# BCNF – Boyce Codd Normal Form in DBMS

## 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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