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## Chapter 2: Chapter 2

Problem 13

# \\} ;$$and$$Ris 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)$$ is not regular. ### Short Answer Expert verified The language $$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)$$ is not regular, as applying the Pumping Lemma leads to a contradiction. See the step by step solution ### Step by step solution ## Unlock all solutions Get unlimited access to millions of textbook solutions with Vaia Premium Over 22 million students worldwide already upgrade their learning with Vaia! ## Step 1: Understanding the Rules First, let's go through the rules one by one: Rule 1: $$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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