Thursday, 13 June 2013

FINITE AUTOTMATA QUESTIONS

This post gives the question and answer for the subject FINITE AUTOMATA from all the previous year question papers of UGC NET.
First the year in which the question appears would be mentioned followed by the question and its answer.

DECEMBER 2004

1. The context-free languages are closed for :
(i) Intersection    (ii) Union
(iii) Complementation   (iv) Kleene Star
then
(A) (i) and (iv)    (B) (i) and (iii)
(C ) (ii) and (iv)    (D) (ii) and (iii)

Ans:-C

Explanation:- Context-free languages are closed under union,intersection and star-closure. Context-free languages are not closed under intersection and complementation.


June - 2005

2. Which of the following is not true ?
(A) Power of deterministic automata is equivalent to power of non-deterministic automata.
(B) Power of deterministic pushdown automata is equivalent to power of non-deterministic pushdown automata.
(C) Power of deterministic turing machine is equivalent to power of non-deterministic
turing machine. (D) All the above

Ans:-B

Explanation:- Deterministic and nondeterministic finite automata have equivalent computational capabilities.
Deterministic and nondeterministic Turing machines have the same computational capabilities.
However, the computational capabilities of deterministic push-down automata are less than those of nondeterministic push-down automata. So, the answer is B.


3. Identify the language which is not context - free.
(A) L={wwR|w is a member of {0,1}*}
(B) L={anbn|n>=0}
(C )L={ww|w is a member of {0,1}*}
(D) L={anbmcmdn|n,m>=0}

Ans:-B


December 2005

4. Which sentence can be generated by S → d/bA, A → d/ccA :
(A) bccddd
(B) aabccd
(C) ababccd
(D) abbbd

Ans:-A

Explanation:- The most closest answer seems to be A only. The terminals in the grammar are d,b and c. There is no terminal 'a' at all. So the options B,C and D are ruled out. option A can also be derived at something like this.
S->bA
->bccA
->bccccA
->bccccd
So the sentence generated should be bccccd. If there are any other explanations for the same please post them.


5. Regular expression a+b denotes the set :
(A) {a}
(B) {Epsilon, a, b}
(C) {a, b}
(D) None of these

Ans:-C


June 2006

6. Which of the following strings is in the language defined by grammar
S → OA, A → 1A/0A/1
(A) 01100
(B) 00101
(C ) 10011
(D) 11111

Ans:-B

Explanation:- Non terminals in the above grammar are S, and A. Terminals are 0 and 1. The start symbol of the grammar is S. S->0A. So rule out strings beginning with 1. so that leaves us with two options A and B. S->0A
->00A
->001A
->0010A
->00101
So the option B is correct. If you try to derive the string for option A you would not be able to get it. So, the correct answer is option A.


7. The logic of pumping lemma is a good example of :
(A) pigeon hole principle
(B) recursion
(C) divide and conquer technique
(D) iteration

Ans:-A


December - 2006

8. Which of the regular expressions corresponds to this grammar ? S → AB/AS, A → a/aA, B → b
(A) aa*b+
(B) aa*b
(C ) (ab)*
(D) a(ab)*

Ans:-B


JUNE 2007

9. The regular expression given below describes :
r=(1+01)*(0+λ)
(A) Set of all string not containing '11'
(B) Set of all string not containing '00'
(C) Set of all string containing '01'
(D) Set of all string ending in '0'

Ans:-B

Explanation:-
The meaning of (1+01)* means that the set of strings of 1 and 01 of any length including the NULL string. (0+λ) means either 0 or λ. So the full regular expression stands for the set of strings of 1's and 01's of any length ending with 0 or λ. So we cannot say the string will only end in '0'. It could end in λ also. The string can begin with 1 or 01 and multiple times it can get repeated.But the string '11' cannot be together. So the option could be B.


