with two or more vertices), there are atleast two pendant vertices
Distance, d(vi , vj) between any two vertices vi and vj is the length of the shortest path
(i.e., number of edgesin the shortest path) between them.
Metric: A function that satisifies the following three conditions is called a metric
 Nonnegativity: 𝑓 𝑥, 𝑦 > 0 𝑎𝑛𝑑 𝑓 𝑥, 𝑦 = 0 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑥 = 𝑦
 Symmetry: 𝑓 𝑥, 𝑦 = 𝑓 𝑦, 𝑥
 Triangle inequality: 𝑓 𝑥, 𝑦 ≤ 𝑓 𝑥, 𝑧 + 𝑓 𝑧, 𝑦 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑧
The distance between the vertices of a connected graph is a metric
The eccentricity of a vertex 𝑬(𝒗) in a graph G is the distance from 𝒗 to the vertex
farthest from 𝒗 in G
Centre of G: A vertex with minimum eccentricity in graph G is called a centre of G
Every tree has either one or two centres
If a tree has two centres , the two centres must be adjacent
The eccentricity of a centre in a tree is defined as the radius of the tree. The diameter
of a tree T is defined as the length of the longest path in T
A tree in which one vertex (called the root) is distinguished from all the other is
called a rooted tree.
A binary tree is defined as a tree in which there is exactly one vertex (root) is of
degree two, and each of the remaining vertices is of degree one or three.
The number of vertices n in a binary tree is always odd
𝑛 +1
Number of pendant vertices in a binary tree T is 𝑝 = 2
A non pendant vertex in a tree is called an internal vertex of a tree
In a binary tree a vertex vi is said to be at level li if vi is at a distance of li from the root
Root is at level 0
The maximum level, lmax, of any vertex in a binary tree is called the height of the tree
Path length can be defined as the sum of the path lengths from root to all pendant
vertices (or sum of the levels of all pendant vertices)
A tree T is said to be a spanning tree of a connected graph G if Tis a sub graph of G
and T contains all vertices of G
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PROBLEMS
1. Cite three different situations that can be represented by trees.
2. Show a tree in which its diameter is not equal to twice the radius. Under what
condition does this inequality hold? Elaborate.
3. Suppose you are given eight coins and are told that seven of them are of equal
weight, and one coin is either heavier
or lighter than the rest. You are provided
with an equal arm balance, which you may use only three times, for comparing coins.
Sketch a strategy in the form of decision trees for identifying the nonconforming coin,
as well as for finding out whether it is heavier or lighter than the rest.
4. Show that a path is its own spanning tree
5. Prove that a pendant edge in a connected graph G is conatine din every spanning tree
of G
6. What is the nullity of a complete graph of n vertices?
PREVIOUS YEAR UNIVERSITY QUESTIONS
1. List down any two properties of trees and also prove the theorem: A graph is a
tree if and only if it is a minimally connected
2. Define spanning tree and distance between two spanning trees.
3. Consider the tree T, given below
Label the vertices of T appropriately and find the center and diameter of T
4. What is eccentricity of a node? How is it used in finding the center of a graph?
5. Let G = (V,E) be a connected graph, and let T = (V, S) be a spanning tree of G. Let e = (a, b)
be an edge of G not in T. Prove that, for any edge f on the path from a to b in T, (V, (S∪{e})
−{f}) is another spanning tree for G
6. Define spanning trees. Consider the graph G given below and obtain any three
spanning trees from G. Calculate the number of distinct spanning trees possible from
a complete graph with n vertices.
7. Define rank and nullity of a graph G
8. Define a Metric and prove that distance is a metric
9. Explain Center and radius of a tree
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10. Prove that a tree with 𝑛 vertices has 𝑛 − 1 edges
11. Show that the distance between vertices of a connected graph is a metric.
12. Discuss the center of a tree with suitable example.
13. Write notes on the fundamental circuit.
14. Prove that in a tree T(V,E),|V|=|E|+1.
15. Define spanning tree with example.
16. Explain the difference between a general tree and a binary tree
17. Show a tree in which its diameter is not equal to twice the radius. Under what
conditions does this inequality hold? Elaborate.
18. Prove that every connected graph has at least one spanning tree
19. Prove that every tree has either one or two centers
20. Prove that an arborescence is a tree in which every vertex other than the root has an
in-degree of exactly one
21. Show that the number of labeled trees with n vertices n n-2
22. Define spanning tree. Show that the distance between the spanning tree of a graph is a
metric
1.
2.
3.
4.
5.
6.
7.
WORK SHEET
Write notes on the fundamental circuit.
Prove that in a tree T(V,E),|V|=|E|+1.
Define spanning tree with example.
Show that the distance between vertices of a connected graph is a metric.
Discuss the center of a tree with suitable example.
List down any two properties of trees and also prove the theorem: A graph is a tree if
and only if it is a minimally connected.
Consider the tree T, given below
Label the vertices of T appropriately and find the center and diameter of T.
8. Define spanning trees. Consider the graph G given below and obtain any three
spanning trees from G. Calculate the number of distinct spanning trees possible from
a complete graph with n vertices.
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9. Let G = (V,E) be a connected graph, and let T = (V, S) be a spanning tree of G. Let e = (a, b)
be an edge of G not in T. Prove that, for any edge f on the path from a to b in T, (V, (S∪{e})
−{f}) is another spanning tree for G
10. Find all spanning trees in the following graph:
11. Draw two different binary tree with 5 vertices having maximum number of leaves
12. A tree has five vertices of degree 2, three vertices of degree 3 and four vertices of
degree 4. How many vertices of degree one does it have?
13. Define binary trees. Draw binary trees with 11 vertices:
a. having maximum height
b. with minimum height
14. How many labeled trees can be constructed using 4 vertices? Show all the labeled
trees with four vertices
15. Sketch all (unlabeled) binary trees with six pendant vertices. Find the path length of
each.
[Hint: Distribute the 11 vertices (because n=6+5) among different levels. Observe
that at level 0 has exactly one vertex, level 1 has exactl two vertices; level 2 can have
either two or four vertices; and so on. There are six such trees, and two of them are
shown in the following figure.]
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16. Draw all trees of n labeled vertices for n=1, 2, 3, 4 and 5
17. Draw all trees of n unlabeled vertices for n=1, 2, 3, 4 and 5
18. Draw all unlabeled rooted trees of n vertices for n=1, 2, 3, 4 and 5
19. It can be shown that there are only six different trees of six vertices. Two such trees
are given in the figure. Draw the other four.
20. Sketch all spanning trees of the graphs in the figure
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MODULE IV
Cut Set and Cut Vertices
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In a coonected graph G, a cut-set a set of edges whose removal from G leaves G
disconnected, provided removal of no proper subset of these edges disconnects G
Cut-set is also referred to as minimal cut-set, proper cut-set, simple cut-set or cocycle
From the figure, 𝑎, 𝑐, 𝑑, 𝑓 , 𝑎, 𝑏, 𝑔 , 𝑎, 𝑏, 𝑒, 𝑓 , 𝑑, 𝑕, 𝑓 , 𝑘 are set of cut-sets. But ,
𝑎, 𝑐, 𝑕, 𝑑 is not a cut-set because one of its proper subsets , 𝑎, 𝑐, 𝑕 is a cut-set
A cut-set always “cuts” a graph into two. Therefore , a cut-set can also be defined as
a minimal set of edges in a connected graph whose removal reduces the rank of the
graph by one.
From the figure , the rank of the graph in figure (b) is four one less than that of the
graph in figure (a)
If we partition all the vertices of a connected graph G into two mutually exclusive
subsets , a cut-set is a minimal number of edges whose removal from G destroys all
paths between these two sets of vertices
From the above figure, cut set , 𝑎, 𝑐, 𝑑, 𝑓 connects vertex set , 𝑣1 , 𝑣2 , 𝑣6 with
𝑣3 , 𝑣4 , 𝑣5
Since removal of any edge from a tree breaks the tree into two parts, every edge
of a tree is a cut-set
Properties of Cut-Set
Theorem:
Every cut-set in a connected graph G must contain at least one branch of every
spanning tree of G
Proof:
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In a given connected graph G, let Q be a minimal set of edges containing atleast one
branch of every spanning tree of G. Consider G-Q , the sub graph that remains after
removing the edges in Q from G.
Since the sub graph G-Q contains no spanning tree of G, G-Q is disconnected.
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Also, since Q is a minimal set of edges with this property, any edge e from Q returned
to G-Q will create at least one spanning tree. Thus the sub graph G-Q+e will be a
connected graph.
Therefore, Q is a minimal set of edges whose removal from G disconnects G. This, by
definition , is a cut set. Hence the theorem.
Theorem:
In a connected graph G, any minimal set of edges containing at least one branch of
every spanning tree of G is a cut set
Theorem:
Every circuit has even number of edges in common with any cut set
Proof:
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Consider a cut set S in a graph G. Let the removal of S partition the vertices of G into
two (mutually exclusive or disjoint)subsets V1 and V2.
Consider a circuit Γ in G. If all the vertices in Γ are entirely within vertex set V1(or
V2), the number of edges common to S and Γ is zero; that is 𝑁 𝑆 ∩ Γ = 0, an even
number
If, on the other hand, some vertices in Γ are in V1 and some in V2, we traverse back
and forth between the sets V1 and V2 as we traverse the circuit. Because of the closed
nature of a circuit, the number of edges we traverse between V 1 and V2 must be even.
And since every edge in S has one end in V1 and the other in V2, and no other edge in
G has this property (of separating sets V1 and V2), the number of edges common to S
and Γ is even.
Fundamental Cut Sets
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A cut set S containing exactly one branch of a tree T is called a Fundamental cut-set
with respect to T
Consider a spanning tree T of a connected graph G. Take any branch b in T. Since
𝑏 is a cut set in T, 𝑏 partitions all vertices of T into two disjoint sets- one at each
end of b
Consider the same partition of vertices in G, and the cut set S in G that corresponds to
this partition. Cut set S will contain only one branch b of T, and the rest of the edges
in S are chords with respect to T.
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
Such a cut set is called Fundamental cut set or basic cut set.
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In the above figure, a spanning tree T and all five of the fundamental cut-sets with
respect to T are shown(broken lines “cutting” through each cut set).
The ring sum of any two cut sets in a graph is either a third cut set or an edge
