IS CS-2018S1-02

题目来源：Problem 2 日期：2024-08-07 题目主题：CS-形式语言与自动机-上下文无关文法

解题思路

本题涉及对上下文无关文法（Context-Free Grammar, CFG）的理解和操作，包括生成语言的描述、推导树的绘制、以及通过抽象结果证明某些语言不可由上下文无关文法生成。我们将依次回答每个小问题，提供详细的推导和证明。

Solution

Question 1

To derive the word $aacbbc$ in the grammar $G_0$, we can construct the following syntax tree:

graph TB
    S --- A1[A] & A2[A]
    A1 --- a1[a] & A3[A] & b1[b]
    A3 --- a2[a] & A4[A] & b2[b]
    A4 --- c1[c]
    A2 --- c2[c]

Explanation:

• Start from $S$ and use the rule $S\to AA$.
• For the left $A$, we use $A\to aAb$ followed by $A\to aAb$ and then $A\to c$.
• For the right $A$, we use $A\to c$.

Question 2

To find the words $u,v,w,x,y$ such that:

1. $uvwxy=acbc$
2. $u{v}^{n}w{x}^{n}y\in \mathcal{L}(G_0)$ for every $n\ge 0$
3. $|vx|>0$

Let’s consider $acbc$:

• $u=ϵ$, $v=a$, $w=c$, $x=b$, $y=c$

This satisfies all the conditions:

• $uvwxy=acbc$
• For any $n\ge 0$, $u{v}^{n}w{x}^{n}y=a^{n}c{b}^{n}c$ which is in $\mathcal{L}(G_0)$ because $a^{n}c{b}^{n}c$ can be derived by $n$ applications of $A\to aAb$ and then $A\to c$.
• $|vx|=|a|+|b|=2>0$

Question 3

To prove that for every context-free grammar $G$ there exists an integer $N$ such that for every word $z\in \mathcal{L}(G)$ with $|z|>N$, there exist words $u,v,w,x,y$ satisfying:

1. $z=uvwxy$
2. $u{v}^{n}w{x}^{n}y\in \mathcal{L}(G)$ for every $n\ge 0$
3. $|vx|>0$
4. $|wvx|\le N$

Proof Outline

1. Chomsky Normal Form: Any CFG can be converted to Chomsky Normal Form (CNF) where each production is of the form $A\to BC$ or $A\to a$.
2. Pumping Lemma for CFLs: If $z$ is derived by a CFG with more than $N$ derivations, some non-terminal must repeat due to the pigeonhole principle.
3. Pumping Length $N$: Choose $N$ as the maximum number of unique derivations before a non-terminal repeats.

Detailed Proof

1. Derivation Tree: Consider a derivation tree for a word $z\in \mathcal{L}(G)$. In CNF, each internal node has two children, resulting in a binary tree structure.
2. Repetition of Non-Terminals: If the length of $z$ exceeds $N$, due to CNF and finite number of non-terminals, there must be a path from the root to a leaf with a repeated non-terminal.
3. Decomposition into $uvwxy$: The repeated non-terminal defines the segments $v$ and $x$ such that the string can be decomposed into $uvwxy$.
4. Pumping Condition: Repeating the repeated portion ($v$ and $x$) any number of times results in words still derivable by $G$, i.e., $u{v}^{n}w{x}^{n}y\in \mathcal{L}(G)$.

Question 4

1. Setup

Let’s assume, for the sake of contradiction, that $L=\{ww\mid w\in \{a,b{\}}^{\ast }\}$ is context-free.

2. Applying the Pumping Lemma

By the pumping lemma for context-free languages, there exists a pumping length $p>0$ such that for any string $s\in L$ with $|s|\ge p$, we can write $s=uvxyz$ where:

1. $|vxy|\le p$
2. $|vy|\ge 1$
3. $u{v}^{i}x{y}^{i}z\in L$ for all $i\ge 0$

3. Choosing a String

Let’s choose the string $s=a^p{b}^p{a}^p{b}^p$. Clearly, $s\in L$ and $|s|=4p\ge p$.

4. Analyzing the Decomposition

Now, $s=uvxyz$. Due to condition 1, $vxy$ must be entirely within the first half of $s$ (i.e., within $a^p{b}^p$) or entirely within the second half (i.e., within $a^p{b}^p$), or it must straddle the middle.

5. Case Analysis

Case 1: $vxy$ is entirely within $a^p{b}^p$

In this case, pumping $v$ and $y$ will change the first half of the string but not the second half. Thus, $u{v}^{2}x{y}^{2}z\notin L$ because it’s no longer of the form $ww$.

Case 2: $vxy$ straddles the middle

In this case, $v$ or $y$ (or both) must contain both $a$‘s and $b$‘s. Pumping will change the number of $a$‘s and $b$‘s in a way that can’t be matched in both halves of the string. Thus, $u{v}^{2}x{y}^{2}z\notin L$.

Case 3: $vxy$ is entirely within the second $a^p{b}^p$

This case is similar to Case 1. Pumping will change the second half of the string but not the first half, resulting in $u{v}^{2}x{y}^{2}z\notin L$.

6. Conclusion

In all cases, we can find an $i$ (specifically, $i=2$) such that $u{v}^{i}x{y}^{i}z\notin L$. This contradicts condition 3 of the pumping lemma.

Therefore, our initial assumption must be false, and $L$ is not context-free.

知识点

上下文无关文法Chomsky范式泵引理

解题技巧和信息

1. 上下文无关文法的定义：包括推导树和语言生成的规则。
2. Chomsky 范式：将 CFG 转换为 CNF，使每个产生式最多有两个非终结符，或一个终结符。
3. Pump Lemma：关键工具，用于证明某些语言不是上下文无关语言。主要思想是超长的字符串可以被”泵出”，并且结果仍然在语言中。

重点词汇

1. Context-Free Grammar 上下文无关文法
2. Chomsky Normal Form Chomsky 范式
3. Pumping Lemma Pumping 引理

参考资料

1. Michael Sipser, Introduction to the Theory of Computation, Chapter 2: Context-Free Grammars and Languages.