Graph-Theory

Question 1
Consider a simple undirected graph of 10 vertices. If the graph is disconnected, then the maximum number of edges it can have is ____________.
A
36
Question 1 Explanation: 
We obtain maximum number of edges in a disconnected graph, with one isolated vertex and remaining completely connected.
Given 10 vertices,
Then we can have a complete graph with 9 vertices and one isolated verted.
Number of edges in complete graph with 9 edges is n(n-1)/2 = 9*8/2 = 36
Question 2
Consider a simple undirected unweighted graph with at least three vertices. If A is the adjacency matrix of the graph, then the number of 3-cycles in the graph is given by the trace of
A
A ^3
B
A^ 3 divided by 2
C
A ^3 divided by 3
D
A ^3 divided by 6
Question 3
The following simple undirected graph is referred to as the Peterson graph.

Which of the following statements is/are TRUE?
A
The chromatic number of the graph is 3.
B
The graph has a Hamiltonian path
C
The following graph is isomorphic to the Peterson graph.
D
The size of the largest independent set of the given graph is 3. (A subset of vertices of a graph form an independent set if no two vertices of the subset are adjacent.)
Question 3 Explanation: 
Chromatic number is 3.

Peterson graph ihas hamiltonian path but not hamiltonian cycle
Given graph in option C is isomorphic as the given has same number of vertice, edges, degree sequence and cycles.
LArgest independent set can be more than 3
Question 4
Which of the properties hold for the adjacency matrix A of a simple undirected unweighted graph having n vertices?
A
The diagonal entries of A 2 are the degrees of the vertices of the graph.
B
If the graph is connected, then none of the entries of A^ n + 1 + I n can be zero.
C
If the sum of all the elements of A is at most 2( n- 1), then the graph must be acyclic.
D
If there is at least a 1 in each of A ’s rows and columns, then the graph must be Connected.
Question 4 Explanation: 
IF A is adjacency matrix, then in the matrix A*A = A^2 we have
The entries aii show the number of 2-length paths between the nodes i and j. Why this happens is easy to see: if there is an edge ij and an edge jk, then there will be a path ik through j. The entries ii are the degrees of the nodes i.
Similarly in A^3 we have the entries aii that show the number of 3-length paths between the nodes i and j.
In A^n-1 + I n, we will have at least n-1 length paths, so there is no possibility of zero entires
Question 5

The number of distinct simple graphs with upto three nodes is

A
15
B
10
C
7
D
9
Question 5 Explanation: 
Question 6

The number of edges in a regular graph of degree d and n vertices is _________.

A
d*n/2
Question 6 Explanation: 
Sum of degree of vertices = 2 × no. of edges
d * n = 2 * |E|
∴ |E| = d*n/2
Question 7

Graph G is obtained by adding vertex s to K3,4 and making s adjacent to every vertex of K3,4. The minimum number of colours required to edge-colour G is _____.

A
7
Question 7 Explanation: 
In k3x4 there are two sets with sizes 3,4. (it is a complete bipartite graph).
The vertex in the set of size 3 has 4 edges connected to 4 vertices on other set. So, edge color of G is max(3,4) i.e. 4.
When a vertex is added to the graph with 7 vertices ( K3x4 has 7 vertices), there would be 7 edges associated to that new vertex. As per the edge coloring “no two adjacent edges have same color).
As the new vertex with 7 edges need to be colored with 7 colors, the edge color of graph G is 7.
Question 8

Let G be a graph with 100 vertices numbered 1 to 100. Two vertices i and j are adjacent if |i - j| = 8 or |i - j| = 12. The number of connected components in G is

A
8
B
4
C
12
D
25
Question 8 Explanation: 
From the description, it is clear that vertices are connected as follows:
1 — 9 — 17 — ......... — 97
2 — 10 — 18 — ......... — 98
3 — 11 — 19 — ......... — 99
4 — 12 — 20 — ......... — 100
5 — 13 — 21 — ......... — 93
6 — 14 — 22 — ......... — 94
7 — 15 — 23 — ......... — 95
8 — 16 — 24 — ......... — 96
We have covered all vertices using 8 vertex sets considering only |i - j| = 8. Using |i - j| = 12 we can see the vertex 1 is connected to 13, 2 to 14, 3 to 15 and 4 to 16. So the top 4 vertex sets are infact connected to the bottom 4 sets, thus reducing the connected components to 4.
Question 9

(a) Prove by induction that the expression for the number of diagonals in a polygon of n sides is n(n-3)/2.

