Question 1 |

Consider an undirected random graph of eight vertices. The probability that there is an edge between a pair of vertices is 1/2. What is the expected number of unordered cycles of length three?

1/8 | |

1 | |

7 | |

8 |

**GATE CS 2013**

**Graph Theory**

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Question 1 Explanation:

Question 2 |

Which of the following statements is/are TRUE for undirected graphs?

P: Number of odd degree vertices is even. Q: Sum of degrees of all vertices is even.

P only | |

Q only | |

Both P and Q | |

Neither P nor Q |

**GATE CS 2013**

**Graph Theory**

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Question 2 Explanation:

Question 3 |

The line graph L(G) of a simple graph G is defined as follows:
· There is exactly one vertex v(e) in L(G) for each edge e in G.
· For any two edges e and e' in G, L(G) has an edge between v(e) and v(e'), if and only if e and e'are incident with the same vertex in G.
Which of the following statements is/are TRUE?

(P) The line graph of a cycle is a cycle. (Q) The line graph of a clique is a clique. (R) The line graph of a planar graph is planar. (S) The line graph of a tree is a tree.

P only | |

P and R only | |

R only | |

P, Q and S only |

**GATE CS 2013**

**Graph Theory**

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Question 3 Explanation:

Question 4 |

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

3 | |

4 | |

5 | |

6 |

**GATE CS 2012**

**Graph Theory**

**Discuss it**

Question 4 Explanation:

If the graph is planar, then it must follow below Euler's Formula for planar graphs

```
v - e + f = 2
v is number of vertices
e is number of edges
f is number of faces including bounded and unbounded
10 - 15 + f = 2
f = 7
There is always one unbounded face, so the number of bounded faces = 6
```

Question 6 |

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

15 | |

30 | |

45 | |

360 |

**GATE CS 2012**

**Graph Theory**

**Discuss it**

Question 6 Explanation:

There can be total

^{6}C_{4}ways to pick 4 vertices from 6. The value of^{6}C_{4}is 15. Note that the given graph is complete so any 4 vertices can form a cycle. There can be 6 different cycle with 4 vertices. For example, consider 4 vertices as a, b, c and d. The three distinct cycles are cycles should be like this (a, b, c, d,a) (a, b, d, c,a) (a, c, b, d,a) (a, c, d, b,a) (a, d, b, c,a) (a, d, c, b,a) and (a, b, c, d,a) and (a, d, c, b,a) (a, b, d, c,a) and (a, c, d, b,a) (a, c, b, d,a) and (a, d, b, c,a) are same cycles. So total number of distinct cycles is (15*3) = 45. **NOTE**: In original GATE question paper 45 was not an option. In place of 45, there was 90.Question 7 |

K4 is planar while Q3 is not | |

Both K4 and Q3 are planar | |

Q3 is planar while K4 is not | |

Neither K4 nor Q3 are planar |

**GATE CS 2011**

**Graph Theory**

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Question 7 Explanation:

A Graph is said to be planar if it can be drawn in a plane without any edges crossing each other.
Following are planar embedding of the given two graphs (Source: http://clweb.csa.iisc.ernet.in/rahulsharma/gate2011key.html)

Question 8 |

Let G = (V,E) be a graph. Define ξ(G) = Σd id x d, where id is the number of vertices of degree d in G. If S and T are two different trees with ξ(S) = ξ(T),then

|S| = 2|T| | |

|S| = |T|-1 | |

|S| = |T| | |

|S| = |T|+1 |

**GATE CS 2010**

**Graph Theory**

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Question 8 Explanation:

The expression ξ(G) is basically sum of all degrees in a tree. For example, in the following tree, the sum is 3 + 1 + 1 + 1.

a / | \ b c dNow the questions is, if sum of degrees in trees are same, then what is the relationship between number of vertices present in both trees? The answer is, ξ(G) and ξ(T) is same for two trees, then the trees have same number of vertices. It can be proved by induction. Let it be true for n vertices. If we add a vertex, then the new vertex (if it is not the first node) increases degree by 2, it doesn't matter where we add it. For example, try to add a new vertex say 'e' at different places in above example tee.

