edited by
783 views

2 Answers

Best answer
3 votes
3 votes

Given Language is accepting all the strings which is not start with 0 and not having 2 consecutive 0's.

So, it can ∈ and 1 because both are not starting with 0.

L={∈,1,10,101,11,111................}

Here any string which is start with 0 and having consecutive will lead to dead state q3

edited by
0 votes
0 votes

(1+10) means we will reach final state either by 1 or by 10 now its star closure is given which includes empty string means initial state is also final state . and any amount of repetition of either 1 or 10 will also take to final state and even 1 followed by 10 or 10 followed by1 (any no of times) will also take us to final state hence its DFA should be like 

Related questions

1 votes
1 votes
2 answers
1
0 votes
0 votes
1 answer
2
0 votes
0 votes
2 answers
3
rohankrishan asked Jun 29, 2022
251 views
Example: 11110100000111 should be accepted. There are 6 zeros. 6 is divisble by 2 and 3. This machine required at least six states.
3 votes
3 votes
1 answer
4
all mix asked Jan 29, 2018
812 views
create a dfa with odd occurrence of sub string "abb" .and short cut for finding no.of states