Question:
Consider the following two languages over the alphabet \( \{a, b\} \):
\[
L_1 = \{ \alpha \beta \alpha \mid \alpha \in \{a, b\}^+ { and } \beta \in \{a, b\}^+ \}
\]
\[
L_2 = \{ \alpha \beta \alpha \mid \alpha \in \{a\}^+ { and } \beta \in \{a, b\}^+ \}
\]
Which ONE of the following statements is CORRECT?
Consider the following two languages over the alphabet \( \{a, b\} \):
\[
L_1 = \{ \alpha \beta \alpha \mid \alpha \in \{a, b\}^+ { and } \beta \in \{a, b\}^+ \}
\]
\[
L_2 = \{ \alpha \beta \alpha \mid \alpha \in \{a\}^+ { and } \beta \in \{a, b\}^+ \}
\]
Which ONE of the following statements is CORRECT?
Show Hint
A language where the prefix must match the suffix typically requires memory beyond what a finite automaton can provide, making it non-regular. However, languages with more restrictive patterns, like \( L_2 \), can often be regular.
Updated On: Jan 30, 2026
- Both \( L_1 \) and \( L_2 \) are regular languages.
- \( L_1 \) is a regular language but \( L_2 \) is not a regular language.
- \( L_1 \) is not a regular language but \( L_2 \) is a regular language.
- Neither \( L_1 \) nor \( L_2 \) is a regular language.
Show Solution
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The Correct Option is C
Solution and Explanation
We are given two languages, \( L_1 \) and \( L_2 \), and need to determine which statement is correct regarding their regularity.
1. For \( L_1 \):
The language \( L_1 \) consists of strings of the form \( \alpha \beta \alpha \), where both \( \alpha \) and \( \beta \) can be any non-empty strings over \( \{a, b\} \).
Since the string \( \alpha \) appears twice (once as a prefix and once as a suffix), this language is not regular, as it requires the ability to "remember" the prefix to match it with the suffix, which a finite automaton cannot do.
2. For \( L_2 \):
The language \( L_2 \) consists of strings of the form \( \alpha \beta \alpha \), where \( \alpha \) consists only of \( a \)'s and \( \beta \) can be any string over \( \{a, b\} \).
This restriction makes the language \( L_2 \) regular, as a finite automaton can track the repetition of the prefix \( \alpha \) and match it with the suffix.
Thus, the correct answer is \( \boxed{C} \).
1. For \( L_1 \):
The language \( L_1 \) consists of strings of the form \( \alpha \beta \alpha \), where both \( \alpha \) and \( \beta \) can be any non-empty strings over \( \{a, b\} \).
Since the string \( \alpha \) appears twice (once as a prefix and once as a suffix), this language is not regular, as it requires the ability to "remember" the prefix to match it with the suffix, which a finite automaton cannot do.
2. For \( L_2 \):
The language \( L_2 \) consists of strings of the form \( \alpha \beta \alpha \), where \( \alpha \) consists only of \( a \)'s and \( \beta \) can be any string over \( \{a, b\} \).
This restriction makes the language \( L_2 \) regular, as a finite automaton can track the repetition of the prefix \( \alpha \) and match it with the suffix.
Thus, the correct answer is \( \boxed{C} \).
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