Consider a relational schema \(π‘‘π‘’π‘Žπ‘š(π‘›π‘Žπ‘šπ‘’, 𝑐𝑖𝑑𝑦, π‘œπ‘€π‘›π‘’π‘Ÿ)\),…

2025

Consider a relational schema \(π‘‘π‘’π‘Žπ‘š(π‘›π‘Žπ‘šπ‘’, 𝑐𝑖𝑑𝑦, π‘œπ‘€π‘›π‘’π‘Ÿ)\), with functional dependencies \(\{π‘›π‘Žπ‘šπ‘’ β†’ 𝑐𝑖𝑑𝑦, π‘›π‘Žπ‘šπ‘’ β†’ π‘œπ‘€π‘›π‘’π‘Ÿ\}\).

The relationΒ \(π‘‘π‘’π‘Žπ‘š\) is decomposed into two relations,Β \(𝑑1(π‘›π‘Žπ‘šπ‘’, 𝑐𝑖𝑑𝑦)\) and \(𝑑2(π‘›π‘Žπ‘šπ‘’, π‘œπ‘€π‘›π‘’π‘Ÿ)\). Which of the following statement(s) is/are TRUE?

  1. A.

    The relationΒ \(π‘‘π‘’π‘Žπ‘š\) is NOT in BCNF.

  2. B.

    The relationsΒ \(t1\) andΒ \(t2\) are in BCNF.

  3. C.

    The decomposition constitutes a lossless join.

  4. D.

    The relationΒ \(π‘‘π‘’π‘Žπ‘š \) is NOT in 3NF.

Attempted by 160 students.

Show answer & explanation

Correct answer: B, C

Identify the candidate key: name determines both city and owner, so name is a candidate key for the relation.

BCNF for the original relation: Every non‑trivial functional dependency (name -> city, name -> owner) has the left side equal to a key, so the original relation is in BCNF.

  • t1 (name, city): name -> city, so name is the key for t1; therefore t1 is in BCNF.

  • t2 (name, owner): name -> owner, so name is the key for t2; therefore t2 is in BCNF.

Lossless join: The common attribute between the decomposed relations is name, which is a key for the relations. Since the common attribute is a key in at least one relation (here in both), the decomposition is lossless.

3NF: Because the relation is in BCNF, it also satisfies 3NF.

Conclusion: The statements that the decomposed relations are in BCNF and that the decomposition is lossless are true. The statements claiming the original relation is not in BCNF or not in 3NF are false.

A video solution is available for this question β€” log in and enroll to watch it.

Explore the full course: Gate Guidance By Sanchit Sir