extented entity relationship model

Moussa Demba-Aljouf univer

Advanced Database

What you learn today, prepares you for tomorrow!

Moussa

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Outline

rsity 1

Advanced Database-CSC426

Fall 2016


a Demba-Aljouf university

� Entity, Attribute, Relationship, Record

� 1-1, 1-n, n-m

� Primary key, Foreign key

� Subclass, Superclass and Inheritance

� Specialization and generalization

� Constraints and Characteristics

� Relationalschema

� Weakentity, partial key

utline

1


Moussa Demba-Aljouf university 2

Chap.1: Enhanced ER Model: Review

� The Entity-Relationship model models an enterprise as a collection of

entities and relationships:

� Entity: a “thing” in the real world that is distinguishable from all

other objects

Example: Student, Course, Professor, Account, Employee, Room,

Customer,…

� Described by a set of properties: attributes

Example: name, salary, grade, account_id, street,…

� Relationship: an association among several entities

,…: Takes, Teaches, DepositorExample

� ER Diagram Symbols and Notations:

Type of Relationships: Cardinality

� One-to-one (1-1):each entity is related to at most one entity.

With x,y=0, 1

E F R

(x,1) (y,1)


An instanceofE is associated with at most oneinstanceofF, and aninstanceofF is

associated with at most oneinstanceofE.

� one-to-many:1-n or n-1

n: no limit

An instanceofE is associated with at most oneinstanceofF, and an instance ofF is

associated with any numberinstancesofE.

� many-to-many:n-m

n,m: no limit

E F R (x,1) (y,n)

E F R (x,n) (y,m)

1-n Relationship


An instanceofE is associated with any numberinstancesofF, and an instance ofF

is associated with any numberinstancesofE.

Primary Key:is an attribute(or set of attributes) that is selected

to uniquely identify each instance of an entity.

the primary key. underlineewIn Design:

Example :Bank

1,n

PlaceOfBirth

1,1

n-m Relationship


Primary Key in SQL:InPractice

Create table Customer(

Customer_idvarchar2(50),

Cnamevarchar2(20),

Street varchar2(15),

POB varchar2(15),

Primary key(Customer_id));

Foreign Key: a field that links one table to another table. A table can

have an unlimited number of foreign keys linking to other tables. We

use a sharp(#) for foreign keys.

Create table Customer(


Cnamevarchar2(20),

Street varchar2(15),

POB varchar2(15),

Primary key(Customer_id));

CREATE TABLE Account(

Account_Numbervarchar2(10),

Balance numeric(6,3),


Primary key(Account_Number),

FOREIGN KEY(Customer_id) references Customer)


1,1

1,1

0,1

0,1

1,1 0,1

1,1 0,1

Mapping an ER Model to theRelational Schema

• One-to-one

With x=0,1

Relational Schema:

The FK must be in the relation with min=1.

Exercise

E(e1,e2)

F(f1, f2, e1#)

E F R (x,1) (1,1)

e1 e2 f1 f2

E

F

G

R1

R2

R3 R4

f1 f2

e1

e2

g1

g2


1,1

1,1

0,1

1,n 0,1

1,1 1,1

1,1

• One-to-many

Relational Schema:

The FK must be in the relation with max=1.

Exercise

E(e1,e2)

F(f1, f2, e1#)

E F R (x,n) (1,1)

e1 e2 f1 f2

E

G

R1

R2

R3 R4

e1

e2

g1

g2

F

f1 f2


• many-to-many

Relational Schema:

The FKs must be in a new relation.

Composite, derived and multivalued attributes

• address is a composite attribute.

• Years-of- services is a derivedattribute, it canbecalculated from Start-Date

andcurrent date.

• Skills is a multivalued attribute: an employee can have many skills.

E F R (x,n) (y,n)

E(e1,e2)

R(e1#,f1#,r1,r2)

F(f1,f2)

e1 e2 f1 f2 r1 r2

Employee

@

Box

street

city

eid Start date

Years of services skills


In relational schema,

� Composite attributes are represented by components.

� Derived attributes are ignored

� Multi-valued attributes are represented by a separated table

Employee(eid, Box, street, city, start-date)

Emp-Skills(eid#, skills)

Reflexive relationships

Person(pid, husband#,…)

Here is another example:

Emp(eid, guarantor#,…)

wife

husband

0,1

0,1 or 0,n

Person Married-to pid

guarantor

beneficiary

n

1

Emp guarantor-of eid


1 ,n

Course(cid,…)

Prerequisite(cdi1#, cid2#)

Weak entity � In a relational database, a weak entity is an entity that cannot be

uniquely identified by its attributes alone; � Therefore, it must use a foreign key in conjunction with its attributes to

create a primary key. � The foreign key is typically a primary key of an entity it is related to.

