Net June 2012 Computer science Questions and solutions

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Net June 2012 Computer science and applicationts Question paper with solutions
( UGC Net answer key for June 24, 2012 )


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(1)  Granulairity refers to

(a) Size of a database

(b) Size of a data item

(c) Size of a record

(d) .........

Answer: (a) size of a data item

(2) R = (A, B, C, D). We decompose it into R1 = (A, B), R2 = (C, D). The set of

functional dependencies is:  A → B,  C → D, Then decomposition is a

(a) Dependency preserving but lossy

(b)Dependency preserving and lossless

(c) lossless

(d) Neither lossy nor dependency preserving

Answer: (b) Dependency preserving but lossy

(3) if A(x) = {0.1,0.2,0.3, .....}

        B(x) = {0.3,0.4,0.5,......}

then find out Complement of A ∩ B

Solution:

(1)find out Intersection first using below rule

Standard intersection (A ∩ B)(x) = min [A(x), B(x)]

(2) Find out complement using complement rule

Standard complement cA(x) = 1 − A(x)

(4) Find out which of the following grammer is in GNF(Greibach Normal Form )

Options: given 4 grammers and we need to find out the grammers which are in GNF

(5) Find out the Regular expression produced by the following DFA

(a) b(b*+a*b)

(b) a*b(b*+aa*b)*

Answer: (B)

(6) (1) Every context-sensitive language is recursive

     (2) There exist recursive languages that are not context-sensitive

which of the following are correct statements

(a) (1) only

(b) (2) only

(c) Both (1) and (2)

(d) Neither (1) nor (2)

Answer: (C)

Explanation:

Every regular language is context-free, every context-free language is context-sensitive

and every context-sensitive language is recursive and every recursive language is

recursively enumerable. These are all proper inclusions, meaning that there exist

recursively enumerable languages which are not recursive, recursive languages that are

not context-sensitive, context-sensitive languages which are not context-free and contextfree languages which are not regular.

(7)  One Question on the closure properties of Recursively enumarable languages, Recursive languages and context free languages

Closure properties

Recursively enumerable languages are closed under the following operations. That is, if L and P are two recursively enumerable languages, then the following languages are recursively enumerable as well:

• the Kleene star  of L
• the concatenation  of L and P
• the union 
• the intersection 

Note that recursively enumerable languages are not closed under set difference or complementation. The set difference L - P may or may not be recursively enumerable. If L is recursively enumerable, then the complement of L is recursively enumerable if and only if L is also recursive.

Recursive languages are closed under the following operations. That is, if L and P are two recursive languages, then the following languages are recursive as well:

• The Kleene star 
• The image φ(L) under an e-free homomorphism φ
• The concatenation 
• The union 
• The intersection 
• The complement of 
• The set difference 

The last property follows from the fact that the set difference can be expressed in terms of intersection and complement.

Closure properties

Context-free languages are closed under the following operations. That is, if L and P are context-free languages, the following languages are context-free as well:

• the union  of L and P
• the reversal of L
• the concatenation  of L and P
• the Kleene star  of L
• the image  of L under a homomorphism 
• the image  of L under an inverse homomorphism 
• the cyclic shift of L (the language )

Context-free languages are not closed under complement, intersection, or difference. However, if L is a context-free language and D is a regular language then both their intersection  and their difference  are context-free languages.

[edit]Nonclosure under intersection and complement

The context-free languages are not closed under intersection. This can be seen by taking the languages  and , which are both context-free. Their intersection is , which can be shown to be non-context-free by the pumping lemma for context-free languages.

Context-free languages are also not closed under complementation, as for any languages A and B: .

Closure properties

The regular languages are closed under the various operations, that is, if the languages K and L are regular, so is the result of the following operations:

• the set theoretic Boolean operations: union , intersection , and complement . From this also difference  follows.
• the regular operations: union , concatenation , and Kleene star .
• the trio operations: string homomorphism, inverse string homomorphism, and intersection with regular languages. As a consequence they are closed under arbitrary finite state transductions, like quotient  with a regular language. Even more, regular languages are closed under quotients with arbitrary languages: If L is regular then L/K is regular for any K.
• the reverse (or mirror image) .

