Institute of Southern Punjab, MultanNormalization (Brief Overview) The main goal of Database...

Institute of Southern Punjab, Multan

Mr. Muhammad Nouman Farooq

BSC-H (Computer Science)

MS (Telecomm. and Networks)

Honors:

Magna Cumm Laude Honors Degree

Gold Medalist!

Blog Url: noumanfarooqatisp.wordpress.com

E-Mail: [email protected]

Advance Database Systems

Lecture# 6

Advanced Normalization

Lecture 6: Advanced Normalization

Normalization (Brief Overview)

Functional Dependencies and Keys

1st Normal Form

2nd Normal Form

3rd Normal Form

3.5 Normal Form (Boyce Codd Normal Form-BCNF)

4th Normal Form

5th Normal Form (Project-Join Normal Form-PJNF)

Domain Key Normal Form (DKNF)

6th Normal Form

Class Activity

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4



The main goal of Database Normalization is to restructure the

logical data model of a database to:

1. Eliminate Redundancy.

2. Organize Data Efficiently.

3. Reduce the possibility of Data Anomalies/Irregularities.

5

Continued…

Data anomalies are inconsistencies in the data stored in a database as a result of an operation such as update, insertion, and deletion.

Such inconsistencies may arise when have a particular record stored in multiple locations and not all of the copies are updated.

We can prevent such anomalies by implementing 7 different level of normalization called Normal Forms (NF)

We’ll only look at the first Three.

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Continued…

Database Normalization was first proposed by Edgar F. Codd.

Codd defined the first three Normal Forms, which we’ll look into,of the 7 known Normal Forms.

In order to do normalization we must know what the requirementsare for each of the three Normal Forms that we’ll go over.

One of the key requirements to remember is that Normal Formsare progressive. That is, in order to achieve 3rd NF we must have2nd NF and in order to have 2nd NF we must have 1st NF.

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Levels of Normalization

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Levels of normalization based on the amount of redundancy

in the database.

Various levels of normalization are:

First Normal Form (1NF)

Second Normal Form (2NF)

Third Normal Form (3NF)

Boyce-Codd Normal Form (BCNF) or 3.5 NF

Fourth Normal Form (4NF)

Fifth Normal Form (Project-Join Normal Form-PJNF)


Sixth Normal Form (6NF)

Redundancy

Decre

ase

s

Num

ber o

f Table

s Incre

ase

s

Data Anomalies

1. Insertion Anomaly:

Cannot make a record of Jone’s address because he is not taking any classes.

10

Continued…

2. Update Anomaly:

Clearly, Name and Address are redundant (larger relation and we have to update 3 rows to update the Address)

11

Continued…

3. Delete Anomaly:

Cannot delete Jones’ enrolment without loosing his address as well

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. Example is shown on next slide

Prime and Non-Prime Attributes

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In Normalization Terminology, Any attribute that is completely or at least

part is a member of a Primary Key is known as a Prime Attribute instead of

the more common term Key Attribute.

So, a Non-Prime attribute or Non-Key attribute, is not part of any

Candidate Key.

Closure Set of Attributes

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Trivial means D and E contains only its own value

Finding Candidate Keys using Closure Set of Attributes

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So:

There are 15 possible Keys;

We have to find Candidate

key from these possible keys

Continued..

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There are 15 possible Keys;

We have to find Candidate

key from these possible keys

Continued..

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Finding Candidate Keys using Closure Set of Attributes

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23

Continued..

1st Normal Form

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1st Normal Form

The requirements to satisfy the 1st NF are:

1) Each table has a Primary Key: minimal set of attributes which

can uniquely identify a record

2) The values in each column of a table are Atomic (No multi-

value attributes allowed eg. stdCellNo; stdEmailAddress)

3) There are no repeating groups: two columns do not store

similar information in the same table.

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2nd Normal Form

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2nd Normal Form

The requirements to satisfy the 2nd NF:

All requirements for 1st NF must be met

Any partial functional dependencies have been removed (i.e., non-keys are identified by the Whole Primary Key).

bookISBN --> bookTitle, bookFirstAuthorName, bookPublisher.

In the given above example; The title of a book, the name of the firstauthor, and the publisher are functionally dependent on the book'sInternational Standard Book Number (ISBN). In other words all listedattributes are identified uniquely by the whole Primary Key which is“bookISBN” and this relation/file/table/ENTITY is in 2nd Normal Form.

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3rd Normal Form

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Continued…

We have seen how Database Normalization can decrease redundancy,

increase efficiency and reduce anomalies by implementing three of

seven different levels of normalization called Normal Forms.

The first Three Normal Forms (3-NF) are usually sufficient for small to

medium size applications.

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3.5 Normal Form (Boyce Codd Normal

Form-BCNF)

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4th Normal Form

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Conditions that Satisfies a Relation R is in 4-NF

A relation R is in 4-NF if and only if the following Conditions satisfied:

1) R must be in 3.5 NF (BCNF).

2) It must not contain Multi Valued Dependencies (MVD’s).

