東京大学 情報理工学系研究科 コンピュータ科学専攻 2016年8月実施 専門科目I 問題1

标签：

• Tokyo-University
• Nondeterministic-Finite-Automaton
• Deterministic-Finite-Automaton
• Regular-Language-Closure-Properties
• Regular-Language-Decision-Problems

Author

kainoj

Description

A language $L \subseteq \Sigma^*$ over a finite alphabet $\Sigma$ is said to be regular if there exists a finite automaton $\mathcal{A}$ such that $L = \mathcal{L}(\mathcal{A})$. Here

$$\mathcal{L}(\mathcal{A}) = \{ w \in \Sigma^* \mid w \text{ is accepted by } \mathcal{A} \}.$$

Answer the following questions:

(1) We fix an alphabet $\Sigma$ by $\Sigma = \{a, b\}$. For the language $L_1$ below, present a nondeterministic finite automaton (NFA) $\mathcal{A}_1$ such that: $\mathcal{L}(\mathcal{A}_1) = L_1$, and the number of states of $\mathcal{A}_1$ is not greater than $4$.

$$L_1 = \{ w \in \Sigma^* \mid \text{there is a character } l \in \Sigma \text{ that occurs more than once in } w \}.$$

(2) Assume that $\Sigma$ is a finite alphabet. Prove the following: any finite language $L = \{w_1, \ldots, w_n\} \subseteq \Sigma^*$ is regular. Here $n$ is a nonnegative integer.

(3) We fix an alphabet $\Sigma$ by $\Sigma = \{a, b\}$. For the language $L_1$ in Question (1), present a deterministic finite automaton (DFA) $\mathcal{A}_2$ such that: $\mathcal{L}(\mathcal{A}_2) = \Sigma^* \setminus L_1$, and the number of states of $\mathcal{A}_2$ is not greater than $5$. Here $\Sigma^* \setminus L_1$ denotes the complement of $L_1 \subseteq \Sigma^*$.

(4) Give a decision procedure for the following problem, and explain it briefly.

• Input: Nondeterministic finite automaton $\mathcal{A}$.
• Output: Whether the language $\mathcal{L}(\mathcal{A})$ is an infinite set or not.

题目描述

对有限字母表 $\Sigma$ 上的语言 $L\subseteq\Sigma^*$，若存在有限自动机 $\mathcal{A}$ 使 $L=\mathcal{L}(\mathcal{A})$，则称 $L$ 为正则语言，其中

$$\mathcal{L}(\mathcal{A}) =\{w\in\Sigma^*\mid w\text{ 被 }\mathcal{A}\text{ 接受}\}.$$

回答下列问题。

（1）固定 $\Sigma=\{a,b\}$。对语言

$$L_1=\{w\in\Sigma^*\mid w\text{ 中存在某个字符出现至少两次}\},$$

构造一个状态数不超过 $4$ 的 NFA $\mathcal{A}_1$，满足 $\mathcal{L}(\mathcal{A}_1)=L_1$。

（2）设 $\Sigma$ 为有限字母表。证明任意有限语言 $L=\{w_1,\ldots,w_n\}\subseteq\Sigma^*$ 都是正则语言，其中 $n$ 可以为非负整数。

（3）仍令 $\Sigma=\{a,b\}$。针对第（1）问的 $L_1$，构造一个状态数不超过 $5$ 的 DFA $\mathcal{A}_2$，使

$$\mathcal{L}(\mathcal{A}_2)=\Sigma^*\setminus L_1.$$

（4）给出并简要说明一个判定过程：输入一个 NFA $\mathcal{A}$，输出其语言 $\mathcal{L}(\mathcal{A})$ 是否为无限集。

Kai

(1)

$\Sigma = \{a,b\}$. Give NFA $\mathcal{A}_1$ with no more than $4$ states recognizing $L_1 = \{w\in \Sigma \:|\: \exists l\in \Sigma \: |w|_l > 1\}$.

$$\begin{array}{ll||l|l} & & a & b \\ \hline s & q_0 & q_0,q_1 & q_0,q_2 \\ & q_1 & q_3 & q_1 \\ & q_2 & q_2 & q_3 \\ * & q_3 & q_3 & q_3 \end{array}$$

(2)

Prove: if $\Sigma$ is finite alphabet, then any finite language $L = \{w_1,\cdots,w_n\} \subseteq \Sigma^*$ is regular, $n\in \mathbb{N}$.

That is we should construct a finite automaton accepting $L$. We can construct a $\epsilon$-NFA containing $n$ "branches", each recognizing $w_i$. Now, for every $\epsilon$-NFA, there must exist equivalent DFA recognizing the same language.

We can also construct the DFA explicitly. Start with automaton (NFA) recognizing $w_1$: there are $|w_1|+1$ states, the last one is accepting and transitions are labeled with next letters of $w_1$. For $w_i$ ($i=2,\cdots,n$) try to traverse the automaton as far as you can, i.e. until transition for symbol $w_{ij}$ exist. If we can go no further, that is we read the longest common prefix of $w_i$ and some $w_k$, $k<i$, then we make a new branch from a current state. This branch consist of states and transitions labeled $w_{i,j+1}\cdots w_{i, |w_i|}$.

Now we need to assure that we constructed a DFA. For every state missing some transitions on some letters, add those transition leading to a "dead state", i.e. nonaccepting state with a self-loop labeled $\Sigma$.

(3)

Give DFA recognizing complement of $L_1$ from (Q1), i.e $L_2 = \Sigma^* \setminus L_1$.

Obviously $L_2 = \{w\in \Sigma \:|\: \forall l\in \Sigma \: |w|_l \leq 1\} = \{\epsilon, a, b, ab, ba\}$.

$$\begin{array}{ll||l|l} & & a & b \\ \hline s,* & q_1 & q_2 & q_3 \\ * & q_2 & q_5 & q_4 \\ * & q_3 & q_4 & q_5 \\ * & q_4 & q_5 & q_5 \\ & q_5 & q_5 & q_5 \end{array}$$

(4)

Given NFA $\mathcal{A}$, decide whether $\mathcal{L(A)}$ is empty or not. % For every NFA $\mathcal{A}$, there exist equivalent DFA $\mathcal{D}$, that is, $\mathcal{L(A)} = \mathcal{L(D)}$ (subset construction). % Let's examine such DFA $\mathcal{D}$. Since language of $\mathcal{A}$ is regular, then from pumping lemma we can "pump" words longer than some $N$ – pumping lemma constant. That is, if $\mathcal{L(A)}$ has some word longer than $N$, then $\mathcal{L(A)}$ is infinite. We just need to check every possible word $w$: $N < w \leq 2N$. If any such word is accepted by $\mathcal{A}$, then $\mathcal{L(A)}$ is infinite. We don't need to check words longer of $2N$: if a word is longer than $2N$, then from PL, we can iterative reduce its length, so $w$ has length shorter than $2N$.

How to choose $N$? We know that for every NFA $\mathcal{A}$, there exist equivalent DFA $\mathcal{D}$, that is, $\mathcal{L(A)} = \mathcal{L(D)}$ (subset construction). We don't need to construct such DFA. All we know is that, $\mathcal{D}$ might have exponentially more states than $\mathcal{A}$. Take $N = |\Sigma|^{|Q_D|} + 42$, where $Q_D$ is set of $\mathcal{D}$'s states.