Noida Institute of Engineering and Technology,

Greater Noida

Database Design &

Normalization

Unit: 3

Database Management System

(New Code)
Manu Sharma
Course Details Assistant
(B Tech 5th Sem)
Professor
(CSE)
 Brief Introduction of Faculty
member

Name: Ms. Manu Sharma

Designation: Assistant Professor, CSE Department

Qualification: M.Tech From Shobhit University, Meerut

B.Tech From UPTU, Lucknow

Teaching Experience: 8 Years

Manu Sharma DBMS

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Unit-1
 Content
• Brief Introduction about me
• Evaluation Scheme
• Syllabus
• Introduction to the subject
• Branchwise Application
• Course Objective
• Course Outcomes
• Program Outcome
• Program Specific Outcomes
• CO-PO Mapping
• CO-PSO Mapping
• CO-PEO Mapping
• Previous Result Analysis
• Question Paper Template
• Unit Content with objective and Mapping with Cos
• Faculty Video Links, Youtube & NPTEL Video Links and Online Courses Details
• Daily Quiz
• Weekly Assignments
• Old university papers
• Sessional paper template
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 Evaluation Scheme

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 Subject Syllabus

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 Subject Syllabus

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 Course Objective

Present an introduction to database management systems,

with an emphasis on how to organize, maintain and retrieve -
efficiently, and effectively - information from a DBMS.

Differentiate between database system and file system.

Knowledge of the different models like database modeling,

relational, hierarchical and network models.

Explain basic issues of transaction processing and

concurrency control.

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Unit-3
 Brief Introduction about
Subject
A database management system (DBMS) refers to the technology for creating and managing
databases. DBMS is a software tool to organize (create, retrieve, update, and manage) data in
a database.

The main aim of a DBMS is to supply a way to store up and retrieve database information that
is both convenient and efficient. By data, we mean known facts that can be recorded and that
have embedded meaning. Usually, people use software such as DBASE IV or V, Microsoft
ACCESS, or EXCEL to store data in the form of a database. A datum is a unit of data.
Meaningful data combined to form information. Hence, information is interpreted data - data
provided with semantics. MS. ACCESS is one of the most common examples of database
management software.

Video link: https://www.youtube.com/watch?

v=kBdlM6hNDAE&list=PLxCzCOWd7aiFAN6I8CuViBuCdJgiOkT2Y

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 Branch wise Applications

There are various application of DBMS in different fields like Railway

Reservation System, Library Management System, Banking,
Universities and colleges, Credit card transactions etc

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 Course Outcome
S. NO. Course Outcomes Blooms’
Taxonom
y
K3
Apply knowledge of database for real life
.1
applications.
Apply query processing techniques to K3, K4
.2 automate the real time problems of
databases.
K2, K3
Identify and solve the redundancy problem
.3
in database tables using normalization

Understand the concepts of transactions, K2, K4

their processing so they will familiar with
.4 broad range of database management
issues including data integrity, security and
recovery.
09/24/22 Design, develop and DBMS
Manu Sharma implementUnit-1
a small K3, K610
.5
 Program Outcomes (POs)
Engineering Graduates will be able to:

1. Engineering knowledge: Apply the knowledge of

mathematics, science, engineering fundamentals, and an
engineering specialization to the solution of complex engineering
problems.
2. Problem analysis: Identify, formulate, review research
literature, and analyze complex engineering problems reaching
substantiated conclusions using first principles of mathematics,
natural sciences, and engineering sciences.
3. Design/development of solutions: Design solutions for
complex engineering problems and design system components or
processes that meet the specified needs with appropriate
consideration for the public health and safety, and the cultural,
societal, and environmental considerations.
4. Conduct investigations of complex problems: Use
research-based knowledge and research methods including design
of experiments, analysis
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and
Manu interpretation
Sharma DBMS of data, and synthesis11
Unit-3
of the information to provide valid conclusions.
 Program Outcomes (POs)

Contd..

5. Modern tool usage: Create, select, and apply

appropriate techniques, resources, and modern engineering
and IT tools including prediction and modeling to complex
engineering activities with an understanding of the
limitations.
6. The engineer and society: Apply reasoning informed
by the contextual knowledge to assess societal, health,
safety, legal and cultural issues and the consequent
responsibilities relevant to the professional engineering
practice.
7. Environment and sustainability: Understand the
impact of the professional engineering solutions in societal
and environmental contexts, and demonstrate the
knowledge of, and need for sustainable development.
8. Ethics: Apply ethical
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Manu Sharma
and commit to
DBMS
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12
 Program Outcomes (POs)

Contd..

9. Individual and team work: Function effectively as an

individual, and as a member or leader in diverse teams, and in
multidisciplinary settings.
10. Communication: Communicate effectively on complex
engineering activities with the engineering community and with
society at large, such as, being able to comprehend and write
effective reports and design documentation, make effective
presentations, and give and receive clear instructions.
11. Project management and finance: Demonstrate
knowledge and understanding of the engineering and
management principles and apply these to one’s own work, as a
member and leader in a team, to manage projects and in
multidisciplinary environments.
12. Life-long learning: Recognize the need for, and have the
preparation and ability
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to Sharma
Manu engage in independent and life-long
DBMS
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Unit-3
learning in the broadest context of technological change.
 Program Specific Outcomes
On successful completion of graduation degree the Computer Science
& Engineering graduates will be able to:

Work as a software developer, database administrator,

PSO1: tester or networking engineer for providing solutions to the
real world and industrial problems. ​
Apply core subjects of information technology related to
data structure and algorithm, software engineering, web
PSO2: technology, operating system, database and networking to
solve complex IT problems. ​
Practice multi-disciplinary and modern computing
PSO3: techniques by lifelong learning to establish innovative
career.
Work in a team or individual to manage projects with
PSO4: ethical concern to be a successful employee or employer in
IT industry.

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Unit-3
 Program Educational Objectives
(PEOs)

PEO1: Able to apply sound knowledge in the field of information technology to fulfill the
needs of IT industry.
PEO2: Able to design innovative and interdisciplinary systems through latest digital
technologies.
PEO3: Able to inculcate professional ethics, team work and leadership for serving the
society.
PEO4: Able to inculcate lifelong learning in the field of computing for successful career
in organizations and R&D sectors.

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 COs and POs Mapping

PO1 PO1
PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO11
0 2

.1
2 2 3 3 3 2 3 2 2 2 2 3
.2
3 3 3 2 2 2 2 2 2 2 2 3
.3
2 3 3 3 3 2 2 2 2 2 2 2
.4
2 3 2 2 2 2 2 2 2 3 2 2
.5
2 3 2 2 2 3 2 2 3 2 2 2

AVG
2.20 2.80 2.60 2.40 2.40 2.20 2.20 2.00 2.20 2.20 2.00 2.40

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 COs and PSOs Mapping

Program Specific Outcomes

PSO1 PSO2 PSO3 PSO4

.1
3 1 3 1

.2
3 1 3 1
.3
3 1 3 1
.4
3 1 3 1
.5
3 1 3 1

AVG
3.00 1.00 3.00 1.00

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 COs and PEOs Mapping

Program Specific Outcomes

PEO1 PEO2 PEO3 PEO4

.1
3 1 3 1

.2
3 1 3 1
.3
3 1 3 1
.4
3 1 3 1
.5
3 1 3 1

AVG
3.00 1.00 3.00 1.00

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 Result Analysis

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 Question Paper Template

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Unit-1
 Question Paper Template

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 Syllabus of Unit 3

Database Design & Normalization:

Functional dependencies, normal forms, first, second, third

normal forms, BCNF, inclusion dependence, loss less join
decompositions, normalization using FD, MVD, and JDs,
alternative approaches to database design.

Book References:
1. Korth, Silbertz, Sudarshan,” Database Concepts”, McGraw Hill
2. Date C J, “An Introduction to Database Systems”, Addision
Wesley
3. Elmasri, Navathe, “ Fundamentals of Database Systems”,
Addision Wesley
4.Bipin C. Desai, “ An Introduction to Database Systems”, Galgotia
Publications

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 Unit Objective

 Students will be able to learn Functional dependency

that is a relationship that exists when one attribute
uniquely determines another attribute.

 Normalization to minimizing redundancy from a relation

or set of relations. It deals with insertion, deletion, and
update anomalies. So, it helps to minimize the
redundancy in relations that are used to eliminate or
reduce redundancy in database tables to create a good
database.

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Unit-3
 Prerequisite and Recap
• A good background in DBMS fundamentals is required.
• Students should be comfortable with the relational
model, SQL, and the basic functions of database
systems.
• The proper understanding of data structures and
algorithms will help you to understand the DBMS
quickly.

