Muaz I. answered 08/22/24
Hello, My name is Muaz. I am a student. If you want I can help yo
### (1) Automata Theory for the Subway Gate
#### (i) Transition Table
The transition table describes the behavior of the gate system based on its current state and inputs.
| Current State | Input | Next State |
|---------------|-------|------------|
| LOCKED | TOKEN | UNLOCKED |
| LOCKED | PUSH | LOCKED |
| UNLOCKED | TOKEN | UNLOCKED |
| UNLOCKED | PUSH | LOCKED |
#### (ii) Transition Diagram
To represent this in a transition diagram:
1. **States**: Represent the states LOCKED and UNLOCKED as circles.
2. **Transitions**:
- An arrow from LOCKED to UNLOCKED labeled "TOKEN".
- An arrow from UNLOCKED to LOCKED labeled "PUSH".
- Self-loops on LOCKED labeled "PUSH" and on UNLOCKED labeled "TOKEN".
Here is a textual representation of the transition diagram:
```
LOCKED --TOKEN--> UNLOCKED
^ |
| |
+--PUSH--<------------+
UNLOCKED --PUSH--> LOCKED
^ |
| |
+--TOKEN--<-----------+
```
(2) Context-Free Grammar and Derivation
(i) Derive the string (p + q) X p – r X p/(q + q)
To derive the expression `(p + q) X p – r X p/(q + q)`, follow these steps:
1. Start with `E` and apply the production rules to build up the expression.
```
E → E – E (Rule 5)
```
2. For the first `E`, derive `(p + q) X p` and for the second `E`, derive `r X p/(q + q)`.
```
E → (E) X E (Rule 6)
```
3. Derive `(p + q)` and `p` for the first `E`:
```
E → (E + E) (Rule 4)
```
4. Derive `p` and `q`:
```
E → p (Rule 1)
E → q (Rule 2)
```
5. Derive `r X p/(q + q)` for the second `E`:
```
E → E X E (Rule 6)
E → E / E (Rule 7)
```
6. Derive `r` and `p` for the first `E` and `(q + q)` for the second `E`:
```
E → p (Rule 1)
E → q (Rule 2)
E → (E + E) (Rule 4)
```
7. Combine all derivations:
E → (E + E) X p – E / E
E → (p + q) X p – r X p / (q + q)
(ii) Parse Tree
A parse tree visualizes the structure of the expression according to the grammar.
```
E
/|\
/ | \
/ | \
E - E
/|\ /|\
/ | \ / | \
( E ) / E \
| / | \
E X E / \
/|\ / E \
/ | \ / | \
p + q X p r / \
/ \
E E
/ \ / \
( + q q
/ \
E E
/ \
q q
```
In the parse tree:
- The root node represents the entire expression.
- Each internal node represents an operator or parenthesis.
- Leaf nodes represent the variables `p`, `q`, and `r` and the operators.
This parse tree reflects the order of operations as specified by the context-free grammar rules.