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Welcome to the Theory of Computation(TOC) MCQs Page

Dive deep into the fascinating world of Theory of Computation(TOC) with our comprehensive set of Multiple-Choice Questions (MCQs). This page is dedicated to exploring the fundamental concepts and intricacies of Theory of Computation(TOC), a crucial aspect of UGC CBSE NET Exam. In this section, you will encounter a diverse range of MCQs that cover various aspects of Theory of Computation(TOC), from the basic principles to advanced topics. Each question is thoughtfully crafted to challenge your knowledge and deepen your understanding of this critical subcategory within UGC CBSE NET Exam.

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Check out the MCQs below to embark on an enriching journey through Theory of Computation(TOC). Test your knowledge, expand your horizons, and solidify your grasp on this vital area of UGC CBSE NET Exam.

Note: Each MCQ comes with multiple answer choices. Select the most appropriate option and test your understanding of Theory of Computation(TOC). You can click on an option to test your knowledge before viewing the solution for a MCQ. Happy learning!

Theory of Computation(TOC) MCQs | Page 1 of 16

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Q1.
Given the following expressions of a grammar

E --> E * F / F + E / F
F --> F - F / id

Which of the following is true ?

a.

* has higher precedence than +

b.

- has higher precedence than *

c.

+ and  - have same precedence

d.

+ has higher precedence than *

Discuss
Answer: (b).- has higher precedence than *
Q2.
Which of the following is true while converting CFG to LL(1) grammar ?

a.

Remove left recursion alone

b.

Factoring grammar alone

c.

Both of the above

d.

None of the above

Discuss
Answer: (c).Both of the above
Q3.
Let L be a set accepted by a non deterministic finite automaton. The number of states in non-deterministic finite automaton is |Q|. The maximum number of states in equivalent finite automaton that accepts L is

a.

|Q|

b.

2|Q|

c.

2^|Q| – 1

d.

2^|Q|

Discuss
Answer: (d).2^|Q|
Q4.
The grammar ‘G1’

S ---> OSO| ISI | 0|1|∈

and the grammar  ‘G2’ is
S ---> as |asb| X,
X -----> Xa | a.

Which is the correct statement ?

a.

G1 is ambiguous, G2 is unambiguous

b.

G1 is unambiguous, G2 is ambiguous

c.

Both G1 and G2 are ambiguous

d.

Both G1 and G2 are unambiguous

Discuss
Answer: (b).G1 is unambiguous, G2 is ambiguous
Q5.
Which of the following regular expression identities are true ?

a.

(r + s)* = r* s*

b.

(r + s)* = r* + s*

c.

(r + s)* = (r*s*)*

d.

r* s* = r* + s*

Discuss
Answer: (c).(r + s)* = (r*s*)*
Q6.
The minimum number of states of the non-deterministic finite automation which accepts the language
{a b a bn| n  ≥  0} ∪ {a b an|n  ≥ 0} is 

a.

3

b.

4

c.

5

d.

6

Discuss
Answer: (c).5
Q7.
Which of the following definitions generates the same Language as L, where
L = {WWR | W ∈ {a, b}*} 

a.

S ∈ asb|bsa|∈

b.

S ∈ asa|bsb|∈

c.

S ∈ asb|bsa|asa|bsb|∈

d.

S ∈ asb|bsa|asa|bsb

Discuss
Answer: (b).S ∈ asa|bsb|∈
Q8.
If the parse tree of a word w generated by a Chomsky normal form grammar has no path of length greater than i, then the word w is of length 

a.

no greater than 2^(i+1) 

b.

no greater than 2^i

c.

no greater than 2^(i–1)

d.

no greater than i

Discuss
Answer: (c).no greater than 2^(i–1)
Q9.
Given the following statements :

S1: SLR uses follow information to guide reductions.In case of LR and LALR parsers, the look-aheads are associated with the  Items and they make use of the left context available to the parser.
S2: LR grammar is a larger sub-class of context free grammar as compared to that SLR and LALR grammars. 

Which of the following is true ? 

a.

S1 is not correct and S2 is not correct

b.

S1 is not correct and S2  is correct

c.

S1 is correct and S2 is not correct

d.

SI is correct and S2 is correct

Discuss
Answer: (c).S1 is correct and S2 is not correct
Q10.
The context free grammar for the language

L= {anbm  | n ≤ m+3,n ≥ 0 ,m≥ 0 } : is

a.

S → aaa; A → aAb | B, B →  Bb |  λ 

b.

S → aaaA | λ ; A → aAb | B; B →  Bb |  λ ;

c.

S → aaaA | aaA | λ ; A → aAb | B; ​B →  Bb |  λ ;

d.

S → aaaA | aaA | aA |  λ ; A → aAb | B; ​B →  Bb |  λ ;

Discuss
Answer: (c).S → aaaA | aaA | λ ; A → aAb | B; ​B →  Bb |  λ ;
Page 1 of 16

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