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Thursday, 21 April 2022

4NF and 5NF

Fourth normal form (4NF)

Multivalue dependency occurs when two attributes (or columns) in a table are independent of one another, but both depend on a third attribute.

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 multivalued dependency.

  • 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

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.

Decompose above table into two tables to make it in 4NF:

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)

Join dependency is a constraint which is similar to functional dependency or multivalued dependency. 

A relation that has join dependency cannot be divided into two(or more) relations such that the resulting tables can be combined to form the original table.

A relation is in 5NF if it is in 4NF and do not contains any join dependency and joining should be lossless.

5NF is also known as Project-join normal form (PJ/NF).

consider example:

SUBJECT

LECTURER

SEMESTER

Computer

Anshika

Semester 1

Computer

John

Semester 1

Math

John

Semester 1

Math

Akash

Semester 2

Chemistry

Praveen

Semester 1

 

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

 

The natural Join of all the three relations will give original relation. So original relation has no join dependency. Three decomposed relations are in 4NF and have no join dependency.


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