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6. Using the following grammar, show a parse tree and rightmost derivation for A= (B+ (A+ (CA)))

6. Using the following grammar, show a parse tree and rightmost derivation for A= (B+ (A+ (CA)))

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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### Grammar and Parse Tree in Context of Simple Assignment Statements #### Problem Statement: Using the following grammar, show a parse tree and rightmost derivation for \[ A = ( B + ( A + ( C * A ) ) ) \] #### A Grammar for Simple Assignment Statements ``` <assign> → <id> = <expr> <id> → A | B | C <expr> → <id> + <expr> | <id> * <expr> | ( <expr> ) | <id> ``` --- ### Explanation #### Rightmost Derivation 1. Start with: \[ \langle \text{assign} \rangle \] 2. Replaced by: \[ \langle \text{id} \rangle = \langle \text{expr} \rangle \] 3. Substitute \[ \langle \text{id} \rangle \] with A: \[ A = \langle \text{expr} \rangle \] 4. Derive \[ \langle \text{expr} \rangle \] as a plus operation: \[ A = \langle \text{id} \rangle + \langle \text{expr} \rangle \] 5. Substitute \[ \langle \text{id} \rangle \] with B: \[ A = B + \langle \text{expr} \rangle \] 6. Derive the next expression with parentheses: \[ A = B + ( \langle \text{expr} \rangle ) \] 7. Inside parentheses, apply plus operation to \[ \langle \text{expr} \rangle \]: \[ A = B + ( \langle \text{id} \rangle + \langle \text{expr} \rangle ) \] 8. Substitute \[ \langle \text{id} \rangle \] inside parentheses with A: \[ A = B + ( A + \langle \text{expr} \rangle ) \] 9. Derive the next expression with parentheses and multiplication: \[ A = B + ( A + ( \langle \text{expr} \rangle \langle \text{expr} \rangle \) ) \] 10.
Transcribed Image Text:### Grammar and Parse Tree in Context of Simple Assignment Statements #### Problem Statement: Using the following grammar, show a parse tree and rightmost derivation for \[ A = ( B + ( A + ( C * A ) ) ) \] #### A Grammar for Simple Assignment Statements ``` <assign> → <id> = <expr> <id> → A | B | C <expr> → <id> + <expr> | <id> * <expr> | ( <expr> ) | <id> ``` --- ### Explanation #### Rightmost Derivation 1. Start with: \[ \langle \text{assign} \rangle \] 2. Replaced by: \[ \langle \text{id} \rangle = \langle \text{expr} \rangle \] 3. Substitute \[ \langle \text{id} \rangle \] with A: \[ A = \langle \text{expr} \rangle \] 4. Derive \[ \langle \text{expr} \rangle \] as a plus operation: \[ A = \langle \text{id} \rangle + \langle \text{expr} \rangle \] 5. Substitute \[ \langle \text{id} \rangle \] with B: \[ A = B + \langle \text{expr} \rangle \] 6. Derive the next expression with parentheses: \[ A = B + ( \langle \text{expr} \rangle ) \] 7. Inside parentheses, apply plus operation to \[ \langle \text{expr} \rangle \]: \[ A = B + ( \langle \text{id} \rangle + \langle \text{expr} \rangle ) \] 8. Substitute \[ \langle \text{id} \rangle \] inside parentheses with A: \[ A = B + ( A + \langle \text{expr} \rangle ) \] 9. Derive the next expression with parentheses and multiplication: \[ A = B + ( A + ( \langle \text{expr} \rangle \langle \text{expr} \rangle \) ) \] 10.
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