Chapter4 - Schema Refinement and Normalisation

7/28/2019 Chapter4 - Schema Refinement and Normalisation

1/24

SCHEMA REFINEMENT

AND

NORMALIZATION


2/24

Objectives

After completing this chapter, you will be able to:

Define Normalization

List reason for Normalization

Refining a database

Defining Normal Form

Describe Types of Normal Forms


3/24

Normalization is a step-by-step decomposition of complex

records into simple records.It results in the formation of

tables that satisfy certain specified constraints, and

represent certain normal forms.

Normalization reduces redundancy using the principle of non-

loss decomposition. A fully normalized record consists of:

A primary key that identifies an entity

A set of attributes that describe the entity

Normalization


4/24

Does the design ensure that all database operations will be

efficiently performed and that the design does not make

the DBMS perform expensive consistency checks which

could be avoided? Is the information unnecessarily replicated?

Why Normalization?

Unless these issues are properly handled, several difficulties like

redundancy and loss of information may arise. There are several

methods to avoid the above mentioned problems. One such

method is database decomposition throughnormalization.


5/24

Redundancy is at the root of several problems associated

with relational schemas

More seriously, data redundancy causes several anomalies:

Insert Update

Delete

Wastage of storage. Main refinement technique: decomposition (replacing

ABCD with, say, AB and BCD, or ACD and ABD).

The Evils of Redundancy


6/24

Refining an E-R Diagram

Department

did

dname

budget

since

Works_inEmployee

ssn

name

lot

Before


7/24

Works_in

since

Employee

ssnname

lot

Department

did

dname

budget

After

Refining continued...


8/24

Normal Forms

Forms are designed to logically address potential problems

in referring to and working with information stored in a

database.

A database is said to be in one of the Normal Forms, if it

satisfies the rules required by that Form.

It also will not suffer from any of the problems addressed by

the Form.


9/24

Types of Normal Forms

Several normal forms have been identified, the most

important and widely used of which are:

First normal form (1NF)

Second normal form (2NF)

Third normal form (3NF)

Boyce-Codd normal form (BCNF)

Fourth normal form (4NF)


10/24

First Normal Form (1NF)

A table is in 1NF, if every row contains exactly one value foreach attribute.

Disallow multivalued attributes, composite attributes and their

combinations.

1NF states that :

Domains of attributes must include only atomic (simple,

indivisible) values and that value of any attribute in a

tuple must be a single value from the domain of thatattribute.

By definition, any relational table must be in 1NF.


11/24

Functional Dependency (FD)

They provide a formal mechanism to express

constraints between attributes.

Given a relation R, attribute Y of R isfunctionally dependent on the attribute X of R

if and only if each X-value in R has associated

with it precisely one Y-value in R.The dependency is represented asY X.


12/24

Full Dependency

An FD X Y is a full functional dependency if

removal of any attribute A from X means that

the dependency does not hold any more.

An attribute B of a relation R is fully functional

dependent on attribute A of R if it is

functionally dependent on A and notfunctionally dependent on any proper subset of

A.


13/24

Partial Dependency

A FD X Y is a partial dependency if there is

some attribute A X that can be removed from Xand the dependency will still hold.


14/24

123- 22- 3666 Attishoo

231- 31- 5368

131- 24- 3650

434- 26- 3751612- 67- 4134

Smiley

Smethurst

GulduMadayan

48

22

35

3535

8

8

5

58

10

10

7

710

40

30

30

3240

S N L R W H

5

8

7

10

R W

123- 22- 3666 Attishoo

231- 31- 5368131- 24- 3650

434- 26- 3751

612- 67- 4134

SmileySmethurst

Guldu

Madayan

48

2235

35

35

S N L

40

3030

32

40

H

8

R

85

5

8

Constraints on Entity Set


15/24

Second Normal Form (2NF)

A relation schema R is in 2NF if:

It is in 1NF

Every non-prime attribute A in R is fully

functionally dependent on the primary keyof R.

2NF prohibitspartial dependencies.

If a relation is not in 2NF, it can be further normalized

into a number of 2NF relations.


16/24

Converting a Database to 2NF

Find and remove attributes that are related to only a

part of the key.

Group the removed items in another table.

Assign the new table a key that consists of that part of

the old composite key.


17/24

Transitive Dependency

An FD X Y in a relation schema R is a transitive

dependency if:

There is a set of attributes Z that is not a

subset of any key of R.

Both X Z and Z Y hold.


18/24

Emp{Eno, Dept, ProjCode, Hours}

Primary key: {Eno, ProjCode}

{Eno} -> {Dept}, {Eno, ProjCode} -> {Hours}

Test of 2NF:

{Eno} -> {Dept}:partial dependency.

Emp is in 1NF, but not in 2NF.

Decomposition:

Emp {Eno, Dept}

Proj {Eno, ProjCode, Hours}

Example


19/24

Third Normal Form (3NF)

A relation schema R is in 3NF if whenever a functional

dependency X A hold in R, then either:

X is a super key of R

A is a prime attribute of R.

If 3NF is violated by X A, one of the following holds:

X is a subset of some key K

We store (X, A) pairs redundantly

X is not a proper subset of any key


20/24

Emp{Eno, Dept, Dept_Head}Primary key: {Eno}

{Eno} -> {Dept}, {Dept} -> {Dept_Head}

Test of 3NF

{Eno} -> {Dept} -> {Dept_Head}: Transitive

dependency.

Emp is in 2NF, but not in 3NF.

Decomposition:

Emp {Eno, Dept}

Dept {Dept, Dept_Head}

Example


21/24

Boyce-Codd Normal Form (BCNF)

A relation R is said to be in BCNF, if and only if

every determinant is a candidate key.

3NF does not satisfactorily handle the case of a

relation processing two or more composite or

overlapping candidate keys.


22/24

Suppose that relation R contains attributes A1... An.

A decomposition of R consists of replacing R by two

or more relations such that:

Each new relation scheme contains a subset of

the attributes of R (and no attributes that do

not appear in R).

Every attribute of R appears as an attribute of

one of the new relations.

Decomposition of a Relation Schema


23/24

Fourth Normal Form (4NF)

Under fourth normal form (4NF):

A record type should not contain two or more

independent multi-valued facts about an entity.

The record must satisfy 3NF.


24/24

Summary

You now should be able to:

Define Normalization

List reason for Normalization

Refining a database

Defining Normal Form

Describe Types of Normal Forms

Documents

Chapter4 - Schema Refinement and Normalisation