In the picture below, I'm trying to figure out what exactly this NFA is accepting. What's confusing me is the $\epsilon$ jump at $q_0$. If a $0$ is entered, does the system move to both $q_0$ and $q_1$ (the accept state)? If a $1$ is entered, does the system move to both $q_1$ and $q_2$? Does the system only move to $q_1$ (accept state), if no input is given (empty string)?
How does an NFA use epsilon transitions?
automata;finite automata;nondeterminism
Every time you are in a state which has a $\epsilon$ transition, it means you automatically are in BOTH states, to simplify this to you:If the string is $\epsilon$ then your automata ends both in $q_0$ and $q_1$If your string is '0' it'll be again in $q_0$ and $q_1$If your string is '1', it'll be only in $q_2$, because if you look from the point of $q_0$, you have a '1' transition to $q_2$, but you have also to look at case you're in $q_1$(if you were in $q_0$ you always were in $q_1$ also) then there is no '1' transition, so this alternative path just dies.Just by looking at these cases its easy to see that your automata accepts $\epsilon$, $0^*$, and going from $q_0$ to $q_1$, the only way to reach $q_2$ is $0^*11^*1$, so, this resumes your automata to $\epsilon$, $0^*$, $0^*11^*1$Hope this helped you, if you have any further doubts, just ask!