(b) Let R be a binary relation on A = {a, b, c, d, e, f, g, h} represented by following two component digraph. Find the smallest integers m and n such that mm = Rn.

A
Theory Explanation.
Question 10

Let (A, *) be a semi group. Furthermore, for every a and b in A, if a ≠ b, then a*b ≠ b*a.

(a) Show that for every a in A

 a*a = a 

(b) Show that for every a, b in A

 a*b*a = a  

(c) Show that for every a, b, c in A

 a*b*c = a*c 

A
Theory Explanation.
Question 11

Let G be a connected, undirected graph. A cut in G is a set of edges whose removal results in 0 being broken into two or more components which are not connected with each other. The size of a cut is called its cardinality. A men-cut of G is a cut in G of minimum cardinality. Consider the following graph.

(a) Which of the following sets of edges is a cut?
(i) {(A,B), (E,F), (B,D), (A,E), (A,D)}
(ii) {(B,D), (C,F), (A,B)}

(b) What is the cardinality of a min-cut in the graph?

(c) Prove that if a connected undirected graph G with n vertices has a min-cut of cardinality K, then G has atleast (nk/2) edges.

A
Theory Explanation.
Question 12

Let S be a set of n elements {1, 2, …., n} and G a graph with 2n vertices, each vertex corresponding to a distinct subset of S. Two vertices are adjacent iff the symmetric difference of the corresponding sets has exactly 2 elements. Note: The symmetric difference of two sets R1 and R2 is defined as (R1/R2)∪(R2/R1)
(a) Every vertex in G has the same degree. What is the degree of a vertex in G?
(b) How many connected components does G have?

A
Theory Explanation is given below.
Question 13

How many undirected graphs (not necessarily connected) can be constructed out of a given set V = {v1, v2, ...,vn} of n vertices?

A
n(n-1)/2
B
2n
C
n!
D
2n(n-1)/2
Question 13 Explanation: 
With n vertices no. of possible edges = n C 2
Each subset of these edges will be form a graph.
No. of possible undirected graphs is 2(n C 2)
⇒ 2(n(n-1)/2)
Question 14

The minimum number of colours required to colour the vertices of a cycle with η nodes in such a way that no two adjacent nodes have the same colour is

A
2
B
3
C
4
D
n - 2[n/2] + 2
Question 14 Explanation: 
We need 2 colours to colour even cycle and 3 colours to colour odd cycle.
Question 15

Maximum number of edges in a n-node undirected graph without self loops is

A
n2
B
n(n-1)/2
C
n-1
D
(n+1)(n)/2
Question 15 Explanation: 
The set of vertices has size n, the number of such subsets is given by the binomial coefficient C(n, 2)⋅ C(n, 2) = n(n-1)/2.
Question 16

Let G be an arbitrary graph with n nodes and k components. If a vertex is removed from G, the number of components in the resultant graph must necessarily lie between

A
k and n
B
k – 1 and k + 1
C
k – 1 and n – 1
D
k + 1 and n – k
Question 16 Explanation: 
While a vertex is removed from a graph then that can be itself be forms a new component. The minimum number of components is k-1.
If a vertex is removed then it results that all the components are also be disconnected. So removal can create (n-1) components.
Question 17

How many perfect matching are there in a complete graph of 6 vertices ?

