Programa para construir DFA para expresiones regulares C( A + B)+

Dada una string S , la tarea es diseñar un autómata finito determinista (DFA) para aceptar el lenguaje L = C (A + B)+ . Si DFA acepta la string proporcionada, imprima «Sí» . De lo contrario, escriba “No” .

Ejemplos:

Entrada: S = “CABABABAB”
Salida: Sí
Explicación: La string dada tiene la forma C(A + B)+ ya que el primer carácter es C y es seguido por A o B.

Entrada: S = “ABAB”
Salida: No

Enfoque: La idea es interpretar el lenguaje dado L = C (A + B)+ y para la expresión regular de la forma C(A + B)+, el siguiente es el Diagrama de Transición de Estado DFA :

Siga los pasos a continuación para resolver el problema:

• Si la string dada tiene una longitud menor que igual a 1 , imprima «No» .
• Si el primer carácter siempre es C , recorra la string restante y verifique si alguno de los caracteres es A o B .
• Si existe algún carácter que no sea A o B mientras se desplaza en el paso anterior, imprima «No» .
• De lo contrario, escriba «Sí» .

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find whether the given
// string is Accepted by the DFA
void DFA(string str, int N)
{
    // If n <= 1, then print No
    if (N <= 1) {
        cout << "No";
        return;
    }
 
    // To count the matched characters
    int count = 0;
 
    // Check if the first character is C
    if (str[0] == 'C') {
        count++;
 
        // Traverse the rest of string
        for (int i = 1; i < N; i++) {
 
            // If character is A or B,
            // increment count by 1
            if (str[i] == 'A' || str[i] == 'B')
                count++;
            else
                break;
        }
    }
    else {
 
        // If the first character
        // is not C, print -1
        cout << "No";
        return;
    }
 
    // If all characters matches
    if (count == N)
        cout << "Yes";
    else
        cout << "No";
}
 
// Driver Code
int main()
{
    string str = "CAABBAAB";
    int N = str.size();
    DFA(str, N);
 
    return 0;
}

// Java program for the above approach
import java.util.*;
class GFG
{
 
  // Function to find whether the given
  // String is Accepted by the DFA
  static void DFA(String str, int N)
  {
 
    // If n <= 1, then print No
    if (N <= 1)
    {
      System.out.print("No");
      return;
    }
 
    // To count the matched characters
    int count = 0;
 
    // Check if the first character is C
    if (str.charAt(0) == 'C')
    {
      count++;
 
      // Traverse the rest of String
      for (int i = 1; i < N; i++)
      {
 
        // If character is A or B,
        // increment count by 1
        if (str.charAt(i) == 'A' ||
            str.charAt(i) == 'B')
          count++;
        else
          break;
      }
    }
    else
    {
 
      // If the first character
      // is not C, print -1
      System.out.print("No");
      return;
    }
 
    // If all characters matches
    if (count == N)
      System.out.print("Yes");
    else
      System.out.print("No");
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    String str = "CAABBAAB";
    int N = str.length();
    DFA(str, N);
  }
}
 
// This code is contributed by 29AjayKumar

# Python3 program for the above approach
 
# Function to find whether the given
# is Accepted by the DFA
def DFA(str, N):
     
    # If n <= 1, then print No
    if (N <= 1):
        print("No")
        return
 
    # To count the matched characters
    count = 0
 
    # Check if the first character is C
    if (str[0] == 'C'):
        count += 1
 
        # Traverse the rest of string
        for i in range(1, N):
 
            # If character is A or B,
            # increment count by 1
            if (str[i] == 'A' or str[i] == 'B'):
                count += 1
            else:
                break
    else:
        # If the first character
        # is not C, print -1
        print("No")
        return
 
    # If all characters matches
    if (count == N):
        print("Yes")
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
    str = "CAABBAAB"
    N = len(str)
    DFA(str, N)
 
# This code is contributed by mohit kumar 29.

// C# program for the above approach
using System;
class GFG
{
 
  // Function to find whether the given
  // String is Accepted by the DFA
  static void DFA(string str, int N)
  {
 
    // If n <= 1, then print No
    if (N <= 1)
    {
      Console.Write("No");
      return;
    }
 
    // To count the matched characters
    int count = 0;
 
    // Check if the first character is C
    if (str[0] == 'C') {
      count++;
 
      // Traverse the rest of String
      for (int i = 1; i < N; i++) {
 
        // If character is A or B,
        // increment count by 1
        if (str[i] == 'A'
            || str[i] == 'B')
          count++;
        else
          break;
      }
    }
    else {
 
      // If the first character
      // is not C, print -1
      Console.Write("No");
      return;
    }
 
    // If all characters matches
    if (count == N)
      Console.Write("Yes");
    else
      Console.Write("No");
  }
 
  // Driver Code
  static public void Main()
  {
 
    string str = "CAABBAAB";
    int N = str.Length;
    DFA(str, N);
  }
}
 
// This code is contributed by Dharanendra L V

<script>
    // Javascript program for the above approach
     
      // Function to find whether the given
      // String is Accepted by the DFA
function  DFA(str,N) {
    // If n <= 1, then print No
    if (N <= 1)
    {
          document.write("No");
          return;
    }
  
    // To count the matched characters
    let count = 0;
  
    // Check if the first character is C
    if (str[0] == 'C')
    {
      count++;
  
      // Traverse the rest of String
      for (let i = 1; i < N; i++)
      {
  
        // If character is A or B,
        // increment count by 1
        if (str[i] == 'A' ||
            str[i] == 'B')
          count++;
        else
          break;
      }
    }
    else
    {
  
      // If the first character
      // is not C, print -1
      document.write("No");
      return;
    }
  
    // If all characters matches
    if (count == N)
      document.write("Yes");
    else
      document.write("No");
}
     
     
    // Driver Code
    let str = "CAABBAAB";
    let N = str.length;
    DFA(str, N);
 
 
// This code is contributed by patel2127
</script>

Yes

Complejidad temporal: O(N)
Espacio auxiliar: O(1)