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To solve the problem of constructing the regular expression representing the given automata using Arden's Theorem, let's proceed step by step.

### Step 1: Identify the states and transitions

The given automaton has four states: \(q_0\), \(q_1\), \(q_2\) (which appears to be a mistake since there's only a state \(q_3\) visible), and \(q_3\). The transitions are:

- \(q_0 \) has a self-loop with input \( a \).
- \(q_0 \) transitions to \(q_1\) on input \( a \) or \( b \).
- \(q_1 \) transitions to \(q_3\) on input \( b \).
- \(q_1 \) transitions to \(q_1\) (self-loop) on input \( a \) or \( b \).
- \(q_3 \) transitions to \(q_0\) on input \( a \) or \( b \).

### Step 2: Define the regular expression equations based on transitions

Let the regular expressions for the states \(q_0\), \(q_1\), and \(q_3\) be \(R_0\), \(R_1\), and \(R_3\) respectively. According to the transitions, the equations for the states are:

1. \( R_0 = aR_0 + aR_1 + bR_1 \)  \(\rightarrow \) (From \(q_0\) with loop on \(a\) and transition to \(q_1\) on \(a\) or \(b\))
2. \( R_1 = (a + b)R_1 + bR_3 \) \(\rightarrow \) (From \(q_1\) with self-loop on \(a\) and \(b\) and transition to \(q_3\) on \(b\))
3. \( R_3 = (a + b)R_0 \) \(\rightarrow \) (From \(q_3\) with transition to \(q_0\) on \(a\) or \(b\))

### Step 3: Solve the equations using Arden's Theorem

Arden's Theorem states that if \(R = Q + RP\), then the solution is \(R = QP^*\).

#### Solving for \(R_1\):

The equation for \(R_1\) is:
\[ R_1 = (a + b)R_1 + bR_3 \]

Using Arden's Theorem:
\[ R_1 = bR_3(a + b)^* \]

#### Substituting \(R_1\) in \(R_0\):

Now, substitute \(R_1\) in the equation for \(R_0\):

\[ R_0 = aR_0 + a(bR_3(a + b)^*) + b(bR_3(a + b)^*) \]
\[ R_0 = aR_0 + (a + b)bR_3(a + b)^* \]

Again, using Arden's Theorem:
\[ R_0 = (a + b)bR_3(a + b)^*a^* \]

#### Substituting \(R_3\) in \(R_0\):

Substitute \(R_3 = (a + b)R_0\) in the expression for \(R_0\):
\[ R_0 = (a + b)b((a + b)R_0)(a + b)^*a^* \]

This leads to a recursive structure, which simplifies as:
\[ R_0 = (a + b)b(a + b)^*(a + b)b(a + b)^*a^* \]

Thus, the final regular expression for the automaton is:
\[ R = (a + b)b(a + b)^*(a + b)b(a + b)^*a^* \]

This expression represents the language accepted by the automaton.

### Do you want more details or have any questions?

Here are five follow-up questions you might be interested in:

1. How would you modify the regular expression if the automaton had a different final state?
2. Can you explain the role of self-loops in the regular expression calculation?
3. How does Arden's Theorem simplify the process of finding regular expressions for automata?
4. What would the regular expression look like if the automaton had no loops?
5. How would the regular expression change if transitions were added or removed?

**Tip:** Arden's Theorem is very useful when working with linear equations derived from regular automata, especially when self-loops are present.

Learn how to construct regular expressions from automata using Arden's Theorem. This detailed solution covers the application of Arden's Theorem in transforming automaton transitions into regular expressions. Suitable for graduate-level study in automata theory and formal languages.

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