A
15
B
24
C
30
D
60
Question 17 Explanation: 
We have formula to find no. of perfect matching in complete graphs of 2n vertices,
(2n)!/n!×2n
Given, 2n = 6 ⇒ n = 3
So, finally, 6!/3!×23 = 15
Question 18

A graph G = (V,E) satisfies |E|≤ 3|V|-6. The min-degree of G is defined as . Therefore, min-degree of G cannot be

A
3
B
4
C
5
D
6
Question 18 Explanation: 
The minimum degree of G = minv∈V {degree(v)}
|E| ≤ 3|v| - 6
Based on handshaking lemma, the minimum degree is (min×|v|)/2
⇒ (min×|v|)/2 ≤ 3|v| - 6
Checking the options lets take min=6
(6×|v|)/2 ≤ 3|v| - 6
0 ≤ -6 (Not satisfied)
And which is inconsistent.
Question 19

The minimum number of colours required to colour the following graph, such that no two adjacent vertices are assigned the same colour, is

A
2
B
3
C
4
D
5
Question 19 Explanation: 

→ a, b, c, d = 4
→ The minimum no. of colours required to colour a graph = 4 (no two adjacent vertices have same colours)
Question 20

How many graphs on n labeled vertices exist which have at least (n2 - 3n)/2 edges?

A
B
C
D
Question 20 Explanation: 
No. of atleast edges = (n2-3n)/2 = e
Maximum no. of vertices = n(n-1)/2 = v
No. of graphs with minimum b edges is
= C(v,e) + C(v,e+1) + C(v,e+2) + ... + C(v,v)
= C((v,v-e) + C(v,v-(e+1)) + C(v,v-(e+2)) + ... + C(v,0)
= C(a,n) + C(a,n-1) + C(a,n-2) + ... + C(a,0) (since a-b=n)
= C(n(n-1)/2,n) + C(n(n-1)/2,n-1) + ... + C(n(n-1)/2,0)
Question 21

Let G1 = (V,E1) and G2 = (V,E2) be connected graphs on the same vertex set V with more than two vertices. If G1 ∩ G2 = (V, E1 ∩ E2) is not a connected graph, then the graph G1 U G2 = (V, E1 U E2)

A
cannot have a cut vertex
B
must have a cycle
C
must have a cut-edge (bridge)
D
has chromatic number strictly greater than those of G1 and G2
Question 21 Explanation: 
Lets try to take counter example for each of them,
(A)

False, since in G1∪G2 'C' is a cut vertex.
(B) True, for all conditions.
(C)

False. G1∪G2 has no bridge.
D)

False. G1∪G2, G1, G2 all the three graphs have chromatic number of 2.
Question 22

Maximum number of edges in a planar graph with n vertices is ________

A
3n - 6
Question 22 Explanation: 
The maximum is 3(n - 8) for every n>2.
⇒ (3n - 2) = 3n - 6
Question 23

A non-planar graph with minimum number of vertices has

A
9 edges, 6 vertices
B
6 edges, 4 vertices
C
10 edges, 5 vertices
D
9 edges, 5 vertices
Question 23 Explanation: 

The above graph with 5 vertices and 10 edges is non-planar.
Question 24

Let T be a Depth First Tree of a undirected graph G. An array P indexed by vertices of G is given. P[V] is the parent of vertex V, in T. Parent of the root is the root itself.

Give a method for finding and printing the cycle formed if the edge (u,v) of G not in T (i.e., e ∈ G − T) is now added to T.

Time taken by your method must be proportional to the length of the cycle.

Describe the algorithm in a PASCAL – like language. Assume that the variables have been suitably declared.

A
Theory Explanation.
Question 25

(a) If G is a group of even order, then show that there exists an element a ≠ e, the identity in g, such that a2 = e

(b) Consider the set of integers {1,2,3, 4,6,8,12,24} together with the two binary operations LCM (lowest common multiple) and GCD (greatest common divisor). Which of the following algebraic structures does this represent?

 (i) group            (ii) ring
 (iii) field          (iv) lattice  
A
Theory Explanation.
Question 26

What is the maximum number of edges in an acyclic undirected graph with n vertices?

A
n - 1
B
n
C
n + 1
D
2n - 1
Question 26 Explanation: 
Maximum number of edges in an acyclic undirected graph = No. of vertices - 1
= n - 1
Question 27

What is the number of vertices in an undirected connected graph with 27 edges, 6 vertices of degree 2, 3 vertices of degree 4 and remaining of degree 3?

A
10
B
11
C
18
D
19
Question 27 Explanation: 
Let x = Total no. of vertices
By Handshaking Lemma,
6 * 2 + 3 * 4 + (x - 9) * 3 = 27 * 2
24 + (x - 9) * 3 = 54
x = 19
Question 28

Let G be a weighted undirected graph and e be an edge with maximum weight in G. Suppose there is a minimum weight spanning tree in G containing the edge e. Which of the following statements is always TRUE?

