How to transform a non-deterministic finite automaton (with or without empty moves) in a deterministic finite automaton?
Please, you have an example
Sorry, I am not entirely sure what you mean by movements then. An automata consists of a 5-tuple, $\displaystyle (Q, \Sigma, T, \iota, F)$ where $\displaystyle Q$ is the set of states, $\displaystyle \Sigma$ is your alphabet, $\displaystyle T: Q \times \Sigma \rightarrow \mathcal{P}(Q)$ is your transition function (lines), $\displaystyle \iota \in Q$ is your start state and $\displaystyle F \subseteq Q$ is the set of terminating sets.
Which of these are you wanting to be empty, if any?
This should be in your lecture notes:
Let $\displaystyle EM = (Q, \Sigma, q_0, \delta _{EM}, F_{EM})$ be a non-deterministic automaton with epsilon-travels ("empty movements")
Then the equivalent non-deterministic automaton without epsilon travels is $\displaystyle ND = (Q, \Sigma, q_0, \delta _{ND}, F_{ND})$ where:
$\displaystyle \forall q \in Q, \sigma \in \Sigma, \ \delta_{ND}(q, \sigma) = \delta_{EM}(q, \sigma)$ (this means that any state that was reachable from $\displaystyle q$ with epsilon-travels and input $\displaystyle \sigma$, is now reachable from $\displaystyle q$ only with input $\displaystyle \sigma$)
and
if $\displaystyle \epsilon \notin L(EM)$ then $\displaystyle F_{ND} = F_{EM}$
if $\displaystyle \epsilon \in L(EM)$, then $\displaystyle F_{ND} = F_{EM} \cup \{q_0 \}$
Sure:
Take a look at the images I attached. The one with EM is the original automaton, and the one with ND is what it should look like after the 'conversion'.
http://img821.imageshack.us/i/28510076.jpg/
http://img202.imageshack.us/i/51132577.jpg/