Can´t find anything affirmative about it. And a NFA with any epsilon transition is a epsilon-NFA ?
Thanks.
Can´t find anything affirmative about it. And a NFA with any epsilon transition is
Share
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
Can´t find anything affirmative about it. And a NFA with any epsilon transition is a epsilon-NFA ?
Thanks.
DFA doesn’t have epsilon transitions.If it had it, it could transit from current state to other state without any input i.e. with nothing , not even {} or phi. And as definition , we know that the input must be from the input set.
Hope this cleared your doubt …