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Depar tment o f Mathemat i cs Babu BanarAsi Das National Institute of Technology & Management, Lucknow

Class Room Notes: Graph Theory (ECS 505)

Math Notes/GT/ECS 505/Unit-1

NOTE: This handout is not a replacement of the prescribed textbooks. You are recommended to refer the following

prescribed textbooks for further reading. You may find good number of examples for practice in these textbooks

together with applications of graph theory in engineering problems and computer science.

1. Graph Theory With Applications to Engineering & Computer Science, Narsingh Deo, Prentice Hall ofIndia

2. Graph Theory, Frank Harary, Narosa Publishing House3. Graph Theory & Applications, Bondy & Murthy, Addison Wesley

The matter contained in will be discussed/explained in classroom lectures. I will appreciate your active

participation. Wish you all success. Your course begins here!

1. GRAPH THEORY BASICS: DEFINITIONS & TYPES OF GRAPHS

The word graph refers to a specific mathematical structure usually represented as diagram

consisting of points joined by lines. In applications the points may, for instance, correspond to

chemical atoms, towns, electrical terminals or anything that can be connected in pairs. The lines may be

chemical bonds, roads, wires or other connections. Applications of graph theory are found in

communications, structures and mechanisms, electrical networks, transport systems, social

networks and computer science etc.

A graph is a mathematical structure comprising a set ofvertices, V, and a set ofedges, E, which

connect the vertices. It is usually represented as a diagram consisting of points, representing the vertices

(or nodes), joined by lines, representing the edges (figure 1.1). It is formally denoted by G = (V, E).

If in the graph the vertices are connected by directed lines, the graph is called directed graph or a

digraph (figure 1.4).The graph is called labeled graph, if the vertices have labels or names (figure 1.2).

If each edge has a weightassociated with it, it is then called a weighted graph,(figure 1.3).

The two vertices joined by an edge are said to be adjacent. The vertices are said to be incident

with the edge that joins them and an edge is said to be incident with the vertices it joins. The two or

more edges joining the same pair of vertices are called multiple/ parallel edges (figure 1.5). An edgejoining a vertex to itself is called a loop (figure 1.5). A graph with no multiple edges and no loops is a

simple graph(figure 1.1).

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The distance between two vertices a and b, denoted by dist(a, b), is the length of a shortest path

joining them. The diameter of a connected graph, denoted diam (G), is maxdist(a, b).

A subgraph of G is a graph all of whose vertices and edges are vertices and edges of G. If

G1 = (V!, E1) is a subgraph of G = (V, E) then V! Vand E1 E. In figure 1.7 below, G1, G2, G3 and G4are subgraphs ofG.

Any two subgraphs G1 and G2 of the graph G are edge-disjoint if they do not have any edge in

common. The two subgraphs are said to be vertex-disjoint subgraphs if there is no common vertex in the

two subgraphs. Note that edge-disjoint subgraphs may have common vertices but vertex-disjoint

subgraphs cannot have common edges.

A subgraph His aspanning subgraphifV(H) = V(G). In figure 1.7, G3 is a spanning subgraph

ofG. An induced (generated) subgraph is a subset of the vertices of the graph together with all the

edges of the graph between the vertices of this subset. In figure 1.7, G2 is an induced subgraph ofG.

The degree of a vertex v is defined as the number of edges incident with v. A loops count as 2.

A vertex having no edge being incident on it is isolated vertex. The degree of isolated vertex is

0. A vertex of degree 1 is pendent.

Ifv is a vertex in a graph G, then G v denotes a subgraph obtained by deleting the vertex v from

G. By deleting a vertex, all edges incident on that vertex are also deleted. Similarly, G e denoted a

subgraph obtained by deleting the edge e from G. The deletion of an edge does not delete the vertices of

graph.

The degree sequence of a graph G is the sequence obtained by listing, in ascending order with

repeats, the degrees of the vertices of G. For example, in figure 1.7 the degree sequence of G is

(1, 2, 2, 3, 4). Try to write the degree sequence of graphs in figures 1.5 and 1.6 .

In a digraph, indegree of a vertex is the number of edges incident on it and the outdegree of a

vertex is the number of edges incident from it.