Wednesday, 15 May 2013

JUNE 2012 - PAPER III

21. A* algorithm uses f' = g + h' to estimate the cost of getting from the initial state to the goal state, where g is a measure of the cost of getting from initial state to the current node and the function h' is an estimate of the cost of getting from the current node to the goal state. To find a path involving the fewest number of steps, we should set
(A) g=1
(B) g=0
(C) h'=0
(D) h'=1

Ans:-A


22. The transform which possesses the highest ‘energy compaction’ property is
(A) Slant transform
(B) Cosine transform
(C) Fourier transform
(D) Karhunen-Loeve transform

Ans:-D


23. Which one of the following prolog programs correctly implement “if G succeeds then execute goal P else execute goal θ ?”
(A) if-else (G, P, θ) :- !, call(G), call(P). if-else (G, P, θ) :- call(θ).
(B) if-else (G, P, θ) :- call(G), !, call(P). if-else (G, P, θ) :- call(θ).
(C) if-else (G, P, θ) :- call(G), call(P), !. if-else (G, P, θ) :- call(θ).
(D) All of the above

Ans:-B

Explanation:- The syntax of If--then---else in prolog goes the following way…
(A->B;C) :-
    call(A),
    !,
    call(B)
(A->B;C) :-
    call(C )
So according to the above syntax option B is correct.


24. The _______ memory allocation function modifies the previous allocated space.
(A) calloc( )
(B) free()
(C) malloc( )
(D) realloc()

Ans:-D


25. Which is not the correct statement(s) ?
(i) Every context sensitive language is recursive.
(ii) There is a recursive language that is not context sensitive.
(A) (i) is true, (ii) is false.
(B) (i) is true and (ii) is true.
(C) (i) is false, (ii) is false.
(D) (i) is false and (ii) is true.

Ans:-B


26. The mechanism that binds code and data together and keeps them secure from outside world is known as
(A) Abstraction
(B) Inheritance
(C) Encapsulation
(D) Polymorphism

Ans:-C


27. Identify the addressing modes of below instructions and match them :
(a) ADI (1) Immediate addressing
(b) STA (2) Direct addressing
(C )CMA (3) Implied addressing
(d) SUB (4) Register addressing
(A) a-1,b-2,c-3,d-4
(B) a-2,b-1,c-4,d-3
(C ) a-3,b-2,c-1,d-4
(D) a-4,b-3,c-2,d-1

Ans:- A

Explanation:- The instruction ADI adds some content to the accumulator. It is an immediate addressing mode instruction.
The instruction STA stores the contents of the accumulator in the particular memory location specified as operand.
CMA instruction takes complement of the contents of the accumulator.
SUB instruction subtracts the contents of the register to the contents of the accumulator.
. So the option is A.


28. Which one of the following is not a Greibach Normal form grammar ?
(i) S → a | bA | aA | bB
A→a B→b
(ii) S→a|aA|AB
A→a
B→b
(iii) S→a|A|aA
A→a
(A) (i) and (ii)
(B) (i) and (iii)
(C) (ii) and (iii)
(D) (i), (ii) and (iii)

Ans:-C

Explanation:- Restriction for GNF:-
The first symbol on the right hand side of the production must be a terminal. It can be followed by zero or more variables. In grammar (ii) of the question, S->AB is a production. AB are two non-terminals and it can be in GNF. In grammar (iii) S->A is given and it is a unit production and that is not allowed in GNF. So the grammar which is not in GNF is (ii) and (iii). So the option is C.


29. Which of the following IP address class is a multicast address ?
(A) Class A
(B) Class B
(C) Class C
(D) Class D

Ans:-D


30. While unit testing a module, it is found that for a set of test data, maximum 90% of the code alone were tested with a 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

Ans:-C


Wednesday, 20 March 2013

JUNE 2012 - PAPER III

18. Consider a schema R(A, B, C, D) and functional dependencies A → B and C → D. Then the decomposition R1(A, B) and R2(C, D) is
(A) Dependency preserving but not lossless join
(B) Dependency preserving and lossless join
(C) Lossless Join but not dependency preserving
(D) Lossless Join

Ans:-A

Explanation:-
I have given the explanation in question no. 17 of December 2010 UGC paper. I am repeating it once again here for dependency preservation and lossless join. First of all let us consider the dependency preservation and how to understand it.
Definition of Dependency preservation decomposition:-

Each FD specified in F either appears directly in one of the relations in the decomposition, or be inferred from FDs that appear in some relation.


