The correct answer is Option 1: b* a b * a b * a b *.

Key Points

• The regular expression b* a b * a b * a b * represents the language accepted by the given finite automata.
• This regular expression can be broken down as follows:
  • b* matches zero or more occurrences of the character 'b'.
  • a matches a single occurrence of the character 'a'.
  • Repeating the pattern b* a three more times ensures that the sequence 'a' appears exactly three times, each possibly preceded by zero or more 'b's.
• This pattern matches any string of 'b's followed by 'a', and this sequence is repeated three times.

Additional Information

• The finite automata can be visualized as having states that transition between each other upon reading specific inputs (characters).
• Understanding the structure of the automata and the transitions helps in deriving the correct regular expression that represents the language it accepts.
• Finite automata are used in various fields such as text processing, compiler design, and network protocols to recognize patterns and validate input sequences.
• Regular expressions are a powerful tool for describing patterns in strings and are widely used in search algorithms, text processing, and input validation.

The correct answer is It applies to all regular languages.

Key Points

• The Pumping Lemma is a property that applies to all regular languages, and it is used to prove whether a language is not regular.
  • If a language is regular, there exists some length (p) such that any string longer than p can be divided into three parts, xyz, satisfying certain conditions.
  • The conditions are: for the string xyz, the length of xy is at most p, y is not an empty string, and the string xy^iz (i ≥ 0) is still in the language.
  • The Pumping Lemma helps in identifying strings that cannot be pumped, thereby proving that the language is not regular.

Additional Information

• The Pumping Lemma does not apply to finite languages, as there is no need to pump strings in such cases.
• While the Pumping Lemma provides a necessary condition for regularity, it is not a sufficient condition; that is, some non-regular languages may satisfy the lemma.
• The lemma is a fundamental concept in the theory of computation and automata theory, helping in understanding the limitations of regular languages.
• It is an essential tool in formal language theory for distinguishing between regular and non-regular languages.

The correct answer is {\(\varepsilon\)} 

Concepts

\({{\rm{L}}^{\rm{*}}} = {{\rm{L}}^0} \cup {{\rm{L}}^1} \cup {{\rm{L}}^2} \cup {{\rm{L}}^3} \cup {{\rm{L}}^4} \ldots \ldots .{{\rm{L}}^{\rm{k}}} \cup {{\rm{L}}^{{\rm{k}} + 1}} \ldots \ldots \ldots \)

For any language \({{\rm{L}}^0} = \epsilon\)

Concatenation of \({\rm{\Phi }}\) with any other language is \({\rm{\Phi }}\). It works as 0 in multiplication.

And for \({{\rm{L}}^{\rm{k}}} = {\rm{L}}.{\rm{L}}.{\rm{L}}.{\rm{L}}.{\rm{L}}.{\rm{L}}.{\rm{L}} \ldots \ldots \ldots {\rm{\;L}}\) language is concatenated K- times.

Data:

 L1 = Φ, L2 = {a}

Calculation:

\({L_1}L_2^* \cup L_1^* = \;?\)

\({{\rm{L}}_1}\) concatenated with \(L_2^*\) and whole thing union with \(L_1^*\).

First \({\rm{\;L}}_1^{\rm{*}} = {\rm{\;\;}}{\phi ^0} \cup {\phi ^1} \cup {\phi ^2} \ldots \ldots {\rm{\;}}\)  \(= \epsilon \cup \phi \cup \phi \ldots \ldots = \left\{ \epsilon \right\}.\)

Similarly, \({\rm{L}}_2^{\rm{*}} = {\left\{ {\rm{a}} \right\}^0} \cup {\left\{ {\rm{a}} \right\}^1} \cup {\left\{ {\rm{a}} \right\}^2} \ldots \ldots {\rm{\;}} = \epsilon \cup \left\{ {\rm{a}} \right\} \cup \left\{ {{\rm{aa}}} \right\} \ldots \ldots .. = {{\rm{a}}^{\rm{*}}}\)

\({{\rm{L}}_1}{\rm{L}}_2^{\rm{*}} \cup {\rm{L}}_1^{\rm{*}} = {\rm{\Phi \;}}{{\rm{a}}^{\rm{*}}} \cup \left\{ \epsilon \right\} = \left\{ \epsilon \right\}\)

The correct answer is x+1

Key Points

• The pumping length needs to be greater than the length of the string itself (x) to guarantee the existence of two separate substrings (x and y).
• These two substrings should have a combined minimum length that exceeds p.
• This minimum combined length (x + 1) then becomes the pumping length (p).

