The best normal form of relation scheme R(A, B, C, D) along with the set of…

Answer: Third Normal Form

Compute candidate keys by closure calculations:

• AB+ = {A, B, C, D} (from AB → C and AB → D), so AB is a candidate key.

• C+ = {C, A} (from C → A), not a key.

• D+ = {D, B} (from D → B), not a key.

• CD+ = {C, D, A, B} (C → A, D → B), so CD is a candidate key.

• AD+ = {A, D, B, C} (D → B then AB → C), so AD is a candidate key.

• BC+ = {B, C, A, D} (C → A then AB → D), so BC is a candidate key.

Thus candidate keys include AB, AD, BC, and CD; the prime attributes are A, B, C, and D (each appears in some candidate key).

Check Boyce-Codd Normal Form (BCNF):

• C → A: the determinant C is not a superkey, so this dependency violates BCNF.

• D → B: the determinant D is not a superkey, so this dependency violates BCNF.

• AB → C and AB → D: AB is a candidate key, so these do not violate BCNF.

Because there exist nontrivial dependencies with non-superkey determinants, the relation is not in BCNF.

Check Third Normal Form (3NF):

• 3NF allows a non-superkey determinant if the right-hand side is a prime attribute.

• The violating dependencies C → A and D → B have right-hand sides A and B, which are prime attributes.

Therefore all functional dependencies satisfy the conditions for Third Normal Form, so the highest normal form of the relation is Third Normal Form.