Program to build a DFA that accepts strings starting and ending with different character


Prerequisite: Deterministic Finite Automata
Given a string str consists of characters ‘a’ & ‘b’. The task is to check whether string str starts and ends with different characters or not. If it does, print ‘YES’ with state transitions, else print ‘NO’.
Examples:

Input: ababab
Output: YES
Explanation:
The string “ababab” is starting with ‘a’ and ends with ‘b’

Input : ababa
Output : NO
Explanation:
The string “ababab” is starting with ‘a’ and ends with ‘a’



In DFA, there is no concept of memory, therefore we have to check the string character by character, beginning with the 0th character. The input set of characters for the problem is {a, b}. For a DFA to be valid, there must a transition rule defined for each symbol of the input set at every state to a valid state.

DFA Machine: For the above problem statement build a DFA machine. It is similar to a flowchart with various states and transitions. DFA machine corresponding to the above problem is shown below, Q2 and Q4 are the final states:

Explanation:
Suppose the first character in the input string is ‘a’, then on reading ‘a’, the control will shift to the upper branch of the machine. Now, it is defined that the string must not end with an ‘a’ to be accepted. At state Q1, if again ‘a’ comes, it keeps circling at the same state because for the machine the last read character might be the last character of the string. If it gets a ‘b’, then it can go to the final state, since a string ending in ‘b’ is acceptable in this case, so it moves to state Q2. Here, if it gets an ‘a’, it again enters the non-final state else for consecutive ‘b’s, it keeps circling in the final state.
The same is in the case when the first character is detected as ‘b’.

Approach:

  1. Define the minimum number of states required to make the state diagram. Use functions to various states.
  2. List all the valid transitions. Each state must have a transition for every valid symbol.
  3. Define the final states by applying the base condition.
  4. Define all the state transitions using state function calls.
  5. Define a returning condition for the end of the string.

For the given DFA Machine, the specifications are as follows:

  1. Q0, Q1, Q2, Q3, Q4 are the defined states.
  2. a and b are valid symbols. Each state has a transition defined for a and b.
  3. Q2 and Q4 are defined as the final state. If the string input ends at any of these states, it is accepted else rejected.
  4. Suppose at state Q0, if ‘a’ comes, the function call is made to Q1. If ‘b’ comes, the function call is made to Q3.
  5. If by following the process, the program reaches the end of the string, the output is made according to the state the program is at.

Below is the implementation of the above approach:

CPP

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// CPP Program to DFA that accepts
// string if it starts and end with
// same character
#include <bits/stdc++.h>
using namespace std;
  
// various states of DFA machine
// are defined using functions.
bool q1(string, int);
bool q2(string, int);
bool q3(string, int);
bool q4(string, int);
  
// vector to store state transition
vector<string> state_transition;
  
// end position is checked using string
// length value.
// q0 is the starting state.
// q2 and q4 are intermediate states.
// q1 and q3 are final states.
  
bool q1(string s, int i)
{
    state_transition.push_back("q1");
    if (i == s.length()) {
        return false;
    }
  
    // state transitions
    // a takes to q1, b takes to q2
    if (s[i] == 'a')
        return q1(s, i + 1);
    else
        return q2(s, i + 1);
}
  
bool q2(string s, int i)
{
    state_transition.push_back("q2");
    if (i == s.length()) {
        return true;
    }
  
    // state transitions
    // a takes to q1, b takes to q2
    if (s[i] == 'a')
        return q1(s, i + 1);
    else
        return q2(s, i + 1);
}
  
bool q3(string s, int i)
{
    state_transition.push_back("q3");
    if (i == s.length()) {
        return false;
    }
  
    // state transitions
    // a takes to q4, 1 takes to q3
    if (s[i] == 'a')
        return q4(s, i + 1);
    else
        return q3(s, i + 1);
}
  
bool q4(string s, int i)
{
    state_transition.push_back("q4");
    if (i == s.length()) {
        return true;
    }
  
    // state transitions
    // a takes to q4, b takes to q3
    if (s[i] == 'a')
        return q4(s, i + 1);
    else
        return q3(s, i + 1);
}
  
bool q0(string s, int i)
{
    state_transition.push_back("q0");
    if (i == s.length()) {
        return false;
    }
  
    // state transitions
    // a takes to q1, b takes to q3
    if (s[i] == 'a')
        return q1(s, i + 1);
    else
        return q3(s, i + 1);
}
  
int main()
{
    string s = "ababab";
  
    // all state transitions are printed.
    // if string is acceptable, print YES.
    // else NO is printed
    bool ans = q0(s, 0);
    if (ans) {
        cout << "YES" << endl;
  
        // print transition state of given
        // string str
        for (auto& it : state_transition) {
            cout << it << ' ';
        }
    }
    else
        cout << "NO" << endl;
    return 0;
}

