Relational Algebra

1. RELATIONAL ALGEBRA
Md. Zahid Hasan
Lecturer
Department of Computer
Science and Engineering

2. •Relation schema
–Named relation defined by a set of attribute and
domain name pairs.
•Relational database schema
–Set of relation schemas, each with a distinct
name.
•Each tuple is distinct; there are no duplicate tuples.
•Order of attributes has no significance.
•Order of tuples has no significance, theoretically.
•Relation name is distinct from all other relation
names in relational schema.
•Each cell of relation contains exactly one atomic
(single) value.
•Each attribute has a distinct name.
•Values of an attribute are all from the same domain.
•Each tuple is distinct; there are no duplicate tuples.
•Order of attributes has no significance.
•Order of tuples has no significance, theoretically.

3. Database Scheme
A relational database scheme, or schema,
corresponds to a set of table definitions.
Eg: product(p_id, name, category, description)
supply(p_id, s_id, qnty_per_month)
supplier(s_id, name, address, ph#)
* remember the difference between a DB instance and
a DB scheme.

4. SAMPLE SCHEMAS AND INSTANCES
The Schemas:
Sailors(sid: integer, sname: string, rating: integer, age: real)
Boats(bid: integer, bname: string, color: string)
Reserves(sid: integer, bid: integer, day: date)
The Instances:

6. What is Relational Algebra?
• Relational algebra is a procedural query language.
• It consists of the select, project, union, set
difference, Cartesian product, and rename
operations.
• Set intersection, division, natural join, and
assignment combine the fundamental operations.
• SQL is based on relational algebra

7. •Relational algebra and relational
calculus are formal languages
associated with the relational
model.
•Both are equivalent to one another

8. • It is an abstract language. We use it to express
the set of operations that any relational query
language must perform.
• Two types of operations:
• 1.set-theoretic operations: tables are essentially
sets of rows
• 2.native relational operations: focus on the
structure of the rows Query languages are
specialized languages for asking
questions,or queries,that involve the
data in database.
What are the query languages ?

9. Query languages
• procedural vs. non-procedural
• commercial languages have some of both
• we will study:
– relational algebra (which is procedural, i.e. tells
you how to process a query)
– relational calculus (which is non-procedural i.e.
tells what you want)

10. SEMANTICS OF THE SAMPLE RELATIONS
• Sailors: Entity set; lists the relevant properties of sailors.
• BoatsBoats: Entity set; lists the relevant properties of boats.
• Reserves: Relationship set: links sailors and boats by describing the
boat number and date for which a sailor made a reservation.
Example of the declarative sentences for which rows stand:
Row 1: “Sailor ’22’ reserved boat number ‘101’ on 10/10/98”.

11. Selection and Projection
• Selection Operator: σrating>8 (S2)
Retrieves from the current instance of relation named S2 those rows
where the value of the attribute ‘rating’ is greater than 8.
Applying the above selection operator to the sample instance of S2
shown in figure 4.2 yields the relational instance on figure 4.4 as
shown below:
• π
condition

12. Projection Operator πsname,rating(S2)
Retrieves from the current instance of the relation named S2 those
columns whose names are ‘sname’ and ‘rating’.
Applying the above operator to the sample instance of S2 shown in figure
4.2 yields the relational instance on figure 4.5 as shown below:
N. B.: Note that the projection operator can produce
duplicate rows in the resulting instance.

13. - Projection Operator (cont’d)
Similarly πage(S2) yields the following relational instance
Note here the elimination of
duplicates
SQL would yield
For πage (S2):
age
35.0
55.0
35.0
35.0

14. Introduction
• one of the two formal query languages of the
relational model
• collection of operators for manipulating
relations
• Operators: two types of operators
– Set Operators: Union(∪),Intersection(∩),
Difference(-), Cartesian Product (x)
– New Operators: Select (σ), Project (π), Join (⋈)

15. Introduction – cont’d
• A Relational Algebra Expression: a sequence
of relational algebra operators and operands
(relations), formed according to a set of rules.
• The result of evaluating a relational algebra
expression is a relation.

16. Selection
• Denoted by σc
(R)
• Selects the tuples (rows) from a relation R that
satisfy a certain selection condition c.
• It is a unary operator
• The resulting relation has the same attributes
as those in R.

