R03-C-問2

問2

文脈自由文法を4つ組 $(N,\Sigma ,P,S)$ で表す。ただし、$N,\Sigma ,P,S$ をそれぞれ非終端記号の集合、終端記号の集合、生成規則の集合、開始記号とする。$\Sigma$ 上の文字列 $w$ と終端記号 $x\in \Sigma$ に対して、$|w{|}_x$ を $w$ 中の $x$ の出現回数とする。以下の各問いに答えよ。

(A) 以下に示す言語 $L_i(i=1,2,3)$ を導出する文脈自由文法 $(N_i,{\Sigma }_i,P_i,S)$ の生成規則の集合 $P_i$ を与えよ。

1. $L_1=\{w\in \{a,b{\}}^{\ast }||w{|}_a\ge 3\}$。ただし ${\Sigma }_1=\{a,b\}$ である。答えには $N_1=\{S,X\}$ を用いよ。
2. $L_2=\{w\in \{a,b{\}}^{\ast }||w{|}_a\ge 2|w{|}_b\}$。ただし ${\Sigma }_2=\{a,b\}$ である。答えには $N_2=\{S\}$ を用いよ。
3. $L_3=\{a^i{b}^j{c}^k|i,j,k\ge 0かつ(2i=jまたは3j=k)\}$。ただし ${\Sigma }_3=\{a,b,c\}$ である。答えには $N_3=\{S,V,X,Y,Z\}$ を用いよ。

(B) 以下に示す言語 $L_4,L_5$ は文脈自由言語であるか否か。証明せよ。

4. $L_4=\varnothing$。
5. 有限個の有限長文字列から構成される空でない任意の言語 $L_5$。

Q2

We represent a context-free grammar by a quartet $(N,\Sigma ,P,S)$, where $N,\Sigma ,P$, and $S$ represent the set of non-terminal symbols, the set of terminal symbols, the set of productions, and the start symbol, respectively. For a string over $\Sigma$ and a terminal symbol $x\in \Sigma$, let $|w{|}_x$ denote the number of occurrences of $x$ in $w$. Answer the following questions.

(A) For the following languages $L_i(i=1,2,3)$, give a set $P_i$ of production rules of a context-free grammar $(N_i,{\Sigma }_i,P_i,S)$ which generates $L_i$:

1. $L_1=\{w\in \{a,b{\}}^{\ast }\mid |w{|}_a\ge 3\}$, where ${\Sigma }_1=\{a,b\}$. Use $N_1=\{S,X\}$ in your answer.
2. $L_2=\{w\in \{a,b{\}}^{\ast }\mid |w{|}_a\ge 2|w{|}_b\}$, where ${\Sigma }_2=\{a,b\}$. Use $N_2=\{S\}$ in your answer.
3. $L_3=\{a^i{b}^j{c}^k\mid i,j,k\ge 0\text{ and }(2i=j\text{ or }3j=k)\}$, where ${\Sigma }_3=\{a,b,c\}$. Use $N_3=\{S,V,X,Y,Z\}$ in your answer.

(B) Prove whether each of the languages $L_4$ and $L_5$ is a context-free language, or not:

4. $L_4=\varnothing$.
5. Any non-empty language $L_5$ which consists of a finite number of finite-length strings.

中文翻译

用四元组 $(N,\Sigma ,P,S)$ 表示上下文无关文法，其中 $N,\Sigma ,P$ 和 $S$ 分别表示非终结符集合、终结符集合、产生式集合和开始符号。对于 $\Sigma$ 上的字符串 $w$ 和终结符 $x\in \Sigma$，令 $|w{|}_x$ 表示 $x$ 在 $w$ 中出现的次数。回答以下问题。

(A) 对于以下语言 $L_i(i=1,2,3)$，给出上下文无关文法 $(N_i,{\Sigma }_i,P_i,S)$ 的产生式集合 $P_i$，该文法生成语言 $L_i$：

1. $L_1=\{w\in \{a,b{\}}^{\ast }\mid |w{|}_a\ge 3\}$，其中 ${\Sigma }_1=\{a,b\}$。在你的答案中使用 $N_1=\{S,X\}$。
2. $L_2=\{w\in \{a,b{\}}^{\ast }\mid |w{|}_a\ge 2|w{|}_b\}$，其中 ${\Sigma }_2=\{a,b\}$。在你的答案中使用 $N_2=\{S\}$。
3. $L_3=\{a^i{b}^j{c}^k\mid i,j,k\ge 0\text{ and }(2i=j\text{ or }3j=k)\}$，其中 ${\Sigma }_3=\{a,b,c\}$。在你的答案中使用 $N_3=\{S,V,X,Y,Z\}$。

