What it normalization? Process of Normalization. Functional dependencies,

1. Normalization
by Ms. Amrit Kaur
Normalisation by Ms.Amrit Kaur 1
Source: Google Images

2.  Proposed By E.F. Codd
 Normalization
 tool to validate and improve logical design
 Process of decomposing relation with anomalies and
produces smaller, well structured relation.
What is it?
Normalisation by Ms.Amrit Kaur 2

3.  Organize the attributes (columns) and tables
(relation) of relational database to
 reduce Data redundancy
 improve Data Integrity
 No data inconsistency
 Decide which attributes should be group together
in a table to create an accurate representation of
data, its relationship and constraint
Objective
Normalisation by Ms.Amrit Kaur 3

4. Process of Normalization
Normalisation by Ms.Amrit Kaur 4

5.  Normalization is executed as series of steps. Each
step correspond to specific normal form.
 Normal form is a state of relation(table) generated
by applying rules regarding relationship between
attributes to that relation.
Process
Normalisation by Ms.Amrit Kaur 5

6. Process
Normalisation by Ms.Amrit Kaur 6

7. First step of Normalization
First normal form
Normalisation by Ms.Amrit Kaur 7

8.  First Normal Form
 RULE: A relation is said to be in First Normal Form if each
cell contains one and only one value.
 Converting Unnormalized relation to 1 NF
 Enter data into empty columns of rows
 Create separate row for multivalued column
1NF
Normalisation by Ms.Amrit Kaur 8

9. CustNo CustName ProdNo ProdName Price Qty OrderDate
????
C101
John P001 Pen 10 100 10.Oct.2020
John P002 Pencil 3 200 11.Oct.2020
????
????
C102
Pat P001 Pen 10 130 10.Oct.2020
Pat P1004 Eraser 5 150 12.Oct.2020
Pat P006 Ruler 7 200 11.Oct.2020
Unnormalised Table
Normalisation by Ms.Amrit Kaur 9
CustNo CustName ProdNo ProdName Price Qty OrderDate
C101
C101
John
?????
P001
P002
Pen
Pencil
10
3
100
200
10.Oct.2020
????
C102
Pat P001 Pen 10 130 10.Oct.2020
Pat P1004 Eraser 5 150 12.Oct.2020
Pat P006 Ruler 7 200 11.Oct.2020

10. CustNo CustName ProdNo ProdName Price Qty OrderDate
C101 John P001 Pen 10 100 10.Oct.2020
C101 John P002 Pencil 3 20
0
11.Oct.2020
C102 Pat P001 Pen 10 130 10.Oct.2020
C102 Pat P1004 Eraser 5 150 12.Oct.2020
C102 Pat P006 Ruler 7 200 11.Oct.2020
Table in 1NF
Normalisation by Ms.Amrit Kaur 10

11. Second Normal Form
Normalisation by Ms.Amrit Kaur 11

12.  Second Normal Form
 RULE: A relation is in 2NF if and only if it is in 1NF and
every non key attributes are FULLY functionally
dependent on primary key.
 Procedure: Converting from 1NF to 2NF
 Identify functional dependencies
 Identify primary key
 Identify partial dependencies on primary key,
 If partial dependencies exist, remove it from 1NF relation by
placing them in new relation along with copy of determinant.
2NF
Normalisation by Ms.Amrit Kaur 12

13. Functional
Dependencies
Normalisation by Ms.Amrit Kaur 13

14.  Functional Dependencies defines relation between
attributes(columns) in a relation(table).
 Given a relation R, a set of attributes X in R is said
to be functionally dependent on another set of
attribute Y if and only if, for each one X value in R is
associated with precisely one Y value in R.
What is it?
Normalisation by Ms.Amrit Kaur 14

15.  Y(attribute) functionally dependent on X(attribute)
 It is denoted by X->Y
 The attribute on left hand side of arrow is called
determinant.
 The attribute on right side of arrow is called dependent.
How is it represented?
Normalisation by Ms.Amrit Kaur 15

