Kingdom of Saudi Arabia Ministry of Higher Education Al-Imam Muhammad Ibn Saud Islamic University College of Computer and Information Sciences Normalization.

Kingdom of Saudi ArabiaMinistry of Higher Education

Al-Imam Muhammad Ibn Saud Islamic UniversityCollege of Computer and Information Sciences

Normalization

IS 320: Introduction to Database

Hatoon AlSagri

Introduction to Databases

Informal Design Guidelines for Relational Databases

• Relational database design: The grouping of attributes to form "good" relation schemas

• Two levels of relation schemas:

• The logical "user view" level

• The storage "base relation" level

• Design is concerned mainly with base relations

Introduction to Databases 2

Informal Design Guidelines for Relational Databases

Four informal measures of quality for relation schema design:

1. Semantics of the Relation Attributes

2. Reducing the redundant information in tuples

3. Reducing Null values in tuples

4. Disallowing the possibility of one generating spurious tuples.


1- Semantics of the Relation Attributes

Each tuple in a relation should represent one entity or

relationship instance

Guideline #1: Design a schema that can be explained easily

relation by relation. The semantics of attributes should be

easy to interpret.


2- Redundant Information in Tuples and Update Anomalies

• Mixing attributes of multiple entities may cause problems:

• Information is stored redundantly wasting storage

• Problems with update anomalies:

• Insertion anomalies

• Deletion anomalies

• Modification anomalies

Guideline #2: Design a schema that does not suffer from the insertion,

deletion and update anomalies. If there are any present, then note them so

that applications can be made to take them into account


Base Relations EMP_PROJ with redundant information


3- Null Values in Tuples

Reasons for nulls:

a. attribute not applicable or invalid

b. attribute value unknown (may exist)

c. value known to exist, but unavailable

Guideline #3: Relations should be designed such that their tuples will have as few NULL values as possible

Attributes that are NULL frequently could be placed in separate relations (with the primary key)


4- Spurious Tuples

• Bad designs for a relational database may result in

erroneous results for certain JOIN operations

Guideline #4: The relations should be designed to satisfy

the lossless join condition. No spurious tuples should be

generated by doing join of any relations.


Functional Dependencies

• Functional dependencies (FDs) are used to specify formal

measures of the "goodness" of relational designs

• FDs and keys are used to define normal forms for

relations

• FDs are constraints that are derived from the meaning

and interrelationships of the data attributes


Examples of FD constraints

• Social security number determines employee name

SSN -> ENAME• Project number determines project name and location

PNUMBER -> {PNAME, PLOCATION}• Employee ssn and project number determines the hours per week that

the employee works on the project

{SSN, PNUMBER} -> HOURS




Describes the relationship between attributes in a relation.

If A and B are attributes of relation R,

B is functionally dependent on A, denoted by A B, if each value of A is associated with exactly one value of B. B may have several values of A.

Determinant Dependent

A BB is functionallydependent on A



Example

StaffNo positionPosition is functionallydependent on Staffno

position StaffNoStaffNo is NOT functionallydependent on position

SL21 Manager

Manager SL21 SG5

1:1 or M:1 relationship

between attributes in a

relation

1:M relationship

between attributes in a

relation

Trivial Functional Dependencies


A B is trivial if B A

StaffNo, Sname SName

StaffNo, SName StaffNo

We are not interested in trivial functional dependencies as it provides no genuine integrity constraints on the value held by these attributes.

Question

Find FDs of the relation shown below that lists dentist/patient appointment data; known that:

• A patient is given an appointment at a specific time and date with a dentist located at a particular surgery.

• On each day of patient appointments, a dentist is allocated to a specific surgery for that day.

Dentist-patient (staffNo, aDate, aTime, dentistName, patNo, patName, surgeryNo)


Question


FDs list

FD1: staffNo, aDate, aTime patNo, patName

FD2: staffNo dentistName

FD3: patNo patName, surgeryNo

FD4: staffNo, aDate surgeryNo

FD5: aDate, aTime, patNo dentistName, staffNo


Introduction to Normalization

• Normalization: Process of decomposing unsatisfactory

"bad" relations by breaking up their attributes into smaller

relations

• Normal form: Condition using keys and FDs of a relation to

certify whether a relation schema is in a particular normal

form


Normalization into 1NF


ExamplesFirst Normal Form

• EMP_PROJ (Ssn, Ename, {Phone#}) { } Mulitvalue attribute

EMP_PROJ1 (Ssn, Ename)

EMP_PROJ2 (Ssn, Phone#)

• EMP_PROJ (Ssn, Ename (Fname, Lname)) ( ) composite attribute

EMP_PROJ (Ssn, Fname,Lname)

• EMP_PROJ (Ssn, Ename, {PROJS (Pnamber, Hours)})

EMP_PROJ1 (Ssn, Ename)

EMP_PROJ2 (Ssn, Pnamber, Hours)


Second Normal Form

• Uses the concepts of FDs, primary key

• Definitions:

• Prime attribute - attribute that is member of the primary key K

• Full functional dependency - a FD Y Z where removal of any

attribute from Y means the FD does not hold any more


Full Functional Dependency


If A and B are attributes of a relation.

B is fully functionally dependent on A if B is functionally dependent on A, but not on any proper subset of A.

B is partial functional dependent on A if some attributes can be removed from A & the dependency still holds.

