# Does a NFA/DFA need a failure state?

I'm learning for my exam in theoretical computer science and I'm not sure about it, because I got different sources with different answers. So I wanted to ask here:

If I draw a NFA/DFA and I maybe don't need the transition from a state $z_k \rightarrow z_{k+1}$ with a $0$ above the alphabet $\Sigma = \{0,1\}^*$ so I only need there a $1$, do I always need a failure state $z_F$ so that $\delta: z_k (0) \rightarrow z_F$. Or can I omit that in the drawing? Or is there a difference between DFA and NFA, so that I need to write this failure function if its an NFA but I can omit it if it's an DFA? In my exercises my prof deducted one point because I had no failure state but just one or two times..

Hope somebody can help

• Depends on the definition. Use the one your professor uses. – Raphael Jul 17 '16 at 21:32