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Construct DFA for Regular Expression a+aa*b in C
Design a Deterministic Finite Automata (DFA) for accepting the language L = (a+aa*b)* If the given string is accepted by DFA, then print “string is accepted”. Otherwise, print “string is rejected”.
Example 1
Input: Enter Input String aaaba Output: String Accepted.
Explanation − The given string is of the form (a+aa*b)* as the first character is a and it is followed by a or ab.
Example 2
Input: Enter Input String baabaab Output: String not Accepted.
The DFA for the given regular expression (a+aa*b) is −
Explanation −
If the first character is always a, then traverse the remaining string and check if any of the characters is a or ab.
If the first character in the string is ‘b’ then it prints “string is rejected”.
If there exists any character other than a or b while traversing, then it prints “entered value is wrong”.
Otherwise, print “string is accepted”.
Program
Following is the C program to construct DFA for Regular Expression (a+aa*b)* −
#include<stdio.h> #include<conio.h> #include<strings.h> void main() { int table[2][2],i,j,l,status=0,success; char input[100]; printf("To implementing DFA of language (a+aa*b)*
Enter Input String:”); table[0][0]=1; table[0][1]=-1; table[1][0]=1; table[1][1]=0; scanf("%s",input); l=strlen(input); for (i=0;i<l;i++) { if(input[i]!='a'&&input[i]!='b') { printf("The entered Value is wrong"); getch(); exit(0); } if(input[i]=='a') status=table[status][0]; else status=table[status][1]; if(status==-1) { printf("String not Accepted"); break; } } if(i==l) printf("String Accepted"); getch(); }
Output
When the above program is executed, it gives the following output −
Run 1: To implementing DFA of language (a+aa*b)* Enter Input String:cbsd The entered Value is wrong. Run 2: To implementing DFA of language (a+aa*b)* Enter Input String:abbababa String not Accepted. Run 3: To implementing DFA of language (a+aa*b)* Enter Input String:babbaab String not Accepted.