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Module 1: Introduction & Regular Languages

20232m

If PP and QQ are regular expressions (PP is not null), then R=Q+RPR = Q + RP has the unique solution:

(i) R=QPR = QP
(ii) R=PQ∗R = PQ^*
(iii) R=QP∗R = QP^*
(iv) R=QP∗R = QP^*

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Answer

Correct Option: (iii) R=QP∗R = QP^* (Arden’s Theorem)

Explanation

Arden’s Theorem: If PP does not contain ϵ\epsilon, then equation R=Q+RPR = Q + RP has unique solution R=QP∗R = QP^*.

Proof: R=Q+(Q+RP)P=Q(ϵ+P)+RP2=⋯=Q(ϵ+P+P2+⋯+Pk)+RPk+1R = Q + (Q + RP)P = Q(\epsilon + P) + RP^2 = \dots = Q(\epsilon + P + P^2 + \dots + P^{k}) + RP^{k+1}. As k→∞k \to \infty, RPk+1RP^{k+1} contributes no finite strings if language, so R=QP∗R = QP^*.

  • (i) QPQP missing star → incomplete.
  • (ii) PQ∗PQ^* swaps order (concatenation not commutative).
  • (iv) duplicate of (iii).

Condition: PP not null ($\epsilon
otin P$) ensures uniqueness.

Final Answer

R=QP∗R = QP^* → Option (iii) (also (iv) if duplicate).

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