disjoint union of cut sets
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Fundamental Circuits and Cut Sets
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Consider a spanning tree T in a given connected graph G.. Let ci be a chord with
respect to T, and let the fundamental circuit made by ci be called Γ, consisting of k
branches 𝑏1, 𝑏2, … , 𝑏𝑘 , in addition to the chord ci; that is
𝚪 = 𝒄𝒊 , 𝒃𝟏, 𝒃𝟐, … , 𝒃𝒌, is a fundamental circuit with respect to T.
Every branch of any spanning tree has a fundamental cut set associated with it. Let S i
be the fundamental cut-set associated with bi, consisting of q chords in addition to the
branch b1; that is
𝐒𝐢 = 𝒃𝟏 , 𝒄𝟏, 𝒄𝟐, … , 𝒄𝒒, is a fundamental cut-set with respect to T
With respect to a given spanning tree T, a chord ci that determines a fundamental
circuit 𝚪 occurs in every fundamental cut-set associated with the branches in 𝚪 and in
no other.
With respect to a given spanning tree T, a branch bi that determines a fundamental
cut-set S is contained in every fundamental circuit associated with the chords in S and
in no others.
Edge Connectivity
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The number of edges in the smallest cut set (i.e., cut set with fewest number of edges)
is defined as the edge connectivity of G
Edge connectivity of a connected graph can be defined as the minimum number of
edges whose removal (i.e., deletion ) reduces the rank of the graph by one.
The edge connectivity of a tree is one
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Vertex Connectivity
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Vertex connectivity of a connected graph G is defined as the minimum number of
vertices whose removal from G leaves the remaining graph disconnected.
Vertex connectivity of a tree is one
`
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Separable Graph:
 Aconnected graph is said to be separable if its vertex connectivity is one
 All other connected graphs are called nonseparable
 In a separable graph a vertex whose removal disconnects the graph is called a cutvertex, a cut node or an articulation point
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The maximum vertex connectivity one can achieve with a graph G of n vetices and e
edges (𝑒 ≥ 𝑛 − 1) is the integral part of the number 2𝑒 𝑛; that is 2𝑒 𝑛
Every cut-set in a non separable graph with more than two vertices contains atleast
two edges
A vertex v in a connected graph G is a cut-vertex if and only if there exist two vertices
x and y in G such that every path between x and y passes through v
Theorem:
The edge connectivity of a graph G cannot exceed the degree of the vertex with the
smallest degree in G
Proof:
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Let vertex vi be the vertex with the smallest degree in G. Let 𝑑(𝑣𝑖 ) be the degree of vi.
Vertex vi can be separated from G by removing the 𝑑(𝑣𝑖 ) edges incident on vertex vi.
Hence the theorem.
Theorem:
The vertex connectivity of any graph G can never exceed the edge connectivity of G
Proof:
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Let 𝛼 denote the edge connectivity of G. Therefore, there exists a cut-set S in G with
𝛼 edges. Let S partition the vertices of G into subsets V1 and V2.
By removing atmost 𝛼 vertices from V1 or V2 on which the edges in S are incident,
we can effect the removal of S(together with all other edges incident on these
vertices) from G. Hence the theorem.
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1-Isomorphism
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A separable graph consists of two or more nonseparable subgraphs. Each of the
largest nonseparable subgraphs is called a block.
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Two graphs G1 and G2 are said to be 1-isomorphic if they become isomorphic to
each other under the following operations
o “Split” a cut-vertex into two vertices to produce two disjoint sub graphs
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If G1 and G2 are two 1-isomorphic graphs, the rank of G1 equals the rank of G2 and
the nullity of G1 equals the nullity of G2
2-Isomorphism
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2-Connected Graph: Graph with vertex connectivity of two
In a 2-connected graph G let vertices x and y be a pair of vertices whose removal
from G will leave the remaining graph disconnected. Suppose that we perform the
following operation 2 on G:
Operation 2: “Split” the vertex x into x1 and x2 and the vertex y into y1 and y2 such
that G split into g1 and g2 . Let vertices x1 and y1 go with g1 and x2 and y2 with g2.
Now rejoin the graphs g1 and g2 by merging x1 with y2 and x2 with y1.
Operation 1: “Split” a cut-vertex into two vertices to produce two disjoint sub graphs
Two graphs are said to be 2-ismorphic if they become isomorphic after undergoing
operation 1 or operation 2 or both operation any number of times.
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Circuit Correspondance:
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Two graphs G1 and G2 are said to have circuit correspondence if they meet the
following condition:
There is one-to-one correspondence between the edges of G1 and G2 and a one to
one correspondence between the circuits of G1 and G2, such that a circuit in G1
formed by certain edges of G1 has a corresponding circuit in G2 formed by the
corresponding edges of G2, and vice cersa.
Isomorphic graphs have circuit correspondence
2- isomorphic graph also have circuit correspondance
Planar Graph
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A graph G is said to be planar if there exists some geometric representation of G
which can be drawn on a plane such that no two of its edges intersect.
The geometric represnttaion of planar graph is also called embedding. An embedding
of a planar graph G on a plane is called a plane representation of G
A graph that cannot be drawn on a plane without a crossover between edges is called
non planar
Kuratowski’s Graphs
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Two specific non planar graphs
These are called Kuratowski’s graphs, after the Polish mathematician Kasimir
Kuratowski, who discovered their unique property
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Kuratowski’s Graphs are
 A complete graph with five vertices
 Regular connected graph with six vertices and nine edges
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Two common geometric representation of Kuratowski’s second graph is shown in the
above figure
Properties of Kuratowski’s Graphs
1)
2)
3)
4)
Both are regular graphs
Both are non planar
Removal of one edge or a vertex makes each a planar graph
Kuratowski’s first graph is the non planar graph with the smallest number of
vertices and Kuratowski’s second graph is the non planar graph with the smallest
number of edges. Thus both are the simplest non planar graphs
Different Representation of Planar Graphs
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Straight line representation
Plane representation
Embedding on a sphere
Straight Line Representation
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Any simple planar graph can be embedded in a plane such that every edge is drawn as
a straight line segment
This representation is only used for simple graph because self loop and parallel edges
cannot be represented by straight line
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Plane Representation
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A plane representation of a graph divides the plane into regions (also called windows,
faces, or meshes). A region is characterized by the set of edges (or the set of vertices)
forming its boundary.
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Infinite Region: The portion of the plane lying outside a graph embedded in a plane.
Sucha region is called the infinite, unbounded, outer, or exterior region for that
particular plane representation.
Embedding on a Sphere
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To eliminate the distinction between finite and infinite region , a planar graph is often
embedded in the surface of a sphere. It is accomplished by stereographic projection
of a sphere on the plane
Stereographic Projection
 Put the sphere on the plane and call the point of contact SP(south pole). At point
SP, draw a straight line perpendicular to the plane and let the point where line
intersects the surface of the sphere be called NP( north pole)
 Now corresponding to any point p on the plane, there exists a unique point p’ on
the sphere and vice versa, where p’ is the point at which the straight line from
point p to point NP intersects the surface of the sphere.
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 Thus there is a one-to-one correspondence between the points of the sphere and
the finite points on the plane, and points at infinity in the plane correspond to the
point NP on the sphere.
Theorem:
A graph can be embedded in the surface of a sphere if and only if it can be embedded in
a plane
Proof:
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A planar graph embedded in the surface of a sphere divides the surface into different
regions.
Each region on the sphere is finite, the infinite region on the plane having been
mapped onto the region containing the point NP.
Now it is clear that by suitably rotating the sphere we can make any specified region
map onto the infinite region on the plane.
Euler’s Theorem
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Euler’s formula gives the number of regions in any planar graph
Theorem:
A connected planar graph with n vertices and e edges has e-n+2 regions
Proof:
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We can disregard a self loop or pararllel edges because it simply adds one region to
the graph and simultaneously increases the value of e by one
We can also disregard all edges that do not form boundaries of any region.
Adding of any such edge increases( or decreases) e by one and increases (or
decreases) n by one, keeping the quanitiy e-n unaltered
Since any simple planar graph can have a plane representation such that each edge is a
straight line, any planar graph can be drawn such that each region is a polygon(a
polygonal net)
Let the polygonal net representing the given graph consist of f regions or faces, and
let 𝑘𝑝 be the number of p-sided regions.
Since each edge is on the boundary of exactly two regions,
3. 𝑘3 + 4. 𝑘4 + 5. 𝑘5 + … + 𝑟. 𝑘𝑟 = 2. 𝑒
𝐸𝑞𝑛 1
where 𝑘𝑟 is the number of polygons with maximum edges.
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Also,
𝑘3 + 𝑘4 + 𝑘5 + ⋯ + 𝑘𝑟 = 𝑓
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𝐸𝑞𝑛(2)
Sum of angles subtended at each vertex in the polygonal net = 2𝜋𝑛 𝐸𝑞𝑛(3)
Sum of all interior angles of a p-sided polygon = 𝜋 𝑝 − 2
Sum of exterior angles of a p-sided polygon = 𝜋 𝑝 + 2
𝐸𝑞𝑛 3 𝑏𝑒𝑐𝑜𝑚𝑒𝑠
Sum of angles subtended at each vertex in the polygonal net = Sum of all interior
angles+ Sum of exterior angles of a p-sided
polygon
= 𝜋 3 − 2 . 𝑘3 + 𝜋 4 − 2 . 𝑘4 + 𝜋 5 − 2 . 𝑘5 + … + 𝜋 𝑟 − 2 . 𝑘𝑟 + 4𝜋
= 𝜋 2𝑒 − 2𝑓 + 4𝜋 𝐸𝑞𝑛(4)

