Relational algebra.pptx

©Silberschatz, Korth and Sudarshan3.1Database System Concepts
Chapter 3: Relational Model
 Structure of Relational Databases
 Relational Algebra
 Tuple Relational Calculus
 Domain Relational Calculus
 Extended Relational-Algebra-Operations
 Modification of the Database
 Views

Example of a Relation

Relation Instance
 The current values (relation instance) of a relation are
specified by a table
 An element t of r is a tuple, represented by a row in a table
Jones
Smith
Curry
Lindsay
customer-name
Main
North
North
Park
customer-street
Harrison
Rye
Rye
Pittsfield
customer-city
customer
attributes
(or columns)
tuples
(or rows)

Query Language
 A query language is a language in which a user requests
information from the database.
 These languages are usually on a level higher than that of a
standard programming language.
 Query languages can be categorized as either procedural or
nonprocedural.
 In a procedural language, the user instructs the system to perform a
sequence of operations on the database to compute the desired
result.
 In a nonprocedural language, the user describes the desired
information without giving a specific procedure for obtaining that
information.
 The relational algebra is procedural.
 the tuple relational calculus and domain relational calculus are
nonprocedural.

RELATIONAL ALGEBRA
 The relational algebra is a procedural query language. It consists
of a set of operations that take one or two relations (tables) as
input and produce a new relation, on the request of the user to
retrieve the specific information, as the output.
 The relational algebra contains the following operations

Relational Algebra
 Procedural language
 Six basic operators
 select
 project
 union
 set difference
 Cartesian product
 rename
 The operators take one or more relations as inputs and give a
new relation as a result.

 Unary Relational Operations
 SELECT (symbol: σ)
 PROJECT (symbol: π)
 RENAME (symbol: ρ)
 Relational Algebra Operations From Set Theory
 UNION (υ)
 INTERSECTION ( ),
 DIFFERENCE (-)
 CARTESIAN PRODUCT ( x )
 Binary Relational Operations
 JOIN
 DIVISION

SELECTION (σ)
• The SELECT operation is used for selecting a subset
of the tuples according to a given selection condition.
Sigma(σ)Symbol denotes it.
• Ex:- find all employees born after 1st Jan
1950:
•
dob σ
'01/JAN/1950'(employee)

 σp(r)
 σ is the predicate
 r stands for relation which is the name of the table
 p is prepositional logic

 Example 1
 σ topic = "Database" (Tutorials)
 Output - Selects tuples from Tutorials where topic = 'Database'.
 Example 2
 σ topic = "Database" and author = "guru99"( Tutorials)
 Output - Selects tuples from Tutorials where the topic is
'Database' and 'author' is guru99.
 Example 3
 σ sales > 50000 (Customers)
 Output - Selects tuples from Customers where sales is greater
than 50000

sSalary > 40000 (Employee)
SSN Name Salary
1234545 John 200000
5423341 Smith 600000
4352342 Fred 500000
SSN Name Salary
5423341 Smith 600000
4352342 Fred 500000

σsubject = "database"(Books)
Output − Selects tuples from books where subject is 'database'.
σsubject = "database" and price = "450"(Books)
Output − Selects tuples from books where subject is 'database' and
'price' is 450.
σsubject = "database" and price = "450" or year > "2010"(Books)
Output − Selects tuples from books where subject is 'database' and
'price' is 450 or those books published after 2010.

PROJECTION(∏ )Pi
• ∏ (pi) symbol used to choose attributes
from a relation.
This helps to extract the values of
specified attributes to eliminates
duplicate values. (pi) symbol is used to
choose attributes from a relation.

SELECTION & PROJECTION Example
Id
Person
Name Address Hobby
1123
1123
5556
9876
John
John
Mary
Bart
123 Main
123 Main
7 Lake Dr
5 Pine St
stamps
coins
hiking
stamps
Id Hobby
1123
Name Address
John 123 Main stamps
9876 Bart 5 Pine St stamps
σ Hobby=‘stamps’(Person)
∏Name,Hobby(Person)
Name Hobby
John
John
Mary
Bart
stamps
coins
Hiking
stamps

P Name,Salary (Employee)
SSN Name Salary
1234545 John 200000
5423341 John 600000
4352342 John 200000
Name Salary
John 20000
John 60000

Project Operation – Example
 Relation r: A B C




10
20
30
40
1
1
1
2
A C




1
1
1
2
=
A C



1
1
2
 A,C (r)

