3. 1. Why Normalize?
• If your database is not normalized, it can be inaccurate,
slow, inefficient.
• It might not produce the data you expect, or want
(update and delete anomalies).
• It makes creating queries, forms, and reports are much
easier to design.
4. 4
• Normalization: sequence of steps by which RDB is both created and improved.
Advantages of Normalization:
1. Get relations in more desirable form.
2. Make database more accurate and efficient.
3. Make creating queries easier.
4. Remove duplication.
2. Normalization
5. 5
• Normalization: sequence of steps by which RDB is both created and improved.
2. Normalization (Normalization Flow)
6. • A relation is said to be in first normal form if the data is held in a two-
dimensional table with each attribute represented by an atomic value.
• The intersection of a row and a column must contain an indivisible value.
• Each row and column position in the table there exists one value, never a set
of values.
• All attributes are atomic - any single attribute must not be composed of
multiple attributes.
6
First Normal Form (1NF)
7. The data we would want to store could be expressed as:
Project
No
Project Name Employee No Employee Name Rate
category
Rate
1203 Madagascar
travel site
11 Jessica Brookes A £90
12 Andy Evans B £80
16 Max Fat C £70
1506 Online estate
agency
11 Jessica Brookes A £90
17 Alex Branton B £80
7
First Normal Form (1NF)
8. • The intersection of a row and a column must contain an
indivisible value
Project
No.
Project Name Employee
No.
Employee
Name
Rate
category
Rate
1203 Madagascar
travel site
11 Jessica
Brookes
A £90
1203 Madagascar
travel site
12 Andy Evans B £80
1203 Madagascat
travel site
16 Max Fat C £70
1506 Online estate
agency
11 Jessica
Brookes
A £90
1506 Online estate
agency
17 Alex Branton B £70
8
Project
No.
Project Name Employee
No.
Solution
9. Three problems become apparent with our current model:
Tables in a RDBMS use a simple grid structure
• All tables in an RDBMS need a key
• Data entry should be kept to a minimum
• Redundant data
9
First Normal Form (1NF)
10. 10
1NF:
• Relation is said to be in 1NF if data is held in a table with each attribute is
represented by atomic value.
2NF:
• Relation is said to be in 2NF if:
1) It is in 1NF.
2) Remove partial dependency.
Normalization (Normalization Flow)
11. 11
3NF:
• Relation is said to be in 3NF if:
1) It is in 2NF.
2) Remove transitive dependency.
Normalization (Normalization Flow)
12. 12
Partial dependency: when non-key attribute is determined by a part, but not all
composite P.K.
Transitive dependency: when non-key attribute determines another non-key
attribute.
3. Dependencies
13. • A relation is said to be in second normal form if the relation
(1) is in 1NF and (2) all attributes that are not part of the
primary key are completely functionally dependent on the
primary key. (Partial dependencies must be removed)
• Second Normal Form (2NF) the relation must be in 1NF and each
non key attribute must be fully dependent on the whole key (not a
subset of the key).
13
Second normal form 2NF
14. Third normal form 3NF
• A relation is said to be in 3NF if it (1) is in 2NF and (2) no
attributes that are not part of the primary key are transitively
dependent on the primary key.
• The key then to move 2NF relations into 3NF is removing any
transitive dependencies that may exist in the relations.
14
15. Dependencies: Definitions
• Partial Dependency – when an non-key attribute is determined by a
part, but not the whole, of a COMPOSITE primary key.
15
CUSTOMER
Cust_ID Name Order_ID
101 AT&T 1234
101 AT&T 156
125 Cisco 1250
Partial
Dependency
16. • Transitive Dependency – when a non-key attribute determines
another non-key attribute.
16
EMPLOYEE
Emp_ID F_Name L_Name Dept_ID Dept_Name
111 Mary Jones 1 Acct
122 Sarah Smith 2 Mktg
Transitive
Dependency
Dependencies: Definitions
20. 3 NF
20
.
Service Place
Service Type
Service Date
.
12/2002
3/2000
11/99
Service Type
Service Place
Service Type
Transitive Dependency
21. 21
• What is meant by third normal form (3NF)? Examine the following table to
check if it is in 3NF. If yes, explain your answer. Otherwise convert the table
into 3NF.
4. Normalization Example
Client_no CName PropertyNo Address rent_start rent_end rent ownerNo oName
CR76 John kay
PG4
PG16
6 st.G
5 Novar
1-Jul-00
1-Sep-02
31-Aug-01
1-Sep-02
350
450
C040
C093
Tina
Tony
CR56 Aline Set
PG4
PG36
PG16
6 st.G
2 Manor
5 Novar
1-Sep-99
10-oct-00
1-Nov-02
10-Jun-00
1-Dec-01
1-Aug-03
350
370
450
C040
C093
C093
Tina
Tony
Tony
22. 22
Solution:
3NF: Relation is said to be in 3NF if:
1) It is in 2NF.
2) Remove transitive dependency.
• It is not in 3NF.
4. Normalization Example (Ex.1): Solution
23. 23
1) 1NF: make each attribute is represented by atomic value.
