The number of states in the minimal DFA of the set of strings over {0, 1} which start with 000 or 111 is ________.

Open in App

Solution

the minimal DFA has 7 states, so 7 is the answer.

Suggest Corrections

Similar questions

W e d e f i n e t h e p o w e r s o f a s t r i n g ω o v e r t h e c o n c a t e n a t i o n o p e r a t i o n (*) a s f o l l o w s : ω^{0} = \in ω^{i + 1} = ω^{i} . ω (1^{3} + 1^{2019}) * + (1^{6} + 1^{2016}) * T h e n u m b e r o f s t a t e s i n t h e m i n i m a l D F A (o v e r t h e u n a r y a l p h a b e t (1}) o f t h e l a n g u a g e c o r r e s p o n d i n g t o t h e r e g u l a r e x p r e s s i o n i s e q u a l t o______.

L ((11^{*} 0 + 0) (0 + 1)^{*} 0^{*} 1^{*})

{0, 1}

w^{0} = ϵ

w^{' + 1} - w^{'} . w

The number of states in the minimal DFA (over the unary alphabet {1}) of the language corresponding to the regulare expression (13+12013)∗+(16+12022)∗ is equal to ______.

Join BYJU'S Learning Program