1. Lecture 07: Relational Algebra

2. Outline
Relational Algebra (Section 6.1)

3. Declarative query language
Relational Algebra
Formalism for creating new relations from existing ones
Its place in the big picture:
Declarative query language
Algebra
Implementation
Relational algebra
SQL, relational calculus

4. Relational Algebra Five operators: Derived or auxiliary operators:
Union: 
Difference: -
Selection: s
Projection: P
Cartesian Product: 
Derived or auxiliary operators:
Intersection, complement
Joins (natural,equi-join, theta join, semi-join)
Renaming: r

5. 1. Union and 2. Difference R1  R2 Example: R1 – R2 Example:
ActiveEmployees  RetiredEmployees
R1 – R2 Example:
AllEmployees − RetiredEmployees

6. What about Intersection ?
It is a derived operator R1  R2 = R1 – (R1 – R2)
Also expressed as a join (will see later) Example
UnionizedEmployees  RetiredEmployees

7. 3. Selection Returns all tuples which satisfy a condition
Notation: sc(R)
Examples
sSalary > (Employee)
sname = “Smith” (Employee)
The condition c can be =, <, , >, , <>
[in SQL: SELECT * FROM Employee WHERE Salary > 40000]

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

9. 4. Projection Eliminates columns, then removes duplicates
Notation: P A1,…,An (R)
Example: project to social-security number and names:
P SSN, Name (Employee)
Output schema: Answer(SSN, Name)
[In SQL: SELECT DISTINCT SSN, Name FROM Employee]

10. P SSN, Name (Employee)

11. 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]

13. Relational Algebra Five operators: Derived or auxiliary operators:
Union: 
Difference: -
Selection: s
Projection: P
Cartesian Product: 
Derived or auxiliary operators:
Intersection, complement
Joins (natural,equi-join, theta join, semi-join)
Renaming: r

14. Renaming Changes the schema, not the instance Schema: R(A1, …, An )
Notation: r B1,…,Bn (R)
Example:
rLastName, SocSocNo (Employee)
Output schema: Answer(LastName, SocSocNo)
[in SQL: SELECT Name AS LastName, SSN AS SocSocNo FROM Employee]

15. LastName, SocSocNo (Employee)
Renaming Example
Employee
Name
SSN
John
Tony
LastName, SocSocNo (Employee)
LastName
SocSocNo
John
Tony

16. Natural Join Notation: R1 ⋈ R2 Meaning: R1 ⋈ R2 = PA(sC(R1  R2))
Where:
The selection sC 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)]

17. Natural Join Example
Employee
Name
SSN
John
Tony
Dependents
SSN
Dname
Emily
Joe
Employee Dependents =
PName, SSN, Dname(s SSN=SSN2(Employee x rSSN2, Dname(Dependents))
Name
SSN
Dname
John
Emily
Tony
Joe

18. Natural Join
R= S=
R ⋈ S= A B X Y Z V B C Z U V W A B C X Z U V Y W

19. 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 ?

20. Theta Join A join that involves a predicate R1 ⋈ q R2 = s q (R1  R2)
Here q can be any condition

21. Eq-join A theta join where q is an equality
R1 ⋈A=B R2 = s A=B (R1  R2)
Example:
Employee ⋈SSN=SSN Dependents
Most useful join in practice (difference to natural join?)

22. Semijoin R ⋉ S = P A1,…,An (R ⋈ S)
Where A1, …, An are the attributes in R
Example:
Employee ⋉ Dependents

23. Semijoins in Distributed Databases
Semijoins are used in distributed databases
Dependents
Employee
SSN
Dname
Age
. . .
SSN
Name
. . .
network
Employee ⋈ssn=ssn (s age>71 (Dependents))
T = P SSN s age>71 (Dependents)
R = Employee ⋉ T
Answer = R ⋈ Dependents

24. Complex RA Expressions
P name
buyer-ssn=ssn
pid=pid
seller-ssn=ssn
P ssn
P pid
sname=fred
sname=gizmo
Person Purchase Person Product

25. 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)
temp T2)
- If predicates for selection have significant overlap, then it will require unnecessary computation and memory consumption.
The earlier we process selections, less tuples we need to manipulate
higher up in the tree (predicate pushdown)
Disadvantages?

26. 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
s C AND C’(R) = s C(s C’(R)) = s C(R) ∩ s C’(R)
s C (R S) = s C (R) S
When C involves only attributes of R
Laws involving projections
PM(PN(R)) = PM,N(R)

27. 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}
sC(R): preserve the number of occurrences
PA(R): no duplicate elimination
Cartesian product, join: no duplicate elimination
Important ! Relational Engines work on bags, not sets !

28. 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
Joe
Cousin
Bill
Spouse
Nancy
Lou
Sister

29. 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)

30. 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’

31. Formulating Queries in RA
PCourse# (s Quarter=W99 ((s Name= ‘John Smith’ (STUDENT) ⋈ ENROLL))
OtherDept = PDept ((s Publisher <> ‘PS Publishers’ (BOOK_ADOPTION ⋈ TEXT)) ⋈ COURSE) AllDept = PDept (BOOK_ADOPTION ⋈ COURSE) Answer = AllDept - OtherDept
And how will you express it in SQL?
WHY?