Question 9 |

The degree sequence of a simple graph is the sequence of the degrees of the nodes in the graph in decreasing order. Which of the following sequences can not be the degree sequence of any graph?

I. 7, 6, 5, 4, 4, 3, 2, 1 II. 6, 6, 6, 6, 3, 3, 2, 2 III. 7, 6, 6, 4, 4, 3, 2, 2 IV. 8, 7, 7, 6, 4, 2, 1, 1

I and II | |

III and IV | |

IV only | |

II and IV |

**GATE CS 2010**

**Graph Theory**

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Question 9 Explanation:

A generic algorithm or method to solve this question is
1: procedure isV alidDegreeSequence(L)
2: for n in list L do
3: if L doesn’t have n elements next to the current one then return false
4: decrement next n elements of the list by 1
5: arrange it back as a degree sequence, i.e. in descending order
6: if any element of the list becomes negative then return false
7: return true
Rationale behind this method comes from the properties of simple graph. Enumerating the f alse returns, 1) if L doesn’t have enough elements after the current one or 2) if any element of the list becomes negative, then it means that there aren’t enough nodes to accommodate edges in a simple graph fashion, which will lead to violation of either of the two conditions of the simple graph (no self-loops and no multiple-edges between two nodes), if not others.
See http://www.geeksforgeeks.org/data-structures-and-algorithms-set-25/
This solution is contributed by

**Vineet Purswani.****Another one:**A degree sequence d1,d2,d2. . . dn of non negative integer is graphical if it is a degree sequence of a graph. We now introduce a powerful tool to determine whether a particular sequence is graphical due to Havel and Hakimi**Havel–Hakimi Theorem :**→ According to this theorem, Let D be sequence the d1,d2,d2. . . dn with d1 ≥ d2 ≥ d2 ≥ . . . dn for n≥ 2 and di ≥ 0. → Then D0 be the sequence obtained by: → Discarding d1, and → Subtracting 1 from each of the next d1 entries of D. → That is Degree sequence D0 would be : d2-1, d2-1, d3-1 . . . , dd1+1 -1 . . . , dn → Then, D is graphical if and only if D0 is graphical. Now, we apply this theorem to given sequences: option I) 7,6,5,4,4,3,2,1 → 5,4,3,3,2,1,0 → 3,2,2,1,0,0 → 1,1,0,0,0 → 0,0,0,0 so its graphical. Option II) 6,6,6,6,3,3,2,2 → 5,5,5,2,2,1,2 ( arrange in ascending order) → 5,5,5,2,2,2,1 → 4,4,1,1,1,0 → 3,0,0,0,0 → 2,-1,-1,-1,0 but d (degree of a vertex) is non negative so its not a graphical. Option III) 7,6,6,4,4,3,2,2 → 5,5,3,3,2,1,1 → 4,2,2,1,1,0 → 1,1,0,0,0 → 0,0,0,0 so its graphical. Option IV) 8,7,7,6,4,2,1,1 , here degree of a vertex is 8 and total number of vertices are 8 , so it’s impossible, hence it’s not graphical. Hence only option I) and III) are graphic sequence and answer is option-D This solution is contributed by**Nirmal Bharadwaj.**Question 10 |

What is the chromatic number of an n-vertex simple connected graph which does not contain any odd length cycle? Assume n >= 2.

2 | |

3 | |

n-1 | |

n |

**GATE-CS-2009**

**Graph Theory**

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Question 10 Explanation:

The chromatic number of a graph is the smallest number of colours needed to colour the vertices of so that no two adjacent vertices share the same colour. These types of questions can be solved by substitution with different values of n.
1) n = 2
This simple graph can be coloured with 2 colours.
2) n = 3
Here, in this graph let us suppose vertex A is coloured with C1 and vertices B, C can be coloured with colour C2 => chromatic number is 2 In the same way, you can check with other values, Chromatic number is equals to 2
This solution contributed by

**Anil Saikrishna Devarasetty**//A simple graph with no odd cycles is bipartite graph and a Bipartite graph can be colored using 2 colors (See this)Question 11 |

Which one of the following is TRUE for any simple connected undirected graph with more than 2 vertices?