16.  Partial Functional Dependencies
 A functional dependency in which one or more non key
attributes are functionally dependent on part of primary key.
 Fully Functional Dependencies
 A functional dependency in which one or more non key
attributes are totally depending functionally dependent on
primary key.
Transitivity Dependencies
Types of Functional Dependencies
Normalisation by Ms.Amrit Kaur 16

17. Example:
Converting 1 NF Table to
2NF
Normalisation by Ms.Amrit Kaur 17

18. CourseNo DepartmentName CourseName Room Enrollment
4604 CS Database 201 45
4604 Dance Tree Dancing 302 45
4604 English The Basics of Data 123 45
2604 CS Data Structure 456 100
2604 Physics Dark Matter 302 100
Step 1 ….Identify FD
Normalisation by Ms.Amrit Kaur 18
Relation (Functional Dependency between CourseNo(X) and DepartmentName(Y)
For each X values(CourseNo) 4604 – How many departmentName ???? … 3 (CS,Dance,
English)….all are different…. 3 Y values (DepartmentName) .
Hence, CourseNo and DepartmentName not functionally dependent
Relation (Functional Dependency between CourseNo(X) and CourseName(Y)
For each X values(CourseNo) 4604 – How many courseName ?? … 3 different
courseName. It means 3 Y values (DepartmentName) .
Hence, CourseNo and CourseName not functionally dependent

19. CourseNo DepartmentName CourseName Room Enrollment
4604 CS Database 201 45
4604 Dance Tree Dancing 302 45
4604 English The Basics of Data 123 45
2604 CS Data Structure 456 100
2604 Physics Dark Matter 302 100
[contd.] Step 1
Normalisation by Ms.Amrit Kaur 19
Relation (Functional Dependency between CourseNo(X) and Room(Y)
For each X values(CourseNo) 4604 –3 different rooms; i.e. 3 Y values .
Hence, CourseNo and Room not functionally dependent
Relation (Functional Dependency between CourseNo(X) and EnrollmentY)
For each X values(CourseNo) 4604 – How many enrollment?? … 3 but value is 1; i.e.
45. It means 1 Y Value
Hence, CourseNo and Enrollment are functionally dependent. CourseNo ->Enrollment

20.  Identified Functional Dependencies considering each
attributes individually and then attribute combinations
1. CourseNo -> Enrollment
2. CourseName -> Room
3. CourseName->Enrollment
4. CourseNo+DepartmentName -> CourseName
5. CoursrNo+DepartmentName -> Room
6. CourseNo+ Department -> Enrollment
7. CourseName+DepartmentName -> CourseNo
8. CoursrName+DepartmentName -> Room
9. CourseName+ Department -> Enrollment
[contd] Step 1
Normalisation by Ms.Amrit Kaur 20

21.  Make list of Determinants from identified FDs
 Rules of Choosing Primary Key
 Dependent cannot be primary key
 For Determinant, the column must not contain duplicate
values
Step 2 ….Choose Primary Key
Normalisation by Ms.Amrit Kaur 21

22.  List of Determinants
(left hand side of FDs)
 CourseNo
 CourseName
 CourseNo+DepartmentName
 CourseName+ DepartmentName
 Identified functional dependencies in Step 1
[contd.] Step 2
Normalisation by Ms.Amrit Kaur 22
• Apply Rules
– In FD (4), CourseName is dependent
– In FD (7), CourseNo is dependent
Both Cannot be Primary Key
1. CourseNo -> Enrollment 2. CourseName -> Room
3. CourseName->Enrollment 4. CourseNo+DepartmentName -> CourseName
5.CoursrNo+DepartmentName -> Room 6.CourseNo+ Department -> Enrollment
7.CourseName+DepartmentName -> CourseNo 8. CoursrName+DepartmentName -> Room
9. CourseName+ Department -> Enrollment
• Choose any ONE as Primary Key
– CourseNo + DepartmentName
– CourseName + DepartmentName

