4bb07functional Dependencies

8/8/2019 4bb07functional Dependencies

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Functional Dependencies and

Normalization


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E-R model gives table structure of the database.

Table structure given by E-R model may be or may not

be good. i.e. tables created by E-R model may have

redundancies and various anomalies.

So, how do we recognize a bad database

design or good database design?


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Normalization

Thus, Normalization is the process of refining the data

structure built by E-R model.


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Normalization takes a relation schema through a series of

tests to certify whether it satisfy a certain normal form.

1 NF

2NF

3NF

BOYEE and CODD

4NF

5NF


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Minimizing the redundancy

Removing the insertion, deletion and update anomalies.

Normalization is a formalized procedure to eliminatingredundancy from data by the progressive use of non-

lose decomposition, which involves splitting records

without losing information

Normalization


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(cont.)

Normalization is based on the idea that an attributemay depend on another attribute in some way.

There are 2 different kinds of dependencies involvedup to 5NF

Functional dependency

Multivalued dependence


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Functional Dependence

In a relation including attribute A and B,B is functional dependent on A if,

for every valid occurrence, the value A determinesthe value B

A and B can be composite

If B is Functional Dependent on A, then A is thedeterminant of B

All fields are functionally dependent on the primary key or indeed any candidate key.


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Let X, Y be two subsets of relation schema R then if the given value of X,

there is only one value ofY , then Y is said to Functionally dependent on

X, or

X functionally determines Y.

X Y

i.e. functional dependencies is a constraints between two subsets of

attributes of relation schema R.

If X Y hold in R, we can not say that whetherY X holds in R or

not.


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Interference Rule forFunctional Dependency

Let F be the given set ofFunctional Dependencies specified on

Schema R.

1. Augmentation Rule:

if ZY holds in R and X is a set of attributes from R then XZXY

holds.

2. Transitive Rule:

if XY andY Z holds in R, then X Z also holds.

3. Union or Additive Rule:

if X Y and X Z hold in R , then X YZ holds in R.

4. Decomposition or Projective Rule:

if X YZ hold in R , then X Y or X Z holds in R.


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5. Pseudotransitive Rule:

if X Y and ZY W holds in R then ZX W holds in R.


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Functional Dependence

S# CITY P# QTY

S1 Khon Kaen P1 100

S1 Khon Kaen P2 100

S2 Saraburii P1 200

S2 Saraburii P2 200

S3 Saraburii P2 300

S4 Bangkok P2 400

S4 Bangkok P4 400

S4 Bangkok P5 400

S#,P# QTY QTY is functionally dependent on S#andP#

S# and P# are the determinant of QTY


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First Normal Form

A relation is in First Normal form if, and onlyif, it contains no multi-value or no repeating

groups.


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First NF (cont.)

NO Name Province PayDate1 Amount1 PayDate2 Amount2

E001 Somchai Khon Kaen 15/04/2004 5,000.00 30/04/2004 5,000.00

E002 Sompong Sarakham 15/04/2004 4,500.00 30/04/2004 4,500.00

E003 Somchay Ubon 15/04/2004 5,200.00 30/04/2004 5,200.00

Repeat


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Problem

ENO Name Dno DeptName ProjNo ProjName

E001 Somchai D01 Physic P01, P02 NMR, Laser

E002 Sompong D01 Physic P03 Medical Imageprocessing

E003 Somchay D02 Computer

Science

P04, P05 Voice ordering, Speech

Coding

E004 SomSiri D02 Computer

Science


Synthesis

Multi-value

Problem1. Difficult to manipulate data2. Redundancy

UPDATE ANOMALIES

Staff


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Insert





Science


Coding


Science


Synthesis

We can not insert new project if the project has not assigned toany employee yet.

