GATE CS 2025 Set 2 Q36 — Normal Forms

To determine if the relations are in 3NF, we find their candidate keys and check the 3NF conditions for each functional dependency (FD). A relation is in 3NF if, for every non-trivial FD $X \rightarrow Y$, either $X$ is a superkey or $Y$ is a prime attribute.

Analysis of $R_1(A, B, C, D, E)$ with FDs: $\{D \rightarrow E, EA \rightarrow B, EB \rightarrow C\}$

• The attributes $A$ and $D$ do not appear on the right-hand side of any FD. Thus, they must be part of any candidate key.
• Computing the closure: $\{A, D\}^+ = \{A, D, E, B, C\}$.
• Therefore, $\{A, D\}$ is the unique candidate key.
• Prime attributes: $\{A, D\}$; Non-prime attributes: $\{B, C, E\}$.
• For the FD $D \rightarrow E$, $D$ is not a superkey and $E$ is not a prime attribute. This violates the 3NF condition.
• Hence, $R_1$ is NOT in 3NF (Option C is TRUE).

Analysis of $R_2(A, B, C, D)$ with FDs: $\{A \rightarrow D, A \rightarrow B, C \rightarrow A\}$

• The attribute $C$ does not appear on the right-hand side of any FD. Thus, it must be part of any candidate key.
• Computing the closure: $\{C\}^+ = \{C, A, D, B\}$.
• Therefore, $\{C\}$ is the unique candidate key.
• Prime attributes: $\{C\}$; Non-prime attributes: $\{A, B, D\}$.
• For the FD $A \rightarrow D$, $A$ is not a superkey and $D$ is not a prime attribute. This violates the 3NF condition.
• Hence, $R_2$ is NOT in 3NF (Option D is TRUE).