Normalization

o Normalization is the process of organizing the data in the database.

o Normalization divides the larger table into smaller and links them using relationships.
o The normal form is used to reduce redundancy from the database table.
o It is a technique to remove or reduce redundancy from a table. There are basically two
types of duplicacy in the database. One is Row level another is Column level.

Primary Key
Sid Sname Age
1 Ram 20
2 Varun 25
1 Ram 20
Row Level Redundancy

The First row and the third row is exactly same means both the rows are exactly same, this is
row level duplicacy or redundancy. Row level duplicacy should not be there in a table. Method
to remove row level duplicacy that we use the concept of Primary Key. We set a Primary key in
our table to any attribute. When we made a Sid Primary Key (Unique + Not Null), means neither
value can be repeated in it, nor it can be left empty. So, Primary Key will take care of row level
duplicacy.

Sid Sname Cid Cname Fid Fname Salary

1 Ram C1 DBMS F1 Jass 45,000
2 Sham C2 JAVA F2 Kiran 40,000
3 Twinkle C1 DBMS F1 Jass 45,000
4 Neha C1 DBMS F1 Jass 45,000
Column Level Redundancy

Data modification anomalies can be categorized into three types:

o Insertion Anomaly: Insertion Anomaly refers to when one cannot insert a new tuple into a
relationship due to lack of data.
o Deletion Anomaly: The delete anomaly refers to the situation where the deletion of data results
in the unintended loss of some other important data.
o Updatation Anomaly: The update anomaly is when an update of a single data value requires
multiple rows of data to be updated.

(Anomaly means a kind of problem, but such a problem that occurs on a special occasion. )

Ms. Jasmeen Kaur

Assistant Professor
(Computer Applications)
 Types of Normal Forms:
Normalization works through a series of stages called Normal forms. The normal forms apply to
individual relations. The relation is said to be in particular normal form if it satisfies constraints.

Following are the various types of Normal forms:

First Normal Form (1NF)

o A relation will be 1NF if it contains an atomic value.
o It states that an attribute of a table cannot hold multiple values. It must hold only single-valued attribute.
o First normal form disallows the multi-valued attribute, composite attribute, and their combinations.
o Example: Relation EMPLOYEE is not in 1NF because of multi-valued attribute EMP_PHONE.

EMPLOYEE table:

EMP_ID EMP_NAME EMP_PHONE EMP_STATE

14 John 7272826385, UP
9064738238

20 Harry 8574783832 Bihar

12 Sam 7390372389, Punjab

8589830302

o The decomposition of the EMPLOYEE table into 1NF has been shown below:

Ms. Jasmeen Kaur

Assistant Professor
(Computer Applications)
 EMP_ID EMP_NAME EMP_PHONE EMP_STATE

14 John 7272826385 UP

14 John 9064738238 UP

20 Harry 8574783832 Bihar

12 Sam 7390372389 Punjab

12 Sam 8589830302 Punjab

EMP_ID EMP_NAME EMP_PHONE1 EMP_PHONE2 EMP_STATE

14 John 7272826385 9064738238 UP

20 Harry 8574783832 Null Bihar

12 Sam 7390372389 8589830302 Punjab

Second Normal Form (2NF)

o In the second normal form, all non-key attributes are fully functional dependent on the primary key

o According to the EF codd. In the 2NF, relational must be in 1NF.

o Example: Let's assume, a school can store the data of teachers and the subjects they teach. In a school, a
teacher can teach more than one subject.

TEACHER table

TEACHER_ID SUBJECT TEACHER_AGE

25 Chemistry 30

25 Biology 30

47 English 35

83 Math 38

83 Computer 38

o In the given table, non-prime attribute TEACHER_AGE is dependent on TEACHER_ID which is a proper
subset of a candidate key. That's why it violates the rule for 2NF.
o To convert the given table into 2NF, we decompose it into two tables:

Ms. Jasmeen Kaur

Assistant Professor
(Computer Applications)
 o TEACHER_DETAIL table:

TEACHER_ID TEACHER_AGE

25 30

47 35

83 38

o TEACHER_SUBJECT table:

TEACHER_ID SUBJECT

25 Chemistry

25 Biology

47 English

83 Math

83 Computer

Third Normal Form (3NF)

o A relation will be in 3NF if it is in 2NF and not contain any transitive partial dependency.
o 3NF is used to reduce the data duplication. It is also used to achieve the data integrity.
o If there is no transitive dependency for non-prime attributes, then the relation must be in third normal form.
o Transitive Dependency is a non-prime attribute determining a non-prime attribute.