23. Partial Dependency means
 one or more non key
attributes are functionally
dependent on part of primary
key.
 Identified functional dependencies in Step 1
Step 3…. Identify Partial Dependencies
Normalisation by Ms.Amrit Kaur 23
DepartmentName->Enrollment (FALSE)
It means from FD(1), Enrollment is depending
on primary key as a whole and also ONLY on
CourseNo (part of primary Key) .
1. CourseNo -> Enrollment 2. CourseName -> Room
3. CourseName->Enrollment 4. CourseNo+DepartmentName -> CourseName
5.CoursrNo+DepartmentName -> Room 6.CourseNo+ Department -> Enrollment
7.CourseName+DepartmentName -> CourseNo 8. CoursrName+DepartmentName -> Room
9. CourseName+ Department -> Enrollment
• Identified PRIMARY KEY in Step2 CourseNo + DepartmentName
• Partial Dependencies Exist
CourseNo - Enrollment

24.  Remove partial dependencies Enrollment from Table in 1 NF
 Create separate table with Enrollment
 Copy determinants CourseNo also
Step 4 ….Remove Partial
Dependencies
Normalisation by Ms.Amrit Kaur 24
Course
No
DepartmentName CourseName Room
4604 CS Database 201
4604 Dance Tree Dancing 302
4604 English The Basics of Data 123
2604 CS Data Structure 456
2604 Physics Dark Matter 302
CourseNo Enrollment
4604 45
2604 100
After 2NF, Table Structure

25. CustNo CustName ProdNo ProdName Price Qty Order
No
OrderDate
C101 John P001 Pen 10 100 1 10.Oct.2020
C101 John P002 Pencil 3 200 2 11.Oct.2020
C102 Pat P001 Pen 10 130 3 10.Oct.2020
C102 Pat P1004 Eraser 5 150 4 12.Oct.2020
C102 Pat P006 Ruler 7 200 5 11.Oct.2020
Example 2. Convert Table to 2NF
Normalisation by Ms.Amrit Kaur 25
Step1: Identify Functional Dependency
(CustNo+ProdNo)->Qty
(CustNo+ProdNo)->ProdName
(CustNo+ProdNo)-> Price
(CustNo+ProdNo)->OrderNo
(CustNo+ProdNo)->OrderDate
Step2: Identify Primary Key
(CustNo+ProdNo)
Step3: Partial Dependency
(CustomerName is depending only on CustNo,
not on ProdNo
CustNo->Custname
(ProductName is depending only on ProdNo
not on CustNo
Prodno->ProdName
STEP 4. Remove dependencies CustName and ProdName from Table in 1 NF and create
two separate tables and copy determinants CustNo and ProdNo respectively.

26. Example 2. After 2 NF
Normalisation by Ms.Amrit Kaur 26
CustNo CustName
C101 John
C102 Pat
CustNo ProdNo Price Qty OrderNo OrderDate
C101 P001 10 100 1 10.Oct.2020
C101 P002 3 200 2 11.Oct.2020
C102 P001 10 130 3 10.Oct.2020
C102 P1004 5 150 4 12.Oct.2020
C102 P006 7 200 5 11.Oct.2020
ProdNo ProdName
P001 Pen
P002 Pencil
P1004 Eraser
P006 Ruler
Customer Product
Purchase

27. Third Normal Form
Normalisation by Ms.Amrit Kaur 27

28.  Third Normal Form
 RULE: A relation is in 3NF if and only if it is in
2NF and every nonkey attribute is non
transitively dependent on primary key.
 Procedure: Converting 2NF to 3NF
 Identify Functional Dependencies
 Identify primary key
 If transitive dependency exist on primary key, remove them
by placing them in new relation along with dominant.
3NF
Normalisation by Ms.Amrit Kaur 28

29.  A transitive dependency exists when the following
functional dependency pattern:
 A → B
 B → C ; therefore A → C.
 If any attribute is depending upon primary key and also
depending upon other attribute
What is Transitive Dependency?
Normalisation by Ms.Amrit Kaur 29

30. Example:
Converting 2NF Table to
3NF
Normalisation by Ms.Amrit Kaur 30

31. 2 NF Tables
Normalisation by Ms.Amrit Kaur 31
CustNo CustName
C101 John
C101 John
C102 Pat
C102 Pat
C102 Pat
CustNo ProdNo Price Qty OrderNo OrderDate
C101 P001 10 100 1 10.Oct.2020
C101 P002 3 200 2 11.Oct.2020
C102 P001 10 130 3 10.Oct.2020
C102 P1004 5 150 4 12.Oct.2020
C102 P006 7 200 5 11.Oct.2020
ProdNo ProdName
P001 Pen
P002 Pencil
P001 Pen
P1004 Eraser
P006 Ruler
Customer Product
Purchase