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The Handshaking Theorem states that the sum of the degrees of the vertices of a graph is equal

to twice the number of edges. This follows readily from the fact that each edge joins two vertices (not

necessarily distinct) and so contributes 1 to the degree of each of those vertices. See the systematic proof

in next Section.

In a regular graph,all vertices have the same degree. If the degree is r the graph is r-regular. If

G is r-regular with n vertices, then it has n redges (from the Handshaking Theorem). Verify it!

A complete graph is a graph in which every vertex is joined to every other vertex by exactly one

edge. The complete graph with n vertices is denoted by Kn. Kn is (n 1)-regular and so has

n(n 1) /2 edges. Verify it!

A null graph is a graph with no edges. The null graph with n vertices is denoted Nn. Nn is

0-regular graph.

A cycle graph consists of a single cycle of vertices and edges. The cycle graph with n vertices is

denoted by Cn.

A bipartite graph is a graph whose vertices can be split into two subsets A and B in such a way

that every edge ofG joins a vertex in A with one in B. Figure 1.11 shows some bipartite graphs. Notice

that the allocation of the nodes to the sets A and B can sometimes be done in several ways.

A complete bipartite graph is a bipartite graph in which every vertex in A is joined to every

vertex in B by exactly one edge. The complete bipartite graph with rvertices in A and s vertices in B is

denoted Kr, s. Figure 1.12 shows some complete bipartite graphs.

A graph whose vertex set is and edge set is a finite set is called a finite graph otherwise an

infinite is an infinite graph (figure 1.3).

An infinite graph with finite number of edges has an infinite number of

isolated vertices. (Can u prove it?)

A walk of length kin a graph is a succession ofkedges joining two vertices. In other words, a

walk is an alternating sequence of vertices and edges, such that each edge is incident with the

vertices preceding and following it. Note that an edge cannot occur more than once in a walk.

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However, a vertex may appear more than once in a walk. If a walk begins and ends at same vertex, it is

called a closed walk. A walks, which are not closed, is called an open walk.

A path is a walk in which all the edges and all the vertices are distinct. In other words, a walk

which does not intersects it self is a path.

A circuit is a closed walk in which no vertex appears more that once,

except the initial and final vertex.

In figure 1.14, a b c d a e is a walk of length 5 between a and e. The walk a b c d e is a path of length 4

between a and e. The walkd a b c dis a circuit.

NOTE: For more definitions and graph theory terminology, refer to classroom discussions and prescribed

textbooks.

2. SOME IMPORTANT RESULS

THEOREM 2.1: The maximum number of degree of any vertex in a simple graph of n vertices is 1n .

Prove it yourself.

THEOREM 2.2: (Handshaking Theorem):The sum of degrees of all vertices is two times the number of

edges.

PROOF. Let G = (V, E) be a graph with vertices v1, v2, v3, , vn ,and e edges. Since each edge is

associated with two vertices and hence it contributes two degree in the sum of degree of all vertices.

Thus, if there are e edges in G then sum of degrees of all vertices ofG will be equal to 2e. That is,

deg (v1) + deg (v2) + + deg (vn) = 2e.

Hence the theorem.

THEOREM 2.3: In a digraph,

Sum of even degree vertices = Sum of odd degree vertices = No. of edges in G.

Prove it yourself.

THEOREM 2.4: The sum of all degrees of a graph is even.

Prove it yourself.

THEOREM 2.5: The maximum number of edges in a simple graph with n vertices is 2)1( nn .

PROOF. Let G be a graph with n vertices and e edges. By handshaking theorem

Sum of degrees of all n vertices = 2 e

We know that maximum number of degree of any vertex in a simple graph ofn vertices is 1n ,

therefore 1n + 1n + 1n + . n term = 2 e

n ( 1n ) = 2 e , or e = n ( 1n ) /2.

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THEOREM 2.6: In a graph G, the number of vertices of odd degree is even.

PROOF.. We know that in a graph G

Sum of degrees of all vertices = 2 (no. of edges in G)

Sum of even degree vertices + Sum of odd degree vertices = 2 (no. of edges in G)

Sum of odd degree vertices = 2 (no. of edges in G) - Sum of even degree vertices

This implies that, Sum of odd degree vertices is an even number. Hence, number of odd degree vertices iseven. Hence the theorem.