P06 Speech Corpus

Staff


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UPDATE ANOMALIES





Science


Coding


Science


Synthesis

Change ProjName from Voice Ordering to Speech Orderingneed to change all in Database

Staff


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DELETE Problem





Science


Coding


Science


Synthesis

Delete Employee E002 SompongProject P03 Medical Image Processing was deleted also

Staff


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Solution

Remove the repeating group

In case of multi-valued Create new relation

Columns = Key + multi-valued


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Repeating group

ENO Name Province PayDate1 Amount1 PayDate2 Amount2

E001 Somchai Khon Kaen 15/04/2004 5,000.00 30/04/2004 5,000.00

E002 Sompong Sarakham 15/04/2004 4,500.00 30/04/2004 4,500.00

E003 Somchay Ubon 15/04/2004 5,200.00 30/04/2004 5,200.00

ENO Name Province

E001 Somchai Khon Kaen

E002 Sompon

g

Sarakham

E003 Somcha

y

Ubon

ENO PayDate Amount

E001 15/04/2004 5,000.00

E001 30/04/2004 5,000.00

E002 15/04/2004 4,500.00

E002 30/04/2004 4,500.00

E003 15/04/2004 5,200.00

E003 30/04/2004 5,200.00

Employee

Employee

PayCheck


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Multi-Valued



E002 Sompong D01 Physic P03 Medical Image processing

E003 Somchay D02 Computer Science P04, P05 Voice ordering, Speech Coding

E004 SomSiri D02 Computer Science P04, P06 Voice ordering, Speech Synthesis

Staff


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Multi-Valued


E001 Somchai D01 Physic P01 NMR

E001 SomchaiD01

Physic P02 Laser E002 Sompong D01 Physic P03 Medical Image processing

E003 Somchay D02 Computer Science P05 Voice ordering

E003 Somchay D02 Computer Science P04 Speech Coding

E004 SomSiri D02 Computer Science P04 Voice ordering

E004 SomSiri D02 Computer Science P06 Speech Synthesis

Insert Project still has problem


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Second Normal Form

(2NF) A relation is in first normal form if and only

if

It is in 1NF

Every non-key attribute is dependent on all

parts of the primary key.


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Problem



E001 Somchai D01 Physic P02 Laser






We can not insert Project if have not yetassigned project to any employee


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Solution

Remove the attribute involved

Take its determinant with it


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Normalize



E001 Somchai D01 Physic P02 Laser







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Result

ENO Name Dno DeptNa

me

E001 Somchai D01 Physic

E003 Somchay D02 Computer Science

E004 SomSiri D02 Computer Science

Proj

No

ProjName

P01 NMR

P02 Laser

P03 Medical Image

processing

P04 Speech Coding

P05Voice ordering

P06 Speech Synthesis

ProjectPERSON

ENO ProjN

o

E001 P01

E001 P02

E002 P03

E003 P04

E004 P05

E004 P06

PERSON_Proj

PERSON(ENO,NAME,Dno,DeptName)

PROJECT(ProjNo,ProjName)

PERSON_PROJ(ENO

,P

rojN

o)


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Third Normal Form

A relation is in 3NF if, and only if:

It is in 2NF

Every non-key attribute is functionallydependent upon the key. (No non-keyattribute is functional dependent on another

non-key attribute)

Or non-key attribute no transitive dependenton key


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3NF?

ENO Name Dno DeptNa

me

E001 Somchai D01 Physic

E003 Somchay D02 Computer Science

E004 SomSiri D02 Computer Science

Proj

No

ProjName

P01 NMR

P02 Laser

P03 Medical Image

processing

P04 Speech Coding

P05Voice ordering


ProjectPERSON

ENO ProjN

o

E001 P01

E001 P02

E002 P03

E003 P04

E004 P05

E004 P06

PERSON_Proj

Answer is NoBecause DeptName is dependenton Dno (has transitive dependent

on key)


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Solution

Remove the offending attributes

Take the determinant along


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Result

ENO Name Dno

E001 Somchai D01E003 Somchay D02

E004 SomSiri D02

ProjNo

ProjName

P01 NMR

P02 Laser

P03 Medical Image

processing

P04 Speech Coding

P05 Voice ordering


Project

PERSON

ENO ProjNo

E001 P01

E001 P02

E002 P03

E003 P04E004 P05

E004 P06

PERSON_Proj

Dno DeptName

D01 Physic

D02 Computer

Science

D02 Computer

Science

Department


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Boyce-Codd Normal Form

(BCNF) A relation is in BCNF if, and only if, every

determinant is a candidate key.

BCNF is a refinement to third normal form,

and tightens its duration.

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