Finite automata or finite state machine can be thought of as a severely restricted model of a computer.

Finite state machine that are acceptable only input alphabet**‘0’** and** ‘1’.**

- Determine the initial state.
- The transition occurs on every input alphabet.
- Determine whether the self-loop should apply or not.
- Mark’s final state.

**Designing DFA step by step:**

**Step -1:**

Make initial state “q_{0}” then it is the possibility that there would not be any ‘1’ but have only ‘0’ in the string which is acceptable because 1 is multiple by 3. So, in this case, any number of 0’s can be present here and for this self-loop of ‘0’ on initial state “q_{0}“.

**Step -2:**

Create transition of input alphabet ‘1’ from state “q_{0}” to state “q_{1}“.

**Step -3:**

After one ‘1’ any number of 0’s can be present i.e, no ‘0’ or more than one ‘1’. For this put self-loop ‘0’ on state “q_{1}“.

**Step -4:**

Now create transition of input alphabet ‘1’ from state “q_{1}” to state “q_{2}“.

**Step -5:**

After that two 1’s any number of 0’s can be found in the string and for this put self loop of 1’s on initial state “q_{2}“.

**Step -6:**

Before transition of third ‘1’ we need to think about the logic so that after the transition the machine will accept string the number of ones multiple by 3. For this transit ‘1’ from state “q_{2}” to state “q_{0}“. As third one is reach to state “q_{0}” so make state “q_{0}” be final state.

Transition table of above DFA:

In above table -> represents initial state and * represents final state. In this post, initial and final state is same which is final state.

**Code:**

**Main.java**

import java.util.Scanner; public class main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); String str = sc.nextLine(); checkStateA(str); } private static void checkStateA(String str) { if (str.length() == 0){ System.out.println("String Accepted"); } else { if(str.charAt(0)=='0') checkStateA(str.substring(1)); else stateB(str.substring(1)); } } private static void stateB(String str) { if(str.length() == 0){ System.out.println("string not accepted"); } else { if(str.charAt(0)=='0') stateB(str.substring(1)); else stateC(str.substring(1)); } } private static void stateC(String str) { if(str.length() == 0){ System.out.println("string not accepted"); } else { if(str.charAt(0)=='0') stateC(str.substring(1)); else checkStateA(str.substring(1)); } } }

**Output:**

Enter the String 10101 String Accepted

**Recommended:**

- Introduction to Finite Automata
- Deterministic Finite Automata (DFA)
- Number of 1’s is not multiple of 3 on {0,1} using DFA
- Automata Theory and Formal Languages
- Kleene Closure
- Recursive Definition of a Language
- Finite Representation of language