𝐹𝑟𝑜𝑚 𝐸𝑞𝑛 3 𝑎𝑛𝑑 𝐸𝑞𝑛 4
𝜋 2𝑒 − 2𝑓 + 4𝜋 = 2𝜋𝑛
2𝜋 𝑒 − 𝑓 + 2 = 2𝜋𝑛
𝑒−𝑓+2 =𝑛

Therefore number of regions is
𝑓 =𝑒−𝑛+2
In any simple graph, connected planar graph with f regions, n vertices, and e edges
𝒆 > 2 , the following inequality holds:
𝟑
𝒆≥ 𝒇,
𝟐
𝒆 ≤ 𝟑𝒏 − 𝟔
Proof:

Since each region is bounded by at least three edges and each edge belongs to exactly
two regions,
2𝑒 ≥ 3𝑓 ,
𝑒≥

3
𝑓
2
Substituting for f from Euler’s formula in in the above inequality
𝑒≥
3
𝑒−𝑛+2
2
𝑒 ≤ 3𝑛 − 6
The inequality 𝒆 ≤ 𝟑𝒏 − 𝟔 is often useful in finding out of a graph is non planar
Question:
Check whether the graph 𝑲𝟓 , a complete graph of five vertices is planar
Ans:
𝒏 = 𝟓, 𝒆 = 𝟏𝟎, 𝟑𝒏 − 𝟔 = 𝟗 < 𝑒
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Thus the graph violates the inequality, and hence it is not planar
Prove that Kuratowski’s first graph is non planar
Ans:
 Kuratowski’s first graph is a complete graph with five vertices
𝒏 = 𝟓, 𝒆 = 𝟏𝟎, 𝟑𝒏 − 𝟔 = 𝟗 < 𝑒
Thus the graph violates the inequality, and hence it is not planar
Prove that Kuratowski’s second graph is non planar
Ans:
 Kuratowski’s second graph is a regular connected graph with 6 vertices and 9
edges.
 To prove the non planarity of Kuratowski’s second graph , we can make use of
the additional fact that no region in this graph can be bounded with fewer than
four edges.
 Hence, if this graph were planar, we would have

𝟐𝒆 ≥ 𝟒𝒇,
And substituting for f from Euler’s formula
𝟐𝒆 ≥ 𝟒 𝒆 − 𝒏 + 𝟐
𝟐×𝟗 ≥ 𝟒 𝟗−𝟔+𝟐
𝟏𝟖 ≥ 𝟐𝟎
Thus the graph violates the inequality, and hence it is not planar
Unique Embedding
 If all possible embeddings on a sphere no two are distinct, the graph is said to have a
unique embedding on a sphere. The following figure shows unique embedding
Detection of Planarity

We can check the planarity of a graph through Elementary Reduction. It consists of the
following steps
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Step 1: Since a disconnected graph is planar if and only if each of its components is planar,
we need consider only one component at a time. Also a separable graph is planar
if and only if each of its block is planar. Therefore, for the given arbitrary graph
G, determine the set
𝐺 = 𝐺1 , 𝐺2 , … , 𝐺𝑘
𝑤𝑕𝑒𝑟𝑒 𝑒𝑎𝑐𝑕𝐺𝑖 𝑖𝑠 𝑎 𝑛𝑜𝑛 𝑠𝑒𝑝𝑎𝑟𝑎𝑏𝑙𝑒 𝑏𝑙𝑜𝑐𝑘 𝑜𝑓 𝐺
Thus we have to check each 𝐺𝑖 for planarity
Step 2: Elimination of a vertex of degree two by merging two edges in series doesnot affect
planarity. Therefore, eliminate all edges in series .
[In a graph two edges are said to be in series if they have exactly one vertex in
common and if this vertex is of degree two)
Step 3: Since addition or removal of self loops doesnot affect planarity, remove all self loops
Step 4: Since parallel edges also do not affect planarity, eliminate edges in paralle by
removing all but one edge between every pair of vertices
Repeated application of step2, 3 and 4 will reduce the graph into a new graph which has any
one of the following property:
1. Graph has a single edge or
2. A complete graph of four vertices
3. A nonseparable, simple graph with 𝒏 ≥ 𝟓 and 𝒆 ≥ 𝟕
For the first two cases, the graph is planar. For the third case we need to use Euler’s theorem
and planarity inequality for planarity detection
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Homomorphic Graphs

Two graphs are said to be homomorphic if one graph can be obtained from the other
by the creation of edges in series( i.e., by insertion of vertices of degree two) or by the
merger of edges in series.
Geometric Dual

Consider the plane representation of a graph in Fig (a), with six regions or faces
𝐹1 , 𝐹2 , 𝐹3 , 𝐹4 , 𝐹5 and 𝐹6 . Let us place six points 𝑝1 , 𝑝2 , … , 𝑝6 one in each of the regions.
Next let us join these six points according to the following procedure

If two regions 𝐹𝑖 and 𝐹𝑗 are adjacent(i.e., have a common edge), draw a line joining
points 𝑝𝑖 and 𝑝𝑗 that intersects the common edge between 𝐹𝑖 and 𝐹𝑗 exactly once.
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
If there is more than one edge common between 𝐹𝑖 and 𝐹𝑗 , draw one line between
points 𝑝𝑖 and 𝑝𝑗 for each of the common edges.

