Can DFA have epsilon / lambda transitions?

Question

I can not find anything positive. And the NFA with any epsilon transition is epsilon-NFA? Thanks.

Answer 1

DFA does not have epsilon transitions. If he had been with him, he could have passed from the current state to another state without any input data, i.e. nothing, even {} or phi. And as a definition, we know that the input must be from an input set. Hope this cleared your doubts ...

Answer 2

DFA must have a specific input character to transition from one state to another. Epsilon migration is not allowed in the DFA because it will change the DFA to NFA. For example, suppose you are in state Q1 and you have the transition (Q1, e) = Q2, in which case you can directly go to Q2 without applying any input or stay in state Q1, so that you have two choices in state Q1. In the case of DFA, you do not have to select selection criteria. This is why DFA does not have epsilon movements.

Answer 3

From the DFA definition, “Deterministic state machines are a machine that cannot move in a different state without receiving any input.” And since epsilon means nothing. DFA cannot move on epsilon moves.

While from the NFA definition "Non-deterministic finite state machines is a machine that can move through a different state without any input data." So the NFA can move along the epsilon path.