Functional Dependency

The functional dependency is a relationship that exists between two

attributes. It typically exists between the primary key and non-key
attribute within a table.

1. 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.

 Types of Functional dependency

1. Trivial functional dependency

o A → B has trivial functional dependency if B is a subset of A.
o The following dependencies are also trivial like: A → A, B → B

Example:

1. Consider a table with two columns Employee_Id and Employee_Name.

2. {Employee_id, Employee_Name} → Employee_Id is a trivial function

al dependency as
3. Employee_Id is a subset of {Employee_Id, Employee_Name}.
4. Also, Employee_Id → Employee_Id and Employee_Name → Employe
e_Name are trivial dependencies too.
 2. Non-trivial functional dependency
o A → B has a non-trivial functional dependency if B is not a subset
of A.
o When A intersection B is NULL, then A → B is called as complete
non-trivial.

Example:

1. ID → Name,
2. Name → DOB

Inference Rule (IR):

o The Armstrong's axioms are the basic inference rule.
o Armstrong's axioms are used to conclude functional
dependencies on a relational database.
o The inference rule is a type of assertion. It can apply to a set of
FD(functional dependency) to derive other FD.
o 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)
Normalization
A large database defined as a single relation may result in data
duplication. This repetition of data may result in:

o Making relations very large.

o It isn't easy to maintain and update data as it would involve
searching many records in relation.
o Wastage and poor utilization of disk space and resources.
o The likelihood of errors and inconsistencies increases.

So to handle these problems, we should analyze and decompose the

relations with redundant data into smaller, simpler, and well-
structured relations that are satisfy desirable properties. Normalization
is a process of decomposing the relations into relations with fewer
attributes.

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

Why do we need Normalization?

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

Data modification anomalies can be categorized into three types:

o Insertion Anomaly: Insertion Anomaly refers to when one

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

Types of Normal Forms:

Normalization works through a series of stages called Normal forms.
The normal forms apply to individual relations. The relation is said to
be in particular normal form if it satisfies constraints.

Following are the various types of Normal forms:

Normal Description
Form

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

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

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

BCNF A stronger definition of 3NF is known as Boyce Codd's normal form.

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

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

Advantages of Normalization
o Normalization helps to minimize data redundancy.
o Greater overall database organization.
o Data consistency within the database.
o Much more flexible database design.
o Enforces the concept of relational integrity.

Disadvantages of Normalization
o You cannot start building the database before knowing what the
user needs.
o The performance degrades when normalizing the relations to
higher normal forms, i.e., 4NF, 5NF.
o It is very time-consuming and difficult to normalize relations of a
higher degree.
o Careless decomposition may lead to a bad database design,
leading to serious problems.
First Normal Form (1NF)
o A relation will be 1NF if it contains an atomic value.
o It states that an attribute of a table cannot hold multiple values.
It must hold only single-valued attribute.
o First normal form disallows the multi-valued attribute, composite
attribute, and their combinations.

Example: Relation EMPLOYEE is not in 1NF because of multi-valued

attribute EMP_PHONE.

EMPLOYEE table:

EMP_ID EMP_NAME EMP_PHONE EMP

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

14 John 7272826385 UP

14 John 9064738238 UP
20 Harry 8574783832 Bihar

12 Sam 7390372389 Punjab

12 Sam 8589830302 Punjab

Second Normal Form (2NF)

o In the 2NF, relational must be in 1NF.
o In the second normal form, all non-key attributes are fully
functional dependent on the primary key

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

TEACHER table

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

A relation is in third normal form if it holds atleast one of the following

conditions for every non-trivial function dependency X → Y.

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

Example:

EMPLOYEE_DETAIL table:

EMP_ID EMP_NAME EMP_ZIP EMP_STATE

222 Harry 201010 UP

333 Stephan 02228 US

444 Lan 60007 US

555 Katharine 06389 UK

666 John 462007 MP

 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_ZI

222 Harry 201010

333 Stephan 02228

444 Lan 60007

555 Katharine 06389

666 John 462007

EMPLOYEE_ZIP table:
 EMP_ZIP EMP_STATE EMP_CIT

201010 UP Noida

02228 US Boston

60007 US Chicago

06389 UK Norwich

462007 MP Bhopal

Boyce Codd normal form (BCNF)

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

Example: Let's assume there is a company where employees work in

more than one department.

