Following questions have been asked in GATE CS 2013 exam.

**1) Which of the following statements is/are TRUE for an undirected graph?P: Number of odd degree vertices is evenQ: Sum of degrees of all vertices is even**

A) P Only

B) Q Only

C) Both P and Q

D) Neither P nor Q

Answer (C)

Q is true: Since the graph is undirected, every edge increases the sum of degrees by 2.

P is true: If we consider sum of degrees and subtract all even degrees, we get an even number (because Q is true). So total number of odd degree vertices must be even.

**2) 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?**

(A) 1/8

(B) 1

(C) 7

(D) 8

Answer (C)

A cycle of length 3 can be formed with 3 vertices. There can be total 8C3 ways to pick 3 vertices from 8. The probability that there is an edge between two vertices is 1/2. So expected number of unordered cycles of length 3 = (8C3)*(1/2)^3 = 7

**3) What is the time complexity of Bellman-Ford single-source shortest path algorithm on a complete graph of n vertices?**

(A) Θ(n^{2})

(B) Θ(n^{2} Logn)

(C) Θ(n^{3})

(D) Θ(n^{3} Logn)

Answer (C).

Time complexity of Bellman-Ford algorithm is Θ(VE) where V is number of vertices and E is number edges (See this). If the graph is complete, the value of E becomes Θ(V^{2}). So overall time complexity becomes Θ(V^{3})

**4) Which of the following statements are TRUE?****(1) The problem of determining whether there exists a cycle in an undirected graph is in P.(2) The problem of determining whether there exists a cycle in an undirected graph is in NP.(3) If a problem A is NP-Complete, there exists a non-deterministic polynomial time algorithm to solve A.**

(A) 1,2 and 3

(B) 1 and 2 only

(C) 2 and 3 only

(D) 1 and 3 only

Answer (A)**1 **is true because cycle detection can be done in polynomial time using DFS (See this).**2 **is true because P is a subset of NP.**3 **is true because NP complete is also a subset of NP and NP means **N**on-deterministic **P**olynomial time solution exists. (See this)

**5) Which one of the following is the tightest upper bound that represents the time complexity of inserting an object into a binary search tree of n nodes?**

(A) O(1)

(B) O(log n)

(C) O(n)

(D) O(n log n)

Answer (C)

The worst case occurs for a skewed tree. In a skewed tree, when a new node is inserted as a child of bottommost node, the time for insertion requires traversal of all node. For example, consider the following tree and the case when something smaller than 70 is inserted.

100 / 90 / 80 / 70

**6) Which one of the following is the tightest upper bound that represents the number of swaps required to sort n numbers using selection sort?**

(A) O(log n)

(B) O(n)

(C) O(n log n)

(D) O(n^2)

Answer (B)

Selection sort requires only O(n) swaps. See this for details.

**7) Consider the following operation along with Enqueue and Dequeue operations onqueues, where k is a global parameter**

MultiDequeue(Q){ m = k while (Q is not empty and m > 0) { Dequeue(Q) m = m - 1 } }

**What is the worst case time complexity of a sequence of n MultiDequeue() operations on an initially empty queue?**

(A) Θ(n)

(B) Θ(n + k)

(C) Θ(nk)

(D) Θ(n^{2})

Answer (A)

Since the queue is empty initially, the condition of while loop never becomes true. So the time complexity is Θ(n)

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