GATE CS 2012

Question 1

Which of the followings is/are automatically added to every class, if we do not write our own.

• A. Copy Constructor
• B. Assignment Operator
• C. A constructor without any parameter
• D. All of the above

Answer: (D)

In C++, if we do not write our own, then compiler automatically creates a default constructor, a copy constructor and a assignment operator for every class.

Question 2

Consider the following logical inferences.

I1: If it rains then the cricket match will not be played.

The cricket match was played.

Inference: There was no rain.

I2: If it rains then the cricket match will not be played.

It did not rain.

Inference: The cricket match was played.

Which of the following is TRUE?

• A. Both I1 and I2 are correct inferences
• B. I1 is correct but I2 is not a correct inference
• C. I1 is not correct but I2 is a correct inference
• D. Both I1 and I2 are not correct inferences

Answer: (B)

The cricket match may not be played even if doesn't rain.

Question 3

Which of the following is TRUE?

• A. Every relation in 3NF is also in BCNF
• B. A relation R is in 3NF if every non-prime attribute of R is fully functionally dependent on every key of R
• C. Every relation in BCNF is also in 3NF
• D. No relation can be in both BCNF and 3NF

Answer: (C)

BCNF is a stronger version 3NF. So every relation in BCNF will also be in 3NF.

Question 4

What will be the output of the following C program segment?

char inchar = 'A';
switch (inchar)
{
case 'A' :
    printf ("choice A n") ;
case 'B' :
    printf ("choice B ") ;
case 'C' :
case 'D' :
case 'E' :
default:
    printf ("No Choice") ;
}

• A.

  No choice

• B.

  Choice A

• C.

  Choice A
  Choice B No choice

• D.

  Program gives no output as it is erroneous

Answer: (C)

There is no break statement in case ‘A’. If a case is executed and it doesn’t contain break, then all the subsequent cases are executed until a break statement is found. That is why everything inside the switch is printed. Try following program as an exercise.

int main()
{
    char inchar = 'A';
    switch (inchar)
    {
    case 'A' :
        printf ("choice A \n") ;
    case 'B' :
    {
        printf ("choice B") ;
        break;
    }
    case 'C' :
    case 'D' :
    case 'E' :
    default:
        printf ("No Choice") ;
    }
}

Question 5

Assuming P != NP, which of the following is true ?
(A) NP-complete = NP
(B) NP-complete  P = 
(C) NP-hard = NP
(D) P = NP-complete

• A. A
• B. B
• C. C
• D. D

Answer: (B)

The answer is B (no NP-Complete problem can be solved in polynomial time). Because, if one NP-Complete problem can be solved in polynomial time, then all NP problems can solved in polynomial time. If that is the case, then NP and P set become same which contradicts the given condition.

(Wikipedia)

Question 6

The worst case running time to search for an element in a balanced in a binary search tree with n2^n elements is

(A) 

(B) 

(C) 

(D)  

• A. A
• B. B
• C. C
• D. D

Answer: (C)

-> The search time in a binary search tree depends on the form of the tree, that is on the order in which its nodes were inserted. A pathological case: The n nodes were inserted by increasing order of the keys, yielding something like a linear list (but with a worse space consumption), with O(n) search time(in the case of skew tree). -> A balanced tree is a tree where every leaf is “not more than a certain distance” away from the root than any other leaf.So in balanced tree, the height of the tree is balanced to make distance between root and leafs nodes a low as possible. In a balanced tree, the height of tree is log₂(n). -> So , if a Balanced Binary Search Tree contains n2n elements then Time complexity to search an item: Time Complexity = log(n2ⁿ) = log (n) + log(2ⁿ) = log (n) +n = O(n) So Answer is C. See http://www.geeksforgeeks.org/data-structures-and-algorithms-set-28/ This solution is contributed by Nirmal Bharadwaj

Question 7

The truth table represents the Boolean function

• A. X
• B. X+Y
• C. X xor Y
• D. Y

Answer: (A)

The value of f(X, Y) is same as X for all input pairs. Also sum of product form of expression we get,

= XY’+XY
= X(Y’+Y)
= X *1
= X 

We see from truth table –

Column x = f(x,y) 
So , 
f(x,y)=x 

Option (A) is correct.

Question 8

The decimal value 0.5 in IEEE single precision floating point representation has

• A. fraction bits of 000…000 and exponent value of 0
• B. fraction bits of 000…000 and exponent value of −1
• C. fraction bits of 100…000 and exponent value of 0
• D. no exact representation

Answer: (B)

The IEEE 754 standard specifies following distribution of bits: Sign bit: 1 bit Exponent width: 8 bits Significand or Fraction: 24 (23 explicitly stored) 0.5 in base 10 means 1 X 2^-1 in base 2. So exponent bits have value -1 and all fraction bits are 0.

Question 9

A process executes the code

fork();
fork();
fork(); 

The total number of child processes created is

• A. 3
• B. 4
• C. 7
• D. 8

Answer: (C)

Let us put some label names for the three lines

  fork ();    // Line 1
  fork ();   // Line 2
  fork ();   // Line 3

       L1       // There will be 1 child process created by line 1
    /     \
  L2      L2    // There will be 2 child processes created by line 2
 /  \    /  \
L3  L3  L3  L3  // There will be 4 child processes created by line 3

We can also use direct formula to get the number of child processes. With n fork statements, there are always 2^n – 1 child processes. Also see this post for more details.

Question 10

Consider the function f(x) = sin(x) in the interval [π/4, 7π/4]. The number and location(s) of the local minima of this function are

• A. One, at π/2
• B. One, at 3π/2
• C. Two, at π/2 and 3π/2
• D. Two, at π/4 and 3π/2

Answer: (D)

Sine function increases till π/2 and so for the considered interval π/4 would be a local minimum. From π/2, value of sine keeps on decreasing till 3π/2 and hence 3π/2 would be another local minima.