Q19,p2,d14. The best normal form of relation scheme R(A, B, C, D) along with the set of functional dependencies F = {AB →C, AB →D, C→A, D →B} is

(A) Boyce-Codd Normal form

(B) Third Normal form

(C) Second Normal form

(D) First Normal form .

 

Answer B.

 Explanation .

A relation is in 3NF if for every functional dependency X->Y in a set of functional dependency either of them hold.

·         X is super key  of R.

Or

·         Y is a subset of X..

Or

·         Y is a subset of K for some key of R.

Repeat steps for all dependency.

Step 1 check whether a key exists.(AB->C,D) hence AB->c,d qualifies for bcnf.

Take C->A,D->B.

Step 2 check for trivial dependency(C-> A, D->B no trivial dependency.

Then check for relaxed 3 nf.

In dependency  type x->y, y is part of any key.

A and B are part of key AB  hence it is in 3nf.