For an edge e lying entirely in one region, say 𝐹𝑘 , draw a self-loop at points 𝑝𝑘
intersecting e exactly once.
By this procedure we obtain a new graph 𝐺 ∗ (broken lines in the above figure)
consisting of six vertices 𝑝1 , 𝑝2 , … , 𝑝6 and edges joining these vertices. Such graph 𝐺 ∗
is called a dual or geometric dual of G

Relationship between G and its dual 𝑮∗
An edge forming a self loop in G yields a pendant edge in 𝑮∗
A pendant edge in G yields a self loop in 𝑮∗
Edges that are in series in G produce parallel edges in 𝑮∗
Paralle edges in G produce edges in series in 𝑮∗
Number of edges constituting the boundary of a region𝐹𝑖 in G is equal to the degree of
the corresponding 𝑝𝑖 in 𝑮∗ and vice versa
6. Graph 𝑮∗ is also embedded in a plane, so it is planar
7. G and 𝑮∗ are called self dual graphs
8. If 𝑛, 𝑒, 𝑓, 𝑟, 𝜇 denotes the number of vertices, edges, regions, rank and nullity of a
connected planar graph G and if 𝑛∗ , 𝑒 ∗ , 𝑓 ∗ . 𝑟 ∗ , 𝜇 ∗ are the corresponding numbers in
dual graph 𝑮∗ , then
𝑛∗ = 𝑓
𝑒∗ = 𝑒
𝑓∗ = 𝑛
𝑟∗ = 𝜇
𝜇∗ = 𝑟
1.
2.
3.
4.
5.
Combinatorial Dual


Two planar graphs 𝐺 and 𝑮∗ are said to be combinatorial duals of each other if there
is one to one correspondence between the edges in G and 𝑮∗ such that a set of edges
in 𝐺 forms a circuit if and only if the corresponding set in 𝑮∗ forms a cut set
Two planar graphs 𝐺 and 𝑮∗ are said to be combinatorial duals of each other if there
is one to one correspondence between the edges in 𝐺 and 𝑮∗ such that if g is a ny sub
graph of 𝐺1 and h is the corresponding subgraph in 𝑮∗ then
𝒓𝒂𝒏𝒌 𝒐𝒇 𝑮∗ − 𝒉 = 𝒓𝒂𝒏𝒌 𝒐𝒇𝑮∗ − 𝒏𝒖𝒍𝒍𝒊𝒕𝒚 𝒐𝒇 𝒈
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Theorem:
There is a one to one correspondence between the edges in G and G* such that a
set of edges in G forms a circuit if and only if the corresponding set in G*forms a
cut set
Proof:
 Let us consider planar representation of a planar graph. Let us also draw dual
G* of G. Then consider an arbitrary circuit Γ in G. Clearly, Γ will form some
closed simple curve in the plane representation of G- dividing the plane into
two areas
 Thus the vertices of G * are partitioned into two non empty, mutually exclusive
subsets- one inside Γ and the other outside
 That is set of edges Γ* in G* corresponding to the set Γ in G is a cut set in G *
 Likewise it is apparent that corresponding to a cut-set S* in G*, there is a
unique circuit consisting of the corresponding edge-set S in G such that S is a
circuit. This proves the theorem.
Theorem:
A graph has a dual if and only if it is planar
Proof:
 Use proof by contradiction. Inorder to prove the theorem, we have only to
prove that a nonplanar graph does not have a dual.
 Let G be a nonplanar graph. Then according to Kuratowski’s theorem, G
contains K5 or K3,3 or a graph homomorphic to either of these.
 Thus if we can show that neither K5 nor K3,3 has a dual, we have proved the
theorem. This we shall prove by contradiction as follows:
 Suppose that K3,3 has adual D. Observe that the cut-sets in K3,3 corresponding
to circuits in D and vice versa. Since K3,3 has no cut-set consisting of two
edges, D has no circuit consisting of two edges. That is , D contains no pair of
parallel edges,Since every circuit in K3,3 is of length four or six, D has no cutsets with less than four edges. Therefore, the degree of every vertex in Dis at
least four.As D has no parallel edges and the degree of every vertex is atleast
four, D must have atleast five verticeseach of degree four or more. That is D
must have atleast (5 × 4)/10 = 10 edges. This is a contradiction, because
K3,3 has 9 edges and so must its dual. Thus K 3,3 cannot have a dual
 Suppose that the graph K5 has a dual H. Note that K5 has
 10 edges
 no pair of parallel edges
 no cut-set with two nedges
 cut set with only four or six edges
 Consequently, graph H must have
 10 edges
 no vertex with degree less than three
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

 no pair of parallel edges
 circuits of length four and six only
Now graph H contains a hexagon(a circuit of length six) and no more than
three edges can be added to a hexagon without creating a circuit of length
three or a pair of parallel edges
Since both of these are forbidden in H and H has 10 edges, there must be
atleast seven vertices in H. the degree of each of these vertices is atleast three..
This leads to H having atleast 11 edges. a contradiction
Dual of Sub Graph





Let G be a planar graph and G* be its dual. Let a be an edge in G, and the
corresponding edge in G* be a*.
Suppose that we delete an edge a from G and then try to find the dual of G –a. If edge
a was on the boundary of two regions, removal of a would merge these two regions
into one.
Thus the dual of 𝐺 − 𝑎 * can be obtained from G * by deleting the corresponding
edge a* and then fusing the two end vertices of a* in G*- a*
If the edge a is not the boundary, a* forms a self loop.In that case G*- a* is the same
as 𝐺 − 𝑎 *
Thus if a graph G has a dual G*, the dual of any subgraph of G can be obtained
by successive application of this procedure
Self Dual Graphs

If a planar graph G is isomorphic to its own dual , it is called a self dual graphs
Thickness and Crossing of Planar graphs
Thickness:



Least number of planar subgraphs in G is called thickness of G
Thickness of planar graph is one
Thickness of Kuratowski’s graph is two
Crossings




Least number of crossing or intersectionsnecesssary inorder to draw the graph in the
plane
Crossing number of planar graph is zero
Crossing number of Kuratowski’s graph is one
Points to Remember
Cut Set: In a coonected graph G, a cut-set a set of edges whose removal from G
leaves G disconnected, provided removal of no proper subset of these edges
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


HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
disconnects G
Every edge of a tree is a cut-set
Every cut-set in a connected graph G must contain at least one branch of every
spanning tree of G
Every circuit has even number of edges in common with any cut set

