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. $L_1 = \{w \in \{a, b\}^* | |w|_a \geq 3\}$。ただし $\Sigma_1 = \{a, b\}$ である。答えには $N_1 = \{S, X\}$ を用いよ。
2. $L_2 = \{w \in \{a, b\}^* | |w|_a \geq 2|w|_b\}$。ただし $\Sigma_2 = \{a, b\}$ である。答えには $N_2 = \{S\}$ を用いよ。
3. $L_3 = \{a^i b^j c^k | i, j, k \geq 0 \ かつ \ (2i = j \ または \ 3j = k)\}$。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. $L_4 = \emptyset$。
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. $L_1 = \{w \in \{a, b\}^* \mid |w|_a \geq 3\}$, where $\Sigma_1 = \{a, b\}$. Use $N_1 = \{S, X\}$ in your answer.
2. $L_2 = \{w \in \{a, b\}^* \mid |w|_a \geq 2|w|_b\}$, where $\Sigma_2 = \{a, b\}$. Use $N_2 = \{S\}$ in your answer.
3. $L_3 = \{a^i b^j c^k \mid i, j, k \geq 0 \text{ and } (2i = j \text{ or } 3j = k)\}$, 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. $L_4 = \emptyset$.
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. $L_1 = \{w \in \{a, b\}^* \mid |w|_a \geq 3\}$，其中 $\Sigma_1 = \{a, b\}$。在你的答案中使用 $N_1 = \{S, X\}$。
2. $L_2 = \{w \in \{a, b\}^* \mid |w|_a \geq 2|w|_b\}$，其中 $\Sigma_2 = \{a, b\}$。在你的答案中使用 $N_2 = \{S\}$。
3. $L_3 = \{a^i b^j c^k \mid i, j, k \geq 0 \text{ and } (2i = j \text{ or } 3j = k)\}$，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. $L_4 = \emptyset$。
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 $L_1 = \{ w \in \{a, b\}^* \mid |w|_a \geq 3 \}$, where $N_1 = \{S, X\}$ and $\Sigma_1 = \{a, b\}$, we can use the following productions:

• $S \rightarrow aA$
• $A \rightarrow aB$
• $B \rightarrow aC$
• $C \rightarrow bC \mid aC \mid \epsilon$

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 $L_2 = \{ w \in \{a, b\}^* \mid |w|_a \geq 2|w|_b \}$, where $N_2 = \{S\}$ and $\Sigma_2 = \{a, b\}$, we can use the following productions:

• $S \rightarrow aS$
• $S \rightarrow aS a$
• $S \rightarrow aS b$
• $S \rightarrow \epsilon$

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 $L_3 = \{ a^i b^j c^k \mid i, j, k \geq 0 \text{ and } (2i = j \text{ or } 3j = k) \}$, where $N_3 = \{S, V, X, Y, Z\}$ and $\Sigma_3 = \{a, b, c\}$, we can use the following productions:

• $S \rightarrow V \mid Z$
• $V \rightarrow a V b b \mid \epsilon$
• $Z \rightarrow b Z c c c \mid \epsilon$
• $V \rightarrow a V \mid a V c$
• $Z \rightarrow a Z \mid a Z c$

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:

$L_4 = \emptyset$ 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 $L_3 = \{ a^i b^j c^k \mid i, j, k \geq 0 \text{ and } (2i = j \text{ or } 3j = k) \}$ 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):

$$\begin{align*} S &\rightarrow V \mid Z \\ V &\rightarrow a V b b \mid X \\ X &\rightarrow c X \mid \epsilon \\ Z &\rightarrow b Z c c c \mid Y \\ Y &\rightarrow a Y \mid \epsilon \\ \end{align*}$$

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.
   • $V \rightarrow a V b b$ ensures that for every a, there are exactly two b's.
   • $V \rightarrow X$ allows for the addition of any number of c's.

2. Non-terminal $X$:

   • Generates any number of c's.
   • $X \rightarrow c X \mid \epsilon$

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.
   • $Z \rightarrow b Z c c c$ ensures that for every b, there are exactly three c's.
   • $Z \rightarrow Y$ allows for the addition of any number of a's.

4. Non-terminal $Y$:

   • Generates any number of a's.
   • $Y \rightarrow a Y \mid \epsilon$

5. Start symbol $S$:

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

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.