Question

Let $\Sigma = \{a, b, c\}$. For $x \in \Sigma^*$, and $\alpha \in \Sigma$, let $\#\alpha(x)$ denote the number of occurrences of $\alpha$ in $x$. Which one or more of the following option(s) define(s) regular language(s)?

• A. $\{ a^m b^n \mid m, n \geq 0 \}$
• B. $\{a, b\}^* \cap \{ a^m b^n c^{m-n} \mid m \geq n \geq 0 \}$
• C. $\{ w \mid w \in \{a, b\}^*, \#a(w) \equiv 2 \pmod{7}, { and } \#b(w) \equiv 3 \pmod{9} \}$
• D. $\{ w \mid w \in \{a, b\}^*, \#a(w) \equiv 2 \pmod{7}, { and } \#a(w) = \#b(w) \}$

Hint:

Regular languages are closed under union, intersection, and modulo counting but not under constraints like equality of counts of two symbols.

Answer 1

The Correct Option is A, C

A language is regular if it can be accepted by a finite automaton.

• (A) The language \( \{ a^m b^n \mid m, n \geq 0 \} \) is regular because it consists of any number of \(a\)'s followed by any number of \(b\)'s, which can be accepted by a finite state machine. This language is simple and can be recognized by a finite automaton that transitions through states as it reads each symbol.
• (B) The language \( \{ a^m b^n c^{m-n} \mid m \geq n \geq 0 \} \) is not regular because it requires counting the difference between \( m \) and \( n \), which is a context-free property. A finite automaton cannot keep track of this type of relationship between different parts of the string, so this language is not regular.
• (C) The language \( \{ w \mid w \in \{a, b\}^*, \#a(w) \equiv 2 \pmod{7}, \#b(w) \equiv 3 \pmod{9} \} \) is regular because modular counting of occurrences can be done using a finite-state automaton. This involves checking the remainder of the count of \( a \)'s modulo 7 and \( b \)'s modulo 9, which is a task that finite automata can handle using state transitions to track these counts.
• (D) The language \( \{ w \mid w \in \{a, b\}^*, \#a(w) \equiv 2 \pmod{7}, \#a(w) = \#b(w) \} \) is not regular because enforcing \( \#a(w) = \#b(w) \) requires a counter, which is a non-regular property. The condition \( \#a(w) = \#b(w) \) implies that the automaton must be able to compare the number of \( a \)'s with the number of \( b \)'s, which is not possible for finite state machines without additional memory.

Thus, the correct options are (A) and (C).