1. Chapter 3 CIS 458 Sungchul Hong
The Relational Model
Chapter 3
CIS 458
Sungchul Hong

2. Last Class ERD Entities, Relations, Attributes
It is not model specific

3. This Class Relational Model & Schema Conversion Cartesian product
Cf)
Hierarchical model
Network model
Object Oriented Model
Cartesian product
Integrity
Relational integrity
Entity integrity

4. Relational Database Management Systems
Dominant data processing software
The estimates sales is between $15 billion - $20 billion per year.
Relational model
In the relational model, all data is logically structured within relations (tables).
Each relation has a name and is made up of named attributes (columns) of data.
Each tuple (row) contains one value per attribute.

6. Relational Data Structure
The relational model is based on the mathematical concept of a relation.
Relation is a table with columns and rows.
Attribute: An attribute is a named column of a relation
Domain: A domain is the set of allowable values for one or more attributes.
Tuple: a tuple is a row of a relation

7. Relations

8. Relational Data Structure (2)
Degree: The degree of a relation is the number of attributes it contains.
n-ary
Cardinality: The cardinality of a relation is the number of tuples it contains.
Relational database: A collection of normalized relations with distinct relation names.

9. Mathematical Relations
Sets D1 and D2 where D1 = {2, 4} and D2={1, 3, 5}.
Cartesian product D1  D2, is the set of all ordered pairs such that the first element is a member of D1 and the second element is a member of D2.
D1  D2 ={(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5)}

10. Exercise E1 = {a, b, c} E2 = {3, 4, 5} E1  E2 = {(a,3), (a, 4), (a,5)
(b,3), (b, 4), (b,5)
(c,3), (c, 4), (c,5)}

11. Mathematical Relations (2)
Any subset of this Cartesian product is a relation. D1= {2, 4} , D2={1, 3, 5}
Relation R = {(2, 1), (4, 1)}
R = { (x, y) | x  D1, y  D2, and y = 1}
You want to find out some records not entire table.
S = {(x, y) | x  D1, y  D2, and x= 2y}
S = { (2,1) }

12. Mathematical Relations (3)
D1 = {1, 3}, D2 = {2, 4}, D3 = {5, 6}
D1  D2  D3 =
{(1,2,5), (1,2,6), (1,4,5), (1,4,6)
(3,2,5), (3,2,6), (3,4,5), (3,4,6)}

13. Database Relations
Relation Schema: A named relation defined by a set of attribute and domain name pairs.
Let A1, A2, …, An be attributes with domain D1, D2, …, Dn. Then the set {A1:D1, A2:D2, …, An:Dn} is a relation schema.
A relation R defined by a relation schema S is a set of mappings from the attribute names to their corresponding domains.
(A1:d1, A2:d2, … , An:dn) such that d1  D1, d2  D2, … , dn  Dn

14. Database Relations (2) Example
Branch table
{branchNo: B005, street: 22Deer Rd, city: London, postcode: SW1 4EH}
Relational instance
Relational database schema: A set of relation schemas each with a distinct name.
If R1, R2, … , Rn are a set of relation schemas, then we can write the relational database schema or relational schema R, as: R= {R1, R2, …, Rn}

15. Branch A3 A4 A1 A2 D1 D2 Domain D3 D4 BranchNo street city postcode
22 Deer Rd
London
SW14EH
B007
16 Argyll St.
Aberdeen
AB2 3SU
B003
163 Main St
Glasgow
G11 9QX
B004
32 Manse Rd
Bristol
BS99 1NZ
B002
56 Clover Dr.
NW10 6EU D1 D2
Domain D3 D4

16. R1 R2 R3 R4 R5 R6
R={R1, R2, R3, R4, R5, R6}

17. Properties of Relations (1/2)
The relation has a name that is distinct from all other relation names in the relational schema.
Each cell of the relation contains exactly one atomic value.
Each attribute has a distinct name.
The values of an attribute are all from the same domain.