1 ,1 Building Room Has

box street

bid

rid area

cost

0,n

0,n

Course Prerequisite

s

cid


� rid is a partial key of the weak entity Rooms.

� (rid+bid) is the primary key of Rooms

� bid is a foreign key in Rooms

in schema:

Building(bid, box, street)

Room(rid, bid#, area, cost)

Exercise

Give a relational schema that best describes the ER diagram:

1,1 0,n

1,1

0,n

1,1

1,n

1,n

1,n

1,n

0,nD

B

C

EF

R6

R3

R4

R5

R2

A R6

a1a2

b1

b2

c1

c2

d1

d2r21

r22

f1

f2

e1

e2

0,1

0,1

0,1 f3

d22d21

c3


Subclass, Superclass and Inheritance

� As in C++, or other programming languages, attributes are inherited.

� If we declare A ISA B then every A entity is also considered to be a B

entity.

� And, every attribute of B is also considered to be an attribute of A.

� For example

Camel IS-A Animal Chicken IS-A Animal

Cow IS-A Animal Lion IS-A Animal

Lion

Cow Chicken

Animals

Camel

superclass

subclasses

Inheritance(IS A)

B

A

b1

b2

a1

C R


� As well, a subclass inherits all the relationships in which the super class

participates.

Example 1:A student and a professor are both person.

Example 2:

Bottom-Up: generalization

Top-Down: specialization

Student pid

DOB name

level

Professor

salary

pid

DOB

name

Person

Student Professor

pid

DOB name

salary level

1,1 0,n lives Person

Student Professor

pid

DOB

salary

Building

level


� EMPLOYEE may be further grouped into:

� SECRETARY, ENGINEER, TECHNICIAN, …

� Based on the EMPLOYEE’s Job

superclass

…subclasses

We read: SECRETARY IS-A EMPLOYEE

ENGINEER IS-A EMPLOYEE

TECHNICIAN IS-A EMPLOYEE

Employee

Secretary Engineer Technician

Inheritance(Is-A)

Car

VehiculeID Price

MaxSpeed LPlateNumbe

Nb_Of_Passengers

Truck

VehiculeID Price

Tonnage LPlateNumbe

Nb_Of_Axles

Ge

ne

raliza

tion

Sp

ecia

lizatio

n

Vehicle VehiculeID LPlateNumbe

Price

Car MaxSpeed

Nb_Of_Passengers

Tonnage

Nb_Of_Axles

Truck


� MANAGER

� EMPLOYEEs who are managers

superclass

subclass

MANAGERIS-AEMPLOYEE

� SALARIED_EMPLOYEE, HOURLY_EMPLOYEE

� Based on the EMPLOYEE’s method of pay

SALARIED_EMPLOYEE IS-A EMPLOYEE

HOURLY_EMPLOYEE IS-A EMPLOYEE

� An Entity CANNOT exist in the database merely by being a member of a

subclass; it must also be a member of the superclass.

� Every Manager or Secretary or Salaried_employee is an employee.

� A member of the superclass can be OPTIONALLY included as a member of

any number of its subclasses

� An employee may not be a Manager.

� An employee may be a secretary and salaried_employee.

Employee

Salaried_Employee Hourly_Employee

Employee

Manager

IS-A


1,1 0,n

0,n

1,n

0,n

1,n

1,1

0,n

0,n

1,n

0,n

1,n

1,1

0,n

We have also:

lives

takes

Course

Person

Student Professor

pid

DOB

salary level

Building

teaches

Generalization

lives1

takes

Course

Student Professor pid

DOB

salary

pid

Building

teaches

Specialization

lives2

DOB

level


Constraints on Specialization and Generalization

Two other conditions apply to a specialization/generalization:

1. Disjointness Constraint: – Specifies that an entity can be a memberofat most one of the subclasses. – Specified by “d” in EER diagram – If not disjointed, overlap; that is the same entity may be a member of

morethan one subclass of the specialization – Specified by “o” in EER diagram

2. Completeness Constraint: – Total specialization: specifies that every entity in the superclass must be a

member of some subclass. Shown in EER diagrams by a double line – Partial specialization:allows an entity not to belong to any of the subclasses.

