## Theory-of-Computation

 Question 1
Suppose that L1is a regular language and L2is a context free language. Which one of the following languages is NOT necessarily context free?
 A L1. L2 B L1 ∪ L2 C L1 ∩ L2 D L1 − L2
Question 1 Explanation:

L1. L2 =>  regular . CFL  == CFL. CFL  (as every regular is CFL so we can assume regular as CFL)

Since CFL is closed under concatenation so

CFL. CFL= CFL

Hence

Regular . CFL = CFL is true

L1 ∪ L2 => Regular ∪ CFL = CFL

Regular ∪ CFL = CFL ∪ CFL  (as every regular is CFL so we can assume regular as CFL)

Since CFL is closed under union

Hence Regular ∪ CFL = CFL   is true

L1 ∩ L2 => Regular ∩ CFL = CFL

Regular languages are closed under intersection with any language

I.e,

Regular ∩ L = L  (where L is any language such as CFL, CSL etc)

Hence Regular ∩ CFL = CFL  is true

Please note this is a special property of regular languages so we will not upgrade regular into CFL (as we did in S1 and S2). We can directly use these closure properties.

L1 − L2 => Regular − CFL = CFL

=> Regular − CFL = regular ∩ CFL (complement)

Since CFL is not closed under complement so complement of CFL may or may not be CFL

Hence Regular − CFL need not be CFL

For ex:

R= (a+b+c)*  and L= {am bn ck | m ≠ n or n ≠ k} which is CFL.

The complement of L = {an bn cn | n>0} which is CSL but not CFL.

So

R L = (a+b+c)* {am bn ck | m ≠ n or n ≠ k}

=> (a+b+c)* ∩  L (complement)

=> (a+b+c)* ∩  {an bn cn | n>0}

=> {an bn cn | n>0}

Which is CSL. Hence Regular − CFL need not be CFL.

 Question 2
Let <M> denote an encoding of an automaton M. Suppose that Σ = {0,1}. Which of the following languages is/are NOT recursive?
 A L = { | M is a DFA such that L(M) = Σ*} B L = { | M is a DFA such that L(M) = ∅} C L = { | M is a PDA such that L(M) = Σ*} D L = { | M is a PDA such that L(M) = ∅}
Question 2 Explanation:
 Question 3
Consider the following language.

 A 2
Question 3 Explanation:

Option A accepts string “01111” which does not end with 011 hence wrong.

Option C accepts string “0111” which does not end with 011 hence wrong.

Option D accepts string “0110” which does not end with 011 hence wrong.

Option B is correct.

The NFA for language in which all strings ends with “011”

">
 Question 4
For a Turing machine M, <M> denotes an encoding of M. Consider the following two languages.
L1=〈M〉|M takes more than 2021 steps on all inputs
L2=〈M〉|M takes more than 2021 steps on some input
Which one of the following options is correct?
 A Both L1and L2 are undecidable. B L1is undecidable and L2is decidable. C L1is decidable and L2is undecidable. D Both L1and L2 are decidable.
Question 4 Explanation:

L1 is decidable.

We can take all strings of length zero to length 2021.

If TM takes more than 2021 steps on above inputs then definitely it will take more than 2021 steps on all input greater than length 2021.

If TM takes less than 2021 steps:

In such a case suppose TM  takes less than 2021 steps (let's say  2020 steps ) for string length 2021, 2022, 2023, then definitely TM  will take 2020 steps for all input greater than 2023. Hence in both cases it is decidable.

Similarly L2 is also decidable. If we can decide for all inputs then we can decide for some inputs also.

 Question 5
 A 50
Question 5 Explanation:

 Question 6

(a) Given a set
S = {x| there is an x-block of 5's in the decimal expansion of π}
(Note: x-block is a maximal block of x successive 5’s)
Which of the following statements is true with respect to S? No reasons need to be given for the answer.

(i) S is regular
(ii) S is recursively enumerable
(iii) S is not recursively enumerable
(iv) S is recursive

(b) Given that a language L1 is regular and that the language L1 ∪ L2 is regular, is the language L2 always regular? Prove your answer.

