The document discusses various types of normal forms in database normalization including 1NF, 2NF, 3NF, BCNF, and 4NF. It defines each normal form and provides examples to illustrate situations that violate the given normal form and how to normalize the data to satisfy that normal form. The goal of normalization is to organize data to eliminate issues like data redundancy, insertion anomalies, update anomalies, and deletion anomalies.

1. NORMALIZATION
Sadik & Shruti

2. AGENDA
Functional Dependency
Armstrong axioms & Inference rules
​Normalization

3. Functional Dependency: 3
The functional dependency is a relationship that exist between two
attributes. It typically exists between the primary key and non-key attribute
within a table.
The left side of FD is known as a determinant, and
The right side of the production is known as a dependent
X → 𝑌
Determinant
Dependent

4. Types of Functional Dependency:
4
Trivial FD
Non-Trivial FD
Multivalued FD
Transitive FD

5. 5
Trivial Functional Dependency
If 𝐴 → 𝐵 is a Functional Dependency and ‘B’ is a subset of ‘A’ then
this FD is called as Trivial Functional Dependency.
Example
S_roll S_name age
3 a 20
4 b 19
5 c 20
Here, {S_roll, S_name} → S_name is a trivial FD, since name is a subset of
{S_roll, S_name}.
S_roll → S_roll is also trivial FD
• AB→A
• A → A

6. 6
Non-Trivial Functional
Dependency
In Non-trivial FD, the dependent is strictly not a subset of the
determinant.
i.e. If X→ Y and Y is not a subset of X, then it is called Non-trivial
FD
Example S_roll S_name age
3 a 20
4 b 19
5 c 20
Here, S_roll→ S_name is Non-trivial FD, since name is not a subset of S_roll.
{S_roll, S_name} → age is also Non-trivial FD
• A→B
• AB→ C

7. 7
Multivalued Functional
Dependency
• In Multivalued functional dependency, entities of dependent set
are not dependent on each other.
• When two attributes in a table are independent of one another
but both rely on a third attribute, this is known as multivalued
dependency.
• A multivalued dependency includes at least three attributes since
it consists of at least two attributes that are based on a third
attribute.
• It is represented by double arrow: →→
• For example: P→→ Q
P →→ R

8. 8
Multivalued Functional
Dependency
Example S_name Course Activities
A Mathematics Singing
A Mathematics Dancing
B Computer Cricket
C Literature Dancing
C Literature Cricket
C Literature Singing
In the above table, we can see Students A and C have interest in more than one activity.
This is multivalued dependency because Course of a student are independent of Activities but
are dependent of the student.
Therefore multivalued dependency:
S_name → Course
S_name → Activities

9. 9
Transitive Functional Dependency
• In transitive functional dependency, dependent is indirectly
dependent on determinant
• i.e. If A→B & B→C, then according to axiom of transitivity, A→C.
This is a transitive functional dependency
Example Roll_no Name Dept Dept_loc
1 A CSE SB1
2 B ECE SB1
3 C IT SB2
4 D EEE SB3
Here, Roll_no → Dept and Dept →Dept_loc
Hence, according to the axiom of transitivity, Roll_no → Dept_loc is a valid functional
dependency.

10. 10
Armstrong’s Axioms in FD
These are the rules to find logically implied functional
dependencies.
W, X, Y, Z some attribute set over R.
1. Reflexivity:
In the reflexive rule, if Y is a subset of X, then X determines Y.
if Y ⊆ X then X → Y.
2. Transitivity:
In the transitive rule, if X determines Y and Y determines Z, then
X
must also determine Z.
if X → Y and Y → Z then X → Z
3. Augmentation:
In augmentation, If X determines Y, then XZ determines YZ for
any
Z.

11. 11
Armstrong’s Axioms in FD
Secondary Rules-
These rules can be derived from the above axioms.
1. Union Rule:
If A →B holds and A → C holds, then A → BC holds.
2. Composition:
If X → Y and P → Q then XP → YQ always true.
3. Decomposition:
If X →YZ is a FD then X →Y and X →Z always true.
4. Pseudo transitive Rule:
If X →Y and YZ → P are two FDs then XZ → P is always true.

12. 12
Normalization
Normalization is a process of organizing the data in database to avoid:
i. Data redundancy
ii. Insertion anomaly
iii. Update anomaly
iv. Deletion anomaly

13. 13
Data redundancy in DBMS
Redundancy in DBMS is the problem that arises when the database is
not normalized. It is the concept of storing multiple copies of the same
data in different parts of the database.
Example
student_id student_name student_age dept_id dept_name dept_head
1 A 18 100
Computer
Science
F1
2 B 18 100
Computer
Science
F1
In this student table, we have repeated the same department details,
dept_id, dept_name, and dept_head in every student record. This
causes redundancy in the student table.

14. 14
Insertion Anomaly in
DBMS
The term ‘insertion anomaly’ is used to describe when a new row is
added to a table and it causes an inconsistency.
Example
S_id S_name C_id C_name Fac_id Fac_name Fac_sal.
1 A C1 DBMS F1 Bob 40K
2 B C2 C++ F2 Doe 30K
3 C C1 DBMS F1 Bob 40K
4 D C1 DBMS F1 Bob 40K
In this table S_id is PK. Suppose, we want to add a student in this table
it will not give any error but if we add new course or faculty then error
occur.
Because, S_id is empty for this case. This is Insertion Anomaly.

