0

Let's suppose we have this PA:

-> q0 (e, e -> $) --> q1

Where:

q0 is a final and initial state; e is epsilon (empty); and q1 is another state.

If the automaton were to read the e word, it could either make the transition to q1 or stop in q0.

So, would this PA be Non-Deterministic?

My teacher says it wouldn't because, in reality, there's only one path for the automaton to follow: since the word is empty and all symbols had already been consumed in q0, it would make no transition whatsoever; however, we're not sure if he's right (by the way, he says that in order for a PA to recognize a word it needs not only to be in a final state but also all the word's symbols must have been consumed).

andre_ss6
  • 1,195
  • 1
  • 13
  • 34

1 Answers1

1

For a PA to be deterministic it must follow at least the following rule:

If there is an epsilon transition from a state q, there must not be any alphabet transition from that state.

So in your case, if there isn't any other rule, the PA is deterministic.

Hassan Abbasi
  • 272
  • 1
  • 3
  • Even in that case I specified? Note that the q0 state is not only the initial state, but also a final one. – andre_ss6 Oct 13 '15 at 12:46