relational algebra and it's implementation

RELATIONAL ALGEBRA
Shyama Barna Bhattacharya
Lecturer
Department of Computer
Science and Engineering

•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.

Database Scheme/schema
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.

Difference Between Schema and
Instance
Schema Instance
It is the overall description of the database.
It is the collection of information stored in a database at a
particular moment.
The schema is same for the whole database.
Data in instances can be changed using addition, deletion, and
updation.
Does not change Frequently. Changes Frequently.
Defines the basic structure of the database i.e. how the data will
be stored in the database.
It is the set of Information stored at a particular time.

“Schema” and “Instance” are key ideas in a database management system (DBMS) that
help organize and manage data. Let’s begin by examining their distinctions from one
another.
Instances
An Instance is the state of an operational database with data at any given time. It contains
a snapshot of the database. The instances can be changed by certain CRUD operations,
such as like addition, and deletion of data. It may be noted that any search query will not
make any kind of changes in the instances.
Example of instances:
Let’s say a table teacher in our database whose name is School, suppose the table has 50
records so the instance of the database has 50 records for now and tomorrow we are
going to add another fifty records so tomorrow the instance has a total of 100 records.
This is called an instance.
Schema
Schema is the overall description of the database. The basic structure of how the data will
be stored in the database is called schema.

SCHEMA (**Already discussed in
class**)
Example:
Let’s say a table teacher in our database named school, the teacher table requires the name, dob, and doj in their table
so we design a structure as:
Teacher table
name: String
doj: date
dob: date
Above given is the schema of the table teacher.
Schema
Schema is the overall description of the database. The basic structure of how the
data will be stored in the database is called schema.
SCHEMA
Schema is of three types: Logical Schema,
Physical Schema and view Schema.
Logical Schema – It describes the database
designed at a logical level.
Physical Schema – It describes the database
designed at the physical level.
View Schema – It defines the design of the
database at the view level.

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:

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 (), Project (), Join
(⋈)

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.

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

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

• 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 ?

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)

Selection Operator
•The select operation selects tuples that
satisfy a given predicate.
•It is denoted by sigma (σ).
Notation: σ p(r)
Where:
σ is used for selection prediction
r is used for relation
p is used as a propositional logic formula which may use connectors
like: AND OR and NOT. These relational can use as relational operators
like =, ≠, ≥, <, >, ≤.

Example of Selection
Input:
1.σ BRANCH_NAME="perryride" (LOAN)
BRANCH_NAME LOAN_NO AMOUNT
Downtown L-17 1000
Redwood L-23 2000
Perryride L-15 1500
Downtown L-14 1500
Mianus L-13 500
Roundhill L-11 900
Perryride L-16 1300

Example of Selection
Output:
BRANCH_NAME LOAN_NO AMOUNT
Perryride L-15 1500
Perryride L-16 1300

SEMANTICS OF THE SAMPLE RELATIONS
• Sailors: Entity set; lists the relevant properties of sailors.
• Boats: 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”.

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:
Select operators only select rows from relation.
• π
condition

Projection operator
•This operator shows the list of those
attributes that we wish to appear in the
result. Rest of the attributes are
eliminated from the table.
•It is denoted by ∏(pie).
•It is used to choose all columns from
relation, deletes unwanted columns for
relation.

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:

- 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

Selection
• Denoted by c(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.

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:
 state=‘IL’(S)

Example 2:
CNO CNAME CREDIT DEPT
C1
Databas
e
3 CS
C2
Statistic
s
3 MATH
C3 Tennis 1 SPORTS
C4 Violin 4 MUSIC
C5 Golf 2 SPORTS
C6 Piano 5 MUSIC
C:
 CREDIT  3(C)

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:
SNO=‘S1’and CNO=‘C1’(E)

Selection - Properties
• Selection Operator is commutative
C1(C2 (R)) = C2(C1 (R))
• The Selection is an unary operator, it
cannot be used to select tuples from more
than one relations.

Projection
• Denoted by L(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.

Projection - Example
• Example 1: STATE (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

Example 2: CNAME, DEPT(C)
CNO CNAME CREDIT DEPT
C1
Databas
e
3 CS
C2
Statistic
s
3 MATH
C3 Tennis 1 SPORTS
C4 Violin 4 MUSIC
C5 Golf 2 SPORTS
C6 Piano 5 MUSIC
CNAME DEPT
Databas
e
CS
Statistic
s
MATH
Tennis SPORTS
Violin MUSIC
Golf SPORTS
Piano MUSIC

Example 3: S#(STATE=‘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

SET Operations
• UNION: R1  R2
• INTERSECTION: R1  R2
• DIFFERENCE: R1 - R2
• CARTESIAN PRODUCT: R1  R2

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).