(8) (1) Deterministic and non-deterministic DFA's are equivalent

      (2) Deterministic and non-deterministic PDA's are equivalent

which of the following are correct statements

(a) (1) only

(b) (2) only

(c) Both (1) and (2)

(d) Neither (1) nor (2)

Answer: (A)

(9)

which of the above graphs are planar.

Options

(a) G1 only

(b) G2 only

(c) Both G1 and G2

(d) Neither G1 nor G2

Answer (b)

Explanation:

Graph G2 is planar because we can redraw the same graph with out any crossovers.

(10) (a+b) (a+b) (a+b)  ..... (a+b) n times ... minimum number of states required to implement using DFA

Options:

(a) n   (b) n+1 (c) n+2 (d) None

Answer: (c)

Explanation:

if n=0,  then DFA should accept only epsilon .....

total 2 states are reuired for n=0


if n=1,  then DFA should accept { a,b}

total 3 states are reuired for n=1


if n=2,  then DFA should accept { aa,ab,ba,bb}

total 4 states are reuired for n=2

(11) Length of the IPV4 header field

Answer: 4 bits

(12) Transaction manager functionality is

(13) Match the following

OLAP                     Datawarehouse

OLTP                      RDBMS

...............................

(14) Which of the trees needs to have all leaves in the same level

Answer: B trees

(15) which of the following tree gives sorted list during traversal 

Answer: BST (Binary Search Tree )

(16)  two questions from software validation and verification... 

• Software verification asks the question, "Are we building the product right?"; that is, does the software conform to its specification.
• Software validation asks the question, "Are we building the right product?"; that is, is the software doing what the user really requires.

(17)  CMM level 4 also included in 


(a) CMM Level 2
(b) CMM level 3
(c) CMM level 5
(d) none


Answer: (C)




(18) A grammer has given  ... four grammers given as four options . we need to find the equivalent grammer to the given grammer




(19)On a disk with 1000 cylinders, numbers 0 to 999, compute the number of tracks

    the disk arm must move to satisfy all the requests in the disk queue. Assume the

    last request serviced was at track 345 and the head is moving toward track 0.The

    queue in FIFO order contains requests for the following tracks :

    123,874,692,475,105,376.Perform the computation for SCAN scheduling

    algorithm :

Answer: 1219

(20) Question on Amdahl's law on parallel processing


It is often advised to focus system design on hardware scalability rather than on capacity. It is typically cheaper to add a new node to a system in order to achieve improved performance than to partake in performance tuning to improve the capacity that each node can handle. But this approach can have diminishing returns (as discussed in performance engineering). For example: suppose 70% of a program can be sped up if parallelized and run on multiple CPUs instead of one. If  is the fraction of a calculation that is sequential, and  is the fraction that can be parallelized, the maximum speedup that can be achieved by using P processors is given according to Amdahl's Law: . Substituting the value for this example, using 4 processors we get . If we double the compute power to 8 processors we get . Doubling the processing power has only improved the speedup by roughly one-fifth. If the whole problem was parallelizable, we would, of course, expect the speed up to double also. Therefore, throwing in more hardware is not necessarily the optimal approach.








(21) java.util.* 


Match the following


Calendar               getTimeZone()
Random                getNumber()
Timezone             setId()
...................


I think...


Random   getNumber()

.......................................

(22)  level of abstraction, describes what data are stored in database

(a) view

(b) abstraction

(c) physical

(d) logical

Answer: Logical

Explanation

The are three levels of abstraction:

1. Physical level: The lowest level of abstraction describes how data are stored.
2. Logical level: The next higher level of abstraction, describes what data are stored in database and what relationship among those data.
3. View level: The highest level of abstraction describes only part of entire database.

(23)  hiding data and code ..........

Answer: Encapsulation

(24) Which of the following is linear data type

(a) Strings   (b) lists  (c) Queues (d) All the above

Answer: All of the above

(25) optimal binary search tree if probability of successful and unsuccessful search are same

(26) Computational Complexity for graph coloring problem

(27) Deadlock is

(a) Timeout

(b) Timein

(c) Both a and b

(d)None

Answer:  (a) Timeout 


Explanation : 
DB2 allows you to put a limit on the amount of time you’ll wait at a database level using the LOCKTIMEOUT configuration parameter.

(28) Software is

(a) schedule with in the budget ....

