Skip to content
Related Articles
Get the best out of our app
GeeksforGeeks App
Open App

Related Articles

Normal Forms in DBMS

Improve Article
Save Article
Like Article
Improve Article
Save Article
Like Article

Prerequisite – Database normalization and functional dependency concept. Normalization is the process of minimizing redundancy from a relation or set of relations. Redundancy in relation may cause insertion, deletion, and update anomalies. So, it helps to minimize the redundancy in relations. Normal forms are used to eliminate or reduce redundancy in database tables.


In database management systems (DBMS), normal forms are a series of guidelines that help to ensure that the design of a database is efficient, organized, and free from data anomalies. There are several levels of normalization, each with its own set of guidelines, known as normal forms.

Here are the important points regarding normal forms in DBMS:

  1. First Normal Form (1NF): This is the most basic level of normalization. In 1NF, each table cell should contain only a single value, and each column should have a unique name. The first normal form helps to eliminate duplicate data and simplify queries.
  2. Second Normal Form (2NF): 2NF eliminates redundant data by requiring that each non-key attribute be dependent on the primary key. This means that each column should be directly related to the primary key, and not to other columns.
  3. Third Normal Form (3NF): 3NF builds on 2NF by requiring that all non-key attributes are independent of each other. This means that each column should be directly related to the primary key, and not to any other columns in the same table.
  4. Boyce-Codd Normal Form (BCNF): BCNF is a stricter form of 3NF that ensures that each determinant in a table is a candidate key. In other words, BCNF ensures that each non-key attribute is dependent only on the candidate key.
  5. Fourth Normal Form (4NF): 4NF is a further refinement of BCNF that ensures that a table does not contain any multi-valued dependencies.
  6. Fifth Normal Form (5NF): 5NF is the highest level of normalization and involves decomposing a table into smaller tables to remove data redundancy and improve data integrity.

Normal forms help to reduce data redundancy, increase data consistency, and improve database performance. However, higher levels of normalization can lead to more complex database designs and queries. It is important to strike a balance between normalization and practicality when designing a database

The advantages of using normal forms in DBMS include:

  • Reduced data redundancy: Normalization helps to eliminate duplicate data in tables, reducing the amount of storage space needed and improving database efficiency.
  • Improved data consistency: Normalization ensures that data is stored in a consistent and organized manner, reducing the risk of data inconsistencies and errors.
  • Simplified database design: Normalization provides guidelines for organizing tables and data relationships, making it easier to design and maintain a database.
  • Improved query performance: Normalized tables are typically easier to search and retrieve data from, resulting in faster query performance.
  • Easier database maintenance: Normalization reduces the complexity of a database by breaking it down into smaller, more manageable tables, making it easier to add, modify, and delete data.

Overall, using normal forms in DBMS helps to improve data quality, increase database efficiency, and simplify database design and maintenance.

1. First Normal Form –

If a relation contain composite or multi-valued attribute, it violates first normal form or a relation is in first normal form if it does not contain any composite or multi-valued attribute. A relation is in first normal form if every attribute in that relation is singled valued attribute.

  • Example 1 – Relation STUDENT in table 1 is not in 1NF because of multi-valued attribute STUD_PHONE. Its decomposition into 1NF has been shown in table 2.
  • Example 2 –
ID   Name   Courses
1    A      c1, c2
2    E      c3
3    M      C2, c3
  • In the above table Course is a multi-valued attribute so it is not in 1NF. Below Table is in 1NF as there is no multi-valued attribute
ID   Name   Course
1    A       c1
1    A       c2
2    E       c3
3    M       c2
3    M       c3

2. Second Normal Form –

To be in second normal form, a relation must be in first normal form and relation must not contain any partial dependency. A relation is in 2NF if it has No Partial Dependency, i.e., no non-prime attribute (attributes which are not part of any candidate key) is dependent on any proper subset of any candidate key of the table. Partial Dependency – If the proper subset of candidate key determines non-prime attribute, it is called partial dependency.

