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 even
Q: 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) [Tex]\Theta(n^2)[/Tex]
(B) [Tex]\Theta(n^2 Logn)[/Tex]
(C) [Tex]\Theta(n^3)[/Tex]
(D) [Tex]\Theta(n^3 Logn)[/Tex]

Answer (C).
Time complexity of Bellman-Ford algorithm is [Tex]\Theta(VE)[/Tex] where V is number of vertices and E is number edges (See this). If the graph is complete, the value of E becomes [Tex]\Theta(V^2)[/Tex]. So overall time complexity becomes [Tex]\Theta(V^3)[/Tex]



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 Non-deterministic Polynomial 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 on
queues, 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) [Tex]\Theta(n)[/Tex]
(B) [Tex]\Theta(n + k)[/Tex]
(C) [Tex]\Theta(nk)[/Tex]
(D) [Tex]\Theta(n^2)[/Tex]

Answer (A)
Since the queue is empty initially, the condition of while loop never becomes true. So the time complexity is [Tex]\Theta(n)[/Tex]



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