# GATE CS 2012

Question 1 |

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**?Both I1 and I2 are correct inferences | |

I1 is correct but I2 is not a correct inference | |

I1 is not correct but I2 is a correct inference | |

Both I1 and I2 are not correct inferences |

**GATE CS 2012**

**General Aptitude**

**Discuss it**

Question 1 Explanation:

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

Question 2 |

Which of the following is

**TRUE**?Every relation in 3NF is also in BCNF | |

A relation R is in 3NF if every non-prime attribute of R is fully functionally dependent on every key of R | |

Every relation in BCNF is also in 3NF | |

No relation can be in both BCNF and 3NF |

**GATE CS 2012**

**Database Design(Normal Forms)**

**Discuss it**

Question 3 |

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") ; }

No choice | |

Choice A | |

Choice A Choice B No choice | |

Program gives no output as it is erroneous |

**GATE CS 2012**

**Discuss it**

Question 3 Explanation:

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 4 |

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 | |

B | |

C | |

D |

**GATE CS 2012**

**Discuss it**

Question 4 Explanation:

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.

**Related Article:**NP-Completeness | Set 1 (Introduction) P versus NP problem (Wikipedia)

Question 5 |

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 | |

B | |

C | |

D |

**GATE CS 2012**

**Discuss it**

Question 5 Explanation:

-> 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 log

_{2}(n). -> So , if a Balanced Binary Search Tree contains n2n elements then Time complexity to search an item: Time Complexity = log(n2^{n}) = log (n) + log(2^{n}) = 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 6 |

X | |

X+Y | |

X xor Y | |

Y |

**GATE CS 2012**

**Digital Logic & Number representation**

**Discuss it**

Question 6 Explanation:

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 = XWe see from truth table –

Column x = f(x,y) So , f(x,y)=xOption (A) is correct.

Question 7 |

T he decimal value 0.5 in IEEE single precision floating point representation has

fraction bits of 000…000 and exponent value of 0 | |

fraction bits of 000…000 and exponent value of −1 | |

fraction bits of 100…000 and exponent value of 0 | |

no exact representation |

**GATE CS 2012**

**Number Representation**

**Discuss it**

Question 7 Explanation:

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 0Question 8 |

A process executes the code

fork(); fork(); fork();The total number of child processes created is

3 | |

4 | |

7 | |

8 |

**GATE CS 2012**

**OS Process Management**

**Discuss it**

Question 8 Explanation:

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 3We 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 9 |

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

One, at π/2 | |

One, at 3π/2 | |

Two, at π/2 and 3π/2 | |

Two, at π/4 and 3π/2 |

**GATE CS 2012**

**Numerical Methods and Calculus**

**Discuss it**

Question 9 Explanation:

Question 10 |

Let A be the 2 × 2 matrix with elements a11 = a12 = a21 = +1 and a22 = −1. Then the eigenvalues of the matrix A

^{19}areA | |

B | |

C | |

D |

**GATE CS 2012**

**Linear Algebra**

**Discuss it**

Question 10 Explanation:

A = 1 1 1 -1 A^{2}= 2 0 0 2 A^{4}= A^{2}X A^{2}A^{4}= 4 0 0 4 A^{8}= 16 0 0 16 A^{16}= 256 0 0 256 A^{18}= A^{16}X A^{2}A^{18}= 512 0 0 512 A^{19}= 512 512 512 -512 Applying Characteristic polynomial 512-lamda 512 512 -(512+lamda) = 0 -(512-lamda)(512+lamda) - 512 x 512 = 0 lamda^{2 = 2 x 5122 }

**Alternative solution:**

det(A) = -2. det(A^19) = (det(A))^19 = -2^19 = lambda1*lambda2. The only viable option is D.Thanks to Matan Mandelbrod for suggesting this solution.

There are 59 questions to complete.