(29)A common property of logic programming languages and functional languages is:

(a) both are procedural languages 
(b) both are based on λ-calculus
(c) both are declarative 
(d) both use Horn-clauses

Answer: C

Explanation:

Both are declarative ( http://en.wikipedia.org/wiki/Declarative_programming )

(30) A* Algorithm ..heauristic function = g+h ......................

....................................................

(a) g=0  (b) g=1

(c)h=0 (d) h=1

(31) Question on GPS ( GLOBAL POSITIONING SYSTEM )

(32) In unit testing of a module, it is found for a set of test data, at the maximum 90% of the code alone were tested with the probability of success 0.9. the reliability of the module is


a. Atleast greater than 0.9
b. Equal to 0.9
c. Atmost 0.81
d. Atleast 1/0.81

Answer: (C)

(33) which of the following software metric does not depend on programming language

(a) LOC ( Lines of code )

(b) Function Point

(c) member of token

(d) None

(34) Reliability of a software depend on

(a) Number of errors present

(35) There is an edge between u and v.. (u,v). shortest path from s to u is 53 and shortest path from s to v is 65 then what can you say about (u,v)

(a) (u,v) =12

(b) (u,v) >= 12

(c) (u,v) <=12

(d) (u,v) > 12

Answer(b) (u,v)>=12, using Floyd Warshall algorithm

(36) which of the following is not UNIX shell

(a) Bourne Shell

(b) C Shell

(c) Net Shell

(d) Korne Shell

Answer: (c) Net Shell

(37) CTRL+C Unix .....

(a) user mode

(b) kernel mode

(c) super user mode

(38) two lines .....

(a) Histogram

(b) covariance

....................

...............

(39) Printf("%c",100)

Answer: Ascii values corresponds to 100 ( that is d )

(40) Match the following

Microcontroller          8051

..............................

..............................

(41) Match the following

Instructions        addressing modes

..........................................................

...........................................................

(42)  Match the following

Regular languages                 DFA

CSL                                     LBA

CFL                                     PDA

Recursive languages              Turing Machine

(43)  Question on Prolog ... If --else

(44) Pipeline processing .....

instruction decoding .... instruction execution ....

(45) Microcomputer consits of

(a) perpherials

(b)Micro processor

(c)

(46) Color magneta ............

Red and white

Red and black

(47) In pre-emtive scheduling algorithm if time quantum increases, effective turn around time

Answer: I think Decreases

(48) making capabilities

(a) Conditional transfer

(b) Unconditional transfer

(49) non-sharable resources

(a) Mutual exclusion

...............

.............

(50) if an integer takes 2 bytes.. what is the maximum value can be represented

(51) interrupts

(a) software

(b) Hardware

(c) internal

(d) External

(52) Relationship among claases and objects .... can be represented

(a) class diagram

(b) Object diagram

(53) Which of the following deletes the structure of the data

(a) Erase
(b) Delete
(c) ZAP
(d) Pack 

reference: http://msdn.microsoft.com/en-US/library/00c5e99f(v=vs.80)

Command:Visual Studio 2005: Removes all records from a table, leaving just the table structure.

(54) when one transaction updates a database item and then 
the transaction fails for some reason is called  

(a) Dirty read problem

(b) temporary update problem

(55) amplification, modulation and ....................

(56) Number of binary trees with 5 nodes

(a) 32 (b) 36 (c) 120 (d) 42

answ: 42

reference: http://theory.cs.uvic.ca/inf/tree/BinaryTrees.html

(57)  Given post order traversal ... need find out the preorder traversal\

(58)  Consider the methods used by processes P1 and P2 for accessing their critical sections whenever needed, as given below. The initial values of shared boolean variables S1 and S2 are randomly assigned.

Method Used by P1
while (S1 == S2) ;
Critica1 Section
S1 = S2;

Method Used by P2
while (S1 == S2) ;
Critica1 Section
S2 = not (S1);

Which one of the following statements describes the properties achieved? (GATE CS 2010)
(A) Mutual exclusion but not progress
(B) Progress but not mutual exclusion
(C) Neither mutual exclusion nor progress
(D) Both mutual exclusion and progress

Progress Requirement: If no process is executing in its critical section and there exist some processes that wishes to enter their critical section, then the selection of the processes that will enter the critical section next cannot be postponed indefinitely.

