Question

Consider the following context-free grammar \(G\), where \(S\), \(A\), and \(B\) are the variables (non-terminals), \(a\) and \(b\) are the terminal symbols, \(S\) is the start variable, and the rules of \(G\) are described as:
\[ S \to aaB \mid Abb \] \[ A \to a \mid AaA \] \[ B \to b \mid bB \] Which ONE of the languages \(L(G)\) is accepted by \(G\)?

• A. \( L(G) = \{a^2 b^n \mid n \geq 1\} \cup \{a^n b^2 \mid n \geq 1\} \)
• B. \( L(G) = \{a^n b^{2n} \mid n \geq 1\} \cup \{a^{2n} b^n \mid n \geq 1\} \)
• C. \( L(G) = \{a^n b^n \mid n \geq 1\} \)
• D. \( L(G) = \{a^{2n} b^{2n} \mid n \geq 1\} \)

Hint:

In context-free grammar problems, carefully analyze the production rules step by step and look at how the rules combine to form strings. Always check for patterns in the generated strings to match the language correctly.

Answer 1

The Correct Option is A

Let’s examine the production rules and how they generate strings:

1. Production Rule 1: \( S \to aaB \mid Abb \)
- The rule \(S \to aaB\) produces strings that start with "aa" followed by a string generated by \(B\). Since \(B\) generates any number of \(b\)'s, this can result in strings like \(aa b^n\), where \(n \geq 1\).
- The rule \(S \to Abb\) generates strings starting with "A" followed by two \(b\)'s. Since \(A\) generates strings of 'a's (from \(A \to a \mid AaA\)), this results in strings like \(a^n b^2\), where \(n \geq 1\).

2. Production Rule 2: \( A \to a \mid AaA \)
- The production \(A \to a\) generates a single \(a\).
- The recursive production \(A \to AaA\) means that \(A\) can generate any number of \(a\)'s, leading to strings like \(a^n\).

3. Production Rule 3: \( B \to b \mid bB \)
- This generates any number of \(b\)'s, starting with one \(b\), as \(B\) recursively produces \(b\)'s.

Derivation of Strings:

- From \( S \to aaB \), we generate strings of the form \( aa b^n \), where \( n \geq 1 \). For example:
\( aa b \), \( aa bb \), \( aa bbb \), etc. These strings match the pattern \( a^2 b^n \), where \( n \geq 1 \).

- From \( S \to Abb \), we generate strings of the form \( a^n b^2 \), where \( n \geq 1 \). For example:
\( ab^2 \), \( a^2 b^2 \), \( a^3 b^2 \), etc. These strings match the pattern \( a^n b^2 \), where \( n \geq 1 \).

Conclusion:
Thus, the correct language \( L(G) \) is the union of the two patterns:
\[ L(G) = \{a^2 b^n \mid n \geq 1\} \cup \{a^n b^2 \mid n \geq 1\}. \]
Therefore, the correct answer is (A).