No two vertices have the same degree. | |

At least two vertices have the same degree. | |

At least three vertices have the same degree. | |

All vertices have the same degree. |

**GATE-CS-2009**

**Graph Theory**

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Question 11 Explanation:

Since the graph is simple, there must not be any self loop and parallel edges.
Since the graph is connected, the degree of any vertex cannot be 0.
Therefore, degree of all vertices should be be from 1 to n-1. So the degree of at least two vertices must be same.

Question 12 |

Which of the following statements is true for every planar graph on n vertices?

The graph is connected | |

The graph is Eulerian | |

The graph has a vertex-cover of size at most 3n/4 | |

The graph has an independent set of size at least n/3 |

**Graph Theory**

**GATE CS 2008**

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Question 12 Explanation:

A planar graph is a graph which can drawn on a plan without any pair of edges crossing each other.
A) FALSE: A disconnected graph can be planar as it can be drawn on a plane without crossing edges.
B) FALSE: An Eulerian Graph may or may not be planar. An undirected graph is eulerian if all vertices have even degree.
For example, the following graph is Eulerian, but not planar
C) TRUE:
D) FALSE:

Question 13 |

G is a graph on n vertices and 2n - 2 edges. The edges of G can be partitioned into two edge-disjoint spanning trees. Which of the following is NOT true for G?

For every subset of k vertices, the induced subgraph has at most 2k-2 edges | |

The minimum cut in G has at least two edges | |

There are two edge-disjoint paths between every pair to vertices | |

There are two vertex-disjoint paths between every pair of vertices |

**Graph Theory**

**GATE CS 2008**

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Question 13 Explanation:

Counter for option D is as follows. Take two copies of K4(complete graph on 4 vertices), G1 and G2. Let V(G1)={1,2,3,4} and V(G2)={5,6,7,8}. Construct a new graph G3 by using these two graphs G1 and G2 by merging at a vertex, say merge (4,5). The resultant graph is two edge connected, and of minimum degree 2 but there exist a cut vertex, the merged vertex.
Thanks to

**Renjith P**for providing above explanation.Question 14 |

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

9 edges and 5 vertices | |

9 edges and 6 vertices | |

10 edges and 5 vertices | |

10 edges and 6 vertices |

**Graph Theory**

**GATE-CS-2007**

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Question 14 Explanation:

For a simple, connected, planar graph with v vertices and e edges, the following simple conditions hold:
If v ≥ 3 then e ≤ 3v − 6;
Note that the question is about non-planar graph G. Only option C doesn't follow above.

**Alternate Explanation:**We know that K5K5 (which has 10 edges and 5 vertices) and K3,3K3,3 (which has 9 edges and 6 vertices) are non-planar graphs. Since we are asked about minimum number of edges, answer should be k3,3k3,3 i.e. option**(B)**is correct. Source: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2007.htmlQuestion 15 |

Which of the following graphs has an Eulerian circuit?

Any k-regular graph where kis an even number. | |

A complete graph on 90 vertices | |

The complement of a cycle on 25 vertices | |

None of the above |

**Graph Theory**

**GATE-CS-2007**

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Question 15 Explanation:

A graph has Eulerian Circuit if following conditions are true.
….a) All vertices with non-zero degree are connected. We don’t care about vertices with zero degree because they don’t belong to Eulerian Cycle or Path (we only consider all edges).
….b) All vertices have even degree.
Let us analyze all options.

**A) Any k-regular graph where k is an even number**. is not Eulerian as a k regular graph may not be connected (property b is true, but a may not)**B) A complete graph on 90 vertices**is not Eulerian because all vertices have degree as 89 (property b is false)**C) The complement of a cycle on 25 vertices**is Eulerian. In a cycle of 25 vertices, all vertices have degree as 2. In complement graph, all vertices would have degree as 22 and graph would be connected.Question 16 |

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 | |

B | |

C | |

D |

**Graph Theory**

**GATE-CS-2014-(Set-1)**

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Question 16 Explanation:

If we reverse directions of all arcs in a graph, the new graph has same set of strongly connected components as the original graph. Se http://www.geeksforgeeks.org/strongly-connected-components/ for more details.

