## Regular-Language

 Question 1

If L is a regular language over Σ = {a,b}, which one of the following languages is NOT regular?

 A Suffix (L) = {y ∈ Σ* such that xy ∈ L} B {wwR │w ∈ L} C Prefix (L) = {x ∈ Σ*│∃y ∈ Σ* such that xy ∈ L} D L ∙ LR = {xy │ x ∈ L, yR ∈ L}
Theory-of-Computation       Regular-Language       GATE 2019       Video-Explanation
Question 1 Explanation:
wwR cannot be recognized without using stack, so it cannot be regular.
 Question 2

For Σ = {a,b}, let us consider the regular language L = {x|x = a2+3k or x = b10+12k, k ≥ 0}. Which one of the following can be a pumping length (the constant guaranteed by the pumping lemma) for L?

 A 3 B 9 C 5 D 24
Theory-of-Computation       Regular-Language       GATE 2019       Video-Explanation
Question 2 Explanation:
Pumping Lemma for Regular Languages:
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
(1) |uv| ≤ n
(2) |v| ≥ 1
(3) for all i ≥ 0: uviw ∈ L
We have to find "n" which satisfies for all the strings in L.
Considering strings derived by b10+12k.
The minimum string in L = "bbbbbbbbbb" but this string b10 cannot be broken in uvw.
So, pumping length 3, 9 and 5 cannot be the correct answer.
So, the minimum pumping length, such that any string in L can be divided into three parts "uvw" must be greater than 10.
 Question 3

Which of the following languages is generated by the given grammar?

S→ aS|bS|ε
 A {anbm |n,m ≥ 0} B {w ∈ {a,b}* | w has equal number of a’s and b’s} C {an |n ≥ 0}∪{bn |n ≥ 0}∪{an b(sup>n|n ≥ 0} D {a,b}*
Theory-of-Computation       Regular-Language       GATE 2016 [Set-1]       Video-Explanation
Question 3 Explanation:
From the given grammar we can draw the DFA, Question 4
Language L1 is deﬁned by the grammar: S1 → aS1b|ε
Language L2 is deﬁned by the grammar: S2 → abS2
Consider the following statements:
P: L1is regular
Q: L2is regular
Which one of the following is TRUE?
 A Both P and Q are true B P is true and Q is false C P is false and Q is true D Both P and Q are false
Theory-of-Computation       Regular-Language       GATE 2016 [Set-2]       Video-Explanation
Question 4 Explanation:
The language L1 generated by the grammar contains equal number of a’s and b’s, but b’s comes after a’s.
So, in order to compare equality between a’s and b’s memory (stack) is required.
Hence, L1 is not regular.
Moreover, L1 = {an bn | n ≥ 0} which is DCFL.
The language L2 generated by grammar contains repetition of “ab” i.e. L2 = (ab)* which is clearly a regular language.
 Question 5

Which of the following language is/are regular ?

L1: {wxwR ⎪ w, x ∈ {a, b}* and ⎪w⎪, ⎪x⎪ >0} wR is the reverse of string w
L2: {anbm ⎪m ≠ n and m, n≥0}
L3: {apbqcr ⎪ p, q, r ≥ 0}
 A L1 and L3 only B L2 only C L2 and L3 only D L3 only
Theory-of-Computation       Regular-Language       GATE 2015 [Set-2]
Question 5 Explanation:
L1: All strings of length 3 or more with same start and end symbol, as everything in middle is consumed by x as per the definition.
L2: In this number of a's is dependent on number of b's. So PDA is needed.
L3: Any number of a's followed by any number of b's followed by any number of c's. Hence Regular.
 Question 6

Let P be a regular language and Q be a context-free language such that Q ⊆ P. (For example, let P be the language represented by the regular expression p*q* and Q be [pnqn | n ∈ N]). Then which of the following is ALWAYS regular?

 A P ∩ Q B P – Q C Σ* – P D Σ* – Q
Theory-of-Computation       Regular-Language       GATE 2011
Question 6 Explanation:
Exp: Σ* - P is the complement of P so it is always regular,
since regular languages are closed under complementation.
 Question 7

Which of the following is TRUE?

 A Every subset of a regular set is regular. B Every finite subset of a non-regular set is regular. C The union of two non-regular sets is not regular. D Infinite union of finite sets is regular.
Theory-of-Computation       Regular-Language       GATE 2007
Question 7 Explanation:
If a set is finite then it must be regular , as every language which contains finite elements is regular. Hence, every finite subset of a non-regular set is regular.
Every subset of regular set is regular, is false. For example L = {an bn | n ≥ 0} is subset of ∑* and L is CFL, whereas ∑* is regular. Hence, every subset of regular set need not be regular.
The union of two non-regular sets is not regular, is also a false statement.
For example, consider two CFL’s.
L = {an bn | n ≥ 0} and its complement Lc = {am bn | m ≠ n } U b*a*.
If we take UNION of L and Lc , we will get ∑*, which is regular. Hence the UNION of two non-regular set may or may not be regular.
The statement, Infinite union of finite sets is regular is also a false statement.
 Question 8

Which of the following languages is regular?

