WEEK—4 Lecture-3 hr

Functional dependency:
Functional dependency: overview

➢ Functional Dependency is a relationship that exists between multiple attributes of a relation.

➢ This concept is given by E. F. Codd.
➢ It is a type of constraint existing between various attributes of a relation.
➢ It is used to define various normal forms.
➢ These dependencies are restrictions imposed on the data in database.
➢ The attributes of a table is said to be dependent on each other when an attribute of a table
uniquely identifies another attribute of the same table.
➢ It typically exists between the primary key and non-key attribute within a table.
X → Y
The left side of FD is known as a determinant, the right side of the production is known as a
dependent.
For example:
Assume we have an employee table with attributes: Emp_Id, Emp_Name, Emp_Address.
Here Emp_Id attribute can uniquely identify the Emp_Name attribute of employee table because if
we know the Emp_Id, we can tell that employee name associated with it.
Functional dependency can be written as:
1. Emp_Id → Emp_Name
We can say that Emp_Name is functionally dependent on Emp_Id.
Importance of functional dependency
➢ Functional Dependency avoids data redundancy. Therefore same data do not repeat at
multiple locations in that database.
➢ It helps you to maintain the quality of data in the database.
➢ It helps you to defined meanings and constraints of databases.
➢ It helps you to identify bad designs.
➢ It helps you to find the facts regarding the database design.
Rules of Functional Dependencies
Inference Rule (IR):
• The Armstrong's axioms are the basic inference rule.
• Armstrong's axioms are used to conclude functional dependencies on a relational database.
• The inference rule is a type of assertion. It can apply to a set of FD(functional dependency)
to derive other FD.
• Using the inference rule, we can derive additional functional dependency from the initial
set.
The Functional dependency has 6 types of inference rule:
1. Reflexive Rule (IR1)
In the reflexive rule, if Y is a subset of X, then X determines Y.
1. If X ⊇ Y then X → Y
Example:
1. X = {a, b, c, d, e}
2. Y = {a, b, c}
2. Augmentation Rule (IR2)
The augmentation is also called as a partial dependency. In augmentation, if X determines Y, then
XZ determines YZ for any Z.
1. If X → Y then XZ → YZ
Example:
 1. For R(ABCD), if A → B then AC → BC
3. Transitive Rule (IR3)
In the transitive rule, if X determines Y and Y determine Z, then X must also determine Z.
1. If X → Y and Y → Z then X → Z
4. Union Rule (IR4)
Union rule says, if X determines Y and X determines Z, then X must also determine Y and Z.
1. If X → Y and X → Z then X → YZ
Proof:
1. X → Y (given)
2. X → Z (given)
3. X → XY (using IR2 on 1 by augmentation with X. Where XX = X)
4. XY → YZ (using IR2 on 2 by augmentation with Y)
5. X → YZ (using IR3 on 3 and 4)
5. Decomposition Rule (IR5)
Decomposition rule is also known as project rule. It is the reverse of union rule.
This Rule says, if X determines Y and Z, then X determines Y and X determines Z separately.
1. If X → YZ then X → Y and X → Z
Proof:
1. X → YZ (given)
2. YZ → Y (using IR1 Rule)
3. X → Y (using IR3 on 1 and 2)
6. Pseudo transitive Rule (IR6)
In Pseudo transitive Rule, if X determines Y and YZ determines W, then XZ determines W.
1. If X→ Y and YZ → W then XZ → W
Proof:
1. X → Y (given)
2. WY → Z (given)
3. WX → WY (using IR2 on 1 by augmenting with W)
4. WX → Z (using IR3 on 3 and 2)

Types Functional Dependencies

There are mainly four types of Functional Dependency in DBMS. Following are the types of
Functional Dependencies in DBMS:
1. Multivalued Dependency
2. Trivial Functional Dependency
3. Non-Trivial Functional Dependency
4. Transitive Dependency
Multivalued Dependency in DBMS
➢ Multivalued dependency occurs in the situation where there are multiple independent
multivalued attributes in a single table.
➢ A multivalued dependency is a complete constraint between two sets of attributes in a
relation.
➢ It requires that certain tuples be present in a relation.
➢ Consider the following Multivalued Dependency Example to understand.
Example:
Car_model Maf_year Color
H001 2017 Metallic
H001 2017 Green
H005 2018 Metallic
H005 2018 Blue
H010 2015 Metallic
H033 2012 Gray
In this example, Maf_year and Color are independent of each other but dependent on car_model.
In this example, these two columns are said to be multivalue dependent on car_model.
This dependence can be represented like this:
car_model -> Maf_year
car_model-> Color
Trivial Functional Dependency in DBMS
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.
For example:

Emp_id Emp_name
AS555 Harry
AS811 George
AS999 Kevin
Consider this table of with two columns Emp_id and Emp_name.
{Emp_id, Emp_name} -> Emp_id is a trivial functional dependency as Emp_id is a subset of
{Emp_id,Emp_name}.
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

Microsoft Satya Nadella 51
Google Sundar Pichai 46
Apple Tim Cook 57
Example:

(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.
Transitive Dependency in DBMS
A Transitive Dependency is a type of functional dependency which happens when “t” is indirectly
formed by two functional dependencies.
Let’s understand with the following Transitive Dependency Example.

