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Normalization and three normal forms.pptx

Normalization is a process of organizing data in a database to minimize redundancy and eliminate undesirable anomalies like insertion, update, and deletion anomalies. Various normal forms (1NF, 2NF, 3NF, BCNF, 4NF, 5NF) are defined, each addressing specific types of data dependency and redundancy issues. While normalization promotes data consistency and integrity, it can lead to performance degradation and is time-consuming, particularly at higher normal forms.

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Normalization
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.
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.
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.
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.
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: • {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_COUN TRY 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}
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.
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) • 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