A relation is in third normal form if it holds atleast one of the following conditions for every non-trivial function
dependency X → Y.

1. X is a super key.
2. Y is a prime attribute, i.e., each element of Y is part of some candidate key.

Example:

EMPLOYEE_DETAIL table:

EMP_ID EMP_NAME EMP_ZIP EMP_STATE EMP_CITY

222 Harry 201010 UP Noida

333 Jass 02228 PUNJAB Mohali

444 Kiran 60007 HARYANA Ambala

555 Rahul 06389 UK Norwich

666 Priya 462007 MP Bhopal

Ms. Jasmeen Kaur

Assistant Professor
(Computer Applications)
 Super key in the table above:

1. {EMP_ID}, {EMP_ID, EMP_NAME}, {EMP_ID, EMP_NAME, EMP_ZIP}....so on

Candidate key: {EMP_ID}

Non-prime attributes: In the given table, all attributes except EMP_ID are non-prime.

Here, EMP_STATE & EMP_CITY dependent on EMP_ZIP and EMP_ZIP dependent on EMP_ID. The non-prime
attributes (EMP_STATE, EMP_CITY) transitively dependent on super key(EMP_ID). It violates the rule of third
normal form.

That's why we need to move the EMP_CITY and EMP_STATE to the new <EMPLOYEE_ZIP> table, with EMP_ZIP
as a Primary key.

EMPLOYEE table:

EMP_ID EMP_NAME EMP_ZIP

222 Harry 201010

333 Jass 02228

444 Kiran 60007

555 Rahul 06389

666 Priya 462007

EMPLOYEE_ZIP table:

EMP_ZIP EMP_STATE EMP_CITY

201010 UP Noida

02228 PUNJAB Mohali

60007 HARYANA Ambala

06389 UK Norwich

462007 MP Bhopal

Boyce Codd normal form (BCNF)

o BCNF is the advance version of 3NF. It is stricter than 3NF.
o A table is in BCNF if every functional dependency X → Y, X is the super key of the table.
o For BCNF, the table should be in 3NF, and for every FD, LHS is super key.

Example: Let's assume there is a company where employees work in more than one department.

Ms. Jasmeen Kaur

Assistant Professor
(Computer Applications)
 EMPLOYEE table:

EMP_ID EMP_COUNTRY EMP_DEPT DEPT_TYPE EMP_DEPT_NO

264 India Designing D394 283

264 India Testing D394 300

364 UK Stores D283 232

364 UK Developing D283 549

In the above table Functional dependencies are as follows:

1. EMP_ID → EMP_COUNTRY
2. EMP_DEPT → {DEPT_TYPE, EMP_DEPT_NO}

Candidate key: {EMP-ID, EMP-DEPT}

The table is not in BCNF because neither EMP_DEPT nor EMP_ID alone are keys.

To convert the given table into BCNF, we decompose it into three tables:

EMP_COUNTRY table:

EMP_ID EMP_COUNTRY

264 India

264 India

EMP_DEPT table:

EMP_DEPT DEPT_TYPE EMP_DEPT_NO

Designing D394 283

Testing D394 300

Stores D283 232

Developing D283 549

Ms. Jasmeen Kaur

Assistant Professor
(Computer Applications)
 EMP_DEPT_MAPPING table:

EMP_ID EMP_DEPT

D394 283

D394 300

D283 232

D283 549

Functional dependencies:

1. EMP_ID → EMP_COUNTRY
2. EMP_DEPT → {DEPT_TYPE, EMP_DEPT_NO}

Candidate keys:

For the first table: EMP_ID

For the second table: EMP_DEPT

For the third table: {EMP_ID, EMP_DEPT}

Fourth normal form (4NF)

o A relation will be in 4NF if it is in Boyce Codd normal form and has no multi-valued dependency.
o For a dependency A → B, if for a single value of A, multiple values of B exists, then the relation will be a
multi-valued dependency.