15. 15
Update Anomaly in DBMS
If there are some changes in the database, we have to apply that
change in all the rows. And if we miss any row, we will have one more
field, creating an update anomaly in the database.
Example
S_id S_name C_id C_name Fac_id Fac_name Fac_sal.
1 A C1 DBMS F1 Bob 40K
2 B C2 C++ F2 Doe 30K
3 C C1 DBMS F1 Bob 40K
4 D C1 DBMS F1 Bob 40K
5 E C3 Java F1 Bob 40K
In the given table, we have two rows for a faculty named Bob, and he
taught two different courses. If we need to update Bob’s salary, we must
update the same in all rows. Otherwise, the data will become
inconsistent.

16. 16
Normal Forms in DBMS
First Normal Form (1NF)
Second Normal Form (2NF)
Third Normal Form (3NF)
Boyce-Codd Normal Form (BCNF)
Fourth Normal Form (4NF)
Fifth Normal Form (5NF)

17. 17
First Normal Form (1NF)
For a table to be in the First Normal Form, it should follow the following 4 rules:
1. Values stored in a column should be of the same domain
2. All the columns in a table should have unique names.
3. It should only have single valued attributes/columns.
4. And the order in which data is stored, does not matter.
5. After converting ER model into relational model, all the relations are in 1st
Normal form.
EXAMPLE:
Relation Employee is not in 1NF because of multi-valued attribute e_ph
E_id E_name E_ph E_state
14 John
7728263853,
9062978898
WB
20 Harry 9088989763 Bihar
11 Tony
6291359760,
9800146215
Punjab

18. 18
First Normal Form (1NF)
Decomposed Relations:
Now ‘Employee’ Table would be decomposed as given below:
Solution 1
E_id E_name E_ph E_state
14 John 7728263853 WB
14 John 9062978898 WB
20 Harry 9088989763 Bihar
11 Tony 6291359760 Punjab
11 Tony 9800146215 Punjab
Solution 2
E_id
E_nam
e
E_ph1 E_ph2 E_state
14 John 7728263853 9062978898 WB
20 Harry 9088989763 - Bihar
11 Tony 6291359760 9800146215 Punjab
Solution 3
E_id E_name E_state
14 John WB
20 Harry Bihar
11 Tony Punjab
Solution 3
E_id E_ph
14
772826385
3
14
906297889
8
20
908898976
3

19. 19
Second Normal Form (2NF)
• In the 2NF, relation must be in 1NF
• In the second normal form, all non-key attributes are fully functional dependent
on the primary key.
• There should be non 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_ID SUBJECT TEACHER_AGE
25 Chemistry 30
25 Biology 30
47 English 35
83 Math 38
83 Computer 38

20. 20
Second Normal Form (2NF)
In the given table {TEACHER_ID, SUBJECT} is primary key.
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 it into 2NF, we decompose it into two tables.
TEACHER_ID TEACHER_AGE
25 30
47 35
83 38
TEACHER_ID SUBJECT
25 Chemistry
25 Biology
47 English
83 Math
83 Computer

21. 21
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).
STUD_NO STUD_NAME STUD_STATE STUD_COUNTRY STUD_AGE
1 A Punjab India 20
2 B WB India 20
3 C WB India 21
In relation STUDENT
FD set: {STUD_NO -> STUD_NAME, STUD_NO -> STUD_STATE, STUD_STATE ->
STUD_COUNTRY, STUD_NO -> STUD_AGE}
Candidate Key: {STUD_NO}

22. 22
Third Normal Form (3NF)
For this relation, 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 as:
STUD_NO STUD_NAME STUD_STATE STUD_AGE
1 A Punjab 20
2 B WB 20
3 C WB 21
STUD_STATE STUD_COUNTRY
Punjab India
WB India

23. 23
Boyce Codd Normal Form
A Relation Scheme ‘R’ is in BCNF if it is in 3NF and for every dependency X→Y, the
determinant must be a super key.
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:
EMP_ID → EMP_COUNTRY
EMP_DEPT → {DEPT_TYPE, EMP_DEPT_NO}
Candidate key: {EMP-ID, EMP-DEPT}

24. 24
Boyce Codd Normal Form
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
EMP_ID EMP_COUNTRY
264 India
264 India
EMP_DEPT
EMP_DEPT DEPT_TYPE EMP_DEPT_NO
Designing D394 283
Testing D394 300
Stores D283 232
Developing D283 549
EMP_DEPT_MAP
EMP_ID EMP_DEP
T
D394 283
D394 300
D283 232
D283 549

25. 25
Fourth Normal Form (4NF)
A relation will be in 4NF if it is in BCNF and has no multi-valued dependency.
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.

26. 26
Fourth Normal Form (4NF)
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
59 Physics
STUDENT_HOBBY
STU_ID HOBBY
21 Dancing
21 Singing
34 Dancing
59 Hockey

27. 27
Fifth Normal Form (5NF)
• A relation is in 5NF if it is in 4NF and not contains any join dependency and
joining should be lossless.
• 5NF is satisfied when all the tables are broken into as many tables as possible in
order to avoid redundancy.
• 5NF is also known as Project-join normal form (PJ/NF).
SUBJECT LECTURER SEMESTER
Computer Anshika Semester 1
Computer John Semester 1
Math John Semester 1
Math Akash Semester 2
Chemistry Praveen Semester 1

28. 28
Fifth Normal Form (5NF)
• 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:

29. 29
Fifth Normal Form (5NF)
SUBJECT LECTURER SEMESTER
Computer Anshika Semester 1
Computer John Semester 1
Math John Semester 1
Math Akash Semester 2
Chemistry Praveen Semester 1
SEMESTER SUBJECT
Semester 1 Computer
Semester 1 Math
Semester 1 Chemistry
Semester 2 Math
SUBJECT LECTURER
Computer Anshika
Computer John
Math John
Math Akash
Chemistry Praveen
SEMSTER LECTURER
Semester 1 Anshika
Semester 1 John
Semester 2 Akash
Semester 1 Praveen