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Lecture 07:Relational Algebra

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Outline

• Relational Algebra (Section 6.1)

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Relational Algebra

• Formalism for creating new relations from existing ones

• Its place in the big picture:

Declarativequery

language

Declarativequery

languageAlgebraAlgebra ImplementationImplementation

SQL,relational calculus

Relational algebra

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Relational Algebra• Five operators:

– Union: – Difference: -– Selection:– Projection: – Cartesian Product:

• Derived or auxiliary operators:– Intersection, complement– Joins (natural,equi-join, theta join, semi-join)– Renaming:

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1. Union and 2. Difference

• R1 R2Example: – ActiveEmployees RetiredEmployees

• R1 – R2Example:– AllEmployees − RetiredEmployees

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What about Intersection ?

• It is a derived operatorR1 R2 = R1 – (R1 – R2)

• Also expressed as a join (will see later)Example– UnionizedEmployees RetiredEmployees

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3. Selection

• Returns all tuples which satisfy a condition

• Notation: c(R)

• Examples– Salary > 40000 (Employee)

– name = “Smith” (Employee)

• The condition c can be =, <, , >, , <>

[in SQL: SELECT * FROM Employee WHERE Salary > 40000]

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Selection Example

EmployeeSSN Name DepartmentID Salary999999999 John 1 30,000777777777 Tony 1 32,000888888888 Alice 2 45,000

SSN Name DepartmentID Salary888888888 Alice 2 45,000

Find all employees with salary more than $40,000.Salary > 40000 (Employee)

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4. Projection• Eliminates columns, then removes duplicates

• Notation: A1,…,An (R)

• Example: project to social-security number and names:– SSN, Name (Employee)

– Output schema: Answer(SSN, Name)

[In SQL: SELECT DISTINCT SSN, Name FROM Employee]

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Projection Example

EmployeeSSN Name DepartmentID Salary999999999 John 1 30,000777777777 Tony 1 32,000888888888 Alice 2 45,000

SSN Name999999999 John777777777 Tony888888888 Alice

SSN, Name (Employee)

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5. Cartesian Product

• Combine each tuple in R1 with each tuple in R2• Notation: R1 R2• Example:

– Employee Dependents

• Very rare in practice; mainly used to express joins

[In SQL: SELECT * FROM R1, R2]

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Cartesian Product Example Employee Name SSN John 999999999 Tony 777777777 Dependents EmployeeSSN Dname 999999999 Emily 777777777 Joe Employee x Dependents Name SSN EmployeeSSN Dname John 999999999 999999999 Emily John 999999999 777777777 Joe Tony 777777777 999999999 Emily Tony 777777777 777777777 Joe

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Relational Algebra• Five operators:

– Union: – Difference: -– Selection:– Projection: – Cartesian Product:

• Derived or auxiliary operators:– Intersection, complement– Joins (natural,equi-join, theta join, semi-join)– Renaming:

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Renaming

• Changes the schema, not the instance

• Schema: R(A1, …, An )

• Notation: B1,…,Bn (R)

• Example:– LastName, SocSocNo (Employee)

– Output schema: Answer(LastName, SocSocNo)

[in SQL: SELECT Name AS LastName, SSN AS SocSocNo FROM Employee]

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Renaming Example

EmployeeName SSNJohn 999999999Tony 777777777

LastName SocSocNoJohn 999999999Tony 777777777

LastName, SocSocNo (Employee)

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Natural Join• Notation: R1 R2⋈• Meaning: R1 R2 = ⋈ A(C(R1 R2))

• Where:– The selection C checks equality of all common attributes– The projection eliminates the duplicate common attributes

[in SQL: SELECT DISTINCT R1.A, R1. B, R2.C FROM R1, R2

WHERE R1.B = R2.B

Schema: R1(A,B), R2(B,C)]

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Natural Join Example

EmployeeName SSNJohn 999999999Tony 777777777

DependentsSSN Dname999999999 Emily777777777 Joe

Name SSN DnameJohn 999999999 EmilyTony 777777777 Joe

Employee Dependents = Name, SSN, Dname( SSN=SSN2(Employee x SSN2, Dname(Dependents))

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Natural Join

• R= S=

• R ⋈ S=

A B

X Y

X Z

Y Z

Z V

B C

Z U

V W

Z V

A B C

X Z U

X Z V

Y Z U

Y Z V

Z V W

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Natural Join

• Given the schemas R(A, B, C, D), S(A, C, E), what is the schema of R ⋈ S ?

