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Chapter 8: Problem 5

Consider the set \(P\) of parenthesis structures defined in Example 8.4.4. Give derivations showing that each of the following is in \(P\). a. ()\((())\) b. \((())(())\)

Short Answer

Expert verified
Using the grammar rules (P.1, P.2, and P.3) in Example 8.4.4, we derived the following expressions: a. ()(()) is derived as: \(P \Rightarrow (P) \Rightarrow () \Rightarrow PP \Rightarrow ()(P) \Rightarrow ()(P(P)) \Rightarrow ()(()) \) b. (())(()) is derived as: \(P \Rightarrow (P) \Rightarrow (P(P)) \Rightarrow (()) \Rightarrow PP \Rightarrow (())(P) \Rightarrow (())(P(P)) \Rightarrow (())(()) \) Thus, both ()(()) and (())(()) belong to the set P.

Step by step solution

01

Understanding the rules of set P

The set P in the Example 8.4.4 is defined by the grammars and rules as follows: 1. Rule P.1: \(P \to \epsilon\) 2. Rule P.2: \(P \to (P)\) 3. Rule P.3: \(P \to PP\)
02

Deriving ()(()()) for set P

To derive the expression ()(()) by following rules: a. Apply Rule P.2 on the first P: \(P \Rightarrow (P) \) b. Apply Rule P.1 on the inner P: \((P) \Rightarrow () \) c. Apply Rule P.2 again for the second part of the expression (()): \(P \Rightarrow (P) \) d. Apply Rule P.2 on the inner P once more: \((P) \Rightarrow (P(P)) \) e. Apply Rule P.1 on the center P: \((P(P)) \Rightarrow (()) \) f. Finally, using Rule P.3, join the first part () with the second part (()): \(P \Rightarrow PP \Rightarrow ()(())\) Thus, the expression ()(()) is derived using rules and grammars, so it belongs to set P.
03

Deriving (())(()) for set P

To derive the expression (())(()) by following rules: a. Apply Rule P.2 on the first P: \(P \Rightarrow (P) \) b. Apply Rule P.2 on the inner P for the first part of the expression: \((P) \Rightarrow (P(P)) \) c. Apply Rule P.1 on the center P: \((P(P)) \Rightarrow (()) \) d. Apply Rule P.2 on the second part of the expression: \(P \Rightarrow (P) \) e. Apply Rule P.2 on the inner P once more: \((P) \Rightarrow (P(P)) \) f. Apply Rule P.1 on the center P: \((P(P)) \Rightarrow (()) \) g. Finally, using Rule P.3, join the first part (()) with the second part (()): \(P \Rightarrow PP \Rightarrow (())(())\) Thus, the expression (())(()) is derived using rules and grammars, so it belongs to set P.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Grammar Rules
In the context of context-free grammars, grammar rules are essential for defining the structure of languages. These rules consist of a set of production rules that describe how to generate strings within a language. In this specific case, we are discussing the grammar rules for constructing correct parenthesis structures, set P, that ensures all parentheses are well-balanced.

The rules for set P are as follows:
  • **Rule P.1:** Allows an empty string, denoted by \(P \to \epsilon\). This rule essentially means that we can have a structure with no parentheses at all.
  • **Rule P.2:** Describes forming a pair of parentheses around another structure, represented as \(P \to (P)\). This rule ensures that any valid structure can be enclosed in another set of parentheses.
  • **Rule P.3:** Allows the concatenation of two valid structures, shown as \(P \to PP\). This rule helps in combining two separate valid structures into one larger structure.
These grammar rules collectively ensure that any sequence or combination derived is a valid parenthesis structure within the set P.
Derivation
Derivation is a process through which we can repeatedly apply the grammar rules to develop or validate a structure in a language. It involves step-by-step transformations of non-terminal symbols into terminal ones, meaning actual strings, using a series of production rules.

To derive a structure in parenthesis grammar, you start with a non-terminal symbol \(P\) and progressively apply the appropriate rules:
  • When deriving an expression such as \(()(())\), the transformation begins with Rule P.2 to form external parentheses and then uses subsequent rules to handle inner components.
  • For complex structures like \((())(())\), derivation involves repeated application of rules. External structures use Rule P.2, while Rule P.1 helps conclude the innermost structures, representing empty strings within the nesting.
The process continues until all non-terminals are resolved into terminal structures, and the expression becomes a complete and valid instance of the structure.
Parenthesis Structures
Parenthesis structures form a unique subset of strings where each opening parenthesis must be matched by a closing one, and these must be properly nested. Such structures can become complex, and understanding their construction is crucial for various applications, such as parsing expressions in programming languages or managing mathematical order of operations.

Key concepts of these structures include:
  • **Balance:** For any valid parenthesis structure, the number of opening and closing parentheses must be equal.
  • **Nesting:** Parentheses must be correctly nested, i.e., every opening parenthesis should have a corresponding closing parenthesis at the right position.
  • **Generation through Grammar:** Utilizing context-free grammar rules allows for systematic and correct generation of these structures. By adhering to set rules such as enclosing or combining substructures, correct parenthesis sequences are ensured.
By understanding these concepts, one can derive or verify whether complex parenthesis expressions are valid and belong to the defined set, such as set P in this context.

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