Which one or more of the following option(s) define(s) regular language(s)?
To determine which of the given options define regular languages, we analyze each option using the properties of regular languages.
- \(\{a^m b^n \mid m,n > 0\}\):
- This language consists of strings comprising of 'a's followed by 'b's with both \(m\) and \(n\) greater than zero.
- It can be expressed by the regular expression \(a^+b^+\), which is a regular language since it can be accepted by a finite automaton.
- \(\{a,b\}^* \cap \{a^m b^n c^{m-n} \mid m > n > 0\}\):
- This language describes strings containing 'a's, 'b's, and 'c's such that the number of 'c's is the difference between the number of 'a's and 'b's.
- Non-equal counting constraints and arithmetic conditions like \(c^{m-n}\) imply the language is context-free or even more complex, but not regular.
- \(\{w \mid w \in \{a,b\}^*, \#_a(w) = 2 \pmod 7, \text{ and } \#_b(w) = 3 \pmod 9\}\):
- This language imposes modular arithmetic conditions on the count of 'a's and 'b's in the strings.
- It is possible to construct a finite automaton with states representing remainders when the count of 'a' and 'b' is divided by 7 and 9, respectively.
- \(\{w \mid w \in \{a,b\}^*, \#_a(w) = 2 \pmod 7, \text{ and } \#_a(w) = \#_b(w)\}\):
- While modular arithmetic condition alone can be handled by finite automata, equating counts of symbols generally cannot, especially mixed with modulus constraints.
- Such a condition usually requires more computational power than that of finite automata, indicating it's not regular.
In conclusion, the correct regular languages are:
- \(\{a^m b^n \mid m,n > 0\}\)
- \(\{w \mid w \in \{a,b\}^*, \#_a(w) = 2 \pmod 7, \text{ and } \#_b(w) = 3 \pmod 9\}\)
Important Questions from Regular Languages
Which of the following are correct on regular expressions?
A. φ + L = L + φ = L
B. εL = Lε = L
C. φL = Lφ = φ
D. φL = Lφ = L
Choose the correct answer from the options given below:
- Let $L_1$ and $L_2$ be two languages over a finite alphabet, such that $L_1 \cap L_2$ and $L_2$ are regular languages.
Which of the following statements is/are always true? Regular expression for the complement of language \(L = \{a^n b^m \mid n \ge 4,\ m \le 3\}\) is
Given two languages :
\(L_1 = \{(ab)^n a^k \mid n \gt k,\ k \ge 0\}\)
\(L_2 = \{a^n b^m \mid n \ne m\}\)
Using pumping lemma for regular language, it can be shown that
Which of the following statements is false ?