• Recap
In last unit we studied about
• Relational Data Model
– Relational Algebra Query
– Relational Calculus Query
• Structured Query Language
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Unit-3
 Content
Unit-3

•Functional Dependencies
•Normal forms
– 1NF
– 2NF
– 3NF
– BCNF
•Loss less join decompositions
•Multivalued dependencies
•Join dependencies

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 Topic 1 objective

 Informal Design Guidelines for Relational

Databases
1.1Semantics of the Relation Attributes
1.2 Redundant Information in Tuples and Update
Anomalies
1.3 Null Values in Tuples
1.4 Spurious Tuples
 Summary of Informal database Design

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 1.1Normalization (CO3)

• Normalization is the process of structuring and handling the

relationship between data to minimize redundancy in the
relational table.
• It avoids the unnecessary anomalies from the database like
insertion, updation and deletion.
• It helps to divide large database tables into smaller tables and
make a relationship between them.
• It can remove the redundant data and ease to add, manipulate
or delete table fields.

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 1.1Normalization (CO3)

• A normalization defines rules for the relational table as to

whether it satisfies the normal form.
• A normal form is a process that evaluates each relation
against defined criteria and removes the multi-valued, joins,
functional and trivial dependency from a relation.
• If any data is updated, deleted or inserted, it does not cause
any problem for database tables and help to improve the
relational table' integrity and efficiency.

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 1.1Purpose of Normalization
(CO3)

• It is used to remove the duplicate data and database anomalies

from the relational table.
• Normalization helps to reduce redundancy and complexity by
examining new data types used in the table.
• It is helpful to divide the large database table into smaller
tables and link them using relationship.
• It avoids duplicate data or no repeating groups into a table.
• It reduces the chances for anomalies to occur in a database.

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 1.1Anomalies (CO3)

• Student Table:
StudReg. CourseID StudName Address Course
205 6204 James Los Angeles Economics
205 6247 James Los Angeles Economics
224 6247 TrentBolt New York Mathematics
230 6204 Ritchie Rich Egypt Computer
230 6208 Ritchie Rich Egypt Accounts

There are two students in the above table, 'James' and 'Ritchie Rich', whose
records are repetitive when we enter a new CourseID. Hence it repeats the
studRegistration, StudName and address attributes.

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 1.1Anomalies (CO3)

• Insert Anomaly: An insert anomaly occurs in the relational database when

some attributes or data items are to be inserted into the database without
existence of other attributes. For example, In the Student table, if we want
to insert a new courseID, we need to wait until the student enrolled in a
course. In this way, it is difficult to insert new record in the table. Hence, it
is called insertion anomalies.
• Update Anomalies: The anomaly occurs when duplicate data is updated
only in one place and not in all instances. Hence, it makes our data or table
inconsistent state. For example, suppose there is a student 'James' who
belongs to Student table. If we want to update the course in the Student, we
need to update the same in the course table; otherwise, the data can
be inconsistent. And it reflects the changes in a table with updated values
where some of them will not.

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 1.1Anomalies (CO3)
• Delete Anomalies: An anomaly occurs in a database table
when some records are lost or deleted from the database table
due to the deletion of other records. For example, if we want
to remove Trent Bolt from the Student table, it also removes
his address, course and other details from the Student table.
Therefore, we can say that deleting some attributes can
remove other attributes of the database table.
• So, we need to avoid these types of anomalies from the tables
and maintain the integrity, accuracy of the database table.
Therefore, we use the normalization concept in the database
management system.

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 A simplified COMPANY relational
database schema (CO3)

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 Examples of Violating Guideline 1
(CO3)

 Attributes of different entities should not be mixed in

the same relation
 Informally, each tuple in a relation should represent
one entity or relationship instance.
 Design a schema that can be explained easily relation
by relation. The semantics of attributes should be
easy to interpret.
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 1.2 Redundant Information in Tuples and
Update Anomalies (CO3)
1.2. A :- Redundant Information in Tuples

One goal of schema design is to minimize the

storage space used by the base relations (and
hence the corresponding files). Grouping
attributes into relation schemas has a significant
effect on storage space.

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 Redundant Information in Tuples
(CO3)
Mixing attributes of multiple entities may cause
problems.
1. Information is stored redundantly wasting
storage, compare the space used by the two base
relations EMPLOYEE and DEPARTMENT in with that
for an EMP_DEPT base relation . (Wastage of Space
by Emp_Dept)

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 1.2 A. Redundant Information in Tuples
(CO3)
Example :- Good database Design Approach

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 Problems with update
anomalies (CO3)
Update Anomaly:- An update anomaly is a data
inconsistency that results from data redundancy and
a partial update.
Consider the relation:

Update Anomaly: Changing the name of project number 1

from “ProductX” to “ProductABC” may cause this update
to be made for all n employees working on project 1.
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 Problems with update anomalies
(CO3)
Problems with update anomalies

– Insertion anomalies
– Deletion anomalies
– Modification anomalies

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 Insertion
anomalies (CO3)
1. To insert a new employee tuple into EMP_DEPT, we
must include either the attribute values for the
department that the employee works for, or NULLs (if
the employee does not work for a department as yet).
Example

2. But, It is difficult to insert a new department that has

no employees as yet in the EMP_DEPT relation. The
only way to do this is to place NULL values in the
attributes for employee. This violates the entity
integrity for EMP_DEPT because its primary key Ssn
cannot be null.
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 Deletion Anomalies (CO3)
• The problem of deletion anomalies is related to the second
insertion anomaly situation just discussed. If we delete from
table EMP_DEPT an employee tuple that happens to
represent the last employee working for a particular
department, the information concerning that department is
lost accidentally from the database.
Example :- Emp_Dept

• When a department is deleted, it will result in deleting all the

employees who work only on that project. Alternately, if an
employee is the sole employee on a department, deleting that
employee would result in deleting the corresponding project.
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 Modification anomalies
(CO3)
In EMP_DEPT, if we change the value of one of the
attributes of a particular department—say, the
manager of department 5—we must update the tuples
of all employees who work in that department;
otherwise, the database will become inconsistent. If we
fail to update some tuples, the same department will
be shown to have two different values for manager in
different employee tuples, which would be wrong.
Example :- Emp_Dept

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 Guideline to Redundant Information in Tuples and
Update Anomalies(CO3)

• It is easy to see that these three anomalies are

undesirable and cause difficulties to maintain
consistency of data as well as require
unnecessary updates that can be avoided; hence,
we can state the next guideline as follows:-

Guideline 2. Design the base relation schemas so

that no insertion, deletion, or modification
anomalies are present in the relations. If any
anomalies are present, note them clearly and
make sure that the programs that update the
database will operate correctly.

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 No Anomalies & No redundancy
Exists (CO3)

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 1.3 Null Values in Tuples(CO3)
In some schema designs we may group many attributes
together into a “fat” relation. If many of the attributes
do not apply to all tuples in the relation, we end up
with many NULLs in those tuples.
Example

 This can waste space at the storage level and may

also lead to problems with understanding the
meaning of the attributes and with specifying JOIN
operations at the logical level.
 Another problem with NULLs is how to account for
them when aggregate operations such as COUNT or
SUM are applied
Note :- If NULL values are present, the results may
become unpredictable.
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 Guideline Null Values in
Tuples(CO3)
GUIDELINE 3: As far as possible, avoid placing
attributes in a base relation whose values may
frequently be NULL. If NULLs are unavoidable,
make sure that they apply in exceptional cases
only and do not apply to a majority of tuples in the
relation.

 Relations should be designed such that their tuples

will have as few NULL values as possible (wastage
of space in storage level and difficulties in JOIN
operation with NULL values).

 Attributes that are NULL frequently could be

placed in separate relations (with the primary key).

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 Summary of Informal database
Design(CO3)
The problems we pointed out, which can be detected
without additional tools of analysis, are as follows:-

 Anomalies that cause redundant work to be done

during insertion into and modification of a relation,
and that may cause accidental loss of information
during a deletion from a relation.
 Waste of storage space due to NULLs and the
difficulty of performing selections, aggregation
operations, and joins due to NULL values.
 Generation of invalid and spurious data during joins
on base relations with matched attributes that may
not represent a proper (foreign key, primary key)
relationship
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 Summary of Informal database
Design(CO3)
we present informal concepts and theory that may
be used to define
the goodness and badness of individual relation
schemas more precisely.

Now we discuss functional dependency as a tool for

analysis.

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 Faculty Video Links, Youtube & NPTEL Video
Links and Online Courses Details

You tube /other Video Links

•https://nptel.ac.in/courses/106/104/106104135/
•https://nptel.ac.in/courses/106106220/
•https://www.youtube.com/watch?v=1057YmExS-I

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Unit-3
 Weekly Assignment 1.1

1. What do you mean by BCNF ? Why it is used and how it differ from
3 NF ? CO3

2. Discuss the various normal forms in normalization with suitable

examples?