 A {wwR|w ∈ {0,1}+} B {wwRx|x, w ∈ {0,1}+} C {wxwR|x, w ∈ {0,1}+} D {xwwR|x, w ∈ {0,1}+}
Theory-of-Computation       Regular-Language       GATE 2007
Question 8 Explanation:
The regular expression corresponding to option C is:
0 (0+1)+ 0 + 1 (0+1)+ 1
Any string which begins and ends with same symbol, can be written in form of “wxwr
For example consider a string: 10010111, in this string “w=1” , “x= 001011” and wr = 1.
Hence any string which begins and ends with either “0” or with “1” can be written in form of “wxwr” and L = {wxwr | x,w ϵ {0,1}+} is a regular language.
 Question 9

If s is a string over (0 + 1)* then let n0(s) denote the number of 0’s in s and n1(s) the number of 1’s in s. Which one of the following languages is not regular?

 A L = {s ∈ (0+1)* | n0(s) is a 3-digit prime} B L = {s ∈ (0+1)* | for every prefix s' of s,|n0(s') - n1(s')| ≤ 2} C L = {s ∈ (0+1)* |n0(s) - n1(s)| ≤ 4} D L = {s ∈ (0+1)* | n0(s) mod 7 = n1(s) mod 5 = 0}
Theory-of-Computation       Regular-Language       GATE 2006
Question 9 Explanation:
Since 3-digit prime numbers are finite so language in option A is finite, hence it is regular.
Option B: The DFA contains 6 states
State 1: n0(s') - n1(s') = 0
State 2: n0(s') - n1(s') = 1
State 3: n0(s') - n1(s') = 2
State 4: n0(s') - n1(s') = -1
State 5: n0(s') - n1(s') = -2
State 6: Dead state (trap state)
Hence it is regular.
Option D: Product automata concept is used to construct the DFA.
mod 7 has remainders {0,1,2,3,4,5,6} and mod 5 remainders {0,1,2,3,4}
So product automata will have 35 states.
But option C has infinite comparisons between number of 0’s and 1’s.
For ex: n0(s) = 5 and n1(s) = 1 then n0(s) - n1(s) = 4 and if n0(s) = 15 and n1(s) = 11 then n0(s) - n1(s) = 4.
Hence this is CFL.
 Question 10

Consider the following languages:

L1 = {ww|w ∈ {a,b}*}
L2 = {wwR|w ∈ {a,b}*, wR is the reverse of w}
L3 = {02i|i is an integer}
L3 = {0i2|i is an integer}

Which of the languages are regular?

 A Only L1 and L2 B Only L2, L3 and L4 C Only L3 and L4 D Only L3
Theory-of-Computation       Regular-Language       GATE 2001
Question 10 Explanation:
L1 = {ww|w∈{a,b}*}
⇒ This is not regular language. We can't be able to identify where the 'w' will ends and where the next 'w' starts.
L2 = {wwR|w∈{a,b}*, wR is the reverse of w}
⇒ This also not a regular language. We can't identify where 'w' ends.
L4 = {0i2|i is an integer}
= {0i*0i|i is an integer}
⇒ This is also not a regular language. We can't identify where 0i ends.
L3 = {02i|i is an integer}
⇒ This is regular. We can easily find whether a string is even or not.
 Question 11

(a) Given that A is regular and A∪B is regular, does it follow that B is necessarily regular? Justify your answer.
(b) Given two finite automata M1, M2, outline an algorithm to decide if L(M1)⊆L(M2). (note: strict subset)

 A Theory Explanation.
Theory-of-Computation       Regular-Language       GATE 1999
 Question 12

Which of the following statements is false?

 A Every finite subset of a non-regular set is regular B Every subset of a regular set is regular C Every finite subset of a regular set is regular D The intersection of two regular sets is regular
Theory-of-Computation       Regular-Language       GATE 1998
Question 12 Explanation:
Let regular language L = a*b* and subset of L is anbn, n ≥ 0, which is not regular. Hence option (B) is false.
 Question 13

Consider the following statements.

I. If L1 ∪ L2 is regular, then both L1 and L2 must be regular.
II. The class of regular languages is closed under infinite union.

Which of the above statements is/are TRUE?

 A Both I and II B II only C Neither I nor II D I only
Theory-of-Computation       Regular-Language       GATE 2020       Video-Explanation
Question 13 Explanation:
Statement I is wrong.
Assume L1 = {an bn | n>0} and L2 = complement of L1
L1 and L2 both are DCFL but not regular, but L1 U L2 = (a+b)* which is regular.
Hence even though L1 U L2 is regular, L1 and L2 need not be always regular.
Statement II is wrong.
Assume the following finite (hence regular) languages.
L1 = {ab}
L2 = {aabb}
L3 = {aaabbb}
.
.
.
L100 = {a100 b100}
.
.
.
If we take infinite union of all above languages i.e,
{L1 U L2 U ……….L100 U ……}
then we will get a new language L = {an bn | n>0}, which is not regular.
Hence regular languages are not closed under infinite UNION.
 Question 14

Choose the correct alternatives (More than one may be correct). Let R1 and R2 be regular sets defined over the alphabet Σ Then:

 A R1 ∩ R2 is not regular. B R1 ∪ R2 is regular. C Σ* − R1 is regular. D R1* is not regular. E Both (B) and (C).
Theory-of-Computation       Regular-Language       GATE 1990       Video-Explanation
Question 14 Explanation:
Regular languages are closed under,
1) Intersection
2) Union
3) Complement
4) Kleen-closure
Σ* - R1 is the complement of R1.
Hence, (B) and (C) are true.
 Question 15

Regularity is preserved under the operation of string reversal.

 A True B False
Theory-of-Computation       Regular-Language       GATE 1987       Video-Explanation
Question 15 Explanation:
Regular language is closed under reversal.
There are 15 questions to complete.

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