EMP_ID EMP_COUNTRY EMP_DEPT DEPT_TYPE EMP_DEPT

264 India Designing D394 283

264 India Testing D394 300

364 UK Stores D283 232

364 UK Developing D283 549

EMPLOYEE table:

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

Example
STUDENT

STU_ID COURSE HOBBY

21 Computer Dancing

21 Math Singing

34 Chemistry Dancing

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)

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

Example
SUBJECT LECTURER SEMESTER

Computer Anshika Semester 1

Computer John Semester 1

Math John Semester 1

Math Akash Semester 2

Chemistry Praveen Semester 1

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

Suppose we add a new Semester as Semester 3 but do not know

about the subject and who will be taking that subject so we leave
Lecturer and Subject as NULL. But all three columns together acts as a
primary key, so we can't leave other two columns blank.

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

P1

SEMESTER SUBJECT

Semester 1 Computer

Semester 1 Math

Semester 1 Chemistry
Semester 2 Math

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

Relational Decomposition
o When a relation in the relational model is not in appropriate
normal form then the decomposition of a relation is required.
o In a database, it breaks the table into multiple tables.
o If the relation has no proper decomposition, then it may lead to
problems like loss of information.
o Decomposition is used to eliminate some of the problems of bad
design like anomalies, inconsistencies, and redundancy.
Types of Decomposition

Lossless Decomposition
o If the information is not lost from the relation that is
decomposed, then the decomposition will be lossless.
o The lossless decomposition guarantees that the join of relations
will result in the same relation as it was decomposed.
o The relation is said to be lossless decomposition if natural joins
of all the decomposition give the original relation.

Example:

EMPLOYEE_DEPARTMENT table:

EMP_ID EMP_NAME EMP_AGE EMP_CITY DEPT_ID

22 Denim 28 Mumbai 827

33 Alina 25 Delhi 438

46 Stephan 30 Bangalore 869

52 Katherine 36 Mumbai 575

60 Jack 40 Noida 678

The above relation is decomposed into two relations EMPLOYEE and

DEPARTMENT

EMPLOYEE table:

EMP_ID EMP_NAME EMP_AGE EMP

22 Denim 28 Mum

33 Alina 25 Delh

46 Stephan 30 Bang

52 Katherine 36 Mum

60 Jack 40 Noid

DEPARTMENT table
DEPT_ID EMP_ID DEPT_NAME

827 22 Sales

438 33 Marketing

869 46 Finance

575 52 Production

678 60 Testing

Now, when these two relations are joined on the common column
"EMP_ID", then the resultant relation will look like:

Employee ⋈ Department

EMP_ID EMP_NAME EMP_AGE EMP_CITY DEPT_ID

22 Denim 28 Mumbai 827

33 Alina 25 Delhi 438

46 Stephan 30 Bangalore 869

52 Katherine 36 Mumbai 575

60 Jack 40 Noida 678

Hence, the decomposition is Lossless join decomposition.

Dependency Preserving
o It is an important constraint of the database.
o In the dependency preservation, at least one decomposed table
must satisfy every dependency.
o If a relation R is decomposed into relation R1 and R2, then the
dependencies of R either must be a part of R1 or R2 or must be
derivable from the combination of functional dependencies of R1
and R2.
o For example, suppose there is a relation R (A, B, C, D) with
functional dependency set (A->BC). The relational R is
decomposed into R1(ABC) and R2(AD) which is dependency
preserving because FD A->BC is a part of relation R1(ABC).

Multivalued Dependency
o Multivalued dependency occurs when two attributes in a table
are independent of each other but, both depend on a third
attribute.
 o A multivalued dependency consists of at least two attributes that
are dependent on a third attribute that's why it always requires at
least three attributes.

Example: Suppose there is a bike manufacturer company which

produces two colors(white and black) of each model every year.

BIKE_MODEL MANUF_YEAR C

M2011 2008 W

M2001 2008 Bla

M3001 2013 W

M3001 2013 Bla

M4006 2017 W

M4006 2017 Bla

Here columns COLOR and MANUF_YEAR are dependent on

BIKE_MODEL and independent of each other.