StaffNo, Sname BranchNo Partial dependency

ClientNo, PropertyNo RentDate Full dependency

1NF 2NF


1. Start with 1NF relation.

2. Find the FDs of a relation.

3. Test the FDs whose determinant attribute is part of the PK.

ExamplesSecond Normal Form

• {SSN, PNUMBER} HOURS is a full FD since neither

SSN HOURS nor PNUMBER HOURS hold

• {SSN, PNUMBER} ENAME is not a full FD (it is called a partial

dependency ) since SSN ENAME also holds

• A relation schema R is in second normal form (2NF) if every non-

prime attribute A in R is fully functionally dependent on the

primary key

• R can be decomposed into 2NF relations via the process of 2NF

normalization


ExamplesSecond Normal Form


Second Normal Form

Note: The test for 2NF involves testing for functional

dependencies whose left-hand side attributes are part of

the primary key. If the primary key contains a single

attribute, the test need not be applied at all.


Third Normal Form

• Definition

• Transitive functional dependency – a FD X Y in R is a transitive

dependency if there is a set of attributes Z that are neither a primary

or candidate key and both X Z and Z Y holds.

• Examples:

• SSN DMGRSSN is a transitive FD since

SSN DNUMBER and DNUMBER DMGRSSN hold

• SSN ENAME is non-transitive since there is no set of

attributes X where SSN X and X ENAME


3rd Normal Form

A relation schema R is in third normal form (3NF) if it is

in 2NF and no non-prime attribute A in R is transitively

dependent on the primary key


ExamplesThird Normal Form


SUMMARY OF NORMAL FORMS based on Primary Keys


BCNF (Boyce-Codd Normal Form)

• A relation schema R is in Boyce-Codd Normal Form (BCNF) if

whenever an FD X A holds in R, then X is a superkey of R

• Each normal form is strictly stronger than the previous one:

• Every 2NF relation is in 1NF

• Every 3NF relation is in 2NF

• Every BCNF relation is in 3NF

• There exist relations that are in 3NF but not in BCNF

• The goal is to have each relation in BCNF (or 3NF)


BCNF

R1(A,C)R2(C,B)



BCNF

FDs:

• {Student,course} Instructor

• Instructor Course

It is in 3NF not in BCNF

• Decomposing into 2 schemas

{Student, Instructor} {Instructor, Course}


ExamplesBCNF

R ( Client#, Problem, Consultant _name)

R1 (Client#, Consultant _name)

R2 (Consultant _name, Problem)

■ R (Stud#, Class#, Instructor, Grade)

R1 (Stud#, Instructor, Grade)

R2 (Instructor, Class#)


Example

Consider the following relation for published books:

BOOK (Book_title, Author_name, Book_type, Listprice,

Author_affil, Publisher)

- Author_affil referes to the affiliation of the author.

Suppose thefollowing dependencies exist:

Book_title -> Publisher, Book_type

Book_type -> Listprice

Author_name -> Author-affil

(a) What normal form is the relation in? Explain your answer.

(b) Apply normalization until you cannot decompose the relations further.

State the reasons behind each decomposition.


Answer

BOOK (Book_title, Authorname, Book_type, Listprice, Author_affil, Publisher)

(a) The key for this relation is (Book_title, Authorname). This relation is in 1NF and not in 2NF as no attributes are Full FD on the key. It is also not in 3NF.

(b) 2NF decomposition:Book0(Book_title, Authorname)Book1(Book_title, Publisher, Book_type, Listprice)Book2(Authorname, Author_affil)

This decomposition eliminates the partial dependencies.3NF decomposition:

Book0(Book_title, Authorname)Book1-1(Book_title, Publisher, Book_type)Book1-2(Book_type, Listprice)Book2(Authorname, Author_affil)

This decomposition eliminates the transitive dependency of ListpriceIntroduction to Databases 35

Example

Given the relation schemaCar_Sale (Car#, Salesman#, Date_sold, Commission%, Discount_amt)with the functional dependenciesDate_sold -> Discount_amtSalesman# -> Commission%Car# -> Date_sold

This relation satisfies 1NF but not 2NF (Car# -> Date_sold and Salesman# -> Commission%)so these two attributes are not Full FD on the primary key and not 3NF


To normalize,2NF:Car_Sale1 (Car#, Salesman#)Car_Sale2 (Car#, Date_sold, Discount_amt)Car_Sale3 (Salesman#,Commission%)3NF:Car_Sale1(Car#, Salesman#)Car_Sale2-1(Car#, Date_sold)Car_Sale2-2(Date_sold, Discount_amt)Car_Sale3(Salesman#,Commission%)


Answer

Question

Given the following Dentist-patient database schema:


Normalize the above relation, showing appropriate dependency diagrams to justify decomposition.


FDs List:

FD1: staffNo, aDate, aTime patNo, patName

FD2: staffNo dentistName

FD3: patNo patName, surgeryNo

FD4: staffNo, aDate surgeryNo

FD5: aDate, aTime, patNo dentistName, staffNo

Answer1NF


2NF (fd2 and fd4 violates 2NF)

Dentist-patient (staffNo, aDate, aTime, patNo, patName)

Surgery (staffNo, aDate, surgeryNo)

Dentist (staffNo, dentistName)

3NF (Fd3’ violates 3NF)

Dentist-patient (staffNo, aDate, aTime, patNo)



Patient (patNo, patName)Introduction to Databases 39

Answer

BCNF (No violation)

Dentist-patient (staffNo, aDate, aTime, patNo)



Patient (patNo, patName)


Kingdom of Saudi Arabia Ministry of Higher Education Al-Imam Muhammad Ibn Saud Islamic University College of Computer and Information Sciences Normalization.

Documents

relation attributes

base relations introduction

databases slide

hours introduction

data attributes introduction

relation schema design

account introduction

databases9 slide