Normalization in rdbms types and examples

DATABASE
Normalization
UNIT III
RDBMS

Normalization
 Normalization is the process of organizing the data in the database.
 Normalization is used to minimize the redundancy from a relation or set of
relations. It also eliminates undesirable characteristics like Insertion, Update,
and Deletion Anomalies.
 Normalization divides the larger table into smaller ones and links them
using relationships.
 The normal form is used to reduce redundancy from the database table.

Why do we need Normalization?
 The main reason for normalizing the relations is removing these anomalies.
 Failure to eliminate anomalies leads to data redundancy and can cause data
integrity and other problems as the database grows.
 Normalization consists of a series of guidelines that helps to guide you in creating a
good database structure.

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

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.

Normal Form Description
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.
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
 Normalization helps to minimize data redundancy.
 Greater overall database organization.
 Data consistency within the database.
 Much more flexible database design.
 Enforces the concept of relational integrity.
Disadvantages of Normalization
 You cannot start building the database before knowing what the user needs.
 The performance degrades when normalizing the relations to higher normal forms, i.e., 4NF,
5NF.
 It is very time-consuming and difficult to normalize relations of a higher degree.
 Careless decomposition may lead to a bad database design, leading to serious problems.

First Normal Form (1NF)
EMP_ID EMP_NAME EMP_PHONE EMP_STATE
14 John
7272826385,
9064738238
UP
20 Harry 8574783832 Bihar
12 Sam
7390372389,
8589830302
Punjab
 A relation will be 1NF if it contains an atomic value.
 It states that an attribute of a table cannot hold multiple values. It must hold only single-valued
attribute.
 First normal form disallows the multi-valued attribute, composite attribute, and their
combinations.
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
14 John 9064738238 UP
20 Harry 8574783832 Bihar
12 Sam 7390372389 Punjab
12 Sam 8589830302 Punjab
The decomposition of the EMPLOYEE table into 1NF has been shown below:

TEACHER_ID SUBJECT TEACHER_AGE
25 Chemistry 30
25 Biology 30
47 English 35
83 Math 38
83 Computer 38
Second Normal Form (2NF)
 In the 2NF, relational must be in 1NF.
 In the second normal form, all non-key attributes are fully functional dependent on the primary key
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
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.

To convert the given table into 2NF, we decompose it into two
tables
TEACHER_ID TEACHER_AGE
25 30
47 35
83 38
TEACHER_ID SUBJECT
25 Chemistry
25 Biology
47 English
83 Math
83 Computer
TEACHER_DETAIL
table
TEACHER_SUBJECT table

Third Normal Form (3NF)
EMP_ID EMP_NAME EMP_ZIP EMP_STATE EMP_CITY
222 Harry 201010 UP Noida
333 Stephan 02228 US Boston
444 Lan 60007 US Chicago
555 Katharine 06389 UK Norwich
666 John 462007 MP Bhopal
 A relation will be in 3NF if it is in 2NF and not contain any transitive partial dependency.
 3NF is used to reduce the data duplication. It is also used to achieve the data integrity.
 If there is no transitive dependency for non-prime attributes, then the relation must be in third normal
form.
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
Super key in the table {EMP_ID}, {EMP_ID, EMP_NAME}, {EMP_ID, EMP_NAME, EMP_ZIP}....so on
Candidate key: {EMP_ID}

EMP_ID EMP_NAME EMP_ZIP
222 Harry 201010
333 Stephan 02228
444 Lan 60007
555 Katharine 06389
666 John 462007
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_ZIP EMP_STATE EMP_CITY
201010 UP Noida
02228 US Boston
60007 US Chicago
06389 UK Norwich
462007 MP Bhopal
EMPLOYEE_ZIP
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
 BCNF is the advance version of 3NF. It is stricter than 3NF.
 A table is in BCNF if every functional dependency X Y, X is the super key of the table.
→
 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.
EMPLOYEE table:
In the above table Functional dependencies are as follows:
1.EMP_ID EMP_COUNTRY
→
2.EMP_DEPT {DEPT_TYPE, EMP_DEPT_NO}
→
Boyce Codd normal form (BCNF)