A
There exists a cutset in G having all edges of maximum weight
B
There exists a cycle in G having all edges of maximum weight
C
Edge e cannot be contained in a cycle
D
All edges in G have the same weight
Question 28 Explanation: 
(A) True, because if there is heaviest edge in MST, then there exist a cut with all edges with weight equal to heaviest edge.

(B) False, because the cutset of heaviest edge may contain only one edge.
(C) False. The cutset may form cycle with other edge.
(D) False. Not always true.
Question 29

Let G be a directed graph whose vertex set is the set of numbers from 1 to 100. There is an edge from a vertex i to a vertex j iff either j = i + 1 or j = 3i. The minimum number of edges in a path in G from vertex 1 to vertex 100 is

A
4
B
7
C
23
D
99
Question 29 Explanation: 
Edge set consists of edges from i to j, using either
j = i +1
(or)
j = 3i
Second option will help us reach from 1 to 100 rapidly. The trick to solve this question is to think in reverse way. Instead of finding a path from 1 to 100, try to find a path from 100 to 1.
So, the edge sequence with minimum number of edges is
1 → 3 → 9 → 10 → 11 → 33 → 99 → 100
which consists of 7 edges.
Question 30

Consider the undirected graph G defined as follows. The vertices of G are bit strings of length n. We have an edge between vertex u and vertex v if and only if u and v differ in exactly one bit position (in other words, v can be obtained from u by flipping a single bit). The ratio of the chromatic number of G to the diameter of G is

A
1/(2n-1)
B
1/n
C
2/n
D
3/n
Question 30 Explanation: 
For the given condition we can simply design a K-map and mark an edge between every two adjacent cells in K-map.
That will give us a bipartite graph, with chromatic number = 2
Also from the same we can conclude that we need for a 'n' bit string, to traverse no more than (n-1) edges or 'n' vertices to get a path between two arbitrary points. So the ratio is (2/n).
Question 31

What is the chromatic number of the following graph?

A
2
B
3
C
4
D
5
Question 31 Explanation: 
Chromatic number = 3
→ Chromatic number of a graph is the smallest number of colours needed to colour the vertices so that no two adjacent vertices share the same colour.
Question 32

What is the size of the smallest MIS(Maximal Independent Set) of a chain of nine nodes?

A
5
B
4
C
3
D
2
Question 32 Explanation: 
1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9
(2, 5, 8) is the maximal independent set for a chain of 9 nodes. If we add any start node to the set then it will not be MIS.
Independent set:
A set of vertices is called independent set such that no two vertices in the set are adjacent.
Question 33

G is a simple undirected graph. Some vertices of G are of odd degree. Add a node v to G and make it adjacent to each odd degree vertex of G. The resultant graph is sure to be

A
Regular
B
Complete
C
Hamiltonian
D
Euler
Question 33 Explanation: 
Euler graph theory is a trail in a finite graph which visits every edge exactly once and which is a undirected graph.
→ In Euler graph all degrees must be even for all nodes. And number of odd degree vertices should be even.
→ So, degree of this new node will be even and as a new edge is formed between this new node and all other nodes of odd degree hence here is not a single node exists with degree odd so this is Euler graph.
Question 34

Choose the correct alternatives (More than one may be correct).

A graph is planar if and only if,
A
It does not contain subgraphs homeomorphic to k5 and k3,3.
B
It does not contain subgraphs isomorphic to k5 or k3,3.
C
It does not contain a subgraph isomorphic to k5 or k3,3.
D
It does not contain a subgraph homeomorphic to k5 or k3,3.
Question 34 Explanation: 
A graph is non-planar if and only if it contains a subgraph which is homomorphic to k5or k3,3. This is kuratowshi theorem.
Question 35

Which of the following graphs is/are planner?