17. Example 1:
SNO SNAME AGE STATE
S1 MIKE 21 IL
S2 STEVE 20 LA
S3 MARY 18 CA
S4 MING 19 NY
S5 OLGA 21 NY
S:
 σstate=‘IL’(S)

18. Example 2:
CNO
CNAM
E
CREDI
T
DEPT
C1
Databas
e
3 CS
C2
Statistic
s
3 MATH
C3 Tennis 1
SPORT
S
C4 Violin 4 MUSIC
C5 Golf 2
SPORT
S
C6 Piano 5 MUSIC
C:
 σCREDIT ≥ 3(C)

19. Example 3
SNO CNO Grade
S1 C1 90
S1 C2 80
S1 C3 75
S1 C4 70
S1 C5 100
S1 C6 60
S2 C1 90
S2 C2 80
S3 C2 90
S4 C2 80
S4 C4 85
S4 C5 100
E:
σSNO=‘S1’and CNO=‘C1’(E)

21. Selection - Properties
• Selection Operator is commutative
σC1(σC2(R)) = σC2(σC1(R))
• The Selection is an unary operator, it cannot
be used to select tuples from more than one
relations.

22. Projection
• Denoted by πL
(R), where L is list of attribute names and R is a
relation name or some other relational algebra expression.
• The resulting relation has only those attributes of R specified
in L.
• The projection is also an unary operation.
• Duplicate rows are not permitted in relational algebra.
Duplication is removed from the result.
• Duplicate rows can occur in SQL, though they may be
controlled by explicit keywords.

23. Projection - Example
• Example 1: πSTATE
(S)
SNO SNAME AGE STATE
S1 MIKE 21 IL
S2 STEVE 20 LA
S3 MARY 18 CA
S4 MING 19 NY
S5 OLGA 21 NY
STATE
IL
LA
CA
NY

24. Projection - Example
Example 2: πCNAME,DEPT
(C)
CNO
CNAM
E
CREDI
T
DEPT
C1
Databas
e
3 CS
C2
Statistic
s
3 MATH
C3 Tennis 1
SPORT
S
C4 Violin 4 MUSIC
C5 Golf 2
SPORT
S
CNAM
E
DEPT
Databas
e
CS
Statistic
s
MATH
Tennis
SPORT
S
Violin MUSIC
Golf
SPORT
S

25. Projection - Example
Example 3: πS#
(σSTATE=‘NY'
(S))
SNO SNAME AGE STATE
S1 MIKE 21 IL
S2 STEVE 20 LA
S3 MARY 18 CA
S4 MING 19 NY
S5 OLGA 21 NY
SNO
S4
S5

26. SET Operations
• UNION: R1
∪ R2
• INTERSECTION: R1
∩ R2
• DIFFERENCE: R1
- R2
• CARTESIAN PRODUCT: R1
× R2

27. Union Compatibility
• For operators ∪, ∩, -, the operand relations
R1(A1, A2, ..., An) and R2(B1, B2, ..., Bn)
must have the same number of attributes, and
the domains of the corresponding attributes
must be compatible; that is, dom(Ai)=dom(Bi)
for i=1,2,...,n.
• The resulting relation for ∪, ∩, or - has the
same attribute names as the first operand
relation R1 (by convention).

28. Union Compatibility - Examples
• Are S(SNO, SNAME, AGE, STATE) and
C(CNO, CNAME, CREDIT, DEPT) union
compatible?
• Are S(S#, SNAME, AGE, STATE) and
C(CNO, CNAME, CREDIT_HOURS,
DEPT_NAME) union compatible?

30. UNION, SET DIFFERENCE & SET
INTERSECT
• Union puts all tuples of two relations in one relation. To use this operator,
two conditions must hold:
1. The two relations must be of the same arity.
2. The domain of ith
attribute of the two participating relation must be
the same.
• Set difference operator computes tuples that are in one relation, but not
in another.
• Set intersect operator computes tuples that are common in two relations:
• The five fundamental operations of the relational algebra are: select,
project, cartesian product, Union, and set difference
• All other operators can be constructed using these operators

31. EXAMPLE
• Assume a database with the following three
relations:
Sailors (sid, sname, rating)
Boats (bid, bname, color)
Reserve (sid, bid, date)
• Query 1: Find the bid of red colored boats:

32. EXAMPLE
• Assume a database with the following three
relations:
Sailors (sid, sname, rating)
Boats (bid, bname, color)
Reserve (sid, bid, date)
• Query 1: Find the bid of red colored boats:
– ∏bid(бcolor=red(Boats))

33. EXAMPLE
• Assume a database with the following three
relations:
Sailors (sid, sname, rating)
Boats (bid, bname, color)
Reserve (sid, bid, date)
• Query 1: Find the name of sailors who have
reserved Boat number 2.