Question 17 |

Consider an undirected graph G 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 __________.

500 | |

502 | |

506 | |

510 |

**Graph Theory**

**GATE-CS-2014-(Set-1)**

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Question 17 Explanation:

Given: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. There can be total 12*12 possible vertices. The vertices are (1, 1), (1, 2) ....(1, 12) (2, 1), (2, 2), ....The number of edges in this graph?Number of edges is equal to number of pairs of vertices that satisfy above conditions. For example, vertex pair {(1, 1), (1, 2)} satisfy above condition. For (1, 1), there can be an edge to (1, 2), (2, 1), (2, 2). Note that there can be self-loop as mentioned in the question. Same is count for (12, 12), (1, 12) and (12, 1) For (1, 2), there can be an edge to (1, 1), (2, 1), (2, 2), (2, 3), (1, 3) Same is count for (1, 3), (1, 4)....(1, 11), (12, 2), ....(12, 11) For (2, 2), there can be an edge to (1, 1), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2), (3, 3) Same is count for remaining vertices. For all pairs (i, j) there can total 8 vertices connected to them if i and j are not in {1, 12} There are total 100 vertices without a 1 or 12. So total 800 edges. For vertices with 1, total edges = (Edges where 1 is first part) + (Edges where 1 is second part and not first part) = (3 + 5*10 + 3) + (5*10) edges Same is count for vertices with 12 Total number of edges: = 800 + [(3 + 5*10 + 3) + 5*10] + [(3 + 5*10 + 3) + 5*10] = 800 + 106 + 106 = 1012 Since graph is undirected, two edges from v1 to v2 and v2 to v1 should be counted as one. So total number of undirected edges = 1012/2 = 506.

Question 18 |

An ordered n-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?

(1, 1, 1, 1, 1, 1) | |

(2, 2, 2, 2, 2, 2) | |

(3, 3, 3, 1, 0, 0) | |

(3, 2, 1, 1, 1, 0) |

**Graph Theory**

**GATE-CS-2014-(Set-1)**

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Question 18 Explanation:

The required graph is not possible with the given degree set of (3, 3, 3, 1, 0, 0). Using this 6-tuple the graph formed will be a Disjoint undirected graph, where the two vertices of the graph should not be connected to any other vertex ( i.e. degree will be 0 for both the vertices ) of the graph. And for the remaining 4 vertices the graph need to satisfy the degrees of (3, 3, 3, 1).

Let's see this with the help of a logical structure of the graph :

Let's say vertices labelled as <ABCDEF> should have their degree as <3, 3, 3, 1, 0, 0> respectively.

Now E and F should not be connected to any vertex in the graph. And A, B, C and D should have their degree as <3, 3, 3, 1> respectively. Now to fulfill the requirement of A, B and C, the node D will never be able to get its degree as 1. It's degree will also become as 3. This is shown in the above diagram.

Hence tuple <3, 3, 3, 1, 0, 0> is not graphic.

Question 19 |

The maximum number of edges in a bipartite graph on 12 vertices is __________________________.

36 |

**Graph Theory**

**GATE-CS-2014-(Set-2)**

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Question 19 Explanation:

Number of edges would be maximum when there are 6 edges on each side and every vertex is connected to all 6 vertices of the other side.

Question 20 |

A cycle on n vertices is isomorphic to its complement. The value of n is _____.

2 | |

4 | |

6 | |

5 |

**Graph Theory**

**GATE-CS-2014-(Set-2)**

**Discuss it**

Question 21 |

If G is a forest with n vertices and k connected components, how many edges does G have?

floor(n/k) | |

ceil(n/k) | |

n-k | |

n-k+1 |

**Graph Theory**

**GATE-CS-2014-(Set-3)**

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Question 21 Explanation:

Each component will have n/k vertices (pigeonhole principle). Hence, for each component there will be (n/k)-1 edges.
Since there are k components, total number of edges= k*((n/k)-1) = n-k.

Question 22 |

Let d denote the minimum degree of a vertex in a graph. For all planar graphs on n vertices with d ≥ 3, which one of the following is TRUE?

