Theory Of Computation

Section B

Arden’s Theorem

In order to find out a regular expression of a Finite Automaton, we use Arden’s Theorem

along with the properties of regular expressions.

Statement:

Let P and Q be two regular expressions.

If P does not contain null string, then R = Q + RP has a unique solution that is R

= QP*

Proof:

R = Q + (Q + RP)P [After putting the value R = Q + RP]

= Q + QP + RPP

When we put the value of R recursively again and again, we get the following equation:

R = Q + QP + QP2 + QP3.....

R = Q (є + P + P2 + P3 + .... )

R = QP* [As P* represents (є + P + P2 + P3 + ....) ]

Hence, proved.

Assumptions for Applying Arden’s Theorem:

1. The transition diagram must not have NULL transitions

2. It must have only one initial state

Method

Step 1: Create equations as the following form for all the states of the DFA having

n states with initial state q1.

q1 = q1R11 + q2R21 + ... + qnRn1 + є

q2 = q1R12 + q2R22 + ... + qnRn2

................................

.................................

.................................

qn = q1R1n + q2R2n + ... + qnRnn

Rij represents the set of labels of edges from qi to qj, if no such edge exists, then Rij = Ø

Step 2: Solve these equations to get the equation for the final state in terms of Rij

Problem

Construct a regular expression corresponding to the automata given below:

Solution

Here the initial state is q2 and the final state is q1.

The equations for the three states q1, q2, and q3 are as follows:

q1 = q1a + q3a + є (є move is because q1 is the initial state0

q2 = q1b + q2b + q3b

q3 = q2a

Now, we will solve these three equations:

q2 = q1b + q2b + q3b

= q1b + q2b + (q2a)b (Substituting value of q3)

= q1b + q2(b + ab)

= q1b (b + ab)* (Applying Arden’s Theorem)

q1 = q1a + q3a + є

= q1a + q2aa + є (Substituting value of q3)

= q1a + q1b(b + ab*)aa + є (Substituting value of q2)

= q1(a + b(b + ab)*aa) + є

= є (a+ b(b + ab)*aa)*

= (a + b(b + ab)*aa)*

Hence, the regular expression is (a + b(b + ab)*aa)*.