ISRO CS 2013

Question 1

Let A(1:8, -5:5, -10:5) be a three dimensional array. How many elements are there in the array A?

• A. 1200
• B. 1408
• C. 33
• D. 1050

Answer: (B)

ISRO CS 2013 C Arrays
Discuss it

Number of elements in first dimension of array = upper limit - lower limit + 1 = 8 - 1 + 1 =8 Number of elements in second dimension of array = 5 - (-5) + 1 = 11 Number of elements in third dimension of array = 15 - (-10) + 1 = 16 total number of elements = 8 * 11 * 16 = 1408 So, Answer (B) is correct.

Question 2

The number of rotations required to insert a sequence of elements 9,6,5,8,7,10 into an empty AVL tree is?

• A. 0
• B. 1
• C. 2
• D. 3

Answer: (D)

ISRO CS 2013 Tree Traversals
Discuss it

The insertion and rotation of the various elements are shown in the following figure: So, the total number of rotations are 3.

Question 3

Opportunistic reasoning is addressed by which of the following knowledge representation

• A. Script
• B. Blackboard
• C. Production Rules
• D. Fuzzy Logic

Answer: (B)

ISRO CS 2013 Arithmetic Aptitude 3 50 Aptitude & Reasoning MCQs with Answers
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Opportunistic reasoning is a method of selecting suitable logical inference strategy from the knowledge base of the AI system. Blackboard is an AI model of opportunistic reasoning where a common knowledge base is updated frequently by diverse knowledge sources. So, (B) is the correct option.

Question 4

The following steps in a linked list

p = getnode()
 info (p) = 10
 next (p) = list
 list = p

result in which type of operation?

• A. pop operation in stack
• B. removal of a node
• C. inserting a node
• D. modifying an existing node

Answer: (C)

ISRO CS 2013 Linked List
Discuss it

p = getnode() // getnode() allocates the space which is equal to the size of the node type structure and returns a pointer. info (p) = 10 // value of the new node is equal to 10 next (p) = list // adding new node to the list. Clearly, through these steps, insertion of a node is performed. Option (C) is correct.

Question 5

Shift reduce parsing belongs to a class of

• A. bottom up parsing
• B. top down parsing
• C. recursive parsing
• D. predictive parsing

Answer: (A)

ISRO CS 2013 Parsing and Syntax directed translation
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Shift reduce are the class of parsers which built a parse tree in bottom-up manner and scans it from left to right.

In these parsers, through shift step, a single character from the token stream is pushed on the parse stack and becomes a new single-node parse tree.

A reduce step applies a complete grammar production rule to the recent parse trees, joining them together as one tree with a new root symbol.

These parsers include LR parsers which are non-backtracking in nature. Examples: SLR, CLR, LALR etc.

Question 6

Which of the following productions eliminate left recursion in the productions given below: S → Aa | b A → Ac | Sd | ε

• A. S → Aa | b A → bdA' A' → A'c | A'ba | A | ε
• B. S → Aa | b A → A' | bdA', A' → cA' | adA' | ε
• C. S → Aa | b A → A'c | A'd A' → bdA' | cA | ε
• D. S → Aa | b A → cA' | adA' | bdA' A' → A | ε

Answer: (B)

ISRO CS 2013 Parsing and Syntax directed translation
Discuss it

To remove left recursion from the grammar of the
form :  A → Aα | β
We rewrite the production rules as:
           A → βA'
           A'→ αA'| ε

Given Grammar: S → Aa | b
               A → Ac | Sd | ε

after finding indirect left recursion, grammar:

               S → Aa | b
               A → Ac | Aad | bd | ε

here, α = c, ad, β = bd

So, Grammar after removing left recursion = 
               S → Aa | b
               A → A' | bdA'
               A'→ CA'| ada'| ε 

So, option (B) is correct.

Question 7

Consider the following psuedocode:

 x : integer := 1
 y : integer := 2
 procedure add
 x := x + y

 procedure second (P: procedure)
 x : integer := 2
 P()
 procedure first
 y : integer := 3
 second(add)
 first()
 write_integer (x)

What does it print if the language uses dynamic scoping with deepbinding?

• A. 2
• B. 3
• C. 4
• D. 5

Answer: (C)

ISRO CS 2013 C Storage Classes and Type Qualifiers
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Question 8

Which logic gate is used to detect overflow in 2's complement arithmetic?

• A. OR gate
• B. AND gate
• C. NAND gate
• D. XOR gate

Answer: (D)

ISRO CS 2013 Digital Logic & Number representation Logic functions and Minimization
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In 2's complement numbers, we can check whether there's an overflow or not by taking the XOR of carry in and carry out of the most significant bit. If XOR value = 0, then no overflow. If XOR value = 1, then overflow.

Question 9

In an array of 2N elements that is both 2-ordered and 3-ordered, what is the maximum number of positions that an element can be from its position if the array were 1-ordered?

• A. 1
• B. 2
• C. N/2
• D. 2N-1

Answer: (A)

ISRO CS 2013 C Arrays
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An array can be termed as 2-ordered array, if it contains an element which is atmost two positions away from its original position in a sorted array. So the maximum number of positions that an element can be from its position if the array were 1-ordered = 1 Option (A) is correct.

Question 10

If the frame buffer has 8 bits per pixel and 8 bits are allocated for each of the R, G, B components, what would be the size of the lookup table?

• A. 24 bytes
• B. 1024 bytes
• C. 768 bytes
• D. 256 bytes

Answer: (D)

ISRO CS 2013 Misc Topics in Computer Networks
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Number of indexes the framebuffer = 2⁸ = 256 Size of each entry = 8 bits for R component + 8 bits for G component + 8 bits for B component = 24 bits = 3 Bytes Size of frame buffer = 256 * 3 = 768 Bytes Option (C) is correct.