In any planar embedding, the number of faces is at least n/2 + 2 | |

In any planar embedding, the number of faces is less than n/2 + 2 | |

There is a planar embedding in which the number of faces is less than n/2 + 2 | |

There is a planar embedding in which the number of faces is at most n/(d+1) |

**Graph Theory**

**GATE-CS-2014-(Set-3)**

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Question 22 Explanation:

Euler's formula for planar graphs: v − e + f = 2. v => Number of vertices e => Number of edges f => Number of faces Since degree of every vertex is at least 3, below is true from handshaking lemma (Sum of degrees is twice the number of edges) 3v >= 2e 3v/2 >= e Putting these values in Euler's formula. v - 3v/2 + f >= 2 f >= v/2 + 2

Question 23 |

The 2

^{n}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:1 | |

n | |

n+1 | |

2 ^{n} |

**Graph Theory**

**GATE-CS-2006**

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Question 23 Explanation:

There are n nodes which are single and 1 node which belong to empty set. And since they are not having 2 or more elements so they won’t be connected to anyone hence total number of nodes with degree 0 are n+1 hence answer should be none. Thanks to roger for the explanation.

Question 24 |

The 2

^{n}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:n | |

n+2 | |

2 ^{n/2} | |

2 ^{n} / n |

**Graph Theory**

**GATE-CS-2006**

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Question 24 Explanation:

n+1 nodes of the graph not connected to anyone as explained in question 70 while others are connected so total number of connected components are n+2 (n+1 connected components by each of the n+1 vertices plus 1 connected component by remaining vertices).

Question 25 |

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

6 | |

8 | |

9 | |

13 |

**Graph Theory**

**GATE-CS-2005**

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Question 25 Explanation:

An undirected graph is called a planar graph if it can be drawn on a paper without having two edges cross and such a drawing is called Planar Embedding. We say that a graph can be embedded in the plane, if it planar. A planar graph divides the plane into regions (bounded by the edges), called faces. Graph K4 is palanar graph, because it has a planar embedding as shown in
figure below.

**Euler's Formula :**For any polyhedron that doesn't intersect itself (Connected Planar Graph),the • Number of Faces(F) • plus the Number of Vertices (corner points) (V) • minus the Number of Edges(E) , always equals 2. This can be written: F + V − E = 2. Solution : Here as given, F=?,V=13 and E=19 -> F+13-19=2 -> F=8 So Answer is (B). This solution is contributed by**Nirmal Bharadwaj**We can apply Euler's Formula of planar graphs. The formula is v − e + f = 2.Question 26 |

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

12 | |

8 | |

Less than 8 | |

More than 12 |

**Graph Theory**

**GATE-CS-2005**

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Question 26 Explanation:

**Background Explanation:**

**Vertex cover**is a set S of vertices of a graph such that each edge of the graph is incident to at least one vertex of S.

**Independent set**of a graph is a set of vertices such that none of the vertices in this set have an edge connecting them i.e. no two are adjacent. A single vertex is an independent set, but we are interested in maximum independent set, that is largest set which is independent set.

**Relation between Independent Set and Vertex Cover :**An interesting fact is, the number of vertices of a graph is equal to its minimum vertex cover number plus the size of a maximum independent set. How? removing all vertices of minimum vertex cover leads to maximum independent set. So if S is the size of minimum vertex cover of G(V,E) then the size of maximum independent set of G is |V| - S.

**Solution:**size of minimum vertex cover = 8 size of maximum independent set = 20 - 8 =12 Therefore, correct answer is (A). References : vertex cover maximum independent set. This solution is contributed by

**Nitika Bansal.**

Question 27 |

G1 | |

G2 | |

G3 | |

G4 |

**Graph Theory**

**GATE-CS-2005**

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Question 27 Explanation:

A graph is planar if it can be redrawn in a plane without any crossing edges. G1 is a typical example of nonplanar graphs.

