DFA that recognizes number of 0 is multiple of 3 on input {0,1}

Finite Automata is known to finite state machine which works with transition from one state to another on every input alphabet, as per the defined rules.
In this article, strings that have number of zeros divisible by 3 are acceptable otherwise not acceptable. on the input alphabet ‘0’ and 1′.

Determine initial state.
Transition occurs on every input alphabets.
Determine weather the self loop should apply or not.
Mark final state.

Designing DFA step by step :

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

Step-2:
Create transition of input alphabet ‘0’ from state “A” to state “B”.

Step-3:
After one ‘0’ any number of 1’s can be present i.e, no ‘1’ or more than one ‘1’.For this put self loop of ‘1’ on state “B”.

Step-4:
Now create transition of input alphabet ‘0’ from state “B” to state “C” and after two 0’s any number of 1’s can be found in the string and for this put self loop of ‘1’ on initial state “C”.

Step-5:
Before transition of third ‘0’ we need to think about the logic so that after this transition the machine will accept string that have number of zeros divisible by 3.For this transit ‘o’ from state “C” to state “A”.AS third zero is reaching to state “A” so make state “A” be final state.

Transition table of above DFA:

States	Input (0)	Input (1)
—> A *	B	A
B	C	B
C	A	C

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

Transition rules of above DFA:

Implementing :

Python3

def checkStateA(n): 
      
    # check length of n  
    # is 0 then print 
    # string accepted 
    if(len(n)== 0): 
        print("string accepted") 
          
    # if 1 is found call function 
    # checkStateA otherwise if 0 
    # is found call function stateB 
    else:     
        if(n[0]=='1'): 
            checkStateA(n[1:]) 
        else: 
            stateB(n[1:]) 
   
   
def stateB(n): 
      
    # check length of n  
    # is 0 then print 
    # string not accepted 
    if(len(n)== 0): 
        print("string not accepted") 
          
    # if 1 is found call function 
    # stateB otherwise if 0 
    # is found call function stateC     
    else:     
        if(n[0]=='1'): 
            stateB(n[1:]) 
        else: 
            stateC(n[1:]) 
         
          
def stateC(n): 
      
    # check length of n  
    # is 0 then print 
    # string not accepted 
    if(len(n)== 0): 
        print("string not accepted") 
          
    # if 1 is found call function 
    # stateC otherwise if 0 
    # is found call function checkStateA  
    else:     
        if(n[0]=='1'): 
            stateC(n[1:]) 
        else: 
            checkStateA(n[1:]) 
          
# take string input 
n = input() 
  
# call checkStateA 
# to check the inputted string 
checkStateA(n) 

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My Personal Notes

Python3

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