THEOREM 2.7. In a simple graph with at least 2 vertices, there are at least 2 vertices of the same degree.

PROOF. Let G be the simple graph with n vertices, n2. The graph G is simple; it has no loops or paralleledges. Hence the degree of each vertex is less than or equal to 1n . Suppose all vertices ofG are of

different degree. Hence possible degrees for n vertices ofG are 0, 1, 2, 3, , 1n . Let v be a vertex of

degree 0. Hence v is an isolated vertex. Let u be a vertex of degree n 1, then u has n 1 adjacent

vertices. Hence v cannot be an isolated vertex; this contradiction proves that a simple graph contains two

vertices of same degree.

PROBLEM 1.1:Find the number of vertices in a graph with 21 edges and 3 vertices of degree 4 and other vertices

each of degree 3.

PROBLEM 1.2: What is the largest number of vertices in a graph with 35 edges if all vertices are of degree 3.

PROBLEM 1.3: Show that there exists no graph corresponding to the following degree sequence:

(i) 0, 2, 2, 3, 4 (ii) 1, 1, 2, 3 (iii) 2,2 2, 3, 4, 5, 5

PROBLEM 1.4: Show that compliment of a bipartite graph need not to be a bipartite.

PROBLEM 1.5: Find the size of the k regular graph.

PROBLEM 1.6: Show that there exists there exists no cycle of odd length in a bipartite graph.

3. OPERATIONS ON GRAPHS

Union: Let ),( 111 EVG = and ),( 222 EVG = be two graphs. The union of two graphs 21 GG is a graph

whose vertex set and edge sets are 21 VV and 21 EE respectively.

Intersection: The intersection of two graphs 21 GG is a graph whose vertex set and edge sets are

21 VV and 21 EE respectively.

Sum: Let ),( 111 EVG = and ),( 222 EVG = be two graphs such that = 21 VV . The sum 21 GG + is

defined as a graph whose vertex set is 21 VV and edge set consists of those edges that are in 1G and in

2G and the edges obtained by joining each vertex of 1G to each vertex of 2G .

Ring Sum: The ring sum 21 GG is defined as a graph whose vertex set is 21 VV and edge set is

)()( 2121 EEEE .

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Compliment: The compliment of a graph G , denoted as G , is a simple graph with same vertex set as

that ofG and in which any two vertices are adjacent if they are not adjacent in G . In other words, the

complement of agraph G = (V, E) is a graph with vertex set Vand edge set E such that eE if and

only ifeE. A graph is self-complimentary if it is isomorphic to its compliment (see the definition of

isomorphism in next section).

Decomposition: A graph G is said to be decomposed in to two graphs 1G and 2G if 21 GG = G and

21 GG = a null graph. It may be noted that ifG is decomposed into two graphs 1G and 2G , then every

edge of G occurs in 1G or in 2G , but not in both. However, some of the vertices may occur in both

1G and 2G . Further, a graph with m edges can be decomposed in 2m1

1 different ways into pair of

subgraphs.

PROBLEM 1.7:Find the union, intersection and ring sum of the following graphs:

PROBLEM 1.8: Find the compliment of the first graph in Problem 1.1.

4. CONNECTED & DISCONNECTED GRAPHS AND COMPONENTS

A graph is said to be connected if its every vertex is reachable from every vertex. A graph that is

not connected is a disconnected graph. In figure 4.1 the graph is a connected graph and that in figure 4.2

is a disconnected graph. It is easy to observe that a disconnected graph is union of two or more connected

graphs, which are called connected components. For example, in figure 4.2, the disconnected graph has

two connected components. Moreover, each connected subgraph of a graph G is a connected component

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of G . In other words, each graph can be expressed as the union of one or morepair-wise disjoint connected graphs.

An edge in a connected graph is called a bridge if its removal leaves a disconnected graph. A

vertex in a connected graph is called a cut-vertex if its removal makes the graph disconnected.

THEOREM 4.1: A graph G is disconnected if and only if its vertex set V can be partitioned into two non-

empty disjoint sets V1 and V2 such that there exists no edge in whose one vertex in V1 and other in V2.

