CSE 480: Database Systems
Lecture 15: Relational Algebra
Reference:
Read Chapter 6.1-6.3.1 of the textbook
‹#›
What is Relational Algebra?

A formal way to express a query in relational model
– A query consists of relational expressions describing the
sequence of operators applied to obtain the query result

Example :
 LName, Sex ( DNo=4
OR Salary>100000 (EMPLOYEE)
)
–  and  are the operators
‹#›
Why Study Relational Algebra?
SELECT A.Name, B.Grade
FROM A, B
WHERE A.Id = B.Id
Query processing
in DBMS
 Name, Grade (Id,Name(A  B))
‹#›
Relational Algebra Operator

Output of a relational algebra expression is a relation

Types of operators in a relational algebra expression
– Unary:
op(Relation)
– Binary: Relation op Relation

Relational algebra operators
– select, project, union, set difference, Cartesian product
‹#›
Select Operator ()

condition (R)

Resulting relation
(NOT THE SAME AS SELECT in SQL)
– contains only tuples in R that satisfy the given condition
– has the same attributes (columns) as the original relation

Example:
Status=‘Junior’(Student)
Student
Id
1123
1234
5556
9876
Name
John
Lee
Mary
Bart
Address
123 Main
123 Main
7 Lake Dr
5 Pine St
Status
Junior
Senior
Freshman
Junior
Id
1123
9876
Name Address
John 123 Main
Bart
5 Pine St
Status
Junior
Junior
‹#›
Selection Condition

select_condition (relation)
– Selection condition acts as a filter to the rows in relation

Selection condition is a Boolean expression
– Simple selection condition:


<attribute> operator <constant>
<attribute> operator <attribute>
where operator: <, , , >, =, 
– <condition> AND <condition>
– <condition> OR <condition>
– NOT <condition>
‹#›
Selection Condition - Examples
Student
Id
1123
1234
5556
9876
Selectivity
Name
John
Lee
Mary
Bart
Address
123 Main
123 Main
7 Lake Dr
5 Pine St
Status
Junior
Senior
Freshman
Junior
Fraction of tuples selected
by a selection condition

 Id>3000
OR Status=‘Freshman’

 Id>3000
AND Id <3999 (Student)

 NOT(Status=‘Senior’) (Student)
selectivity = 3/4

 Status‘Senior’ (Student)
selectivity = 3/4
(Student)
selectivity = 2/4
selectivity = 0/4
‹#›
Selection Condition – Incorrect Queries

Find professors who taught both CS315 and CS305
–  CrsCode=‘CS315’

AND CrsCode=‘CS305’ (TEACHING)
X
Find courses taught by CS professors
–  PROFESSOR.Id=Teaching.ProfId
AND PROFESSOR.DeptId=‘CS’
(TEACHING)
X
‹#›
Properties of SELECT operator

Commutative:
–  <condition1>( < condition2> (R)) =  <condition2> ( < condition1> (R))

Cascade of SELECT operations may be applied in any order:
– <cond1>(<cond2> (<cond3> (R)) = <cond2> (<cond3> (<cond1> ( R)))

Cascade of SELECT operations may be replaced by a single selection
with a conjunction of all the conditions:
– <cond1>(< cond2> (<cond3>(R)) =  <cond1> AND < cond2> AND < cond3>(R)))
‹#›
Project Operator

Produces a relation containing a subset of columns of its
input relation
– attribute list(R)

Example:
Name,Status(Student)
Student
Id
1123
1234
5556
9876
Name
Address
John 123 Main
Lee 123 Main
Mary 7 Lake Dr
Bart 5 Pine St
Status
Name
Junior
Freshman
Senior
Junior
John
Lee
Mary
Bart
Status
Junior
Freshman
Senior
Junior
‹#›
Project Operator

Resulting relation has no duplicates; therefore it can have
fewer tuples than the input relation

Example:
Address(Student)
Student
Id
1123
1234
5556
9876
Name
John
Lee
Mary
Bart
Address
123 Main
123 Main
7 Lake Dr
5 Pine St
Status
Junior
Freshman
Senior
Junior
Address
123 Main
7 Lake Dr
5 Pine St
‹#›
Properties of PROJECT Operator

Number of tuples in the result of projection <list>(R) is
always less or equal to the number of tuples in R
– If list of projected attributes includes a superkey, the resulting
relation has the same number of tuples as the input relation

PROJECT operator is not commutative
–  <list1> ( <list2> (R) )   <list2> ( <list1> (R) )
‹#›
Relational Algebra Expressions

Find the Ids and names of junior undergraduate students
Student
Id
1123
1234
5556
9876
Name
Address
John 123 Main
Lee 123 Main
Mary 7 Lake Dr
Bart 5 Pine St
Status
Junior
Freshman
Senior
Junior
Id
Name
1123 John
9876 Bart
Result
 Id, Name ( Status=’Junior’ (Student) )
‹#›
Relational Algebra Expressions

