GATE CS 1994 Q3.6 — Normal Forms

The statement is False (represented by 0).

While it is always possible to find a decomposition into 3NF that is both lossless and dependency-preserving, the same does not hold for BCNF. We can always find a lossless BCNF decomposition, but it is not always possible to preserve all functional dependencies.

Counterexample:

Consider a relation $R(A, B, C)$ with the following functional dependencies (FDs):

• $AB \rightarrow C$
• $C \rightarrow A$

The candidate keys for $R$ are $AB$ and $BC$.

The dependency $C \rightarrow A$ violates BCNF because $C$ is not a superkey. To decompose $R$ into BCNF losslessly, we must decompose it based on the violating FD $C \rightarrow A$:

• $R_1(A, C)$ with FD $C \rightarrow A$ (which is in BCNF)
• $R_2(B, C)$ with no non-trivial FDs (which is in BCNF)

This decomposition is lossless because $R_1 \cap R_2 = \{C\}$, which is a key for $R_1$. However, the functional dependency $AB \rightarrow C$ is lost because the attributes $A$ and $B$ do not appear together in any of the decomposed relations. Thus, a lossless and dependency-preserving BCNF decomposition is not possible for this relation.