Procedural and Non Procedural Language

Procedural Language (DBMS)

• User specifies what data is required and how to get it (sequence of steps)

• Query execution path is defined by user

• Define Steps, More control, less abstraction

Examples:

1. Relational Algebra
2. Tuple Relational Calculus (considered weakly procedural in exams)

eg:- Relational Algebra (RA)

π name (σ marks > 80 (STUDENT))

• First select tuples

• Then project attributes

eg:- Tuple Relational Calculus (TRC)

{ t.name | STUDENT(t) ∧ t.marks > 80 }

• Specify desired attribute in result
• Give condition on tuples (no explicit operation order)

1. Relational Algebra Operations ⭐

Primitive (Basic) Operations:

• Selection (σ)
• Projection (π)
• Cartesian Product (×)
• Union (∪)
• Set Difference (−)
• Rename (ρ)

Derived Operations:

• Intersection (∩)
• Join (⨝)
• Division (÷)

Unary Operations

Operation	Symbol	Purpose	Example
Selection	σ	Select rows satisfying condition	σ_Age>20(Student)
Projection	π	Select specific columns	π_Name, Age(Student)
Rename	ρ	Rename relation or attributes	ρ_Stud(Student)

Binary Operations

Operation	Symbol	Purpose	Example
Union	∪	Tuples in either relation	Student ∪ Alumni
Set Difference	-	Tuples in first, not in second	Student - Alumni
Intersection	∩	Tuples common to both	Student ∩ Scholarship
Cartesian Product	×	All combinations of tuples	Student × Course
Join	⨝	Combine tuples based on condition	Student ⨝_ID=Enroll.ID Enroll
Division	÷	Tuples related to all in second relation	StudentID ÷ CourseID

• Unary: operate on one relation
• Binary: operate on two relations
• Rename (ρ) helps in queries and avoiding conflicts.

Examples :

1. Find all students older than 20 σ_Age>20 (Student)
2. Find all students enrolled in a 4-credit course σ_Credit=4 (Student ⨝ Student.CourseID = Course.ID Course)
3. Find all students with marks > 80 σ_Marks>80 (Student)
4. Find names of students in “DBMS” course π_Name (σ_CourseName='DBMS' (Student ⨝ Student.CourseID = Course.ID Course))
5. Find students who have taken all courses offered π_StudentID (Enroll) ÷ π_CourseID (Course)

2. Tuple Relational Calculus (TRC)

Feature	Description
Definition	Non-procedural query language using tuples as variables
Tuple Variable	Represents a tuple from a relation
Syntax	{t | P(t)}
Predicate	Condition on tuple t (comparisons, logical operators)
Operators	AND (∧), OR (∨), NOT (¬), =, ≠, <, >, ≤, ≥
Quantifiers	==∃ (exists)==, ==∀ (for all)==
Characteristics	Non-procedural; specifies what to retrieve, not how
Difference from SQL	TRC is theoretical; SQL is practical
Examples :

1. Find all students older than 20 { t | Student(t) ∧ t.Age > 20 }
2. Find all students enrolled in a 4-credit course { t | ∃c (Course(c) ∧ t.CourseID = c.ID ∧ c.Credit = 4) }
3. Find all students with marks > 80 { t | Student(t) ∧ t.Marks > 80 }
4. Find names of students in “DBMS” course { t.Name | ∃c (Course(c) ∧ t.CourseID = c.ID ∧ c.CourseName = 'DBMS') }
5. Find students who have taken all courses offered { s | Student(s) ∧ ∀c (Course(c) → ∃e (Enroll(e) ∧ e.StudentID = s.ID ∧ e.CourseID = c.ID)) }

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Non-Procedural Language (DBMS)

• User specifies what data is required, not how to retrieve

• DBMS decides execution plan
• Define Result Only, High-level, declarative, high abstraction

Examples:

1. SQL
2. Query By Example
3. Domain Relational Calculus (QBE)

eg:- SQL

SELECT name FROM STUDENT WHERE marks > 80;

eg:- Query By Example (QBE)

• Put condition >80 under marks
• Put P. under name

eg:- Domain Relational Calculus (DRC)

{ <n> | ∃ r (STUDENT(n, r) ∧ r > 80) }

• Specify required domain variable in result
• Use logical condition on domains (no execution order)

3. SQL (Structured Query Language)

Feature	Description
Definition	Declarative query language for relational databases
Nature	Non-procedural (user specifies what, not how)
Basis	Based on Relational Algebra & Relational Calculus
Core Clauses	SELECT, FROM, WHERE, GROUP BY, HAVING, ORDER BY
Joins	INNER, LEFT, RIGHT, FULL
Quantification	EXISTS, NOT EXISTS, IN, ALL, ANY
Characteristics	Practical, optimized by DBMS
Difference from TRC	SQL is executable; TRC is mathematical

Examples :

1. Find all students older than 20 SELECT * FROM Student WHERE Age > 20;
2. Find all students enrolled in a 4-credit course SELECT DISTINCT s.* FROM Student s, Course c WHERE s.CourseID = c.ID AND c.Credit = 4;
3. Find all students with marks > 80 SELECT * FROM Student WHERE Marks > 80;
4. Find names of students in “DBMS” course SELECT s.Name FROM Student s, Course c WHERE s.CourseID = c.ID AND c.CourseName = 'DBMS';

2. QBE (Query By Example)

Feature	Description
Definition	Visual, example-based query language
Nature	Non-procedural
Basis	Based on Domain Relational Calculus
Query Style	Conditions filled in table skeletons
Output Indication	P. (print)
Usage	Educational, conceptual (rare in real systems)
Difference from SQL	QBE is graphical; SQL is textual

Examples :

1. Find all students older than 20
   • Put >20 under Age
   • Put P. under required attributes
2. Find all students enrolled in a 4-credit course
   • Course table: put 4 under Credit
   • Match CourseID
   • Put P. under Student attributes
3. Find all students with marks > 80
   • Put >80 under Marks
   • Put P. under Student attributes
4. Find names of students in “DBMS” course
   • Course table: put DBMS under CourseName
   • Match CourseID
   • Put P. under Name
5. Find students who have taken all courses offered
   • Use ALL rows of Course table
   • Match StudentID and CourseID via Enroll
   • Put P. under Student attributes

3. Domain Relational Calculus (DRC)

Feature	Description
Definition	Non-procedural query language using domain variables
Domain Variable	Represents attribute values
Syntax	{ <x1, x2, …> | P(x1, x2, …) }
Predicate	Conditions on domains using logic
Operators	∧, ∨, ¬, =, <, >
Quantifiers	==∃, ∀==
Characteristics	Mathematical, declarative
Difference from TRC	DRC uses domains; TRC uses tuples

Examples :

1. Find all students older than 20 { <n, a> | Student(n, a) ∧ a > 20 }
2. Find all students enrolled in a 4-credit course { <s> | ∃c (Student(s, c) ∧ Course(c, 4)) }
3. Find all students with marks > 80 { <n> | ∃m (Student(n, m) ∧ m > 80) }
4. Find names of students in “DBMS” course { <n> | ∃c (Student(n, c) ∧ Course(c, 'DBMS')) }
5. Find students who have taken all courses offered { <s> | ∀c (Course(c) → ∃e (Enroll(s, c))) }

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