4. Normalization Example (Ex.1): Solution
Client_no CName PropertyNo Address rent_start rent_end rent ownerNo oName
CR76
CR76
John kay
John kay
PG4
PG16
6 st.G
5 Novar
1-Jul-00
1-Sep-02
31-Aug-01
1-Sep-02
350
450
C040
C093
Tina
Tony
CR56
CR56
CR56
Aline Set
Aline Set
Aline Set
PG4
PG36
PG16
6 st.G
2 Manor
5 Novar
1-Sep-99
10-oct-00
1-Nov-02
10-Jun-00
1-Dec-01
1-Aug-03
350
370
450
C040
C093
C093
Tina
Tony
Tony
24. 24
Choose Client_no + PropertyNo composite primary key
4. Normalization Example (Ex.1): Solution
Client_no PropertyNo CName Address rent_start rent_end rent ownerNo oName
CR76
CR76
PG4
PG16
John kay
John kay
6 st.G
5 Novar
1-Jul-00
1-Sep-02
31-Aug-01
1-Sep-02
350
450
C040
C093
Tina
Tony
CR56
CR56
CR56
PG4
PG36
PG16
Aline Set
Aline Set
Aline Set
6 st.G
2 Manor
5 Novar
1-Sep-99
10-oct-00
1-Nov-02
10-Jun-00
1-Dec-01
1-Aug-03
350
370
450
C040
C093
C093
Tina
Tony
Tony
27. 27
• Apply the various normalization steps to convert the following table into a
normal form.
4. Normalization Example (Ex.2)
Invoice
No.
Date
Cust.
No.
Cust.
Name
Cust.
Address
Cust.
City
Cust.
State
ItemID
Item
Description
Item.Q
ty
Item
Price
Item
Total
Order
total
price
125 9/13/2002 56 Foo, Inc.
23 Main
St,thorpleb
urg
thorpleb
urg
TX 563 56"Blue Fre 4 3.50 $ 14.00 $ 82.00 $
851 Spline End I 32 0.25 $ 8.00 $ 82.00 $
652 3' Red Fre 5 12.00 $ 60.00 $ 82.00 $
126 9/14/2002 2
Freens R
Us
1600
Pennsylva
nia
Washing
ton
DC 563 56"Blue Fre 500 3.50 $
1750.00
$
10750.0
0 $
652 3' Red Fre 750 12.00 $
9000.00
$
10750.0
0 $
28. 28
Solution:
• To be in 1NF
1) 1NF: make each attribute is represented by atomic value.
4. Normalization Example (Ex.2): Solution
Invoic
e No.
Date
Cust.No
.
Cust. Name
Cust.
Address
Cust. City
Cust.
State
ItemID
Item
Description
Item.Qt
y
Item Price Item Total
Order
total price
125 9/13/2002 56 Foo, Inc.
23 Main
St,thorplebur
g
thorplebur
g
TX 563 56"Blue Fre 4 3.50 $ 14.00 $ 82.00 $
125 9/13/2002 56 Foo, Inc.
23 Main
St,thorplebur
g
thorplebur
g
TX 851 Spline End I 32 0.25 $ 8.00 $ 82.00 $
125 9/13/2002 56 Foo, Inc.
23 Main
St,thorplebur
g
thorplebur
g
TX 652 3' Red Fre 5 12.00 $ 60.00 $ 82.00 $
126 9/14/2002 2
Freens R
Us
1600
Pennsylvani
a
Washingto
n
DC 563 56"Blue Fre 500 3.50 $ 1750.00 $
10750.00
$
126 9/14/2002 2
Freens R
Us
1600
Pennsylvani
a
Washingto
n
DC 652 3' Red Fre 750 12.00 $ 9000.00 $
10750.00
$
29. 29
Choose Invoice No. + ItemID composite primary key
4. Normalization Example (Ex.2): Solution
Invoice
No.
ItemID Date Cust.No. Cust. Name Cust. Address Cust. City
Cust.
State
Item
Description
Item.Qt
y
Item
Price
Item
Total
Order total price
125 563 9/13/2002 56 Foo, Inc.
23 Main
St,thorplebur
g
thorplebu
rg
TX 56"Blue Fre 4 3.50 $ 14.00 $ 82.00 $
125 851 9/13/2002 56 Foo, Inc.
23 Main
St,thorplebur
g
thorplebu
rg
TX Spline End I 32 0.25 $ 8.00 $ 82.00 $
125 652 9/13/2002 56 Foo, Inc.
23 Main
St,thorplebur
g
thorplebu
rg
TX 3' Red Fre 5 12.00 $ 60.00 $ 82.00 $
126 563 9/14/2002 2
Freens R
Us
1600
Pennsylvania
Washingt
on
DC 56"Blue Fre 500 3.50 $
1750.0
0 $
10750.00 $
126 652 9/14/2002 2
Freens R
Us
1600
Pennsylvania
Washingt
on
DC 3' Red Fre 750 12.00 $
9000.0
0 $
10750.00 $
30. 30
2) 2NF: remove partial dependency
Invoice No. Date, Cust.No., Cust. Name, Cust. Address, Cust. City, Cust. State, Order
total price
ItemID Item Description, Item Price
Invoice No. + ItemID Item.Qty, Item Total
4. Normalization Example (Ex.2): Solution
Invoice No. ItemID Item.Qty Item Total
125 563 4 14.00 $
125 851 32 8.00 $
125 652 5 60.00 $
126 563 500 1750.00 $
126 652 750 9000.00 $
Invoice
No.
Date
Cust.N
o.
Cust.
Name
Cust.
Address
Cust.
City
Cust.
State
Order total
price
125 9/13/2002 56
Foo,
Inc.
23 Main
St,thorple
burg
thorple
burg
TX 82.00 $
126 9/14/2002 2
Freens
R Us
1600
Pennsylv
ania
Washi
ngton
DC 10750.00 $
ItemID Item Description Item Price
563 56"Blue Fre 3.50 $
851 Spline End I 0.25 $
652 3' Red Fre 12.00 $
31. 31
3) 3NF: Tables are in 3 NF as there is no transitive dependency
4. Normalization Example (Ex.2): Solution
