### Context-Free Grammar (CFG) for Generating Strings in the Form \(0^n 1^n\)**Problem Statement:**2. Build a CFG to generate all strings of the form \(0^n 1^n\), where \(n\) is greater than or equal to 0. Please demonstrate the derivation of 11100000111.**Detailed Derivation Explanation:**In order to generate all strings of the form \(0^n 1^n\) using a Context-Free Grammar, we define a grammar with variables, terminals, a start symbol, and production rules. For generating strings where the number of 0s is equal to the number of 1s, the variable should be designed to add pairs of 0s and 1s progressively.#### CFG Definition:- **Variables**: S (Start symbol) - **Terminals**: 0, 1 - **Production Rules**: 1. \( S \rightarrow 0S1 \) 2. \( S \rightarrow \epsilon \) (where \(\epsilon\) represents the empty string)This set of production rules implies that the variable \(S\) can either produce an empty string (to stop the recursion) or add a 0 at the beginning and a 1 at the end of the string contained within \(S\).**Demonstration of the Derivation for the String "11100000111":**The demonstration shows the derivation process of the string following the CFG rules, but note that based on the given rules, generating the string "11100000111" seems incorrect since it does not conform to the form \(0^n 1^n\). The expected derivation should lead from \(S\) to a symmetric string of zeros followed by ones.The string "11100000111" does not follow the rule \(0^n 1^n\), hence it cannot be derived from the given CFG. Here is the correct derivation for a string that follows \(0^n1^n\):Let's demonstrate the derivation for \(S \Rightarrow 000111\) instead:1. Start with S.2. Apply the production \(S \rightarrow 0S1\): \( S \Rightarrow 0S1 \)3. Apply the production \(S \rightarrow 0S1\) again: \(

We can find the character unit by repeatedly changing the non-terminal area on the right side of the…

Answered: 2. Build CFG to generate all strings of…

2. Build CFG to generate all strings of the form 0n 1 n, where n is greater or equal to 0. Please demonstrate the derivation of 111000000111

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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Question

### Context-Free Grammar (CFG) for Generating Strings in the Form \(0^n 1^n\)

**Problem Statement:**

2. Build a CFG to generate all strings of the form \(0^n 1^n\), where \(n\) is greater than or equal to 0. Please demonstrate the derivation of 11100000111.

**Detailed Derivation Explanation:**

In order to generate all strings of the form \(0^n 1^n\) using a Context-Free Grammar, we define a grammar with variables, terminals, a start symbol, and production rules. For generating strings where the number of 0s is equal to the number of 1s, the variable should be designed to add pairs of 0s and 1s progressively.

#### CFG Definition:

- **Variables**: S (Start symbol)

- **Terminals**: 0, 1

- **Production Rules**:
1. \( S \rightarrow 0S1 \)
2. \( S \rightarrow \epsilon \) (where \(\epsilon\) represents the empty string)

This set of production rules implies that the variable \(S\) can either produce an empty string (to stop the recursion) or add a 0 at the beginning and a 1 at the end of the string contained within \(S\).

**Demonstration of the Derivation for the String "11100000111":**

The demonstration shows the derivation process of the string following the CFG rules, but note that based on the given rules, generating the string "11100000111" seems incorrect since it does not conform to the form \(0^n 1^n\). The expected derivation should lead from \(S\) to a symmetric string of zeros followed by ones.

The string "11100000111" does not follow the rule \(0^n 1^n\), hence it cannot be derived from the given CFG. Here is the correct derivation for a string that follows \(0^n1^n\):

Let's demonstrate the derivation for \(S \Rightarrow 000111\) instead:

1. Start with S.
2. Apply the production \(S \rightarrow 0S1\):
\( S \Rightarrow 0S1 \)
3. Apply the production \(S \rightarrow 0S1\) again:
\(

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