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Java

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// Java Program to DFA that accepts 
// string if it starts and end with 
// same character 
import java.util.*;
  
class GFG 
{
      
    // vector to store state transition 
    static Vector state_transition = new Vector(); 
      
    // end position is checked using string 
    // length value. 
    // q0 is the starting state. 
    // q2 and q4 are intermediate states. 
    // q1 and q3 are final states. 
      
    static boolean q1(String s, int i) 
    
        state_transition.add("q1"); 
        if (i == s.length()) 
        
            return false
        
      
        // state transitions 
        // a takes to q1, b takes to q2 
        if (s.charAt(i) == 'a'
            return q1(s, i + 1); 
        else
            return q2(s, i + 1); 
    
      
    static boolean q2(String s, int i) 
    
        state_transition.add("q2"); 
        if (i == s.length()) 
        
            return true
        
      
        // state transitions 
        // a takes to q1, b takes to q2 
        if (s.charAt(i) == 'a'
            return q1(s, i + 1); 
        else
            return q2(s, i + 1); 
    
      
    static boolean q3(String s, int i) 
    
        state_transition.add("q3"); 
        if (i == s.length()) 
        
            return false
        
      
        // state transitions 
        // a takes to q4, 1 takes to q3 
        if (s.charAt(i) == 'a'
            return q4(s, i + 1); 
        else
            return q3(s, i + 1); 
    
      
    static boolean q4(String s, int i) 
    
        state_transition.add("q4"); 
        if (i == s.length())
        
            return true
        
      
        // state transitions 
        // a takes to q4, b takes to q3 
        if (s.charAt(i) == 'a'
            return q4(s, i + 1); 
        else
            return q3(s, i + 1); 
    
      
    static boolean q0(String s, int i) 
    
        state_transition.add("q0"); 
        if (i == s.length()) 
        
            return false
        
      
        // state transitions 
        // a takes to q1, b takes to q3 
        if (s.charAt(i) == 'a'
            return q1(s, i + 1); 
        else
            return q3(s, i + 1); 
    
      
    // Driver code
    public static void main (String[] args)
    
        String s = "ababab"
      
        // all state transitions are printed. 
        // if string is acceptable, print YES. 
        // else NO is printed 
        boolean ans = q0(s, 0); 
        if (ans == true)
        
            System.out.println("YES"); 
      
            // print transition state of given 
            // string str 
            for(int index = 0; index < state_transition.size(); index++) 
            { //(auto& it : ) { 
                System.out.print((String)state_transition.get(index) + ' '); 
            
        
        else
            System.out.println("NO"); 
    
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 Program to DFA that accepts
# if it starts and end with
# same character
  
  
# vector to store state transition
state_transition = []
  
# end position is checked using string
# length value.
# q0 is the starting state.
# q2 and q4 are intermediate states.
# q1 and q3 are final states.
def q1(s, i):
  
    state_transition.append("q1")
    if (i == len(s)):
        return False
  
    # state transitions
    # a takes to q1, b takes to q2
    if (s[i] == 'a'):
        return q1(s, i + 1)
    else:
        return q2(s, i + 1)
  
def q2(s, i):
  
    state_transition.append("q2")
    if (i == len(s)):
        return True
  
    # state transitions
    # a takes to q1, b takes to q2
    if (s[i] == 'a'):
        return q1(s, i + 1)
    else:
        return q2(s, i + 1)
  
def q3(s, i):
  
    state_transition.append("q3")
    if (i == len(s)):
        return False
  
    # state transitions
    # a takes to q4, 1 takes to q3
    if (s[i] == 'a'):
        return q4(s, i + 1)
    else:
        return q3(s, i + 1)
  
def q4(s, i):
  
    state_transition.append("q4")
    if (i == len(s)):
        return True
  
    # state transitions
    # a takes to q4, b takes to q3
    if (s[i] == 'a'):
        return q4(s, i + 1)
    else:
        return q3(s, i + 1)
  
def q0(s, i):
  
    state_transition.append("q0")
    if (i == len(s)):
        return False
  
    # state transitions
    # a takes to q1, b takes to q3
    if (s[i] == 'a'):
        return q1(s, i + 1)
    else:
        return q3(s, i + 1)
  
s = "ababab"
  
# all state transitions are printed.
# if is acceptable, print YES.
# else NO is printed
ans = q0(s, 0)
if (ans):
    print("YES")
  
    # print transition state of given
    # str
    for it in state_transition:
        print(it, end = " ")
  
else:
    print("NO")
  
# This code is contributed by mohit kumar 29

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Output:

YES
q0 q1 q2 q1 q2 q1 q2

Time Complexity: O(n) where a string of length n requires traversal through n states.



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Improved By : mohit kumar 29, AnkitRai01