Construct The Minimal Finite Automata That Accepts All Base 8 Numbers Which Are Divisible By ‘6’
(Binary Means Base 2, i.e Σ(0,1) Input Symbols Will Be 2 Only 0 & 1)
(Integer Means Base 10,i.e (0,1,2,3,4,5,6,7,8,9) Input Symbols Will Be 10 Only 0&1&2&3&4&5&6&7&8&9)
Here Given Input Symbols Are Base ‘8’, It Means OCTAT.
My Input Symbols Are Σ={0,1,2,3,4,5,6,7}
Example :- (22)8
= (2×81+2×80)10
=(16+2)10
=(18)
After Conversion Into Decimal (22)8 Is Divisible By ‘6’
Possible Remainders For ‘6’
Question Is Divisible By ‘6’
Means Remainder Should Be ‘0’
For ‘0’ We Gave ‘q0‘As State , ‘q0‘ Is Final State.
q1= q4
q2=q5
q1& q4 Are Same, I’ll Merge Both.
q1= q4
q2 & q5
Are Same, I’ll Merge Both.
q2=q5
No Commons States In This Table.
MAKE DFA USING THIS TABLE.
DFA Is Ready.
Now Take Any Number Which Has Base 8, Which Is Divisible By ‘6’
Example:- (22)8
If I Give ‘2’ To ‘q0‘ , I’ll Go To ‘q2‘
If I Give ‘2’ To ‘q2‘, I’ll Go To ‘q0‘(Final State)
Hence Satisfied.
TYPES OF PROBLEMS WE SOLVED TILL NOW.
- BINARY(INPUT SYMBOL) – (0,1)
- INTEGER (INPUT SYMBOL) – (0,9)
- BASE 8, OCTAT (INPUT SYMBOL) – (0-7)