34. EXAMPLE
• Assume a database with the following three
relations:
Sailors (sid, sname, rating)
Boats (bid, bname, color)
Reserve (sid, bid, date)
• Query 1: Find the name of sailors who have reserved
Boat number 2.
– ∏sname(бbid=2(Sailors (sid)Reserve))

35. EXAMPLE
• Assume a database with the following three
relations:
Sailors (sid, sname, rating)
Boats (bid, bname, color)
Reserve (sid, bid, date)
• Query 1: Find the name of sailors who have reserved
both a red and a green boat.

36. Union, Intersection, Difference
• T= R U S : A tuple t is in relation T if and only
if t is in relation R or t is in relation S
• T = R ∩ S: A tuple t is in relation T if and only
if t is in both relations R and S
• T= R - S :A tuple t is in relation T if and only
if t is in R but not in S

37. Set-Intersection
• Denoted by the symbol .
• Results in a relation that contains only the
tuples that appear in both relations.
• R  S = R – (R – S)
• Since set-intersection can be written in terms
of set-difference, it is not a fundamental
operation.

38. Examples
A1 A2
1 Red
3 White
4 green
B1 B2
3 White
2 Blue
R S

39. Examples
A1 A2
1 Red
3 White
4 Green
2 Blue
A1 A2
3 White
R ∩S
R ∪ S
S - R
B1 B2
2 Blue
A1 A2
1 Red
4 Green
R - S

40. RENAME OPERATOR
• Rename operator changes the name of its
input table to its subscript,
– ρe2(Emp)
– Changes the name of Emp table to e2

41. RELATIONAL ALGEBRA INTRODUCTION
• Assume the following two relations:
Emp (SS#, name, age, salary, dno)
Dept (dno, dname, floor, mgrSS#)
• Relational algebra is a procedural query language, i.e., user must define
both “how” and “what” to retrieve.
• Relational algebra consists of a set of operators that consume either one
or two relations as input. An operator produces one relation as its output.
• Unary operators include: select, project, and rename
• Binary operators include: cartesian product, equality join, natural join,
join, semi-join, division, union, and set difference.

42. SELECT OPERATOR
• Select (б): selects tuples that satisfy a predicate; e.g., retrieve the
employees whose salary is 30,000
бSalary=30,000(Employee)
• Conjunctive ( ) and disjunctive ( ) selection predicates are allowed;
e.g., retrieve employees whose salary is higher than 30,000 and are
younger than 25 years old:
бSalary>30,000 age<25(Employee)
• Note that only selection predicates are allowed. A selection predicate
is either (1) a comparison (=, ≠, ≤, ≥, <, >) between an attribute and a
constant (e.g., salary = 30,000) or (2) a comparison between two
different attributes of the same relation (e.g., salary = age × 100).
• Note: This operator is different than the SELECT command of SQL.
<
<
<

43. EXAMPLE
• Emp table:
SS# Name Age Salary dno
1 Joe 24 20000 2
2 Mary 20 25000 3
3 Bob 22 27000 4
4 Kathy 30 30000 5
5 Shideh 4 4000 1

44. EXAMPLE
• Emp table:
• бSalary=30,000(Employee)
SS# Name Age Salary dno
1 Joe 24 20000 2
2 Mary 20 25000 3
3 Bob 22 27000 4
4 Kathy 30 30000 5
5 Shideh 4 4000 1

45. EXAMPLE
• Emp table:
• бSalary=30,000(Employee)
SS# Name Age Salary dno
1 Joe 24 20000 2
2 Mary 20 25000 3
3 Bob 22 27000 4
4 Kathy 30 30000 5
5 Shideh 4 4000 1
SS# Name Age Salary dno
4 Kathy 30 30000 5