3. Why is concurrency control needed? Explain lost update, Inconsistent

retrievals and uncommitted dependency anomalies. CO3

4. Explain the Codd’s Rule in detail. CO3

5. Explain Normalization with example. CO3

6. What are the rules of 1NF,2NF,3NF. CO3

7. Discuss Boyce Codd Normalization Form. CO3

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Unit-3
 Recap

1.1Semantics of the Relation Attributes

1.2 Redundant Information in Tuples and Update
Anomalies
1.3 Null Values in Tuples
1.4 Spurious Tuples

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Unit-3
 Topic 2 Objective

 Functional dependency
 Types of Functional dependency

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 2.1 Functional
Dependencies(CO3)
We now introduce a formal tool for analysis of relational
schemas that enables us to detect and describe some of
the above-mentioned problems in precise terms.
The single most important concept in relational schema
design theory is that of a functional dependency.

Important points for FD are

1. Functional dependencies is a constraint between 2 sets
of attributes
2. Functional dependencies (FDs) are used to specify
formal measures of the "goodness" of relational
schema designs
3. FDs and keys are used to define normal forms for
relations
4. FDs are constraints that are derived from the meaning
and interrelationships of the data attributes.
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 Definition of Functional
Dependency(CO3)
A functional dependency, denoted by X → Y, between
two sets of attributes X and Y that are subsets of R
specifies a constraint on the possible tuples that can
form a relation state r of R.
The constraint is that, for any two tuples t1 and t2 in r
that have t1[X] = t2[X], they must also have t1[Y] =
t2[Y].

This means that the values of the Y component of a

tuple in r depend on, or are determined by, the
values of the X component;

Alternatively, the values of the X Component of a tuple

uniquely (or functionally) determine the values of the Y
component. We also say that there is a functional
dependency from X to Y, or that Y is functionally
dependent on X.
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 FD on Key and Non key
Attribute(CO3)
Functional dependency exists between the primary key
and non-key attribute within a table.
FD on Non Key Attribute
Thus, X functionally determines Y in a relation schema R if,
and only if, whenever two tuples of r(R) agree on their X-
value, they must necessarily agree on their Y-value.
FD on Key Attribute
 If a constraint on R states that there cannot be more
than one tuple with a given X-value in any relation
instance r(R)—that is, X is a candidate key of R—this
implies that X → Y for any subset of attributes Y of R
(because the key constraint implies that no two tuples in
any legal state r(R) will have the same value of X). If X is
a candidate key of R, then X → R.
 If X → Y in R, this does not say whether or not Y → X in R.
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 Example

Rollno Name Marks

1 A 10
2 B 20
1 A 10
4 B 20
5 C 30
6 D 30

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 Example
Let us Consider the following table of data r(R) of the relation
schema R(ABCDE) shown in

Since the values of A are unique (a1, a2, a3, etc.), it

follows from the FD definition that:-
A → B, A → C, A → D, A → E,
Means A attribute uniquely identifies the B,C,D,E
attribute of R relation because if we know the A we
can tell the B,C,D,E associated with it.
This can be also write as A →BCDE.
From our understanding of primary keys, A is a primary
key.
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 Diagrammatic notation for
displaying FD(CO3)
• Consider the relation schema EMP_PROJ from the
semantics of the attributes and the relation, we know
that the following functional dependencies should
hold:-

a. Pnumber → {Pname, Plocation}

b. Ssn → Ename
c. {Ssn, Pnumber} → Hours
 Pnumber uniquely determines Pname and Plocation.
 Ssn of an employee uniquely determines the Ename.
 Combination of Ssn & Pnumber uniquely determines
the Hours
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 Types of Functional
dependency(CO3)
There are mainly two types of Functional
Dependency in DBMS.

Following are the types of Functional

Dependencies in DBMS:

 Trivial Functional Dependency

 Non-Trivial Functional Dependency

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 Trivial Functional
Dependency(CO3)
The Trivial dependency is a set of attributes which are called a
trivial if the set of attributes are included in that attribute.
So, X -> Y is a trivial functional dependency if Y is a subset of X.

Let’s understand with a Trivial Functional Dependency

Example.

Emp_id Emp_name
For example:
AS555 Harry
AS811 George
AS999 Kevin
{Emp_id, Emp_name} -> Emp_id is a trivial functional
dependency as Emp_id is a subset of {Emp_id,Emp_name}.

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 Non-Trivial Functional
Dependency(CO3)
Non Trivial Functional Dependency in DBMS
Functional dependency which also known as a nontrivial
dependency occurs when A->B holds true where B is not
a subset of A. In a relationship, if attribute B is not a
subset of attribute A, then it is considered as a non-trivial
dependency.
Company CEO Age
Example:
Microsoft Satya Nadella 51
Google Sundar Pichai 46
Apple Tim Cook 57

(Company} -> {CEO} (if we know the Company, we knows

the CEO name)
But CEO is not a subset of Company, and hence it’s non-
trivial functional dependency.

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 Recap

 Functional dependency
 Types of Functional dependency

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Unit-3
 Topic 3 Objective

Inference rules

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 Inference Rules for Functional
Dependencies(CO3)
An inference rule is an assertion that we
can apply to a set of functional
dependencies to derive other functional
dependencies.

 We denote by F the set of functional dependencies

that are specified on relation schema R.
 Inference rules are sound meaning that they are an
immediate consequence of the definition of FD and
that only FD that can be derived from a relation from
a given set of FD using them to true.
 So, inference rules can be used to find all the derived
FD’s logically by a set of FD is true and complete.
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 Definition(CO3)
Definition:
An FD X → Y is inferred from or implied by a set of
dependencies F specified on R if X → Y holds in every
legal relation state r of R; that is, whenever r satisfies all
the dependencies in F, X → Y also holds in r.
Inference rules basically based on Armstrong axiom.

Armstrong Axiom basic inference rules are :-

1. IR1 (reflexive rule) : If X ⊇ Y, then X →Y.
2. IR2 (augmentation rule) : {X → Y} |= XZ → YZ.
3. IR3 (transitive rule): {X → Y, Y → Z} |= X → Z.

IR1, IR2, IR3 form a sound and complete set of inference

rules

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 Armstrong Axiom Rules(CO3)

There are three other inference rules that follow from IR1,
IR2 and IR3.
They are as follows:
4. IR4 (decomposition, or projective, rule): {X → YZ} |=X →
Y,
X →Z
5. IR5 (union, or additive, rule): {X → Y, X → Z} |=X → YZ.
6 . IR6 (pseudotransitive rule): {X → Y, WY → Z} |=WX → Z.

Note:- Rules from IR1 to IR3 are known as Armstrong

Axiom rules. These rules are mainly used in FD to find
complete set of all possible dependencies.

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 Exercise 3.1

Prove or disprove the following

inference rules:-

1. {WY,XZ} |= {WXY}

2. {XZ,Y Z} |= {XY}

3. {XY,Y W} |= {XZYW}

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 Exercise 3.2

Prove or disprove the following

inference rules

1. {WY,XZ} |= {WXY}
2. {XY,XW,WYZ} |= {XZ}
3. {XY} |= {XYZ}
4. {XY, Z Y} |= {XZY}

09/24/22 Manu Sharma DBMS Unit-3 68

 Question(CO3)
Let us consider the Universal relation schema R =
( A, B, C, G, H, I ) and the set of functional
dependencies F ={ A → B, A → C, CG → H, CG → I , B
→ H}.
Determines the other FD’s from given set of FD F.
Answer:-
Given F= { A → B, A → C, CG → H, CG → I , B → H}.
1. Since A → B and B → H hold, we apply the transitivity rule that A → H
2. Since CG → H and CG → I , the union rule implies that CG → HI.
3. Since A → C and CG → I , the pseudo-transitivity rule implies that AG
→ I holds
4. Since A → B and A→ C hold, we apply the union rules that A → BC
All possible FD are:-
F’= { A → B, A → C, CG → H, CG → I , B → H, A → H, CG → HI ,A
→BC, AG → I }.

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 Conti………… (CO3)

Note:-
Armstrong axioms are complete. As for a given set
F of functional dependencies, all FD implied by F
can be inferred by using rules IR1 through IR3.

Armstrong axiomS are also sound as to no

additional FD or incorrect FD’s can be deduced
from F by using inference rules.

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 Recap

 Inference rules

Manu Sharma DBMS

09/24/22 71
Unit-3
 Topic 4,5 Objective

 Closure of Attribute set.

 Definitions of Keys
 Find the super key and candidate key from
FD

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 Closure of a Attribute set(CO3)

 After finding a set of FD’s that are hold on a

relation the next step is to find super key and
candidate key for a relation and,

 Also to test whether a set X is a super key, we

must devise an algorithm for computing the set of
attributes functionally determined by X.

