GATE CS 2025 Set 1 Q.47: Functional Dependencies and Normalization

Consider a relational schema $\textit{team}(\textit{name}, \textit{city}, \textit{owner})$ , with functional dependencies $\{\textit{name} \to \textit{city}, \; \textit{name} \to \textit{owner}\}$ .

The relation $\textit{team}$ is decomposed into two relations, $t_1(\textit{name}, \textit{city})$ and $t_2(\textit{name}, \textit{owner})$ . Which of the following statement(s) is/are TRUE?

The relation $\textit{team}$ is NOT in BCNF.

The relations $t_1$ and $t_2$ are in BCNF.

The decomposition constitutes a lossless join.

The relation $\textit{team}$ is NOT in 3NF.

Official answer and solution

Official answer: B and C
From: Official GATE 2025 answer key

Solution

Find the candidate key of team, t1, and t2, and check the BCNF and lossless-join conditions.

In $\textit{team}(\textit{name}, \textit{city}, \textit{owner})$ , the closure of $\{\textit{name}\}$ is $\{\textit{name}, \textit{city}, \textit{owner}\}$ , so $\textit{name}$ is the sole candidate key. Since both FDs $\textit{name} \to \textit{city}$ and $\textit{name} \to \textit{owner}$ have a superkey on the left-hand side, $\textit{team}$ is already in BCNF (and hence in 3NF). Thus (A) and (D) are false.
In $t_1(\textit{name}, \textit{city})$ with $\textit{name} \to \textit{city}$ and $t_2(\textit{name}, \textit{owner})$ with $\textit{name} \to \textit{owner}$ , $\textit{name}$ is a superkey in both relations, so both $t_1$ and $t_2$ are in BCNF ((B) is TRUE).
The common attribute is $t_1 \cap t_2 = \{\textit{name}\}$ , which is a key for both $t_1$ and $t_2$ , so the decomposition is a lossless join ((C) is TRUE).

Final answer

Options (B), (C)

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2025 Computer Science and Information Technology - Session 1 (CS1)
Question: 47 of 65
Section: Computer Science and Information Technology

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