We may want to apply several relational algebra
operations one after the other
– We can write the operations as a single relational algebra
expression by nesting the operations, or
– We can apply one operation at a time and create intermediate
result relations.
– In the latter case, we must give names to the relations that hold
the intermediate results.
‹#›
Single versus Sequence of operations

Query: Find the first name, last name, and salary of all
employees who work in department number 5

We can write a single relational algebra expression:
– FNAME, LNAME, SALARY( DNO=5(EMPLOYEE))

OR We can explicitly show the sequence of operations,
giving a name to each intermediate relation:
– DEP5_EMPS   DNO=5(EMPLOYEE)
– RESULT   FNAME, LNAME, SALARY (DEP5_EMPS)
‹#›
RENAME operator

The RENAME operator is denoted by 

We may rename the attributes of a relation or the relation
name or both
– S (B1, B2, …, Bn )(R)

Change the relation name from R to S, and

Change column (attribute) names to B1, B2, …..Bn
– S(R) :

Change the relation name only to S
– (B1, B2, …, Bn )(R) :

Change the column (attribute) names only to B1, B2, …..Bn
‹#›
Example
 Change relation name to Emp and attributes to First_name,
Last_name, and Salary
Emp(First_name, Last_name, Salary ) (  FNAME,LNAME,SALARY (EMPLOYEE))
or
Emp(First_name, Last_name, Salary)   FNAME,LNAME,SALARY (EMPLOYEE)
‹#›
Set Operators

A relation is a set of tuples; therefore set operations are
applicable:
– Intersection: 
– Union: 
– Set difference (Minus): 

Set operators are binary operators
– Operator: Relation  Relation  Relation

Result of set operation is a relation that has the same
schema as the combining relations
‹#›
Examples: Set Operations
X
Y
A
B
A
B
x1
x2
x1
x2
x3
x4
x5
x6
XY
XY
X–Y
A
B
A
B
A
B
x1
x2
x1
x2
x3
x4
x3
x4
x5
x6
‹#›
Example
STUDENT

INSTRUCTOR
STUDENT – INSTRUCTOR
STUDENT

INSTRUCTOR – STUDENT
INSTRUCTOR
‹#›
Union Compatible Relations

Set operations are limited to union compatible relations
– Two relations are union compatible if:


they have the same number of attributes, and

the domains of corresponding attributes are type compatible (i.e.
dom(Ai)=dom(Bi) for i=1, 2, ..., n).
The resulting relation has the same attribute names as
the first operand relation
‹#›
Example
Tables:
Student (SSN, Name, Address, Status)
Professor (Id, Name, Office, Phone)
are not union compatible.
But
 Name (Student) and  Name (Professor)
are union compatible
‹#›
Exercise

Relations:
– Employee (SSN, Name, Address)
– Professor (SSN, Office, Phone)
– Student (SSN, Status)

Find the SSN of student employees
 SSN (Employee)   SSN (Student)

Find the SSN of employees who are not professors
 SSN (Employee) –  SSN (Professor)

Find the SSN of employees who are neither professors
nor students
 SSN (Employee) – ( SSN (Student)   SSN (Professor))
‹#›
Cartesian Product

A B
x1 x2
x3 x4
C D
y1 y2
y3 y4
R
S
A
x1
x1
x3
x3
B C
x2 y1
x2 y3
x4 y1
x4 y3
R S
D
y2
y4
y2
y4
R  S is expensive to compute:
– Number of columns = degree(R) + degree(S)
– Number of rows = number of rows (R) × number of rows (S)
‹#›
Example

Broker (BrokerId, BrokerName)
BrokerID
Client (ClientId, ClientName)
1
Merrill Lynch
2
Morgan Stanley
3
Salomon Smith Barney
List all the possible Broker-Client pairs
– Broker  Client
BrokerId
BrokerName
ClientId
ClientName
1
Merrill Lynch
A11
Bill Gates
1
Merrill Lynch
B11
Steve Jobs
2
Morgan Stanley
A11
Bill Gates
2
Morgan Stanley
B11
Steve Jobs
3
Salomon Smith Barney
A11
Bill Gates
3
Salomon Smith Barney
B11
Steve Jobs
BrokerName
ClientId
ClientName
A11
Bill Gates
B11
Steve Jobs
‹#›
More Complex Examples
Find the names of employees who are not supervisors
Nonsup   SSN Employee    SuperSSN Employee 
Answer   Fname , MInit , Lname  Nonsup. SSN Employee. SSN Nonsup  Employee 
‹#›
Example
Find the names of employees who earn more than their supervisors
Emp (SSN1, Sal)   SSN , SalaryEmployee 
Answer   Fname, Lname  SuperSSN SSN1 ANDSalarySal Employee  Emp 
‹#›
Summary (Relational Algebra Operators)


Unary operators
– SELECT
condition(R)
– PROJECT
Attribute-List(R)
– RENAME
S(A1,A2,…Ak)(R)
Set operators
– RS
– RS
– R – S or S – R

Cartesian product
– RS
‹#›