Therefore, the correct option is C. (x + 1).

The correct answer is L1 ∩ L2 is Recursively Enumerable

Concept:

To determine the properties of the intersection of two context free languages L1 and L2 , we need to analyze each option provided.

Context-Free Languages

1. L1 and L2 are Context-Free Languages (CFLs): 
    Context-free languages are closed under union and concatenation but not closed under intersection.

Explanation:

1. L1 ∩ L2 is context-free:  
    False. The intersection of two context-free languages is not guaranteed to be context free. For example, if \(L_1 = \{ a^n b^n | n \geq 0 \} and L_2 = \{ b^n c^n | n \geq 0 \} , then L_1 \cap L_2 = \{ b^n | n \geq 0 \}\) , which is not context-free in this context.

2. L1 ∩ L2 is regular:  
    False. The intersection of two context-free languages is not necessarily regular. Regular languages are a subset of context free languages, but the intersection of two CFLs can produce a language that is not regular.

3. L1 ∩ L2 is recursively enumerable:  
    True. Context-free languages are a subset of recursively enumerable languages. Since both L1 and L2 are context-free, their intersection L1 ∩ L2 is recursively enumerable.

4. L1 ∩ L2 is context-sensitive:  
    True, but it is a broader statement. All context-free languages are context-sensitive, and since the intersection can produce a language that fits within the context-sensitive category, this statement is also valid, but it doesn't specifically apply only to CFLs.

Conclusion: The correct answer is: 3) L1 ∩ L2 is Recursively Enumerable

Grammar:

S → XY

X → aX | a

Y → aYb | ϵ

Explanation:

X generates at least one a. Generates a string of type {a, aa, aaa, aaaa……}

a^mbⁿ, if n = 0 and m = 0 ∴ string = ϵ which cannot be generated from the given grammar and hence option 2 is incorrect.

Y generates an equal number of a and b { ϵ, ab, aabb, ……} with a comes before b.

Since at least one a is generated by X and an equal number of a’s and b’s generated by Y ∴ m> n and hence option 1 is incorrect

The smallest length string generated by Y is ϵ. In a^mbⁿ, n ≥ 0 and hence option 4 is incorrect.

{ a, aab, aaabb,………….} which is equivalent to :

L = {a^mbⁿ | m > n, n ≥ 0}

S → aS | bS | ϵ 

This grammar results in (a + b)*

DFA for this grammar is:

Diagram

Now, consider the options one by one:

Option 1:

{aⁿb^m | n, m ≥ 0}

Here order is fixed, means first any number of a will come then any number of b. It is incorrect.

Option 2: 

{w ϵ {a, b} * | w has an equal number of a’s and b’s}

As given grammar is not generating the expression which generates only equal number of a and b. So,

not correct.

Option 3:

{aⁿ | n ≥ 0} {bⁿ | n ≥ 0} {aⁿbⁿ | n ≥ 0}

Here, also order is fixed. So, it is incorrect.

Option 4:

{a, b}*

This is equivalent to (a + b)*. So, it is correct.

Suppose grammar G1:

S₁ → aS₁b|ϵ

It generates language in which number of a’s are equal to number of b’s and all a’s are followed by b’s

L₁ = { aⁿ bⁿ | n ≥ 0}.

It is deterministic context free language. Extra memory is required for this. So, it can’t be accepted by finite state automata. L₁ is not a regular language.

Now, another grammar let G2: S₂ → abS₂|ϵ

This grammar also generates language in which number of a’s are equal to number of b’s. But it generates language L₂ = { (ab)ⁿ | n ≥ 0 }

Regular expression for this = (ab)*

• Non-deterministic push down automata (NPDA) determines the middle position of the string in the language and start popping until Z(end of stack elements) is meet to tell the acceptance of it .
• As all the string present in the language ww^R accepted by NPDA Hence it is a context free language (CFL).
• From the properties of regular language: Reverse, Suffix, Prefix, Concatenation of Regular language is Regular.
• Every regular language is context free language, but every context free language is not a regular language also ww^R it is not possible to give, DFA, NFA or ϵ - NFA accepting the language. Therefore, it is not Regular language.