A cut set S containing exactly one branch of a tree T is called a Fundamental cut-set
with respect to T
 The number of edges in the smallest cut set (i.e., cut set with fewest number of edges)
is defined as the edge connectivity of G
 The edge connectivity of a tree is one
 Vertex connectivity of a connected graph G is defined as the minimum number of
vertices whose removal from G leaves the remaining graph disconnected.
 Vertex connectivity of a tree is one
 Aconnected graph is said to be separable if its vertex connectivity is one. All other
connected graphs are called nonseparable
 In a separable graph a vertex whose removal disconnects the graph is called a cutvertex, a cut node or an articulation point
 Two types of Isomorphism: 1 isomorphism, 2-Isomorphism
 2-Connected Graph: Graph with vertex connectivity of two
 n-Connected Graph: Graph with vertex connectivity of n
 A graph G is said to be planar if there exists some geometric representation of G
which can be drawn on a plane such that no two of its edges intersect.
 The geometric represnttaion of plananr graph is also called embedding.
Kuratowski’s Graphs
 A complete graph with five vertices
 Regular connected graph with six vertices and nine edges
Different Representation of Planar Graphs
 Straight line representation
 Plane representation
 Embedding on a sphere
Euler’s Theorem
 A connected planar graph with n vertices and e edges has e-n+2 regions
 In any simple graph, connected planar graph with f regions, n vertices, and e
edges 𝒆 > 2 , the following inequality holds:
𝟑
𝒆≥ 𝒇,
𝟐
𝒆 ≤ 𝟑𝒏 − 𝟔
 If all possible embeddings on a sphere no two are distinct, the graph is said to have a
unique embedding on a sphere. The following figure shows unique embedding
Homomorphic Graphs
 Two graphs are said to be homomorphic if one graph can be obtained from the other
by the creation of edges in series( i.e., by insertion of vertices of degree two) or by the
merger of edges in series.
Combinatorial Dual
 Two planar graphs 𝐺 and 𝑮∗ are said to be combinatorial duals of each other if there
is one to one correspondence between the edges in G and 𝑮∗ such that a set of edges
in 𝐺 forms a circuit if and only if the corresponding set in 𝑮∗ forms a cut set
 Two planar graphs 𝐺 and 𝑮∗ are said to be combinatorial duals of each other if there
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is one to one correspondence between the edges in 𝐺 and 𝑮∗ such that if g is a ny sub
graph of 𝐺1 and h is the corresponding subgraph in 𝑮∗ then
𝒓𝒂𝒏𝒌 𝒐𝒇 𝑮∗ − 𝒉 = 𝒓𝒂𝒏𝒌 𝒐𝒇𝑮∗ − 𝒏𝒖𝒍𝒍𝒊𝒕𝒚 𝒐𝒇 𝒈
Self Dual Graphs
 If a planar graph G is isomorphic to its own dual , it is called a self dual graphs
Thickness:
 Least number of planar subgraphs in G is called thickness of G
 Thickness of planar graph is one
 Thickness of Kuratowski’s graph is two
Crossings
 Least number of crossing or intersectionsnecesssary inorder to draw the graph in the
plane
 Crossing number of planar graph is zero
 Crossing number of Kuratowski’s graph is one
PREVIOUS YEAR UNIVERSITY QUESTIONS
1. Explain self dual with an example
2. Prove the statement:
Every cut-set in a connected graph G must also contain at least one branch of every
spanning tree of G
3. List down the properties stating the relationship between the edges of graph G and its
dual G*
4. Define cut set. Find any four cut sets from the graph G given below and also find the
edge connectivity of G.
5. Define vertex connectivity and draw a graph with an articulation point.
6. State Euler’s Theorem (formula).
7. Draw two Kuratowski’s graphs and also prove that Kuratowsk’s first graph is non
planar using appropriate inequality.
8. Draw the geometric dual (G*) of the graph G given below and also check whether G
and G* are self dual or not, substantiate your answer clearly?
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9. Prove that the ring sum of any two cut-sets in a graph is either a third cut-set or an
edge-disjoint union of cut-sets.
10. Prove that a connected planar graph with n vertices and e edges has e-n+2 regions.
11. Consider the following graph G and any one of its spanning trees, T.List all
fundamental circuits and fundamental cut-sets with respect to T.
12. State and prove Euler’s theorem involving number of regions, edges and vertices in a
planar graph
13. A connected planar graph has 10 vertices each of degree 3. Into how many regions
does the representation of planar graph split the plane
14. Define the terms thickness and crossings with an example
15. Explain Planar Graph and show that K5 is not planar
16. Show that a graph can be embedded in the surface of a sphere if and only if it can be
embedded in a plane
17. Let G be a planar graph with𝑣 ≥ 3 vertices and 𝑒 edges. Then prove that 𝑒 ≤ 3𝑣 − 6
18. What are the relationship between a planar graph G and its dual G *?Find the
geometric dual of the following graph
9. Prove that a complete graph of five vertices is non planar
10. Explain the procedure to obtain the geometric dual of a given graph
11. Prove that every planar graph contain at least one vertex of degree ≤ 5
12. Prove that a graph has a dual if and only if it is planar
13. State and prove Euler Theorem
14. What do you mean by geometric dual of a graph? Illustrate with examples
15. A necessary and sufficient condition for two planar graphs G1 and G2 to be duals of
each other is as follows: There is one-to-one correspondence between the edges in G1
forms a circuit if and only if the corresponding set in G2 forms a cut –set. Prove it
geometrically. (Consider a graph with atleast 6 vertices and 6 faces)
16. Differentiate Geometric dual with Combinatorial dual
17. Prove that the complete graph K5 is non planar
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18. If a connected planar graph G has n vertices, e edges and r regions, then prove that
𝑛−𝑒+𝑟 = 2
19. Describe the methods used to determine the planarity of a graph
20. What is the necessary condition for a graph to have geometric dual? List the
characteristics of graphs that are geometric duals of each other. Give two examples of
graphs having geometric duals.
21. Using Euler’s formula show that Kwatowski’s two graphs are not planar
22. Prove that a graph has a dual if and only if it is planar
WORKSHEET
1. Explain self dual with an example
2. Prove the statement:
Every cut-set in a connected graph G must also contain at least one branch of every
spanning tree of G
3. List down the properties stating the relationship between the edges of graph G and its
dual G*
4. Define cut set. Find any four cut sets from the graph G given below and also find the
edge connectivity of G.
5. Define vertex connectivity and draw a graph with an articulation point.
6. State Euler’s Theorem (formula).
7. Draw two Kuratowski’s graphs and also prove that Kuratowsk’s first graph is non
planar using appropriate inequality.
8. Draw the geometric dual (G*) of the graph G given below and also check whether G
and G* are self dual or not, substantiate your answer clearly?
9. Prove that the ring sum of any two cut-sets in a graph is either a third cut-set or an
edge-disjoint union of cut-sets.
10. Prove that a connected planar graph with n vertices and e edges has e-n+2 regions.
11. Consider the following graph G and any one of its spanning trees, T.List all
fundamental circuits and fundamental cut-sets with respect to T.
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MODULE V
MATRIX REPRESENTATION OF GRAPHS
Incidence Graph

Let G be a graph with n vertices, e edges, and no self loops. Define an n by e matrix
𝐴 = [𝑎𝑖𝑗 ],whose n rows correspond to the n vertices and the e columns correspond to
the e edges, as follows:
𝑎𝑖𝑗 = 1,
𝑖𝑓 𝑗𝑡𝑕 𝑒𝑑𝑔𝑒 𝑒𝑗 𝑖𝑠 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑜𝑛 𝑡𝑕𝑒 𝑖 𝑡𝑕 𝑣𝑒𝑟𝑡𝑒𝑥 𝑣𝑖, and
𝑎𝑖𝑗 = 0,


𝑜𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒
Such a matrix A is called the vertex-edge incidence matrix, or simply incidence
matrix. Matrix A for a graph G is sometimes also written as A(G)
The matrix contains only two elements, 0 and 1. Such a matrix is called a binary
matrix or a (0,1) -matrix
Properties of Incidence Matrix
1) Since every edge is incident on exactly two vertices, each column of A has exactlt two
1’s.
2) The number of 1’s in each row equals the degree of the corresponding vertex.
3) A row with all 0’s represents an isolated vertex.
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4) Parallel edges in a graph produce identical columns in its incidence matrix
5) If a graph G is disconnected and consists of two components 𝑔1 and 𝑔2 , the incidence
matrix A(G) of graph G can be written in a block diagonal form as
𝐴(𝑔1 )
0
𝐴 𝐺 =
0
𝐴(𝑔2 )
where 𝐴 𝑔1 and 𝐴(𝑔2 ) are the incidence matrices of components 𝑔1 and 𝑔2 .
6) Permutation of any two rows or columns in an incidence matrix simply corresponds to
relabeling the vertices and edges of the same graph.
Theorem:
Two graphs G1 and G2 are isomorphic if and only if their incidence matrices
A(G1) and A(G2) differ only by permutations of rows and columns
Rank of the Incidence Matrix


Let G be a graph and let A(G) be its incidence matrix. Now each row in A(G) is a
vector over GF(2) in the vector space of graph G. Let the row vectors be denoted by
A1, A2, . . ., An. Then,
𝐴1
𝐴2
.
𝐴 𝐺 =
.
.
𝐴𝑛
Since there are exactly two ones in every column of A, the sum of all these vectors is
0 (this being a modulo 2 sum of the corresponding entries). Thus vectors A1, A2, . . .,
An are linearly dependent. Therefore, rank A < n. Hence, rank A ≤ n−1.
Theorem:
If A(G) is an incidence matrix of a connected graph G with n vertices, the rank
of A(G) is n-1
Pro
of:


Let G be a connected graph with n vertices and let the number of edges in G be m. Let
A(G) be the incidence matrix and let A1, A2, . . ., An be the row vector of A(G).Then,
𝐴1
𝐴2
.
𝐴 𝐺 =
.
.
𝐴𝑛
Clearly, rank A(G) ≤ n−1.
(1)
Consider the sum of any m of these row vectors, m ≤ n−1. Since G is connected,
A(G) cannot be partitioned in the form
𝐴(𝐺1 )
0
𝐴 𝐺 =
0
𝐴(𝐺2 )
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such that 𝐴(𝐺1 ) has m rows and 𝐴(𝐺2 ) has n−m rows.
Thus there exists no m×m submatrix of A(G) for m≤ n−1, such that the modulo 2
sum of these m rows is equal to zero.
As there are only two elements 0 and 1 in this field, the additions of all vectors taken
m at a time for m = 1, 2, . . ., n−1 gives all possible linear combinations of n−1 row
vectors. Thus no linear combinations of m row vectors of A, for m ≤ n−1, is zero.
Therefore, rank A(G) ≤ n−1.
(2)
Combining (1) and (2), it follows that rank A(G) = n−1.
Theorem:
Rank of A(G) is n-k, if G is disconnected graph with n vertices and k components
Reduced Incidence Matrix



