Found 691 Articles for Computer Science

Kleene's Theorem states the equivalence of the following three statements −A language accepted by Finite Automata can also be accepted by a Transition graph.A language accepted by a Transition graph can also be accepted by Regular Expression.A language accepted by Regular Expression can also be accepted by finite Automata.Kleene’s theorem proof part-1A language accepted by Finite Automata can also be accepted by a Transition graph.Consider an example Let L=aba over an alphabet {a, b}Third part of Kleene’s theoremA language accepted by Regular Expression can also be accepted by finite Automata.TheoremAny language that can be defined with RE can be accepted ... Read More

The complementation process in the deterministic finite automata (DFA) is explained below −Let’s take a DFA which is defined by (Q, Σ, δ, q0, F) and it accepts the language L1. Now, the DFA that accepts the language L2, where L2 = ̅L1, is defined as follows −         (Q, Σ, δ, q0, Q-F)The complement of a DFA is obtained by making the non-final states as final states and final states as non-final states.The language which is accepted by the complemented DFA L2 is the complement of language L1.ExamplesLet’s consider some examples to get the clarity on the ... Read More

The cross product method process in the deterministic finite automata (DFA) is explained below −Let a's DFA diagram has m number of states and b's DFA diagram has n number of states the cross product m x n will have mxn states.Languages represented by even number of ‘a’ and even number of ‘b’ are given below −L1 = {ε, baa, aa, aba, aab, aaaa, ... }L2 = {ε bb, abb, bab, bba, ...}After cross product we will find the DFA as mentioned below −As, L = {ab, aab, abb, aaab, ...}ExampleLet’s taken two DFAsEven number of a'sEven number of b'sThe ... Read More

The concatenation process in the deterministic finite automata (DFA) is explained below −If L1 and If L2 are two regular languages, their union L1 ∩ L2 will also be regular.For example, L1 = {an | n > O} and L2 = {bn | n > O}L3 = L1 ∩ L2 = {an ∩ bn | n > O} is also regular.In this property, we say that concatenation of two DFAs is also DFA.ProblemDesign a DFA over an alphabet {a, b} where the string starts with ‘a’ and ends with ‘b’.SolutionThere are two different types of languages are formed for a ... Read More

The union process in the deterministic finite automata (DFA) is explained below −If L1 and If L2 are two regular languages, their union L1 U L2 will also be regular.For example, L1 = {an | n > O} and L2 = {bn | n > O}L3 = L1 U L2 = {an U bn | n > O} is also regular.ProblemDesign a DFA over an alphabet {a, b} where the start and end are of different symbols.SolutionThere are two different types of languages are formed for a given condition −L1={ab, aab, abab, abb, …….}L1={ab, aab, abab, abb, …….}Here, L1= starts ... Read More

A Deterministic Finite automaton (DFA) is a 5-tuplesM=(Q, Σ, δ, q0, F)Where, Q : Finite set called states.Σ : Finite set called alphabets.δ : Q × Σ → Q is the transition function.q0 ϵ Q is the start or initial state.F : Final or accept state.ProblemDesign a finite automaton where the second symbol from the right hand side is ‘a’.SolutionThe language for a given string over an alphabet {a, b} is −L={aa, abaa, abbab, aaabab, ………}ExampleInput − aabaOutput − Not acceptedBecause the second letter from right hand side is not aInput − aabbabOutput − AcceptedIt is complicated to directly construct ... Read More

A Deterministic Finite automaton (DFA) is a 5-tuplesM=(Q, Σ, δ, q0, F)Where, Q : Finite set called states.Σ : Finite set called alphabets.δ : Q × Σ → Q is the transition function.q0 ϵ Q is the start or initial state.F : Final or accept state.ProblemConstruct a DFA machine accepting odd numbers of 1’s and even number of 0’s.SolutionDesign two separate machines for the two conditions over an alphabet Σ={0, 1}:DFA accepts only an odd number of 1’s.DFA accepts only even number of 0’s.Here, s1 = starts2=odd 1 or start 11s3= starts 11 accepted and stay theres4 = accept even ... Read More

Non-deterministic finite automata (NFA) also have five states which are same as DFA, but with different transition function, as shown follows −δ: Q X Σ → 2QWhere, Q : Finite set called states.Σ : Finite set called alphabets.δ : Q × Σ → Q is the transition function.q0 ϵ Q is the start or initial state.F : Final or accept state.ProblemConstruct NFA over an alphabet Σ={0, 1}.SolutionDesign two separate machines for the two conditions, as given below −NFA accepting only odd number of 1’sNFA accepting only even number of 0’sNFA accepting only odd number of 1’s over an alphabet Σ= ... Read More

A Deterministic Finite automaton (DFA) is a five tuplesM=(Q, Σ, δ, q0, F)Where, Q : Finite set called states.Σ : Finite set called alphabets.δ : Q × Σ → Q is the transition function.q0 ϵ Q is the start or initial state.F : Final or accept state.ProblemConstruct a DFA machine accepting odd numbers of 0’s or even numbers of 1’s.SolutionDesign two separate machines for the two conditions over an alphabet Σ={0, 1} −DFA accepting only odd number of 0’sDFA accepting only even number of 1’sDFA accepting only odd number of 1’s over an alphabet Σ={0, 1}    The language L= ... Read More

A Deterministic Finite automaton (DFA) is a 5-tuplesM=(Q, Σ, δ, q0, F)Where, Q : Finite set called states.Σ : Finite set called alphabets.δ : Q × Σ → Q is the transition function.q0 ϵ Q is the start or initial state.F : Final or accept state.ProblemConstruct DFA of a string with at least two 0’s and at least two 1’s.SolutionThe language generated based on the given condition over the alphabet Σ ={0, 1) is −L={0011, 001011, 0001010, 0011001, 010101, ……}The given language accepts at least two zero’s means it can accept two or more than two zero’s and at least ... Read More