In this case, these two columns can be called as multivalued

dependent on BIKE_MODEL. The representation of these dependencies
is shown below:

1. BIKE_MODEL → → MANUF_YEAR
2. BIKE_MODEL → → COLOR

This can be read as "BIKE_MODEL multidetermined MANUF_YEAR" and

"BIKE_MODEL multidetermined COLOR".
Join Dependency
o Join decomposition is a further generalization of Multivalued
dependencies.
o If the join of R1 and R2 over C is equal to relation R, then we can
say that a join dependency (JD) exists.
o Where R1 and R2 are the decompositions R1(A, B, C) and R2(C,
D) of a given relations R (A, B, C, D).
o Alternatively, R1 and R2 are a lossless decomposition of R.
o A JD ⋈ {R1, R2,..., Rn} is said to hold over a relation R if R1, R2,.....,
Rn is a lossless-join decomposition.
o The *(A, B, C, D), (C, D) will be a JD of R if the join of join's
attribute is equal to the relation R.
o Here, *(R1, R2, R3) is used to indicate that relation R1, R2, R3 and
so on are a JD of R.
Inclusion Dependency
o Multivalued dependency and join dependency can be used to
guide database design although they both are less common than
functional dependencies.
o Inclusion dependencies are quite common. They typically show
little influence on designing of the database.
o The inclusion dependency is a statement in which some columns
of a relation are contained in other columns.
o The example of inclusion dependency is a foreign key. In one
relation, the referring relation is contained in the primary key
column(s) of the referenced relation.
o Suppose we have two relations R and S which was obtained by
translating two entity sets such that every R entity is also an S
entity.
o Inclusion dependency would be happen if projecting R on its key
attributes yields a relation that is contained in the relation
obtained by projecting S on its key attributes.
o In inclusion dependency, we should not split groups of attributes
that participate in an inclusion dependency.
o In practice, most inclusion dependencies are key-based that is
involved only keys.
Canonical Cover
In the case of updating the database, the responsibility of the system
is to check whether the existing functional dependencies are getting
violated during the process of updating. In case of a violation of
functional dependencies in the new database state, the rollback of the
system must take place.

A canonical cover or irreducible a set of functional dependencies FD is

a simplified set of FD that has a similar closure as the original set FD.

Extraneous attributes
An attribute of an FD is said to be extraneous if we can remove it
without changing the closure of the set of FD.

Example: Given a relational Schema R( A, B, C, D) and set of Function

Dependency FD = { B → A, AD → BC, C → ABD }. Find the canonical
cover?

Solution: Given FD = { B → A, AD → BC, C → ABD }, now decompose

the FD using decomposition rule( Armstrong Axiom ).

1. B → A
2. AD → B ( using decomposition inference rule on AD → BC)
3. AD → C ( using decomposition inference rule on AD → BC)
4. C → A ( using decomposition inference rule on C → ABD)
5. C → B ( using decomposition inference rule on C → ABD)
6. C → D ( using decomposition inference rule on C → ABD)

Now set of FD = { B → A, AD → B, AD → C, C → A, C → B, C → D }
The next step is to find closure of the left side of each of the given FD
by including that FD and excluding that FD, if closure in both cases are
same then that FD is redundant and we remove that FD from the given
set, otherwise if both the closures are different then we do not exclude
that FD.

Calculating closure of all FD { B → A, AD → B, AD → C, C → A, C →

B, C → D }

1a. Closure B+ = BA using FD = { B → A, AD → B, AD → C, C → A, C →

B, C → D }

1b. Closure B+ = B using FD = { AD → B, AD → C, C → A, C → B, C → D

}

From 1 a and 1 b, we found that both the Closure( by including B →

A and excluding B → A ) are not equivalent, hence FD B → A is
important and cannot be removed from the set of FD.

2 a. Closure AD+ = ADBC using FD = { B →A, AD → B, AD → C, C → A,

C → B, C → D }

2 b. Closure AD+ = ADCB using FD = { B → A, AD → C, C → A, C → B, C

→D}

From 2 a and 2 b, we found that both the Closure (by including AD →

B and excluding AD → B) are equivalent, hence FD AD → B is not
important and can be removed from the set of FD.

Hence resultant FD = { B → A, AD → C, C → A, C → B, C → D }

3 a. Closure AD+ = ADCB using FD = { B →A, AD → C, C → A, C → B, C

→D}

3 b. Closure AD+ = AD using FD = { B → A, C → A, C → B, C → D }

From 3 a and 3 b, we found that both the Closure (by including AD →
C and excluding AD → C ) are not equivalent, hence FD AD → C is
important and cannot be removed from the set of FD.