 CustomerID CustomerName Status
 1 Google Active
 2 Amazon Active
 3 Apple Inactive
 4 Alibaba Active
 Here, the projection of CustomerName and status will give
 Π CustomerName, Status (Customers)
 CustomerName Status
 Google Active
 Amazon Active
 Apple Inactive
 Alibaba Active

 The results of relational algebra are also relations but without
any name. The rename operation allows us to rename the output
relation. 'rename' operation is denoted with small Greek letter rho
ρ.
 Notation − ρ x (E)
 Where the result of expression E is saved with name of x
'rename'

UNION
 UNION is symbolized by ∪ symbol. It includes all tuples that are
in tables A or in B. It also eliminates duplicate tuples. So, set A
UNION set B would be expressed as:
 The result <- A ∪ B
 For a union operation to be valid, the following conditions must
hold -
 R and S must be the same number of attributes.
 Attribute domains need to be compatible.
 Duplicate tuples should be automatically removed.

Union Operation – Example
 Relations r, s:
r  s:
A B



1
2
1
A B


2
3
r
s
A B




1
2
1
3

 TABLE A Table B
 column 1 column 2 column 1 column 2
 1 1 1 1
 1 2 1 3
 A ∪ B gives
 Table A ∪ B
 column 1 column 2
 1 1
 1 2
 1 3

Set Difference (-)
 - Symbol denotes it. The result of A - B, is a relation which
includes all tuples that are in A but not in B.
 The attribute name of A has to match with the attribute name in
B.

Set Difference Operation – Example
 Relations r, s:
r – s:
A B



1
2
1
A B


2
3
r
s
A B


1
1

What about Intersection ?
 An intersection is defined by the symbol ∩
 A ∩ B
 Defines a relation consisting of a set of all tuple that are in both
A and B. However, A and B must be union-compatible.

 Table A ∩ B
 TABLE A Table B
 column 1 column 2 column 1 column 2
 1 1 1 1
 1 2 1 3
 column 1 column 2
 1 1

Cartesian-Product Operation
 artesian Product in DBMS is an operation used to merge
columns from two relations.
 Notation − r Χ s
 Where r and s are relations and their output will be defined as −
 σauthor = 'tutorialspoint'(Books Χ Articles)
 Output − Yields a relation, which shows all the books and articles
written by tutorialspoint.

Cartesian-Product Operation-Example
Relations r, s:
r x s:
A B


1
2
A B








1
1
1
1
2
2
2
2
C D








10
10
20
10
10
10
20
10
E
a
a
b
b
a
a
b
b
C D




10
10
20
10
E
a
a
b
br
s

Join Operations

NATURAL JOIN (⋈)
Natural join can only be performed if there is a common attribute
(column) between the relations. The name and type of the attribute
must be same.

 Example: Find the employee names and city who have salary
details.
semp_name, salary, city ( employee ⋈ employee_works )
 The join operation selects all employees with salary details, from
where we can easily project the employee names, cities and
salaries. Natural Join operation results in some loss of
information

Right Outer Join:

Outer Join – Example
 Relation loan
 Relation borrower
customer-name loan-number
Jones
Smith
Hayes
L-170
L-230
L-155
3000
4000
1700
loan-number amount
L-170
L-230
L-260
branch-name
Downtown
Redwood
Perryridge

 Inner Join
loan Borrower
loan-number amount
L-170
L-230
3000
4000
customer-name
Jones
Smith
branch-name
Downtown
Redwood
Jones
Smith
null
loan-number amount
L-170
L-230
L-260
3000
4000
1700
customer-namebranch-name
Downtown
Redwood
Perryridge
 Left Outer Join
loan Borrower

 Right Outer Join
loan borrower
loan borrower
 Full Outer Join
loan-number amount
L-170
L-230
L-155
3000
4000
null
customer-name
Jones
Smith
Hayes
branch-name
Downtown
Redwood
null
loan-number amount
L-170
L-230
L-260
L-155
3000
4000
1700
null
customer-name
Jones
Smith
null
Hayes
branch-name
Downtown
Redwood
Perryridge
null

Banking Example
branch (branch-name, branch-city, assets)
customer (customer-name, customer-street, customer-only)
account (account-number, branch-name, balance)
loan (loan-number, branch-name, amount)
depositor (customer-name, account-number)
borrower (customer-name, loan-number)