 A Theory Explanation.
 Question 7

A grammar G is in Chomsky-Normal Form (CNF) if all its productions are of the form A → BC or A → a, where A, B and C, are non-terminals and a is a terminal. Suppose G is a CFG in CNF and w is a string in L(G) of length, then how long is a derivation of w in G?

 A Theory Explanation.
 Question 8

Every subset of a countable set is countable.
State whether the above statement is true or false with reason.

 A True B False
Question 8 Explanation:
Because if a set itself is countable then the subset of set is definitely countable.
 Question 9

The regular expression for the language recognized by the finite state automaton of figure is __________

 A L = 0*1*
Question 9 Explanation:
L = 0*1*
L contains all binary strings where a 1 is not followed by a 0.
 Question 10

Which of the following features cannot be captured by context-free grammars?

 A Syntax of if-then-else statements B Syntax of recursive procedures C Whether a variable has been declared before its use D Variable names of arbitrary length
Question 10 Explanation:
Context free grammars are used to represent syntactic rules while designing a compiler.
Syntactic rules not checking the meaningful things such as if a variable is declared before it use (or) not.
Like this, things are handled by semantic analysis phase.
 Question 11

Which of the following conversions is not possible (algorithmically)?

 A Regular grammar to context free grammar B Non-deterministic FSA to deterministic FSA C Non-deterministic PDA to deterministic PDA D Non-deterministic Turing machine to deterministic Turing machine
Question 11 Explanation:
NPDA to DPDA conversion is not possible. They have different powers.
 Question 12

Let L be a language over ∑ i.e., *L ≤ ∑ . Suppose L satisfies the two conditions given below
(i) L is in NP and
(ii) For every n, there is exactly one string of length n that belongs to L.
Let Lc be the complement of L over ∑*. Show that Lc is also in NP.

 A Theory Explanation.
 Question 13

Let Σ = {0,1}, L = Σ* and R = {0n1n such that n >0} then the languages L ∪ R and R are respectively

 A regular, regular B not regular, regular C regular, not regular D not regular, no regular
Question 13 Explanation:
L∪R is nothing but L itself. Because R is subset of L and hence regular. R is deterministic context free but not regular as we require a stack to keep the count of 0's to make that of 1's.
 Question 14

A finite state machine with the following state table has a single input x and a single out z.

If the initial state is unknown, then the shortest input sequence to reach the final state C is:

 A 01 B 10 C 101 D 110
Question 14 Explanation:

If A is the start state, shortest sequence is 10 'or' 00 to reach C.
If B is the start state, shortest sequence is 0 to reach C.
If C is the start state, shortest sequence is 10 or 00 to reach C.
If D is the start state, shortest sequence is 0 to reach C.
∴ (B) is correct.
 Question 15

Which of the following definitions below generates the same language as L, where L = {xnyn such that n >= 1}?

``` I. E → xEy|xy
II. xy|(x+xyy+)
III. x+y+ ```
 A I only B I and II C II and III D II only
Question 15 Explanation:
(I) is the correct definition and the other two is wrong because the other two can have any no. of x and y. There is no such restriction over the number of both being equal.
 Question 16

In some programming languages, an identifier is permitted to be a letter following by any number of letters or digits. If L and D denote the sets of letters and digits respectively, which of the following expressions defines an identifier?

 A (L ∪ D)+ B L(L ∪ D)* C (L⋅D)* D L⋅(L⋅D)*
Question 16 Explanation:
Which is to be letter followed by any number of letters (or) digits
L(L ∪ D)*
 Question 17

Consider a grammar with the following productions

```S → a∝b|b∝c| aB
S → ∝S|b
S → ∝b b|ab
S ∝ → bd b|b ```

The above grammar is:

 A Context free B Regular C Context sensitive D LR(k)
Question 17 Explanation:
S ∝→ [violates context free]
Because LHS must be single non-terminal symbol.
S ∝→ b [violates CSG]
→ Length of RHS production must be atleast same as that of LHS.
Extra information is added to the state by redefining iteams to include a terminal symbol as second component in this type of grammar.
Ex: [A → αβa]
A → αβ is a production, a is a terminal (or) right end marker \$, such an object is called LR(k).
So, answer is (D) i.e., LR(k).
 Question 18

Consider the following language.

