IS CS-2022W-03

题目来源：Problem 3 日期：2024-07-19 题目主题：CS-Formal Languages-Finite Automata and Context-Free Languages

具体题目

Let $\Sigma =\{a,b\}$. Answer the following questions.

1. Give a non-deterministic finite state automaton with 3 states that accepts the following language.

$$\{wba\mid w\in {\Sigma }^{\ast }\}$$

1. Show the minimal deterministic finite state automaton that accepts the language given in question (1).

2. Prove that the following language $L$ over $\Sigma$ is not regular. You may use the pumping lemma for regular languages.

$$L=\{w^{\mathbf{R}}baw\mid w\in {\Sigma }^{\ast }\}$$

Here, $w^{\mathbf{R}}$ denotes the reverse of $w$. For example, $(abb{)}^{\mathbf{R}}=bba$.

1. Is the language $L$ in question (3) a context-free language? If it is, construct a context-free grammar that generates $L$. If not, prove that $L$ is not a context-free language.

正确解答

Question 1

A non-deterministic finite state automaton (NFA) with 3 states that accepts the language $\{wba\mid w\in {\Sigma }^{\ast }\}$ can be constructed as follows:

• States: $q_0,q_1,q_2$
• Alphabet: $\{a,b\}$
• Start state: $q_0$
• Accept state: $q_2$
• Transition function:
  • $q_0\stackrel{a,b}{\to }{q}_0$
  • $q_0\stackrel{b}{\to }{q}_1$
  • $q_1\stackrel{a}{\to }{q}_2$
  • $q_1\stackrel{b}{\to }{q}_0$
  • $q_2\stackrel{a}{\to }{q}_0$
  • $q_2\stackrel{b}{\to }{q}_1$

The state diagram for the NFA is as follows:

graph LR
    q0 -->|a,b| q0
    q0 -->|b| q1
    q1 -->|a| q2
    q1 -->|b| q0
    q2 -->|a| q0
    q2 -->|b| q1

Question 2

To construct the minimal deterministic finite state automaton (DFA) for the language $\{wba\mid w\in {\Sigma }^{\ast }\}$, we can use the subset construction method to convert the NFA to a DFA.

We can represent the NFA as a DFA with the following states:

• $\delta (q_0,a)=q_0$
• $\delta (q_0,b)=\{q_0,q_1\}$
• $\delta (\{q_0,q_1\},a)=\{q_0,q_2\}$
• $\delta (\{q_0,q_1\},b)=\{q_0,q_1\}$
• $\delta (\{q_0,q_2\},a)=q_0$
• $\delta (\{q_0,q_2\},b)=\{q_0,q_1\}$

This DFA is minimal and accepts the language $\{wba\mid w\in {\Sigma }^{\ast }\}$.

The transition table for the DFA is as follows:

So we can construct the minimal DFA as follows:

• States: $\{Q_0,Q_1,Q_2\}$
• Alphabet: $\{a,b\}$
• Start state: $Q_0$
• Accept state: $Q_2$
• Transition function:
  • $\delta (Q_0,a)=Q_0$
  • $\delta (Q_0,b)=Q_1$
  • $\delta (Q_1,a)=Q_2$
  • $\delta (Q_1,b)=Q_1$
  • $\delta (Q_2,a)=Q_0$
  • $\delta (Q_2,b)=Q_1$

The state diagram for the minimal DFA is as follows:

graph LR
    Q0 -->|a| Q0
    Q0 -->|b| Q1
    Q1 -->|a| Q2
    Q1 -->|b| Q1
    Q2 -->|a| Q0
    Q2 -->|b| Q1

Question 3

To prove that the language $L=\{w^{\mathbf{R}}baw\mid w\in {\Sigma }^{\ast }\}$ is not regular, we will use the pumping lemma for regular languages.

Pumping Lemma for Regular Languages

If $L$ is a regular language, then there exists a pumping length $p$ such that any string $s\in L$ with $|s|\ge p$ can be split into three parts $s=xyz$ satisfying the following conditions:

1. $|xy|\le p$
2. $|y|>0$
3. $x{y}^{i}z\in L$ for all $i\ge 0$

Let $w=a^p$, and consider the string $s=w^{\mathbf{R}}baw=a^{p}ba{a}^p$. According to the pumping lemma, $s$ can be decomposed into $xyz$ such that the conditions are met.

• Let $x=a^i$, $y=a^j$, $z=a^{p-i-j}ba{a}^p$ where $1\le i+j\le p$ and $j>0$.

If we pump $y$, the resulting string is $x{y}^{2}z=a^i{a}^{2j}{a}^{p-i-j}ba{a}^p=a^{p+j}ba{a}^p$.

This string is not in $L$ because it does not match the form $w^{\mathbf{R}}baw$ where $w=a^p$. Thus, $L$ is not a regular language.

Question 4

To determine whether $L$ is a context-free language, we can use the context-free grammar (CFG) approach.

Constructing a CFG for $L$

We need to create a CFG that generates strings of the form $w^{\mathbf{R}}baw$ where $w\in {\Sigma }^{\ast }$.

A CFG that generates this language can be defined as follows:

• Non-terminals: $S,A$
• Terminals: $a,b$
• Start symbol: $S$
• Production rules:
  • $S\to ba\mid aSa\mid bSb$

This CFG generates strings of the form $w^{\mathbf{R}}baw$ by recursively placing matching symbols $a$ and $b$ around the center substring $b$.

Therefore, $L$ is a context-free language.

知识点

NFADFA正则语言泵引理上下文无关语言

难点解题思路

对于问题 3，使用抽象的语言对泵引理的证明往往会遇到困难。我们选择具体的字符串 $w=a^p$ 并展示在不同情况下的行为以证明该语言非正则性。问题 4 则是运用构造法，结合递归的生成规则，清晰地表达语言的上下文无关性质。

解题技巧和信息

1. NFA 转换为 DFA 时，确保考虑每种状态转换的可能性，尽量简化状态数量。
2. 利用泵引理证明非正则性时，选择一个能覆盖所有可能情况下的字符串。
3. 构造 CFG 时，利用递归规则可以简洁地表达嵌套的结构。

重点词汇

• Non-deterministic finite state automaton (NFA) 非确定性有限状态自动机
• Deterministic finite state automaton (DFA) 确定性有限状态自动机
• Pumping Lemma 泵引理
• Context-free grammar (CFG) 上下文无关文法
• Regular language 正则语言
• Context-free language 上下文无关语言

参考资料

1. “Introduction to the Theory of Computation” by Michael Sipser, Chapters 1-2 for Automata and Regular Languages, Chapter 4 for Context-Free Grammars.