## Regular-Language

 Question 1

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       2016 set-01
Question 1 Explanation:
From the given grammar we can draw the DFA,
 Question 2

Language L1 is deﬁned by the grammar: S1 → aS1b|ε
Language L2 is deﬁned by the grammar: S2 → abS2

Consider the following statements:

P: L1 is regular
Q: L2 is 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
Question 2 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 3

 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 3 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 4
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 4 Explanation:
Exp: Σ* - P is the complement of P so it is always regular,
since regular languages are closed under complementation
 Question 5
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 5 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 6
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 6 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 7
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 7 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
State1: n0(s') - n1(s') = 0
State2: n0(s') - n1(s') = 1
State3: n0(s') - n1(s') = 2
State4: n0(s') - n1(s') = -1
State5: n0(s') - n1(s') = -2
State6: 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 8

 A Both I and IV B Only I C Only IV D Both II and III
Theory-of-Computation       Regular-Language       Gate 2006-IT
Question 8 Explanation:
Repeat(L) = {ww|w ∈ L} is non-regular language.
Half (L), Suffix (L) and Prefix (L) are regular languages.
 Question 9

 A (a + b)* B {ϵ, a, ab, bab} C (ab)* D {anbn | n ≥ 0}
Theory-of-Computation       Regular-Language       Gate 2006-IT
Question 9 Explanation:
A counter example which proves all the conclusions of the last question in one go should have the following properties:
1) L should be regular due to demand of question.
2) L should be an infinite set of strings.
3) L should have more than one alphabet in its grammar, otherwise repeat(L) would be regular.
∴ (a + b)* is the perfect example to support the conclusions of last questions.
 Question 10
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 10 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 11

 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 11 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 12

 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
Question 12 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 13
Regularity is preserved under the operation of string reversal.
 A True B False
Theory-of-Computation       Regular-Language       GATE-1987
Question 13 Explanation:
Regular language is closed under reversal.
 Question 14
All subsets of regular sets are regular.
 A True B False
Theory-of-Computation       Regular-Language       GATE-1987
Question 14 Explanation:
a*b* is regular but its subset anbn is not regular.
There are 14 questions to complete.