Example Queries
 Find all loans of over $1200
Find the loan number for each loan of an amount greater than
$1200
samount > 1200 (loan)
loan-number (samount > 1200 (loan))

Example Queries
 Find the names of all customers who have a loan, an account, or
both, from the bank
Find the names of all customers who have a loan and an
account at bank.
customer-name (borrower)  customer-name (depositor)
customer-name (borrower)  customer-name (depositor)

53
Examples
Sailors (sid, name,
rating, age) bid name color
101 Interlake blue
102 Interlake red
103 Clipper green
104 Marine red
Boats (bid, name,
color) sid bid day
1 101 10/10/12
1 102 10/10/12
1 101 10/7/12
2 102 11/9/12
2 102 7/11/12
3 101 7/11/12
3 102 7/8/12
4 103 19/9/12
Reserves (sid, bid,
day)
List names of boats.
List ratings and ages sailors.
List names of sailors who are over 21 years old.
List names of red boats.
πname (
Boats)
πrating, age
(Sailors)
πname ( σage>21
(Sailors))πname ( σcolor=red
(Boats))
sid name rating age
1 Dustin 7 45
2 Rusty 10 35
3 Horatio 5 35
4 Zorba 8 18
5 Julius 25

54
Examples
Sailors (sid, name,
101 Interlake blue
102 Interlake red
103 Clipper green
104 Marine red
Boats (bid, name,
color) sid bid day
1 101 10/10/12
1 102 10/10/12
1 101 10/7/12
2 102 11/9/12
2 102 7/11/12
3 101 7/11/12
3 102 7/8/12
4 103 19/9/12
Reserves (sid, bid,
day)
List ids of boats named Interlake
List ids of boats reserved on 10/10/12
sid name rating age
1 Dustin 7 45
2 Rusty 10 35
3 Horatio 5 35
4 Zorba 8 18
5 Julius 25

55
Examples (solution)
Sailors (sid, name,rating, age) Boats (bid, name,
color) Reserves (sid, bid,day)bid name color
101 Interlake blue
102 Interlake red
103 Clipper green
104 Marine red
sid bid day
1 101 10/10/12
1 102 10/10/12
1 101 10/7/12
2 102 11/9/12
2 102 7/11/12
3 101 7/11/12
3 102 7/8/12
4 103 19/9/12List ids of boats named Interlake
bid (snameInterlake(Boats))
List ids of boats reserved on 10/10/12
bid (sday10/10/12 (Reserves))
sid name rating age
1 Dustin 7 45
2 Rusty 10 35
3 Horatio 5 35
4 Zorba 8 18
5 Julius 25

56
Examples
Sailors (sid, name,
101 Interlake blue
102 Interlake red
103 Clipper green
104 Marine red
Boats (bid, name,
color) sid bid day
1 101 10/10/12
1 102 10/10/12
1 101 10/7/12
2 102 11/9/12
2 102 7/11/12
3 101 7/11/12
3 102 7/8/12
4 103 19/9/12
Reserves (sid, bid,
day)
List ids of sailors who reserved boat 102
πsid ( σbid=102 Reserves)
List names of sailors who reserved boat 102
πname (Sailors ⨝ ( σbid=102
Reserves))
πname (σbid=102 (Sailors ⨝
Reserves))
both are correct!
which is better?
sid name rating age
1 Dustin 7 45
2 Rusty 10 35
3 Horatio 5 35
4 Zorba 8 18
5 Julius 25

Example Queries
 Find the names of all customers having a loan at the Perryridge
branch
{t | s  borrower( t[customer-name] = s[customer-name]
 u  loan(u[branch-name] = “Perryridge”
 u[loan-number] = s[loan-number]))
 not v  depositor (v[customer-name] =
t[customer-name]) }
 Find the names of all customers who have a loan at the
Perryridge branch, but no account at any branch of the bank
{t | s  borrower(t[customer-name] = s[customer-name]
 u  loan(u[branch-name] = “Perryridge”
 u[loan-number] = s[loan-number]))}

Example Queries
 Find the names of all customers having a loan from the
Perryridge branch, and the cities they live in
{t | s  loan(s[branch-name] = “Perryridge”
 u  borrower (u[loan-number] = s[loan-number]
 t [customer-name] = u[customer-name])
  v  customer (u[customer-name] = v[customer-name]
 t[customer-city] = v[customer-city])))}

