Program to construct a DFA which accepts the language having all ‘a’ before all ‘b’

Last Updated : 14 Dec, 2021

Given a string S, the task is to design a Deterministic Finite Automata (DFA) for accepting the language L = {a^Nb^M| N ≥ 0, M ≥ 0, N+M ≥ 1}. , i.e., a regular language L such that all ‘a’ occur before the first occurrence of ‘b’ {a, ab, aab, bb…, }. If the given string follows the given language L, then print “Accepted”. Otherwise, print “Not Accepted”.

Examples

Input: S = “aabbb”
Output: Accepted
Explanation: All the ‘a’ come before ‘b’ s.

Input: S = “ba”
Output: Not Accepted
Explanation: ‘b’ comes before ‘a’.

Input: S = “aaa”
Output: Accepted
Explanation: Note that ‘b’ does not need to occur in S

Input: S = “b”
Output: Accepted
Explanation: Note that ‘a’ does not need to occur in S

Approach: The problem can be accepted only when the following cases are met:

All the characters can be ‘a’.
All the characters can be ‘b’.
All the ‘b’ come occur after all the ‘a’.
There is at least one character in the string.

This can be better visualized with the help of the state transition diagram of the DFA

State Transition Diagram:

State Transition Diagram of the above DFA

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;  
 
// Function for state zero Q0
int startStateQ0(char s) {
     
    int state;
    if (s == 'a')
        state = 1;
    else if (s == 'b')
        state = 2;
    else
        state = -1;
 
    return state;
}
         
// Function for first state Q1
int firstStateQ1(char s) {
     
    int state;
    if (s == 'a')
        state = 1;
    else if (s == 'b')
        state = 2;
    else
        state = -1;
         
    return state;
}
         
// Function for second state Q2
int secondStateQ2(char s) {
     
    int state;
    if (s == 'b')
        state = 2;
    else if (s == 'a')
        state = 3;
    else
        state = -1;
         
    return state;
}
         
// Function for third state Q3
int thirdStateQ3(char s) {
     
    int state = 3;
    return state;
}
         
// Function to check 
// if the string is accepted or not
int isAcceptedString(string String) {
     
    int l = String.length();
 
    // dfa tells the number associated
    // with the present dfa = state
    int state = 0;
    for (int i = 0; i < l; i++) {
        if (state == 0)
            state = startStateQ0(String[i]);
        else if (state == 1)
            state = firstStateQ1(String[i]);
        else if (state == 2)
            state = secondStateQ2(String[i]);
        else if (state == 3)
            state = thirdStateQ3(String[i]);
        else
            return 0;
    }
    if (state == 1 || state == 2)
        return 1;
    else
        return 0;
}
 
int main() {
     
    string String = "ba";
    if (isAcceptedString(String))
        cout << "ACCEPTED";
    else
        cout << "NOT ACCEPTED";
}     
 
// This code is contributed by Samim Hossain Mondal.

Java

// Java Program to implement
// the above approach
import java.util.*;
public class GFG {
 
    // Function for state zero Q0
    static int startStateQ0(char s)
    {
 
        int state;
        if (s == 'a')
            state = 1;
        else if (s == 'b')
            state = 2;
        else
            state = -1;
 
        return state;
    }
 
    // Function for first state Q1
    static int firstStateQ1(char s)
    {
 
        int state;
        if (s == 'a')
            state = 1;
        else if (s == 'b')
            state = 2;
        else
            state = -1;
 
        return state;
    }
 
    // Function for second state Q2
    static int secondStateQ2(char s)
    {
 
        int state;
        if (s == 'b')
            state = 2;
        else if (s == 'a')
            state = 3;
        else
            state = -1;
 
        return state;
    }
 
    // Function for third state Q3
    static int thirdStateQ3(char s)
    {
 
        int state = 3;
        return state;
    }
 
    // Function to check
    // if the string is accepted or not
    static int isAcceptedString(String Str)
    {
 
        int l = Str.length();
 
        // dfa tells the number associated
        // with the present dfa = state
        int state = 0;
        for (int i = 0; i < l; i++) {
            if (state == 0)
                state = startStateQ0(Str.charAt(i));
            else if (state == 1)
                state = firstStateQ1(Str.charAt(i));
            else if (state == 2)
                state = secondStateQ2(Str.charAt(i));
            else if (state == 3)
                state = thirdStateQ3(Str.charAt(i));
            else
                return 0;
        }
        if (state == 1 || state == 2)
            return 1;
        else
            return 0;
    }
 
    public static void main(String args[])
    {
 
        String Str = "ba";
        if (isAcceptedString(Str) != 0)
            System.out.println("ACCEPTED");
        else
            System.out.println("NOT ACCEPTED");
    }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Function for state zero Q0
def startStateQ0(s):
    if (s == 'a'):
        state = 1
    elif (s == 'b'):
        state = 2
    else:
        state = -1
    return state
 
