2. Relational Algebra
Relational algebra and relational calculus are
formal languages associated with the relational
model.
The Relational Algebra is used to define the ways
in which relations (tables) can be operated to
manipulate their data.
It is used as the basis of SQL for relational
databases, and illustrates the basic operations
required of any DML.
2
3. 3
Relational Algebra
Relational algebra operations work on one or
more relations to define another relation
without changing the original relations.
Both operands and results are relations, so
output from one operation can become input to
another operation.
Allows expressions to be nested, just as in
arithmetic. This property is called closure.
4. 4
Relational algebra VS Relational Calculus
Informally, relational algebra is a (high-level)
procedural language and relational calculus a
non-procedural language. – Difference ??
However, formally both are equivalent to one
another.
A language that produces a relation that can be
derived using relational calculus is relationally
complete.
What & How
5. 5
Relational Algebra
This Algebra is composed of Unary operations
(involving a single table) and Binary operations
(involving multiple tables).
Five basic operations in relational algebra:
Selection, Projection, Cartesian product,
Union, and Set Difference.
These perform most of the data retrieval
operations needed.
Also have Join, Intersection, and Division
operations, which can be expressed in terms of
5 basic operations.
8. 8
Selection (or Restriction)
σpredicate (R)
– Works on a single relation R and defines a relation that
contains only those tuples (rows) of R that satisfy the
specified condition (predicate).
– Unary Operation
σ <condition> < tablename >
Conditions in Selection:
Simple Condition: (attribute)(comparison)(attribute)
(attribute)(comparison)(constant)
Comparison: =,≠,≤,≥,<,>
9. Select Operator Example
Name Age Weight
Harry 34 80
Sally 28 64
George 29 70
Helena 54 54
Peter 34 80
Name Age Weight
Harry 34 80
Helena 54 54
Peter 34 80
Person бAge≥34(Person)
Name Age Weight
Helena 54 54
бAge=Weight(Person)
10. 10
Example - Selection (or Restriction)
List all staff with a salary greater than £10,000.
σsalary >10000 (Staff)
11. 11
Projection
Πcol1, .. . ,coln(R)
– Works on a single relation R and defines a
relation that contains a vertical subset of R,
extracting the values of specified attributes and
eliminating duplicates.
π <columnlist > < tablename >
e.g., name of employees:
∏ name(Employee)
e.g., name of employees earning more than 80,000:
∏ name(бSalary>80,000(Employee))
12. Project Operator Example
Name Age Salary
Harry 34 80,000
Sally 28 90,000
George 29 70,000
Helena 54 54,280
Peter 34 40,000
Name
Harry
Sally
George
Helena
Peter
Employee
∏ name(Employee)
13. Project Operator Example
Name Age Salary
Harry 34 80,000
Sally 28 90,000
George 29 70,000
Helena 54 54,280
Peter 34 40,000
Name
Sally
Employee бSalary>80,000(Employee)
Name Age Salary
Sally 28 90,000
∏ name(бSalary>80,000(Employee))
14. 14
Example - Projection
Produce a list of salaries for all staff, showing only
staffNo, fName, lName, and salary details.
ΠstaffNo, fName,lName, salary(Staff)
15. Union, Intersection, Set-Difference
All of these operations take two input relations, which
must be union-compatible:
– Same number of fields.
– `Corresponding’ fields have the same type.
15
16. 16
Union
R ∪ S
– Union of two relations R and S defines a relation
that contains all the tuples of R, or S, or both R
and S, duplicate tuples being eliminated.
– R and S must be union-compatible.
If R and S have I and J tuples, respectively, union
is obtained by concatenating them into one relation
with a maximum of (I + J) tuples.
17. Union Operator Example
FN LN
Susan Yao
Ramesh Shah
Barbara Jones
Amy Ford
Jimmy Wang
FN LN
John Smith
Ricardo Brown
Susan Yao
Francis Johnson
Ramesh Shah
Student Professor
FN LN
Susan Yao
Ramesh Shah
Barbara Jones
Amy Ford
Jimmy Wang
John Smith
Ricardo Brown
Francis Johnson
Student U Professor
18. 18
Example - Union
List all cities where there is either a branch office
or a property for rent.
Πcity(Branch) ∪ Πcity(PropertyForRent)
19. 19
Set Difference
R – S
– Defines a relation consisting of the tuples that
are in relation R, but not in S.
– R and S must be union-compatible.
20. Set Difference Operator Example
FN LN
Susan Yao
Ramesh Shah
Barbara Jones
Amy Ford
Jimmy Wang
FN LN
John Smith
Ricardo Brown
Susan Yao
Francis Johnson
Ramesh Shah
Student
Professor
FN LN
Barbara Jones
Amy Ford
Jimmy Wang
Student - Professor
FN LN
John Smith
Ricardo Brown
Francis Johnson
Professor - Student
21. 21
Example - Set Difference
List all cities where there is a branch office but no
properties for rent.
Πcity(Branch) – Πcity(PropertyForRent)
22. 22
Intersection
R ∩ S
– Defines a relation consisting of the set of all
tuples that are in both R and S.