Candidate key: {EMP-ID, EMP-DEPT}
EMP_ID EMP_COUNTRY
264 India
264 India
EMP_DEPT DEPT_TYPE EMP_DEPT_NO
Designing D394 283
Testing D394 300
Stores D283 232
Developing D283 549
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_DEPT table:

EMP_ID EMP_DEPT
D394 283
D394 300
D283 232
D283 549
EMP_DEPT_MAPPING table:
Now, this is in BCNF because left side part of both the functional dependencies is a key.
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}

 A relation will be in 4NF if it is in Boyce Codd normal form and has no multi-valued
dependency.
 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.
Fourth normal form (4NF)
STU_ID COURSE HOBBY
21 Computer Dancing
21 Math Singing
34 Chemistry Dancing
74 Biology Cricket
59 Physics Hockey
STUDENT
 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:
STU_ID COURSE
21 Computer
21 Math
34 Chemistry
74 Biology
59 Physics
STU_ID HOBBY
21 Dancing
21 Singing
34 Dancing
74 Cricket
59 Hockey
STUDENT_COURSE STUDENT_HOBB
Y

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

SEMESTER SUBJECT
Semester 1 Computer
Semester 1 Math
Semester 1 Chemistry
Semester 2 Math
So to make the above table into 5NF, we can decompose it into three relations P1, P2 & P3:
P1
SUBJECT LECTURER
Computer Anshika
Computer John
Math John
Math Akash
Chemistry Praveen
P2

SEMSTER LECTURER
Semester 1 Anshika
Semester 1 John
Semester 1 John
Semester 2 Akash
Semester 1 Praveen
P3

 Domain/key normal form (DKNF) is a normal form used in database
normalization which requires that the database contains no constraints other
than domain constraints and key constraints.
 A domain constraint specifies the permissible values for a given attribute, while
a key constraint specifies the attributes that uniquely identify a row in a given
table.
 The domain/key normal form is achieved when every constraint on the relation
is a logical consequence of the definition of keys and domains, and enforcing
key and domain restraints and conditions causes all constraints to be met. Thus,
it avoids all non-temporal anomalies.
DOMAIN KEY NORMAL FORM

 The reason to use domain/key normal form is to avoid having general
constraints in the database.
 Most databases can easily test domain and key constraints on attributes.
General constraints however would normally require special database
programming in the form of stored procedures that are expensive to maintain
and expensive for the database to execute. Therefore general constraints are
split into domain and key constraints.
 It's much easier to build a database in domain/key normal form .
 While the domain/key normal form eliminates the problems found in most
databases, it tends to be the most costly normal form to achieve.

 The third normal form, Boyce–Codd normal form, fourth normal form and
fifth normal form are special cases of the domain/key normal form. All have
either functional, multi-valued or join dependencies that can be converted
into (super)keys.
 The domains on those normal forms were unconstrained so all domain
constraints are satisfied. However, transforming a higher normal form into
domain/key normal form is not always a dependency-preserving
transformation and therefore not always possible.

A table is in domain key normal form if:
 The table has domain and key constraint
 It should not have any other general constraint

Normalization in rdbms types and examples

Denormalization in DBMS
 Denormalization is a database optimization technique where we add redundant data in the
database to get rid of the complex join operations.
 This is done to speed up database access speed.
 Denormalization is done after normalization for improving the performance of the database.
 The data from one table is included in another table to reduce the number of joins in the query
and hence helps in speeding up the performance.

Example: Suppose after normalization we have two tables first, Student
table and second, Branch table. The student has the attributes
as Roll_no , Student-name , Age , and Branch_id .

The branch table is related to the Student table with Branch_id as the foreign key in the Student
table.
 If we want the name of students along with the name of the branch name then we need to
perform a join operation.
 The problem here is that if the table is large we need a lot of time to perform the join
operations.
 So, we can add the data of Branch_name from Branch table to the Student table and this will
help in reducing the time that would have been used in join operation and thus optimize the

Advantages of Denormalization
1. Query execution is fast since we have to join fewer tables.
Disadvantages of Denormalization
2. As data redundancy is there, update and insert operations are more expensive and take
more time. Since we are not performing normalization, so this will result in redundant
data.
3. Data Integrity is not maintained in denormalization. As there is redundancy so data can
be inconsistent.

Normalization in rdbms types and examples

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