BCNF – Boyce Codd Normal Form in DBMS

Let R be the relational schema, R is in BCNF only if :

  • R should be in 3NF.
  • Every Functional Dependency will have a Superkey on the LHS or all determinants are the superkeys.

Example :

Consider the following relationship R(ABCD) having following functional dependencies

F = {A → BCD, BC → AD, D → B}
Candidate Keys are :
  (A)+ = {ABCD}
(BC)+ = {BCAD}
(DC)+ = {DCBA}
Functional Dependency Is FD in BCNF or not ? Reason ?
A  → BCD Yes A is a super key
BC  → AD Yes BC is also a super key
D  → A No D is not super key, it is part of key

Solution : Decomposition in BCNF

The relation R(ABCD) is decomposed into two relations R1 and R2 such that :

R1(A,D,C)                         R2(D,B)

The above two relations R1 and R2

  1. Lossless Join
  2. BCNF Decomposition
  3. But Not Dependency Preserving
Redundancy in BCNF

0% redundancy, Because of Single Valued Functional Dependency.
Redundancy may exist because of Multivalued Dependency.

Some Notes Regarding BCNF
  • There is sometimes more than one BCNF decomposition of a given schema.
  • Some of the BCNF decompositions may also yield dependency preservation, while others may not.

Difference between 3NF and BCNF –

S.NO 3NF BCNF
1. It concentrates on Primary Key It concentrates on Candidate Key.
2. Redundancy is high as compared to BCNF 0% redundancy
3. It may preserve all the dependencies It may not preserve the dependencies.
4. A dependency X → Y is allowed in 3NF if X is a super key or Y is a part of some key. A dependency X → Y is allowed if X is a super key

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