PROOF. Let there exists partition V1and V2.of the vertex set Vof the graph G. Let aV1 and bV2

such that no edge exists that connects a and b. If such a partition exists then G is disconnected.

Conversely, let us now suppose that G is disconnected. Let a be any vertex ofG. Consider now

the set V1 of all vertices ofG that are joined by a path to a. Obviously, the set V1 does not contain all

vertices ofG as G is disconnected. Thus the remaining vertices will be contained in another set, say V2..

Thus there exists a partition. This completes the proof.

THEOREM 4.2: If a graph (connected or disconnected) has exactly two vertices of odd degree, there must

be a path joining two vertices.

PROOF. IfG is a connected graph, i.e., in G there exists a path between every two vertices. The theorem

is proved.

Let us now suppose that G is a disconnected graph. Let v1 and v2 be the vertices of odd degree

and all other vertices are of even degree. Since the number of odd degree vertices in a graph is always

even, therefore v1 and v2 must belong to the same component. As each component is a connected graph,

there exists a path between v1 and v2. This completes the proof.

THEOREM 4.3: A simple graph with n vertices and k components can have at most

2)1)(( + knkn edges.

PROOF. Let G be the graph with n vertices and k components. Letk

nnn ,.....,, 21 be the number vertices

in kcomponents. Therefore, nnnn k =,.....,, 21

We know that the maximum number of edges in i th component can be 2)1( nn , therefore maximum

number of edges in G is

== ==

==k

i

i

k

i

k

i

ii

k

i

ii

nnnnnn

1

2

1 1

2

1 22

1

2

1

2

1)1(

2

1

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Using the following inequality

)2)(1(2

1

2 knknnk

i

i

=

,

We get )1)((2

1)1(

2

1

1

+=

knknnnk

i

ii.

PROBLEM 1.9:Show that if G is not connected then G is connected.

PROBLEM 1.10: Show that if a graph (connected or disconnected) has exactly two vertices of odd degree, then there

must be a path joining two vertices.

PROBLEM 1.11: Show that if the intersection of two paths is a disconnected graph, then the union of two paths has

atleast one circuit.

PROBLEM 1.12: What is the total number of subgraphs and spanning subgraphs ofK6.

5. GRAPH ISOMOSRPHISM

The graphs G1 and G2 are said to be isomorphic if there exists a one-to-one correspondence

between vertices in G1 and vertices in G2 such that there is an edge between two vertices in G1 if and onlyif there is an edge between the two corresponding vertices in G2. The correspondence itself is called an

isomorphism.

Consider the following correspondence between vertices

in the two graphs shown in figure 5.1. A corresponds to 1,

B corresponds to 2, D corresponds to 4, C corresponds to 3.

Since there is an edge between two vertices in the first graph if

and only if there is an edge between the two corresponding vertices

in the second graph on the right. Therefore, the two graphs are

isomorphic.

Further, two isomorphic graphs may be drawn to look quite

different. For example, here are two different ways of drawing C5

(figure 5.2). The isomorphic graphs share a great many properties,

such as the number of vertices, number of edges, and the pattern

of vertex degrees. Thus, two graphs can be proved to be

non-isomorphicby identifying some property that one possesses

and the other does not. For example, if one graph has two vertices

of degree 5 and another has three vertices of degree 5, and then the two graphs can not be isomorphic.

More precisely, two graphs are isomorphic then (i) they have same number of edges, (ii) they have

same number of vertices, and (iii) their degree sequence are same. It is important to note that the

converse is not true, i.e., if two graphs are isomorphic then all above conditions are satisfied, but if two

graphs satisfy these conditions they are not necessarily be isomorphic. The graphs in figure 5.3 further

illustrate this idea. The two graphs are isomorphic under the correspondence a A, b B, c C,

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d D, e E, fF. (also check their number of vertices, number of edges and their degree

sequences).

PROBLEM 1.13:Which of the following graphs are isomorphic:

(i) (ii)

PROBLEM 1.14: Show that the following graphs are isomorphic:

PROBLEM 1.15: Construct three non-isomorphic spanning subgraphs of the following graph

PROBLEM 1.16: Show that the two simple connected graphs with n vertices each of degree 2 are isomorphic.

