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Introduction to the Theory of Computation

Sipser Michael

Chapter 2: Chapter 2

Problem 13

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\\} ;$ and $R$ is the set of rules: $$ \begin{aligned} &S \rightarrow T T \mid U \\ &T … # Let $G=(V, \Sigma, R, S)$ be the following grammar. $V=\\{S, T, U\\} ; \Sigma=\\{0, \\#\\} ;$ and $R$ is the set of rules: $$ \begin{aligned} &S \rightarrow T T \mid U \\ &T \rightarrow 0 T|T 0| \\ &U \rightarrow 0 U 00 \mid \\# \end{aligned} $$ a. Describe $L(G)$ in English. b. Prove that $L(G)$ is not regular.

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The language $L(G)$ contains strings that either have an even number of consecutive 0s or have half as many 0s on the left side of the symbol '\\#' as on the right side. The language $L(G)$ is not regular, as applying the Pumping Lemma leads to a contradiction.

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Understanding the Rules

First, let's go through the rules one by one: Rule 1: $S \rightarrow T T \mid U$ S denotes the starting symbol, which will either produce two Ts or one U. Rule 2: $T \rightarrow 0 T|T 0|$ T produces any combination of 0s, which will always have an even total number of 0s. Rule 3: $U \rightarrow 0 U 00 \mid \\#$ U produces any combination of 0s, with every 0 in the middle section doubled in the right section. The doubled 0s are denoted with the symbol '\\#'.

Step 2: Describing L(G)

Combining the understanding of these rules, we can describe $L(G)$ in English as follows: The language $L(G)$ contains strings that either have an even number of consecutive 0s or have half as many 0s on the left side of the symbol '\\#' as on the right side. b. Prove that $L(G)$ is not regular.

Step 1: Assuming L(G) is regular

To prove that $L(G)$ is not regular, we will use the Pumping Lemma for regular languages. First, we assume that $L(G)$ is regular.

Step 2: Applying the Pumping Lemma

According to the Pumping Lemma, given any regular language $L$ and sufficiently long string $w$, there exist substrings $x, y, z$ such that: 1. $w = xyz$ 2. $|xy| \leq p$ 3. $y ≠ ε$ 4. $xy^iz$ is in $L$ for every $i \geq 0$ Now let's find a string $w$ that, when pumped, leads to a contradiction. We choose the string $w = 0^p\#0^{2p}$, where $p$ is the pumping length. The string $w$ ∉ L(G) if $L(G)$ is regular.

Step 3: Finding Contradiction

We can represent the substrings $x, y, z$ from the Pumping Lemma as $w = 0^a0^b0^c\#0^{2p}$, where $a+b+c = p$ and $b > 0$. Then, the pumped string for any $i \geq 1$ is: $$w' = xy^iz = 0^a0^{bi}0^c\#0^{2p}$$ By simple observation, we can see that as long as $b > 0$, the total number of 0s on the left side of '\\#' in the pumped string $w'$ will be greater than $p$. But, on the right side, we still have only $2p$ 0s. Thus, the pumped string $w'$ does not maintain the property of half as many 0s on the left side of '\\#' as on the right side. Therefore, $w'$ ∉ L(G), contradicting the conditions of the Pumping Lemma, proving that our initial assumption of $L(G)$ being regular is false. Thus, we conclude that the language $L(G)$ is not regular.

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Chapter 2: Chapter 2

Short Answer

Step by step solution

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Step 1: Understanding the Rules

Step 2: Describing L(G)

Step 1: Assuming L(G) is regular

Step 2: Applying the Pumping Lemma

Step 3: Finding Contradiction

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More chapters from the book ‘Introduction to the Theory of Computation’

Chapter 0

Chapter 1

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

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