32. Steps…Converting 2NF to 3NF
Normalisation by Ms.Amrit Kaur 32
Step1: Identify Functional Dependency
Customer Relation
1. CustNo -> CustName
Product Relation
2. ProdNo -> ProdName
Purchase Relation
3. (CustNo+ProdNo)->Qty
4. (CustNo+ProdNo)-> Price
5. (CustNo+ProdNo)->OrderNo
6. (CustNo+ProdNo)->OrderDate
7. OrderNo -> OrderDate
Step2: Identify Primary Key
Customer Relation – CustNo
Product Relation – ProdNo
Purchase – CustNo+ ProdNO
Step3: Transitive Dependency
Non key attribute is depending upon other
attribute that is not primary key then
transitive dependencies exist. Here
(CustNo+ProdNo)->OrderNo
OrderNo -> OrderDate
(CustNo+ProdNo)->OrderDate
STEP 4. Remove transitive
dependencies OrderDate from Table in
2 NF and create separate tables and
copy OrderNo

33. After 3 NF
Normalisation by Ms.Amrit Kaur 33
CustNo CustName
C101 John
C101 John
C102 Pat
C102 Pat
C102 Pat
CustNo ProdNo Price Qty OrderNo
C101 P001 10 100 1
C101 P002 3 200 2
C102 P001 10 130 3
C102 P1004 5 150 4
C102 P006 7 200 5
ProdNo ProdName
P001 Pen
P002 Pencil
P001 Pen
P1004 Eraser
P006 Ruler
Customer Product
Purchase
OrderNo OrderDate
1 10.Oct.2020
2 11.Oct.2020
3 10.Oct.2020
4 12.Oct.2020
5 11.Oct.2020
Order

34. Boyce Codd Normal Form(BCNF)
Normalisation by Ms.Amrit Kaur 34

35.  Relations in 3NF still contain redundancy when
 There are two or more candidate key
 Candidate Keys are composite key
 Candidate Key overlaps (one attribute common)
 BCNF
 RULE: A relation is in BCNF if and only if every
determinant is a candidate key
BCNF
Normalisation by Ms.Amrit Kaur 35

36.  Procedure: Converting to BCNF
 Identify all candidate keys
 Indentify all functional dependencies
 If functional dependencies exist where determinants are
not candidate key, remove functional dependencies by
placing them in new relation along with determinant
BCNF
Normalisation by Ms.Amrit Kaur 36

37. Example: Apply BCNF
Normalisation by Ms.Amrit Kaur 37
ClientNo InterviewDate Interview Time Staff No Room No
C176 20-Oct-2020 10:30 S5 101
C156 20-Oct-2020 12:00 S5 101
C174 20-Oct-2020 12:00 S37 102
C156 22-Oct-2020 10:30 S5 102
Primary Key (ClientNo,InterviewDate)->(InterviewTime,StaffNo,RoomNo)
Candidate Keys
1. (StaffNo,InterviewDat,InterviewTime) -> (ClientNo, RoomNo)
2. (RoomNo,InterviewDate,InterviewTime)-> (StaffNo, ClientNo)
Other Functional Dependency – Determinant not Candidate Key
Remove and create another relation and copy determinant
1. (StaffNo, InterviewDate) -> RoomNo

38. After BCNF
Normalisation by Ms.Amrit Kaur 38
ClientNo InterviewDate Interview Time Staff No
C176 20-Oct-2020 10:30 S5
C156 20-Oct-2020 12:00 S5
C174 20-Oct-2020 12:00 S37
C156 22-Oct-2020 10:30 S5
Staff No InterviewDate Room No
S5 20-Oct-2020 101
S37 20-Oct-2020 101
S5 22-Oct-2020 102

39. Questions
Normalisation by Ms.Amrit Kaur 39

40. Thank You
Normalisation by Ms.Amrit Kaur 40