A
Theory Explanation is given below.
Question 35 Explanation: 
(i) G1 is K33 which is planar graph with the minimum number of edges.
→ Let us assume K33is a planar graph. Then it satisfy the useful corollary. As there is no triangle in K33.
Let G be a connected planar graph with n vertices and m edges, and no triangles. Then m≤2n-4.
Where m=9, n=6
⇒ 9 ≤ 12 - 4
⇒ 9 ≤ 8, which is to be false, then K33 is non-planar graph.
(ii) G2 is a planar graph. Because it can be redrawn like as below.

(iii) Let us assume G3 be a planar graph then it also be satisfy useful corollary.
Where m=9, n=6
then 9 ≤ 12-4
9 ≤ 8 is False
So, G3 is non-planar graph.
Answer: Only G2 is planar graph.
Question 36

The minimum number of colours that is sufficient to vertex-colour any planar graph is ________.

A
4
B
5
C
6
D
7
Question 36 Explanation: 
The 4-colour theorem of the planar graph describes that any planar can atmost be colored with 4 colors.
Here it is asked about the sufficient number of colors, so with the worst case of 4 colors we can color any planar graph.
Question 37

The maximum number of possible edges in an undirected graph with a vertices and k components is ________.

A
(n-k)(n-k+1)/2
Question 37 Explanation: 
N vertices and K components.
To get maximum, take one vertex each for each component, except last component.
Now, k-1 components have 1 vertex each and so on edges.
The last component has (n-(k-1)) vertices. So make the last component complete, i.e., it has n-(n-1)C2
= (n-k)(n-k+1)/2
= maximum no. of edges
Question 38

Let G be a simple connected planar graph with 13 vertices and 19 edges. Then, the number of faces in the planar embedding of the graph is:

A
6
B
8
C
9
D
13
Question 38 Explanation: 
No. of faces in a planar embedding of a graph is
F = E - V + 2 [From Euler's formula i.e., F + V - E = 2]
F = 19 - 13 +2
F = 8
Question 39

Let G be a simple graph with 20 vertices and 100 edges. The size of the minimum vertex cover of G is 8. Then, the size of the maximum independent set of G is

A
12
B
8
C
Less than 8
D
More than 12
Question 39 Explanation: 
No. of vertices = 20
Edges = 100
Minimum cover of vertex G is = 8
Maximum Independent set of G = No. of vertices - Minimum cover of vertex G
= 20 - 8
= 12
Question 40

Which one of the following graphs is NOT planar?

A
G1
B
G2
C
G3
D
G4
Question 40 Explanation: 
G2 can also be drawn as

which is planar
G3 can also be drawn as

which is planar
G4 can also be drawn as

which is planar
But G1 cannot be drawn as planar graph.
Hence, option (A) is the answer.
Question 41

The 2n vertices of a graph G corresponds to all subsets of a set of size n, for n ≥ 6. Two vertices of G are adjacent if and only if the corresponding sets intersect in exactly two elements.

The number of vertices of degree zero in G is:

A
1
B
n
C
n+1
D
2n
Question 41 Explanation: 
No. of vertices with degree zero is
= no. of subsets with size less than or equal to 1
= n+1, because in question it is given that the two vertices are connected if and only if the corresponding sets intersect in exactly two elements.
Question 42

The 2n vertices of a graph G corresponds to all subsets of a set of size n, for n ≥ 6. Two vertices of G are adjacent if and only if the corresponding sets intersect in exactly two elements.

The maximum degree of a vertex in G is:

A
B
2n-2
C
2n-3 × 3
D
2n-1
Question 42 Explanation: 
The degree of each subset with k elements will be
(k(c))2 2n-k
∴ We need to find 'k' value such that, the value will be maximum.[k should be an integer].
If you differentiate (k(c))2 2n-k w.r.t. k and equal to 0.
You will get k = 2/(loge)2 which is not an integer.
So you can see it like

∴ The maximum degree 3⋅2n-3 at k=3 or k=4.
Question 43

The 2n vertices of a graph G corresponds to all subsets of a set of size n, for n ≥ 6. Two vertices of G are adjacent if and only if the corresponding sets intersect in exactly two elements.

