Pumping lemma is based on
Principle of mathematical induction
Pigeon hole principle
Principle of countable and uncountable sets.
None of these
Difficulty Level: 1
Positive Marks: 1.00
Negative Marks: 0.33
Consider the following statements about regular expression

S1: φ.ε = ε

S2: φ+ε = φ

Select the correct option

Both S1 and S2 are correct
Both S1 and S2 are false
S1 is correct but S2 is false
S1 is false but S2 is correct
Difficulty Level: 1
Positive Marks: 1.00
Negative Marks: 0.33
Consider the following statements about regular expression

S1: (φ.ε)* = ε

S2: (φ+ε)* = φ*

Select the correct option

Both S1 and S2 are correct
Both S1 and S2 are false
S1 is correct but S2 is false
S1 is false but S2 is correct
Difficulty Level: 1
Positive Marks: 1.00
Negative Marks: 0.33
Consider the following Finite automata

Select the correct regular expression

0*1*0*1*0*
0*10*1*0*
0*10*10*
0*110*
Difficulty Level: 1
Positive Marks: 1.00
Negative Marks: 0.33
Consider the statements, which are given below.

S1: We can use pumping lemma to prove that language generated by regular expression R= a*b* is a regular language.

S2: We can use pumping lemma to prove that language L = { an bn | n>0} is a non-regular language.

Select the correct option

Both S1 and S2 are correct
Both S1 and S2 are false
S1 is correct but S2 is false
S1 is false but S2 is correct
Difficulty Level: 1
Positive Marks: 1.00
Negative Marks: 0.33
Consider the following grammar, whose productions are.

S→ 0P0 | 1P1 | ε

P→ P0 | P1 |ε

Select the correct option

The given grammar is a regular grammar
The given grammar is left linear grammar but not regular.
The given grammar generates a regular language
None of these
Difficulty Level: 1
Positive Marks: 2.00
Negative Marks: 0.66
Consider the following regular expressions

R1: (a+b)* a (a+b)*

R2: b*a (a+b)*

Select the correct option

L(R1) = L(R2)
L(R1) is proper subset of L(R2)
L(R2) is proper subset of L(R1)
L(R1) ∩ L(R2) =Ø
Difficulty Level: 1
Positive Marks: 2.00
Negative Marks: 0.66
Consider the following grammars

G1: S→ aSb| bSa | aSa | bSb |a | b | ε

G2: S→ aS | bS | ε

Select the correct option.

Complement of (L(G1) ∩ L(G2)) =φ
L(G1) ⸦ L(G2)
L(G2) ⸦ L(G1)
L(G1) ∩ L(G2) = φ
Difficulty Level: 1
Positive Marks: 2.00
Negative Marks: 0.66
Consider the following regular expressions.

R1: (a*b*b*)*

R2: (a*b*+b*)*

Select the correct option:

R1=R2 but both R1 and R2 ≠(a+b)*
R1≠R2 but R1=(a+b)*
R1=R2=(a+b)*
None of these
Difficulty Level: 1
Positive Marks: 2.00
Negative Marks: 0.66
Consider the following language L such that every string ends with “0” or contains “1”. Consider the below regular expressions

R1: (0+1)* 1 (0+1)* 0

R2 : (0+1)+

Select the correct option.

Both R1 and R2 are correct regular expressions for L.
Both R1 and R2 are not correct regular expressions for L.
R1 generates L while R2 doesn’t generate L
R2 generates L while R1 doesn’t generate L.
Difficulty Level: 1
Positive Marks: 2.00
Negative Marks: 0.66
Pumping Lemma for regular language is defined as: For any language L, there exists an integer n, such that for all x ∈ L with |x| ≥ n, there exists u,v, w ∈ Σ*, such that x = uvw, and the three conditions. Select the option which represents the three conditions correctly.
(1) |uv| ≤ n

(2) |v| ≥ 1

(3) for all i ≥ 0: uviw ∈ L

(1) |uv| < n

(2) |v| ≥ 1

(3) for all i ≥ 0: uviw ∈ L

(1) |uv| ≤ n

(2) |v| > 1

(3) for all i ≥ 0: uviw ∈ L

(1) |uv| ≤ n

(2) |v| ≥ 1

(3) for all i > 0: uviw ∈ L

Difficulty Level: 1
Positive Marks: 2.00
Negative Marks: 0.66
Consider the language L=a(a+b)*b. Select the correct option which generates language L.
S→aAb & A→aA | bA | a | b | ε
S→aAb & A→aA | bA | ε
S→aAb & A→aAa | bAb | aAb | bAa | a | b | ε
All of these
Difficulty Level: 1
Positive Marks: 2.00
Negative Marks: 0.66
Consider the following grammar

S ⟶ aSb|bSa|∈

Select the correct option

The given grammar is a regular grammar
The grammar generates all strings which start and end with different symbols.
The language generated by the grammar can be accepted by Finite automata
None of these
Difficulty Level: 1
Positive Marks: 2.00
Negative Marks: 0.66
Consider the following equality statements between regular expressions.

(i) (a+b)* = (a*+b)+

(ii) (a+b)*= (a+ + b+ )*

Select the correct option.

Both (i) and (ii) are correct.
Both (i) and (ii) are false.
(i) is correct but (ii) is false.
(i) is false but (ii) is true.
Difficulty Level: 1
Positive Marks: 2.00
Negative Marks: 0.66
Which one of the following regular expressions represents the set of all strings with an odd number of b’s.
a*b (a+ba*b)*
(a*ba*ba*)*ba*
ba*(a*ba*ba*)*
(a*ba*ba*)*a*b
Difficulty Level: 1
Positive Marks: 2.00
Negative Marks: 0.66