UNIT-III Data Base Design & Normalization

 

UNIT-III

Data Base Design & Normalization

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

• A → B has trivial functional dependency if B is a subset of A.
• 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 functional dependency as   

3.      Employee_Id is a subset of {Employee_Id, Employee_Name}.  

4.      Also, Employee_Id → Employee_Id and Employee_Name   →    Employee_Name are trivial dependencies too.  

2. Non-trivial functional dependency

• A → B has a non-trivial functional dependency if B is not a subset of A.
• 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):

• 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)

 

Normalization

A large database defined as a single relation may result in data duplication. This repetition of data may result in:

• Making relations very large.
• It isn't easy to maintain and update data as it would involve searching many records in relation.
• Wastage and poor utilization of disk space and resources.
• 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?

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

Why do we need Normalization?

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

Data modification anomalies can be categorized into three types:

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

Types of Normal Forms:

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

Following are the various types of Normal forms:

Normal Form	Description
1NF	A relation is in 1NF if it contains an atomic value.
2NF	A relation will be in 2NF if it is in 1NF and all non-key attributes are fully functional dependent on the primary key.
3NF	A relation will be in 3NF if it is in 2NF and no transition dependency exists.
BCNF	A stronger definition of 3NF is known as Boyce Codd's normal form.
4NF	A relation will be in 4NF if it is in Boyce Codd's normal form and has no multi-valued dependency.
5NF	A relation is in 5NF. If it is in 4NF and does not contain any join dependency, joining should be lossless.

Advantages of Normalization

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

Disadvantages of Normalization

• You cannot start building the database before knowing what the user needs.
• The performance degrades when normalizing the relations to higher normal forms, i.e., 4NF, 5NF.
• It is very time-consuming and difficult to normalize relations of a higher degree.
• Careless decomposition may lead to a bad database design, leading to serious problems.

First Normal Form (1NF)

• A relation will be 1NF if it contains an atomic value.
• It states that an attribute of a table cannot hold multiple values. It must hold only single-valued attribute.
• First normal form disallows the multi-valued attribute, composite attribute, and their combinations.

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

EMPLOYEE table:

EMP_ID	EMP_NAME	EMP_PHONE	EMP_STATE
14	John	7272826385, 9064738238	UP
20	Harry	8574783832	Bihar
12	Sam	7390372389, 8589830302	Punjab

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

Second Normal Form (2NF)

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

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

TEACHER table

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)

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

A relation is in third normal form if it holds atleast one of the following conditions for every non-trivial function dependency X → Y.

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

Example:

EMPLOYEE_DETAIL table:

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)

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

Example: Let's assume there is a company where employees work in more than one department.

EMPLOYEE table:

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}

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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)

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

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:

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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)

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• 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).

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

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

Types of Decomposition

Lossless Decomposition

• If the information is not lost from the relation that is decomposed, then the decomposition will be lossless.
• The lossless decomposition guarantees that the join of relations will result in the same relation as it was decomposed.
• 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	DEPT_NAME
22	Denim	28	Mumbai	827	Sales
33	Alina	25	Delhi	438	Marketing
46	Stephan	30	Bangalore	869	Finance
52	Katherine	36	Mumbai	575	Production
60	Jack	40	Noida	678	Testing

The above relation is decomposed into two relations EMPLOYEE and DEPARTMENT

EMPLOYEE table:

EMP_ID	EMP_NAME	EMP_AGE	EMP_CITY
22	Denim	28	Mumbai
33	Alina	25	Delhi
46	Stephan	30	Bangalore
52	Katherine	36	Mumbai
60	Jack	40	Noida

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	DEPT_NAME
22	Denim	28	Mumbai	827	Sales
33	Alina	25	Delhi	438	Marketing
46	Stephan	30	Bangalore	869	Finance
52	Katherine	36	Mumbai	575	Production
60	Jack	40	Noida	678	Testing

Hence, the decomposition is Lossless join decomposition.

Dependency Preserving

• It is an important constraint of the database.
• In the dependency preservation, at least one decomposed table must satisfy every dependency.
• 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.
• 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

• Multivalued dependency occurs when two attributes in a table are independent of each other but, both depend on a third attribute.
• 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	COLOR
M2011	2008	White
M2001	2008	Black
M3001	2013	White
M3001	2013	Black
M4006	2017	White
M4006	2017	Black

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

• Join decomposition is a further generalization of Multivalued dependencies.
• If the join of R1 and R2 over C is equal to relation R, then we can say that a join dependency (JD) exists.
• Where R1 and R2 are the decompositions R1(A, B, C) and R2(C, D) of a given relations R (A, B, C, D).
• Alternatively, R1 and R2 are a lossless decomposition of R.
• A JD ⋈ {R1, R2,..., Rn} is said to hold over a relation R if R1, R2,....., Rn is a lossless-join decomposition.
• 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.
• Here, *(R1, R2, R3) is used to indicate that relation R1, R2, R3 and so on are a JD of R.

Inclusion Dependency

• Multivalued dependency and join dependency can be used to guide database design although they both are less common than functional dependencies.
• Inclusion dependencies are quite common. They typically show little influence on designing of the database.
• The inclusion dependency is a statement in which some columns of a relation are contained in other columns.
• 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.
• 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.
• 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.
• In inclusion dependency, we should not split groups of attributes that participate in an inclusion dependency.
• 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 }

• 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.