Hence resultant FD = { B → A, AD → C, C → A, C → B, C → D }

4 a. Closure C+ = CABD using FD = { B →A, AD → C, C → A, C → B, C →

D}

4 b. Closure C+ = CBDA using FD = { B → A, AD → C, C → B, C → D }

From 4 a and 4 b, we found that both the Closure (by including C →

A and excluding C → A) are equivalent, hence FD C → A is not
important and can be removed from the set of FD.

Hence resultant FD = { B → A, AD → C, C → B, C → D }

5 a. Closure C+ = CBDA using FD = { B →A, AD → C, C → B, C → D }

5 b. Closure C+ = CD using FD = { B → A, AD → C, C → D }

From 5 a and 5 b, we found that both the Closure (by including C →

B and excluding C → B) are not equivalent, hence FD C → B is
important and cannot be removed from the set of FD.

Hence resultant FD = { B → A, AD → C, C → B, C → D }

6 a. Closure C+ = CDBA using FD = { B →A, AD → C, C → B, C → D }

6 b. Closure C+ = CBA using FD = { B → A, AD → C, C → B }

From 6 a and 6 b, we found that both the Closure( by including C →

D and excluding C → D) are not equivalent, hence FD C → D is
important and cannot be removed from the set of FD.

Hence resultant FD = { B → A, AD → C, C → B, C → D }
 o Since FD = { B → A, AD → C, C → B, C → D } is resultant FD, now
we have checked the redundancy of attribute, since the left side
of FD AD → C has two attributes, let's check their importance, i.e.
whether they both are important or only one.

Closure AD+ = ADCB using FD = { B →A, AD → C, C → B, C → D }

Closure A+ = A using FD = { B →A, AD → C, C → B, C → D }

Closure D+ = D using FD = { B →A, AD → C, C → B, C → D }

Since the closure of AD+, A+, D+ that we found are not all equivalent,
hence in FD AD → C, both A and D are important attributes and
cannot be removed.

Hence resultant FD = { B → A, AD → C, C → B, C → D } and we can

rewrite as

FD = { B → A, AD → C, C → BD } is Canonical Cover of FD = { B → A,
AD → BC, C → ABD }.

Example 2: Given a relational Schema R( W, X, Y, Z) and set of

Function Dependency FD = { W → X, Y → X, Z → WXY, WY → Z }. Find
the canonical cover?

Solution: Given FD = { W → X, Y → X, Z → WXY, WY → Z }, now

decompose the FD using decomposition rule( Armstrong Axiom ).

1. W → X
2. Y → X
3. Z → W ( using decomposition inference rule on Z → WXY )
4. Z → X ( using decomposition inference rule on Z → WXY )
 5. Z → Y ( using decomposition inference rule on Z → WXY )
6. WY → Z

Now set of FD = { W → X, Y → X, WY → Z, Z → W, Z → X, Z → Y }

The next step is to find closure of the left side of each of the given FD
by including that FD and excluding that FD, if closure in both cases are
same then that FD is redundant and we remove that FD from the given
set, otherwise if both the closures are different then we do not exclude
that FD.

Calculating closure of all FD { W → X, Y → X, Z → W, Z → X, Z → Y,

WY → Z }

1 a. Closure W+ = WX using FD = { W → X, Y → X, Z → W, Z → X, Z →
Y, WY → Z }

1 b. Closure W+ = W using FD = { Y → X, Z → W, Z → X, Z → Y, WY →
Z}

From 1 a and 1 b, we found that both the Closure (by including W →

X and excluding W → X ) are not equivalent, hence FD W → X is
important and cannot be removed from the set of FD.

Hence resultant FD = { W → X, Y → X, Z → W, Z → X, Z → Y, WY →
Z}

2 a. Closure Y+ = YX using FD = { W → X, Y → X, Z → W, Z → X, Z → Y,
WY → Z }

2 b. Closure Y+ = Y using FD = { W → X, Z → W, Z → X, Z → Y, WY → Z
}
From 2 a and 2 b we found that both the Closure (by including Y →
X and excluding Y → X ) are not equivalent, hence FD Y → X is
important and cannot be removed from the set of FD.

