GATE CS 2025 Set 1 Q37 — Normal Forms

Let us analyze the given relational schema $\text{team}(\text{name}, \text{city}, \text{owner})$ and its functional dependencies (FDs):

1. Candidate Key of the relation team

Using the attribute closure of $\text{name}$: $$\text{name}^+ = \{\text{name}, \text{city}, \text{owner}\}$$ Since $\text{name}$ determines all attributes of the relation, the candidate key is $\{\text{name}\}$.

2. Normal Form of the relation team

• For the FD $\text{name} \rightarrow \text{city}$, the left-hand side ($\text{name}$) is a superkey.
• For the FD $\text{name} \rightarrow \text{owner}$, the left-hand side ($\text{name}$) is a superkey.

Since the left-hand side of every non-trivial functional dependency is a superkey, the relation $\text{team}$ is in BCNF (and consequently also in 3NF).

• Statement A is FALSE: The relation $\text{team}$ is in BCNF.
• Statement D is FALSE: The relation $\text{team}$ is in 3NF.

3. Normal Form of decomposed relations $t1$ and $t2$

• For $t1(\text{name}, \text{city})$: The projected FD is $\text{name} \rightarrow \text{city}$. The candidate key is $\{\text{name}\}$. Since the LHS is a superkey, $t1$ is in BCNF.
• For $t2(\text{name}, \text{owner})$: The projected FD is $\text{name} \rightarrow \text{owner}$. The candidate key is $\{\text{name}\}$. Since the LHS is a superkey, $t2$ is in BCNF.

Therefore, Statement B is TRUE.

4. Lossless Join Property

A decomposition of $R$ into $R_1$ and $R_2$ is a lossless join if and only if the common attribute(s) form a superkey in at least one of the decomposed relations:

Here, $t1 \cap t2 = \{\text{name}\}$. Since $\text{name}$ is a candidate key for both $t1$ and $t2$:

Thus, the decomposition is guaranteed to be a lossless join. Statement C is TRUE.