Note: Multi Valued Attributes and Multi Valued Dependencies are different concepts.

Multi-Valued Dependency (MVD)

It is the dependency where one attribute value

is potentially a ‘Multi Valued fact’ about

another:

Important Points:

1) There must be three or more attributes exists

in a Relation (In example three Attributes

exists which are: Person, Mobile and

Food_Like)

2) Attributes or Subset of Attributes must be

independent of each other (In example Mobile

and Food_Like attributes are Independent of

each other).

Overview of Definitions of 4-NF and Multivalued Dependencies (MVD)

Example:

Anomalies in a Relation that is not in 4NF

Class Activity:

a) If not in 4NF than Decompose the given Relation into 4NF and Design an ERD

in Information Engineering Standard.

b) Write a complete Code of an ERD of part a) in MySQL using XAMPP as

Simulator and Show Relational Schema Diagram having Database Name: 4NF

5th Normal Form (Project-Join Normal

Form-PJNF)

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Why we Required Decomposition of a Relation

Redundancy can be reduced

Data Independence can be maintained

Anomalies can be reduced

In Natural Join Operation; After Joining tables; Repeated Records inrows are Eliminated

In Additive Lossy Join; spurious RECORD/TUPPLE/INSTANCE createdafter Natural Joining on two or more than two tables (It can be called asInformation/record retrieve that was missing in original table)

In Non-Additive Lossless Join; We are not losing any Information andnot creating any SPURIOUS/FALSE RECORDS/INSTANCES/DATA/TUPPLES ina Relation/Table/File but repetition in Records must be removed

Additive Lossy Join means EXTRA

RECORD/TUPPLE/INSTANCE created

after Natural Joining on two or more

than two tables

In Natural Join; After Joining

multiple tables; Repeated values

Eliminated as shown in figures

Projection of R1 Natural Join Projection of R2 Natural Join Projection of R3 ….. Natural Join Projection of Rn = R

R = R1 (R) ⋈ R2 (R) ⋈ …… ⋈ Rn(R)

Conditions that Satisfies a Relation R is in 5-NF A relation R is in 5-NF if and only if:

1) R must be in 4-NF.

2) If Join Dependency (JD) not exists.

a) If Non-Additive Lossless Join exists under one condition that is;decomposition of main table/relation into smaller relations that musteliminate redundant records with out losing original information and without creating false/spurious records.

b) If Additive Lossy Join exists than it will also be in 5-NF under onecondition that is; gaining meaningful information that was missing inoriginal relation (not called as spurious record in certain situation)

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Lossy and Lossless Join Examples under Two Conditions

Condition 1: When Data is given in a Relation

Condition 2: When Data is Not given in a Relation

The following given Relation is Lossy Join or Lossless Join?Condition 1: When Data is Given in a Relation

R1 is a Super Set of R2:Condition 1: Example 2

R1 ⋈ R2 ⋈ R3

Continued..

R1 ⋈ R2 ⋈ R3

Continued..

R1 (R) ⋈ R2 (R) ⋈ R3 (R)

Condition 1: Example 3

Condition 2: When Data is not given in a Relation

Solution: -

Step 1: Write data in Tabular Form: -

Step 2:

1. Apply Functional Dependencies as provided in question:

2. We will select column of these determinants containing minimum of

two stars.

Step 2 (Continued..)

Now, We will apply functional dependency where b c and c d

Step 3:

Check number of Stars in each Rows. If a Row contains stars in all

Columns than Lossless Join otherwise Lossy Join.


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6th Normal Form

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Class Activity

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Class Activity 1: Answer the following Questions

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Class Activity 2: Answer the following Questions

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a) Indicate the Level of Normal Form for the given below Relation.

b) If the Relation is not in 4-NF, Than decompose it into 4-NF Relations.

c) Design an ERD of 4-NF Relations in Information Engineering Standard.

Recommended Readings

Chapter 5 from:

Modern Database Management-8th Edition by Jeffrey

A. Hoffer, Mary B. Presscott & Fred R. McFadden

(Page No. 211-219)

Advanced Normal Forms from:

Modern Database Management-8th Edition by Jeffrey

A. Hoffer, Mary B. Presscott & Fred R. McFadden

(Page No. 605-610)


Chapter 14 from:

Database Systems-A Practical Approach to Design,

Implementation and Management by Thomas Connolly

and Carolyn Begg, 4th Edition (Page No. 481-485)

Chapter 4 from:

Database Systems-A Pragmatic Approach by Elvis C.

Foster, Shirpad V. Godbole (Page 72-79)


Chapter 6 from:

Database Systems - Design, Implementation and

Management by Carlos Coronel, Steven Morris and Peter

Rob, 9th Edition(Page No. 204-246)

Summary of Lecture

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Lecture 6➦



1st Normal Form

2nd Normal Form

3rd Normal Form

3.5 Normal Form (Boyce Codd Normal Form-BCNF)

4th Normal Form

5th Normal Form (Project-Join Normal Form-PJNF)


6th Normal Form

Class Activity

END OF LECTURE 6

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