46. EXAMPLE
• Emp table:
• бAge>22(Employee)
SS# Name Age Salary dno
1 Joe 24 20000 2
2 Mary 20 25000 3
3 Bob 22 27000 4
4 Kathy 30 30000 5
5 Shideh 4 4000 1

47. EXAMPLE• Emp table:
• бAge>22(Employee)
SS# Name Age Salary dno
1 Joe 24 20000 2
2 Mary 20 25000 3
3 Bob 22 27000 4
4 Kathy 30 30000 5
5 Shideh 4 4000 1
SS# Name Age Salary dno
1 Joe 24 20000 2
4 Kathy 30 30000 5

48. PROJECT OPERATOR
• Project (∏) retrieves a column. It is a unary operator
that eliminate duplicates.
e.g., name of employees:
∏ name(Employee)
e.g., name of employees earning more than 30,000:
∏ name(бSalary>30,000(Employee))

49. EXAMPLE
• Emp table:
SS# Name Age Salary dno
1 Joe 24 20000 2
2 Mary 20 25000 3
3 Bob 22 27000 4
4 Kathy 30 30000 5
5 Shideh 4 4000 1

50. EXAMPLE
• Emp table:
• ∏ age(Emp)
SS# Name Age Salary dno
1 Joe 24 20000 2
2 Mary 20 25000 3
3 Bob 22 27000 4
4 Kathy 30 30000 5
5 Shideh 4 4000 1
Age
24
20
22
30
4

51. EXAMPLE
• Emp table:
• ∏ name,age(бSalary=4000 (Emp) )
SS# Name Age Salary dno
1 Joe 24 20000 2
2 Mary 20 25000 3
3 Bob 22 27000 4
4 Kathy 30 30000 5
5 Shideh 4 4000 1

52. EXAMPLE
• Emp table:
• ∏ name,age(бSalary=4000 (Emp) )
SS# Name Age Salary dno
1 Joe 24 20000 2
2 Mary 20 25000 3
3 Bob 22 27000 4
4 Kathy 30 30000 5
5 Shideh 4 4000 1
SS# Name Age Salary dno
5 Shideh 4 4000 1

53. EXAMPLE
• Emp table:
• ∏ name,age(бSalary=4000 (Emp) )
SS# Name Age Salary dno
1 Joe 24 20000 2
2 Mary 20 25000 3
3 Bob 22 27000 4
4 Kathy 30 30000 5
5 Shideh 4 4000 1
Name Age
Shideh 4

54. CARTESIAN PRODUCT
• Cartesian Product (R1×R2) combines two relations by
concatenating their tuples together, evaluating all
possible combinations. If the name of a column is
identical for two relations, this ambiguity is resolved
by attaching the name of each relation to a column.
e.g., Emp × Dept
– (SS#, name, age, salary, Emp.dno, Dept.dno, dname, floor,
mgrSS#)
• If t(Emp) and t(Dept) is the cardinality of the
Employee and Dept relations respectively, then the
cardinality of Emp × Dept is: t(Emp) × t(Dept)

55. CARTESIAN PRODUCT (Cont…)
• Example:
Emp table:
Dept table:
345 John Doe 23 25,000 1
943 Jane Java 25 28,000 2
876 Joe SQL 22 32,000 1
SS# Name age salary dno
1 Toy 1 345
2 Sho 2 943
dno dname floor mgrSS#

56. CARTESIAN PRODUCT (Cont…)
• Cartesian product of Emp and Dept: Emp × Dept:
34
5
John
Doe
23 25,00
0
1 1 Toy 1 345
94
3
Jane
Java
25 28,00
0
2 1 Toy 1 345
87
6
Joe SQL 22 32,00
0
1 1 Toy 1 345
34
5
John
Doe
23 25,00
0
1 2 Shoe 2 943
94
3
Jane
Java
25 28,00
0
2 2 Shoe 2 943
87 Joe SQL 22 32,00 1 2 Shoe 2 943
SS# Name age salary Emp.dno Dept.dno dname floor mgrSS#

57. CARTESIAN PRODUCT
• Example: retrieve the name of employees
that work in the toy department:

58. CARTESIAN PRODUCT
• Example: retrieve the name of employees
that work in the toy department:
– ∏name(бEmp.dno=Dept.dno(Emp × бdname=‘toy’(Dept)))

