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Boyce-Codd Normal Form (BCNF)

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Pre-Requisite: First Normal Form, Second Normal Form, Third Normal Form

Application of the general definitions of 2NF and 3NF may identify additional redundancy caused by dependencies that violate one or more candidate keys. However, despite these additional constraints, dependencies can still exist that will cause redundancy to be present in 3NF relations. This weakness in 3NF resulted in the presentation of a stronger normal form called the Boyce-Codd Normal Form (Codd, 1974).

Although, 3NF is an adequate normal form for relational databases, still, this (3NF) normal form may not remove 100% redundancy because of X−>Y functional dependency if X is not a candidate key of the given relation. This can be solved by Boyce-Codd Normal Form (BCNF). 

Boyce-Codd Normal Form (BCNF)

Boyce–Codd Normal Form (BCNF) is based on functional dependencies that take into account all candidate keys in a relation; however, BCNF also has additional constraints compared with the general definition of 3NF.

Rules for BCNF

Rule 1: The table should be in the 3rd Normal Form.

Rule 2: X should be a superkey for every functional dependency (FD) X−>Y in a given relation. 

Note: To test whether a relation is in BCNF, we identify all the determinants and make sure that they are candidate keys.



You came across a similar hierarchy known as the Chomsky Normal Form in the Theory of Computation. Now, carefully study the hierarchy above. It can be inferred that every relation in BCNF is also in 3NF. To put it another way, a relation in 3NF need not be in BCNF. Ponder over this statement for a while. 

To determine the highest normal form of a given relation R with functional dependencies, the first step is to check whether the BCNF condition holds. If R is found to be in BCNF, it can be safely deduced that the relation is also in 3NF, 2NF, and 1NF as the hierarchy shows. The 1NF has the least restrictive constraint – it only requires a relation R to have atomic values in each tuple. The 2NF has a slightly more restrictive constraint. 

The 3NF has a more restrictive constraint than the first two normal forms but is less restrictive than the BCNF. In this manner, the restriction increases as we traverse down the hierarchy. 


Here, we are going to discuss some basic examples which let you understand the properties of BCNF. We will discuss multiple examples here.

Example 1

Let us consider the student database, in which data of the student are mentioned.

101Computer Science & EngineeringDBMSB_001201
101Computer Science & EngineeringComputer NetworksB_001202
102Electronics & Communication EngineeringVLSI TechnologyB_003401
102Electronics & Communication EngineeringMobile CommunicationB_003402

Functional Dependency of the above is as mentioned:

Stu_ID −> Stu_Branch
Stu_Course −> {Branch_Number, Stu_Course_No}

Candidate Keys of the above table are: {Stu_ID, Stu_Course}

Why this Table is Not in BCNF?

The table present above is not in BCNF, because as we can see that neither Stu_ID nor Stu_Course is a Super Key. As the rules mentioned above clearly tell that for a table to be in BCNF, it must follow the property that for functional dependency X−>Y, X must be in Super Key and here this property fails, that’s why this table is not in BCNF.

How to Satisfy BCNF?

For satisfying this table in BCNF, we have to decompose it into further tables. Here is the full procedure through which we transform this table into BCNF. Let us first divide this main table into two tables Stu_Branch and Stu_Course Table.

Stu_Branch Table

101Computer Science & Engineering
102Electronics & Communication Engineering

Candidate Key for this table: Stu_ID.

Stu_Course Table

Computer NetworksB_001202
VLSI TechnologyB_003401
Mobile CommunicationB_003402

Candidate Key for this table: Stu_Course.

Stu_ID to Stu_Course_No Table


Candidate Key for this table: {Stu_ID, Stu_Course_No}.

After decomposing into further tables, now it is in BCNF, as it is passing the condition of Super Key, that in functional dependency X−>Y, X is a Super Key.

Example 2

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 subsets can determine all attributes of the relation, So AC will be the 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 attributes.

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 a prime attribute. So the highest normal form of relation will be the 2nd Normal form. 

Note: A prime attribute cannot be transitively dependent on a key in BCNF relation. 

Consider these functional dependencies of some relation R

AB ->C 
C ->B 
AB ->B 

Suppose, it is known that the only candidate key of R is AB. A careful observation is required to conclude that the above dependency is a Transitive Dependency as the prime attribute B transitively depends on the key AB through C. Now, the first and the third FD are in BCNF as they both contain the candidate key (or simply KEY) on their left sides. The second dependency, however, is not in BCNF but is definitely in 3NF due to the presence of the prime attribute on the right side. So, the highest normal form of R is 3NF as all three FDs satisfy the necessary conditions to be in 3NF. 

Example 3

For example consider relation R(A, B, C)

A -> BC, 
B -> A

A and B both are super keys so the above relation is in BCNF. 

Note: BCNF decomposition may always not be possible with dependency preserving, however, it always satisfies the lossless join condition. For example, relation R (V, W, X, Y, Z), with functional dependencies:

V, W -> X
Y, Z -> X
W -> Y 

It would not satisfy dependency preserving BCNF decomposition. 

Note: Redundancies are sometimes still present in a BCNF relation as it is not always possible to eliminate them completely.

There are also some higher-order normal forms, like the 4th Normal Form and the 5th Normal Form.

For more, refer to the 4th and 5th Normal Forms.

Last Updated : 22 May, 2023
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