Hence resultant FD = { W → X, Y → X, Z → W, Z → X, Z → Y, WY →
Z}

3 a. Closure Z+ = ZWXY using FD = { W → X, Y → X, Z → W, Z → X, Z

→ Y, WY → Z }

3 b. Closure Z+ = ZXY using FD = { W → X, Y → X, Z → X, Z → Y, WY →

Z}

From 3 a and 3 b, we found that both the Closure (by including Z →

W and excluding Z → W ) are not equivalent, hence FD Z → W is
important and cannot be removed from the set of FD.

Hence resultant FD = { W → X, Y → X, Z → W, Z → X, Z → Y, WY →
Z}

4 a. Closure Z+ = ZXWY using FD = { W → X, Y → X, Z → W, Z → X, Z

→ Y, WY → Z }

4 b. Closure Z+ = ZWYX using FD = { W → X, Y → X, Z → W, Z → Y, WY

→Z}

From 4 a and 4 b, we found that both the Closure (by including Z →

X and excluding Z → X ) are equivalent, hence FD Z → X
is not important and can be removed from the set of FD.

Hence resultant FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

5 a. Closure Z+ = ZYWX using FD = { W → X, Y → X, Z → W, Z → Y, WY

→Z}
5 b. Closure Z+ = ZWX using FD = { W → X, Y → X, Z → W, WY → Z }

From 5 a and 5 b, we found that both the Closure (by including Z →

Y and excluding Z → Y ) are not equivalent, hence FD Z → X is
important and cannot be removed from the set of FD.

Hence resultant FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

6 a. Closure WY+ = WYZX using FD = { W → X, Y → X, Z → W, Z →

Y, WY → Z }

6 b. Closure WY+ = WYX using FD = { W → X, Y → X, Z → W, Z → Y }

From 6 a and 6 b, we found that both the Closure (by including WY →

Z and excluding WY → Z) are not equivalent, hence FD WY → Z is
important and cannot be removed from the set of FD.

Hence resultant FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

Since FD = { W → X, Y → X, Z → W, Z → Y, WY → Z } is resultant FD
now, we have checked the redundancy of attribute, since the left side
of FD WY → Z has two attributes at its left, let's check their
importance, i.e. whether they both are important or only one.

Closure WY+ = WYZX using FD = { W → X, Y → X, Z → W, Z → Y, WY →

Z}

Closure W+ = WX using FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

Closure Y+ = YX using FD = { W → X, Y → X, Z → W, Z → Y, WY → Z }

Since the closure of WY+, W+, Y+ that we found are not all equivalent,
hence in FD WY → Z, both W and Y are important attributes and
cannot be removed.
Hence resultant FD = { W → X, Y → X, Z → W, Z → Y, WY → Z } and we
can rewrite as:

FD = { W → X, Y → X, Z → WY, WY → Z } is Canonical Cover of FD =

{ W → X, Y → X, Z → WXY, WY → Z }.

Example 3: Given a relational Schema R( V, W, X, Y, Z) and set of

Function Dependency FD = { V → W, VW → X, Y → VXZ }. Find the
canonical cover?

Solution: Given FD = { V → W, VW → X, Y → VXZ }. now decompose

the FD using decomposition rule (Armstrong Axiom).

1. V → W
2. VW → X
3. Y → V ( using decomposition inference rule on Y → VXZ )
4. Y → X ( using decomposition inference rule on Y → VXZ )
5. Y → Z ( using decomposition inference rule on Y → VXZ )

Now set of FD = { V → W, VW → X, Y → V, Y → X, Y → Z }.

The next step is to find closure of the left side of each of the given FD
by including that FD and excluding that FD, if closure in both cases are
same then that FD is redundant and we remove that FD from the given
set, otherwise if both the closures are different then we do not exclude
that FD.

Calculating closure of all FD { V → W, VW → X, Y → V, Y → X, Y → Z

}.

1 a. Closure V+ = VWX using FD = {V → W, VW → X, Y → V, Y → X, Y

→ Z}
1 b. Closure V+ = V using FD = {VW → X, Y → V, Y → X, Y → Z }

From 1 a and 1 b, we found that both the Closure( by including V →

W and excluding V → W ) are not equivalent, hence FD V → W is
important and cannot be removed from the set of FD.