59. CARTESIAN PRODUCT (Cont…)
• ∏name(бdname=‘toy’(бEmp.dno=Dept.dno(Emp × Dept)))
345 John Doe 23 25,000 1 1 Toy 1 345
943 Jane Java 25 28,000 2 1 Toy 1 345
876 Joe SQL 22 32,000 1 1 Toy 1 345
345 John Doe 23 25,000 1 2 Shoe 2 943
943 Jane Java 25 28,000 2 2 Shoe 2 943
876 Joe SQL 22 32,000 1 2 Shoe 2 943
SS# Name age salary Emp.dno Dept.dno dname floor mgrSS#

60. CARTESIAN PRODUCT (Cont…)
• ∏name(бdname=‘toy’(бEmp.dno=Dept.dno(Emp × Dept)))
345 John
Doe
23 25,000 1 1 Toy 1 345
876 Joe
SQL
22 32,000 1 1 Toy 1 345
943 Jane
Java
25 28,000 2 2 Shoe 2 943
SS# Name age salary Emp.dno Dept.dno dname floor mgrSS#

61. CARTESIAN PRODUCT (Cont…)
• ∏name(бdname=‘toy’(бEmp.dno=Dept.dno(Emp × Dept)))
345 John
Doe
23 25,000 1 1 Toy 1 345
876 Joe SQL 22 32,000 1 1 Toy 1 345
SS# Name age salary Emp.dno Dept.dno dname floor mgrSS#

62. CARTESIAN PRODUCT (Cont…)
• ∏name(бdname=‘toy’(бEmp.dno=Dept.dno(Emp × Dept)))
John Doe
Joe SQL
Name

64. •Join is a derivative of Cartesian product.
•Equivalent to performing a Selection, using join
predicate as selection formula, over Cartesian
product of the two operand relations.
•One of the most difficult operations to implement
efficiently in an RDBMS and one reason why
RDBMSs have intrinsic performance problems.
•Various forms of join operation
–Natural join (defined by Codd)
–Outer join
–Theta join
–Equijoin (a particular type of Theta join)
–Semijoin

65. EQUALITY JOIN, NATURAL JOIN,
JOIN, SEMI-JOIN
• Equality join connects tuples from two relations that match on certain
attributes. The specified joining columns are kept in the resulting relation.
– ∏name(бdname=‘toy’(Emp Dept)))
• Natural join connects tuples from two relations that match on the
specified common attributes
– ∏name(бdname=‘toy’(Emp Dept)))
• How is an equality join between Emp and Dept using dno different than a
natural join between Emp and Dept using dno?
– Equality join: SS#, name, age, salary, Emp.dno, Dept.dno, …
– Natural join: SS#, name, age, salary, dno, dname, …
• Join is similar to equality join using different comparison operators
– A S op = {=, ≠, ≤, ≥, <, >}
att op att
(dno)
(dno)

66. EXAMPLE JOIN
• Equality Join, (Emp Dept)))
SS# Name Age Salary dno
1 Joe 24 20000 2
2 Mary 20 25000 1
3 Bob 22 27000 1
4 Kathy 30 30000 2
5 Shideh 4 4000 1
EMP
dno dname floor mgrss#
1 Toy 1 5
2 Shoe 2 1
Dept
(dno)
SS# Name Age Salary EMP.dn
o
Dept.dn
o
dnam
e
floor mgrss
#
1 Joe 24 20000 2 2 Shoe 2 1
2 Mary 20 25000 1 1 Toy 1 5
3 Bob 22 27000 1 1 Toy 1 5
4 Kathy 30 30000 2 2 Shoe 2 1
5 Shideh 4 4000 1 1 Toy 1 5

67. EXAMPLE JOIN
• Natural Join, (Emp Dept)))
SS# Name Age Salary dno
1 Joe 24 20000 2
2 Mary 20 25000 1
3 Bob 22 27000 1
4 Kathy 30 30000 2
5 Shideh 4 4000 1
EMP
dno dname floor mgrss#
1 Toy 1 5
2 Shoe 2 1
Dept
(dno)
SS# Name Age Salary dno dname floor mgrss#
1 Joe 24 20000 2 Shoe 2 1
2 Mary 20 25000 1 Toy 1 5
3 Bob 22 27000 1 Toy 1 5
4 Kathy 30 30000 2 Shoe 2 1
5 Shideh 4 4000 1 Toy 1 5