Example

STUDENT

STU_ID COURSE HOBBY

21 Computer Dancing

21 Math Singing

34 Chemistry Dancing

Ms. Jasmeen Kaur

Assistant Professor
(Computer Applications)
74 Biology Cricket

59 Physics Hockey

The given STUDENT table is in 3NF, but the COURSE and HOBBY are two independent entity. Hence, there is no
relationship between COURSE and HOBBY.

In the STUDENT relation, a student with STU_ID, 21 contains two courses, Computer and Math and two
hobbies, Dancing and Singing. So there is a Multi-valued dependency on STU_ID, which leads to unnecessary
repetition of data.

So to make the above table into 4NF, we can decompose it into two tables:

STUDENT_COURSE

STU_ID COURSE

21 Computer

21 Math

34 Chemistry

74 Biology

59 Physics

STUDENT_HOBBY

STU_ID HOBBY

21 Dancing

21 Singing

34 Dancing

74 Cricket

59 Hockey

Ms. Jasmeen Kaur

Assistant Professor
(Computer Applications)
Fifth normal form (5NF)
o A relation is in 5NF if it is in 4NF and not contains any join dependency and joining should be lossless.
o 5NF is satisfied when all the tables are broken into as many tables as possible in order to avoid redundancy.
o 5NF is also known as Project-join normal form (PJ/NF).

Example
SUBJECT LECTURER SEMESTER

Computer Anshika Semester 1

Computer John Semester 1

Math John Semester 1

Math Akash Semester 2

Chemistry Praveen Semester 1

In the above table, John takes both Computer and Math class for Semester 1 but he doesn't take Math class for
Semester 2. In this case, combination of all these fields required to identify a valid data.

Suppose we add a new Semester as Semester 3 but do not know about the subject and who will be taking that
subject so we leave Lecturer and Subject as NULL. But all three columns together acts as a primary key, so we can't
leave other two columns blank.

So to make the above table into 5NF, we can decompose it into three relations P1, P2 & P3:

P1

SEMESTER SUBJECT

Semester 1 Computer

Semester 1 Math

Semester 1 Chemistry

Semester 2 Math

Ms. Jasmeen Kaur

Assistant Professor
(Computer Applications)
P2

SUBJECT LECTURER

Computer Anshika

Computer John

Math John

Math Akash

Chemistry Praveen

P3

SEMSTER LECTURER

Semester 1 Anshika

Semester 1 John

Semester 1 John

Semester 2 Akash

Semester 1 Praveen

Normal Description
Form

1NF A relation is in 1NF if it contains an atomic value.

2NF A relation will be in 2NF if it is in 1NF and all non-key attributes are fully
functional dependent on the primary key.

3NF A relation will be in 3NF if it is in 2NF and no transition dependency exists.

Ms. Jasmeen Kaur

Assistant Professor
(Computer Applications)
BCNF A stronger definition of 3NF is known as Boyce Codd's normal form.

4NF A relation will be in 4NF if it is in Boyce Codd's normal form and has no multi-
valued dependency.

5NF A relation is in 5NF. If it is in 4NF and does not contain any join dependency,
joining should be lossless.

Advantages of Normalization

o Normalization helps to minimize data redundancy.

o Greater overall database organization.
o Data consistency within the database.
o Much more flexible database design.
o Enforces the concept of relational integrity.

Disadvantages of Normalization

o You cannot start building the database before knowing what the user needs.
o The performance degrades when normalizing the relations to higher normal forms, i.e., 4NF, 5NF.
o It is very time-consuming and difficult to normalize relations of a higher degree.
o Careless decomposition may lead to a bad database design, leading to serious problems.

Ms. Jasmeen Kaur

Assistant Professor
(Computer Applications)