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 Algorithm for Closure of a
Attribute set(CO3)
The set of attribute that are functionally dependent on the
attribute X is called attribute closure of X and it can be
represented by X+.
Or
For each such set of attributes X, we determine the set X+ of
attributes that are functionally determined by X based on F;
X+ is called the closure of X under F.
Algorithm :- Determining X+, the Closure of X under F .
Input: A set F of FDs on a relation schema R, and a set of attributes X,
which is a subset of R.
X+ := X;
repeat
old X+ := X+;
for each functional dependency Y → Z in F do
if X+ ⊇ Y then X+ := X+ ∪ Z;
until (X+ = old X+);

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 Example

Question 1:-
To compute the closure for relation schema R
={A,B,C,G,H,I} and F= {A → B,A → C,CG → H, CG →
I,B → H,C → G).
Find the closure of A under F . Or {A+ =}

Question 2:-
To compute the closure for relation schema R
={A,B,C,D,E} and F= {A → BC,CD → E,B → D, E → A}.
Find the closure of A and CD under F . or {A+ =} and
{ CD + =}

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 Definitions of Keys and Attributes Participating in Keys

 A super key of a relation schema R = {A1, A2, … , An} is a

set of attributes S ⊆ R with the property that no two tuples
t1 and t2 in any legal relation state r of R will have t1[S] =
t2[S].
 If a relation schema has more than one key, each is called a
candidate key.
 One of the candidate keys is designated to be the primary
key, and the others are called secondary keys. In a practical
relational database, each relation schema must have a
primary key.
 If no candidate key is known for a relation, the entire
relation can be treated as a default super key.
 An attribute of relation schema R is called a prime attribute
of R if it is a member of some candidate key of R.
 An attribute is called nonprime if it is not a prime attribute—
that is, if it is not a member of any candidate key.

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 Find the super key and candidate key
from FD(CO3)
Definition :- The set of attribute whose closure of
attribute is set of all attribute of a relation called
super key of relation and if this is set of minimal of it
is called of candidate key.

Formally say ,
• Let R is the relation and X is the set of attribute over
R.
• If X+ determines all the attributes of R, then X is said
to be super key, or candidate key of R.
• To find the candidate key first find all the super key of
a relation. (because the candidate key is a minimal
set of super key).

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 Step to find the super key and
candidate Key(CO3)
So To identify super key and candidate key, We need to follow
some steps,
1. Compute the closure for the attribute that combination of
attribute on The LHS of FD,
2. If any closure includes all the attribute of given relation, then
that can be denoted as a key for the relation(This would be
the set of super key and candidate key).
3. When we indentify the super key using FD’s and it has
extraneous attribute in the super key then,
4. Now applying the given FD and reduced the set of LHS
elements
5. Repeat Step-4,Apply till the of element of the super key when
it is not functionally determines all FD, If fail Stop, otherwise
6. We determines all FD from a exclusive attribute of super key
is true is called candidate key

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 Example

Question 1-
To find the Keys of relation R={A,B,C,D,E} with
FD’s F={A →BC,CD →E,E → A,B → D}.

Question 2-
To find the Keys of relation R={A,B,C,D,E,H} with
FD’s F={A →BC,CD →E,E → C,C → AEH,AH →D,DH →
BC}.

09/24/22 Manu Sharma DBMS Unit-3 79

 Recap

 Closure of Attribute set.

 Definitions of Keys
 Find the super key and candidate key
from FD

Manu Sharma DBMS

09/24/22 80
Unit-3
 Topic 6 Objective

Equivalence set of FD
Minimal cover, Canonical cover of FD

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 Equivalence Sets of FDs(CO3)

Definition.
A set of functional dependencies F is said to cover
another set of functional dependencies E if every
FD in E is also in F+; that is, if every dependency
in E can be inferred from F; alternatively, we can
say that E is covered by F.
Means,
Two sets of FDs F and E are equivalent if:
- every FD in F can be inferred from E, and
- every FD in E can be inferred from F
• Hence, F and E are equivalent if F + = E +
• F and E are equivalent if F covers E and E covers F

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 Step to Equivalence Sets of
FDs(CO3)
I. We can determine whether F covers E by calculating X+
with respect to F for each FD X → Y in E,
II. And then checking whether this X+ includes the
attributes in Y. If this is the case for every FD in E, then
F covers E. We determine whether E and F are
equivalent by checking that E covers F and F covers E.
Note
If F and G are the two sets of functional dependencies,
then following 3 cases are possible-
1. If all FDs of E can be derived from FDs present in F, we
can say that F ⊃ E.
2. If all FDs of F can be derived from FDs present in E, we
can say that E ⊃ F.
3. If 1 and 2 both are true, F=E.

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 Example(CO3)

Q 1: Given a relational schema R( X, Y, Z, W, V ) set

of functional dependencies P and Q such that:
P = { X → Y, XY → Z, W → XZ, W → V} and Q = { X
→ YZ, W → XV }
using FD sets P and Q which of the following
options are
correct?
a) P is a subset of Q
b) Q is a subset of P
c) P = Q
d) P ≠ Q

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 Example(CO3)
• Using definition of equivalence of FD set, let us determine
the right-hand side of the FD set of P using FD set Q.
• Given P = { X → Y, XY → Z, W → XZ, W → V} and Q = { X →
YZ, W → XV }
• Let's find closure of the left side of each FD of P using FD Q.
• X+ = XYZ (using X → YZ)
• XY+ = XYZ (using X → YZ)
• W+ = WXVYZ (using W → XV and X → YZ)
• W+ = WXVYZ (using W → XV and X → YZ)
Now compare closure of each X, XY, W and W calculated using
FD
Q with the right-hand side of FD P. Closure of each X, XY, W
and W
has all the attributes which are on the right-hand side of each
FD of P. Hence, we can say P is a subset of Q----------1

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 Example(CO3)
• Using definition of equivalence of FD set, let us determine
the right-hand side of the FD set of Q using FD set P.
• Given P = { X → Y, XY → Z, W → XZ, W → V} and Q = { X
→ YZ, W → XV }
• Let us find closure of the left side of each FD of Q using
FD P.
• X+ = XYZ (using X → Y and XY → Z)
• W+ = WXZVY (using W → XZ, W → V, and X → Y)
• Now compare closure of each X, W calculated using FD P
with the right-hand side of FD Q. Closure of each X and W
has all the attributes which are on the right-hand side of
each FD of Q. Hence, we can say Q is a subset of
P-----------2
• From 1 and 2 we can say that P = Q, hence option
c) is correct.

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 Example(CO3)

Q 2: Given a relational schema R( A, B, C, D ) set of

functional dependencies P and Q such that:
P = { A → B, B → C, C → D } and Q = { A → BC, C →
D } using FD
sets P and Q which of the following options are
correct?

a) P is a subset of Q
b) Q is a subset of P
c) P = Q
d) P ≠ Q

09/24/22 Manu Sharma DBMS Unit-3 87

 Example(CO3)

Question 1:-
Let us consider a relation schema R ={A,B,C,D,E} having
two functional dependency(FD) set E and F,
E= {A → B, AB → C, D → AC, D → E} and
F = {A → BC, D → AE}
Check wheteher two set are equivalent or not.
Question 2:-
Let us consider a relation schema R ={A,B,C,D,E,H}
having two functional dependency(FD) set F and G,
F = {A → C, AC → D, E → AD, E → H}
G = {A → CD, E → AH}
Check wheteher two set are equivalent or not.
09/24/22 Manu Sharma DBMS Unit-3 88
 Minimal cover, Canonical
cover of FD

09/24/22 Manu Sharma DBMS Unit-3 89

 Need of minimal cover of Functional
Dependencies(CO3)
 Whenever a user updates database the system must check
whether any FD are getting violated in the process, if there
is a violation of dependencies in the new database state
the system must rollback. If working with a huge set of FD
can cause unnecessary added computational time .

 Remove the extraneous attribute in FD set and also

remove the redundant FD in F and get standard from of FD.

So we can reduce the effort spent in checking for violation by

testing a simplified set of FDs, canonical cover comes to
play big role for remove redundant FD and find the
standard form of FD.

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 Continue…… (CO3)

A minimal cover of a set of functional dependencies

E is a set of functional dependencies F that
satisfies the property that every dependency in E
is in the closure F + of F.

In addition, this property is lost if any dependency

from the set F is removed; F must have no
redundancies in it, and the dependencies in F are
in a standard form.

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 Definition of minimal cover(CO3)
Definition:- An attribute in a functional dependency is considered
an extraneous attribute if we can remove it without changing
the closure of the set of dependencies.
Formally, given F, the set of functional dependencies, and a
functional dependency X → A in F, attribute Y is extraneous in X if
Y ⊂ X, and F logically implies (F − (X → A) ∪ { (X − Y) → A } ).

