Lecture 2

Non-deterministic finite automata

Definition

A non-deterministic finite automata (NFA) is $M = (K, \Sigma, \Delta, s, F)$, where

• $K$ is a finite set of states.
• $\Sigma$ is the input alphabet.
• $s \in k$ is initial state.
• $F \subseteq k$ is the set of final states.

• $\Delta$ is a transition relation.

  $$\Delta \subseteq (K \times (\Sigma \cup \{e\})) \times K$$

Configurations and yields are defined similarly to DFA.

$M$ accepts $w \in \Sigma^*$ if $(s, w) \vdash_M^* (q, e)$ for some $q \in F$.

For $L(M) = \{ w \in \Sigma^*: M \text{ accepts } w \}$, we say $M$ accepts $L(M)$.

The difference between DFA and NFA is that NFA allows

• more than one choice for next state.
• e-transition.

state diagram of NFA

Insight | A Magic

We assume that NFA always makes the right guess.

Example

Construct an FA that accepts

$$L = \{w \in \{a, b\}^* | \text{ the second symbol from the end of $w$ is $b$.}\}$$

Theorem

DFA is equvilent to NFA.

• $\forall \text{ DFA } M', \exist \text{ NFA } M,\ s.t.\ L(M) = L(M').$
• $\forall \text{ NFA } M, \exist \text{ DFA } M',\ s.t.\ L(M) = L(M').$

Proof

The first claim is trivial. For the second claim,

Idea DFA $M'$ simulates "tree-like computation" of NFA $M$.

$\forall \text{ NFA } M' = (K, \Sigma, \Delta, s, F)$, we construct a DFA $M' = (K', \Sigma, \delta, s', F')$, where

• $K' = 2^K = \{Q | Q \subseteq K\}$.
• $F' = \{Q \subseteq K | Q \cap F \neq \emptyset\}$.
• $s' = E(s)$, where $E(q) = \{p \in K | (q, e) \vdash_M^* (p, e) \}$, $\forall q \in K$.
• $\forall Q \in K', a \in \Sigma,\ \ \delta(Q, a) = \bigcup\limits_{q \in Q} \bigcup\limits_{p : (q, a, p) \in \Delta} E(p)$.

Then we claim that $\forall p, q \in K, w \in \Sigma^*, (p, w) \vdash_M^* (q, e) \text{ iff } (E(p), w) \vdash_{M'}^* (Q, e), \text{ where } q \in Q$. (We can prove it by induction on $|w|$)

Thus, DFA $M'$ accepts $w$ $\Leftrightarrow$ NFA $M$ accepts $w$.

Example

NFA

DFA

Closure Property (Cont.)

Since the equivalence of NFA and DFA, so a language is regular if it's accepted by some NFA. Now we can prove the following theorem by NFA.

Theorem

If $A$ and $B$ are regular, so is $A \cdot B$.

Proof

Since $A$ and $B$ are regular, thus there exist NFAs $M_A = (K_A, \Sigma, \Delta_A, s_A, F_A)$ and $M_B = (K_B, \Sigma, \Delta_B, s_B, F_B)$ which accepts $A$ and $B$ respectively.

We construct an NFA $M_\circ = (K_\circ, \Sigma, \Delta_\circ, s_\circ, F_\circ)$, where

• $K_\circ = K_A \cup K_B$.
• $s_\circ = s_A$.
• $F_\circ = F_B$.
• $\Delta_\circ = \Delta_A \cup \Delta_B \cup \{(q, e, s_B) | q \in F_A\}$.

Theorem

If $A$ is regular, so is $A^*$.

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最后更新: 2024.01.29 15:36:20 CST
创建日期: 2023.10.01 15:23:00 CST