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a) S0S1 | 0 1 with string 000111. b) S → + SS SS a with string +* aaa.

a) S0S1 | 0 1 with string 000111. b) S → + SS SS a with string +* aaa.

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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Consider the context-free grammar:

a) Give a leftmost derivation for the string.

b) Give a rightmost derivation for the string.

c) Give a parse tree for the string.

### Grammar Derivation Examples #### a) Production Rule **Production:** \( S \rightarrow 0 \, S \, 1 \mid 0 \, 1 \) **String Example:** The string derived from this production is `000111`. **Explanation:** This production rule describes how to derive strings. The non-terminal symbol \( S \) can be replaced with either of the following sequences: - `0 S 1`: This means a `0` followed by another \( S \) and then a `1`. - `0 1`: This direct substitution skips further recursion and places `0` followed by `1`. For the string `000111`: 1. Start with \( S \). 2. Apply rule \( S \rightarrow 0 \, S \, 1 \) to get `0 S 1`. 3. Replace the new \( S \) with `0 S 1` to continue getting `00 S 11`. 4. Finally, replace the last \( S \) with `0 1` to complete the string: `000111`. #### b) Production Rule **Production:** \( S \rightarrow + \, S \, S \mid * \, S \, S \mid a \) **String Example:** The string derived from this production is `+*aaa`. **Explanation:** This rule shows how to form a string using a combination of operations and terminals. \( S \) can be replaced by: - `+ S S`: Starts with a plus sign, followed by two \( S \). - `* S S`: Starts with an asterisk, followed by two \( S \). - `a`: Acts as a terminal. For the string `+*aaa`: 1. Begin with \( S \). 2. Use rule \( S \rightarrow + \, S \, S \) for `+ S S`. 3. For the first \( S \), apply \( S \rightarrow * \, S \, S \) to get `+ * S S`. 4. Replace both \( S \) with `a` according to \( S \rightarrow a \). 5. The resulting string becomes `+*aaa`.
Transcribed Image Text:### Grammar Derivation Examples #### a) Production Rule **Production:** \( S \rightarrow 0 \, S \, 1 \mid 0 \, 1 \) **String Example:** The string derived from this production is `000111`. **Explanation:** This production rule describes how to derive strings. The non-terminal symbol \( S \) can be replaced with either of the following sequences: - `0 S 1`: This means a `0` followed by another \( S \) and then a `1`. - `0 1`: This direct substitution skips further recursion and places `0` followed by `1`. For the string `000111`: 1. Start with \( S \). 2. Apply rule \( S \rightarrow 0 \, S \, 1 \) to get `0 S 1`. 3. Replace the new \( S \) with `0 S 1` to continue getting `00 S 11`. 4. Finally, replace the last \( S \) with `0 1` to complete the string: `000111`. #### b) Production Rule **Production:** \( S \rightarrow + \, S \, S \mid * \, S \, S \mid a \) **String Example:** The string derived from this production is `+*aaa`. **Explanation:** This rule shows how to form a string using a combination of operations and terminals. \( S \) can be replaced by: - `+ S S`: Starts with a plus sign, followed by two \( S \). - `* S S`: Starts with an asterisk, followed by two \( S \). - `a`: Acts as a terminal. For the string `+*aaa`: 1. Begin with \( S \). 2. Use rule \( S \rightarrow + \, S \, S \) for `+ S S`. 3. For the first \( S \), apply \( S \rightarrow * \, S \, S \) to get `+ * S S`. 4. Replace both \( S \) with `a` according to \( S \rightarrow a \). 5. The resulting string becomes `+*aaa`.
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