PROBLEM 1.17: If a graph with n vertices is isomorphic to its compliment G, show that n or (n -1) must be

divisible by 4.

PROBLEM 1.18:Show that there are 11 non-isomorphic graphs with 4 vertices.

6. EULER GRAPH

A closed walk in a graph that contains all edges of the graph exactly once is called an Euler line.

A graph that containsan Euler line is called an Euler graph.

It is important to note that an Euler graph is always connected. For example, the graphs in figure

6.1 are both Euler graphs. The close walk containing all edges (Euler line) in two graphs are a b c d e a

and A a B d C e D f E c B b E g D h A respectively. The graphs in figure 6.2 are not Euler graphs

because no Euler line exists.

The following are the important results for Euler graphs.

THEOREM 6.1:A connected graph is an Euler graph if and only if its all vertices are of even degree.

PROOF. Let G be an Euler graph. Therefore, an Euler line exist in G. Let the Euler line starting from the

vertex v1 and traversing all edges of graph be v1, e1, v2, e2, v3, . , vn, en, v1. The vertices in this circuit

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may not be all distinct but all edges are distinct. In this circuit every pair of edge ei and ei+1 contributes 2

to the degree of vertex vi+1.. Thus v2, v3, . , vn are all of even degree, and v1 gets 2 degree from e1 and

en. Hence all vertices are of even degree.

Conversely, let connected graph G has all vertices of even degree. Consider v be any vertex. Let

us start constructing a circuit from v and going through edges ofG so that no edge is traversed twice.

Since every vertex is even, so we can always exit from every vertex we enter. The vertex v is also even sowe can arrive at v so as to complete the circuit. If this circuit contains all edges, it is an Euler line. Thus,

G is an Euler graph. If this circuit does not contain all edges of G, the remove the edges included in this

circuit from G, we get a subgraph, say H, of G. Again all vertices in H are even. Moreover, G is

connected; Hmust touch G Hat one vertex, say u. Now, start making a circuit from u. Since all vertices

in Hare even, the circuit must terminate at u. We repeat this process until all edges are covered, i.e., we

get an Euler line. This completes the proof of the theorem.

THEOREM 6.2: The connected graph G is an Euler graph if and only if G can be decomposed into

circuits.

PROOF. Let the connected graph G is decomposed in to circuits. Therefore, G is the union of edge-disjoint

circuits. Since the degree of each vertex of the circuit is 2, hence the degree of each vertex of G is even.

Hence G is an Euler graph.

Conversely, suppose that G is an Euler graph. Consider a vertex u. Obliviously, there are atleast

two edges incident on u as G is Euler graph and all vertices are of even degree. Let other end of the edge

incident on u is v. Since the vertex v is also of even degree, it must have another edge incident ofv, with

w as the vertex on its other end.. If we continue this process we arrive at the vertex already traversed,

thus forming a circuit. Now remove this circuit from G. All vertices in the remaining graph are of even

degree, repeat the above process and remove another circuit formed from G until no edge is left. Thus the

graph is decomposed into circuits.

PROBLEM 1.19: Which of the following graphs are Euler graph;

(i) the complete graph K5, (ii) the complete bipartite graph K2, 3

PROBLEM 1.20:Which of the following graphs are Euler graphs, also find the Euler line if it is Eulerian:

(i) (ii)

PROBLEM 1.21: Show that if a graph had a vertex of odd degree, then it cannot have an Euler circuit.

7. KONIGSBERG BRIDGE PROBLEM

The Seven Bridges of Knigsberg is a notable historical problem in mathematics. Its negative

resolution by Leonhard Euler in 1735 laid the foundations of graph theory. The city of Knigsberg in

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Prussia (now Kaliningrad, Russia) was set on both sides of the Pregel river, and included two large

islands (A and D) which were connected to each other and the mainland (B and C) by seven bridges.

(namely, a, b, c, d, e, f, g) as shown in figure 7.1. The citizens of Knigsberg used to amuse themselves,

during their Sunday afternoon strolls, trying to devise a route which crossed every bridge once only and

returned to its starting point, but no one could find a way to do it.