Hence resultant FD = { V → W, VW → X, Y → V, Y → X, Y → Z }.

2 a. Closure VW+ = VWX using FD = { V → W, VW → X, Y → V, Y → X,

Y→Z}

2 b. Closure VW+ = VW using FD = { V → W, Y → V, Y → X, Y → Z }

From 2 a and 2 b, we found that both the Closure( by including VW →

X and excluding VW → X ) are not equivalent, hence FD VW → X is
important and cannot be removed from the set of FD.

Hence resultant FD = { V → W, VW → X, Y → V, Y → X, Y → Z }.

3 a. Closure Y+ = YVXZW using FD = { V → W, VW → X, Y → V, Y → X,

Y→Z}

3 b. Closure Y+ = YXZ using FD = { V → W, VW → X, Y → X, Y → Z }

From 3 a and 3 b, we found that both the Closure( by including Y →

V and excluding Y → V ) are not equivalent, hence FD Y → V is
important and cannot be removed from the set of FD.

Hence resultant FD = { V → W, VW → X, Y → V, Y → X, Y → Z }.

4 a. Closure Y+ = YXVZW using FD = { V → W, VW → X, Y → V, Y → X,

Y→Z}

4 b. Closure Y+ = YVZWX using FD = { V → W, VW → X, Y → V, Y → Z }

From 4 a and 4 b, we found that both the Closure( by including Y →
X and excluding Y → X ) are equivalent, hence FD Y → X
is not important and can be removed from the set of FD.

Hence resultant FD = { V → W, VW → X, Y → V, Y → Z }.

5 a. Closure Y+ = YZVWX using FD = { V → W, VW → X, Y → V, Y → Z }

5 b. Closure Y+ = YVWX using FD = { V → W, VW → X, Y → V }

From 5 a and 5 b, we found that both the Closure( by including Y →

Z and excluding Y → Z ) are not equivalent, hence FD Y → Z is
important and cannot be removed from the set of FD.

Hence resultant FD = { V → W, VW → X, Y → V, Y → Z }.

Since FD = { V → W, VW → X, Y → V, Y → Z } is resultant FD now, we

have checked the redundancy of attribute, since the left side of FD VW
→ X has two attributes at its left, let's check their importance, i.e.
whether they both are important or only one.

Closure VW+ = VWX using FD = { V → W, VW → X, Y → V, Y → Z }

Closure V+ = VWX using FD = { V → W, VW → X, Y → V, Y → Z }

Closure W+ = W using FD = { V → W, VW → X, Y → V, Y → Z }

Since the closure of VW+, V+, W+ we found that all the Closures of
VW and V are equivalent, hence in FD VW → X, W is not at all an
important attribute and can be removed.

Hence resultant FD = { V → W, V → X, Y → V, Y → Z } and we can

rewrite as

FD = { V → WX, Y → VZ } is Canonical Cover of FD = { V → W, VW

→ X, Y → VXZ }.
CONCLUSION: From the above three examples we conclude that
canonical cover / irreducible set of functional dependency follows the
following steps, which we need to follow while calculating Canonical
Cover.

STEP 1: For a given set of FD, decompose each FD using

decomposition rule (Armstrong Axiom) if the right side of any FD has
more than one attribute.

STEP 2: Now make a new set of FD having all decomposed FD.

STEP 3: Find closure of the left side of each of the given FD by

including that FD and excluding that FD, if closure in both cases are
same then that FD is redundant and we remove that FD from the given
set, otherwise if both the closures are different then we do not exclude
that FD.

STEP 4: Repeat step 4 till all the FDs in FD set are complete.

STEP 5: After STEP 4, find resultant FD = { B → A, AD → C, C → B, C →

D } which are not redundant.

STEP 6: Check redundancy of attribute, by selecting those FD's from

FD sets which are having more than one attribute on its left, let's an
FD AD → C has two attributes at its left, let's check their importance,
i.e. whether they both are important or only one.

STEP 6 a: Find Closure AD+

STEP 6 b: Find Closure A+

STEP 6 c: Find Closure D+

Compare Closure of STEP (6a, 6b, 6c) if the closure of AD+, A+, D+ are
not equivalent, hence in FD AD → C, both A and D are important
attributes and cannot be removed, otherwise, we remove the
redundant attribute.