`   L = {x ∈ {a,b}* | number of a’s in x is divisible by 2 but not divisible by 3} `

The minimum number of states in a DFA that accepts L is ______.

 A 6
Question 18 Explanation:
DFA 1: No. of a’s divisible by 2.

DFA 1: No. of a’s not divisible by 3

Using product automata:
 Question 19

Consider the following languages.

L1 = {wxyx | w,x,y ∈ (0 + 1)+}
L2 = {xy | x,y ∈ (a + b)*, |x| = |y|, x ≠ y}

Which one of the following is TRUE?

 A L1 is context-free but not regular and L2 is context-free. B Neither L1 nor L2 is context-free. C L1 is regular and L2 is context-free. D L1 is context-free but L2 is not context-free.
Question 19 Explanation:
L1 is regular. y can be expanded and w can also expanded. So x can be either "a" or "b".
So it is equivalent to
(a+b)+ a (a+b)+ a + (a+b)+ b (a+b)+ b
L2 is CFL since it is equivalent to complement of L=ww.
Complement of L=ww is CFL.
 Question 20

Which of the following languages are undecidable? Note that indicates encoding of the Turing machine M.

L1 = | L(M) = Φ}
L2 = {| M on input w reaches state q in exactly 100 steps}
L3 = {| L(M) is not recursive}
L4 = {| L(M) contains at least 21 members}
 A L2 and L3 only B L1 and L3 only C L2, L3 and L4 only D L1, L3 and L4 only
Question 20 Explanation:
L1 is undecidable as emptiness problem of Turing machine is undecidable. L3 is undecidable since there is no algorithm to check whether a given TM accept recursive language. L4 is undecidable as it is similar to membership problem.
Only L3 is decidable. We can check whether a given TM reach state q in exactly 100 steps or not. Here we have to check only upto 100 steps, so here is not any case of going to infinite loop.
 Question 21

Which one of the following regular expressions represents the set of all binary strings with an odd number of 1’s?

 A 10*(0*10*10*)* B ((0 + 1)*1(0 + 1)*1)*10* C (0*10*10*)*10* D (0*10*10*)*0*1
Question 21 Explanation:
The regular expression 10*(0*10*10*)* always generate string begin with 1 and thus does not generate string “01110” hence wrong option.
The regular expression ((0+1)*1(0+1)*1)*10* generate string “11110” which is not having odd number of 1’s , hence wrong option.
The regular expression (0*10*10*)10* is not a generating string “01”. Hence this is also wrong . It seems none of them is correct.
NOTE: Option 3 is most appropriate option as it generates the max number of strings with odd 1’s.
But option 3 is not generating odd strings. So, still it is not completely correct.
The regular expression (0*10*10*)*0*1 always generates all string ends with “1” and thus does not generate string “01110” hence wrong option.
 Question 22

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
Question 22 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 23

Consider the language L = {an| n≥0} ∪ {anbn| n≥0} and the following statements.

I. L is deterministic context-free.
II. L is context-free but not deterministic context-free.
III. L is not LL(k) for any k.

Which of the above statements is/are TRUE?

 A II only B III only C I only D I and III only
Question 23 Explanation:
L is DCFL.
We can make DPDA for this.

L is not LL(k) for any “k” look aheads. The reason is the language is a union of two languages which have common prefixes. For example strings {aa, aabb, aaa, aaabbb,….} present in language. Hence the LL(k) parser cannot parse it by using any lookahead “k” symbols.
 Question 24

Consider the set of strings on {0,1} in which, every substring of 3 symbols has at most two zeros. For example, 001110 and 011001 are in the language, but 100010 is not. All strings of length less than 3 are also in the language. A partially completed DFA that accepts this language is shown below.

The missing arcs in the DFA are

 A B C D
Question 24 Explanation:
All states are final states except “q” which is trap state. The strings in language are such that every substring of 3 symbol has at most two zeros. It means that we cannot have 3 consecutive zeros anywhere in string. In the given DFA total four transition is missing, so we have to create the missing transition keeping the criteria in mind that “three consecutive zeros” will lead to trap state “q” as after 3 consecutive zeros whatever comes after that in the string, the string is going to be rejected by DFA.
From the state “00” it is clear that if another “0” comes then the string is going to be rejected, so from state “00” the transition with input “0” will lead to state “q”. So option A and B are eliminated.
Now option C has the self loop of “0” on state “10” which will accept any number of zeros (including greater than three zeros), hence the C option is also wrong. We left with only option D which is correct option.
 Question 25

Given the language L ={ab,aa,baa}, which of the following strings are in L *?