18. Properties of Relations (2/2)
Each tuple is distinct: there are no duplicate tuples.
The order of attributes has no significance.
Practically has significance.
The order of tuples has no significance, theoretically.
Normalized or first normal form

19. Relational Keys
Superkey: An attribute, or set of attributes, that uniquely identifies a tuple within a relation.
Candidate key: A superkey such that no proper subset is a superkey within the relation.
Uniqueness (need semantic information)
Irreducibility

20. Example: The CAR relation schema:
CAR(State, Reg#, SerialNo, Make, Model, Year)
has two keys
Key1 = {State, Reg#}, Key2 = {SerialNo},
which are also superkeys. {SerialNo, Make} is a superkey but not a key.

21. Relational Keys (2)
Primary Key: The candidate key that is selected to identify tuples uniquely within the relation.
The candidate keys that are not selected to be the primary key are called alternate keys.
Foreign key: An attribute, or set of attributes, within one relation that matches the candidate key of some relation.
When an attribute appears in more than one relation, its appearance usually represents a relationship between tuples of the two relations.

23. Schema Conversion Create a table for each entity Relation Attribute
1:1 choose one side and put a foreign key
1:* put a foreign key in many side
*:* create a table for the relation
Relation
*:* create a table
Ternary or higher degree
Attribute
Multi-valued attribute

24. (1..1)
(0..1)
Staff(staffNo, name, position, salary, branchNo)
Branch(branchNo, street, city, postcode)
Telephone (branchNo, telNo) OR
Staff(staffNo, name, position, salary,)
Branch(branchNo, street, city, postcode, staffNo)
Telephone (branchNo, telNo)

25. (1..*)
(1..1)
Staff(staffNo, name, position, salary, manages, belongs)
Branch(branchNo, street, city, postcode)
Telephone (branchNo, telNo)
Domain [manages]= Branch.branchNo
Domain [belongs] = Branch.branchNo

26. Staff (1..*) Branch (1..1) staffNo fName lName manages sex DOB salary
branchNo
SL21
John
White
B005 M
Oct/1/45
30000
SG37
Ann
Beech F
Nov/10/60
12000
B003
SG14
David
Ford
Mar/24/58
18000
SA9
Mary
Howe
19/Feb/70
9000
B007
SG5
Susan
Brand
Jun/3/40
24000
SL41
Julie
Lee
Jun/13/65
(1..*)
Branch
BranchNo
street
city
postcode
B005
22 Deer Rd
London
SW14EH
B007
16 Argyll St.
Aberdeen
AB2 3SU
B003
163 Main St
Glasgow
G11 9QX
B004
32 Manse Rd
Bristol
BS99 1NZ
B002
56 Clover Dr.
NW10 6EU
(1..1)

27. Newspaper(newspaperName)
PropertyForRent
NewsPaper
◄Advertises
newsPaperName {PK}
propertyNo{PK}
Street
City
Postcode
Rooms
Rent
privateOwnerNo …
(1..*)
(0..*)
dateAdvert
Cost
PropertyForRent (propertyNo, street, city, postcode, type, rooms, rent, privateOwnerNo,
businessOwnerNo, staffNo, branchNo)
Newspaper(newspaperName)
Advertisement (newspaperName, PropertyNo, dateAdvert, cost)

28. Exercise Employee Department Project Dependent ◄Supervises Supervisor
0..1
Manages ►
Employee
SSN{pk}
FName
MINIT
LName
BDate
Address
Sex
Salary
Department
Dnumber{pk}
Dname
Mgrstartdate
Location[1..n]
1..1
1..0
◄Has
1..*
1..1
1..*
1..1
In Charge of►
Supervisee
1..*
Works_on ►
1..1
◄Depend On
0..*
0..*
0..*
Project
Pnumber{pk}
Pname
Plocation
Dependent
Dependent_Name
Sex
Bdate
Relationship

29. R1 R2 R3 R4 R5 R6
R={R1, R2, R3, R4, R5, R6}

30. Exercise

31. Representing Relational Database Schemas
DreamHome case study
Branch (branchNo, street, city, postcode, manager)
Staff (staffNo, fName, lName, position, sex, DOB salary, branchNo, supervisor)
PropertyForRent (propertyNo, street, city, postcode, type, rooms, rent, privateOwnerNo, businessOwnerNo, staffNo, branchNo)
Client (clientNo, fName, lName, telNo, prefType, maxRent)
Registration (clientNo, branchNo, staffNo, dateJoined)
Manager(staffNo, branchNo, MgrStartDate, bonus)

