I assume your question is from Chapter 7 Problem 14(iv). Constructing an NFA does not have as many restriction as constructing an FA, i.e. there is no need to have an outgoing edge for each letter of the alphabet from each state when constructing an NFA. Thus, an dead-end state can be eliminated when converting an FA into an NFA.
kiolb if you look at the NFA only the final state lacks outgoing edges ( a or b ). In the transition table you'll see there is no situation where there is a z state that is only equal to this final x state, so theres always somewhere to go! (cause the other two states both have a and b edges)
Oops. Meant ch7 prob14(ix). Did not have to do 14(iv).
Thanks for the replies.
The question asks to change NFA to FA.
In the example on pg138. The final state on the NFA has no outgoing edges, but the equivalent FA has a dead-end state.
The answer to 14(v) has a dead-end state.
Back to 14(ix). State z2, a or b could be in x2, which should then go to a dead-end state, according to the algorithm and the example on pg138.
The answer in the study guide, without using a dead-end, or null state, looks reasonable. But if the algorithm is followed, where is a or b in x2 going to go?
Is it because a or b in z2 has other choices that the dead-end state can be ignored? If so, which rule determines this? This was my original question.