1) abaabaaabaa
2) aaaabaaaa
3) baaaaabaaaab
4) baaaaabaa
 A 1, 2 and 3 B 2, 3 and 4 C 1, 2 and 4 D 1, 3 and 4
Question 25 Explanation:
L* will contain all those strings which can be obtained by any combination (and repetition) of the strings in language i,e, from L= {ab, aa, baa}
String 1: abaabaaabaa : ab aa baa ab aa
String 2: aaaabaaaa : aa aa baa aa
String 3: baaaaabaaaab: baa aa ab aa aa b, because of the last “b” the string cannot belong to L*.
String 4: baaaaabaa : baa aa ab aa
 Question 26

Which of the following problems are decidable?

1) Does a given program ever produce an output?

2) If L is a context-free language, then, is also context-free?

3) If L is a regular language, then, is also regular?

4) If L is a recursive language, then, is also recursive?

 A 1, 2, 3, 4 B 1, 2 C 2, 3, 4 D 3, 4
Question 26 Explanation:
The statement “Does a given program ever produce an output?” is same as the statement “Does a Turing Machine will halt for any arbitrary string?”, which is nothing but the “halting problem of Turing Machine”, hence statement 1 is undecidable.
Context free languages are not closed under complement operation, so compliment of CFL may or may not be CFL. Hence statement 2 is also undecidable.
Complement of Regular languages is also regular. Since a DFA that accepts the complement of L, i.e. ∑* – L, can be obtained by swapping its final states with its non-final states and vice-versa. Hence it is decidable and if L is a regular language, then, L must also be regular.
Recursive languages are closed under complement, so if L is a recursive language then L must also be recursive, hence it is decidable.
 Question 27

What is the complement of the language accepted by the NFA shown below?

Assume Σ={a} and ε is the empty string.

 A ∅ B {ε} C a* D {a ,ε}
Question 27 Explanation:
The Σ= {a} and the given NFA accepts the strings {a, aa, aaa, aaaa, ……….} i.e. the language accepted by the NFA can be represented by the regular expression: {a+}
Hence the complement of language is: {a* − a+} = {ϵ}
 Question 28

Let Q = ({q1,q2}, {a,b}, {a,b,Z}, δ, Z, ϕ) be a pushdown automaton accepting by empty stack for the language which is the set of all non empty even palindromes over the set {a,b}. Below is an incomplete specification of the transitions δ. Complete the specification. The top of the stack is assumed to be at the right end of the string representing stack contents.

```(1) δ(q1,a,Z) = {(q1,Za)}
(2) δ(q1,b,Z) = {(q1,Zb)}
(3) δ(q1,a,a) = {(.....,.....)}
(4) δ(q1,b,b) = {(.....,.....)}
(5) δ(q2,a,a) = {(q2,ϵ)}
(6) δ(q2,b,b) = {(q2,ϵ)}
(7) δ(q2,ϵ,Z) = {(q2,ϵ)} ```
 A Theory Explanation.
 Question 29

Let G be a context-free grammar where G = ({S, A, B, C},{a,b,d},P,S) with the productions in P given below.

```   S → ABAC
A → aA ∣ ε
B → bB ∣ ε
C → d  ```

(ε denotes null string). Transform the grammar G to an equivalent context-free grammar G' that has no ε productions and no unit productions. (A unit production is of the form x → y, and x and y are non terminals.)

 A Theory Explanation.
 Question 30

Consider the given figure of state table for a sequential machine. The number of states in the minimized machine will be

 A 4 B 3 C 2 D 1
Question 30 Explanation:
3 states are required in the minimized machine states B and C can be combined as follows:
 Question 31

If L1 and L2 are context free languages and R a regular set, one of the languages below is not necessarily a context free language. Which one?

