exercise 1.14

a. Theorem

Show that if M is a DFA that recognizes language B, swapping the accept and nonaccept states in M yields a new DFA recognizing the complement of B. Conclude that the class of regular languages is closed under complement.

a. Proof

This proof is based on induction.

Base case: a DFA with a single state, don't care what is the alphabet. Every letter brings to the only state. Obviously, if the state is an accepted state (F = {q₀}), the DFA accepts every string; if the state is a non-accepted state (F = ∅), the DFA doesn't accept strings.

Inductive step: Let N a DFA that respect the theorem. If we add a new state F_new, we must add an outgoing arrow for each letter and we can add some incoming arrow, moving them from other states.

Now, there are three cases: if the string ends before reaching F_new, the theorem is proven, because the existence of F_new isn't relevant. If the string ends at F_new, the change of the acceptance of F_new inverts the behavior of the DFA.

If the string ends at the state after F_new, that we can name F_new+1, the change of the acceptance of F_new+1 inverts the behavior of the DFA.

After F_new+1, we are again inside N, so the theorem is proved.

a. Proof (2)

For each string s, the membership of each state at F don't change the path that s follows. So, if s ends in an accepted state, the string is accepted, and changing this state the string isn't accepted anymore; and vice versa.

a. Conclusion

Let S a class of strings s composed by letters of Σ. Each string of s must belong to a regular language recognized by N or ¬N, since every state is accepted in N or ¬N.

Since "A language is called regular language if some finite automaton recognizes it" (Sipser, definition 1.16), all the languages that recognize every string s must be recognized by N or ¬N.

b. Theorem

Show by giving an example that if M is an NFA that recognizes language C, swapping the accept and nonaccept states in M doesn’t necessarily yield a new NFA that recognizes the complement of C. Is the class of languages recognized by NFAs closed under complement? Explain your answer.

b. Proof

A simple example is:

N = ({q₀, q₁}, {A, B}, δ, q₀, {q₀, q₁})
δ = (q₀, A) = q₁

the string "B" is not recognized both by N and ¬N

Since every NFA have an equivalent DFA (Sipser, theorem 1.39), and the NFA is a special class of DFAs, the class of NFA is closed under complement.