BCNF – Boyce Codd Normal Form

BCNF – Boyce Codd Normal Form 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.

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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