We can formally define a set of functional dependencies

F to be minimal if it satisfies the following conditions: -
1. Every dependency in F has a single attribute for its right-
hand side.
2. We cannot replace any dependency X → A in F with a
dependency Y → A, where Y is a proper subset of X, and still
have a set of dependencies that is equivalent to F.
3. We cannot remove any dependency from F and still have a
set of dependencies that is equivalent to F.
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 Algorithm for Minimal
cover(CO3)
Algorithm:- Finding a Minimal Cover F for a Set of Functional
Dependencies E
Input: A set of functional dependencies E.
1. Set F := E.
2. Replace each functional dependency X → {A1, A2, … , An} in
F by the n functional dependencies X →A1, X →A2, … , X → An.
3. For each functional dependency X → A in F for each attribute
B that is an element of X if { {F − {X → A} } ∪ { (X − {B} ) →
A} } is equivalent to F then replace X → A with (X − {B} ) → A
in F.
(* The above constitutes a removal of the extraneous attribute B from X *)
4. For each remaining functional dependency X → A in F if {F −
{X → A} } is equivalent to F, then remove X → A from F.
(*This constitutes removal of a redundant functional dependency X → A from
F when possible*)

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 Example(CO3)

Question 1:-
Given a relation schema R = {A,B,C,D,E,F}and a set
of functional dependencies F= {AB → C, C → AB,
B → C, ABC → AC, A→ C, AC → B }
To find the minimal cover for above given FD’s.

Sol.
Step 1: { AB → C, C → A, C → B, B → C, ABC→ A, ABC
→ C, A → C, AC → B }
Step 2: { B → C, C → A, C → B, B → C, A → C, A → B}
Step 3: {C → A, B → C, A → B }

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 Example(CO3)

Question 1:-
Given a relation schema R = {A,B,C,D,E,F}and a set
of functional dependencies F= {AB → C, C → A, BC
→ D, ACD → B, BE → C, EC → FA,CF → BD, D → E }
To find the minimal cover for above given FD’s.

Question 2:-
Given a relation schema R = {A,B,C,D,E}and a set of
functional dependencies F= {A → BC, CD → E,B →
D, E → A }
To find the minimal cover for above given FD’s.

09/24/22 Manu Sharma DBMS Unit-3 95

 Recap

 Equivalence set of FD
 Minimal cover, Canonical cover of FD

Manu Sharma DBMS

09/24/22 96
Unit-3
 Topic 7 Objective
 Definition of Normalization and Normal form
 Need of normalization
 Practical Use of Normal Forms
 Types of Normalization
1. First Normal Form(INF)
2. Second Normal Form (2NF)
3. Third Normal Form (3NF)
4. BCNF (Boyce Codd Normal Form)
5. Fourth normal Form (4NF)
6. Fifth Normal form (5NF)

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 Definition of Normalization and
Normal form(CO3)
Definition :- Normalization is a process of organization
the data in database with ensuring well-formed.

Or

The process of decomposing unsatisfactory "bad"

relations by breaking up their attributes into
smaller relations.

Normal form: Condition using keys and FDs of

a relation to certify whether a relation schema is in a
particular normal form .

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 Need of Normalization(CO3)

It is the process of analyzing the given relation

schemas based on the FD’s and primary key to achieve
the desirable properties are :-

I.Minimizing redundancy
II.Eliminates the anomalies(for insuring the integrity
and consistency of the data).
III.Ensuring data dependencies make sense i.e. data
is logically stored (all prime attribute in a relation )
IV.Dependent on the primary key (Normalization
generally involving splitting existing table into
multiple ones).

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 Practical Use of Normal
Forms(CO3)
1. Normalization is carried out in practice so that
the resulting designs are of high quality and meet
the desirable properties.

2. The practical utility of these normal forms becomes

questionable when the constraints on which they
are based are hard to understand or to detect.

3. The database designers need not normalize to the

highest possible normal form. (usually up to 3NF,
BCNF or 4NF).

De normalization: the process of storing the join

of higher normal form relations as a base relation—
which is in a lower normal form

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 Content

Normal form
Types of Normalization
1. First Normal Form(1NF)
2. Second Normal Form (2NF)
3. Third Normal Form (3NF)
4. BCNF (Boyce Codd Normal Form)
5. Fourth normal Form (4NF)
6. Fifth Normal form (5NF)

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 Procedure of
Normalization(CO3)
In normalization we use the degree of the normal forms in a
relation.
The normal form of a relation refer highest normal form condition
that it meets, and hence indicate the degree to which it has
been normalized.
 The top-down process evaluates each relation against the
criteria for normal forms and decomposing relations as
necessary - relational design by analysis.
 Codd proposed 3 normal forms - first, second, and third. A
stronger definition of 3NF—called Boyce-Codd normal form
(BCNF) - proposed by Boyce and Codd.
 All these normal forms are based on a single analytical tool:
the functional dependencies among the attributes of a relation.
Later, a fourth normal form (4NF) and a fifth normal form (5NF)
were proposed, based on the concepts of multivalued
dependencies and join dependencies

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 Procedure of
Normalization(CO3)
Design the good database relation also consider
two other properties.
1. Lossless join or additive join property – which is
guarantee that no spurious tuples are generated after
the decomposition to the relation reconstructed.
2. Dependency preservation property :- it is ensure
that each FD’s represented in some individual relation
resulting after decomposition.

Note :- In practice INF,2NF and 3NF are enough for

database normalization.

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 1. First Normal Form (1NF)
(CO3)
• 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.
• EMPLOYEE table:
• EMP_ID EMP_NAME EMP_PHONE EMP_STATE
• 14 John 7272826385, UP
9064738238
• 20 Harry 8574783832 Bihar
• 12 Sam 7390372389, Punjab
8589830302

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 1. First Normal Form (1NF)
(CO3)
• The decomposition of the EMPLOYEE table into
1NF has been shown below:
• 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
09/24/22 Manu Sharma DBMS Unit-3 105
 Normalization nested relations
into 1NF(CO3)
1. Let us Consider following relation Schema EMP_PROJ.

2. Sample State of relation schema EMP_PROJ.

3. Solution 1NF version relation EMP_PROJ

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 2. Second Normal Form (2NF)
(CO3)
• In the 2NF, relation must be in 1NF.
• There should not be any partial dependency present.
• In the second normal form, all non-key attributes are fully
functional dependent on the primary key.
• TEACHER table
• TEACHER_ID SUBJECT TEACHER_AGE
• 25 Chemistry 30
• 25 Biology 30
• 47 English 35
• 83 Math 38
• 83 Computer 38
• 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
09/24/22 it Sharma
Manu violates the
DBMSrule forUnit-3
2NF. 107
 2. Second Normal Form (2NF)
(CO3)
• To convert the given table into 2NF, we decompose it into two tables:
• TEACHER_DETAIL table:
• TEACHER_ID TEACHER_AGE
• 25 30
• 47 35
• 83 38
• TEACHER_SUBJECT table:
• TEACHER_ID SUBJECT
• 25 Chemistry
• 25 Biology
• 47 English
• 83 Math
• 83 Computer

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 2. Second Normal Form (2NF)
(CO3)
• 1NF meets the relationship, and no partial functional dependencies are
present.
• Partial Dependency occurs when a non-prime attribute is functionally
dependent on part of a candidate key.
• The 2nd Normal Form (2NF) eliminates the Partial Dependency.
• Example
• <StudentProject>
• StudentID ProjectNo StudentName ProjectName
S01 199 Katie Geo Location
S02 120 Ollie Cluster Exploration
In the above table, we have partial dependency; let us see how −
• The prime key attributes are StudentID and ProjectNo, and
• StudentID = Unique ID of the student
StudentName = Name of the student
ProjectNo = Unique ID of the project
ProjectName = Name of the project

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 2. Second Normal Form (2NF)
(CO3)
• As stated, the non-prime attributes i.e. StudentName and ProjectName should be
functionally dependent on part of a candidate key, to be Partial Dependent.
• The StudentName can be determined by StudentID, which makes the relation Partial
Dependent.
• The ProjectName can be determined by ProjectNo, which makes the relation Partial
Dependent.
• Therefore, the <StudentProject> relation violates the 2NF in Normalization and is
considered a bad database design.
• To remove Partial Dependency and violation on 2NF, decompose the tables −
• <StudentInfo>
• StudentID ProjectNo StudentName
S01 199 Katie
S02 120 Ollie
<ProjectInfo>
• ProjectNo ProjectName
199 Geo Location
120 Cluster Exploration
Now the relation is in 2nd Normal form of Database Normalization.

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 2. Second Normal Form (2NF)
(CO3)
Definition :-
A FD X -> Y is full functional dependency, if removal
of any attribute A from X, the dependency does not
hold any more.
That is for any attribute A ∈ X,(X-{A}) does not
functionally determines Y, else it is partial FD. :- (X-
{A}) -> Y(NOT hold )
Examples:-
{SSN, PNUMBER} -> HOURS is a full FD since neither
SSN -> HOURS nor PNUMBER -> HOURS hold .

