**DFA to Regular Expression-**

The two popular methods for converting a given DFA to its regular expression are-

- Arden’s Method
- State Elimination Method

In this article, we will discuss Arden’s Theorem.

**Arden’s Theorem-**

Arden’s Theorem is popularly used to convert a given DFA to its regular expression.

It states that-

Let P and Q be two regular expressions over ∑.

If P does not contain a null string ∈, then-

**R = Q + RP has a unique solution i.e. R = QP***

**Conditions-**

To use Arden’s Theorem, following conditions must be satisfied-

- The transition diagram must not have any ∈ transitions.
- There must be only a single initial state.

**Steps-**

To convert a given DFA to its regular expression using Arden’s Theorem, following steps are followed-

**Step-01:**

- Form a equation for each state considering the transitions which comes towards that state.
- Add ‘∈’ in the equation of initial state.

**Step-02:**

Bring final state in the form R = Q + RP to get the required regular expression.

**Important Notes-**

**Note-01:**

Arden’s Theorem can be used to find a regular expression for both DFA and NFA.

**Note-02:**

If there exists multiple final states, then-

- Write a regular expression for each final state separately.
- Add all the regular expressions to get the final regular expression.

**Also Read-****State Elimination Method**

**PRACTICE PROBLEMS BASED ON CONVERTING DFA TO REGULAR EXPRESSION-**

**Problem-01:**

Find regular expression for the following DFA using Arden’s Theorem-

**Solution-**

**Step-01:**

Form a equation for each state-

- A = ∈ + B.1 ……(1)
- B = A.0 ……(2)

**Step-02:**

Bring final state in the form R = Q + RP.

Using (1) in (2), we get-

B = (∈ + B.1).0

B = ∈.0 + B.1.0

B = 0 + B.(1.0) ……(3)

Using Arden’s Theorem in (3), we get-

B = 0.(1.0)*

Thus, Regular Expression for the given DFA = 0(10)*

**Problem-02:**

Find regular expression for the following DFA using Arden’s Theorem-

**Solution-**

**Step-01:**

Form a equation for each state-

- q1 = ∈ ……(1)
- q2 = q1.a ……(2)
- q3 = q1.b + q2.a + q3.a …….(3)

**Step-02:**

Bring final state in the form R = Q + RP.

Using (1) in (2), we get-

q2 = ∈.a

q2 = a …….(4)

Using (1) and (4) in (3), we get-

q3 = q1.b + q2.a + q3.a

q3 = ∈.b + a.a + q3.a

q3 = (b + a.a) + q3.a …….(5)

Using Arden’s Theorem in (5), we get-

q3 = (b + a.a)a*

Thus, Regular Expression for the given DFA = (b + aa)a*

**Problem-03:**

Find regular expression for the following DFA using Arden’s Theorem-

**Solution-**

**Step-01:**

Form a equation for each state-

- q1 = ∈ + q1.b + q2.a ……(1)
- q2 = q1.a + q2.b ……(2)

**Step-02:**

Bring final state in the form R = Q + RP.

Using Arden’s Theorem in (2), we get-

q2 = q1.a.b* …….(3)

Using (3) in (1), we get-

q1 = ∈ + q1.b + q1.a.b*.a

q1 = ∈ + q1.(b + a.b*.a) …….(4)

Using Arden’s Theorem in (4), we get-

q1 = ∈.(b + a.b*.a)*

q1 = (b + a.b*.a)*

Thus, Regular Expression for the given DFA = (b + a.b*.a)*

**Problem-04:**

Find regular expression for the following DFA using Arden’s Theorem-

**Solution-**

**Step-01:**

Form a equation for each state-

- q1 = ∈ + q1.a + q3.a ……(1)
- q2 = q1.b + q2.b + q3.b ……(2)
- q3 = q2.a …….(3)

**Step-02:**

Bring final state in the form R = Q + RP.

Using (3) in (2), we get-

q2 = q1.b + q2.b + q2.a.b

q2 = q1.b + q2.(b + a.b) …….(4)

Using Arden’s Theorem in (4), we get-

q2 = q1.b.(b + a.b)* …….(5)

Using (5) in (3), we get-

q3 = q1.b.(b + a.b)*.a …….(6)

Using (6) in (1), we get-

q1 = ∈ + q1.a + q1.b.(b + a.b)*.a.a

q1 = ∈ + q1.(a + b.(b + a.b)*.a.a) …….(7)

Using Arden’s Theorem in (7), we get-

q1 = ∈.(a + b.(b + a.b)*.a.a)*

q1 = (a + b.(b + a.b)*.a.a)*

Thus, Regular Expression for the given DFA = (a + b(b + ab)*aa)*

**Also Read-****Minimization of DFA**

To gain better understanding about Arden’s Theorem,

**Next Article-****Non-Deterministic Finite Automata**

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