Consider the following two languages over the alphabet \(\{𝑎, 𝑏\}\): \(L_1 =…
2025
Consider the following two languages over the alphabet \(\{𝑎, 𝑏\}\):
\(L_1 = \{ \alpha \beta \alpha \mid \alpha \in \{a, b\}^+ \text{ AND } \beta \in \{a, b\}^+ \} \\ L_2 = \{ \alpha \beta \alpha \mid \alpha \in \{a\}^+ \text{ AND } \beta \in \{a, b\}^+ \}\)
Which ONE of the following statements is CORRECT ?
- A.
Both
\(L_1\)and\(L_2\)are regular languages. - B.
\(L_1\)is a regular language but\(L_2\)is not a regular language. - C.
\(L_1\)is not a regular language but\(L_2\)is a regular language. - D.
Neither
\(L_1\)nor\(L_2\)is a regular language.
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Correct answer: C
Answer: L1 is a not regular language and L2 is a regular language.
L1 is not regular as we have to match starting and ending part(not alphabet) of string.
NOTE: this is not your typical "starting and ending with same symbol" string, if it is, try to match this string using DFA/NFA - "abbaab"
L2 is regular. here, we have to only check "starting and ending with a" strings which is regular
Analysis of L₁
The language L₁ consists of all strings that can be written as αβα, where both α and β are nonempty strings over a, b. Hence, the prefix and suffix must be identical and nonempty.
To verify this property, a finite automaton would need to “remember” an arbitrary-length prefix α to compare it with the suffix, which requires unbounded memory. This strongly suggests non-regularity.
The language L₂ can be described by the regular expression
a(a+b)+a.
This clearly defines a regular language, since the class of regular languages is closed under concatenation and the Kleene plus operation.
A simple DFA can recognize L₂ by:
Checking that the first symbol is a,
Reading one or more middle symbols,
Accepting only if the final symbol is a.
Thus, L₂ is regular.
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