ARDEN'S THEOREM FOR REGULAR EXPRESSIONS

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    Identities for Regular Expression:- Given R, P, L, Q as regular expressions, the following identities hold − ∅* = ε ε* = ε RR* = R*R R*R* = R* (R*)* = R* RR* = R*R (PQ)*P =P(QP)* (a+b)* = (a*b*)* = (a*+b*)* = (a+b*)* = a*(ba*)* R + ∅ = ∅ + R = R   (The identity for union) R ε = ε R = R   (The identity for concatenation) ∅ L = L ∅ = ∅   (The annihilator for concatenation) R + R = R   (Idempotent law) L (M + N) = LM + LN   (Left distributive law) (M + N) L = ML + NL   (Right distributive law) ε + RR* = ε + R*R = R* 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 equatio
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ARDEN'S THEOREM FOR REGULAR EXPRESSIONS

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 Identities for Regular Expression:-

Given R, P, L, Q as regular expressions, the following identities hold −

  • ∅* = ε
  • ε* = ε
  • RR* = R*R
  • R*R* = R*
  • (R*)* = R*
  • RR* = R*R
  • (PQ)*P =P(QP)*
  • (a+b)* = (a*b*)* = (a*+b*)* = (a+b*)* = a*(ba*)*
  • R + ∅ = ∅ + R = R (The identity for union)
  • R ε = ε R = R (The identity for concatenation)
  • ∅ L = L ∅ = ∅ (The annihilator for concatenation)
  • R + R = R (Idempotent law)
  • L (M + N) = LM + LN (Left distributive law)
  • (M + N) L = ML + NL (Right distributive law)
  • ε + RR* = ε + R*R = R*
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 (ε + P + P2 + P3 + …. )

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

Hence, proved.
Assumptions for Applying Arden’s Theorem

  • The transition diagram must not have NULL transitions
  • 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 + ε

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 = ∅

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

EXAMPLE:-

Construct a regular expression corresponding to the automata given below −

Finite Automata

Solution −

Here the initial state and final state is q1.

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

q1 = q1a + q3a + ε (ε 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 + ε

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

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

= q1(a + b(b + ab)*aa) + ε

= ε (a+ b(b + ab)*aa)*

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

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

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