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 or initial statement. The initial statement is called Hypothesis.Suppose there exists a k > 0 such that for any regular expression r where 0 < OP(r) < k, there exists an NFA- s such that L(M) = L(r). Furthermore, suppose that M has exactly one final state.Inductive StepLet r be ... 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 ε to NFA without ε.Step 3 − Convert the NFA to the equivalent Deterministic Finite Automata (DFA).Some basic RE are as follows −Case 1 − For a regular expression ‘a’, we can construct FA as shown below −Case 2 − For a regular expression ‘ab’ we can construct FA, as ... 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 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 a regular expression which denotes the empty set.ε is a regular expression and denotes the set { ε} and it ... 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 a regular expression which denotes the empty set.ε is a regular expression and denotes the set { ε} and it is called a null string.For each ‘a’ in Σ ‘a’ is a regular expression and denotes the set {a}.If r and s regular expressions denoting the language.L1 and l2 respectively ... 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 a regular expression which denotes the empty set.ε is a regular expression and denotes the set { ε} and it is called a null string.For each ‘a’ in Σ ‘a’ is a regular expression and denotes the set {a}.If r and s regular expressions denoting the language.L1 and l2 respectively ... Read More

Problem 1Write the regular expression for the language accepting all the strings containing any number of a's and b's.SolutionThe regular expression will be −r.e. = (a + b)*This will give the set as L = {E, a, aa, b, bb, ab, ba, aba, bab, .....}, any combination of a and b.The (a + b)* shows any combination with a and b even a null string.Problem 2Write the regular expression for the language starting with a but not having consecutive b's.SolutionThe regular expression has to be built for the language: L = {a, aba, aab, aba, aaa, abab, .....}The regular expression ... Read More

ProblemConstruct a minimum state DFA for the following automata −SolutionWe first construct a transition table for the given finite automata −States\inputs01q0q1q5q1q6q2*q2q0q2q3q2q6q4q7q5q5q2q6q6q6q4q7q6q2Q={q0, q1, q2, q3, q4, q5, q6, q7}Q01={q2} and Q02={q0, q1, q2, q3, q4, q5, q6, q7}S0={{q2} {q0, q1, q2, q3, q4, q5, q6, q7}}Consider the set {q0, q1, q2, q3, q4, q5, q6, q7}{q2} {q0, q1, q3, q5, q6, q7}{q2} {q0, q4, q6} {q1, q3, q5, q7}{q2} {q0, q4} {q6} {q1, q3, q5, q7}{q2}{q0, q4}{q6}{q1, q7}{q3, q5}The minimized state is as follows −M1=(Q1, Σ, δ1, q01, F1)Q1= {[q2], [q0, q4], [q6], [q1, q7], [q3, q5]}qo1= {[q0, q4]}F1= {[q2]}Transition TableNow ... Read More

A finite state machine (FSM) which has a set of states and two functions called the next-state and output function.The set of states correspond to all the possible combinations of the internal storage.If there are n bits of storage, there are 2n possible states.The next state function is a combinational logic function that given the inputs and the current state, determines the next state of the systemA Finite State Machine consists of the following −K states: S = {s1, s2, ... ,sk}, s1 is initial stateN inputs: I = {h, i2, ... ,in}M outputs: O = {o1, o2, . ,om}Next-state ... Read More

In this method, we try to remove all the ε-transitions from the given Non-deterministic finite automata (NFA) −The method is mentioned below stepwise −Step 1 − Find out all the ε-transitions from each state from Q. That will be called as ε-closure(qi) where, qi ∈Q.Step 2 − Then, 𝛿1 transitions can be obtained. The 𝛿1 transitions means an ε-closure on 𝛿 moves.Step 3 − Step 2 is repeated for each input symbol and for each state of given NFA.Step 4 − By using the resultant status, the transition table for equivalent NFA without ε can be built.NFA with ε to ... Read More

The ε closure(P) is a set of states which are reachable from state P on ε-transitions.The epsilon closure is as mentioned below −ε-closure (P) = P, where P ∈ QIf there exists ε-closure (P) = {q} and 𝛿(q, ε) =r then, ε-closure (P) = {q, r}ExampleFind ε-closure for the following Non-deterministic finite automata (NFA) with epsilon.Solutionε-closure (q0)= {q0, q1, q2}self state+ ε-reachable states.ε-closure (q1)= { q1, q2}q1 is self-state and q2 is a state obtained from q1 with epsilon input.ε-closure (q2)= {q2}Lets us consider an example to understand more clear about epsilon closure −Problem − find the number of epsilon ... Read More