32. Schema Lease (leaseNo, PropertyNo, clientNo)
Preference (clientNo, preference)
Newspaper(newspaperName)
Advertisement (newspaperName, PropertyNo, dateAdvert, cost)
PrivateOwner (privateOwnerNo, fName, lName, address, telNo)
BusinessOwner (businessOwnerNo, businessName)

33. Relational Integrity Data Model
Definition, manipulative part
Domain constraints form restrictions on the set of values allowed for the attributes of relation.
Integrity rules
Constraints or restrictions that apply to all instances of the database.
Entity integrity, referential integrity

34. Relational Integrity (2)
Null represents a value for an attribute that is currently unknown or is not applicable for this tuple.
Unknown, incomplete or exceptional data
Null is not zero nor white space.

35. Entity Integrity
Base Relation: A named relation corresponding to an entity in the conceptual schema, whose tuples are physically stored in the database.
View (Virtual relation)
Entity integrity: In a base relation, no attribute of a primary key can be null.

36. Referential Integrity
Referential Integrity: If a foreign key exists in a relation, either the foreign key value must match a candidate key value of some tuple in its home relation or the foreign key value must be wholly null.

37. Enterprise Constraints
Additional rules specified by the users or database administrators of a database.
A employee can not work more than 65 hours per week.
Salary value must be a positive number.

38. Views A view is a virtual or derived relation.
A relation that does not necessarily exist in its own right, but may be dynamically derived from one or more base relations.
Provide a powerful and flexible security mechanism.
Customized user data access
It can simplify complex operations.
Updates are allowed on simple query with a single base relation.
Not allowed with multiple base relations.
Not allowed with aggregation or grouping operations.

39. Exercise
Chairs►
PROFESSOR
DEPARTMENT
1..1
0..1
1..1
1..N
IS ASSIGNED TO►
PROFESSOR(PID, FName, MINIT, LName, DOB, Rank)
DEPARTMENT (DID, DName)

40. Exercise
COURSE
1..1
Has▼
o..*
ENROLL►
SECTION
STUDENTS
0..35
1..6
STUDENTS (SID, FName, LName, DOB, Major[1..3], Minor)
COURSE (COURSE#, C_Name, Description)
SECTION (Section_Num, Max_size)

41. Painter (Painter_ID, FName, LName) Paining (Painting_ID, Title, Date)
Exercise
Paint ►
(0..*)
Displayed
(1..1) ►
Painter
(1,1)
(0..*)
Painting
Gallery
Painter (Painter_ID, FName, LName)
Paining (Painting_ID, Title, Date)
Gallery (Gallery_Name, Location)

42. Employee (EID, FName, LName) Skill (Skill_Name)
Exercise
Learn ►
0..*
0..*
Employee
Skills
Expert Level
Employee (EID, FName, LName)
Skill (Skill_Name)

43. STUDENTS (SID, FName, LName, DOB) CLASS (CLASS#, C_Name, Description)
Exercise
Take ►
1..*
0..*
Student
Classes
Grade
STUDENTS (SID, FName, LName, DOB)
CLASS (CLASS#, C_Name, Description)

44. Exercise (Recursive and many-to-many)
STUDENTS (SID, FName, LName, DOB)
COURSE (COURSE#, C_Name, Description)
SECTION (Section_Num, Max_size)
Has ►
1..1
0..*
Course
Section
0..*
0..*
0..*
Take ▼
Grade
◄Prerequisite
1..*
Student

45. Grade Course# Cname Dept Student Enroll Course Adopt Quarter1 N
Student
SSN{pk}
Name
Major
Bdate
Enroll
Course 1 1
Adopt
Quarter
Quarter{pk} 1 N
Text
ISBN{pk}
Publisher
Author

46. Exercise
Consider the following relations for a database that keeps track of student enrollment in courses and the books adopted for each course:
STUDENT(SSN, Name, Major, Bdate)
COURSE(Course#, Cname, Dept)
ENROLL(SSN, Course#, Quarter, Grade)
BOOK_ADOPTION(Course#, Quarter, Book_ISBN)
TEXT(Book_ISBN, Book_Title, Publisher, Author)
Draw a relational schema diagram specifying the foreign keys for this schema.