Example Queries
 Find the names of all customers who have an account at all
branches located in Brooklyn:
{t |  c  customer (t[customer.name] = c[customer-name]) 
 s  branch(s[branch-city] = “Brooklyn” 
 u  account ( s[branch-name] = u[branch-name]
  s  depositor ( t[customer-name] = s[customer-name]
 s[account-number] = u[account-number] )) )}

Safety of Expressions
 It is possible to write tuple calculus expressions that generate
infinite relations.
 For example, {t |  t r} results in an infinite relation if the
domain of any attribute of relation r is infinite
 To guard against the problem, we restrict the set of allowable
expressions to safe expressions.
 An expression {t | P(t)} in the tuple relational calculus is safe if
every component of t appears in one of the relations, tuples, or
constants that appear in P
 NOTE: this is more than just a syntax condition.
 E.g. { t | t[A]=5  true } is not safe --- it defines an infinite set with
attribute values that do not appear in any relation or tuples or
constants in P.

Domain Relational Calculus
 A nonprocedural query language equivalent in power to the tuple
relational calculus
 Each query is an expression of the form:
{  x1, x2, …, xn  | P(x1, x2, …, xn)}
 x1, x2, …, xn represent domain variables
 P represents a formula similar to that of the predicate calculus

Example Queries
 Find the loan-number, branch-name, and amount for loans of over
$1200
{ c, a  |  l ( c, l   borrower  b( l, b, a   loan 
b = “Perryridge”))}
or { c, a  |  l ( c, l   borrower   l, “Perryridge”, a   loan)}
 Find the names of all customers who have a loan from the
Perryridge branch and the loan amount:
{ c  |  l, b, a ( c, l   borrower   l, b, a   loan  a > 1200)}
 Find the names of all customers who have a loan of over $1200
{ l, b, a  |  l, b, a   loan  a > 1200}

Example Queries
 Find the names of all customers having a loan, an account, or
both at the Perryridge branch:
{ c  |  s, n ( c, s, n   customer) 
 x,y,z( x, y, z   branch  y = “Brooklyn”) 
 a,b( x, y, z   account   c,a   depositor)}
 Find the names of all customers who have an account at all
branches located in Brooklyn:
{ c  |  l ({ c, l   borrower
  b,a( l, b, a   loan  b = “Perryridge”))
  a( c, a   depositor
  b,n( a, b, n   account  b = “Perryridge”))}

Safety of Expressions
{  x1, x2, …, xn  | P(x1, x2, …, xn)}
is safe if all of the following hold:
1.All values that appear in tuples of the expression are values
from dom(P) (that is, the values appear either in P or in a tuple
of a relation mentioned in P).
2.For every “there exists” subformula of the form  x (P1(x)), the
subformula is true if and only if there is a value of x in dom(P1)
such that P1(x) is true.
3. For every “for all” subformula of the form x (P1 (x)), the
subformula is true if and only if P1(x) is true for all values x
from dom (P1).

Result of s branch-name = “Perryridge” (loan)

Loan Number and the Amount of the Loan

Names of All Customers Who Have
Either a Loan or an Account

Customers With An Account But No Loan

Result of borrower  loan

Result of s branch-name = “Perryridge” (borrower  loan)

Result of Pcustomer-name

Result of the Subexpression

Largest Account Balance in the Bank

Customers Who Live on the Same Street and In the
Same City as Smith

Customers With Both an Account and a Loan
at the Bank

Result of Pcustomer-name, loan-number, amount
(borrower loan)

Result of Pbranch-name(scustomer-city =
“Harrison”(customer account depositor))

Result of Pbranch-name(sbranch-city =
“Brooklyn”(branch))

Result of Pcustomer-name, branch-name(depositor account)

The credit-info Relation

Result of Pcustomer-name, (limit – credit-balance) as
credit-available(credit-info).

The pt-works Relation

The pt-works Relation After Grouping

Result of branch-name  sum(salary) (pt-works)

Result of branch-name  sum salary, max(salary) as
max-salary (pt-works)

The employee and ft-works Relations

The Result of employee ft-works

Result of employee ft-works

Tuples Inserted Into loan and borrower

Names of All Customers Who Have a
Loan at the Perryridge Branch

E-R Diagram

The branch Relation

The loan Relation

The borrower Relation

Relational algebra.pptx

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