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Which of the following properties correctly describe a Regular Grammar ?

(A) All production rules are of the form A→xB or A→x, where A and B are non terminal symbols and x is a terminal symbol.

(B) Regular grammars are more powerful than context-free grammars and can express any type of language.

(C) There is a direct correspondence between regular grammar and finite automata.

(D) Regular grammars can generate languages that are not recognised by any type of automata.

Choose the correct answer from the options given below :

This question was previously asked in
UGC NET 2024 Management Question Paper (02-Sep-2024) (Shift 2)
The correct answer is

(A) and (C) Only

A regular grammar is a Type-3 grammar in the Chomsky hierarchy. Its languages are exactly the regular languages, which are also the languages accepted by finite automata and described by regular expressions.

(A) is true: a right-linear regular grammar has every production of the form A → xB or A → x, with A, B non-terminals and x a terminal (an ε-production for the start symbol may also be allowed).

(B) is false: regular grammars are strictly less powerful than context-free grammars; for example \(\{a^nb^n \mid n \ge 0\}\) is context-free but not regular. They certainly cannot express every language.

(C) is true: a state of a finite automaton corresponds to a non-terminal, and a transition on x from A to B corresponds to the production A → xB; this gives a direct two-way conversion.

(D) is false: every language a regular grammar generates is accepted by a finite automaton.

Only (A) and (C) describe regular grammars correctly.

(A) and (C) Only option is correct.

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Similar Questions

  1. Regular expression for the complement of language \(L = \{a^n b^m \mid n \ge 4,\ m \le 3\}\) is

  2. 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

  3. Which of the following statements is false ?

  4. The regular expression corresponding to the language L where

    \(L = \{ x \in \{0, 1\}^{*} \mid x\) ends with 1 and does not contain substring 00 \(\}\) is :

  5. Let L be any language. Define even (W) as the strings obtained by extracting from W the letters in the even-numbered positions and even(L) = {even (W) | W ∈ L}. We define another language Chop (L) by removing the two leftmost symbols of every string in L given by \(\text{Chop}(L) = \{W \mid \nu W \in L, \text{ with } |\nu| = 2\}\).

    If L is regular language then

  6. For the above specified passage, which of the following represents the regular expression?

  7. The precedence of regular expression operators from lowest to highest precedence is:

  8. Which of the following is TRUE about the Pumping Lemma for regular language?

  9. Consider the following two languages :

    L1 = {0i1j| gcd (i, j) = 1}

    L2 is any subset of 0*.

    Which of the following is correct ?

  10. The regular expression for the complement of the language L = {anbm|n ≥ 4, m ≤ 3} is :


Important Questions from Regular Languages

  1. 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: 

  2. 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?
  3. Regular expression for the complement of language \(L = \{a^n b^m \mid n \ge 4,\ m \le 3\}\) is

  4. 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

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