# Function for first state Q1
def firstStateQ1(s):
    if (s == 'a'):
        state = 1
    elif (s == 'b'):
        state = 2
    else:
        state = -1
    return state
 
# Function for second state Q2
def secondStateQ2(s):
    if (s == 'b'):
        state = 2
    elif (s == 'a'):
        state = 3
    else:
        state = -1
    return state
 
# Function for third state Q3
def thirdStateQ3(s):
    state = 3
    return state
 
#Function to check 
#if the string is accepted or not
def isAcceptedString(String):
    l = len(String)
 
    # dfa tells the number associated
    # with the present dfa = state
    state = 0
    for i in range(l):
        if (state == 0):
            state = startStateQ0(String[i])
        elif (state == 1):
            state = firstStateQ1(String[i])
        elif (state == 2):
            state = secondStateQ2(String[i])
        elif (state == 3):
            state = thirdStateQ3(String[i])
        else:
            return 0
    if(state == 1 or state == 2):
        return 1
    else:
        return 0
 
# Driver code
if __name__ == "__main__":
 
    String = "ba"
    if (isAcceptedString(String)):
        print("ACCEPTED")
    else:
        print("NOT ACCEPTED")

C#

// C# Program to implement
// the above approach
using System;
class GFG {
 
    // Function for state zero Q0
    static int startStateQ0(char s)
    {
 
        int state;
        if (s == 'a')
            state = 1;
        else if (s == 'b')
            state = 2;
        else
            state = -1;
 
        return state;
    }
 
    // Function for first state Q1
    static int firstStateQ1(char s)
    {
 
        int state;
        if (s == 'a')
            state = 1;
        else if (s == 'b')
            state = 2;
        else
            state = -1;
 
        return state;
    }
 
    // Function for second state Q2
    static int secondStateQ2(char s)
    {
 
        int state;
        if (s == 'b')
            state = 2;
        else if (s == 'a')
            state = 3;
        else
            state = -1;
 
        return state;
    }
 
    // Function for third state Q3
    static int thirdStateQ3(char s)
    {
 
        int state = 3;
        return state;
    }
 
    // Function to check
    // if the string is accepted or not
    static int isAcceptedString(string Str)
    {
 
        int l = Str.Length;
 
        // dfa tells the number associated
        // with the present dfa = state
        int state = 0;
        for (int i = 0; i < l; i++) {
            if (state == 0)
                state = startStateQ0(Str[i]);
            else if (state == 1)
                state = firstStateQ1(Str[i]);
            else if (state == 2)
                state = secondStateQ2(Str[i]);
            else if (state == 3)
                state = thirdStateQ3(Str[i]);
            else
                return 0;
        }
        if (state == 1 || state == 2)
            return 1;
        else
            return 0;
    }
 
    public static void Main()
    {
 
        string Str = "ba";
        if (isAcceptedString(Str) != 0)
            Console.Write("ACCEPTED");
        else
            Console.Write("NOT ACCEPTED");
    }
}
 
// This code is contributed by ukasp.

Javascript

<script>
 
       // JavaScript Program to implement
       // the above approach 
 
       // Function for state zero Q0
       function startStateQ0(s) {
           if (s == 'a')
               state = 1
           else if (s == 'b')
               state = 2
           else
               state = -1
           return state
       }
        
       // Function for first state Q1
       function firstStateQ1(s) {
           if (s == 'a')
               state = 1
           else if (s == 'b')
               state = 2
           else
               state = -1
           return state
       }
        
       // Function for second state Q2
       function secondStateQ2(s) {
           if (s == 'b')
               state = 2
           else if (s == 'a')
               state = 3
           else
               state = -1
           return state
       }
        
       // Function for third state Q3
       function thirdStateQ3(s) {
           state = 3
           return state
       }
        
       // Function to check 
       // if the string is accepted or not
       function isAcceptedString(String) {
           l = String.length;
 
           // dfa tells the number associated
           // with the present dfa = state
           state = 0
           for (let i = 0; i < l; i++) {
               if (state == 0)
                   state = startStateQ0(String[i])
               else if (state == 1)
                   state = firstStateQ1(String[i])
               else if (state == 2)
                   state = secondStateQ2(String[i])
               else if (state == 3)
                   state = thirdStateQ3(String[i])
               else
                   return 0
           }
           if (state == 1 || state == 2)
               return 1
           else
               return 0
       }
 
       let String = "ba"
       if (isAcceptedString(String))
           document.write("ACCEPTED")
       else
           document.write("NOT ACCEPTED")
 
   // This code is contributed by Potta Lokesh
   </script>

Output

NOT ACCEPTED

Time Complexity: O(N) where N is the length of the string
Auxiliary Space: O(1)

Construct DFA with Σ = {a, b} and Accept All String of Length at Least 2

ashutosh deshwal

Improve

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Practice Tags :

Program to construct a DFA which accepts the language having all ‘a’ before all ‘b’

C++

Java

Python3

C#

Javascript

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