2. Consider the following grammar: <assign> à <id> = <expr>
2. Consider the following grammar:
<assign> à <id> = <expr>
<id> à A | B | C
<expr> à <id> + <expr>
| <id> * <expr>
| ( <expr> )
| <id>
Show a parse tree and leftmost derivation for the following statements:
(a) A = ( A + B ) * C
(b) A = A * ( B + C )
3. [10 Points] Show that the following grammar is ambiguous.
<S> à <A>
<A> à <A> + <A> | <id>
<id> à a | b | c
4. [10 Points] Compute the weakest precondition for each of the following assignment statements with their given postconditions.
(a) a = 2 * (b – 1) – 1 {a > 0}
(b) a = a + 2 * b – 1 {a > 1}
Homework Answers
Question 3: Given the following grammar: assign → id := expr expr → expr + term \ term term -term *factor lfactor factor-(expr) id Using the above grammar, show a leftmost derivation (first five step...
Question 3: Given the following grammar: assign → id := expr expr → expr + term \ term term -term *factor lfactor factor-(expr) id Using the above grammar, show a leftmost derivation (first five steps) for the following assignment statement: A ((A B)+ C) a. [3 marks] b. Using the above grammar, show a rightmost derivation (first five steps) for the following assignment statement: A:-A+B+C)+A [3 marks] Draw the abstract syntax tree for each of the above statements [4 marks]...
Consider the following grammar (G1) for simple assignment statements. (The symbols in double quotation marks are...
Consider the following grammar (G1) for simple assignment statements. (The symbols in double quotation marks are terminal symbols.) assign → id “ = ” expr id → “A” | “B” | “C” expr → expr “ + ” expr | expr “ ∗ ” expr | “(” expr “)” | id a) Give a (leftmost) derivation for string A = B ∗ A + C. b) Give the parse tree for string A = B ∗ A + C. c)...
6. (8 pts) Using grammar below show a Parse tree and leftmost derivation for a). A...
6. (8 pts) Using grammar below show a Parse tree and leftmost derivation for a). A = A * (B+C) <assign> à<id> = <expr> <id> à A | B|C <expr>à <expr> + <term> | <term> <term> à <term> * <factor> |<factor> <factor> à ( <expr> ) |<id>
3. Using the grammar below, show a parse tree and a leftmost derivation for the statement....
3. Using the grammar below, show a parse tree and a leftmost derivation for the statement. A = ( A + (B)) * C assign <idxpr expr>? <expr> <term> term <term factor factor (<expr>) l <term I <factor l <id> 4. Prove that the following grammar is ambiguous (Give sentence that has two parse trees, and show the parse trees):
1.a Consider the following Grammar,
1.a Consider the following Grammar, <assign> à <id> = <expr> <id> à A | B |C <expr> à < expr> + <expr> | <expr> * <expr> | ( <expr> ) | <id> Derive the following statement using leftmost derivation. A = A * (B*(C+ A)) b. 2 a}. Consider the following grammar that expresses parenthesized expressions of digits, including both addition and multiplication. <expr> := <expr> + <expr> | <expr> * <expr> | (expr>) | <digit> <digit> := 0 | 1 | 2 |...
- Using the grammar in Example 3.2, show a parse tree and a leftmost derivation for...
- Using the grammar in Example 3.2, show a parse tree and a leftmost derivation for the following statement: B = C * (A * (B + C)). EXAMPLE 3.2 A Grammar for Simple Assignment Statements <assign> → <id> = <expr> <id> → A | B | C <expr> → <id> + <expr> | <id> * <expr> | ( <expr> ) | <id>
The questions in this section are based on the grammar given as the following: prog ->...
The questions in this section are based on the grammar given as the following: prog -> assign | expr assign -> id = expr expr -> expr + term | expr - term | term term -> factor | factor * term factor -> ( expr ) | id | num id -> A | B | C num -> 0 | 1 | 2 | 3 (2a) What is the associativity of the * operator? (5 points) (2b) What...
The questions in this section are based on the grammar given as the following: prog ->...
The questions in this section are based on the grammar given as the following: prog -> assign | expr assign -> id = expr expr -> expr + term | expr - term | term term -> factor | factor * term factor -> ( expr ) | id | num id -> A | B | C num -> 0 | 1 | 2 | 3 (2a) What is the associativity of the * operator? (5 points) (2b) What...
Compute the weakest precondition for each of the following assignment statements and postconditions: a = 2...
Compute the weakest precondition for each of the following assignment statements and postconditions: a = 2 * (b - 1) - 1 {a > 0} b = (c + 10) / 3 {b > 6} Prove that the following grammar is ambiguous (show 2 trees): <S> → <A> <A> → <A> + <A> | <id> <id> → a | b | c
Use the grammar given below and show a parse tree and a leftmost derivation for each...
Use the grammar given below and show a parse tree and a leftmost
derivation for each of
the following statements.
1. A = A * (B + (C * A))
2. B = C * (A * C + B)
3. A = A * (B + (C))
<assign> → <id> <expr> = <expr> → <id> + <expr> kid<expr> <expr>) ids
Question 3: Given the following grammar: assign → id := expr expr → expr + term \ term term -term *factor lfactor factor-(expr) id Using the above grammar, show a leftmost derivation (first five steps) for the following assignment statement: A ((A B)+ C) a. [3 marks] b. Using the above grammar, show a rightmost derivation (first five steps) for the following assignment statement: A:-A+B+C)+A [3 marks] Draw the abstract syntax tree for each of the above statements [4 marks]...
3. Using the grammar below, show a parse tree and a leftmost derivation for the statement. A = ( A + (B)) * C assign <idxpr expr>? <expr> <term> term <term factor factor (<expr>) l <term I <factor l <id> 4. Prove that the following grammar is ambiguous (Give sentence that has two parse trees, and show the parse trees):
- Using the grammar in Example 3.2, show a parse tree and a leftmost derivation for the following statement: B = C * (A * (B + C)). EXAMPLE 3.2 A Grammar for Simple Assignment Statements <assign> → <id> = <expr> <id> → A | B | C <expr> → <id> + <expr> | <id> * <expr> | ( <expr> ) | <id>
Use the grammar given below and show a parse tree and a leftmost
derivation for each of
the following statements.
1. A = A * (B + (C * A))
2. B = C * (A * C + B)
3. A = A * (B + (C))
<assign> → <id> <expr> = <expr> → <id> + <expr> kid<expr> <expr>) ids
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