Statement I: FALSE

If L₁ ∪ L₂ is regular, then neither L₁ nor L₂ needs necessarily be regular.

Example:

Assume L₁= {aⁿ bⁿ, n ≥ 0} over the alphabet {a, b} and L₂ be the complement of L₁.

Neither L₁ nor L₂ is regular (both are DCFL) but L₁ ∪ L₂= {aⁿ bⁿ} ∪ {aⁿ bⁿ}^c = (a + b)^* is regular.

Statement II: FALSE. The infinite Union of regular languages is not regular.

Example:

Given alphabet {a, b}.

L₁= {ε}

L₂= {ab}

L₃= {aabb}

L₄= {aaabbb}

:

:

L = L₁ ∪ L₂ ∪ L₃ ∪ L₄ …

Each of the above are regular but their infinite Union gives L₁= {aⁿ bⁿ, n ≥ 0} which is not regular but DCFL.

Note:

Regular language L = {x|x = a^2+3k or x = b^10+12k k ≥ 0}.

L = {a², a⁵, a⁸, a¹¹ … ∪ b¹⁰, b²², b³⁴…}

Pumping Lemma:

Let L be an infinite regular language. Then there exists some positive integer m such that any w ϵ L with |w|≥ m can be decomposed as

w = xyz

with |xy|≤ m such that w_i = xyⁱz

is also L for all i = 0, 1, 2, …

Therefore, minimum Pumping Length should be 11, because string with length 10 (i.e., w = b¹⁰) does not repeat anything, but string with length 11 (i.e., w = b¹¹) will repeat states.

Hence option 1, 2, and 3 are eliminated.

Consider all the options one by one:

1) (0 + 1) *0011 (0 + 1)* + (0 + 1) *1100 (0 + 1)*

It consider only those string which have either 0011 or 1100 as a substring. So, it is wrong.

2) (0 + 1) *(00(0 + 1)*11 + 11 (0 + 1)*00) (0 + 1)*

It contains set of all the binary strings that contain two consecutive 0’s and 1’s.

3) (0 + 1) *00 (0 + 1)* + (0 + 1) *11 (0 + 1)*

This set contains only those string which have either 00 or 11 as substring but not does not contain all the strings that contain two consecutive 0’s and 1’s.

4) 00 (0 + 1) *11 + 11 (0 + 1) *00

It represents those strings which start with 00 or 11 and end with 11 or 00 respectively. So, it is wrong.

Lexical analysis: Lexical analysis is the first phase of compiler also known as scanner. It converts the high-level input program into a sequence of tokens. Lexical analysis can be implemented by deterministic finite automata.

Parsing: Parser constructs a production tree. Parse tree is a hierarchical structure which represents the derivation of the programmer to yield input strings.

Register allocation: Register allocation reduces to graph coloring problem in which colors (registers) are assigned to the nodes such that two nodes connected by an edge do not receive the same color.

Expression evaluation: Expression evaluation is done using post order traversal.

The correct answer is option 4.

Key Points

The pumping lemma is often used to prove that a particular language is non-regular. Pumping lemma for regular language is generally used for proving a given grammar is not regular.

Hence the correct answer is a given grammar is not regular.

Additional Information

Pumping Lemma is to be applied to show that certain languages are not regular. It should never be used to show a language is regular.

If L is regular, it satisfies Pumping Lemma.

If L does not satisfy Pumping Lemma, it is non-regular.

Concept:

The regular expression should generate all possible binary strings with odd number of 1’s and doesn’t generate the other strings.

Option 1: Incorrect

Regular Expression: ((0 + 1)*1(0 + 1)*1)*10*

String: 11110

It can generate strings (11110) with even number of 1’s also

Option 2: Incorrect

Regular Expression: (0*10*10*)*0*1

It will generate string only ending with ‘1’

String: 1110

It cannot generate ‘1110’ substring.

Option 3: Incorrect

Regular Expression: 10*(0*10*10*)*

It will generate string only starting with ‘1’

String: 0111

It cannot generate ‘0111’ substring.

Option 4: Correct

Regular expression: 0*(10*10*)*10*

It will generate all possible binary strings with odd number of 1’s and doesn’t generate the other strings.

Note:

Option 4 has been changed, since none of the option is was correct in official GATE CS 2020 paper