(59) if virtual address space and logical address space is same then

(60) cache memory and interleaved memory

(61) compiler generates window programming

(a) text file

(b) binary file

(c) obj file

(62) Virtual switching and packet switching

(63) base class derived class... constructor order

(64) additions done in computer

(a) CPU

(b) Memory

(c)

(65) Connect different LANS with different technologies

(a) Bridges

(b) switches

(66) Encryption and Decryption of data is done in which layer?

Answer: (a) presentation layer


Explanation:



Layer 6:Presentation Layer
The presentation layer presents the data into a uniform format and masks the difference of data format between two dissimilar systems. It also translates the data from application to the network format. Presentation layer is also responsible for the protocol conversion, encryption, decryption and data compression. Presentation layer is a best layer for cryptography.
Network Devices: Gateway Redirector is operates on the presentation layer

(67)

What is the output for above circuit

option (a) is correct

(68)

K=0;

for i=1 to n

 for i1=1 to i

   for i2=1 to i1

................

..........

.............

for im=1 to im-1

k=k+1

output for this program?

Answer:  C(n+m-1,m)

example:


#include<stdio.h>
main()
{
int k=0;
int i,i1,i2,i3,i4,i5,n=3,m=2;
for (i1=1;i1<=n;i1++)
for(i2=1;i2<=i1;i2++)
k=k+1;

printf("Values of n= %d, m=%d, k=%d",n,m,k);
}


output:

Values of n= 3, m=2, k=6

if we consider option(a) as answer then
c(n+m-1,m)=c(3+2-1,2)=c(4,2)=6

if we consider option(b) as answer then
c(n-m+1,m)=c(3-2+1,2)=c(2,2)=1

So Option(a) is correct.

(69) E-commerce cannot be used in

(a) House banking

(b) for evaluating employee performance

(c).......

...........

(70) if virtual address space is equal to physical address space then ...

(71)  Question on COSETS

(72)  How many relations are there on a set with n elements that are  symmetric  and How many relations are there on a set with n elements that are reflexive and symmetric ?

Solution: 

Let R be the set with n elements. Then RxR has n^2 elements in it, and the relationson R correspond exactly to the subsets of RxR, giving us 2^(n^2) relations in general.

If the relation is symmetric, we can think of it slightly differently. Let P2(R) be all subsets of R with 2 elements, and P1(R) be all the subsets of R with a single element. Then all symmetric relations will correspond exactly to the subsets of P2(R) U P1(R). Notice that P2(R) is exactly like RxR, except the pairs aren't ordered, and it only considers pairs with distinct x and y (the pairs where they aren't distinct are covered by P1(R)). How many elements does P2(R) have in it? Well, we are looking for all pairs of the form:

{x, y}

where x and y are distinct and in R. They correspond to an unordered selection of 2 objects from n objects, giving us:

n C 2 = (1/2)n(n - 1)

How many elements of P1(R) are there? Well, clearly, there will be n elements. So, the total number of elements in P2(R) U P1(R) will be:

(1/2)n(n - 1) + n

= (1/2)n[(n - 1) + 2]

= (1/2)n(n + 1)

And, we want the number of subsets of this, so we get:

2^[(1/2)n(n + 1)]

As for all relations that are antisymmetric, that's a bit more tricky. I'll have to think about that one. The relations that are neither reflexive nor irreflexive are not too difficult to count. Assuming that n > 0, it's impossible for a relation to be simultaneously reflexive and irreflexive, so if we count the number of reflexive relations, and the number of irreflexive relations, then we will not have counted the same relation twice, and we can just subtract this number from 2^(n^2).

In both the reflexive and irreflexive cases, essentially membership in the relation is decided for all pairs of the form {x, x}. This leaves n^2 - n pairs to decide, giving us, in each case:

2^(n^2 - n)

choices of relation. That is the number of reflexive relations, and also the number of irreflexive relations. The number of relations that are either reflexive or irreflexive will be the sum:

2^(n^2 - n) + 2^(n^2 - n) = 2^(n^2 - n + 1)

If we subtract this from the total number of relations, 2^(n^2), then we get the number of relations that are neither reflexive or irreflexive:

2^(n^2) - 2^(n^2 - n + 1)

Hope that helps!

