3

I realized that if a language L is recognized by a DFA, then flipping its accept and reject states (aka switching all the accept to reject states and all the reject states to accept) works to recognize the complement of L (which wold be L^c).

I was curious if that worked also with NFAs (flipping the accept and reject states to go from L to L complement), but with their additional paths it doesn't seem like it would work..Definitely seems like it would have too many accept states that aren't correct.

Is this correct? Is there more to it? Thank you!

Raphael
  • 73,212
  • 30
  • 182
  • 400
user3295674
  • 383
  • 4
  • 14

1 Answers1

2

Same thing does not hold for an NFA. Example

q0 -> 0 -> q1
q0 -> 0 -> q2

q0 is the start state, q1 is the final state.

Sarvottamananda
  • 4,877
  • 1
  • 14
  • 19