(B) 证明语言 $L_4$ 和 $L_5$ 是否为上下文无关语言：

4. $L_4=\varnothing$。
5. 任何非空语言 $L_5$，其中包含有限数量的有限长度字符串。

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DeepSeekV3

Answer to Question A1:

To define a set of production rules $P_1$ for the grammar $(N_1,{\Sigma }_1,P_1,S)$ that generates the language $L_1=\{w\in \{a,b{\}}^{\ast }\mid |w{|}_a\ge 3\}$, where $N_1=\{S,X\}$ and ${\Sigma }_1=\{a,b\}$, we can use the following productions:

• $S\to aA$
• $A\to aB$
• $B\to aC$
• $C\to bC\mid aC\mid ϵ$

This ensures that every string generated has at least three a’s, with any number of a’s or b’s following.

Answer to Question A2:

To define $P_2$ for the grammar $(N_2,{\Sigma }_2,P_2,S)$ that generates $L_2=\{w\in \{a,b{\}}^{\ast }\mid |w{|}_a\ge 2|w{|}_b\}$, where $N_2=\{S\}$ and ${\Sigma }_2=\{a,b\}$, we can use the following productions:

• $S\to aS$
• $S\to aSa$
• $S\to aSb$
• $S\to ϵ$

This ensures that for every b generated, there are at least two a’s.

Answer to Question A3:

To define $P_3$ for the grammar $(N_3,{\Sigma }_3,P_3,S)$ that generates $L_3=\{a^i{b}^j{c}^k\mid i,j,k\ge 0\text{ and }(2i=j\text{ or }3j=k)\}$, where $N_3=\{S,V,X,Y,Z\}$ and ${\Sigma }_3=\{a,b,c\}$, we can use the following productions:

• $S\to V\mid Z$
• $V\to aVbb\mid ϵ$
• $Z\to bZccc\mid ϵ$
• $V\to aV\mid aVc$
• $Z\to aZ\mid aZc$

This ensures that either the number of b’s is twice the number of a’s, or the number of c’s is three times the number of b’s.

Answer to Question B4:

$L_4=\varnothing$ is a context-free language because the empty set is a regular language, and all regular languages are context-free.

Answer to Question B5:

Any non-empty language $L_5$ consisting of a finite number of finite-length strings is a context-free language because finite languages are regular, and all regular languages are context-free.

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To create a context-free grammar for the language $L_3=\{a^i{b}^j{c}^k\mid i,j,k\ge 0\text{ and }(2i=j\text{ or }3j=k)\}$ using the specified non-terminals $N_3=\{S,V,X,Y,Z\}$, we can define the grammar as follows:

Grammar for $L_3$

• Non-terminal symbols (N): $\{S,V,X,Y,Z\}$
• Terminal symbols (Σ): $\{a,b,c\}$
• Start symbol (S): $S$
• Production rules (P):

$$\begin{array}{rl}S& \to V\mid Z\\ V& \to aVbb\mid X\\ X& \to cX\mid ϵ\\ Z& \to bZccc\mid Y\\ Y& \to aY\mid ϵ\end{array}$$

Explanation

1. Non-terminal $V$:

   • Generates strings where the number of a’s is exactly half the number of b’s ($2i=j$), with any number of c’s.
   • $V\to aVbb$ ensures that for every a, there are exactly two b’s.
   • $V\to X$ allows for the addition of any number of c’s.

2. Non-terminal $X$:

   • Generates any number of c’s.
   • $X\to cX\mid ϵ$

3. Non-terminal $Z$:

   • Generates strings where the number of c’s is exactly three times the number of b’s ($3j=k$), with any number of a’s.
   • $Z\to bZccc$ ensures that for every b, there are exactly three c’s.
   • $Z\to Y$ allows for the addition of any number of a’s.

4. Non-terminal $Y$:

   • Generates any number of a’s.
   • $Y\to aY\mid ϵ$

5. Start symbol $S$:

   • Chooses between generating strings that satisfy $2i=j$ or $3j=k$.
   • $S\to V\mid Z$

Correctness

• The grammar ensures that strings generated either satisfy $2i=j$ with any number of c’s or $3j=k$ with any number of a’s.
• Strings that do not satisfy either condition are not generated by the grammar.

This grammar correctly captures the language $L_3$ by providing the necessary production rules to generate all and only the strings that meet the specified conditions.