– R and S must be union-compatible.
Expressed using basic operations:
R ∩ S = R – (R – S)
23. Intersection Operator Example
FN LN
Susan Yao
Ramesh Shah
Barbara Jones
Amy Ford
Jimmy Wang
FN LN
John Smith
Ricardo Brown
Susan Yao
Francis Johnson
Ramesh Shah
Student Professor
FN LN
Susan Yao
Ramesh Shah
Student ∩ Professor
24. 24
Example - Intersection
List all cities where there is both a branch office
and at least one property for rent.
Πcity(Branch) ∩ Πcity(PropertyForRent)
25. 25
Cartesian product
R X S
– Defines a relation that is the concatenation of
every tuple of relation R with every tuple of
relation S.
26. 26
Example - Cartesian product
List the names and comments of all clients who have
viewed a property for rent.
(ΠclientNo,fName,lName(Client)) X (ΠclientNo,propertyNo,comment(Viewing))
27. 27
Example - Cartesian product and Selection
Use selection operation to extract those tuples where
Client.clientNo = Viewing.clientNo.
σClient.clientNo=Viewing.clientNo((∏clientNo,fName, lName(Client)) Χ (∏clientNo, propertyNo,
comment(Viewing)))
Cartesian product and Selection can be reduced to a single
operation called a Join.
28. 28
Join Operations
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.
29. 29
Join Operations
Various forms of join operation
– Theta join
– Equijoin (a particular type of Theta join)
– Natural join
– Outer join
– Semijoin
30. 30
Theta join (θ-join)
R FS
– Defines a relation that contains tuples
satisfying the predicate F from the Cartesian
product of R and S.
– The predicate F is of the form R.ai θ S.bi
where θ may be one of the comparison
operators (<, ≤, >, ≥, =, ≠).
31. 31
Theta join (θ-join)
Can rewrite Theta join using basic Selection and
Cartesian product operations.
R FS = σF(R Χ S)
Degree of a Theta join is sum of degrees of the
operand relations R and S. If predicate F contains
only equality (=), the term Equijoin is used.
32. 32
Example - Equijoin
List the names and comments of all clients who
have viewed a property for rent.
(ΠclientNo,fName, lName(Client)) Client.clientNo=Viewing.clientNo (ΠclientNo, propertyNo,
comment(Viewing))
33. 33
Natural join
R S
– An Equijoin of the two relations R and S over all
common attributes x. One occurrence of each
common attribute is eliminated from the result.
34. 34
Example - Natural join
List the names and comments of all clients who
have viewed a property for rent.
(ΠclientNo, fName,lName(Client))
(ΠclientNo,propertyNo, comment(Viewing))
35. 35
Outer join
To display rows in the result that do not have
matching values in the join column, use Outer
join.
R S
– (Left) outer join is join in which tuples from
R that do not have matching values in
common columns of S are also included in
result relation.
36. 36
Example - Left Outer join
Produce a status report on property viewings.
ΠpropertyNo, street, city(PropertyForRent)
Viewing
37. 37
Semijoin
R FS
– Defines a relation that contains the tuples of R that
participate in the join of R with S.
– It performs a join on two relations and then project
Over the attributes of first operand.
Can rewrite Semijoin using Projection and Join:
R F S = ΠA(R F S)
38. 38
Example - Semijoin
List complete details of all staff who work at the
branch in Glasgow.
Staff Staff.branchNo=Branch.branchNo(σcity=‘Glasgow’(Branch))
39. 39
Division
R ÷ S
– Defines a relation over the attributes C that consists of
set of tuples from R that match combination of every
tuple in S.
Expressed using basic operations:
T1 ← ΠC(R)
T2 ← ΠC((S X T1) – R)
T ← T1 – T2
40. 40
Example - Division
Identify all clients who have viewed all properties
with three rooms.
(ΠclientNo,propertyNo(Viewing)) ÷
(ΠpropertyNo(σrooms=3 (PropertyForRent)))
41. Relational DBMS
The following tables form part of a database held in a
relational DBMS:
Hotel (hotelNo, hotelName, city)
Room (roomNo, hotelNo, type, price)
Booking (hotelNo, guestNo, dateFrom, dateTo,
roomNo)
Guest (guestNo, guestName, guestAddress)
41
43. Exercise – Gererate Relational algebra
List all hotels.
List all single rooms with a price below £20 per night.
List the names and cities of all guests.
List the price and type of all rooms at the Grosvenor
Hotel.
List all guests currently staying at the Grosvenor Hotel.
List the details of all rooms at the Grosvenor Hotel,
including the name of the guest staying in the room, if
the room is occupied.
List the guest details (guestNo, guestName, and
guestAddress) of all guests staying at the Grosvenor
43
44. SQL
Structured Query Language (SQL)
– Standardised by ANSI
– Supported by modern RDBMSs
Commands fall into three groups
– Data Definition Language (DLL)
» Create tables, etc
– Data Manipulation Language (DML)
» Retrieve and modify data
– Data Control Language
» Control what users can do – grant and revoke privileges
44