DFA_VG

Question

Number of state in DFA accepting a language L={anbn,1<= n <=3 } over {a,b} ……..

Comments

what is mistake in it can u plz tell @Magma @Gupta731 @Shaik Masthan

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Your diagram is not DFA. you have not given the transition on the final state

==>Also your diagram accepting the string ==>>  aababb

==>>>Remember we have a finite number of string just three. there should not be a loop in the DFA.

===>It is not the form of the a^n | n >= 0. here we can put loop.

I think you got my point.

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yes my dfa accept other string too that does not belong to language.

yes u r correct we cant put loop for finite no strings.

i have given transition after final state i.e to dead state its no looking clear,so its dfa.

got it thanks bro.

Answer 1

Qo = input state,   Q6 -- final state

Qo--Q1  using  'a'

Qo -- deadlock using 'b'

Q1 --- Q2 using 'a'

Q1--Q6 using  'b'

Q2 -- Q3 using 'a'

Q2 -- Q5  using 'b'

Q3 --  Q4 using 'b'

Q4 -- Q5 using 'b'

Q5 -- Q6 using 'b'

Follow above step plot the diagram you get 8 state ( Qo  to Q6  and deadlock state )

complete the each state for each and every input alphabet i.e. definition of DFA

Still any confusion comment below !!