Yes. The automaton will stay in the start state, which, for this automaton, is not a final state.
Here is recognize
in Java. Assume that the transition table has already been filled in.
. . . final int reject = -1 ; // number of the reject state int[numStates][128] transition ; // one row per state, // one column per character . . . public boolean recognize(String str) { int state = 0; // the current state int index = 0; // index of the current character char current; // the current character // continue while there are input characters // and the reject state has not been reached while ( index < str.length() && state != reject ) { current = str.charAt( index++ ) ; state = transition[state][current] ; } // if the last character leads to a final state, // the string is accepted. if ( index == str.length() && isFinal( state ) ) return true; else return false; }
States are represented by integers starting at 0.
The reject state is represented by -1.
The input string is contained in a String
object.
What is the value of "abacc".charAt(0)