Posts

Showing posts from October, 2021

ARDEN'S THEOREM FOR REGULAR EXPRESSIONS

Image
    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

SIMPLIFICATION OF CONTEXT-FREE GRAMMAR

  Simplification of Context Free Grammar Context Free Grammar  has recursive structure. The languages that are accepted with Context Free Grammar are called  Context Free Languages . Context Free Grammar has one condition for production rules, i.e.,  on the left-hand side of each rule, there must be only single variable, and on the right-hand side, there may be combination of variables and terminals including  ?. In Context Free Grammar, sometimes all the productions rules and symbols are not needed for the derivation to solve. Some productions rules are never used during derivation of any string. Elimination of these types of productions or symbols is called  Simplification of Context Free Grammar. We will use the various simple methods to simplify the given context free grammar without changing the resulting language. The following types of productions rules are never used during derivation of any string from context free grammar: Useless productions Null productions Unit productions

CONTEXT FREE GRAMMAR

Image
About C.F.G. :-   -> In   formal language   theory, a   context-free grammar   ( CFG ) is a   formal grammar   whose   production rules   are of the form {\displaystyle A\ \to \ \alpha } with  {\displaystyle A}  a  single   nonterminal  symbol, and  {\displaystyle \alpha }  a string of  terminals  and/or nonterminals ( {\displaystyle \alpha }  can be empty). A formal grammar is "context free" if its production rules can be applied regardless of the context of a nonterminal. No matter which symbols surround it, the single nonterminal on the left hand side can always be replaced by the right hand side. This is what distinguishes it from a  context-sensitive grammar . A formal grammar is essentially a set of production rules that describe all possible strings in a given formal language. Production rules are simple replacements. For example, the first rule in the picture, {\displaystyle \langle {\text{Stmt}}\rangle \to \langle {\text{Id}}\rangle =\langle {\text{Expr}}\rangl

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  j  such that a i  and a j  end in the same state. From this state, b i  

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 − 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 each input symbol. Step 2  

ARDEN'S THEOREM FOR REGULAR EXPRESSIONS

Image
  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