FDS

Consider relation R(A,B,C,D,E) with the following functional dependencies: AB -> C, D -> E, DE -> B.
Is R in BCNF? If not, decompose R into a collection of BCNF relations.

To determine if a relation is in Boyce-Codd Normal Form (BCNF), we need to check if it satisfies the following conditions:

1. It is in 1NF (First Normal Form).
2. For every non-trivial functional dependency (X -> Y), X is a superkey.

Let's analyze the given relation R(A, B, C, D, E) with the functional dependencies AB -> C, D -> E, DE -> B:

1. 1NF: Assuming that all attributes (A, B, C, D, E) are atomic, the relation is in 1NF.

2. Superkey analysis:

   • The functional dependency AB -> C implies that AB is a superkey.
   • The functional dependency D -> E implies that D is a superkey.
   • The functional dependency DE -> B implies that DE is a superkey

• AB -> C: AB is a superkey, so this dependency is fine.
• D -> E: D is a superkey, so this dependency is fine.
• DE -> B: DE is a superkey, so this dependency is fine.

Since all the functional dependencies have superkeys on the left side, the relation R is already in BCNF.

Therefore, there is no need for decomposition in this case, as the relation R(A, B, C, D, E) is already in BCNF.