Consider the Relation RA B C D E F and the following set of

Consider the Relation R(A, B, C, D, E, F) and the following set of functional dependencies:

A --> BCD

BC --> DE

B --> D

D --> A

Give a BCNF(Boyce Codd Normal Form) decomposition of R.

Solution

BCNF Relation:

A BCNF relation is defined as, if there are no non-trivial functional dependencies of attributes on anything other than a superset of a candidate key.

Considered the relation R(A, B, C, D, E, F) is not in BCNF, because there is no super key from the given relation that is no dependency holds F.

So, decompose the relation R into R1(A, B, C, D) and R2(A, E, F) accordingly to obtain the relation is in BCNF.

Now, A->E is an FD in closure of F and cause R2 to violate BCNF. So, once again  decompose R2 and gives R1 (A, B, C, D), R2 (A, E) and R3 (A, F).

Now the schema is in BCNF.