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Given the grammar A → AA | ( A ) | e, a. Describe the language it generates. b. Show that it is ambiguous.

Given the grammar A → AA | ( A ) | e, a. Describe the language it generates. b. Show that it is ambiguous.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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### Exercise 3.2 - Context-Free Grammar (CFG) **Given the grammar \( A \rightarrow AA \mid ( A ) \mid \epsilon \),** a. **Describe the language it generates.** The grammar generates strings that are either balanced parentheses or empty strings. Here is an explanation: - **\(A \rightarrow AA\):** The non-terminal \(A\) can be substituted with two consecutive instances of \(A\), allowing for concatenation of balanced strings. - **\(A \rightarrow ( A )\):** The non-terminal \(A\) can be replaced by a pair of parentheses that contain a balanced string within. - **\(A \rightarrow \epsilon\):** The non-terminal \(A\) can be replaced by the empty string, indicating that an \(A\) can represent no characters. This grammar describes the language of strings composed of balanced parentheses, which is a well-known context-free language. b. **Show that it is ambiguous.** To demonstrate that the grammar is ambiguous, one must show that there exists at least one string that can be generated by the grammar in more than one distinct way (i.e., it has more than one parse tree). Let's consider the string "()(())": - One possible derivation: 1. \( A \rightarrow ( A ) \) 2. \( A \rightarrow AA \) 3. \( A \rightarrow \epsilon \, A \) 4. \( A \rightarrow ( A ) \) 5. \( A \rightarrow \epsilon \) - Another possible derivation: 1. \( A \rightarrow AA \) 2. \( A \rightarrow ( A ) \) 3. \( A \rightarrow \epsilon \) 4. \( A \rightarrow ( A ) \) 5. \( A \rightarrow \epsilon \) Since the string "()(())" can be derived in multiple ways using different production rules, this confirms that the given grammar is ambiguous.
Transcribed Image Text:### Exercise 3.2 - Context-Free Grammar (CFG) **Given the grammar \( A \rightarrow AA \mid ( A ) \mid \epsilon \),** a. **Describe the language it generates.** The grammar generates strings that are either balanced parentheses or empty strings. Here is an explanation: - **\(A \rightarrow AA\):** The non-terminal \(A\) can be substituted with two consecutive instances of \(A\), allowing for concatenation of balanced strings. - **\(A \rightarrow ( A )\):** The non-terminal \(A\) can be replaced by a pair of parentheses that contain a balanced string within. - **\(A \rightarrow \epsilon\):** The non-terminal \(A\) can be replaced by the empty string, indicating that an \(A\) can represent no characters. This grammar describes the language of strings composed of balanced parentheses, which is a well-known context-free language. b. **Show that it is ambiguous.** To demonstrate that the grammar is ambiguous, one must show that there exists at least one string that can be generated by the grammar in more than one distinct way (i.e., it has more than one parse tree). Let's consider the string "()(())": - One possible derivation: 1. \( A \rightarrow ( A ) \) 2. \( A \rightarrow AA \) 3. \( A \rightarrow \epsilon \, A \) 4. \( A \rightarrow ( A ) \) 5. \( A \rightarrow \epsilon \) - Another possible derivation: 1. \( A \rightarrow AA \) 2. \( A \rightarrow ( A ) \) 3. \( A \rightarrow \epsilon \) 4. \( A \rightarrow ( A ) \) 5. \( A \rightarrow \epsilon \) Since the string "()(())" can be derived in multiple ways using different production rules, this confirms that the given grammar is ambiguous.
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