GATE CS 2012

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

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

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") ;
    }
}

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)

-> 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

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.

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

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.

See http://softwareeveryday.wordpress.com/2012/08/07/function-minima-and-maxima-recap/

A =  1    1
     1   -1

A2 = 2   0
                0   2

A4 = A2 X A2
A4 = 4   0
                0   4

A8 = 16   0
                0    16


A16 = 256   0
                 0    256

A18 = A16 X A2
A18 = 512   0
                 0    512 

A19 = 512   512
                 512  -512


Applying Characteristic polynomial

512-lamda   512
512       -(512+lamda)  =   0

-(512-lamda)(512+lamda) - 512 x 512 = 0

lamda2 = 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.