Let us consider the above example for Dependency preservation



Let R be a relation R(A B C D)

Let there be 2 functional dependencies.

FD1: A->B

FD2: C->D


Let the relation R be decomposed into two more relations.

R1(A B )  :  R2(C D)

Let us first consider the relation R1(A B ). Here between A and B the functional dependency FD1 is preserved.

Let us now consider the second relation R2(C D). Between C and D the FD, FD2 is preserved. So in the two relations R1 and R2, all the 2 functional dependencies are preserved.

Now for the lossless join.
A decomposition of R into R1 and R2 is lossless join if and only if at least one of the following dependencies is in F*: R1 ∩ R2 -> R1 R1 ∩ R2 -> R2 In the above example, R1 ∩ R2 = { null }. So, it is not a lossless join. So the answer is dependency preserving but not lossless join. So, the answer is option A.


19. The quantiser in an image-compression system is a
(A) lossy element which exploits the psychovisual redundancy
(B) lossless element which exploits the psychovisual redundancy
(C) lossy element which exploits the statistical redundancy
(D) lossless element which exploits the statistical redundancy

Ans:-A


20. Data Warehouse provides
(A) Transaction Responsiveness
(B) Storage, Functionality Responsiveness to queries
© Demand and supply Responsiveness
(D) None of the above

Ans:-B


Tuesday, 19 March 2013

JUNE 2012 - PAPER III

11. X.25 is ________ Network.
(A) Connection Oriented Network
(B) Connection Less Network
(C) Either Connection Oriented or Connection Less
(D) Neither Connection Oriented nor Connection Less

Ans:-A

Explanation:-
X.25 is a connection oriented protocol.


12. Which of the following can be used for clustering of data ?
(A) Single layer perception
(B) Multilayer perception
(C) Self organizing map
(D) Radial basis function

Ans:-C


13. Which of the following is scheme to deal with deadlock ?

(A) Time out
(B) Time in
(C) Both (A) & (B)
(D) None of the above

Ans:-A


14. If the pixels of an image are shuffled then the parameter that may change is
(A) Histogram
(B) Mean
(C) Entropy
(D) Covariance

Ans:-D


15. The common property of functional language and logical programming language :
(A) Both are declarative
(B) Both are based on λ-calculus
(C) Both are procedural
(D) Both are functional

Ans:-A


16. Given the following statements :
(i) The power of deterministic finite state machine and non- deterministic finite state machine are same.
(ii) The power of deterministic pushdown automaton and non- deterministic pushdown automaton are same.
Which of the above is the correct statement(s) ?
(A) Both (i) and (ii)
(B) Only (i)
(C) Only (ii)
(D) Neither (i) nor (ii)

Ans:-B

Explanation:-
The answer is B. But why is it so?. A very good explanation is given in the book "Theory of computation" by A.A.Puntambekar.
We all know that finite machine is of two types. One is deterministic finite state machine and the other one non deterministic finite state machine. Both these machine accept regular language only. So the power of DFA = NFA. So the first statement is true.
Next comes the question of pushdown automaton and the power of deterministic and non-deterministic being the same. PDA has more power than FA because PDA has a memory and so can accept large class of languages than FA. PDA accepts the language of context free grammar. The power of DPDA is less than NPDA because NPDA accepts a larger class of context free language.
Turing machines can accept a more large class of language. Therefore it is the most powerful computational model. The power of deterministic and non deterministic turing machine is the same.


