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.

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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.

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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={aⁿb^mc^mdⁿ|n,m>=0}

Ans:-B

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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.

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

Ans:-C

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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.

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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

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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

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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.

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