Example:

Company CEO Age

Microsoft Satya Nadella 51
Google Sundar Pichai 46
Alibaba Jack Ma 54
{Company} -> {CEO} (if we know the compay, we know its CEO’s name)
{CEO } -> {Age} If we know the CEO, we know the Age
Therefore according to the rule of rule of transitive dependency:
{ Company} -> {Age} should hold, that makes sense because if we know the company name, we
can know his age.
Note: transitive dependency can only occur in a relation of three or more attributes.
Normalization: normalization process
Importance of normalization

❖ Resolving the database anomalies

The forms of Normalization i.e. 1NF, 2NF, 3NF, BCF, 4NF and 5NF remove all the Insert,
Update and Delete anomalies.

Insertion Anomaly occurs when you try to insert data in a record that does not exist.

Deletion Anomaly is when a data is to be deleted and due to the poor deign of database,
other record also deletes.

❖ Eliminate Redundancy of Data

Storing same data item multiple times is known as Data Redundancy. A normalized table do
not have the issue of redundancy of data.

❖ Data Dependency
The data gets stored in the correct table and ensures normalization.

❖ Isolation of Data
A good designed database states that the changes in one table or field do not affect other.
This is achieved through Normalization.

❖ Data Consistency
While updating if a record is left, it can led to inconsistent data, Normalization resolves it
and ensures Data Consistency.

1NF,2NF,3NF
First Normal Form (1NF)
1. A relation will be 1NF if it contains an atomic value.
2. It states that an attribute of a table cannot hold multiple values. It must hold only single-valued
attribute.
3. 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
9064738238

20 Harry 8574783832 Bihar

12 Sam 7390372389, Punjab

8589830302
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

We can observe that although a few values are getting repeated but values for the EMP_PHONE column are
now atomic for each record/row.
Using the First Normal Form, data redundancy increases, as there will be many columns with same data in
multiple rows but each row as a whole will be unique.
Second Normal Form (2NF)
For a table to be in the Second Normal Form, it must satisfy two conditions:
1. The table should be in the First Normal Form.
2. There should be no Partial Dependency.
Example: Let's assume, a school can store the data of teachers and the subjects they teach. In a school, a
teacher can teach more than one subject.
TEACHER table

TEACHER_ID SUBJECT TEACHER_AGE

25 Chemistry 30

25 Biology 30

47 English 35

83 Math 38

83 Computer 38
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_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
NOTE:

1. For a table to be in the Second Normal form, it should be in the First Normal form and it should not
have Partial Dependency.
2. Partial Dependency exists, when for a composite primary key, any attribute in the table depends
only on a part of the primary key and not on the complete primary key.
3. To remove Partial dependency, we can divide the table, remove the attribute which is causing partial
dependency, and move it to some other table where it fits in well.

Third Normal Form (3NF)

1. A relation will be in 3NF if it is in 2NF and not contain any transitive partial dependency.
2. 3NF is used to reduce the data duplication. It is also used to achieve the data integrity.
3. 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:
we have 3 tables, Student, Subject and Score.

Student Table

student_id name reg_no branch address

10 Akon 07-WY CSE Kerala

11 Akon 08-WY IT Gujarat

12 Bkon 09-WY IT Rajasthan
Subject Table

subject_id subject_name teacher

1 Java Java Teacher

2 C++ C++ Teacher

3 Php Php Teacher

Score Table

score_id student_id subject_id marks

1 10 1 70

2 10 2 75
3 11 1 80
In the Score table, we need to store some more information, which is the exam name and total marks.

score_id student_id subject_id marks exam_name total_marks

With exam_name and total_marks added to Score table, it saves more data now.
Primary key for Score table is a composite key, which is made up of two attributes or columns
→ student_id + subject_id.
new column exam_name depends on both student and subject.
For example, a mechanical engineering student will have Workshop exam but a computer science student
won't. And for some subjects you have Prctical exams and for some you don't. So we can say
that exam_name is dependent on both student_id and subject_id.
The column total_marks depends on exam_name as with exam type the total score changes. For example,
practicals are of less marks while theory exams are of more marks.

But, exam_name is just another column in the score table. It is not a primary key or even a part of the
primary key, and total_marks depends on it.

This is Transitive Dependency. When a non-prime attribute depends on other non-prime attributes rather
than depending upon the prime attributes or primary key.

