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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 + ...
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EQUIVALENCE OF TWO FINITE AUTOMATAs

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An Automaton is a machine that has a finite number of states. Any Two Automaton is said to be equivalent if both accept exactly the same set of input strings.  Two Automaton are equivalent if they satisfy the following conditions :  1. The initial and final states of both the automatons must be same. 2. Every pair of states chosen is from a different automaton only. 3. While combining the states with the input alphabets, the pair results must be either both final states or intermediate states.(i.e both should lie either in the final state or in the non-final state). 4. If the resultant pair has different types of states, then it will be non-equivalent. (i.e. One lies in the final state and the other lies in the intermediate state). Method The method for comparing two FA’s is explained below − Let M and M1 be the two FA’s and Σ be a set of input strings. Step 1  − Construct a transition table that has pairwise entries (q, q 1 ) where q ∈ M and q 1  ∈ M 1  for eac...
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