GATE CS 2001 Q2.23 — Normal Forms

To determine which relation does not have a lossless-join, dependency-preserving BCNF decomposition, we analyze the options:

• Option A: $A \rightarrow B, B \rightarrow CD$. The candidate key is $A$. Decomposing into $R_1(B, C, D)$ and $R_2(A, B)$ is lossless (common attribute $B$ is a key of $R_1$) and preserves all dependencies.
• Option B: $A \rightarrow B, B \rightarrow C, C \rightarrow D$. The candidate key is $A$. Decomposing into $R_1(A, B)$, $R_2(B, C)$, and $R_3(C, D)$ is lossless and preserves all dependencies.
• Option D: $A \rightarrow BCD$. The candidate key is $A$. The relation is already in BCNF, so the trivial decomposition is lossless and dependency-preserving.

For Option C ($AB \rightarrow C, C \rightarrow AD$):

• The candidate keys are $AB$ and $BC$ (since $(AB)^+ = ABCD$ and $(BC)^+ = ABCD$).
• The dependency $C \rightarrow AD$ violates BCNF because $C$ is not a superkey.
• If we decompose to resolve this violation, we must split the attributes of $AB \rightarrow C$ across different relations (e.g., $R_1(A, C, D)$ and $R_2(B, C)$). Any such decomposition loses the dependency $AB \rightarrow C$.

Thus, Option C cannot be decomposed into BCNF while being both lossless and dependency-preserving.