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: Option 1

Key Points

To detect “abb”, we need at least 4 states:

1. q0 — Start state
2. q1 — After seeing 'a'
3. q2 — After seeing 'ab'
4. q3 — After seeing 'abb' → Accepting

We also need a dead state to trap wrong transitions. So the **minimum** DFA needs at least 5 states.

Option 1: This DFA uses 5 states correctly and has transitions:

• q1 → q2 (on a)
• q2 → q3 (on b)
• q3 → q4 (on b) → final
• Final state loops on a, b

Option 2: Incorrect transitions. Accepts some but not all “abb” variants. Incorrect

Option 3: Uses 6 states, which is not optimal/minimal. Incorrect

The correct answer is Option 3.

Key Points

• Option 3 specifies the grammar for the language accepted by the machine as follows:
  • S → AabbA
  • A → aA | bA | ε
• This grammar generates strings where:
  • The start symbol S derives a string with 'A', followed by 'aabb', followed by 'A'.
  • The non-terminal A can be replaced by 'aA', 'bA', or ε (the empty string).
• This allows for the generation of strings with 'a' and 'b' characters appropriately placed in the structure defined by the grammar.

Additional Information

• Grammar rules (productions) define how strings in a language can be generated from the start symbol.
• The symbol 'ε' denotes the empty string, which means a non-terminal can be replaced by nothing.
• The use of non-terminals like A allows for recursive definitions, enabling the generation of complex strings.
• Correct grammar formulations are essential in parsing and generating strings in programming languages and compilers.

The correct answer is Option 3.

Key Points

Option 1: The DFA transitions a → b → b directly to final state. But it lacks proper looping logic to verify intermediate characters. No trap state. Incorrect.

Option 2: Accepts strings only if they end with “abb”. But the question states the string can have “abb” anywhere. Incorrect.

Option 3: The DFA clearly detects “abb” and loops over further inputs. So, even strings like “aabb”, “baabbab”, “abba” will be accepted. Correct ✔

The correct answer is (a + b)* abb(a + b)*.

Key Points

• The given regular expression (a + b)* abb(a + b)* matches any string that has "abb" in between any combination of 'a' and 'b' characters.
  • The part (a + b)* indicates any combination (including none) of 'a' and 'b' characters before "abb".
  • The part abb is the fixed substring that must appear in the string.
  • The part (a + b)* after "abb" indicates any combination (including none) of 'a' and 'b' characters after "abb".
• This regular expression ensures that "abb" is always present in the string, surrounded by any number of 'a' and 'b' characters.

Additional Information

• Regular expressions are powerful tools used for pattern matching and string manipulation.
• They are widely used in various programming languages and tools for searching, replacing, and validating text.
• Understanding the components and structure of regular expressions is essential for effectively utilizing them in different applications.
• Common operations using regular expressions include searching for specific patterns, validating input data, and parsing text.

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.

Given language is:  L = {w₁aw₂ | w₁, w₂ ϵ {a, b}*, |w₁| = 2, |w₂| ≥ 3}

Regular expression: (a+b) (a+b) a (a+b) (a+b) (a+b)(a+b)^*

Minimum length string that is possible = (a+b) (a+b) a (a+b) (a+b) (a+b)

This string has minimum length 6. For these 7 states are needed and one trap state. So, total 8 states are needed.

DFA for given language is

Consider DFA M,

It is accepting all the string that ends with a.

Language for DFA M L(M) = {a, aa, ba, aaa, aba, …}

Regular expression = (a + b )*a

DFA N is accepting all the languages ending with b.

Language for DFA N L (n) = {b, ab, bb, aab, abb, bbb,…….}

Regular expression = (a + b)*b

Intersection of both these languages L (M) ꓵ L(N) = { } = Ø i.e. empty language

For this, we just need 1 state with all transitions to itself and no final state.

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.

Concept:

First convert the given regular expression into an NFA and then find out the DFA from NFA.

Explanation: 

Regular expression is (a + b) * b (a + b)

NFA for this expression is given here as:

Diagram: NFA

NFA Table:

DFA Table from the above NFA

Diagram: DFA

This DFA can’t be minimized further. So, for the given regular expression minimum number of states in the DFA is 4.

Concept:

Extended transition function means when we start from a state and follow a sequence of input.

In case of ϵ-NFA, epsilon moves are also considered.

Diagram: NFA for the given NFA table is:

Here, \(\hat \delta \;\left( {{q_2},\;aba} \right)\) 

Explanation:

When input a is applied on q₂ it will move to q₀, q₁ as there is null transition so it will also get to q₂.

Similarly, states after input b => {q₀, q₂, q₃}

States after input a = {q₀, q₁, q₂}

So, final answer is {q₀, q₁, q₂}

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.