Question 28 |

Let s and t be two vertices in a undirected graph G + (V, E) having distinct positive edge weights. Let [X, Y] be a partition of V such that s ∈ X and t ∈ Y. Consider the edge e having the minimum weight amongst all those edges that have one vertex in X and one vertex in Y.
Let the weight of an edge e denote the congestion on that edge. The congestion on a path is defined to be the maximum of the congestions on the edges of the path. We wish to find the path from s to t having minimum congestion. Which one of the following paths is always such a path of minimum congestion?

a path from s to t in the minimum weighted spanning tree | |

a weighted shortest path from s to t | |

an Euler walk from s to t | |

a Hamiltonian path from s to t |

**Graph Theory**

**GATE-CS-2005**

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Question 28 Explanation:

Suppose shortest path from A->B is 6, but in MST, we have A->C->B (A->C = 4, C->B = 3), then along the path in MST, we have minimum congestion, i.e 4

Question 29 |

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

2 | |

3 | |

4 | |

5 |

**Graph Theory**

**GATE-CS-2004**

**Discuss it**

Question 29 Explanation:

Two vertices are said to be adjacent if they are directly connected, i.e., there is a direct edge between them.
So, here, we can assign same color to 1 & 2 (red), 3 & 4 (grey), 5 & 7 (blue) and 6 & 8 (brown).
Therefore, we need a total of 4 distinct colors.
Thus, C is the correct choice.

Please comment below if you find anything wrong in the above post.

Please comment below if you find anything wrong in the above post.

Question 30 |

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

k and n | |

k - 1 and k + 1 | |

k - 1 and n - 1 | |

k + 1 and n - k |

**Graph Theory**

**GATE-CS-2003**

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Question 30 Explanation:

Minimum: The removed vertex itself is a separate connected component. So removal of a vertex creates k-1 components.
Maximum: It may be possible that the removed vertex disconnects all components. For example the removed vertex is center of a star. So removal creates n-1 components.

Question 31 |

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

15 | |

24 | |

30 | |

60 |

**Graph Theory**

**GATE-CS-2003**

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Question 31 Explanation:

A perfect matching, every vertex of the graph is incident to exactly one edge of the matching. A perfect matching is therefore a matching of a graph containing n/2 edges, the largest possible, meaning perfect matchings are only possible on graphs with an even number of vertices. (Source http://mathworld.wolfram.com/PerfectMatching.html)

Question 32 |

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

3 | |

4 | |

5 | |

6 |

**Graph Theory**

**GATE-CS-2003**

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Question 32 Explanation:

Question 33 |

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

2 | |

3 | |

4 | |

n - 2⌊n/2⌋ + 2 |

**Graph Theory**

**GATE-CS-2002**

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Question 33 Explanation:

We need 3 colors to color a odd cycle and 2 colors to color an even cycle.

Question 34 |

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

n ^{2} | |

n(n - 1)/2 | |

n - 1 | |

(n + 1) (n)/2 |

**Graph Theory**

**GATE-CS-2002**

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Question 34 Explanation:

Background required - Basic Combinatorics
Since the given graph is undirected, that means the order of edges doesn't matter.
Since we have to insert an edge between all possible pair of vertices, therefore problem reduces to finding the count of the number of subsets of size 2 chosen from the set of vertices.
Since the set of vertices has size n, the number of such subsets is given by the binomial coefficient C(n,2) (also known as "n choose 2"). Using the formula for binomial coefficients, C(n,2) = n(n-1)/2.e
This explanation has been contributed by

**Pranjul Ahuja.**Question 35 |

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 _______________.

24 | |

20 | |

32 | |

64 |

**Graph Theory**

**GATE-CS-2015 (Set 1)**

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Question 35 Explanation:

Euler's formula states that if a finite, connected, planar graph is drawn in the plane without any edge intersections, then

v − e + f = 2. v -> Number of vertices e -> Number of edges f -> Number of faces As per the question v = 10 And number of edges on each face is three Therefore, 2e = 3f [Note that every edge is shared by 2 faces] Putting above values in v − e + f = 2 10 - e + 2e/3 = 2 e = 3*10 - 6 = 24

Question 36 |

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

A multiple of 4 | |

Even | |

Odd | |

Congruent to 0 mod 4, or 1 mod 4 |

**Graph Theory**

**GATE-CS-2015 (Set 2)**

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Question 36 Explanation:

