Computer Networking: A Top-Down Approach (7th Edition)
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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.
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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