• Given R(A, B, C), S(D, E), what is R ⋈ S ?

• Given R(A, B), S(A, B), what is R ⋈ S ?

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Theta Join

• A join that involves a predicate

• R1 ⋈ R2 = (R1 R2)

• Here can be any condition

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Eq-join

• A theta join where is an equality

R1 ⋈A=B R2 = A=B (R1 R2)

• Example:– Employee ⋈SSN=SSN Dependents

• Most useful join in practice(difference to natural join?)

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Semijoin

• R ⋉ S = A1,…,An (R ⋈ S)

• Where A1, …, An are the attributes in R

• Example:– Employee ⋉ Dependents

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Semijoins in Distributed Databases

• Semijoins are used in distributed databases

SSN Name

. . . . . .

SSN Dname Age

. . . . . .

EmployeeDependents

network

Employee ⋈ssn=ssn (age>71 (Dependents))Employee ⋈ssn=ssn (age>71 (Dependents))

T = SSN age>71 (Dependents)R = Employee T⋉

Answer = R ⋈ Dependents

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Complex RA Expressions

Person Purchase Person Product

name=fred name=gizmo

pid ssn

seller-ssn=ssn

pid=pid

buyer-ssn=ssn

name

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Application: Query Rewriting for Optimization

Reserves Sailors

sid=sid

bid=100 rating > 5

sname

Reserves Sailors

sid=sid

bid=100

sname

rating > 5(Scan;write to temp T1)

(Scan;write totemp T2)

The earlier we process selections, less tuples we need to manipulatehigher up in the tree (predicate pushdown)Disadvantages?

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Algebraic Laws (Examples)

• Commutative and Associative Laws– R ∩ S = S ∩ R, R ∩ (S ∩ T) = (R ∩ S) ∩ T

– R S = S R, R (S T) = (R S) T

• Laws involving selection– C AND C’(R) = C( C’(R)) = C(R) ∩ C’(R)

– C (R S) = C (R) S • When C involves only attributes of R

• Laws involving projections– M(N(R)) = M,N(R)

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Operations on Bags

A bag = a set with repeated elements

All operations need to be defined carefully on bags• {a,b,b,c}{a,b,b,b,e,f,f}={a,a,b,b,b,b,b,c,e,f,f}• {a,b,b,b,c,c} – {b,c,c,c,d} = {a,b,b,d}

• C(R): preserve the number of occurrences

• A(R): no duplicate elimination

• Cartesian product, join: no duplicate elimination

Important ! Relational Engines work on bags, not sets !

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Finally: RA has Limitations !

• Cannot compute “transitive closure”

• Find all direct and indirect relatives of Fred• Cannot express in RA !!! Need to write C program

Name1 Name2 Relationship

Fred Mary Father

Mary Joe Cousin

Mary Bill Spouse

Nancy Lou Sister

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Formulating queries in RA

• Consider a database for student enrollment for courses, and books used in the courses– STUDENT (SSN, Name, Major, Bdate)

– COURSE (Course#, Cname, Dept)

– ENROLL (SSN, Course#, Quarter, Grade)

– BOOK_ADOPTION (Course#, Quarter, Book_ISBN)

– TEXT (Book_ISBN, Book_Title, Publisher, Author)

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Formulating queries in RA

• Specify the following queries in relational algebra– List the number of courses (Course#) taken by all

students named ‘John Smith’ in Winter 1999 (i.e., Quarter = W99)

– List any department which has all its adopted books published by ‘BC Publishing’

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Formulating Queries in RA

Course# (Quarter=W99 ((Name= ‘John Smith’ (STUDENT) ⋈ ENROLL))

OtherDept = Dept ((Publisher <> ‘PS Publishers’

(BOOK_ADOPTION ⋈ TEXT)) ⋈ COURSE) AllDept = Dept (BOOK_ADOPTION ⋈ COURSE)Answer = AllDept - OtherDept

And how will you express it in SQL?

WHY?