  • Example 1 – Consider table-3 as following below.
STUD_NO            COURSE_NO        COURSE_FEE
1                     C1                  1000
2                     C2                  1500
1                     C4                  2000
4                     C3                  1000
4                     C1                  1000
2                     C5                  2000
  • {Note that, there are many courses having the same course fee. } Here, COURSE_FEE cannot alone decide the value of COURSE_NO or STUD_NO; COURSE_FEE together with STUD_NO cannot decide the value of COURSE_NO; COURSE_FEE together with COURSE_NO cannot decide the value of STUD_NO; Hence, COURSE_FEE would be a non-prime attribute, as it does not belong to the one only candidate key {STUD_NO, COURSE_NO} ; But, COURSE_NO -> COURSE_FEE, i.e., COURSE_FEE is dependent on COURSE_NO, which is a proper subset of the candidate key. Non-prime attribute COURSE_FEE is dependent on a proper subset of the candidate key, which is a partial dependency and so this relation is not in 2NF. To convert the above relation to 2NF, we need to split the table into two tables such as : Table 1: STUD_NO, COURSE_NO Table 2: COURSE_NO, COURSE_FEE
       Table 1                                    Table 2
STUD_NO            COURSE_NO          COURSE_NO                COURSE_FEE     
1                 C1                  C1                        1000
2                 C2                  C2                        1500
1                 C4                  C3                        1000
4                 C3                  C4                        2000
4                 C1                  C5                        2000        
  • 2 C5 NOTE: 2NF tries to reduce the redundant data getting stored in memory. For instance, if there are 100 students taking C1 course, we don’t need to store its Fee as 1000 for all the 100 records, instead, once we can store it in the second table as the course fee for C1 is 1000.
  • Example 2 – Consider following functional dependencies in relation  R (A,  B , C,  D )
AB -> C  [A and B together determine C]
BC -> D  [B and C together determine D]
  • In the above relation, AB is the only candidate key and there is no partial dependency, i.e., any proper subset of AB doesn’t determine any non-prime attribute.
    1. X is a super key.
    2. Y is a prime attribute (each element of Y is part of some candidate key).
  • Example 1 – In relation STUDENT given in Table 4, FD set: {STUD_NO -> STUD_NAME, STUD_NO -> STUD_STATE, STUD_STATE -> STUD_COUNTRY, STUD_NO -> STUD_AGE} Candidate Key: {STUD_NO} For this relation in table 4, STUD_NO -> STUD_STATE and STUD_STATE -> STUD_COUNTRY are true. So STUD_COUNTRY is transitively dependent on STUD_NO. It violates the third normal form. To convert it in third normal form, we will decompose the relation STUDENT (STUD_NO, STUD_NAME, STUD_PHONE, STUD_STATE, STUD_COUNTRY_STUD_AGE) as: STUDENT (STUD_NO, STUD_NAME, STUD_PHONE, STUD_STATE, STUD_AGE) STATE_COUNTRY (STATE, COUNTRY)
  • Example 2 – Consider relation R(A, B, C, D, E) A -> BC, CD -> E, B -> D, E -> A All possible candidate keys in above relation are {A, E, CD, BC} All attributes are on right sides of all functional dependencies are prime.
    • Example 1 – Find the highest normal form of a relation R(A,B,C,D,E) with FD set as {BC->D, AC->BE, B->E} Step 1. As we can see, (AC)+ ={A,C,B,E,D} but none of its subset can determine all attribute of relation, So AC will be candidate key. A or C can’t be derived from any other attribute of the relation, so there will be only 1 candidate key {AC}. Step 2. Prime attributes are those attributes that are part of candidate key {A, C} in this example and others will be non-prime {B, D, E} in this example. Step 3. The relation R is in 1st normal form as a relational DBMS does not allow multi-valued or composite attribute. The relation is in 2nd normal form because BC->D is in 2nd normal form (BC is not a proper subset of candidate key AC) and AC->BE is in 2nd normal form (AC is candidate key) and B->E is in 2nd normal form (B is not a proper subset of candidate key AC). The relation is not in 3rd normal form because in BC->D (neither BC is a super key nor D is a prime attribute) and in B->E (neither B is a super key nor E is a prime attribute) but to satisfy 3rd normal for, either LHS of an FD should be super key or RHS should be prime attribute. So the highest normal form of relation will be 2nd Normal form.
    • Example 2 –For example consider relation R(A, B, C) A -> BC, B -> A and B both are super keys so above relation is in BCNF.

 Applications of normal forms in DBMS:

Data consistency: Normal forms ensure that data is consistent and does not contain any redundant information. This helps to prevent inconsistencies and errors in the database.

Data redundancy: Normal forms minimize data redundancy by organizing data into tables that contain only unique data. This reduces the amount of storage space required for the database and makes it easier to manage.

Query performance: Normal forms can improve query performance by reducing the number of joins required to retrieve data. This helps to speed up query processing and improve overall system performance.

Database maintenance: Normal forms make it easier to maintain the database by reducing the amount of redundant data that needs to be updated, deleted, or modified. This helps to improve database management and reduce the risk of errors or inconsistencies.

Database design: Normal forms provide guidelines for designing databases that are efficient, flexible, and scalable. This helps to ensure that the database can be easily modified, updated, or expanded as needed.

  1. BCNF is free from redundancy.
  2. If a relation is in BCNF, then 3NF is also satisfied.
  3.  If all attributes of relation are prime attribute, then the relation is always in 3NF.
  4. A relation in a Relational Database is always and at least in 1NF form.
  5. Every Binary Relation ( a Relation with only 2 attributes ) is always in BCNF.
  6. If a Relation has only singleton candidate keys( i.e. every candidate key consists of only 1 attribute), then the Relation is always in 2NF( because no Partial functional dependency possible).
  7. Sometimes going for BCNF form may not preserve functional dependency. In that case go for BCNF only if the lost FD(s) is not required, else normalize till 3NF only.
  8. There are many more Normal forms that exist after BCNF, like 4NF and more. But in real world database systems it’s generally not required to go beyond BCNF.
  9. GATE CS 2012, Question 2
  10. GATE CS 2013, Question 54
  11. GATE CS 2013, Question 55
  12. GATE CS 2005, Question 29
  13. GATE CS 2002, Question 23
  14. GATE CS 2002, Question 50
  15. GATE CS 2001, Question 48
  16. GATE CS 1999, Question 32
  17. GATE IT 2005, Question 22
  18. GATE IT 2008, Question 60
  19. GATE CS 2016 (Set 1), Question 31

My Personal Notes arrow_drop_up
Last Updated : 08 May, 2023
Like Article
Save Article
Similar Reads