Partial Function Dependency :- A FD X -> Y, is a

partial functional dependency if some attribute A ∈
X, can be removed from X and dependency still hold.
A ∈ X, (X-{A}) -> Y(hold )
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 2. Second Normal Form (2NF)(CO3)
A relation schema R is in second normal form (2NF)
if every non-prime attribute A in R is fully functionally
dependent on the primary key or candidate key of R.
(means all non key attribute cannot be dependent on
a subset of the primary key or candidate key ).
Otherwise ,
R can be decomposed into 2NF relations via the
process of 2NF normalization .
Note :- The test for 2NF involves testing for functional
dependencies whose left-hand side attributes are part
of the primary key. If the primary key contains a
single attribute, the test need not be applied at all.
If a primary key is not a composite primary key, such
case all non key attribute are always fully dependent
on the primary key .
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 Example 2NF (CO3)
Let us Consider the following relation schema
EMP_PROJ
{SSN,Pnumber,hours,ename,pname,plocation}
with FD Set F=
FD1= {SSN,Pnumber} -> hours
FD2= {SSN} -> ename
FD3={Pnumber} -> {Pname,plocation}

Check for 2NF,if it is not in 2NF then converted into

2NF.(Relation have composite primary key {SSN,
Pnumber}).

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 Example 2NF(CO3)

FD1= {SSN,Pnumber} -> hours

FD2= {SSN} -> ename
FD3={Pnumber} -> {Pname,plocation}

FD1= {SSN,Pnumber} -> hours (Fully Functional dependency)

FD2= {SSN} -> ename (Partial Functional dependency)
Therefore ename is partial dependent on SSN, which is only the
part of the primary key. Therefore FD2 is not satisfied second
normal form.
FD3={Pnumber} -> {Pname, plocation} (Partial Functional
dependency)
{Pname,Plocation } is also partial dependent on Pnumber, which
is only the part of the primary key. Therefore FD3 is not satisfied
second normal form

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 Example(CO3)
Solution :-
To bring relation EMP_PROJ into second normal form, We
decompose the relation

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 3. Third Normal Form (3NF)
(CO3)
• Third Normal Form (3NF):
A relation is in third normal form, if there is no
transitive dependency for non-prime attributes as
well as it is in second normal form.
• A relation is in 3NF if at least one of the following
condition holds in every non-trivial function
dependency X –> Y:
• X is a super key.
• Y is a prime attribute (each element of Y is part of
some candidate key).
• In relation STUDENT given in Table 4,

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 3. Third Normal Form (3NF)
(CO3)

FD set:
{STUD_NO -> STUD_NAME, STUD_NO -> STUD_STATE, STUD_STATE -
> STUD_COUNTRY, STUD_NO -> STUD_AGE}
Candidate Key:
{STUD_NO}
For this relation in table 4, STUD_NO -> STUD_STATE and STUD_STATE -
> STUD_COUNTRY are true. So STUD_COUNTRY is transitively
dependent on STUD_NO.
It violates the third normal form. To convert it in third normal form, we will
decompose the relation STUDENT (STUD_NO, STUD_NAME,
STUD_STATE, STUD_COUNTRY_STUD_AGE) as:
STUDENT (STUD_NO, STUD_NAME, STUD_STATE, STUD_AGE)
STATE_COUNTRY (STUD_STATE, STUD_COUNTRY)

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 General Definitions of 1NF,2NF
and 3NF(CO3)

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 4. Boyce Codd Normal Form (BCNF)
OR 3.5NF (CO3)
BCNF (Boyce Codd Normal Form) is the advanced version of 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.
• Consider a relation R with attributes (student, subject, teacher).
• Student Teacher Subject
Jhansi P.Naresh Database
Jhansi K.Das C
Subbu P.Naresh Database
Subbu R.Prasad C

F: { (student, Teacher) -> subject (student, subject) -> Teacher

Teacher -> subject}
• Candidate keys are (student, teacher) and (student, subject).
• The above relation is in 3NF [since there is no transitive
dependency]. A relation R is in BCNF if for every non-trivial FD X-
>Y, X must be a key.
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 4. Boyce Codd Normal Form (BCNF)
OR 3.5NF (CO3)
• The above relation is not in BCNF, because in the FD (teacher-
>subject), teacher is not a key. This relation suffers with anomalies
−
• For example, if we try to delete the student Subbu, we will lose the
information that R. Prasad teaches C. These difficulties are caused
by the fact the teacher is determinant but not a candidate key.
• Decomposition for BCNF
• Teacher-> subject violates BCNF [since teacher is not a candidate
key].
• If X->Y violates BCNF then divide R into R1(X, Y) and R2(R-Y).
• So R is divided into two relations R1(Teacher, subject) and
R2(student, Teacher).
• R1 R2
• Teacher Subject Student Teacher
• P.Naresh database Jhansi P.Naresh
• K.DAS C Jhansi K.DAS
• R.Prasad C Subbu P.Naresh
• Subbu R.Prasad
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 Lossless Join Decomposition (CO3)
• Lossless-join decomposition is a process in which a relation is decomposed
into two or more relations. This property guarantees that the extra or less
tuple generation problem does not occur and no information is lost from
the original relation during the decomposition. It is also known as non-
additive join decomposition.
• When the sub relations combine again then the new relation must be the
same as the original relation was before decomposition.
• Consider a relation R if we decomposed it into sub-parts relation R1 and
relation R2.
• The decomposition is lossless when it satisfies the following statement −
• If we union the sub Relation R1 and R2 then it must contain all the
attributes that are available in the original relation R before decomposition.
• Intersections of R1 and R2 cannot be Null. The sub relation must contain a
common attribute. The common attribute must contain unique data.

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 Lossless Join Decomposition (CO3)
• The common attribute must be a super key of sub relations
either R1 or R2.
• Here,
• R = (A, B, C)
• R1 = (A, B)
• R2 = (B, C)
• The relation R has three attributes A, B, and C. The relation R is
decomposed into two relation R1 and R2. . R1 and R2 both have
2-2 attributes. The common attribute is B.
• The Value in Column B must be unique. if it contains a duplicate
value then the Lossless-join decomposition is not possible.
• R (A, B, C)
• A B C
• 12 25 34
• 10 36 09
• 12 42 30

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 Lossless Join Decomposition (CO3)
• It decomposes into the two sub relations −
• R1 (A, B)
• A B
• 12 25
• 10 36
• 12 42
• R2 (B, C)
• B C
• 25 34
• 36 09
• 42 30
• Now, we can check the first condition for Lossless-join decomposition.
• The union of sub relation R1 and R2 is the same as relation R.
• R1 U R 2 = R
• A B C
• 12 25 34
• 10 36 09
• 12 42 30
09/24/22 Manu Sharma DBMS Unit-3 123
 Continue….

The formal definition of BCNF differs from the

definition of 3NF in that clause (b) of 3NF, which
allows functional dependency having the RHS as
a prime attribute, is absent from BCNF. That
makes BCNF a stronger normal form compared to
3NF.
(LHS of every functional dependency must be
super key or candidate key and RHS prime
attribute condition not allowed in BCNF)

Note:- Every relation in BCNF is in 3NF,but it is not

necessary that every relation in 3NF is in BCNF.

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 Example

Let us consider the following relation TEACH=

{student,Course,Instructor}

with the following dependencies: -

FD1: {Student, Course} → Instructor
FD2: Instructor → Course
{Student, Course}. Check this relation in
BCNF or not. is a candidate key for this
relation
09/24/22 Manu Sharma DBMS Unit-3 125
 Example

Instructor is not super keys but determines

course. So student relation in 3NF.
But student relation is not in BCNF because in
FD Instructor → Course, Instructor is not
super key or candidate key or primary key.
Solution :-
Therefore the relation Teach decompose in
1. Teach1 {Student, Instructor}
2. Teach 2 {Instructor, Course}
Teach1 Teach 2

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 Continue…

All three decompositions will lose functional

dependency . We have to settle for sacrificing the
functional dependency preservation. But we cannot
sacrifice the non-additivity property after
decomposition.
Out of the above three, only the 3rd decomposition
will not generate spurious tuples after join.(and
hence has the non-additivity property).

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 5. Multi-valued
Dependency(CO3)
Multi-valued dependency (MVD)
- Consequence of first normal form (1NF) which
diasallow an attribute in a tuple to have a set of
values.

- A multivalued dependency always requires at least

three attributes because it consists of at least two
attributes that are dependent on a third attribute .