We need to remove Transitive Dependency here.

the solution is very simple. Take out the columns exam_name and total_marks from Score table and put
them in an Exam table and use the exam_id wherever required.
Score Table: In 3rd Normal Form
score_id student_id subject_id marks exam_id

The new Exam table

exam_id exam_name total_marks

1 Workshop 200

2 Mains 70

3 Practicals 30
The advantage of removing transitive dependency is,
• Amount of data duplication is reduced.
• Data integrity achieved.
Consider another example

EMPLOYEE_DETAIL table:

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

Super key in the table above:
1. {EMP_ID}, {EMP_ID, EMP_NAME}, {EMP_ID, EMP_NAME, EMP_ZIP}....so on
Candidate key: {EMP_ID}
Non-prime attributes: In the given table, all attributes except EMP_ID are non-prime.
Here, EMP_STATE & EMP_CITY dependent on EMP_ZIP and EMP_ZIP dependent on EMP_ID.
The non-prime attributes (EMP_STATE, EMP_CITY) transitively dependent on super
key(EMP_ID). It violates the rule of third normal form.
That's why we need to move the EMP_CITY and EMP_STATE to the new <EMPLOYEE_ZIP>
table, with EMP_ZIP as a Primary key.
EMPLOYEE table:

EMP_ID EMP_NAME EMP_ZIP

222 Harry 201010

333 Stephan 02228

444 Lan 60007

 555 Katharine 06389

666 John 462007

EMPLOYEE_ZIP table:

EMP_ZIP EMP_STATE EMP_CITY

201010 UP Noida

02228 US Boston

60007 US Chicago

06389 UK Norwich

462007 MP Bhopal

Boyce Codd normal form (BCNF)

1. BCNF is the advance version of 3NF. It is stricter than 3NF.

2. A table is in BCNF if every functional dependency X → Y, X is the super key of the table.
3. 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:

EMP_ID EMP_COUNTRY EMP_DEPT DEPT_TYPE EMP_DEPT_NO

264 India Designing D394 283

264 India Testing D394 300

364 UK Stores D283 232

364 UK Developing D283 549

In the above table Functional dependencies are as follows:

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

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

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

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

EMP_COUNTRY table:
 EMP_ID EMP_COUNTRY

264 India

264 India

EMP_DEPT table:

EMP_DEPT DEPT_TYPE EMP_DEPT_NO

Designing D394 283

Testing D394 300

Stores D283 232

Developing D283 549

EMP_DEPT_MAPPING table:

EMP_ID EMP_DEPT

D394 283

D394 300

D283 232

D283 549
Functional dependencies:
1. EMP_ID → EMP_COUNTRY
2. EMP_DEPT → {DEPT_TYPE, EMP_DEPT_NO}
Candidate keys:
For the first table: EMP_ID
For the second table: EMP_DEPT
For the third table: {EMP_ID, EMP_DEPT}
Now, this is in BCNF because left side part of both the functional dependencies is a key.

Fourth normal form (4NF)

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

Example

STUDENT

STU_ID COURSE HOBBY

 21 Computer Dancing

21 Math Singing

34 Chemistry Dancing

74 Biology Cricket

59 Physics Hockey

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

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

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

STUDENT_COURSE

STU_ID COURSE

21 Computer

21 Math

34 Chemistry

74 Biology

59 Physics

STUDENT_HOBBY

STU_ID HOBBY

21 Dancing

21 Singing

34 Dancing
 74 Cricket

59 Hockey

Fifth normal form (5NF)

1. A relation is in 5NF if it is in 4NF and not contains any join dependency and joining should be
lossless.
2. 5NF is satisfied when all the tables are broken into as many tables as possible in order to avoid
redundancy.
3. 5NF is also known as Project-join normal form (PJ/NF).

Example

SUBJECT LECTURER SEMESTER

Computer Anshika Semester 1

Computer John Semester 1

Math John Semester 1

Math Akash Semester 2

Chemistry Praveen Semester 1

In the above table, John takes both Computer and Math class for Semester 1 but he doesn't take Math class

for Semester 2. In this case, combination of all these fields required to identify a valid data.

Suppose we add a new Semester as Semester 3 but do not know about the subject and who will be taking

that subject so we leave Lecturer and Subject as NULL. But all three columns together acts as a primary

key, so we can't leave other two columns blank.

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

P1

SEMESTER SUBJECT

Semester 1 Computer
 Semester 1 Math

Semester 1 Chemistry

Semester 2 Math

P2

SUBJECT LECTURER

Computer Anshika

Computer John

Math John

Math Akash

Chemistry Praveen

P3

SEMSTER LECTURER

Semester 1 Anshika

Semester 1 John

Semester 1 John

Semester 2 Akash

Semester 1 Praveen