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?

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

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:

EXAMPLE
• Assume a database with the following three relations:
• Query 1: Find the bid of red colored boats:
– ∏bid(бcolor=red(Boats))

EXAMPLE
• Assume a database with the following three
relations:
• Query 1: Find the name of sailors who have
reserved Boat number 2.

EXAMPLE
• Query 1: Find the name of sailors who have reserved
Boat number 2.
– ∏sname(бbid=2(Sailors (sid)Reserve))

EXAMPLE
• Query 1: Find the name of sailors who have reserved
both a red and a green boat.

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

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.

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

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

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

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.

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.
<
<
<

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

EXAMPLE
• Emp table:
• бSalary=30,000(Employee)
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

EXAMPLE
• Emp table:
• бSalary=30,000(Employee)
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
4 Kathy 30 30000 5

EXAMPLE
• Emp table:
• бAge>22(Employee)
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

EXAMPLE
• Emp table:
• бAge>22(Employee)
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
1 Joe 24 20000 2
4 Kathy 30 30000 5

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))

EXAMPLE
• Emp table:
• ∏ age(Emp)
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

EXAMPLE
• Emp table:
• ∏ name,age(бSalary=4000 (Emp) )
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

EXAMPLE
• Emp table:
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
5 Shideh 4 4000 1

EXAMPLE
• Emp table:
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

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)

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 Shoe 2 943
dno dname floor mgrSS#

• 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
6
Joe SQL 22 32,00
0
1 2 Shoe 2 943
SS# Name age salary Emp.dno Dept.dno dname floor mgrSS#

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

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

• ∏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

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

345 John
Doe
23 25,000 1 1 Toy 1 345
876 Joe SQL 22 32,000 1 1 Toy 1 345

John Doe
Joe SQL
Name

•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

EQUALITY JOIN, NATURAL JOIN,
JOIN, SEMI-JOIN,LEFT JOIN, RIGHT
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)

EXAMPLE JOIN
• Equality Join, (Emp Dept)))
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.dno Dept.dn
o
dname 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

EXAMPLE JOIN
• Natural Join, (Emp Dept)
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

EMP_ID EMP_NA
ME
CITY SALARY AGE
1 Angelina Chicago 200000 30
2 Robert Austin 300000 26
3 Christian Denver 100000 42
4 Kristen Washingt
on
500000 29
5 Russell Los
angels
200000 36
6 Marry Canada 600000 48
EMPLOYEE
PROJECT
NO
EMP_ID DEPARTMENT
101 1 Testing
102 2 Development
103 3 Designing
104 4 Development
PROJECT

INNER JOIN
In SQL, INNER JOIN selects records that have
matching values in both tables as long as the
condition is satisfied. It returns the combination of
all rows from both the tables where the condition
satisfies.
Syntax
SELECT table1.column1, table1.column2, table2.column1,....
FROM table1
INNER JOIN table2
ON table1.matching_column = table2.matching_column;

Query
SELECT EMPLOYEE.EMP_NAME, PROJECT.DEPARTMENT
FROM EMPLOYEE
INNER JOIN PROJECT
ON PROJECT.EMP_ID = EMPLOYEE.EMP_ID

OUTPUT of INNER JOIN
EMP_NAME DEPARTMENT
Angelina Testing
Robert Development
Christian Designing
Kristen Development

LEFT JOIN
SYNTAX
FROM table1
LEFT JOIN table2
QUERY
FROM EMPLOYEE
LEFT JOIN PROJECT
ON PROJECT.EMP_ID = EMPLOYEE.EMP_ID;

OUTPUT OF LEFT JOIN
EMP_NAME DEPARTMEN
T
Angelina Testing
Robert Development
Christian Designing
Kristen Development
Russell NULL
Marry NULL

RIGHT JOIN
SYNTAX
FROM table1
RIGHT JOIN table2
QUERY
FROM EMPLOYEE
RIGHT JOIN PROJECT

OUTPUT OF RIGHT JOIN
EMP_NAME DEPARTMENT
Angelina Testing
Robert Development
Christian Designing
Kristen Development

FULL JOIN
Syntax
FROM table1
FULL JOIN table2
Query
FROM EMPLOYEE
FULL JOIN PROJECT

Output of FULL JOIN
EMP_NAME DEPARTMENT
Angelina Testing
Robert Development
Christian Designing
Kristen Development
Russell NULL
Marry NULL

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