 A L1, L2 B L1 ∩ L2 C L1 ∩ R D L1 ∪ L2
Question 31 Explanation:
Context free languages are not closed under intersection.
 Question 32

Define for a context free language L ⊆ {0,1}*, init(L) = {u ∣ uv ∈ L for some v in {0,1}∗} (in other words, init(L) is the set of prefixes of L)
Let L = {w ∣ w is nonempty and has an equal number of 0’s and 1’s}
Then init(L) is

 A the set of all binary strings with unequal number of 0’s and 1’s B the set of all binary strings including the null string C the set of all binary strings with exactly one more 0’s than the number of 1’s or one more 1 than the number of 0’s D None of the above
Question 32 Explanation:
(B) is the answer. Because for any binary string of 0's and 1's we can append another string to make it contain equal no. of 0's and 1's, i.e., any string over {0,1} is a prefix of a string in L.
 Question 33

The grammar whose productions are

```  → if id then
→ if id then   else
→ id := id ```

is ambiguous because

 A the sentence if a then if b then c:=d B the left most and right most derivations of the sentence if a then if b then c:=d give rise top different parse trees C the sentence if a then if b then c:=d else c:=f has more than two parse trees D the sentence if a then if then c:=d else c:=f has two parse trees
Question 33 Explanation:
We have to generate
"if a then if b then c:=d else c:=f".
Parse tree 1:

Parse tree 2:
 Question 34

Which of the following statements is false?

 A The Halting problem of Turing machines is undecidable. B Determining whether a context-free grammar is ambiguous is undecidbale. C Given two arbitrary context-free grammars G1 and G2 it is undecidable whether L(G1) = L(G2). D Given two regular grammars G1 and G2 it is undecidable whether L(G1) = L(G2).
Question 34 Explanation:
Equivalence of regular languages is decidable under
1) Membership
2) Emtiness
3) Finiteness
4) Equivalence
5) Ambiguity
6) Regularity
7) Everything
8) Disjointness
All are decidable for Regular languages.
→ First 3 for CFL.
→ Only 1st for CSL and REC.
→ None for RE.
 Question 35

Let L ⊆ Σ* where Σ = {a, b}. Which of the following is true?

 A L = {x|x has an equal number of a's and b's } is regular B L = {anbn|n≥1} is regular C L = {x|x has more a's and b's} is regular D L = {ambn|m ≥ 1, n ≥ 1} is regular
Question 35 Explanation:
L = {ambn|m ≥ 1, n ≥ 1}
Here, m and n are independent.
So 'L' Is Regular.
 Question 36

Which two of the following four regular expressions are equivalent? (ε is the empty string).

(i) (00)*(ε+0)
(ii) (00)*
(iii) 0*
(iv) 0(00)*
 A (i) and (ii) B (ii) and (iii) C (i) and (iii) D (iii) and (iv)
Question 36 Explanation:
(00)*(ε+0),0*
In these two, we have any no. of 0's as well as null.
 Question 37

Which of the following languages over {a,b,c} is accepted by a deterministic pushdown automata?

` Note: wR  is the string obtained by reversing 'w'. `
 A {w⊂wR|w ∈ {a,b}*} B {wwR|w ∈ {a,b,c}*} C {anbncn|n ≥ 0} D {w|w is a palindrome over {a,b,c}}
Question 37 Explanation:
(A) w⊂wR, can be realized using DPDA because we know the center of the string that is c here.
(B) wwR, is realized by NPDA because we can't find deterministically the center of palindrome string.
(C) {anbncn | n ≥ 0} is CSL.
(D) {w | w is palindrome over {a,b,c}},
is realized by NPDA because we can't find deterministically the center of palindrome string.
 Question 38

Which one of the following regular expressions over {0,1} denotes the set of all strings not containing 100 as a substring?

 A 0*(1+0)* B 0*1010* C 0*1*01 D 0(10+1)*
Question 38 Explanation:
(A) generates 100.
(B) generates 100 as substring.
(C) doesn't generate 1.
 Question 39

Which one of the following is not decidable?

 A Given a Turing machine M, a stings s and an integer k, M accepts s within k steps B Equivalence of two given Turing machines C Language accepted by a given finite state machine is not empty D Language generated by a context free grammar is non empty
Question 39 Explanation:
(A) It is not halting problem. In halting problem number of steps can go upto infinity and that is the only reason why it becomes undecidable.
In (A) the number of steps is restricted to a finite number 'k' and simulating a TM for 'k' steps is trivially decidable because we just go to step k and output the answer.
(B) Equivalence of two TM's is undecidable.
For options (C) and (D) we do have well defined algorithms making them decidable.
 Question 40

Given Σ = {a,b}, which one of the following sets is not countable?

 A Set of all strings over Σ B Set of all languages over Σ C Set of all regular languages over Σ D Set of all languages over Σ accepted by Turing machines
Question 40 Explanation:
Uncountable: Set of all languages over Σ is uncountable.
 Question 41

(a) An identifier in a programming language consists of upto six letters and digits of which the first character must be a letter. Derive a regular expression for the identifier.

(b) Build an LL(1) parsing table for the language defined by the LL(1) grammar with productions