The problem came to the attention of a Swiss mathematician named Leonhard Euler. Euler

replaced each land mass with an abstract "vertex" or node, and each bridge with an abstract connection,

an "edge", which only serves to record which pair of vertices (land masses) is connected by that bridge

(see figure 7.2). He finally proved (during the 1730s) that the task was impossible.

We can see this readily using the result of theorem 6.1. What the citizens of Knigsberg were

seeking was an Euler line through this graph. Theorem 6.1 tells us that the graph is Eulerian if and only if

every vertex has even degree. The degree sequence of the graph in figure 7.2 is (3,3,3,5) so it is not

Eulerian and no Euler line exists.

Something of your interest: Two of the seven original bridges were destroyed by bombs during World

War II. Two others were later demolished and replaced by a modern highway. The three other bridges

remain, although only two of them are from Euler's time (one was rebuilt in 1935). Thus, there are now

five bridges in Knigsberg (modern name Kaliningrad).

8. HAMILTONIAN GRAPH

A closed walk in a connected graph that traverses every vertex exactly once (except initial and

terminal vertices) is called a Hamiltonian circuit. Thus, a Hamiltonian circuit in a connected graph with

n vertices has exactly n edges. Further, if any one edge from a Hamiltonian circuit is removed, we get a

path that is called a Hamiltonian path. It is obvious that the number of edges in a Hamiltonian path in a

connected graph with n vertices is n 1. It is also important to note that in a connected graph, a

Hamiltonian path is a subgraph of a Hamiltonian circuit. It is because of this fact that a connected graph

with a Hamiltonian circuit has a Hamiltonian path also. But, on the other hand, a connected graph with a

Hamiltonian path may or may not have a Hamiltonian circuit.

A connected graph that has a Hamiltonian circuit is called Hamiltonian graph. The connected

graph is figure 8.1 is a Hamiltonian graph; it has Hamiltonian circuit a b c d e f g h a (note that all

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vertices have been used but not all edges). The connected graph is figure 8.2 has a Hamiltonian path

a b c d but does not have a Hamiltonian circuit; it is not a Hamiltonian graph. A complete graph is

Hamiltonian graph. Total number of Hamiltonian circuits in a complete graph with n vertices is (n-1) !/2.

It may be noted that an Euler line visits each edge of the graph exactly once whereas a

Hamiltonian circuit visits each vertex of the graph exactly once. Also, a connected graph that is an Euler

graph may or may not be a Hamiltonian graph and vice versa. The graph is figure 8.3 is a Hamiltonian

graph as it has a Hamiltonian circuit a b c d e f a, but this graph is not Euler graph because all vertices are

not of even degree. Next, the graph in figure 8.4 is Euler graph but not a Hamiltonian graph. Lastly, the

graph in figure 8.5 is both Euler and Hamiltonian (Can you give the reasons. Try it !).

THEOREM 8.1 (Diracs Theorem): A simple connected graph with n (n3) vertices is Hamiltonian if

deg v n/2 for every vertex v in G.

THEOREM 8.2: A simple connected graph with n vertices and m edges is Hamiltonian if

m1/2 (n - 1) (n - 2) + 2.

PROOF. Let u and v be two non-adjacent vertices in G. Let His a subgraph obtained on removing u and

v from G. The subgraph Hhas n 2 vertices and number of edges in it is m deg u deg v or less than

this. The maximum number of edges in Hmay be

n-2C2 = (n-1) (n - 2) /2 = (n2 5 n + 6) /2.

Therefore, m deg u deg v (n2

5 n + 6) /2

or, deg u + deg v m - (n2

5 n + 6) /2.

By the hypothesis of the theorem, , deg u + deg v (n2

3 n + 6) /2 - (n2

5 n + 6) /2 = n.

This completes the proof.

PROBLEM 1.22. Show that the complete graph Kn, n 3, is Hamiltonian.

PROBLEM 1.23:Draw a graph which is:

(i) Hamiltonian but not Euler, (ii) Euler graph but not Hamiltonian graph, (iii) both Hamiltonian and Euler graph.

PROBLEM 1.24: Which of the following graphs are Hamiltonian graphs, also find the Hamiltonian circuit

(i) (ii)

PROBLEM 1.25: IfG be a graph with n vertices, n 3, then show that G has a Hamiltonian circuit with each vertex

of degree n/2.

