 First Normal Form (1NF)  Second Normal Form (2NF)  Third Normal Form (3NF)  Boyce-Codd Normal Form (BCNF)  Boyce-Codd Normal Form (BCNF)  Fourth Normal Form (4NF)  Fifth Normal Form (5NF)

1. MCA, MSc(CS), PGDCA
Pratibha Rashmi

2.  First Normal Form (1NF)
 Second Normal Form (2NF)
 Third Normal Form (3NF)
 Boyce-Codd Normal Form (BCNF) Boyce-Codd Normal Form (BCNF)
 Fourth Normal Form (4NF)
 Fifth Normal Form (5NF)
Note: The database designers don’t have
to normalize relations to the highest
possible normal form (usually 3NF, BCNF
or 4NF is enough)

3. First normal form enforces these criteria:
 Eliminate repeating groups in individual
tables.
 Create a separate table for each set of
related data.related data.
 Identify each set of related data with a
primary key.
 Each set of column must have a unique
value.
 It contains atomic values because the table
cannot hold multiple values.

4. tb_Product
p_id colour cost
1 Red, yellow 3000
2 Blue 2000
3 Green, Red 2030
This table is not in first normal form because the
“Colour” column contains multiple Values.

5. p_id cost
1 3000
2 2000
3 2030
p_id Colour
1 Red
1 Yellow
2 Blue
3 Green
3 Red

6. A table is said to be in 2NF if both the
following conditions hold:
 Table is in 1NF (First normal form). Table is in 1NF (First normal form).
 No non-prime attribute is dependent on the
proper subset of any candidate key of table.
 Non-prime attribute-An attribute that is not
part of any candidate key is known as non-
prime attribute.

7.  Or in other words A relation R is said to be in
2NF if it is in 1NF and if all nono-prime
attributes are not partially dependent on
candidate key ie all non- prime attribute
must be fully functionally dependent on
candidate key.
must be fully functionally dependent on
candidate key.
 Partially dependency: In a relationa R XY->Z
is said to be partial FD if either X->Z or Y->Z
exists in R.
 On other hand In a relationa R XY->Z is said
to be fully FD if neither X->Z nor Y->Z exists
in R.

8.  We use a teacher table that have teacher id,
subject and age. Since a teacher can teach more
than one subjects, the table can have multiple
rows for a same teacher.
tb_teacher
t_id subject t_aget_id subject t_age
1 DBMS 40
1 Networking 40
2 OS 38
2 DM 38
3 AI 42
* Candidate Keys: {t_id, subject}
* Non prime attribute: t_age

9.  This table is in 1 NF because each attribute
has atomic values.
 But, it is not in 2NF because non prime
attribute t_age is dependent on t_id alone
which is a proper subset of candidate key.
This violates the rule for 2NF as the rule says
“no non-prime attribute is dependent on the
proper subset of any candidate key of theproper subset of any candidate key of the
table”.
 To make the table complies with 2NF we can
break it in two tables , eg. t_details and
t_subject.

10.  Table 1: t_details
t_details
t_id t_age
1 40
2 38
3 42
 Table 2: t_subject
t_subject
t_id subject
1 DBMS
1 Networking
2 OS
2 DM
3 AI
Now the tables execute with Second normal form (2NF).

11.  Third Normal Form (3NF) is used to minimize
the transitive redundancy.
 In 3NF, the table is required in 2NF.
 While using the 2NF table, there should not
be any transitive partial dependency.be any transitive partial dependency.
 3NF reduces the duplication of data and also
achieves the data integrity.
In other words: A table is in 3NF if it is in 2NF
and for each functional dependency X-> Y at
least one of the following conditions hold:
 X is a super key of table
 Y is a prime attribute of table

12.  Let’s take a table Student details with
attributes student id, name, city, district,
state, zip.
tb_Student
s_id s_name city district state zip
1 Jaunty Delhi Delhi Delhi 1100011 Jaunty Delhi Delhi Delhi 110001
2 Arnav Agra Agra UP 282005
3 Ram Aligarh Aligarh UP 200201
4 Reeta Meerut Meerut UP 250001
5 Sita Delhi Delhi Delhi 110013
Super keys: {s_id}, {s_id, s_name}, {s_id, s_name, zip}…....so on
Candidate Keys: {s_id}
Non-prime attributes: all attributes except s_id are non-prime
as they are not part of any candidate keys.

13.  Here, state, city and district dependent on
zip. And, zip is dependent on s_id that makes
non-prime attributes (state, city and district)
transitively dependent on super key (s_id).
This violates the rule of 3NF.
 To make this table complies with 3NF we
have to break the table into two tables to
remove the transitive dependency:
tb_Student
s_id s_name zip
1 Jaunty 110001
2 Arnav 282005
3 Ram 200201
4 Reeta 250001
5 Sita 110013
tb_Zip
zip city district state
110001 Delhi Delhi Delhi
282005 Agra Agra UP
200201 Aligarh Aligarh UP
250001 Meerut Meerut UP
110013 Delhi Delhi Delhi

14.  Even when a database is in 3rd Normal
Form, still there would be anomalies resulted
if it has more than one Candidate Key.
 BCNF is an advance version of 3NF that’s why
it is also referred as 3.5NF.it is also referred as 3.5NF.
 BCNF is stricter than 3NF.
 A table complies with BCNF if it is in 3NF
and for every functional dependency X->Y, X
should be the super key of the table.

15.  Let's assume there is a company where
employees work in more than one
department.
tb_employee
e_id e_country e_dept dept_type e_dept_id
1 India Sales D003 1003
1 India Account D003 10081 India Account D003 1008
2 USA Developing D001 1001
2 USA Designing D001 1006
3 UK Sales D003 1007
In the above table Functional dependencies are as follows:
e_id → e_country
e_dept → {dept_type, e_dept_id}
Candidate key: {e_id, e_dept}
The table is not in BCNF because neither e_dept nor e_id alone
are keys.

16.  To convert the given table into BCNF, we
decompose it into three tables:
tb_emp_country
e_id e_country
1 India
2 USA
3 UK
tb_emp_dept
e_dept dept_type e_dept_id
Sales D003 1003
Table 1:
Table 2:
3 UK Sales D003 1003
Account D003 1008
Developing D001 1001
Designing D001 1006
Sales D003 1007
tb_emp_dept_mapping
e_id e_dept
1 Sales
1 Account
2 Developing
2 Designing
3 Sales
Table 3:

17.  Functional dependencies:
e_id → e_country
e_dept → {dept_type, e_dept_id}
 Candidate keys:
For the first table: e_id
For the second table: e_dept
For the third table: {e_id, e_dept}
 Now, this is in BCNF because left side part of
both the functional dependencies is a key.

19.  Ques 1: What is normalization? What are
different type of normalization?
 Ques 2: Define Prime and Non - prime
attributes.
Ques 3: Why do we use normalized database? Ques 3: Why do we use normalized database?
 Ques 4: Define partial and full dependency.
 Ques 5: Define normalization. Explain INF,
2NF, and 3NF using appropriate example.
 Ques 6: differentiate between 3NF and BCNF.