- Let us consider a relation R =(A,B,C) and value of A

there is a set of value of B & a set of value of C.
However the set of value for B & C are independent of
each other.
We write A →→ B and
A →→ C.
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 Example
Example :- Employee whose name is ename work on
project Pname and has dependent dname. An employee
may work on several projects and have several
dependents & both project and dependents are
independent from each other.

problem
Multivalued dependencies (MVDs) express a
condition among tuples of a relation that exists
when the relation is trying to represent more than
one to many or many to many relationship.
09/24/22 Manu Sharma DBMS Unit-3 129
 Formal Definition of Multivalued
Dependency(CO3)
A multivalued dependency X →→ Y specified on
relation schema R, where X and Y are both subsets of
R, specifies the following constraint on any relation
state r of R:
If two tuples t1 and t2 exist in r such that t1[X] =
t2[X], then two tuples t3 and t4 should also exist in r
with the following properties,
t1[X] = t2[X] = t3[X] = t4[X]
t1[Y] = t3[Y] and t2[Y] = t4[Y]
t1[Z] = t4[Z] and t3[Z] = t2[Z]

Where X →→ Y holds, we say that X multi-determines

Y, or Y is multi-dependent on X.
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 Forth Normal Form (4NF)(CO3)
The official qualification for 4NF are:-
1. A table is already in BCNF.
2. A table contain no multi-valued functional
dependency .
Example :- Let us consider following relation student:-

Key:- {s_id,course,hobby}
MVD, s_id →→ course,hobby
To check relation is in 4NF or not.
09/24/22 Manu Sharma DBMS Unit-3 131
 Example

As you can see in the above table , student with s_id 1 has
opted for two courses, Science and Maths, and has two
hobbies, Cricket and Hockey.
You must be thinking what problem this can lead to, right?
Problem:- Well the two records for student with s_id 1& 2,
will give rise to two more records, because for one student,
two hobbies exists, hence along with both the courses,
these hobbies should be specified.

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 Continue

In the above table , there is no relationship between

the columns course and hobby. They are
independent of each other.

So there is multi-value dependency, which leads to

un-necessary repetition of data and other anomalies
as well.

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 Solution
Solution:- To make the above relation satisfy the 4th
normal form, we can decompose the table into 2 tables.

• Now this relation satisfies the fourth normal form.

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 6. Fifth Normal Form (5NF)(CO3)

The 5NF (Fifth Normal Form) is also known as project-

join normal form (PJNF).
A relation is in Fifth Normal Form (5NF), if
1. it is in 4NF, and
2. won’t have lossless decomposition into smaller tables.
Definition:- A join dependency (JD), denoted by
JD(R1, R2, … , Rn), specified on relation schema R,
specifies a constraint on the states r of R.
The constraint states that every legal state r of R
should have a non-additive join decomposition into R1,
R2, … , Rn.
Hence, for every such r we have * (πR1 (r), πR2 (r), … ,
πRn (r)) = r

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 6. Fifth Normal Form (5NF)(CO3)

5NF is of little practical use to

the database designer, but it is of interest from a
theoretical point of view.

In all of the further normal forms discussed so far,

no loss decomposition was achieved by the
decomposing of a single table into two separate
tables.

Note :- In considering 5NF, consideration must

be given to tables where this non-loss
decomposition can only be achieved by
decomposition into three or more separate tables.
09/24/22 Manu Sharma DBMS Unit-3 136
 Example

Let us consider the following relation person with

attribute ((Agent, Company, Product _Name))

The table is in 4NF because it contains no multi-

valued dependency.
It does, however, contain an element of redundancy
in that it records the fact that Suneet is an agent for
ABC twice. But there is no way of eliminating this
redundancy without losing information.

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 Continue

• Suppose that the Relation Person is decomposed

into two relation P1 and P2.

• The redundancy has been eliminated, but the

information about which companies make which
products and which of these products they supply to
which agents has been lost.

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 Continue..

The natural join of these projections over the ‘agent’

columns is:

The table resulting from this join is spurious, since the

asterisked row of the table contains incorrect information.
Solution :- When we decompose the relation such case
common attribute must be a candidate key.

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 Cont……..

Now suppose that the original table were to be

decomposed into three tables, the two projections,
P1 (agent,company)and P2(agent,procduct) which
have already shown, and the final, possible
projection, P3 (company,product).
P3 Product_na
Company
me
ABC Nut

ABC Scew

CDE Bolt

ABC Bolt

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 Conti….

If a join is taken of all three projections, first of P1 and P2 with the

(spurious) result shown above, and then of this result with P3 over
the ‘Company’ and ‘Product name’ column, the following table is
obtained:-
Agent Company Product_Name
Suneet ABC Nut
Suneet ABC Screw
Suneet ABC Bolt
Suneet CDE Bolt
Raj ABC Bolt
This still contains a spurious row. The order in which the joins are
performed makes no difference to the final result. It is not simply
possible of decompose the ‘AGENT_COMPANY_PRODUCT’ table,
populated as shown, without losing information. Thus, it has to be
accepted that it is not possible to eliminate all redundancies using
normalization techniques, because it cannot be assumed that all
decompositions will be non-loss.
09/24/22 Manu Sharma DBMS Unit-3 141
 Content

Lecture 8:-
Decomposition of relation Schema
Properties of decomposition

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 Decomposition of relation
Schema(CO3)
Decomposition of relation schema R consists of
replacing the relation schema by two or more sub
relations schemas. So that each contain subset of
attributes of relation schema R and which together
include all attribute or R.
Decomposition of relation R= {A1,A2,A3,…….An} is the
replacement of R by a collection D={R1,R2…….Rm}
where Ri is subset of R such that R={ R1 U R2 U…….RN}
Example :- R= { Eid,position,salary}
D1= {Eid,position}
D2= {Postion,Salary}
Using the functional dependencies we can decompose a
given relation schema R= {A1,A2,……AN} into a set of
relation schema represented by D={R1,R2…..RN} to
remove anomalies present in a database(Relations).

09/24/22 Manu Sharma DBMS Unit-3 143

 Properties of
decomposition(CO3)
There are three properties of decomposition of relation
are
1. Attribute Preservation
2. Dependency Preservation
3. Lossless Join decomposition

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 1.Attribute Preservation(CO3)
We must make sure that each attribute in R will
appear in at least one relation schema Ri in the
decomposition so that no attributes are lost; formally,
we say this is called attribute preservation.
Example :- R={A,B,C,D,E,F}
Decompose relation are
R1={A,B,C,D}
R2={A,E,F}
Note :- Union of all decompose relation will give all
attribute of universal relation R.
R1 U R2= R{A,B,C,D,E,F}.

09/24/22 Manu Sharma DBMS Unit-3 145

 2. Dependency
Preservation(CO3)
• It would be useful if each functional dependency X →
Y specified in F either appeared directly in one of the
relation schemas Ri in the decomposition D or could
be inferred from the dependencies that appear in
some Ri. Informally, this is the dependency
preservation condition.

• We want to preserve the dependencies because

each dependency in F represents a constraint
on the database. If one of the dependencies is not
represented in some individual relation Ri of the
decomposition, we cannot enforce this constraint by
dealing with an individual relation.

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 Cont….. (CO3)

If we decompose a relation R into relations R1

and R2, All dependencies of R either must be a
part of R1 or R2 or must be derivable from
combination of FD’s of R1 and R2.
A Decomposition D = { R1, R2, R3….Rn } of R is
dependency preserving write E a set F of
Functional dependency if
(F1 ∪ F2 ∪ … ∪ Fm) + = F+ .

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 Example

Question 1 :- Let a relation R (A, B, C, D ) and

functional dependency {AB –> C, C –> D, D –> A}.
Relation R is decomposed into R1( A, B, C) and R2(C,
D).
Check whether decomposition is dependency
preserving or not.

Question 2:- Let us consider a relation R (A, B, C,

D,E,F ) with functional dependency set F= {D –> AB,
C–> EF,}and relation R is decomposed into R1( A,C,D),
R2(A,D,B), R3(D,E,F), R4(C,E,F).
Check whether decomposition is dependency
preserving or not.

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 Nonadditive (Lossless) Join Property of a
Decomposition(CO3)

 Another property that a decomposition D

should possess is the nonadditive join
property, which ensures that no spurious tuples are
generated when a NATURAL JOIN operation is applied
to the relations resulting from the decomposition.

Definition. Formally, a decomposition D = {R1, R2,

… , Rm} of R has the lossless (nonadditive) join
property with respect to the set of dependencies F
on R if, for every relation state r of R that satisfies F,
the following holds, where * is the NATURAL JOIN of
all the relations in D: *(πR1(r), … , πRm(r)) = r.

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 Procedure(CO3)

• Procedure :- Decomposition of R into R1 and R2 from a

lossless join decomposition if at least one of the
following dependencies in F+
1. First condition) = (ABC) U (AD) = (ABCD) = Att(R). holds
true as Att(R1) U Att(R2
2. R1 ∩ R2= R1 , that is: all attributes common to both R1
and R2 functionally determine ALL the attributes in R1.
OR
R1 ∩ R2= R2 that is: all attributes common to both R1
and R2 functionally determine ALL the attributes in R2

3. In other word R1 ∩ R2 from a super key or candidate

key of either R1 or R2 or both , The decomposition of R
is a lossless join decomposition.
09/24/22 Manu Sharma DBMS Unit-3 150
 Example

Question 1 :- Let a relation R (A, B, C, D ,E) and functional

dependency {A –> BC, CD –> E, B –> D, E> A}.
Relation R is decomposed into R1( A, B, C) and R2(A, D,E).
Check whether decomposition is lossless or lossy
decomposition.

Question 2 :- Let a relation R (A,B,C,D,E,F) and functional

dependency {AB –> C, C –> D, D –> EF, F -> A,D ->B}.
Relation R is decomposed into R1( A,B,C) ,
R2(C,D,E),R3(E,F).
Check whether decomposition is lossless or lossy
decomposition.