17. LetQ(x,y)denote “x+y=0” and let there be two quantifications given as
(i) ∃y∀x Q(x, y)
(ii) ∀x∃y Q(x, y)
where x & y are real numbers. Then which of the following is valid ?
(A) (i) is true & (ii) is false.
(B) (i) is false & (ii) is true.
(C) (i) is false & (ii) is also false.
(D) both (i) & (ii) are true.

Ans:-B

Explanation:-
The symbol ∀ is called the universal quantifier. The universal qualification of P(x) is the statement "P(x) for all values x in the universe" which is written as ∀xP(x).
The symbol ∃ is called the existential quantifier and represents the phrase "there exists" or "for some". The existential quantification of P(x) is the statement "P(x) for some values x in the universe which is written as ∃xP(x).
A nested quantifier is one where two quantifiers are nested if one is within the scope of the other.
The order of nested universal quantifiers in a statement without other quantifiers can be changed without changing the meaning of the quantified statement.
The order of nested existential quantifiers in a statement without other quantifiers can be changed without changing the meaning of the quantified statement.
But the order of nested universal and existential quantifier is important. The order cannot be changed without affecting the meaning.
Let us consider the quantification ∀x∃y Q(x, y). The domain is real numbers. Q(x,y) is x+y=0. The quantification can be explained this way. Since the universal quantification comes first, it has to be understood as, for all real numbers x there is a real number y such that x+y=0, which is actually true. We are saying that y is the additive inverse of x.
Now let us consider the quantification ∃y∀x Q(x, y). Since the existential quantification comes first, it has to be understood as there is a real number y such that for all real numbers x, x+y=0 , which is false.
So, the quantification (i) is false and (ii) is true.


Saturday, 16 March 2013

MUSINGS ON LIFE

I would be continuing with updating my blog with detailed solution of previous UGC NET PAPERS. Apart from that, I am also going to be penning down my thoughts which i have always wanted to share it with the world. Experiences which i have undergone and lessons i have learnt and generally some musings on life. Hope you all enjoy it and learn and grow from it just like me. I hope and pray that the updates are more regular now.

CHANGE IS THE KEY



CHANGE IS THE KEY


Everyone wants their country to be the best. Being an Indian i also want the same. I want the best things to happen in our country. I want everyone of us to be proud of who we are, for what we stand for. But the recent news of rapes and violence against women has tarnished the image of India worldwide. The capital of our country not safe to women at all does not sound good for the image of the country.

I retrospect a lot about the recent developments happening in our country and how everyone of are restless and looking forward to a corrupt free nation. The entire apathy of the nation is blamed on the politicians. They are made wholly responsible for what is happening in our country wrong. But is it the right approach?.

Roughly India has a population of 1 billion. There are around 500 odd representatives in the parliament for this entire population. Although their influence on the overall running of the governance cannot be completely undermined, you can also not blame them only for all the negative things happening around you. The time has come for each and every citizen to take up responsibility for what is happening around them and not blame only the politicians. John F. Kennedy in his world famous speech asked the americans, "Do not ask what the country has done for you, ask what you have done for the country". The timing is no better right to implement it in our country. I think every one of us need to ask what we have done for the country and the apathy surrounding us.

We need to think on what are the positive changes which one can bring in our society. If every individual changes for the better, the entire country changes. If every person thinks positively and hopes for a better tomorrow it is achievable. Actually India bashing is very fashionable among indians. "There is no future for our country", "Politicians have swindled money and left us with nothing..", "It is better to migrate to a developed nation instead of living in our country"…these are some of the statements which you will hear when we hear talk about the state of our country. No doubt things are a bit gloomy. But we need to look at how we, each one can make things better for others and ourselves too in the same order.

It could be as small a step as following traffic rules and not littering in public. But every single act of improvement on our part is going to get reflected in the society. Teach our kids to be honest and instill good values in them. Let each and everyone of us become the citizen of a modern India, a India which everyone want to live in. What is the behavior you would expect from the people of such a country, try to bring it in yourself.