The number of connected components in G is:

A
n
B
n+2
C
2n/2
D
2n / n
Question 43 Explanation: 
Not connected nodes is n+1.
While other nodes are connected so that total number of connected components is (n+1)+1
(here we are adding 1 because it is connected corresponding remaining vertices)
= n+2
Question 44

Let G be the non-planar graph with the minimum possible number of edges. Then G has

A
9 edges and 5 vertices
B
9 edges and 6 vertices
C
10 edges and 5 vertices
D
10 edges and 6 vertices
Question 44 Explanation: 
Using Euler’s formula we know that,
if n ≥ 3 then e ≤ 3n-6 (for planarity)
where n = no. of vertices
e = no. of edges
Now lets check the options.
A) e=9, n=5
9 ≤ 3(5) - 6
9 ≤ 15 - 6
9 ≤ 9
Yes, it is planar.
B) e=9, n=6
9 ≤ 3(6) - 6
9 ≤ 18 - 6 9 ≤ 12 Yes, it is planar.
iii) e=10, n=5
10 ≤ 3(5) - 6
10 ≤ 15 - 6
10 ≤ 9 No, it is not planar.
So, option C is non-planar graph.
iv) e=10, n=6
10 ≤ 3(6) - 6
10 ≤ 18 - 6
10 ≤ 12
Yes, it is planar.
Question 45

Which of the following graphs has an Eulerian circuit?

A
Any k-regular graph where k is an even number.
B
A complete graph on 90 vertices.
C
The complement of a cycle on 25 vertices.
D
None of the above.
Question 45 Explanation: 
Two necessary condition for the existence of Eulerian circuits is
→ all vertices in the graph have an "even degree".
→ And the graph must be corrected.
Now in option (C) it is saying that the complement of a cycle on 25 vertices without complement the degree of each vertex is 2.
Now since there are 25 vertices, so maximum degree of each vertex will be 24 and so in complement of cycle each vertex degree will be 24 - 2 = 22.
There is a theorem which says "G be a graph with n vertices and if every vertex has a degree of atleast n-1/2 then G is connected."
So we can say that complement of cycle with 25 vertices fulfills both the conditions, and hence is Eulerian circuit.
Question 46

The minimum number of edges in a connected cyclic graph on n vertices is:

A
n - 1
B
n
C
n + 1
D
None of the above
Question 46 Explanation: 
In a normal graph number of edges required for n vertices is n-1, and in cyclic graph it is n.
In cyclic graph:
No. of edges = No. of vertices
⇒ n = n
Question 47

Prove that in finite graph, the number of vertices of odd degree is always even.

A
Theory Explanation.
Question 48

In a connected graph, a bridge is an edge whose removal disconnects a graph. Which one of the following statements is true?

A
A tree has no bridges
B
A bridge cannot be part of a simple cycle
C
Every edge of a clique with size 3 is a bridge (A clique is any complete sub graph of a graph)
D
A graph with bridges cannot have a cycle
Question 48 Explanation: 
Since, every edge in a tree is bridge
∴ (A) is false
Since, every edge in a complete graph kn(n≥3) is not a bridge ⇒
(C) is false
Let us consider the following graph G:

This graph has a bridge i.e., edge ‘e’ and a cycle of length ‘3’
∴ (D) is false
Since, in a cycle every edge is not a bridge
∴ (B) is true
Question 49

A graph is self-complementary if it is isomorphic to its complement for all self-complementary graphs on n vertices, n is

A
A multiple of 4
B
Even
C
Odd
D
Congruent to 0 mod 4, or, 1 mod 4
Question 49 Explanation: 
An n vertex self-complementary graph has exactly half number of edges of the complete graph i.e., n(n-1)/4 edges. Since n(n – 1) must be divisible by 4, n must be congruent to 0 or 1 module 4.
Question 50

Let G be an undirected complete graph on n vertices, where n > 2. Then, the number of different Hamiltonian cycles in G is equal to

A
n!
B
1
C
(n-1)!
D
Question 50 Explanation: 
A Hamiltonian cycle is a closed loop on a graph where every node (vertex) is visited exactly once.
The total number of hamiltonian cycles in a complete graph are
(n-1)!/2, where n is number of vertices.
Question 51

The chromatic number of the following graph is _______.

