**Transform the Following Grammar to a Grammar Without Left-Recursive Rules**Given Grammar:1. $ S \to A \mid B $2. $ A \to AAA \mid a \mid B $3. $ B \to BBb \mid b $To convert the given grammar to a form without left recursion, the direct left recursion in the rules for $ A $ and $ B $ must be resolved by introducing new non-terminal symbols.### Explanation of the Original GrammarThe original grammar consists of three production rules. - The start symbol is $ S $, which can be replaced with either $ A $ or $ B $.- The non-terminal $ A $ includes a left-recursive rule $ A \to AAA $, as well as options to replace with $ a $ or $ B $.- The non-terminal $ B $ contains another left-recursive rule $ B \to BBb $, and can also resolve to $ b $.### Steps to Remove Left RecursionFor each non-terminal with left recursion, introduce a new non-terminal:1. **Convert $ A $:** - Identify left-recursive and non-left-recursive parts: - Left-recursive: $ AAA $ - Non-left-recursive: $ a $, $ B $ - Introduce $ A' $ to handle left recursion: - Convert $ A $ to: $ A \to aA' \mid BA' $ - Define $ A' $ with recursive production: $ A' \to AA' \mid \epsilon $2. **Convert $ B $:** - Identify left-recursive and non-left-recursive parts: - Left-recursive: $ BBb $ - Non-left-recursive: $ b $ - Introduce $ B' $ for left recursion: - Convert $ B $ to: $ B \to bB' $ - Define $ B' $ with recursion: $ B' \to BbB' \mid \epsilon $The resulting grammar will have no left recursion:1. $ S \to A \mid B $2. $ A \to aA' \mid BA' $3. $ A' \to AA' \mid \epsilon $4. $ B \to bB' $5. \(

Answered: Transform the following grammar to a…

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

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

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Question

**Transform the Following Grammar to a Grammar Without Left-Recursive Rules**

Given Grammar:

1. $ S \to A \mid B $
2. $ A \to AAA \mid a \mid B $
3. $ B \to BBb \mid b $

To convert the given grammar to a form without left recursion, the direct left recursion in the rules for $ A $ and $ B $ must be resolved by introducing new non-terminal symbols.

### Explanation of the Original Grammar

The original grammar consists of three production rules.

- The start symbol is $ S $, which can be replaced with either $ A $ or $ B $.
- The non-terminal $ A $ includes a left-recursive rule $ A \to AAA $, as well as options to replace with $ a $ or $ B $.
- The non-terminal $ B $ contains another left-recursive rule $ B \to BBb $, and can also resolve to $ b $.

### Steps to Remove Left Recursion

For each non-terminal with left recursion, introduce a new non-terminal:

1. **Convert $ A $:**
- Identify left-recursive and non-left-recursive parts:
- Left-recursive: $ AAA $
- Non-left-recursive: $ a $, $ B $
- Introduce $ A' $ to handle left recursion:
- Convert $ A $ to: $ A \to aA' \mid BA' $
- Define $ A' $ with recursive production: $ A' \to AA' \mid \epsilon $

2. **Convert $ B $:**
- Identify left-recursive and non-left-recursive parts:
- Left-recursive: $ BBb $
- Non-left-recursive: $ b $
- Introduce $ B' $ for left recursion:
- Convert $ B $ to: $ B \to bB' $
- Define $ B' $ with recursion: $ B' \to BbB' \mid \epsilon $

The resulting grammar will have no left recursion:

1. $ S \to A \mid B $
2. $ A \to aA' \mid BA' $
3. $ A' \to AA' \mid \epsilon $
4. $ B \to bB' $
5. \(

Expert Solution

Step 1

Assume that the set of all A-productions has the form

A $\displaystyle \longmapsto$ A $\displaystyle \alpha_{{1}}^{{}}$ | A $\displaystyle \alpha_{{2}}^{{}}$ | ^....A $\displaystyle \alpha_{{m}}^{{}}$ | $\displaystyle \beta_{{1}}^{{}}$ | $\displaystyle \beta_{{2}}^{{}}$ | ^... $\displaystyle \beta_{{n}}^{{}}$

where no $\beta_{{i}}^{{}}$ begins with an A and where $\alpha_{{i}}^{{}}$ $\neq$ $\varepsilon$ holds for every i = 1^...m. Then we replace these A-productions by

A	$\displaystyle \longmapsto$	$\displaystyle \beta_{{1}}^{{}}$ A' \| $\displaystyle \beta_{{2}}^{{}}$ A' \| ^... $\displaystyle \beta_{{n}}^{{}}$ A'
A'	$\displaystyle \longmapsto$	$\displaystyle \alpha_{{1}}^{{}}$ A' \| $\displaystyle \alpha_{{2}}^{{}}$ A' \| ^... $\displaystyle \alpha_{{m}}^{{}}$ A' \| $\displaystyle \varepsilon$