Question

Consider a relational schema team(name, city, owner), with functional dependencies \( \{ {name} \rightarrow {city}, {name} \rightarrow {owner} \} \). The relation team is decomposed into two relations, \( t_1({name}, {city}) \) and \( t_2({name}, {owner}) \). Which of the following statement(s) is/are TRUE?

• A. The relation team is NOT in BCNF.
• B. The relations \( t_1 \) and \( t_2 \) are in BCNF.
• C. The decomposition constitutes a lossless join.
• D. The relation team is NOT in 3NF.

Hint:

A decomposition is lossless if the common attribute in the decomposed relations is a candidate key for at least one of the relations.

Answer 1

The Correct Option is B, C

We are given the relation:
\( \text{team}(\text{name}, \text{city}, \text{owner}) \) with the functional dependencies:
\( \text{name} \rightarrow \text{city} \)
\( \text{name} \rightarrow \text{owner} \)

Step 1: Checking for BCNF
The relation \( \text{team} \) is not in BCNF because \( \text{name} \) is not a superkey for the relation.

Step 2: Checking Relations \( t_1 \) and \( t_2 \)
The decomposed relations are:
\( t_1(\text{name}, \text{city}) \) and \( t_2(\text{name}, \text{owner}) \).
Since \( \text{name} \) is a superkey in both, both \( t_1 \) and \( t_2 \) are in BCNF.

Step 3: Checking for Lossless Join
The decomposition is lossless because the common attribute \( \text{name} \) is a candidate key in both relations, ensuring that no data is lost when joining \( t_1 \) and \( t_2 \).

Step 4: Checking for 3NF
The relation \( \text{team} \) is not in 3NF because the functional dependencies violate 3NF conditions (since \( \text{name} \) is not a superkey).

Final Answer:
The correct answers are \( \boxed{B \text{ and } C} \).