A
1
B
2
C
3
D
4
Question 51 Explanation: 
Chromatic number of the following graph is “3”
Question 52

Let G be a graph with 100! vertices, with each vertex labeled by a distinct permutation of the numbers 1, 2, …, 100. There is an edge between vertices u and v if and only if the label of u can be obtained by swapping two adjacent numbers in the label of v. Let y denote the degree of a vertex in G, and z denote the number of connected components in G.

Then, y + 10z = ___________.

A
109
B
110
C
111
D
112
Question 52 Explanation: 
G is a graph with 100! vertices. Label of each vertex obtains from distinct permutation of numbers “1, 2, … 100”.
There exists edge between two vertices iff label of ‘u’ is obtained by swapping two adjacent numbers in label of ‘v’.
Example:
12 & 21, 23 & 34
The sets of the swapping numbers be (1, 2) (2, 3) (3, 4) … (99).
The no. of such sets are 99 i.e., no. of edges = 99.
As this is regular, each vertex has ‘99’ edges correspond to it.
So degree of each vertex = 99 = y.
As the vertices are connected together, the number of components formed = 1 = z
y + 102 = 99 + 10(1) = 109
Question 53

Let G be a connected planar graph with 10 vertices. If the number of edges on each face is three, then the number of edges in G is _______________.

A
24
B
25
C
26
D
27
Question 53 Explanation: 
By Euler’s formula,
|V| + |R| = |E| + 2 ------(1) where |V|, |E|, |R| are respectively number of vertices, edges and faces (regions)
Given |V| = 10 ------(2) and number of edges on each face is three
∴3|R| = 2|E| ⇒ |R| = 2/3|E| ------(3)
Substituting (2), (3) in (1), we get
10 + 2/3|E| = |E| + 2 ⇒ |E|/3 = 8 ⇒ |E| = 24
Question 54

G is an undirected graph with n vertices and 25 edges such that each vertex of G has degree at least 3. Then the maximum possible value of n is ___________.

A
16
B
17
C
18
D
19
Question 54 Explanation: 
An undirected graph ‘G’ has ‘n’ vertices & 25 edges.
Degree of each vertex ≥ 3

|v| = 2|E|
The relation between max and min degree of graph are
m ≤ 2|E| / |v| ≤ M
Given minimum degree = 3
So, 3 ≤2 |E| / |v|
3|v| ≤ 2|E|
3(n) ≤ 2(25)
n ≤ 50/3
n ≤ 16.6
(n = 16)
Question 55

Let G be a simple undirected planar graph on 10 vertices with 15 edges. If G is a connected graph, then the number of bounded faces in any embedding of G on the plane is equal to

A
3
B
4
C
5
D
6
Question 55 Explanation: 
If the graph is planar, then we have to consider Euler’s formula
v-e+f = 2
Given 10 vertices & 15 edges
10-15+f = 2
f = 2+15-10
f = 7
There will be an unbounded face always. So, number of faces = 6.
Question 56

Which of the following graphs is isomorphic to

A
B
C
D
Question 56 Explanation: 
Original graph:

(A) 3 cycle graph not in original one.

(B) Correct 5 cycles & max degree is 4.
(C) Original graph doesn’t have a degree of 3.

(D) 4 cycles not in original one.
Question 57

Let G be a complete undirected graph on 6 vertices. If vertices of G are labeled, then the number of distinct cycles of length 4 in G is equal to

