ARDEN'S THEOREM FOR REGULAR EXPRESSIONS

Image
    Identities for Regular Expression:- Given R, P, L, Q as regular expressions, the following identities hold βˆ’ βˆ…* = Ξ΅ Ξ΅* = Ξ΅ 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 + βˆ… = βˆ… + R = R   (The identity for union) R Ξ΅ = Ξ΅ R = R   (The identity for concatenation) βˆ… L = L βˆ… = βˆ…   (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) Ξ΅ + RR* = Ξ΅ + 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  βˆ’ 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  βˆ’ R = Q + (Q + RP)P  [After putting the value R = Q + RP] = Q + ...
Post a Comment

About Us

 

Hello Reader, Welcome to my Blog. This Blog is all about Cafe Foods and their types. You will get full insights of benefits and makings of the beverages. I hope you will enjoy the Blog.

Author : Mohit

Contact No. 7976933871

Contact Mail:- mohitbhatia208@gmail.com

Comments

Post a Comment

Popular posts from this blog

PIGEONHOLE PRINCIPLE AND APPLICATIONS OF PUMPING LEMMA

Image
  Pigeonhole principle for regular language:- a hole or small recess for pigeons to nest a small open compartment (as in a desk or cabinet) for keeping letters or documents a neat category which usu. fails to reflect actual complexities. pigeonhole principle:   if  n  objects are put into  m  containers, where  n  >  m , then at least one container must hold more than one object. The pigeonhole can be used to prove that certain infinite languages are not regular. (Remember, any finite language  is  regular.) As we have informally observed, dfas "can't count." This can be shown formally by using the pigeonhole principle. As an example , we show that L = {a b :  n  > 0} is not regular. The proof is by contradiction. Suppose L is regular. There are an infinite number of values of  n  but M(L) has only a finite number of states. By the pigeonhole principle, there must be distinct values of  i  and...
Read more

EQUIVALENCE OF TWO FINITE AUTOMATAs

Image
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 βˆ’ Let M and M1 be the two FA’s and Ξ£ be a set of input strings. Step 1  βˆ’ Construct a transition table that has pairwise entries (q, q 1 ) where q ∈ M and q 1  βˆˆ M 1  for eac...
Read more

ARDEN'S THEOREM FOR REGULAR EXPRESSIONS

Image
    Identities for Regular Expression:- Given R, P, L, Q as regular expressions, the following identities hold βˆ’ βˆ…* = Ξ΅ Ξ΅* = Ξ΅ 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 + βˆ… = βˆ… + R = R   (The identity for union) R Ξ΅ = Ξ΅ R = R   (The identity for concatenation) βˆ… L = L βˆ… = βˆ…   (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) Ξ΅ + RR* = Ξ΅ + 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  βˆ’ 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  βˆ’ R = Q + (Q + RP)P  [After putting the value R = Q + RP] = Q + ...
Read more