The language for the given regular expression (RE) is as follows −L={ ε, aa, ab, ba, aaaa, ………}ExampleLet the regular expression be ((a+b)(a+b))*(a+b).Construct the Finite automata for the given regular expression.First, generate the language for the given Regular Expression −L={a, d, aaa, bbb, abb, bab, bba, ………..}This is the language ... Read More

To convert the regular expression to Finite Automata (FA) we can use the Subset method.Subset method is used to obtain FA from the given regular expression (RE).Step 1 − Construct a Transition diagram for a given RE using Non-deterministic finite automata (NFA) with ε moves.Step 2 − Convert NFA with ... Read More

The regular expression R= a+ba* divided into r1 and r2r1= a and r2= ba*Let us draw Non-deterministic finite automata (NFA) for r1 as given below −Now, we will go for r2 = ba *Divide r2 into r3 and r4, where, r3=b and r4=a*The NFA for r3 is as follows −The ... Read More

Induction is a powerful tool in mathematics. It is a way of proving propositions that hold for all natural numbers.Hypothesis − The formal proof can be using deductive proof and inductive proof. The deductive proof consists of sequence of statements given with logical reasoning in order to prove the first ... Read More

To convert the regular expression (RE) to Finite Automata (FA), we can use the Subset method.Subset method is used to obtain FA from the given RE.Step 1 − Construct a Transition diagram for a given RE by using Non-deterministic finite automata (NFA) with ε moves.Step 2 − Convert NFA with ... Read More

In order to understand the relationship between finite automata (FA) and regular expression (RE), we need to understand these terminologies. Let us begin by understanding what is a regular expression.Regular ExpressionRegular expression is the language which is used to describe the language and is accepted by finite automata. Regular expressions ... Read More

Problem 1Prove that (1+00*1)+(1+00*1)(0+10*1)*(0+10*)=0*1(0+10*1)*SolutionHere, we need to prove LHS=RHS (Left hand side = Right hand side)Let us solve first LHS(1+00*1)+(1+00*1)(0+10*1)*(0+10*)Take (1+00*1) as a common factor(1+00*1)( ε+(0+10*1)*(0+10*1)Where, (0+10*1)*(0+10*1). It is in the form of R*R where R=0+10*1As we know, (ε+R*R)=( ε+RR*)=R*Therefore, (1+00*1)((0+10*1)*)Taking 1 as common factor(ε+00*)1(0+10*1)*Apply ε+00*=0*0*1(0+10*1)*=RHSHence, the two regular expressions ... Read More

Regular expression is the language which is used to describe the language and is accepted by finite automata. Regular expressions are the most effective way to represent any language. Let Σ be an alphabet which denotes the input set.The regular expression over Σ can be defined as follows −Φ is ... Read More

Arden’s theorem helps in checking the equivalence of two regular expressions.Arden’s TheoremLet, P and Q be two regular expressions over the input set Σ. The regular expression R is given as follows −R=Q+RPThis has a unique solution as R=QP*.ProofLet, P and Q be the two regular expressions over the input ... Read More

The two regular expression’s P and Q are equivalent (denoted as P=Q) if and only if P represents the same set of strings as Q does.For showing the equivalence of two regular expressions we need to show some identities of regular expression’sLet P, Q and R be the regular expressions ... Read More