Chapter 2, Problem 15E

Explanation of Solution

a.

Given context-free grammar:

$\begin{array}{c}\text{G }\to \text{ G B}\\ \to \text{ G N}\\ \to \epsilon \end{array}$

$\begin{array}{c}\text{B }\to \left(\text{ E }\right)\\ \to \text{ E }\left(\text{ E }\right)\\ \to \epsilon \end{array}$

Explanation of Solution

b.

Parse tree:

The parse tree for the given string is as follows:

Explanation of Solution

c.

Canonical (right-most) derivation for the given grammar:

$\begin{array}{c}\text{G }\to \text{ G B}\\ \to \text{ G }\left(\text{ E }\right)\\ \to \text{ G }\left(\right)\\ \to \text{ G N }\left(\right)\end{array}$

Explanation of Solution

d.

FIRST (E) in the given grammar:

• In production rule, the FIRST set is applied to the right-hand side (RHS), and tells all the terminal symbols which can start sentences derived from that RHS.
• The definition of FIRST set is FIRST (x) = {x}, here “x” is terminal symbol.
• For any non-terminal “X” with production rules $X\to x_1\left|x_2\right|\mathrm{...}|x_n,FIRST\left(X\right)=FIRST\left(x_1\right)\cup FIRST\left(x_2\right)\cup \mathrm{...}\cup FIRST\left(x_n\right)$

Explanation of Solution

e.

Justification:

“No”, the given grammar does not have an LL (1) grammar because, the intuition is straightforward.

• In an LL (1) parser, the contents of the parse stack denote the predicted remainder of the program, and decisions are always made on the basis of the top-of-stack symbol and the next token of input...