Question

Using the following grammar, show a parse tree and a leftmost derivation for the following sentence (make sure you do not omit parentheses in your derivation):

        Grammar
        <assign> → <id> = <expr>
   <id> → A | B | C
   <expr> → <expr> + <term> | <term>
   <term> → <term> * <factor> | <factor>
   <factor> → (<expr>) | <id>

Derive

C = (A+B)*(C+A)*(C+B)

Answer 1

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ANSWER:

EXPLANATION:

C = (A+B)*(C+A)*(C+B)

<assign>
→ <id> = <expr>
→ C = <expr>
→ C = <term> * <factor>
→ C = <term> * <factor> * <factor>
→ C = <factor> * <factor> * <factor>
→ C = (<expr>) * <factor> * <factor>
→ C = (<expr> + <term>) * <factor> * <factor>
→ C = (<term> + <term>) * <factor> * <factor>
→ C = (<factor> + <term>) * <factor> * <factor>
→ C = (<id> + <term>) * <factor> * <factor>
→ C = (A + <term>) * <factor> * <factor>
→ C = (A + <factor>) * <factor> * <factor>
→ C = (A + <id>) * <factor> * <factor>
→ C = (A + B) * <factor> * <factor>
→ C = (A + B) * (<expr>) * <factor>
→ C = (A + B) * (<expr>+<term>) * <factor>
→ C = (A + B) * (<term>+<term>) * <factor>
→ C = (A + B) * (<factor>+<term>) * <factor>
→ C = (A + B) * (<id>+<term>) * <factor>
→ C = (A + B) * (C + <term>) * <factor>
→ C = (A + B) * (C + <factor>) * <factor>
→ C = (A + B) * (C + <id>) * <factor>
→ C = (A + B) * (C + A) * <factor>   
→ C = (A + B) * (C + A) * (<expr>)   
→ C = (A + B) * (C + A) * (<expr> + <term>)
→ C = (A + B) * (C + A) * (<term> + <term>)
→ C = (A + B) * (C + A) * (<factor> + <term>)
→ C = (A + B) * (C + A) * (<id> + <term>)
→ C = (A + B) * (C + A) * (C + <term>)
→ C = (A + B) * (C + A) * (C + <factor>)
→ C = (A + B) * (C + A) * (C + <id>)
→ C = (A + B) * (C + A) * (C + B)

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