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PROBLEM 1.26:In a complete graph with n vertices there are (n - 1)/2 edge-disjoint Hamiltonian circuits, ifn is an

odd number 3.

9. TRAVELLING SALESMAN PROBLEM

A sales man is required to visit a number of cities during a trip. Given the distance between

cities, in what order should he travel so as to visit every city precisely once and return home covering

shortest distance.

We can try to approach this problem of finding a shortest rout by constructing a graph. Let us

represent cities by vertices in a graph and roads by edges connecting cities. Since we are given lengths of

roads, so this graph is a weighted graph. The traveling salesman has to visit each city (vertex in graph)

exactly once and has to return home (the vertex from where he starts). Therefore, the problem is to find a

shortest Hamiltonian circuit. Let us consider that there exists a road between every two city, i.e., the

graph is a complete graph. We know that in a complete graph of n vertices there exist (n -1)! /2

Hamiltonian circuits, therefore, one possible approach to get the shortest rout is to calculate the weights

of all the (n -1)! /2 Hamiltonian circuits and find the shortest one.

Note. As we can notice that the approach discussed above is practically impossible because of the

cumbersome calculation involved. For example, if n = 10, the number of Hamiltonian circuits are

181440. Unfortunately, we do not know any efficient algorithm for solving this problem.

PROBLEM 1.27: Solve the traveling salesman problem in the following graph:

1.1. 13

1.2. 23

1.3. In (i) and (ii) there are odd number of odd degree vertices, such a graph is not possible. (iii) Thee r four vertices

given and maximum degree is 6, graph is not possible because maximum degree cannot exceed one less than thenumber of vertices.

1.5. Let G be a k regular graph with n vertices and e edges. We know that

Sum of degree of all n vertices = 2e

i.e., k+ k+ k+ .. n times = 2 e, n . k= 2 e, e = n k/ 2

1.9. Given that G is not connected, that is, G is a disconnected graph. This implies that G has more than one

component, which are connected. Let u and v be any two vertices. To show that G is connected, we need to show

that thereexists a path between u and v.

Ifu and v are vertices in different components, then u and v are not adjacent in G. Hence they are adjacent

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ANSWERS & HINTS TO PROBLEMS

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in G. Next, let us suppose that u and v are in the same component of G. Let w be a vertex of G but in other

component of G. Obiviously, u w v make a path u v in G. Hence G is connected.

1.10. Let in a graph G, the vertices u and v are of odd degree and rest of the vertices are of even degree. Since we

know that in any graph the number of odd vertices are always even in number, therefore the two odd vertices of G

must belong to the same component ofG. Hence there is a path between u and v.

1.12. In K6, no. of vertices is 6 and no. of edges is 15, therefore total number of subgraphs is (26

- 1) 215

, and total

number of spanning subgraphs is 215

.

1.13. (i) Isomorphic (ii) Not isomorphic

1.15.

1.17. Since G and G are isomorphic, so they have same number of edges. Also total nuber of edges in G and G

together equals to number of edges in Kn, that is , n (n 1) /2. Thus, each G and G have n (n 1) /4 edges. It now

follows from above that n (n 1) /4 is a positive integer, and therefore, n or (n 1) must be divisible by 4.

1.19. (i) Degree of each vertex in K5 is even, it is Eulerian. (ii) Each vertex in K2, 3 is not even, it is not Eulerian.

1.20. (i) Not Eulerian (ii) Eulerian

1.22. In Kn, the degree of each vertex is n 1. . If n > 2, we have n 2 > 0 or n 2 > 2 n or n 1 > n /2.

Thus, in Kn , n 3, the degree of every vertex is greater than n /2. Hence Kn is Hamiltonian.

1.24. (i) Hamiltonian (ii) Not Hamiltonian

1.25. Let u and v be two vertices ofG. We know that

degu + degv nor, degu + degv

n/2 + n/2

By comparison (comparison test in binomial expressions) for n =1, we at once get degu n/2, degv n/2.1.27. The graph has two Hamiltonian circuits: A B C D E F H G A and A B C D E F G H A . The total weight of the

circuits are 22 and 25 respectively. Therefore, salesman should travel according to the first circuit A B C D E F H

G A.

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