A
15
B
30
C
45
D
360
Question 57 Explanation: 
Complete graph means there exists an edge between every pair of vertices.
It is asked to find the distinct cycle of length 4. As it is complete graph, if we chose any two vertices, there will be an edge.
So, to get a cycle of length 4 (means selecting the 4 edges which can form a cycle) we can select any four vertices.
The number of such selection of 4 vertices from 6 vertices is 6C4 => 15.
From each set of 4 vertices, suppose a set {a, b, c, d} we can have cycles like
a-b-c-d
a-b-d-c
a-c-b-d
a-c-d-b
a-d-b-c
a-d-c-b (Total 6, which is equal to number of cyclic permutations (n-1)! )
As they are labelled you can observe, a-b-c-d and a-d-c-b are same, in different directions.
So, we get only three combinations from the above 6.
So, total number of distinct cycles of length 4 will be 15*3 = 45.
If it is asked about just number of cycles then 15*6 = 90
Question 58
The chromatic number of a graph is the minimum number of colours used in a proper colouring of the graph. Let G be any graph with n vertices and chromatic number k Which of the following statements is/are always TRUE?
A
G contains a complete subgraph with k vertices
B
G contains an independent set of size at least n/k
C
G contains at least k(k−1)/2 edges
D
G contains a vertex of degree at least k
E
Question 58 Explanation: 
The existence of a proper k-coloring implies the existence of color classes (sets of vertices with the same color) that form independent sets. In a worst-case scenario, where all color classes except one have fewer than n/k vertices, the remaining color class must have at least n/k vertices to ensure all vertices are colored.
Dense graphs with a chromatic number k are more likely to have at least k(k-1)/2 edges
Question 59
The number of edges present in the forest generated by the DFS traversal of an undirected graph G with 100 vertices is 40. The number of connected components in G is _________
A
60
B
C
D
E
Question 59 Explanation: 
Let K be the number of connected components (or trees) in the forest. Number of edges=V-K 40=100-K K=60
Question 60

Let G = (V,E) be a directed graph where V is the set of vertices and E the set of edges. Then which one of the following graphs has the same strongly connected components as G?

A
G1=(V,E1) where E1={(u,v)|(u,v)∉E}
B
G2=(V,E2 )where E2={(u,v)│(u,v)∈E}
C
G3=(V,E3) where E3={(u,v)|there is a path of length≤2 from u to v in E}
D
G4=(V4,E) where V4 is the set of vertices in G which are not isolated
Question 60 Explanation: 
G(V, E) is a directed graph.
→ It strongly connected.
(A) G1=(V,E1) where E1={(u,v)|(u,v)∉E}
If (u, v) does not belong to the edge set ‘E’, then it indicates there are no edges. So, it is not connected.
(B) G2=(V,E2 )where E2={(u,v)│(u,v)∈E}
Given that ‘G’ is directed graph, i.e., it has path from each vertex to every other vertex.
Though direction is changed from (u, v) to (v, u), it is still connected component same as ‘G’.
(C) G3=(V,E3) where E3={(u,v)|there is a path of length≤2 from u to v in E}
This can also be true.
eg:

Both from each vertex to other vertex is also exists. So it is also strongly connected graph.
(D) G4=(V4,E) where V4 is the set of vertices in G which are not isolated.
If ‘G’ has same ‘x’ no. of isolated vertices, one strongly connected component
then no. of SCC = x + 1
G4 contain only ‘1’ component, which is not same as G.
Question 61

Consider an undirected graph where self-loops are not allowed. The vertex set of G is {(i,j): 1 ≤ i ≤ 12, 1 ≤ j ≤ 12}. There is an edge between (a,b) and (c,d) if |a - c| ≤ 1 and |b - d| ≤ 1. The number of edges in this graph is __________.

A
506
B
507
C
508
D
509
Question 61 Explanation: 
The total number of vertices in the graph is 12*12 = 144. The vertices are allowed to connect in both horizontal and vertical directions which are separated by at most 1 distance.
If we observe the graph, it looks like a 12 by 12 grid. Each corner vertex has a degree of 3 and we have 4 corner vertices. 40 external vertices of degree 5 and remaining 100 vertices of degree 8.
From Handshaking theorem, sum of the degrees of the vertices is equal to the 2*number of edges in the graph.
⇒ (4*3) + (40*5) + (100*8) = 2*E
⇒ 1012 = 2*E
⇒ E = 506
Question 62

An ordered -tuple (d1, d2, ..., dn) with d1 ≥ d2 ≥ ... dn is called graphic if there exists a simple undirected graph with n vertices having degrees d1, d2, ..., dn respectively. Which of the following 6-tuples is NOT graphic?

A
(1, 1, 1, 1, 1, 1)
B
(2, 2, 2, 2, 2, 2)
C
(3, 3, 3, 1, 0, 0)
D
(3, 2, 1, 1, 1, 0)
Question 62 Explanation: 
This can be checked by Havel-hakimi theorem:
A) (1, 1, 1, 1, 1, 1)

Yes, it is a graph.
We will see that option (C) is not graphic.
There are 62 questions to complete.

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