DBMS UNIT-3 part 2

Relational Database Design:

Features of Good Relational Designs,
Atomic Domains and First Normal Form,
Decomposition Using Functional Dependencies,
Decomposition Using Multi valued Dependencies, More Normal Forms
Features of Good Relational
Designs
• The goal of the relational database design is to generate a set of relations
having
•
1) Minimum redundancy : redundancy means storage of same data in more
than one location
• or 
• we can say  that same data are repeated in multiple tuples in the same relation
or multiple relations
•
disadvantage : 
• 1)wastage of storage space
• 2)insertion , deletion and modification anamolies 
•
so relational DB should be designed in such way that it should have
• minimum redundancy.
•
Features of Good Relational
Designs
• 2)lesser null values in tuple:
• it may happen in relation the values  of the attributes in the tuple are not
known
• for those values null values are stored.
•
Disadvantages
• 1)it  waste storage space
• 2)difficult to intrepret them
• 3)difficult in applying the aggregate function
• So in good database design the attributes whose value frequently be  null
are avoided
•
Decomposition: 
• Decomposition: 
•
1)In a database, it breaks the table into multiple tables. 
• 2)When a relation in the relational model is not in
• appropriate normal form then the decomposition of a relation is required.
• 3)If the relation has no proper decomposition,
• then it may lead to problems like loss of information.
• 4) Decomposition is used to eliminate some of the problems
• of bad design like anomalies, inconsistencies, and redundancy.
•
Decomposition is divided into two types
•
1) lossless decomposition.
• 2)dependency preserving.
•
Lossless Decomposition:

• Lossless Decomposition:
• 1)If the information is not lost from the relation that is decomposed,
• then the decomposition will be lossless.
• 2)The lossless decomposition guarantees that the join of relations
• will result in the same relation as it was decomposed.
• 3)The relation is said to be lossless decomposition
• if natural joins of all the decomposition give the original relation.
•
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
•
 Lossless Decomposition:
The above relation is decomposed into two relations EMPLOYEE and DEPARTMENT
•1) 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
•
2) DEPARTMENT table
•
DEPT_ID EMP_ID DEPT_NAME
•827               22     Sales
•438               33 Marketing
•869               46 Finance
•575               52 Production
•678 60 Testing
•
Lossless Decomposition:
• 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:
•
2)Dependency Preserving:
•
1)It is an important constraint of the database.
• 2)In the dependency preservation, at least one decomposed table must satisfy every dependency.
• 3)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.
• 4)For example, suppose there is a relation R (A, B, C, D)
• with functional dependency set (A->BC)
• The relational R is decomposed into into R1(ABC) and R2(AD)
• which is dependency preserving because FD A->BC is a part of relation R1(ABC).
•
2) R (A, B, C, D,E)
• FD (A->BC)
•
i) R1:ADE R2:ABC
• Ii) R1:ABC R2:AD R3:AE
•
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 the undesirable characteristics like Insertion,
Update and Deletion Anomalies.
Normalization divides the larger table into the smaller table and links them
using relationship.
Types of Normal Forms

 There are the mainly four types of normal forms:

Types of Normal Forms
Normal Form Description
1NF A relation is in 1NF if it contains atomic values.

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. No partial dependencies

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

dependency exists.

4NF A relation will be in 4NF if it is in Boyce Codd normal form

and has no multi-valued dependency.

5NF A relation is in 5NF if it is in 4NF and not contains any join

dependency and joining should be lossless
First Normal Form (1NF)
 A relation will be 1NF if it contains atomic values.
 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.
EMP_ID EMP_NAME EMP_PHONE EMP_STATE

14 John 7272826385, UP
9064738238

20 Harry 8574783832 Bihar

12 Sam 7390372389, Punjab

8589830302
First Normal Form (1NF)

• 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

Functional Dependency

 The functional dependency is a relationship that exists between

two attributes. It typically exists between the primary key an non-
key attribute within a table
X   →   Y  
 The left side of FD is known as a determinant, the right side of
the FD is known as a dependent
For example:
 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:
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 is a trivial functional dependency if B is a subset of A.
 The following dependencies are also trivial like: A → A, B → B

Example: Consider a table with two columns Employee_Id and Employee_Name.  
 {Employee_id, Employee_Name}   →    Employee_Id is a trivial functional depende
ncy as  Employee_Id is a subset of {Employee_Id, Employee_Name}.  
 Also, Employee_Id → Employee_Id  and 
Employee_Name   →    Employee_Name are trivial dependencies too
Types of Functional dependency

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:
• ID →    Name
• Name   →    DOB  
Second Normal Form (2NF)
• In the 2NF, relation must be in 1NF.
• There should not be any partial dependencies (All non-key attributes
are fully functional dependent on the primary key )
Example: 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
Second Normal Form (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 if it doesn’t
contain any transitive dependency(non-prime attribute
should not determine non-prime attribute)
 3NF is used to reduce the data duplication. It is also used to
achieve the data integrity.
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

Third Normal Form (3NF)
 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.
Third Normal Form (3NF)
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 said to be 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
Boyce Codd normal form (BCNF)

Functional dependencies are as follows:

• EMP_ID  →  EMP_COUNTRY  
• 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.
Boyce Codd normal form (BCNF)
To convert the given table into BCNF, we decompose it into three tables:

EMP_COUNTRY table:

EMP_ID EMP_COUNTRY
264 India
364 UK

EMP_DEPT table: EMP_DEPT_MAPPING table:

EMP_DEPT DEPT_TYPE EMP_DEPT_NO EMP_ID EMP_DEPT

Designing D394 283 D394 283
Testing D394 300 D394 300
Stores D283 232 D283 232
Developing D283 549 D283 549
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
value of B exists, then the table may have multi-valued
dependency. The table should have at least 3 attributes and B
and C should be independent for A ->> B multi-valued
dependency  
STUDENT

STU_ID COURSE HOBBY

The given STUDENT table is
21 Computer Dancing in 3NF, but the COURSE and
21 Math Singing HOBBY are two independent
entity. Hence, there is no
34 Chemistry Dancing
relationship between
74 Biology Cricket COURSE and HOBBY
59 Physics Hockey
Fourth normal form (4NF)
• 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

STUDENT_COURS STUDENT_HOBBY
E
STU_ID COURSE STU_ID HOBBY
21 Computer 21 Dancing
21 Math 21 Singing
34 Chemistry 34 Dancing
74 Biology 74 Cricket
59 Physics 59 Hockey
Fifth normal form (5NF)
• A relation is in 5NF, if it is in 4NF and doesn’t not contain 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

• 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.
Fifth normal form (5NF)
• So to make the above table into 5NF, we can decompose it into
three relations P1, P2 & P3:

P2 SUBJECT LECTURER
P1 SEMESTER SUBJECT
Computer Anshika
Semester 1 Computer
Computer John
Semester 1 Math
Math John
Semester 1 Chemistry
Math Akash
Semester 2 Math
Chemistry Praveen
SEMSTER LECTURER
P3 Semester 1 Anshika
Semester 1 John
Semester 1 John
Semester 2 Akash
Semester 1 Praveen
Thank you