Let G be a connected graph with n vertices and m edges. Then the order of the
incidence matrix A(G) is n×m. Now, if we remove any one row from A(G), the
remaining (n−1) by m submatrix is of rank (n−1). Thus the remaining (n−1) row
vectors are linearly independent.
This shows that only (n−1) rows of an incidence matrix are required to specify the
corresponding graph completely, because (n−1) rows contain the same information as
the entire matrix.
Such an (n−1) × m matrix of A is called a reduced incidence matrix and is denoted
by Af. The vertex corresponding to the deleted row in Af is called the reference
vertex.
Theorem: The reduced incidence matrix of a tree is non-singular.
Proof:


A tree with n vertices has n−1 edges and also a tree is connected. Therefore, the
reduced incidence matrix is a square matrix of order n−1, with rank n−1. Thus the
result follows.
Now a graph G with n vertices and n −1 edges which is not a tree is obviously
disconnected. Therefore the rank of the incidence matrix of G is less than (n−1).
Hence the (n−1)×(n−1) reduced incidence matrix of a graph is non-singular if and
only if the graph is a tree.
Circuit Matrix


Let the graph G have m edges and let q be the number of different circuits in G. The
circuit matrix 𝐵 = [𝑏𝑖𝑗 ]𝑞×𝑚 of G is a (0, 1) − matrix of order q × m, with
bij = 1, if the ith cycle includes jth edge
bij = 0, otherwise.
The circuit matrix B of a graph G is denoted by B(G).
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The graph G1 has four different cycles Z1 = {e1, e2}, Z2 = {e3, e5, e7}, Z3 = {e4, e6, e7}
and Z4 = {e3, e4, e6, e5}.
Properties of Circuit Matrix
1) A column of all zeros corresponds to a non circuit edge, that is, an edge which does
not belong to any circuit.
2) Each row of B(G) is a circuit vector.
3) A circuit matrix has the property of representing a self-loop and the corresponding
row has a single one.
4) The number of ones in a row is equal to the number of edges in the corresponding
circuit.
5) If the graph G is separable (or disconnected) and consists of two blocks (or
components) g1 and g2, then the cycle matrix B(G) can be written in a block-diagonal
form as
𝐵(𝑔1)
0
𝐵 𝐺 =
0
𝐵(𝑔2)
where B(g1) and B(g2)are the cycle matrices of g1 and g2
6) Permutation of any two rows or columns in a circuit matrix corresponds to relabeling
the circuits and the edges.
7) Two graphs G1 and G2 have the same circuit matrix if and only if G1 and G2 are 2isomorphic
Theorem:
If G is a graph without self-loops, with incidence matrix A and circuit matrix B
whose columns are arranged using the same order of edges, then every row of B is
orthogonal to every row of A, that is ABT = BAT = 0 (mod2), where AT and BT are
the transposes of A and B respectively.
Proof:
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Consider a vertex v and circuit ℸ in a graph G. Either v is in ℸ or it is not. If v is not
inℸ, there is no edge in the circuit ℸ that is incident on v. On the other hand, if v is inℸ,
the number of edges in the circuit ℸ that are incident on v is exactly two.
Consider the ith row in A and the jth row in B. Since the edges are arranged in the
same order, the non zero entries in the corresponding position occurs only if a
particular edge is incident on the ith vertex and also in the jth circuit.
If the ith vertex is not in the jth circuit, there is no such non zero entry, and the dot
product of the two rows is zero.
If the ith vertex is in the jth circuit, there will be exactly two 1’s in the sum of the
product of two arbitrary rows. Since the dot product of two arbitrary rows one from A
and other from B is zero. Hence the theorem.
Fundamental Circuit Matrix



A sub matrix of a circuit matrix in which all rows correspond to a set of fundamental
circuit is called a fundamental circuit matrix B f
If n is the number of vertices, m the number of edges in a connected graph G, then Bf
is an (m−n+1)×m matrix because the number of fundamental circuitss is m−n+1,
each fundamental circuit being produced by one chord.
A matrix Bf thus arranged has the form
𝐵𝑓 = 𝐼𝜇 : 𝐵𝑡
where Iμ is an identity matrix of order μ = m−n+1 and Bt is the remaining μ ×(n−1)
submatrix, corresponding to the branches of the spanning tree.
𝑍1 = 𝑒1 , 𝑒2 , 𝑒4 , 𝑒7
𝑍2 = 𝑒3 , 𝑒4 , 𝑒7
𝑍3 = 𝑒5 , 𝑒6 , 𝑒7
Theorem: If B is a circuit matrix of a connected graph G with n vertices and m
edges, then rank B = m−n+1.
Proof:


Let A be the incidence matrix of the connected graph G. Then ABT ≡ 0(mod2).
Using Sylvester’s theorem, we have
rank A+ rank BT ≤ m
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so that
rank A+ rank B ≤ m.

Therefore,
rank B ≤ m− rank A.

As rank A = n−1, we get
rank B ≤ m−(n−1)= m−n+1.

But,

rank B ≥ m−n+1.
Combining, we get rank B = m−n+1.
Theorem: If B is a circuit matrix of a disconnected graph G with n vertices, m edges
and k components, then rank B = m−n+k.
Proof:






Let B be the circuit matrix of the disconnected graph G with n vertices, m edges and k
components. Let the k components be G1, G2, ...,Gk with n1, n2, ..., nk vertices and m1,
m2, . . .,mk edges respectively
Then 𝑛1 + 𝑛2 + ⋯ + 𝑛𝑘 = 𝑛 and 𝑚1 + 𝑚2 + ⋯ + 𝑚𝑘 = 𝑚
Let 𝐵1 , 𝐵2 ,…, 𝐵𝑘 be the circuit matrices of G1, G2, ...,Gk
𝐵1 (𝐺1 )
0
0
…
0
0
𝐵1 (𝐺1 )
0
…
0
0
0
𝐵1 (𝐺1 ) …
0
Then 𝐵 𝐺 =
.
.
.
.
.
.
.
.
.
.
.
.
0
0
0
… 𝐵1 (𝐺1 )
We know rank𝐵𝑖 = 𝑚𝑖 − 𝑛𝑖 + 1, 𝑓𝑜𝑟 1 ≤ 𝑖 ≤ 𝑘
Therefore, 𝑟𝑎𝑛𝑘 𝐵 = 𝑟𝑎𝑛𝑘 𝐵1 + ⋯ + 𝑟𝑎𝑛𝑘 𝐵𝑘
= 𝑚1 − 𝑛1 + 1 + ⋯ + (𝑚𝑘 − 𝑛𝑘 + 1)
= 𝑚1 + ⋯ + 𝑚𝑘 + 𝑛1 + ⋯ + 𝑛𝑘 + 𝑘
𝒓𝒂𝒏𝒌 𝑩 = 𝒎 − 𝒏 + 𝒌
Cut Set Matrix

Let G be a graph with m edges and q cutsets. The cut-set matrix C = [cij]q×m of G is a
(0,1)-matrix with
𝟏,
𝒊𝒇 𝒊𝒕𝒉 𝒄𝒖𝒕𝒔𝒆𝒕 𝒄𝒐𝒏𝒕𝒂𝒊𝒏𝒔 𝒋𝒕𝒉 𝒆𝒅𝒈𝒆
𝒄𝒊𝒋 =
𝟎,
𝑶𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆

Example
Consider the graphs shown in Figure. In the graph G1, E = {e1, e2, e3, e4, e5, e6, e7, e8}.
The cut-sets are c1 = {e8}, c2 ={e1, e2}, c3={e3, e5}, c4 = {e5, e6, e7}, c5 = {e3, e6, e7},
c6 = {e4, e6}, c7 = {e3, e4, e7} and c8 = {e4, e5, e7}.
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Thus the cut-set matrices are given by
Properties of Cut Set Matrix
1) The permutation of rows or columns in a cut-set matrix corresponds simply to
renaming of the cut-sets and edges respectively.
2) Each row in C(G) is a cut-set vector.
3) A column with all zeros corresponds to an edge forming a self-loop.
4) Parallel edges form identical columns in the cut-set matrix.
5) In a non-separable graph, since every set of edges incident on a vertex is a cutset,therefore every row of incidence matrix A(G) is included as a row in the cut-set
matrix C(G). That is, for a non-separable graph G, C(G) contains A(G). For a
separable graph,the incidence matrix of each block is contained in the cut-set matrix.
6) rank C(G)≥rank A(G). Therefore, for a connected graph with n vertices, rank C(G)≥
n−1.
Theorem:
If G is a connected graph, then the rank of a cut-set matrix C(G) is equal to the rank of
incidence matrix A(G), which equals the rank of graph G.
Proof:
 Let A(G), B(G) and C(G) be the incidence, circuit and cut-set matrix of the connected
graph G. Then we have rank
C(G) ≥ n−1.
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Since the number of edges common to a cut-set and a cycle is always even, every row
in C is orthogonal to every row in B, provided the edges in both B and C are arranged
in the same order.Thus,
BCT =CBT ≡ 0 (mod 2).
Now, applying Sylvester’s theorem to the above equation, we have
rank B+ rank C ≤ m.
For a connected graph, we have
rank B = m−n+1.
Therefore,
rank C ≤m- rank B
= m−(m-n−1)= n-1
So,
rank C = n−1.
Path Matrix

Let G be a graph with m edges, and u and v be any two vertices in G. The path matrix
for vertices u and v denoted by P(u, v) = [pij]q×m, where q is the number of different
paths between u and v, is defined as
𝒑𝒊𝒋 =




𝟏,
𝟎,
𝒊𝒇 𝒋𝒕𝒉 𝒆𝒅𝒈𝒆 𝒍𝒊𝒆𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒊𝒕𝒉 𝒑𝒂𝒕𝒉
𝑶𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
Clearly, a path matrix is defined for a particular pair of vertices, the rows in P(u, v)
correspond to different paths between u and v, and the columns correspond to
different edges in G.
Example
The different paths between the vertices v3 and v4 are p1 = {e8, e5}, p2 = {e8, e7, e3}
and p3 = {e8, e6, e4, e3}
The path matrix for v3, v4 is given by
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Properties of Path matrix.
1) A column of all zeros corresponds to an edge that does not lie in any path between u
and v.
2) A column of all ones corresponds to an edge that lies in every path between u and v.
3) There is no row with all zeros.
4) The ring sum of any two rows in P(u, v) corresponds to a cycle or an edge-disjoint
union of cycles.
Theorem: If the columns of the incidence matrix A and the path matrix P(u, v) of a
connected graph are arranged in the same order, then under the product (mod 2).
APT (u, v) = M,
where M is a matrix having ones in two rows u and v, and the zeros in the remaining
n−2 rows
Adjacency Matrix (Connection Matrix)
1) Let V =(V, E) be a graph with V ={v1, v2, . . . , vn}, E ={e1, e2, . . . , em} and without
parallel edges. The adjacency matrix of G is an n×n symmetric binary matrix X = [xij]
defined over the ring of integers such that
1,
𝑖𝑓𝑡𝑕𝑒𝑟𝑒 𝑖𝑠 𝑎𝑛 𝑒𝑑𝑔𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑖 𝑡𝑕 𝑎𝑛𝑑 𝑗𝑡𝑕 𝑣𝑒𝑟𝑡𝑒𝑥
𝑥𝑖𝑗 =
0,
𝑂𝑡𝑕𝑒𝑟𝑤𝑖𝑠𝑒
2) Example:
Consider the graph given in the figure
The adjacency matrix of G is given by
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Properties of Adjacency Matrix
1) The entries along the principal diagonal of X are all zeros if and only if the graph has
no self-loops. However, a self-loop at the ith vertex corresponds to xii = 1
2) If the graph has no self-loops, the degree of a vertex equals the number of ones in the
corresponding row or column of X.
3) Permutation of rows and the corresponding columns imply reordering the vertices.
We note that the rows and columns are arranged in the same order. Therefore, when
two rows are interchanged in X, the corresponding columns are also interchanged.
Thus two graphs G1 and G2 without parallel edges are isomorphic if and only if their
adjacency matrices X(G1) and X(G2) are related by
𝑋 𝐺2 = 𝑅−1 𝑋 𝐺1 𝑅
where R is a permutation matrix.
4) A graph G is disconnected having components G1 and G2 if and only if the adjacency
matrix X(G) is partitioned as
𝑋 𝐺1
0
𝑋 𝐺 =
0
𝑋 𝐺2
where 𝑋 𝐺1 and 𝑋 𝐺2 are respectively the adjacency matrices of the components G1
and G2. Obviously, the above partitioning implies that there are no edges between
vertices in G1 and vertices in G2.
5) If any square, symmetric and binary matrix Q of order n is given, then there exists a
graph G with n vertices and without parallel edges whose adjacency matrix is Q.
Edge sequence



An edge sequence is a sequence of edges in which each edge, except the first and the
last, has one vertex in common with the edge preceding it and one vertex in common
with the edge following it.
A walk and a path are the examples of an edge sequence. An edge can appear more
than once in an edge sequence.
In the graph of following Figure , v1e1v2e2v3e3v4e4v2e2v3e5 v5, or e1e2e3e4e2e5 is an
edge sequence.
Powers of Adjacency Matrix (X)

Let us multiply by itself the adjacency matrix of the simple graph. The result is
another symmetric matrix 𝑋 2
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= number of ones in the product of ith row and jth column (or jth row) of X
𝑖𝑗
= number of positions in which both ith and jth rows of X have ones
= number of vertices that are adjacent to both ith and jth vertices
= number of different paths of length two between ith and jth vertices
Also, 𝑋 2
𝑖𝑖
= number of ones in the ith row (or column) of X
= degree of the corresponding vertex.
Theorem: Let X be the adjacency matrix of a simple graph G. Then ij th entry in Xr is the
number of different edge sequence of r edges between vertices vi and vj
Proof:


We prove the result by using induction on k. The result is trivial for k = 0 and 1. Since
X2 = X.X, X2 is a symmetric matrix, as product of symmetric matrices is also
symmetric.
For k = 2, i ≠ j, we have
𝑋 2 𝑖𝑗 = number of ones in the product of ith row and jth column (or jth row) of X
= number of positions in which both ith and jth rows of X have ones
= number of vertices that are adjacent to both ith and jth vertices
= number of different paths of length two between ith and jth vertices
Also, 𝑋 2
𝑖𝑖
= number of ones in the ith row (or column) of X
= degree of the corresponding vertex.


This shows that 𝑋 2 𝑖𝑗 is the number of different paths and therefore different edge
sequences of length 2 between the vertices vi and vj. Thus the result is true for k = 2.
Assume the result to be true for k, so that
𝑋 𝑘 𝑖𝑗 = number of different edge sequences of length k between vi and vj .
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We have,
𝑋 𝑘 +1

𝑖𝑗 j
= 𝑋𝑘
𝑖𝑗
=
𝑛
𝑟=1
𝑋𝑘
𝑖𝑟
𝑋
𝑟𝑗
=
𝑛
𝑟=1
𝑋𝑘
𝑖𝑟 𝑥𝑟𝑗
Now, every vi −vj edge sequence of length k+1 consists of a vi −vr edge sequence of
length k, followed by an edge vtvj. Since there are 𝑋 𝑘 𝑖𝑟 such edge sequences of
length k and xrj such edges for each vertex vr, the total number of all vi −vj edge
sequences of length k+1 is 𝑛𝑟=1 𝑋 𝑘 𝑖𝑟 𝑥𝑟𝑗 . This proves the result for k+1 also.
In a connected graph, the distance between two vertices vi and vj (for i≠ 𝒋) is k, if and
only if k is the smallest integer for which the i,j th entry in xk is nonzero.
If X is the adjacency matrix of a graph G with n vertices, and 𝒀 = 𝑿 + 𝑿𝟐 + 𝑿𝟑 +
⋯ . +𝑿𝒏 , 𝒕𝒉𝒆 𝒓𝒊𝒏𝒈 𝒐𝒇 𝒊𝒏𝒕𝒆𝒈𝒆𝒓𝒔 ∀ 𝒏 ≥ 𝟐 , then G is disconnected if and only if there
exists at least one entry in matrix Y that is zero
Proof:





We have, 𝑦𝑖𝑗 = [𝑌]𝑖𝑗 = 𝑋 𝑖𝑗 + 𝑋 2 𝑖𝑗 + ⋯ + 𝑋 𝑛 −1 𝑖𝑗
Since 𝑋 𝑘 𝑖𝑗 denotes the number of distinct edge-sequences of length k from vi to vj ,
yij = number of different vi −vj edge sequence of length 1 + number of different vi −vj
edge sequences of length 2+. . .+ number of different vi −vj edge sequences of length
n−1.
Therefore, yij = number of different vi −vj edge sequence of length less than n.
Now let G be connected. Then for every pair of distinct i, j there is a path from v i to
vj. Since G has n vertices, this path passes through atmost n vertices and so has length
less than n.
Thus, 𝑦𝑖𝑗 ≠ 0 for each i, j with 𝑖 ≠ 𝑗.
Conversely, for each distinct pair i, j we have 𝑦𝑖𝑗 ≠ 0. Then from above, there is at
least one edge sequence of length less than n from vi to vj. This implies that vi is
connected to vj. Since the distinct pair i, j is chosen arbitrarily, G is connected.
Relationship between A(G) and X(G)

If a graph G has neither self-loops nor parallel edges( Simple graph), both A(G) and
X(G) contains the entire information. ie., X(G)=A(G).
Adjacency Matrix and Isomprphism
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Points to Remember
 Matrix representation of graphs:
 Adjacency Matrix
 Incidence Matrix
 Circuit Matrix
 Fundamental Circuit Matrix
 Path Matrix
 Cut Set Matrix
Incidence Matrix
 Two graphs G1 and G2 are isomorphic if and only if their incidence matrices A(G1)
and A(G2) differ only by permutations of rows and columns
 If A(G) is an incidence matrix of a connected graph G with n vertices, the rank of
A(G) is n-1
 Rank of A(G) is n-k, if G is disconnected graph with n vertices and k components
 The reduced incidence matrix of a tree is non-singular.
Circuit Matrix
 If G is a graph without self-loops, with incidence matrix A and circuit matrix B whose
columns are arranged using the same order of edges, then every row of B is
orthogonal to every row of A, that is ABT = BAT = 0 (mod2), where AT and BT are the
transposes of A and B respectively.
 If B is a circuit matrix of a connected graph G with n vertices and m edges, then rank
B = m−n+1.
 If B is a circuit matrix of a disconnected graph G with n vertices, m edges and k
components, then rank B = m−n+k.
Cut set Matrix
 If G is a connected graph, then the rank of a cut-set matrix C(G) is equal to the rank of
incidence matrix A(G), which equals the rank of graph G.
Path Matrix
 If the columns of the incidence matrix A and the path matrix P(u, v) of a connected
graph are arranged in the same order, then under the product (mod 2).
APT (u, v) = M,
where M is a matrix having ones in two rows u and v, and the zeros in the remaining
n−2 rows
Powers of Adjacency matrix
 Let X be the adjacency matrix of a simple graph G. Then ijth entry in Xr is the number
of different edge sequence of r edges between vertices vi and vj
 In a connected graph, the distance between two vertices vi and vj (for i≠ 𝑗) is k, if and
only if k is the smallest integer for which the i,j th entry in xk is nonzero.
 If X is the adjacency matrix of a graph G with n vertices, and 𝑌 = 𝑋 + 𝑋 2 + 𝑋 3 +
⋯ . +𝑋 𝑛 , 𝑡𝑕𝑒 𝑟𝑖𝑛𝑔 𝑜𝑓 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 ∀ 𝑛 ≥ 2 , then G is disconnected if and only if
there exists at least one entry in matrix Y that is zero
 For a simple graph, A(G)=X(G)
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PROBLEMS
1. Write the incidence matrices for the labelled simple
graphs shown below. Put the incidence matrix of the
graph in the block diagonal form.
2. Show that for a simple disconnected graph of k components, n vertices, and e edges
the ranks of matrices A,B, and C are n-k, e-n+k, and n-k respectively
3. Label the edges of the graph in the figure and write down its circuit matrix B.
4. Write down the path matrix P(v1,v6) for the graph in the figure
PREVIOUS YEAR UNIVERSITY QUESTIONS
1. Write the properties of Incidence matrices.
2. List down any four properties of adjacency matrix
3. Construct an adjancy matrix(X) for the following graph and also mention how the
concept of edge sequencs is described with X3(no need to find X3 from X)
4. Find the incidence matrix of the following graph
5. Define the adjacency matrix and incidence matrix representation of a graph with an
example
6. Prove the theorem:
If A(G) is an incidence matrix of a connected graph G with n vertices, the rank of A(G)
is n-1
7. Describe with examples the usage of incidence matrix to find two graphs (g1 and g2) are
isomorphic.
8. Define cut-set matrix with an example and list down any four properties of cut set matrix
96
Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor
Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
9. If B is a circuit matrix of a connected graph G with e edges and n vertices, then show that
the number of linearly independent rows in B = e-n+1
10. Define the circuit-matrix B(G) of a connected graph G with n vertices and e edges. Prove
that the rank of B(G) is e-n+1.
11. Define the adjacency matrix X(G) of a graph. Let X(G) be adjacency matrix of a simple
graph G, then prove that ijth entry in Xr is the number of different edge sequences of r
edges between vertices vi and vj.
12. Draw the adjacency graph for the following adjacency matrix.
13. Write the fundamental circuit matrix with respect to the spanning tree shown in heavy
lines for the following graph.
14. Define the incidence matrix of a graph G.Prove that the rank of an incidence matrix of a
connected graph with n vertices is n-1.
15. Draw the graph represented by the following incidence matrix.
WORK SHEET
1. Incidence matrix and Adjacency matrix of a graph will always have same dimensions?
a) True
b) False
Ans:
2. The column sum in an incidence matrix for a simple graph is __________
a) depends on number of edges
b) always greater than 2
c) equal to 2
d) equal to the number of edges
Ans:
3. A matrix whose rows are the rows of the unit matrix, but not necessarily in their natural
order, is called:
a) adjacency matrix
b) adjacency matrix of pseudo graph
c) adjacency matrix of simple graph
d) permutation matrix
Ans:
97
Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor
Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
4. An ____________in a graph is a finite alternating sequence of vertices and edges,
beginning and ending with vertices, such that each edge is incident on the vertices
preceding and following it.
Ans:
5. Two graphs are isomorphic, if and only if their vertices can be labelled in such a way that
the corresponding adjacency matrices are ___________.
Ans:
6. Find the adjacency and incidence matrix of the following graph:
7. Represent the
following
graphs using adjacent
matrices:
8. Represent the incidence matrix for the following graphs:
9. Draw the graphs with the given adjacency matrix:
98
Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor
Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
10. Draw the graphs represented by the following incidence matrices:
11. Write the properties of Incidence matrices.
12. List down any four properties of adjacency matrix
13. Construct an adjancy matrix(X) for the following graph and also mention how the
concept of edge sequencs is described with X3(no need to find X3 from X)
14. Find the incidence matrix of the following graph
15. Define the adjacency matrix and incidence matrix representation of a graph with an
example
16. Prove the theorem:
If A(G) is an incidence matrix of a connected graph G with n vertices, the rank of A(G)
is n-1
17. Describe with examples the usage of incidence matrix to find two graphs (g1 and g2) are
isomorphic.
18. Define cut-set matrix with an example and list down any four properties of cut set matrix
19. If B is a circuit matrix of a connected graph G with e edges and n vertices, then show that
the number of linearly independent rows in B = e-n+1
20. Define the circuit-matrix B(G) of a connected graph G with n vertices and e edges. Prove
that the rank of B(G) is e-n+1.
99
Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor
Dept of CSE
HandOut and WorkSheet of CS309 Graph Theory And Combinatorics
21. Define the adjacency matrix X(G) of a graph. Let X(G) be adjacency matrix of a simple
graph G, then prove that ijth entry in Xr is the number of different edge sequences of r
edges between vertices vi and vj.
22. Draw the adjacency graph for the following adjacency matrix.
23. Write the fundamental circuit matrix with respect to the spanning tree shown in heavy
lines for the following graph.
24. Define the incidence matrix of a graph G.Prove that the rank of an incidence matrix of a
connected graph with n vertices is n-1.
25. Draw the graph represented by the following incidence matrix.
s
100
Prepared by Keerthi A S Pillai, Asst. Prof, Dept of CSE, SBCE Pattoor
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