Solution

The correct answer is: Option 3

Concept:

Boyce-Codd Normal Form (BCNF) is a higher version of the Third Normal Form (3NF). A relation is in BCNF if for every non-trivial functional dependency X → Y, X is a super key.

While BCNF decomposition always guarantees a lossless join, it does not always guarantee dependency preservation.

Key Points

Option 1: R(V, W, X, Y, Z) with FDs:

• VW → X
• WX → V
• VX → W
• W → Y
• Y → Z

Observation: This is a known example from textbooks like Navathe and Korth that looks complex, but it can be decomposed into BCNF with both lossless join and dependency preservation.

Why it is NOT the correct answer: Despite multiple dependencies, by carefully selecting decompositions (e.g., starting from W → Y), we can achieve both BCNF and preserve dependencies.

Option 2: R(V, W, X, Y) with FDs:

• VW → W (trivial)
• X → Y

Observation: The FD VW → W is trivial and doesn't affect BCNF. The FD X → Y violates BCNF if X is not a superkey.

Decomposition:

• Decompose on X → Y → R1(X, Y) and R2(V, W, X)

This decomposition is:

• Lossless: Yes (X is common and a key in R1)
• Dependency preserving: Yes (both X → Y and VW → W are preserved)

Why it is NOT the correct answer: It can be decomposed into BCNF with both lossless join and dependency preservation.

Option 3: Correct Answer
R(V, W, X, Y) with FDs:

• VW → X
• X → VY

Candidate Key: VW (since VW⁺ = V, W, X → V, W, X, Y = all attributes)

Violation: FD X → VY is a violation of BCNF as X is not a superkey

Attempted BCNF Decomposition:

• Decompose on X → VY
• R1(X, V, Y), R2(X, W)

Check:

• Lossless join: Yes (X is common and a key in R1)
• Dependency preservation: ❌ No (FD VW → X is lost; we can’t enforce it without recombining the relations)

Why this IS the correct answer: This relation cannot be decomposed into BCNF while preserving both lossless join and dependencies.

Option 4: R(V, W, X, Y, Z) with FDs:

• VW → X
• X → Y
• Y → Z
• Z → V

Observation: This is a cyclic chain of dependencies. These types of chains can be decomposed incrementally in BCNF.

Possible decomposition:

• Start with Z → V → R1(Z, V)
• Then Y → Z → R2(Y, Z)
• Then X → Y → R3(X, Y)
• Then VW → X → R4(V, W, X)

Lossless Join: ✅

Dependency Preservation: ✅ All dependencies are represented in individual decomposed relations

Why it is NOT the correct answer: It can be decomposed into BCNF with both properties preserved.

Final Answer:

Option 3 is the correct answer because it cannot be decomposed into BCNF with both lossless join and dependency preservation.