An n-vertex self-complementary graph has exactly half number of edges of the complete graph, i.e., n(n − 1)/4 edges, and (if there is more than one vertex) it must have diameter either 2 or 3. Since n(n −1) must be divisible by 4, n must be congruent to 0 or 1 mod 4; for instance, a 6-vertex graph cannot be self-complementary.
Source: http://en.wikipedia.org/wiki/Self-complementary_graph

Question 37 |

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

A tree has no bridge
| |

A bridge cannot be part of a simple cycle | |

Every edge of a clique with size ≥ 3 is a bridge (A clique is any complete subgraph of a graph) | |

A graph with bridges cannot have a cycle |

**Graph Theory**

**GATE-CS-2015 (Set 2)**

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Question 37 Explanation:

A bridge in a graph cannot be a part of cycle as removing it will not create a disconnected graph if there is a cycle.

Question 38 |

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?10 | |

11 | |

18 | |

19 |

**Graph Theory**

**GATE-IT-2004**

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Question 38 Explanation:

The idea is to use Handshaking Lemma :- In any graph, the sum of all the vertex-degree is equal to twice the number of edges.

Let x = Number of vertices of degree 3. By Handshaking Lemma 6*2 + 3*4 + (x-9)*3 = 27*2 24 + (x-9)*3 = 54 x-9 = 10 x = 19

Question 39 |

The minimum number of colours that is sufficient to vertex-colour any planar graph is _______________
[This Question was originally a Fill-in-the-blanks Question]

1 | |

2 | |

3 | |

4 |

**Graph Theory**

**GATE-CS-2016 (Set 2)**

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Question 39 Explanation:

A planar graph is a graph on a plane where no two edges are crossing each other.
The set of regions of a map can be represented more abstractly as an undirected graph that has a vertex for each region and an edge for every pair of regions that share a boundary segment. Hence the four color theorem is applied here.
Here is a property of a planar graph that a planar graph does not require more than 4 colors to color its vertices such that no two vertices have same color. This is known four color theorem.

Question 40 |

If all the edge weights of an undirected graph are positive, then any subset of edges that connects all the vertices and has minimum total weight is a

Hamiltonian cycle | |

grid | |

hypercube | |

tree |

**Tree Traversals**

**Graph Theory**

**GATE IT 2006**

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Question 40 Explanation:

As here we want subset of edges that connects all the vertices and has minimum total weight i.e. Minimum Spanning Tree
Option A - includes cycle, so may or may not connect all edges.
Option B - has no relevance to this question.
Option C - includes cycle, so may or may not connect all edges.
Related:
http://www.geeksforgeeks.org/greedy-algorithms-set-2-kruskals-minimum-spanning-tree-mst/
http://www.geeksforgeeks.org/greedy-algorithms-set-5-prims-minimum-spanning-tree-mst-2/
This solution is contributed by

**Mohit Gupta.**Question 41 |

Consider a weighted undirected graph with positive edge weights and let uv be an edge in the graph. It is known that the shortest path from the source vertex s to u has weight 53 and the shortest path from s to v has weight 65. Which one of the following statements is always true?

weight (u, v) < 12 | |

weight (u, v) ≤ 12 | |

weight (u, v) > 12 | |

weight (u, v) ≥ 12 |

**Graph Shortest Paths**

**Graph Theory**

**Gate IT 2007**

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Question 42 |

2 | |

3 | |

4 | |

5 |

**Graph Theory**

**Gate IT 2008**

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Question 42 Explanation:

The chromatic number of a graph is the smallest number of colors needed to color the vertices so that no two adjacent vertices have the same color.
In this graph, minimum number of colors needed to color given graph would be equal to 3.

Question 43 |

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

regular | |

Complete | |

Hamiltonian | |

Euler |

**Graph Theory**

**Gate IT 2008**

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Question 43 Explanation:

For a graph to be Euler graph all the degrees must be Even for all nodes. In any graph all the Odd degree nodes are connected with a node.

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

=> Euler Graph

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

=> Euler Graph

There are 43 questions to complete.