Be proud of your heritage, history, and our unity in diversity. Learn to respect others. Do not judge others. IMG 5215

Society is just a reflection of the collective thoughts of the individuals. A nation reflects the hopes,beliefs and thoughts of the society. Be the change you want to see in others, in the society and the nation. If every single individual changes for the better, society changes for the better and the nation also would reflect it. So stop blaming politicians and start becoming the champion of change. Make an India where no one would want to migrate from.JAI HIND.

JUNE 2012 - PAPER III

10. In an image compression system 16384 bits are used to represent 256 × 256 image with 256 gray levels. What is the compression ratio for this system ?
(A) 1
(B) 2
(C) 4
(D) 8

Ans:- 

Explanation:-
Number of bits required to store a 256 X 256 image with 256 gray levels is.
256 gray levels = 28
=8 bits
Therefore 256 * 256 * 8 = 524,288 bits
The ratio of the original(uncompressed) image to the compressed image is referred to as the Compression Ratio CR.
CR = Uncompressed image size/Compressed image size
Uncompressed image size for the given data above= 524,288 bits
Compressed image size as given in the question = 16384 bits
Therefore, compression ratio = 524,288/16384 = 32
I am not converting the values into bytes.
There is no option called 32 in the answer at all. The solution website for June 2012 says option B which is 2. The same question is found in another place and the solution there is 8 but no explanation provided. If anyone can provide an explanation as to how 8 is the answer it will be great. I have got the value 32. So i am not choosing any option as the answer.


Thursday, 27 September 2012

DECEMBER 2011 - PAPER II


28. On receiving an interrupt from an I/O device, the CPU

(A) halts for predetermined time
(B) branches off to the interrupt service routine after completion of the current instruction
(C) branches off to the interrupt service routine immediately.
(D) hands over control of address bus and data bus to the interrupting device.

Ans:-B


29. The maximum amount of information that is available in one portion of the disk access arm for a removal disk pack (without further movement of the arm with multiple heads)

(A) a plate of data
(B) a cylinder of data
(C) a track of data
(D) a block of data

Ans:-B


30. Consider a logical address space of 8 pages of 1024 words mapped with memory of 32 frames. How many bits are there in the physical address ?

(A) 9 bits
(B) 11 bits
(C) 13 bits
(D) 15 bits

Ans:-D

Explanation:-
The question asks about the number of bits in the physical address. The word size is 1024. So, the number of bits required to calculate the offset into each page = 210=1024=10 bits.
So for 32 frames(25) of 1024 words each,we would require a total of 5+10 bits = 15 bits. Therefore the option is D.


31. CPU does not perform the operation

(A) data transfer
(B) logic operation
(C) arithmetic operation
(D) all of the above

Ans:-D


32. A chip having 150 gates will be classified as

(A) SSI
(B) MSI
(C) LSI
(D) VLSI

Ans:-B

Explanation:- The number of gates for the different classification are as follows.
Small Scale Integration(SSI) - <10
Medium Scale Integration(MSI) - between 10 to 1000
Large Scale Integration(LSI) - >1000
Very Large Scale Integration(VLSI) - >100000
So the answer is option B.


33.If an integer needs two bytes of storage, then the maximum value of unsigned integer is

(A) 216 –1
(B) 215 –1
(C) 216
(D) 215

Ans:-A

Explanation:- The same question is asked in June 2012,paper -II but there they have asked for the maximum value of signed integer.The link to that answer is JUNE 2012 - PAPER II. Here it is the maximum value of unsigned integer and the answer is option A.


34. Negative numbers cannot be represented in

(A) signed magnitude form
(B) '1’ s complement form
(C) '2’ s complement form
(D) none of the above

Ans:-D

Explanation:-
Negative numbers can be represented in signed magnitude form, or '1' complement form or '2''s complement form. So the option is D.


35. The cellular frequency reuse factor for the cluster size N is

(A) N
(B) N2
(C) 1/N
(D) 1/N2

Ans:-C