```Program → begin d semi X end
X → d semi X | sY
Y → semi s Y | ε ```
 A Theory Explanation.
 Question 42

(a) Let G1 = (N, T, P, S1) be a CFG where,
N = {S1, A, B}, T = {a,b} and
P is given by

```         S1 → aS1b             S1 → aBb
S1 → aAb              B → Bb
A → aA                B → b
A → a ```

What is L(G1)?

(b) Use the grammar in part(a) to give a CFG
for L2 = {ai bj ak bl | i, j, k, l ≥ 1, i=j or k=l} by adding not more than 5 production rule.

(c) Is L2 inherently ambiguous?

 A Theory Explanation.
 Question 43

Let M = ({q0, q1}, {0, 1}, {z0, x}, δ, q0, z0, ∅) be a pushdown automaton where δ is given by

δ(q0, 1, z0) = {(q0, xz0)}
δ(q0, ε, z0) = {(q0, ε)}
δ(q0, 1, X) = {(q0, XX)}
δ(q1, 1, X) = {(q1, ε)}
δ(q0, 0, X) = {(q1, X)}
δ(q0, 0, z0) = {(q0, z0)}

(a) What is the language accepted by this PDA by empty stack?

(b) Describe informally the working of the PDA.

 A Theory Explanation.
 Question 44

Design a deterministic finite state automaton (using minimum number of states) that recognizes the following language:
L = {w ∈ {0,1}* | w interpreted as a binary number (ignoring the leading zeros) is divisible by 5}

 A Theory Explanation.
 Question 45

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
Question 45 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 46

Let L be the set of all binary strings whose last two symbols are the same. The number of states in the minimum state deterministic finite 0 state automaton accepting L is

 A 2 B 5 C 8 D 3
Question 46 Explanation:
NFA:

Equivalent DFA:

Hence, 5 states.
 Question 47

How many sub strings of different lengths (non-zero) can be found formed from a character string of length n?

 A n B n2 C 2n D
Question 47 Explanation:
Let us consider an example S = {APB}
Possible sub-strings are = {A, P, B, AP, PB, BA, APB}
Go through the options.
Option D:
n(n+1)/2 = 3(3+1)/2 = 6
 Question 48

Regarding  the power of recognition of languages, which of the following statements is false?

 A The non-deterministic finite-state automata are equivalent to deterministic finite-state automata. B Non-deterministic Push-down automata are equivalent to deterministic Push- down automata. C Non-deterministic Turing machines are equivalent to deterministic Push-down automata. D Both B and C
Question 48 Explanation:
B: No conversion possible from NPDA to DPDA.
C: Power (TM) > NPDA > DPDA.
 Question 49

The string 1101 does not belong to the set represented by

 A 110*(0 + 1) B 1 ( 0 + 1)* 101 C (10)* (01)* (00 + 11)* D Both C and D
Question 49 Explanation:
Options A & B are generates string 1101.
C & D are not generate string 1101.
 Question 50

If the regular set A is represented by A = (01 + 1)* and the regular set ‘B’ is represented by B = ((01)*1*)*, which of the following is true?

 A A ⊂ B B B ⊂ A C A and B are incomparable D A = B
Question 50 Explanation:
Both A and B are equal, which generates strings over {0,1}, while 0 is followed by 1.
There are 50 questions to complete.

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