Let's define the precedence relations among the terminals:
- Precedence of
*
> Precedence of b
modify the production rules to make it an operator grammar:
\[ \begin{aligned} & \text{Para} \rightarrow \text{Sentence RP | Sentence} \\ & \text{RP} \rightarrow \text{b Sentence RP | b Sentence} \\ & \text{Sentence} \rightarrow \text{Word Sentence'} \\ & \text{Sentence'} \rightarrow \varepsilon \,|\, \text{b Sentence'} \\ & \text{Word} \rightarrow \text{letter Word'} \\ & \text{Word'} \rightarrow \varepsilon \,|\, * \, \text{word Word'} \\ & \text{letter} \rightarrow \text{id} \end{aligned} \]
In the modified grammar, each production has at most one non-terminal on the right side, and the terminals are associated with operators.
Now, let's use a stack algorithm to parse the given string \[\text{id * id b id * id}\] based on the defined precedence relations:
- Initialize an empty stack and start reading the input from left to right.
- For each terminal encountered, push it onto the stack.
- When a terminal with lower precedence encounters a terminal with higher precedence on the stack, reduce the stack using the production rules.
Parsing Steps:
- Stack: [id]
- Input: * id b id * id
- Reduce: (Apply Word -> letter)
- Stack: [Word]
- Input: * id b id * id
- Reduce: (Apply Sentence -> Word Sentence')
- Stack: [Sentence']
- Input: * id b id * id
- Shift: (Push b onto the stack)
- Stack: [Sentence' b]
- Input: id b id * id
- Reduce: (Apply Sentence' -> b Sentence')
- Stack: [Sentence']
- Input: id * id
- Reduce: (Apply Sentence -> Word Sentence')
- Stack: [Word Sentence']
- Input: id * id
- Reduce: (Apply Word -> letter Word')
- Stack: [Word' Sentence']
- Input: * id
- Shift: (Push * onto the stack)
- Stack: [Word' Sentence' *]
- Input: id
- Reduce: (Apply Word' -> ε)
- Stack: [Word Sentence']
- Input: id
- Reduce: (Apply Sentence' -> ε)
At this point, the stack contains only one non-terminal, and the input is empty. The parsing is successful