09/24/22 Manu Sharma DBMS Unit-3 151

 Testing for Non additive join
property
A Universal relation R , decomposition D {R1,R2,……
Rm} of R and a set of Functional dependencies
Step 1:- Create a matrix by the given set of FD’s
(initially matrix will one row i for each relation Ri in D,
and one column J for each attribute Aj in R ).
Step 2:- Set s(i,j) = by for all matrix entries:-
– Put the symbol ‘a’ for those element which are present
in the given set D. And
– Put the symbol ‘b’ for those element which are not
present in the given set D.
Setp3:- Now apply the given FD’s and developed
new matrix by changing the symbol ‘b’ to ‘a’ if the
element functionally dependent.
Step 4:- Check whether any raw is filled with all ‘a’.
Step 5:- if any raw is filled with all ‘a’ element then It
is lossless join decomposition otherwise lossy
decomposition.
09/24/22 Manu Sharma DBMS Unit-3 152
 Example

Question 1:- Let us consider a relation Schema R

={SSN, PNumber, PName, PLocation,Hours} and
functional dependency set f=
 SSN –> Ename,
 Pnumber –> {Pname,Plocation}
 {SSN,Number} –> Hours } and
Decompose a relation D = {R1,R2,R3} where R1=
{SSN, Ename}, R2= {Pnumber, Pname,Plocation}
and R3= {SSN, Pnumber, Hours} .
show whether decomposition is lossy or lossless
join decomposition .

09/24/22 Manu Sharma DBMS Unit-3 153

Question 2:- Let us consider a relation Schema R
={A,B,C,D,E,} and functional dependency set f=
 AB –> CD,
 A –> E
 C –> D } and
Decompose a relation D = {R1,R2,R3} where
R1= {A,B,C}, R2= {B,C,D}, and R3= {C,D,E} .
show whether decomposition is lossy or lossless
join decomposition .

09/24/22 Manu Sharma DBMS Unit-3 154

 Inclusion Dependencies(CO3)
• MVDs and JDs can be used to guide database design,
as we have seen, although they are less common than
FDs and harder to recognize and reason about.
• In contrast, inclusion dependencies are very intuitive
and quite common. However, they typically have little
influence on database design
• The main point to bear in mind is that we should not
split groups of attributes that participate in an
inclusion dependency.
• Most inclusion dependencies in practice are key-
based, that is, involve only keys.

09/24/22 Manu Sharma DBMS Unit-3 155

 Faculty Video Links, Youtube & NPTEL Video
Links and Online Courses Details

•Youtube/other Video Links

•https://www.youtube.com/watch?v=1yUkun2r0N4
•https://www.youtube.com/watch?v=EFdvRm5nse0

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 Daily Quiz

• What is functional dependency?

• Explain normal forms.
• What is multi valued dependency?
• Explain BCNF.
• Explain redundancy.

09/24/22 Manu Sharma DBMS Unit-3 157

 Quiz
Question 1:- When you normalize a relation by breaking it into
two smaller relations, what must you do to maintain data
integrity?
Please select all the correct answers.
A. Link the relations by a common field
B.Remove any functional dependencies from both relations
C .Assign both relations the same primary key field(s)
D .Create a primary key(s) for the new relation

Question 2:- A relation is in 1NF if it doesn't contain any

____________?
A. Determinants
B. Repeating groups
C. Null values in primary key fields
D. Functional dependencies
09/24/22 Manu Sharma DBMS Unit-3 158
 Quiz

Question 3:A functional dependency that exist between

two non-key attributes is called _____
(a) Non-transitive dependency (b) Transitive
dependency
(c) Partial transitive dependency (d) None of the above

Question 4:-In the __________ normal form, a composite

attribute is converted to individual attributes.
a) First
b) Second
c) Third
d) Fourth

09/24/22 Manu Sharma DBMS Unit-3 159

 Daily Quiz

• Relation R has eight attributes ABCDEFGH. Fields of R

contain only atomic values. F = {CH -> G, A -> BC, B ->
CFH, E -> A, F -> EG} is a set of functional
dependencies (FDs) so that F+ is exactly the set of FDs
that hold for R. How many candidate keys does the
relation R have? Gate 2013
• Answer: (B)

Explanation: A+ is ABCEFGH which is all attributes

except D.
• B+ is also ABCEFGH which is all attributes except D.
• E+ is also ABCEFGH which is all attributes except D.
• F+ is also ABCEFGH which is all attributes except D.
• So there are total 4 candidate keys AD, BD, ED and FD
09/24/22 Manu Sharma DBMS Unit-3 160
 Glossary Questions

1. What are advantages and disadvantages of Distributed DBMS.

2. What are the features of DDBMS?

3. Explain the basic Timestamp Ordering Algorithm.

4. What are the objectives of Distributed Query Processing?

5. What is horizontal and vertical fragmentation? What are the types of

horizontal fragmentation. Perform horizontal Fragmentation for student
relation as given below.
Also give the corrextness criteria for it.
Students (studentrollno., Student Name, Course Name, Course Name,
Course fees, year)

6. What are the various kinds of transparencies in distributed database

design? Explain each with the help of example.
09/24/22 Manu Sharma DBMS Unit-1 161
 Weekly Assignment

• What do you mean by BCNF ? Why it is used and how

it differ from 3 NF ? CO3
• Discuss the various normal forms in normalization with
suitable examples? Why is concurrency control
needed? Explain lost update, Inconsistent retrievals
and uncommitted dependency anomalies. CO3
• Explain the Codd’s Rule in detail. CO3
• Explain Normalization with example. CO3
• What are the rules of 1NF,2NF,3NF. CO3
• Discuss Boyce Codd Normalization Form. CO3

09/24/22 Manu Sharma DBMS Unit-3 162

 MCQ s

• Which forms simplifies and ensures that there are

minimal data aggregates and repetitive groups:
a) 1NF
b) 2NF
c) 3NF
d) All of the mentioned

• For any pincode, there is only one city and state. Also,
for given street, city and state, there is just one
pincode. In normalization terms, empdt1 is a relation
in
a) 1 NF only
b) 2 NF and hence also in 1 NF
c) 3NF and hence also in 2NF and 1NF
d) BCNF and hence also in 3NF, 2NF and 1NF

09/24/22 Manu Sharma DBMS Unit-3 163

 MCQ s

• In the __________ normal form, a composite attribute is

converted to individual attributes.
a) First
b) Second
c) Third
d) Fourth

• Functional Dependencies are the types of constraints

that are based on______
a) Key
b) Key revisited
c) Superset key
d) None of the mentioned

09/24/22 Manu Sharma DBMS Unit-3 164

 Sessional paper Template

Manu Sharma DBMS

09/24/22 165
Unit-1
 Old Question Papers

• http://www.aktuonline.com/papers/btech-cs-5-sem-
data-base-management-system-rcs501-2020.pdf
• http://www.aktuonline.com/papers/btech-cs-5-sem-
database-management-system-KCS501-2018-19.pdf
• http://www.aktuonline.com/papers/btech-cs-5-sem-
database-management-system-ncs-502-2017-18.pdf
• http://www.aktuonline.com/papers/btech-cs-5-sem-
database-management-system-ncs-502-2016-17.pdf

09/24/22 Manu Sharma DBMS Unit-3 166

 Expected Questions for
University Exam
• Explain normalization. What is normal form?
• Describe the following terms :
(i) Multivalued dependency
(ii) Trigger
• Why do we normalize database?
• Define partial functional dependency. Consider the
following two sets of functional dependencies F= {A -
>C, AC ->D, E ->AD, E ->H} and G = {A ->CD, E -
>AH}. Check whether or not they are equivalent.
• What is Trigger? Explain different trigger with example.
• Write difference between BCNF Vs 3 NF.

09/24/22 Manu Sharma DBMS Unit-3 167

 Summary

• Knowledge of database design.

• Knowledge of functional dependencies.

• Knowledge of Normal forms.

• Knowledge of Loss less join decomposition,

Multivalued dependency and join dependency

09/24/22 Manu Sharma DBMS Unit-3 168

 References

• Korth, Silbertz, Sudarshan,” Database Concepts”, McGraw

Hill
• Date C J, “An Introduction to Database Systems”, Addision
Wesley
• Elmasri, Navathe, “ Fundamentals of Database Systems”,
Addision Wesley
• O’Neil, Databases, Elsevier Pub.
• RAMAKRISHNAN"Database Management Systems",McGraw
Hill
• Leon &Leon,”Database Management Systems”, Vikas
Publishing House
• Bipin C. Desai, “ An Introduction to Database Systems”,
Galgotia Publications
• Majumdar& Bhattacharya, “Database Management
System”, TMH
• R.P. Mahapatra, Database Management System, Khanna
Publishing House

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09/24/22 Manu Sharma DBMS Unit-3 170