Shown in EER diagrams by a single line Note: Generalization usually is total because the superclass is derived from the subclasses. Example of disjointpartialSpecialization:Example of disjointtotal Specialization:

Student

Undergraduate

student

Graduate

student

d

Employee

Secretary Engineer Technician

d

Car

VehiculeID Price

MaxSpeed LPlateNumbe

Nb_Of_Passengers

Truck

VehiculeID Price

Tonnage LPlateNumbe

Nb_Of_Axles

Ge

ne

raliza

tion

Vehicle VehiculeID LPlateNumbe

Price

Car MaxSpeed

Nb_Of_Passengers

Tonnage

Nb_Of_Axles

Truck

d

Sp

ecia

lizatio

n


Example of overlappingpartial Specialization:

Example of overlappingtotal Specialization: (in a university)

Person

Employee Student

O

Person

Teacher Student

O


Translating in schema:

Fig.1

3 possibiliBes:

a) keep only the super-class

we get something like Fig.2.

and the schema is:

E1 (A1, A11, A2, A3, type) Fig.2

type is a discriminator to disFnguish which type each tuple is (E2, E3,…).

- many null columns + always applicable.

b) keep only the sub-classes

we get something like Fig.3.if the generalization is disjoint and complete

Fig.3

and the corresponding schema is:

E2 (A12, A11, A2)A1 is renamed A12 in E2 and A13 in E3.

E3 (A13, A11, A3)

E1

E2 E3

A1 A11

A2 A3

E1

A3

A11

A2

A1

E3

A11

A3

A13

E2

A11

A2

A12


- This solution involves a significant redundancy of the data on the

super-class if there is not exclusion between the sub-classes.

- For an overlapping generalization, some values such as A11 would be

stored twice unnecessarily.

- Similarly, if the generalization were not complete (i.e., partial), that is

if some E1’s were neither E2 nor E3, then such E1’s could not be

represented by this method.

+ One relation per concrete class: the most natural way

+ Not require joins to retrieve information

c) keep all classes

we get something like Fig.4.

and the corresponding schema is:

E1 (A1, A11,type)

E2 (A1#,A2)

E3 (A1#, A3)

- many joins to retrieve information of a subclass

� Attribute type is here optional, but usingitcan minimize complex joins (improve

performance).

+ simple and allows different kind of queries.

Exercise:An engineer is an employee who works in a department, and he is a

person who lives in a building.

E1

E2 E3

A1 A11

A2 A3

Fig.4.


1,n 1,n 1,1 1,n

0,n 0,n 0,n

1,1

1,1

0,n 0,n

1,1

Give the corresponding schema without keeping the entity Employee?

Exercise:Consideringa university database represented by the ERD:

Give a relational schema that best describes the ERD.

Student-

Boys

Student-

Girls

is-brother Course

offered

cid syllabus

Student enrolls

sid name

Course-

offering

Instructor teaches

time secno room iid name grade

Department

study-in has deptnam

1,1 0,n

1,n

1,1

Building

bid

pname Person

pid

lives

salary

Student grant Employee

Department

dname

works

Engineer type Secretary

age


This ERD is designed using draw.io online tool.

Give a Schema that best describes the above ERD:

a. keeping all entities(generalization). b. without keeping the entity A1(specialization).


Exercise:Draw an ER diagram according to the following specification and

transform it to a relational data model.

A hotel would like to make a registration of the expenses of its

guests. In the hotel the guests can stay in rooms for a specific period.

The main properties of the rooms are their size (floor space), the

number of beds in them and their daily cost. Guests can stay in one

or multiple rooms at the same time, while multiple guests can stay

together as well. Relevant information about room usage of guests

are the times of checking in and checking out. Of course there can be

rooms where no guest is staying. Guests can make reservations, the

related information are the planned dates of arrival and departure.

The guests can use various services (internet, meal, sauna …), related

information are the names of such services, their prices and their

description. Guests can use multiple services and there can be

services which have never been used by anyone.

We register the guest by their name (first name, last name) and

address (city, street, Box).

We want also to keep track of the dependents of each guest, if any.

We keep each dependent’s name, sex, birth date, and relationship to

the guest.


ERD – Crow’s Foot Notation

Relationship Types:

� Relationship types are modeled by lines connectingthe entities

� Line is annotated with the name of the relationship which is a verb

� Cardinalities are represented graphically

ATTENTION: Cardinalities are written on the oppositeside of the relationship

(in contrast to “classic ER”)

Example:

� What happens to n-ary relationships or relationship attributes?

� Problem


• N-ary relationship types are not supported bycrow’s foot notation, neither are

relationshipattributes

� Workaround solution:

• Intermediate entities must be used

• N-ary relationships are broken down in a series of binary

relationship types anchoring on the intermediate entity


Is equivalent to

End.

extented entity relationship model

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