GATE-CS-2003

Question 1
Consider the following C function.
float f(float x, int y)
{
  float p, s; int i;
  for (s=1, p=1, i=1; i < y; i ++)
  {
    p*= x/i;
    s+=p;
  }
  return s;
}  
For large values of y, the return value of the function f best approximates
A
x^y
B
e^x
C
ln(1 + x)
D
x^x
GATE-CS-2003    
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Question 1 Explanation: 
Question 2
Assume the following C variable declaration
 int *A [10], B[10][10];  
Of the following expressions I A[2] II A[2][3] III B[1] IV B[2][3] which will not give compile-time errors if used as left hand sides of assignment statements in a C program?
A
I, II, and IV only
B
II, III, and IV only
C
II and IV only
D
IV only
GATE-CS-2003    
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Question 2 Explanation: 
Question 3
Let P(E) denote the probability of the event E. Given P(A) = 1, P(B) = 1/2, the values of P(A | B) and P(B | A) respectively are
A
1/4, 1/2
B
1/2, 1/14
C
1/2, 1
D
1, 1/2
Probability    GATE-CS-2003    
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Question 3 Explanation: 
Conditional probability measures the probability of an event given that another event has occurred. P(A | B) is Probability of A given B has occurred which is 1. P(B | A) is probability of B given that A has occurred which is 1/2
Question 4
Let A be a sequence of 8 distinct integers sorted in ascending order. How many distinct pairs of sequences, B and C are there such that (i) each is sorted in ascending order, (ii) B has 5 and C has 3 elements, and (iii) the result of merging B and C gives A?
A
2
B
30
C
56
D
256
Combinatorics    GATE-CS-2003    
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Question 4 Explanation: 
Suppose you have selected 3 elements from 8 in 8C3 ways, the remaining elements are treated as another array and merging both the arrays gives the sorted array. Here, you can select either 3 or 5. => 8C3 = 8C5 = 8!/(3!5!) = 7*8 = 56 Ways. This solution is contributed by Anil Saikrishna Devarasetty
Question 5
n couples are invited to a party with the condition that every husband should be accompanied by his wife. However, a wife need not be accompanied by her husband. The number of different gatherings possible at the party isGATECS2003Q4
A
A
B
B
C
C
D
D
Combinatorics    GATE-CS-2003    
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Question 5 Explanation: 
There are three options for every couple.
1) Nobody goes to gathering
2) Wife alone goes
2) Both go
Since there are n couples, total possible ways of gathering are 3n
Question 6
Let T(n) be the number of different binary search trees on n distinct elements. Then GATECS2003Q7, where x is
A
n-k+1
B
n-k
C
n-k-1
D
n-k-2
Binary Search Trees    GATE-CS-2003    
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Question 6 Explanation: 
The idea is to make a key root, put (k-1) keys in one subtree and remaining n-k keys in other subtree. A Binary Search Tree (BST) is a tree in which all the nodes follow the below-mentioned properties −
  • The left sub-tree of a node has a key less than or equal to its parent node's key.
  • The right sub-tree of a node has a key greater than to its parent node's key.
Now construction binary search trees from n distinct number- Lets for simplicity consider n distinct numbers as first n natural numbers (starting from 1) If n=1 We have only one possibility, therefore only 1 BST. If n=2 We have 2 possibilities , when smaller number is root and bigger number is the right child or second when the bigger number is root and smaller number as left child.   parul_1 If n=3 We have 5 possibilities. Keeping each number first as root and then arranging the remaining 2 numbers as in case of n=2. parul_2   If n=4 We have 14 possibilities. Taking each number as root and arranging smaal numbers as left subtree and larger numbers as right subtree. parul_4 Thus we can conclude that with n distinct numbers, if we take ‘k’ as root then all the numbers smaller than k will left subtree and numbers larger than k will be right subtree where the the right subtree and left subtree will again be constructed recursively like the root. Therefore, parul5   This solution is contributed by Parul Sharma.
Question 7
Consider the set ∑* of all strings over the alphabet ∑ = {0, 1}. ∑* with the concatenation operator for strings
A
does not form a group
B
forms a non-commutative group
C
does not have a right identity element
D
forms a group if the empty string is removed from ∑*
Set Theory & Algebra    GATE-CS-2003    
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Question 8
Let G be an arbitrary graph with n nodes and k components. If a vertex is removed from G, the number of components in the resultant graph must necessarily lie between
A
k and n
B
k - 1 and k + 1
C
k - 1 and n - 1
D
k + 1 and n - k
Graph Theory    GATE-CS-2003    
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Question 8 Explanation: 
Minimum: The removed vertex itself is a separate connected component. So removal of a vertex creates k-1 components. Maximum: It may be possible that the removed vertex disconnects all components. For example the removed vertex is center of a star. So removal creates n-1 components.
Question 9
Assuming all numbers are in 2's complement representation, which of the following numbers is divisible by 11111011?
A
11100111
B
11100100
C
11010111
D
11011011
Number Representation    GATE-CS-2003    
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Question 9 Explanation: 
Since most significant bit is 1, all numbers are negative. 2's complement of divisor (11111011) = 1's complement + 1 = 00000100 + 1 = 00000101 So the given number is -5 The decimal value of option A is -25
Question 10
For a pipelined CPU with a single ALU, consider the following situations
1. The j + 1-st instruction uses the result of the j-th instruction
    as an operand
2. The execution of a conditional jump instruction
3. The j-th and j + 1-st instructions require the ALU at the same 
   time
Which of the above can cause a hazard ?
A
1 and 2 only
B
2 and 3 only
C
3 only
D
All of above
Computer Organization and Architecture    GATE-CS-2003    
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Question 10 Explanation: 
Case 1: Is of data dependency .this can’t be safe with single ALU so read after write. Case 2:Conditional jumps are always hazardous they create conditional dependency in pipeline. Case 3:This is write after read problem or concurrency dependency so hazardous All the three are hazardous So (D) is correct option.
There are 90 questions to complete.

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