EDIT: (Fixed a small mistake previously)

OK, I've just thought of a way to deal with the antisymmetric case. Again, we will consider P2(R) U P1(R). We can choose freely which pairs of the form (x, x) we want in our relation, so we can choose freely our subset of P1(R), giving us 2^n possible contributions from P1(R). As for our contribution from P2(R), for each {x, y} pair in P2(R), we must have exactly one of the following three possibilities:

1) Neither (x, y) nor (y, x) is in our relation.

2) Only (x, y) is in our relation.

3) Only (y, x) is in our relation.

Each choice for each {x, y} can be made independently of the other choices chosen previously. Also, making two distinct choices will result in two distinct relations, i.e. we are not counting anything twice. Therefore, the number of contributions from P2(R) will be:

3^(n C 2) = 3^[(1/2)n(n - 1)]

The contributions of P2(R) and P1(R) are independent, so the total number of antisymmetric relations will be:

2^n * 3^[(1/2)n(n - 1)]

(73)  Which data structure is used when you do the post order traversal 

answer: stack

(74) what would be the top of elements in stack when u do post order traversal 

23^8/23*

Answeer: 6,1

(75) which class of network provides multicasting

Answer: class D

(76) ICMP protocol presents in ------

Answer: network layer

(77) x.25 is 

(a) connection oriented

(b) connection less

(c) both

(d) neither of them


Answer : (a)

(78)HTML standard

(a)  ISO 8879 

(79) if an instruction takes i nano seconds ... if it takes extra j nano seconds for every k instructions then effective access time is 

Answer: i+j/k

(80) if Q(x,y) represents x+y=0 and x,y are real numbers then

(1) ∃x∀y Q(x,y)


(2) ∀x∃y Q(x,y)

which of the statements are true

Explanation:

Compare these two sentences: "For all x, there exists at least one y such that Q(x,y)" and "there is at least one y such that, for all x, Q(x,y)". Here Q(x,y) is some sentence about x and y, such as "x+y=0". The first of these means that for all x, there is a y WHICH IS ALLOWED TO VARY WITH EACH x which makes Q(x,y) true. So, for example, "for all x, there exists at least one y such that x+y=0" is true because y=-x makes it true.

that means  ∀x∃y Q(x,y) is true 

The sentence "there is at least one x such that, for all y, Q(x,y)" means that there exists one x which does the job for all y of making Q(x,y) true. This is a much tougher statement to make true! Notice that "there is at least one x such that, for all y, x+y=0" is false for the real numbers---there x has to be -y and must vary with y which this sentence does not allow.

that means ∃x∀y Q(x,y) is false

http://www.math.hawaii.edu/~ramsey/Logic/ThereIs.html

(81) which of the following is preferable in software

Answer : High cohesion and low coupling

(82) link state algorithm

(83) 1000 ms then frequency

(a) 1 KHZ (b) 10 khz

(84) 10 base Tx 

(85) which data structure is used for heirarchical

Answer is Tree

(86) Compare B+ tree with normal trees.. why do we use B+trees 

(87)  which logic family is fastest

Answer:  ECL

Explanation:

 ECL (http://en.wikipedia.org/wiki/Emitter-coupled_logic)

(88) To connect diffrent technology networks 


Answer: bridge

(89) Consider unsigned integer representation. How many bits will be required to store a decimal number containing 3 digits

Answer: 10 bits


Explanation: 

we have to find the lowest power of 2 that is higher than that range.

For instance, 3 decimal digits -> 10^3 = 1000 possible numbers so you have to find the lowest power of 2 that is higher than 1000, which in this case is 2^10 = 1024 (10 bits).

(90)Computational Complexity for graph coloring problem

Ans: O(n*m^n)

Explanation:

http://www.scribd.com/doc/22874034/A-Notes-on-Design-Analysis-of-Algorithm

(91)Planner graphs can be _____________colored problems 

(a) 2 b)3 c) 4 d) 5

ANS: By the four color theorem, every planar graph can be 4-colored

Explanation: http://en.wikipedia.org/wiki/Graph_coloring

(92)consider the declaration Base *b = new Derived;
the order of execution of base and derived class constructors?

a) derived class constructor followed by base class

b) base class constructor followed by derived class

*****************************************************************************

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